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# NCERT Solutions for Class 12 Maths Chapter 5 - In Hindi

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## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability In Hindi pdf download

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Competitive Exams after 12th Science

## Access NCERT Solutions for Mathematics Chapter ५ – सांतत्य तथा अवकलनीयता

### प्रश्नावली 5.1

1. सिद्ध  कीजिए कि फलन ${\text{f(x)}}\;{\text{ = }}\;{\text{5x - 3,}}\;{\text{x}}\;{\text{ = }}\;{\text{0,}}\;{\text{x}}\;{\text{ = }}\;{\text{ - 3,}}\;{\text{x}}\;{\text{ = }}\;{\text{5}}$ पर संतत है।

उत्तर: ${\text{f(x)}}\;{\text{ = }}\;{\text{5x - 3}}$

जब ${\text{x}}\;{\text{ = }}\;{\text{0}}$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 0} {\text{f(x)}}\;{\}}\;\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(5x - 3)}}\;{\text{ = }}\;{\text{5}} \times {\text{0 - 3}}\;{\text{ = }}\;{\text{ - 3}} \hfill \\ {\text{f(0)}}\;{\text{ = }}\;{\text{5}} \times {\text{0 - 3}}\;{\text{ = }}\;{\text{ - 3}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 0} {\text{f(x)}}\;{\text{ = }}\;{\text{f(0)}} \hfill \\ \end{align}

अतएव ${\text{x}}\;{\text{ = }}\;{\text{0}}$ पर ${\text{f}}$ संतत है।

जब ${\text{x}}\;{\text{ = }}\;{\text{ - 3}}$ ,

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to - 3} {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to - 3} {\text{(5x - 3)}}\;{\text{ = }}\;{\text{5}} \times ( - 3){\text{ - 3}}\;{\text{ = }}\; - 15{\text{ - 3}}\;{\text{ = }}\;{\text{ - 18}} \hfill \\ {\text{f( - 3)}}\;{\text{ = }}\;{\text{5}} \times ( - 3){\text{ - 3}}\;{\text{ = }}\;{\text{ - 18}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to - 3} {\text{f(x)}}\;{\text{ = }}\;{\text{f( - 3)}} \hfill \\ \end{align}

अतएव  ${\text{x}}\;{\text{ = }}\;{\text{ - 3}}$ पर ${\text{f}}$ संतत है।

जब ${\text{x}}\;{\text{ = }}\;{\text{5}}$ ,

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 5} {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 5} {\text{(5x - 3)}}\;{\text{ = }}\;{\text{5}} \times (5){\text{ - 3}}\;{\text{ = }}\;25{\text{ - 3}}\;{\text{ = }}\;22 \hfill \\ {\text{f(5)}}\;{\text{ = }}\;{\text{5}} \times (5){\text{ - 3}}\;{\text{ = }}\;22 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 5} {\text{f(x)}}\;{\text{ = }}\;{\text{f(5)}} \hfill \\ \end{align}

अतएव ${\text{x}}\;{\text{ = }}\;{\text{5}}$ पर ${\text{f}}$ संतत है।

2. ${\text{x}}\;{\text{ = }}\;{\text{3}}$ पर फलन ${\text{f(x)}}\;{\text{ = }}\;2{{\text{x}}^2}{\text{ - 1}}$ के सांतत्य कि जांच कीजिए।

उत्तर: ${\text{f(x)}}\;{\text{ = }}\;2{{\text{x}}^2}{\text{ - 1}}$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 3} {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 3} {\text{(2}}{{\text{x}}^2}{\text{ - 1)}}\;{\text{ = }}\;2 \times {(3)^2}{\text{ - 1}}\;{\text{ = }}\;18{\text{ - 1}}\;{\text{ = }}\;17 \hfill \\ {\text{f(3)}}\;{\text{ = }}\;2 \times {(3)^2}{\text{ - 1}}\;{\text{ = }}\;17 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 3} {\text{f(x)}}\;{\text{ = }}\;{\text{f(3)}} \hfill \\ \end{align}

अतएव  पर ${\text{x}}\;{\text{ = }}\;{\text{3}}$ ${\text{f}}$ संतत है।

3. निम्नलिखित फलनों का सांतत्य कि जांच कीजिए:

(a) ${\text{f(x)}}\;{\text{ = }}\;{\text{x - 5}}$

उत्तर: ${\text{f(x)}}\;{\text{ = }}\;{\text{x - 5}}$ एक बहुपद फलन है।

इसलिए ${\text{f}}$ एक संतत फलन है।

(b) ${\text{f(x)}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{x - 5}}}}{\text{,}}\;{\text{x}}\; - \;5$ के बराबर नहीं है

उत्तर: ${\text{f(x)}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{x - 5}}}}$ एक परिमय फलन है।

इसलिए ${\text{f}}$ एक संतत फलन है। ${\text{x}}\; - \;5$ के बराबर नहीं है।

(c) ${\text{f(x)}}\;{\text{ = }}\;\dfrac{{{\text{(}}{{\text{x}}^{\text{2}}}{\text{ - 25)}}}}{{{\text{x + 5}}}}{\text{,}}\;{\text{x - 5}}$ के बराबर नहीं है

उत्तर: ${\text{f(x)}}\;{\text{ = }}\;\dfrac{{{\text{(}}{{\text{x}}^{\text{2}}}{\text{ - 25)}}}}{{{\text{x + 5}}}}$ = ${\text{(x + 5)(x - 5)/(x + 5)}}\;{\text{ = }}\;{\text{x - 5}}$

${\text{f(x)}}\;{\text{ = }}\;{\text{x - 5}}$ एक बहुपद फलन है। इसलिए ${\text{f}}$ एक संतत फलन है।

(d) ${\text{f(x)}}\;{\text{ = }}\;\left| {{\text{x - 5}}} \right|$

उत्तर: ${\text{f(x)}} = \{ \begin{array}{*{20}{c}} {{\text{x - 5,}}}&{{\text{x}} \geqslant {\text{5}}} \\ {{\text{5 - x,}}}&{{\text{x < 5}}} \end{array}$ \begin{align} \mathop {\lim }\limits_{{\text{x}} \to 5 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 5 + } {\text{(x - 5)}}\;{\text{ = }}\;5 - 5\;{\text{ = }}\;0 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 5 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 5 - } {\text{(5 - x)}}\;{\text{ = }}\;5 - 5\;{\text{ = }}\;0 \hfill \\ {\text{f(5 )}}\;{\text{ = }}\;{\text{5 - 5}}\;{\text{ = }}\;0 \hfill \\ \end{align}

बाए पक्ष कि सीमा तथा दाए पक्ष की सीमा बराबर है। ${\text{x}}\;{\text{ = }}\;{\text{5}}$ पर ${\text{f}}$ एक संतत फलन है।

4. सिद्ध कीजिए कि फलन ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{n}}}{\text{,}}\;{\text{x}}\;{\text{ = }}\;{\text{n}}$ पर संतत है, जहा ${\text{n}}$ एक धन पूर्णांक है।

उत्तर: ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{n}}}$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to {\text{n}}} {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{n}}} {\text{(}}{{\text{x}}^{\text{n}}}{\text{)}}\;{\text{ = }}\;{{\text{n}}^{\text{n}}} \hfill \\ {\text{f(n)}}\;{\text{ = }}\;{{\text{n}}^{\text{n}}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to {\text{n}}} {\text{f(x)}}\;{\text{ = }}\;{\text{f(n)}} \hfill \\ \end{align}

${\text{x}}\;{\text{ = }}\;{\text{n}}$ पर ${\text{f}}$ संतत है। जहा ${\text{n}}$ एक धन पूर्णांक है।

5. क्या ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{x,}}}&{{\text{x}} \leqslant {\text{1}}} \\ {{\text{x,}}}&{{\text{x > 1}}} \end{array}$ द्वारा परिभाषित फलन ${\text{f}}$ , ${\text{x}}\;{\text{ = }}\;{\text{0,}}\;{\text{x}}\;{\text{ = }}\;{\text{1,}}\;{\text{x}}\;{\text{ = }}\;{\text{2}}$ पर संतत है? ${\text{f}}$ के सभी असांतत्य के बिन्दुओ को ज्ञात कीजिए, जब कि ${\text{f}}$ निम्नलिखित प्रकार से प्रभाषित है। उत्तर: ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{x,}}}&{{\text{x}} \leqslant {\text{1}}} \\ {{\text{x,}}}&{{\text{x > 1}}} \end{array}$

जब ${\text{x}}\;{\text{ = }}\;{\text{0}}$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 0 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{f(0 - h)}}\;{\text{ = }}\;{\text{0 - 0}}\;{\text{ = }}\;0 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{f(0 + h)}}\;{\text{ = }}\;{\text{0 + 0}}\;{\text{ = }}\;0 \hfill \\ {\text{f(0)}}\;{\text{ = }}\;{\text{0}} \hfill \\ \end{align}

अतएव ${\text{x}}\;{\text{ = }}\;{\text{0}}$ पर ${\text{f}}$ संतत है।

जब ${\text{x}}\;{\text{ = }}\;1$ ,

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{f(1 - h)}}\;{\text{ = }}\;1 - 0\; = \;1 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\;{\text{ = }}\;5 \hfill \\ \end{align}

अतएव  ${\text{x}}\;{\text{ = }}\;1$ पर ${\text{f}}$ संतत नहीं है।

जब ${\text{x}}\;{\text{ = }}\;2$ ,

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 2 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 2 + } {\text{f(x)}}\;{\text{ = }}\;{\text{5}} \hfill \\ {\text{f(2)}}\;{\text{ = }}\;{\text{5}} \hfill \\ \end{align}

अतएव ${\text{x}}\;{\text{ = }}\;2$ पर ${\text{f}}$ संतत है।

6. ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{2x + 3,}}}&{{\text{x}} \leqslant 2} \\ {{\text{2x - 3,}}}&{{\text{x > 2 }}} \end{array}$

उत्तर: $\mathop {\lim }\limits_{{\text{x}} \to 2 - } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 2 - } {\text{(2x + 3)}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[2(2 - h) + 3]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[7 - 2h]}} \hfill \\ {\text{ = }}\;{\text{7 - 2}} \times {\text{0}} \hfill \\ {\text{ = }}\;{\text{7}} \hfill \\ \end{align}

$\mathop {\lim }\limits_{{\text{x}} \to 2 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 2 + } {\text{(2x + 3)}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[2(2 + h) - 3]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[1 + 2h]}} \hfill \\ {\text{ = }}\;1 + {\text{2}} \times {\text{0}} \hfill \\ {\text{ = }}\;1 \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;{\text{2}}$ पर ${\text{f}}$ संतत नहीं है।

7. $\text { 7. } f(x)=\left\{\begin{array}{cc} |x|+3, & x {\pounds}-3 \\ -2 x, & -3<x<3 \\ 6 x+2 & x^{3} 3 \end{array}\right.$

उत्तर: ${\text{x < - 3}}$ के लिए ${\text{f(x)}}\;{\text{ = }}\;\left| {\text{x}} \right|{\text{ + 3}}$

${\text{ - 3 < x < 3}}$ के लिए ${\text{f(x)}}\;{\text{ = }}\;{\text{ - 2x}}$

${\text{x > 3}}$ के लिए ${\text{f(x)}}\;{\text{ = }}\;{\text{ - 2x}}$ एक बहुपद फलन है

इसलिए यह फलन है।

जब ${\text{x}}\;{\text{ = }}\;{\text{ - 3}}$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 3 - } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 3 - } {\text{(}}\left| {\text{x}} \right|{\text{ + 3)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[}}\left| {{\text{ - 3 - h}}} \right|{\text{ + 3]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} (6 + {\text{h)}}\;{\text{ = }}\;{\text{6 + 0}}\;{\text{ = }}\;{\text{6}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 3 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 3 + } {\text{( - 2x)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[ - 2( - 3 + h)]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} (6 - 2{\text{h)}}\;{\text{ = }}\;{\text{6 - 2}} \times {\text{0}}\;{\text{ = }}\;{\text{6}} \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;{\text{ - 3}}$ पर ${\text{f}}$ संतत है।

जब ${\text{x}}\;{\text{ = }}\;{\text{3}}$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 3 - } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 3 - } {\text{( - 2x)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[ - 2(3 - h)]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} ( - 6 + 2{\text{h)}}\;{\text{ = }}\; - {\text{6 + 2}} \times {\text{0}}\;{\text{ = }}\; - {\text{6}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 3 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 3 + } {\text{(6x + 2)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[6(3 + h) + 2]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} (18 + 6{\text{h + 2)}}\;{\text{ = }}\;20 + 6 \times {\text{0}}\;{\text{ = }}\;20 \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;{\text{3}}$ पर ${\text{f}}$ संतत नहीं है।

8. ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {\dfrac{{\left| {\text{x}} \right|}}{{\text{x}}}{\text{,}}}&{{\text{x < 0 }}} \\ {{\text{0 ,}}}&{{\text{x = 0}}} \end{array}$के बराबर नहीं है

उत्तर: $\mathop {\lim }\limits_{{\text{x}} \to 0 - } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 0 - } {\text{(}}\left| {\text{x}} \right|{\text{/x)}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[}}\left| {0 - {\text{h}}} \right|{\text{/0 - h]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} ({\text{h)}}\;{\text{ = }}\;{\text{h/ - h}}\;{\text{ = }}\; - 1 \hfill \\ \end{align}

$\mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{(}}\left| {\text{x}} \right|{\text{/x)}}$

$= \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[}}\left| {0 + {\text{h}}} \right|{\text{/0 + h]}}$ = $\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{h/h}}\;{\text{ = }}\;{\text{1}}$

अतः ${\text{x}}\;{\text{ = }}\;0$ पर ${\text{f}}$ संतत नहीं है।

9. ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {\dfrac{{\text{x}}}{{\left| {\text{x}} \right|}}{\text{,}}}&{{\text{x < 0 }}} \\ {{\text{ - 1 ,}}}&{{\text{x}} \geqslant {\text{0}}} \end{array}$

उत्तर: $\mathop {\lim }\limits_{{\text{x}} \to 0 - } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 0 - } {\text{(x/}}\left| {\text{x}} \right|{\text{)}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[(0 - h)/}}\left| {0 - {\text{h}}} \right|{\text{]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} ({\text{h)}}\;{\text{ = }}\; - {\text{h/h}}\;{\text{ = }}\; - 1 \hfill \\ \end{align}

$\mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{(x/}}\left| {\text{x}} \right|{\text{)}}$

${\text{f(0)}}\;{\text{ = }}\;{\text{ - 1}}$

अतः ${\text{x}}\;{\text{ = }}\;0$ पर ${\text{f}}$ संतत  है।

यहा कोई असांतत्य के बिन्दु नहीं है।

10. ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{x + 1,}}}&{{\text{x}} \geqslant 1{\text{ }}} \\ {{{\text{x}}^2}{\text{ + 1,}}}&{{\text{x < 1}}} \end{array}$

उत्तर: ${\text{x > 1}}$ के लिए ${\text{f(x)}}\;{\text{ = }}\;{\text{x + 1}}$

${\text{x < 1}}$ के लिए ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ + 1}}$ एक बहुपद फलन है

इसलिए यह फलन है।

जब ${\text{x}}\;{\text{ = }}\;1$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{(}}{{\text{x}}^2}{\text{ + 1)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[(1 - h}}{{\text{)}}^2}{\text{ + 1]}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[1 + }}{{\text{h}}^{\text{2}}}{\text{ - 2h + 1]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} (2{\text{ + }}{{\text{h}}^{\text{2}}}{\text{ - 2h)}}\;{\text{ = }}\;2 + 0 - {\text{0}}\;{\text{ = }}\;2 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{(x + 1)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[1 + h + 1]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} ({\text{2 + h)}}\;{\text{ = }}\;2 + 0\;{\text{ = }}\;2 \hfill \\ \end{align}

${\text{f(1)}}\;{\text{ = }}\;{\text{1 + 1}}\;{\text{ = }}\;{\text{2}}$

अतः ${\text{x}}\;{\text{ = }}\;1$ पर ${\text{f}}$ फलन है।

यहा कोई असांतत्य के बिन्दु नहीं है।

11. ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{{\text{x}}^3}{\text{ - 3,}}}&{{\text{x}} \leqslant {\text{2}}} \\ {{{\text{x}}^2}{\text{ + 1,}}}&{{\text{x > 2}}} \end{array}$

उत्तर: ${\text{x < 2}}$ के लिए ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^3} - 3$

${\text{x > 2}}$ के लिए ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ + 1}}$ एक बहुपद फलन है

इसलिए यह फलन है।

${\text{x}}\;{\text{ = }}\;{\text{2}}$ ,

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 2 - } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 2 - } {\text{(}}{{\text{x}}^3}{\text{ - 3)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[(2 - h}}{{\text{)}}^3}{\text{ - 3]}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[8 - }}{{\text{h}}^3}{\text{ - 12h + 6}}{{\text{h}}^2}{\text{ - 3]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{(5 - }}{{\text{h}}^{\text{3}}}{\text{ - 12h + 6}}{{\text{h}}^{\text{2}}}{\text{)}}\;{\text{ = }}\;5 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 2 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 2 + } {\text{(}}{{\text{x}}^2}{\text{ + 1)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[(2 + h}}{{\text{)}}^2}{\text{ + 1]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} ({\text{4 + }}{{\text{h}}^{\text{2}}}{\text{ + 4h + 1)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} (5{\text{ + }}{{\text{h}}^{\text{2}}}{\text{ + 4h)}}\;{\text{ = }}\;{\text{5}} \hfill \\ \end{align}

\begin{align} {\text{f(2)}}\;{\text{ = }}\;{{\text{(2)}}^{\text{3}}}{\text{ - 3}} \hfill \\ {\text{ = }}\;{\text{8 - 3}}\;{\text{ = }}\;{\text{5}} \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;2$ पर ${\text{f}}$ फलन है।

यहा कोई असांतत्य के बिन्दु नहीं है।

12.${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{{\text{x}}^{10}}{\text{ - 1,}}}&{{\text{x}} \leqslant 1} \\ {{{\text{x}}^2}{\text{,}}}&{{\text{x > 1}}} \end{array}$

उत्तर: ${\text{x < 1}}$ के लिए ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{10}} - 1$

${\text{x > 1}}$ के लिए ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}$ एक बहुपद फलन है

इसलिए यह फलन है।

${\text{x}}\;{\text{ = }}\;1$ ,

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{(}}{{\text{x}}^{10}}{\text{ - 1)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[(1 - h}}{{\text{)}}^{10}}{\text{ - 1]}} \hfill \\ {\text{ = }}\;{{\text{(1 - 0)}}^{10}} - 1\;{\text{ = }}\;1 - 1\; = \;0 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{(}}{{\text{x}}^2}{\text{)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[(1 + h}}{{\text{)}}^2}{\text{]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} (1{\text{ + }}{{\text{h}}^{\text{2}}}{\text{ + 2h)}} \hfill \\ {\text{ = }}\;1 \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;1$ पर ${\text{f}}$ संतत नहीं है।

13. ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{x + 5,}}}&{{\text{x}} \leqslant 1} \\ {{\text{x - 5,}}}&{{\text{x > 1}}} \end{array}$द्वारा परिभाषित फलन, एक संतत फलन है। फलन ${\text{f}}$ के सांतत्य पर विचार कीजिए, जहा ${\text{f}}$ निम्नलिखित द्वारा परिभाषित है।

उत्तर: ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{x + 5,}}}&{{\text{x}} \leqslant 1} \\ {{\text{x - 5,}}}&{{\text{x > 1}}} \end{array}$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{(x + 5)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[1 - h + 5]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{(6 - h)}} \hfill \\ {\text{ = }}\;6 - 0\; = \;6 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{(x - 5)}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{[1 + h - 5]}} \hfill \\ {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} ({\text{h - 4)}} \hfill \\ {\text{ = }}\;0 - 4\; = \; - 4 \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;1$ पर ${\text{f}}$ संतत नहीं है।

14. $f(x)=\left\{\begin{matrix} 3, &0\leq x\leq 1 \\ 4, & 1< x< 3\\ 5, & 3\leq x\leq 1 \end{matrix}\right.$

उत्तर: ${\text{0}} \leqslant {\text{x}} \leqslant {\text{1,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{3}}$

\begin{align} {\text{1 < x < 3,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{4}} \hfill \\ {\text{3 < x}} \leqslant {\text{10,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{5}} \hfill \\ \end{align} एक सतत फलन है, इसलिए यह फलन है। ${\text{x}}\;{\text{ = }}\;{\text{1}}$ \begin{align} \mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 - } (3)\; = \;3 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 + } (4)\; = \;4 \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;1$ पर ${\text{f}}$ संतत नहीं है।

${\text{x}}\;{\text{ = }}\;3$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 3 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 3 - } (4)\; = \;4 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 3 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 3 + } (5)\; = \;5 \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;3$ पर ${\text{f}}$ संतत नहीं है।

15.$f(x)=\left\{\begin{matrix} 2x, & x< 1 \\ 0, & 0\leq x\leq 1\\ 4x, & x> 1 \end{matrix}\right.$

उत्तर: ${\text{x < 0,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{2x}}$

\begin{align} {\text{0 < x < 1,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{0}} \hfill \\ {\text{x > 1,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{4x}} \hfill \\ \end{align}

एक बहुपद, सतत फलन है, इसलिए यह फलन है।

${\text{x}}\;{\text{ = }}\;{\text{0}}$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 0 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 0 - } {\text{(2x}}) \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} [2(0 - {\text{h)]}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} ( - 2{\text{h)}}\;{\text{ = }}\;{\text{ - 2}} \times {\text{0}}\;{\text{ = }}\;{\text{0}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 0 + } (0)\; = \;0 \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;0$ पर ${\text{f}}$ संतत नहीं है।

${\text{x}}\;{\text{ = }}\;1$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{(0}})\; = \;0 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 + } (4{\text{x}}) \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} [4(1 + {\text{h)]}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} (4 + 4{\text{h)}}\;{\text{ = }}\;4 + 4 \times {\text{0}}\;{\text{ = }}\;4 \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;1$ पर ${\text{f}}$ संतत नहीं है।

16.$f(x)=\left\{\begin{matrix} -2, & x< -1 \\ 2x, & -1< x\leq 1\\ 2, & x> 1 \end{matrix}\right.$

उत्तर: ${\text{x < 1,}}\;{\text{f(x)}}\;{\text{ = }}\; - {\text{2}}$

\begin{align} {\text{ - 1 < x < 1,}}\;{\text{f(x)}}\;{\text{ = }}\;2{\text{x}} \hfill \\ {\text{x > 1,}}\;{\text{f(x)}}\;{\text{ = }}\;2 \hfill \\ \end{align}

एक बहुपद फलन है, इसलिए यह फलन है।

${\text{x}}\;{\text{ = }}\;{\text{ - 1}}$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{( - 2}})\; = \;2 \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 + } (2{\text{x}}) \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} [2( - 1 + {\text{h)]}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} ( - 2 + 2{\text{h)}}\;{\text{ = }}\; - 2 + {\text{0}}\;{\text{ = }}\; - {\text{2}} \hfill \\ {\text{f( - 1)}}\;{\text{ = }}\;{\text{ - 2}} \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\; - 1$ पर ${\text{f}}$ संतत है।

${\text{x}}\;{\text{ = }}\;1$

\begin{align} \mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{(2x}}) \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} [2({\text{1 - h)]}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{(2 - 2h)}} \hfill \\ {\text{ = }}\;{\text{2 - 2}} \times {\text{0}}\;{\text{ = }}\;{\text{2}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 + } (2)\; = \;2 \hfill \\ {\text{f(1)}}\;{\text{ = }}\;2 \times 1\;{\text{ = }}\;2 \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;1$ पर ${\text{f}}$ संतत है।

17. ${\text{a,}}\;{\text{b}}$ के उन मानो को ज्ञात कीजिए जिनके लिए द्वारा परिभाषित फलन ${\text{x = }}\;{\text{3 }}$ पर संतत है। उत्तर: ${\text{f(x)}}\;{\text{ = }}\;{\text{\{ ax}} + 1;\;{\text{x}} \leqslant 3)$

${\text{f(x)}}\;{\text{ = }}\;{\text{\{ bx}} + 3;\;{\text{x > }}3)$

${\text{x}}\;{\text{ = }}\;{\text{3}}$, ${\text{f(x) = ax + 1}}$

LHL, $\mathop {\lim }\limits_{{\text{x}} \to 0 + } ({\text{ax + 1)}}\;{\text{ = }}\;{\text{3a + 1}}$

${\text{f(3)}}\;{\text{ = }}\;{\text{3a + 1}}$

${\text{x}}\; > \;{\text{3}}$, ${\text{f(x) = bx + 3}}$

RHL, $\mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 0 + } ({\text{bx + 3)}}\;{\text{ = }}\;{\text{3b + 3}}$

\begin{align} {\text{3a + 1}}\;{\text{ = }}\;{\text{3b + 3}} \hfill \\ {\text{a}}\;{\text{ = }}\;{\text{b + }}\dfrac{{\text{2}}}{{\text{3}}} \hfill \\ \end{align}

स्वेच्छा से ${\text{b}}$ के मान के लिए ${\text{a}}$ का मान ज्ञात किया जा सकता है।

18. $\lambda$ के किस मान के लिए

\begin{align} {\text{f(x) = \{ }}\begin{array}{*{20}{c}} {\lambda {\text{(}}{{\text{x}}^2}{\text{ - 2x),}}}&{{\text{x}} \leqslant 0} \\ {{\text{4x + 1,}}}&{{\text{x > 0}}} \end{array} \hfill \\ \hfill \\ \end{align} द्वारा परिभाषित फलन ${\text{x = }}\;0$ पर संतत है। ${\text{x = }}\;1$ पर इसके सातत्य पर विचार कीजिए।

उत्तर: यदि ${\text{f(x)}}$, ${\text{x}}\;{\text{ = }}\;{\text{0}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(0)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 0 - } {\text{f(x)}} \hfill \\ \Rightarrow {\text{(}}{{\text{0}}^2} + 2(0))\; = \;(4(0) + 1)\; = \;({0^2} - 2(0)) \hfill \\ \Rightarrow \;0\; = \;1\; = \;0 \hfill \\ \end{align}

जो सत्य नहीं हो सकता, अर्थात $\lambda$ कि किसी भी मात्रा के लिए यह फलन ${\text{x}}\;{\text{ = }}\;{\text{0}}$ पर संतत नहीं है।

यदि ${\text{f(x)}}$, ${\text{x}}\;{\text{ = }}\;1$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(1)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 1 - } {\text{f(x)}} \hfill \\ \Rightarrow {\text{(4(1) + 1}})\; = \;(4(1) + 1)\; = \;(4(1) + 1) \hfill \\ \Rightarrow \;5\; = \;5\; = \;5 \hfill \\ \end{align}

जो हमेशा सत्य है, अर्थात $\lambda$ कि किसी भी मात्रा के लिए यह फलन ${\text{x}}\;{\text{ = }}\;1$ पर संतत है।

19. दर्शाइए की ${\text{g(x)}}\;{\text{ = }}\;{\text{x - }}\left[ {\text{x}} \right]$ द्वारा परिभाषित फलन समस्त पूर्णांक बिन्दुओ पर असंतत है। यहा $\left[ {\text{x}} \right]$ उस महतम पूर्णांक निरूपित करता है। जो ${\text{x}}$ के बराबर या ${\text{x}}$ से कम है।

उत्तर: ${\text{g(x)}}\;{\text{ = }}\;{\text{x - }}\left[ {\text{x}} \right]$

मान लेते है कि ${\text{n}}$ एक पूर्णांक बिन्दु को निरूपित करता है:

यदि ${\text{g(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{n}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{g(n)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{n}} + } {\text{g(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{n}} - } {\text{g(x)}} \hfill \\ \Rightarrow {\text{(n - n)}}\;{\text{ = }}\;{\text{(n - n)}}\;{\text{ = }}\;{\text{(n - (n - 1)}}) \hfill \\ \Rightarrow \;0\; = \;0\; = \;1 \hfill \\ \end{align}

जो सत्य नहीं हो सकता, अर्थात ${\text{g(x)}}$ किसी भी पूर्णांक बिन्दुक पर संतत नहीं है।

20. क्या ${\text{f (x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - sin(x) + 5}}$ द्वारा परिभाषित फलन ${\text{x = }}\;{\text{\pi }}$ पर संतत है।

उत्तर: ${\text{f (x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - sin(x) + 5}}$

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;\pi$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(\pi }})\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{\pi }} + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{\pi - }}} {\text{f(x)}} \hfill \\ \Rightarrow \;{\text{(}}{{\text{\pi }}^{\text{2}}}{\text{ - sin(\pi ) + 5)}}\; = \;{\text{(}}{{\text{\pi }}^{\text{2}}}{\text{ - sin(\pi ) + 5)}}\; = \;{\text{(}}{{\text{\pi }}^{\text{2}}}{\text{ - sin(\pi ) + 5)}} \hfill \\ \Rightarrow \;{{\text{\pi }}^{\text{2}}}{\text{ + 5}}\;{\text{ = }}\;{{\text{\pi }}^{\text{2}}}{\text{ + 5}}\; = \;{{\text{\pi }}^{\text{2}}}{\text{ + 5}} \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;\pi$ पर संतत है।

21. निम्नलिखित स्तनों के सातत्य पर विचार कीजिए:

(a) ${\text{sinx + cosx}}$

उत्तर: ${\text{f(x)}}$ = ${\text{sinx + cosx}}$

मान लेते है कि ${\text{c}}$ एक कोई भी असली अंक है।

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \;{\text{(sin}}\;{\text{c + cos}}\;{\text{c)}}\;{\text{ = }}\;{\text{(sin}}\;{\text{c + cos}}\;{\text{c)}}\; = \;{\text{(sin}}\;{\text{c + cos}}\;{\text{c)}} \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x)}}$ असली अंक रेखा के हर बिन्दु पर संतत है।

(b) ${\text{sinx - cosx}}$

उत्तर: ${\text{f(x)}}$${\text{sinx - cosx}}$

मान लेते है कि ${\text{c}}$ एक कोई भी असली अंक है।

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \;{\text{(sin}}\;{\text{c - cos}}\;{\text{c)}}\;{\text{ = }}\;{\text{(sin}}\;{\text{c - cos}}\;{\text{c)}}\; = \;{\text{(sin}}\;{\text{c - cos}}\;{\text{c)}} \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x)}}$ असली अंक रेखा के हर बिन्दु पर संतत है।

(c) ${\text{sinx}} \cdot {\text{cosx}}$

उत्तर: ${\text{f(x)}}$ = ${\text{sinx}} \cdot {\text{cosx}}$

मान लेते है कि ${\text{c}}$ एक कोई भी असली अंक है।

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \;{\text{(sin}}\;{\text{c}}{\text{.cos}}\;{\text{c)}}\;{\text{ = }}\;{\text{(sin}}\;{\text{c}}{\text{.cos}}\;{\text{c)}}\; = \;{\text{(sin}}\;{\text{c}}{\text{.cos}}\;{\text{c)}} \hfill \\ \end{align}जो सत्य है, अर्थात ${\text{f(x)}}$ असली अंक रेखा के हर बिन्दु पर संतत है।

22. cosine, cosecant, secant, cotangent फलनों के सातत्य पर विचार कीजिए।

उत्तर: (i) ${\text{f(x)}}$ = ${\text{cos}}\;{\text{x}}$

मान लेते है कि ${\text{c}}$ एक कोई भी असली अंक है।

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \;{\text{(cos}}\;{\text{c)}}\;{\text{ = }}\;{\text{(cos}}\;{\text{c)}}\; = \;{\text{(cos}}\;{\text{c)}} \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x)}}$ असली अंक रेखा के हर बिन्दु पर संतत है।

(ii) ${\text{f(x)}}$ = ${\text{cosec x}}$

मान लेते है कि ${\text{c}}$ एक कोई भी असली अंक है।

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \;{\text{(cosec}}\;{\text{c)}}\;{\text{ = }}\;{\text{(cosec}}\;{\text{c)}}\; = \;{\text{(cosec}}\;{\text{c)}} \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x)}}$ असली अंक रेखा के हर बिन्दु पर संतत है।

(iii) ${\text{f(x)}}$ = ${\text{sec}}\;{\text{x}}$

मान लेते है कि ${\text{c}}$ एक कोई भी असली अंक है।

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \;{\text{(sec}}\;{\text{c)}}\;{\text{ = }}\;{\text{(sec}}\;{\text{c)}}\; = \;{\text{(sec}}\;{\text{c)}} \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x)}}$ असली अंक रेखा के हर बिन्दु पर संतत है।

(iv) ${\text{f(x)}}$ = ${\text{cot x}}$

मान लेते है ${\text{c}}$ एक कोई भी असली अंक है इस प्रकार कि  ${\text{(n - 1)\pi < c < n\pi }}$ जहा ${\text{n}}$ एक पूर्णांक बिन्दु को निरूपित करता है।

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \;{\text{(cot}}\;{\text{c)}}\;{\text{ = }}\;{\text{(cot}}\;{\text{c)}}\; = \;{\text{(cot}}\;{\text{c)}} \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x)}}$ असली अंक रेखा के हर बिन्दु पर संतत है जो ${\text{(n - 1)\pi }}$ से ${\text{n\pi }}$ के बीच है।

अब यदि हम माने ${\text{c}}$ इस प्रकार है ${\text{c}}\;{\text{ = }}\;{\text{n\pi }}$ जहा ${\text{n}}$ एक पूर्णांक बिन्दु को निरूपित करता है, तो

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \; + \infty \;{\text{ = }}\; + \infty \; = \; + \infty \hfill \\ \end{align}

अर्थात ${\text{f(x)}}$ असली अंक रेखा के हर बिन्दु पर संतत है सिवाय उनके जो ${\text{n\pi }}$ बिन्दुओ के प्रकार है।

23. ${\text{f}}$ के सभी असातत्यता के बिन्दुओ को ज्ञात कीजिए जहा ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {\dfrac{{{\text{sinx}}}}{{\text{x}}}{\text{,}}}&{{\text{x < 0}}} \\ {{\text{x + 1,}}}&{{\text{x}} \geqslant {\text{0}}} \end{array}$

उत्तर: दृष्टिकोण 1:

मान लेते है कि ${\text{c}}$ एक असली अंक रेखा पर एक बिन्दु है और ${\text{c < 0}}$

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

$\dfrac{{{\text{sin}}\;{\text{c}}}}{{\text{c}}}$ संतत है, अर्थात ${\text{sin}}\;{\text{c}}$ और ${\text{c}}$ संतत फलन है, जोकि सच है।

अर्थात ${\text{f(x),}}\;{\text{c < 0}}$ के लिए संतत है।

दृष्टिकोण 2:

${\text{c}}\;{\text{ = }}\;{\text{0}}$

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \;1\;{\text{ = }}\;1\; = \;\dfrac{{\sin 0}}{0} \hfill \\ \Rightarrow \;1\; = \;1\; = \;1 \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{0}}$ पर संतत है।

दृष्टिकोण 3:

मान लेते है ${\text{c}}$ एक असली अंक रेखा पर एक बिन्दु है और ${\text{c < 0}}$

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

${\text{c + 1}}$ संतत है, जोकि सच है।

अर्थात ${\text{f(x),}}\;{\text{c < 0}}$ के लिए संतत है।

24. निर्धारित कीजिए कि फलन ${\text{f}}$ ,

${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{{\text{x}}^{\text{2}}}\sin \dfrac{1}{{\text{x}}}{\text{,}}}&{{\text{x}} \ne {\text{0}}} \\ {{\text{0 ,}}}&{{\text{x = 0}}} \end{array}$ द्वारा परिभाषित एक संतत फलन है।

उत्तर: दृष्टिकोण 1:

${\text{c}}\;{\text{ = }}\;{\text{0}}$

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \;0\; = \;0\; = \;0 \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{0}}$ पर संतत है।

दृष्टिकोण 2:

${\text{c}}\; \ne \;0$ और ${\text{c}}\; \subset \mathbb{R}$

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

${{\text{c}}^{\text{2}}}{\text{sin}}\dfrac{{\text{1}}}{{\text{c}}}$ संतत है, अर्थात ${\text{sin}}\dfrac{{\text{1}}}{{\text{c}}}$ और ${{\text{c}}^{\text{2}}}$ संतत फलन है, जोकि सच है।

अर्थात  ${\text{f(x),}}\;{\text{x}} \ne {\text{0}}$ पर भी संतत है।

25. ${\text{f}}$ के सातत्य की जांच कीजिए, जहा ${\text{f}}$ निम्नलिखित प्रकार से परिभाषित है

${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{sinx - cosx,}}}&{{\text{x}} \ne {\text{0}}} \\ {{\text{ - 1,}}}&{{\text{x = 0}}} \end{array}$

उत्तर: दृष्टिकोण 1:

${\text{c}}\;{\text{ = }}\;{\text{0}}$

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(c)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c + }}} f(x)\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{c - }}} f(x) \hfill \\ \Rightarrow \; - 1\; = \;\sin 0 - \cos 0\; = \; - \sin 0 - \cos 0 \hfill \\ \Rightarrow \; - 1\; = \; - 1\; = \; - 1 \hfill \\ \end{align}

जो सत्य है, अर्थात ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{0}}$ पर संतत है।

दृष्टिकोण 2:

${\text{c}}\; \ne \;0$ और ${\text{c}}\; \subset \mathbb{R}$

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{c}}$ पर संतत है, इसका तात्पर्य होगा:

${\text{sin}}\;{\text{c - cos}}\;{\text{c}}$ संतत है, अर्थात ${\text{sin}}\;{\text{c}}$ और $\cos \;{\text{c}}$ संतत फलन है, जोकि सच है।

अर्थात  ${\text{f(x),}}\;{\text{x}} \ne {\text{0}}$ पर भी संतत है।

26 से 29 मे ${\text{k}}$ के मानो को ज्ञात कीजिए ताकि प्रदत फलन निर्दिष्ट बिन्दु पर संतत हो:

26.${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {\dfrac{{{\text{k}}\;{\text{cosx}}}}{{{\text{\pi - 2x}}}}{\text{,}}}&{{\text{x}} \ne \dfrac{{\text{\pi }}}{{\text{2}}}} \\ {{\text{3,}}}&{{\text{x = }}\dfrac{{\text{\pi }}}{{\text{2}}}} \end{array}$ परिभाषित फलन ${\text{x = }}\dfrac{{\text{\pi }}}{{\text{2}}}$ पर

उत्तर: यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;\dfrac{{\text{\pi }}}{{\text{2}}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(}}\dfrac{{\text{\pi }}}{{\text{2}}})\; = \;\mathop {\lim }\limits_{{\text{x}} \to \dfrac{{\text{\pi }}}{{\text{2}}} + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to \dfrac{{\text{\pi }}}{{\text{2}}} - } {\text{f(x)}} \hfill \\ \Rightarrow \;{\text{3}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to \dfrac{{\text{\pi }}}{{\text{2}}} + } \dfrac{{{\text{kcosx}}}}{{{\text{\pi - 2x}}}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to \dfrac{{\text{\pi }}}{{\text{2}}} - } \dfrac{{{\text{kcosx}}}}{{{\text{\pi - 2x}}}} \hfill \\ \Rightarrow \;{\text{3}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to \dfrac{{\text{\pi }}}{{\text{2}}}} \dfrac{{{\text{kcosx}}}}{{{\text{\pi - 2x}}}} \hfill \\ \Rightarrow \;{\text{3}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to {\text{0}}} \dfrac{{{\text{kcos(}}\dfrac{\pi }{2} + {\text{h)}}}}{{{\text{\pi - 2(}}\dfrac{\pi }{2} + {\text{h)}}}} \hfill \\ \Rightarrow \;{\text{3}}\; = \;\dfrac{{\text{k}}}{{\text{2}}} \hfill \\ \Rightarrow \;{\text{k}}\;{\text{ = }}\;{\text{6}} \hfill \\ \end{align}

अर्थात ${\text{k}}\;{\text{ = }}\;{\text{6}}$ कि मात्रा के लिए यह फलन  ${\text{x}}\;{\text{ = }}\;\dfrac{{\text{\pi }}}{{\text{2}}}$ पर संतत है।

27. ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{k}}{{\text{x}}^{\text{2}}}{\text{,}}}&{{\text{x}} \leqslant {\text{2}}} \\ {{\text{3,}}}&{{\text{x > 2 }}} \end{array}$परिभाषित फलन ${\text{x = 2}}$ पर

उत्तर: यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{2}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(2)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 2 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 2 - } {\text{f(x)}} \hfill \\ \Rightarrow \;{\text{k(4)}}\;{\text{ = }}\;{\text{3}}\;{\text{ = }}\;{\text{k(4)}} \hfill \\ \Rightarrow \;{\text{3}}\;{\text{ = }}\;{\text{k(4)}} \hfill \\ \Rightarrow \;{\text{k}}\;{\text{ = }}\;\dfrac{{\text{3}}}{{\text{4}}} \hfill \\ \end{align}

अर्थात ${\text{k}}\;{\text{ = }}\;\dfrac{{\text{3}}}{{\text{4}}}$ कि मात्रा के लिए यह फलन ${\text{x}}\;{\text{ = }}\;{\text{2}}$ पर संतत है।

28. ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{kx + 1,}}}&{{\text{x}} \leqslant \pi } \\ {{\text{cosx,}}}&{{\text{x > }}\pi {\text{ }}} \end{array}$द्वारा परिभाषित फलन ${\text{x = }}\pi$ पर

उत्तर: यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{\pi }}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(\pi )}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to \pi + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to \pi - } {\text{f(x)}} \hfill \\ {\text{k(}}\pi {\text{) + 1}}\;{\text{ = }}\;{\text{cos(}}\pi {\text{)}}\;{\text{ = }}\;{\text{k(}}\pi {\text{) + 1}} \hfill \\ \end{align} \begin{align} {\text{k(}}\pi {\text{) + 1}}\;{\text{ = }}\;{\text{cos(}}\pi {\text{)}} \hfill \\ {\text{k(}}\pi {\text{) + 1}}\;{\text{ = }}\; - 1 \hfill \\ \end{align}

${\text{k}}\;{\text{ = }}\;\dfrac{{ - 2}}{\pi }$

अर्थात ${\text{k}}\;{\text{ = }}\;\dfrac{{ - 2}}{\pi }$ कि मात्रा के लिए यह फलन ${\text{x}}\;{\text{ = }}\;{\text{\pi }}$ पर संतत है।

29. ${\text{f(x) = \{ }}\begin{array}{*{20}{c}} {{\text{kx + 1,}}}&{{\text{x}} \leqslant 5} \\ {{\text{3x - 5,}}}&{{\text{x > 5}}} \end{array}$ द्वारा परिभाषित फलन ${\text{x = 5}}$ पर

उत्तर: यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;5$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(5)}}\;{\text{ = }}\;\mathop {{\text{lim}}}\limits_{{\text{x}} \to {\text{5 + }}} {\text{f(x)}}\;{\text{ = }}\;\mathop {{\text{lim}}}\limits_{{\text{x}} \to {\text{5 - }}} {\text{f(x)}} \hfill \\ {\text{k(5) + 1}}\;{\text{ = }}\;3{\text{(5) - 5}}\;{\text{ = }}\;{\text{k(5) + 1}} \hfill \\ {\text{k(5) + 1}}\;{\text{ = }}\;3{\text{(5) - 5}} \hfill \\ {\text{k(5)}}\;{\text{ = }}\;9 \hfill \\ {\text{k}}\;{\text{ = }}\;\dfrac{9}{5} \hfill \\ \end{align}

अर्थात ${\text{k}}\;{\text{ = }}\;\dfrac{9}{5}$ कि मात्रा के लिए यह फलन ${\text{x}}\;{\text{ = }}\;{\text{5}}$ पर संतत है।

30. ${\text{a,}}\;{\text{b}}$ के मानो को ज्ञात कीजिए ताकि

$f(x)=\left\{\begin{matrix} 5, & x\leq 2 \\ ax+b, & 2< x< 10\\ 21, & x\geq 10 \end{matrix}\right.$द्वारा परिभाषित एक संतत फलन हो।

उत्तर: क्योंकि ${\text{f(x)}}\;{\text{ = }}\;{\text{5,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{ax + b,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{21}}$ संतत फलन है,  ${\text{f(x),}}\;{\text{x < 2,}}\;{\text{2 < x < 10,}}\;{\text{x > 10}}$ पर संतत फलन पहले ही है।

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{2}}$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(2)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 2 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 2 - } {\text{f(x)}} \hfill \\ {\text{5 = }}\;{\text{a(2) + b}}\;{\text{ = }}\;{\text{5}} \hfill \\ \end{align}

${\text{a(2) + b}}\;{\text{ = }}\;{\text{5}}$ ……………………….(1)

यदि ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;10$ पर संतत है, इसका तात्पर्य होगा:

\begin{align} {\text{f(10)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 10 + } {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to 10 - } {\text{f(x)}} \hfill \\ {\text{21 = }}\;{\text{21}}\;{\text{ = }}\;{\text{a(10) + b}} \hfill \\ \end{align}

${\text{a(10) + b}}\;{\text{ = }}\;21$ ……………………….(2)

(2) - (1)

\begin{align} {\text{8a}}\;{\text{ = }}\;{\text{16}} \hfill \\ {\text{a}}\;{\text{ = }}\;{\text{2}} \hfill \\ \therefore \;{\text{(2)(2) + b}}\;{\text{ = }}\;{\text{5}} \hfill \\ {\text{b}}\;{\text{ = }}\;{\text{1}} \hfill \\ \end{align}

अर्थात ${\text{a}}\;{\text{ = }}\;{\text{2,}}\;{\text{b}}\;{\text{ = }}\;{\text{1}}$ कि मात्रा के लिए यह फलन ${\text{f(x)}}$ संतत है।

31. दर्शाइए कि ${\text{f(x)}}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}$ द्वारा परिभाषित फलन एक संतत फलन हो।

उत्तर: ज्ञात है ${\text{f(x)}}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}$

\begin{align} {\text{x}}\;{\text{ = }}\;{\text{c}} \in \mathbb{R}{\text{,}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to {\text{c}}} {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c}}} {\text{cos}}\;{{\text{x}}^{\text{2}}}\; = \;{\text{cos}}\;{{\text{c}}^{\text{2}}} \hfill \\ {\text{f(c)}}\;{\text{ = }}\;{\text{cos}}\;{{\text{c}}^{\text{2}}} \hfill \\ \end{align}

अतः ${\text{f(x)}}\;{\text{ = }}\;{\text{cos}}\;{{\text{x}}^{\text{2}}}$ एक संतत फलन है, इति सिद्धम

32. दर्शाइए कि ${\text{f(x)}}\;{\text{ = }}\;\left| {{\text{cosx}}} \right|$ द्वारा परिभाषित फलन एक संतत फलन है।

उत्तर: ज्ञात है ${\text{f(x)}}\;{\text{ = }}\;\left| {{\text{cosx}}} \right|$

\begin{align} {\text{x}}\;{\text{ = }}\;{\text{c}} \in \mathbb{R}{\text{,}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to {\text{c}}} {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c}}} \left| {{\text{cos}}\;{\text{x}}} \right|\; = \;\left| {{\text{cos}}\;{\text{c}}} \right| \hfill \\ {\text{f(c)}}\;{\text{ = }}\;\left| {{\text{cos}}\;{\text{c}}} \right| \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;{\text{c}} \in \mathbb{R}$ पर  ${\text{f}}$ एक संतत फलन है। इति सिद्धम

33. जाँचिए कि क्या $\sin \left| {\text{x}} \right|$ एक संतत फलन है।

उत्तर: माना ${\text{f(x)}}\;{\text{ = }}\;{\text{sin}}\left| {\text{x}} \right|$

\begin{align} {\text{x}}\;{\text{ = }}\;{\text{c}} \in \mathbb{R}{\text{,}} \hfill \\ \mathop {\lim }\limits_{{\text{x}} \to {\text{c}}} {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {\text{c}}} \sin \left| {\text{x}} \right|\; = \;\sin \left| {\;{\text{c}}} \right| \hfill \\ {\text{f(c)}}\;{\text{ = }}\;\sin \left| {\text{c}} \right| \hfill \\ \end{align}

अतः ${\text{x}}\;{\text{ = }}\;{\text{c}} \in \mathbb{R}$ पर ${\text{f}}$ एक संतत फलन है।

34. ${\text{f(x)}}\;{\text{ = }}\;\left| {\text{x}} \right|{\text{ - }}\left| {{\text{x + 1}}} \right|$ द्वारा परिभाषित फलन ${\text{f}}$ के सभी असातत्यता के बिन्दुओ को ज्ञात कीजिए।

उत्तर: ज्ञात है ${\text{f(x)}}\;{\text{ = }}\;\left| {\text{x}} \right|{\text{ - }}\left| {{\text{x + 1}}} \right|$

\begin{align} {\text{f(x)}}\;{\text{ = }}\;{\text{ - x + [ - (x + 1)]}}\;\;\;\;\,{\text{(x < - 1)}} \hfill \\ {\text{ = }}\;{\text{ - x + x + 1}}\;{\text{ = }}\;{\text{1}} \hfill \\ {\text{f(x)}}\;{\text{ = }}\;{\text{ - x - [(x + 1)]}}\;\;\;\;\,({\text{ - 1}} \leqslant {\text{x < 0)}} \hfill \\ {\text{ = }}\;{\text{ - x - x - 1}}\;{\text{ = }}\;{\text{ - 2x - 1}} \hfill \\ {\text{f(x)}}\;{\text{ = }}\;{\text{x - [(x + 1)]}}\;\;\;\;\,{\text{(x}} \geqslant 0) \hfill \\ {\text{ = }}\;{\text{x - x - 1}}\;{\text{ = }}\;{\text{ - 1}} \hfill \\ \end{align}

### प्रश्नावली 5.2

प्रश्न 1 से 8 के निम्नलिखित फलनों का अवकलन कीजिए:

1. ${\text{sin(}}{{\text{x}}^{\text{2}}}{\text{ + 5)}}$

उत्तर: मान लीजिए ${\text{y}}\;{\text{ = }}\;{\text{sin(}}{{\text{x}}^{\text{2}}}{\text{ + 5)}}$ तथा ${\text{u}}\;{\text{ = }}\;{\text{(}}{{\text{x}}^{\text{2}}}{\text{ + 5)}}$

फिर ${\text{y}}\;{\text{ = }}\;{\text{sin}}\;{\text{u}}$

अतः शृंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{dy}}}}{{{\text{du}}}} \times \dfrac{{{\text{du}}}}{{{\text{dx}}}}\; \hfill \\ {\text{ = }}\;\dfrac{{{\text{d(sin}}\;{\text{u)}}}}{{{\text{du}}}} \times \dfrac{{{\text{d(}}{{\text{x}}^{\text{2}}}{\text{ + 5)}}}}{{{\text{dx}}}}\; \hfill \\ {\text{ = }}\;{\text{cosu \times 2x + 0}}\; \hfill \\ {\text{ = }}\;{\text{2xcosu}}\; \hfill \\ {\text{ = }}\;{\text{2xcos}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 5}}} \right) \hfill \\ \end{align}

2. ${\text{cos(sin}}\;{\text{x)}}$

उत्तर: मान लीजिए ${\text{y}}\;{\text{ = }}\;\cos ({\text{sin}}\;{\text{x)}}$ तथा ${\text{u}}\;{\text{ = }}\;{\text{sin}}\;{\text{x}}$

फिर ${\text{y}}\;{\text{ = }}\;{\text{cos}}\;{\text{u}}$

अतः शृंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{dy}}}}{{{\text{du}}}} \times \dfrac{{{\text{du}}}}{{{\text{dx}}}}\; \hfill \\ {\text{ = }}\;\dfrac{{{\text{d(cos}}\;{\text{u)}}}}{{{\text{du}}}} \times \dfrac{{{\text{d(sin}}\;{\text{x)}}}}{{{\text{dx}}}}\; \hfill \\ {\text{ = }}\;{\text{ - sinu \times cos x}}\; \hfill \\ {\text{ = }}\;{\text{ - sin(sinx)cos x}} \hfill \\ {\text{ = }}\;{\text{ - cosx}}\;{\text{sin(sin}}\;{\text{x)}} \hfill \\ \end{align}

3. ${\text{sin(ax + b)}}$

उत्तर: मान लीजिए ${\text{y}}\;{\text{ = }}\;$ ${\text{sin(ax + b)}}$  तथा ${\text{u}}\;{\text{ = }}\;{\text{ax + b}}$

फिर ${\text{y}}\;{\text{ = }}\;\sin \;{\text{u}}$

अतः शृंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{dy}}}}{{{\text{du}}}} \times \dfrac{{{\text{du}}}}{{{\text{dx}}}}\; \hfill \\ {\text{ = }}\;\dfrac{{{\text{d(sin}}\;{\text{u)}}}}{{{\text{du}}}} \times \dfrac{{{\text{d(ax + b)}}}}{{{\text{dx}}}}\; \hfill \\ {\text{ = }}\;{\text{cosu \times a \times 1 + 0}}\; \hfill \\ {\text{ = }}\;{\text{acos}}\;{\text{u}} \hfill \\ {\text{ = }}\;{\text{acos(ax + b)}} \hfill \\ \end{align}

4. ${\text{sec(tan(}}\sqrt {\text{x}} {\text{))}}$

उत्तर: मान लीजिए ${\text{y}}\;{\text{ = }}\;$${\text{sec(tan(}}\sqrt {\text{x}} {\text{))}}$  तथा ${\text{x}}\;{\text{ = }}\;{\text{u,}}\;\sqrt {\text{u}} \;{\text{ = }}\;{\text{t,}}\;{\text{tan}}\;{\text{t}}\;{\text{ = }}\;{\text{v}}$

फिर ${\text{y}}\;{\text{ = }}\;{\text{sec}}\;{\text{v}}$

अतः शृंखला नियम द्वारा

\begin{align} {\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/dv) \times (dv/dt) \times (dt/du) \times (du/dx)}}\; \hfill \\ {\text{ = }}\;{\text{(d(sec}}\;{\text{v)/dv) \times (d(tan}}\;{\text{t)/dt) \times (d}}\sqrt {\text{u}} {\text{/du) \times (dx/dx)}}\; \hfill \\ {\text{ = }}\;{\text{sec}}\;{\text{v}}\;{\text{tan}}\;{\text{v \times se}}{{\text{c}}^{\text{2}}}{\text{t \times 1/2}}\sqrt {\text{u}} {\text{ \times 1}}\; \hfill \\ {\text{ = }}\;{\text{sec(tan}}\;{\text{t)tan(tan}}\;{\text{t) \times se}}{{\text{c}}^{\text{2}}}\sqrt {\text{u}} {\text{ \times 1/2}}\sqrt {\text{x}} \; \hfill \\ {\text{ = }}\;{\text{sec(tan}}\sqrt {\text{x}} {\text{) \times tan(tan}}\sqrt {\text{x}} {\text{) \times se}}{{\text{c}}^{\text{2}}}\sqrt {\text{x}} {\text{/2}}\sqrt {\text{x}} \hfill \\ \end{align}

5. $\dfrac{{{\text{sin(ax + b)}}}}{{{\text{cos(cx + d)}}}}$

उत्तर: मान लीजिए ${\text{y}}\;{\text{ = }}\;$$\dfrac{{{\text{sin(ax + b)}}}}{{{\text{cos(cx + d)}}}}$

भागफल नियम द्वारा

= $\dfrac{{{\text{cos(cx + d)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sin(ax + b)) - sin(ax + b)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cos(cx + d))}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(cx + d)}}}}$

शृंखला नियम द्वारा

\begin{align} \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sin(ax + b)}}\;{\text{ = }}\;{\text{cos(ax + b)}}{\text{.}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(ax + b)}} \hfill \\ {\text{ = }}\;{\text{cos(ax + b)}}{\text{.a}} \hfill \\ {\text{ = }}\;{\text{acos(ax + b)}} \hfill \\ \end{align} \begin{align} \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cos(cx + d)}}\;{\text{ = }}\;{\text{ - sin(cx + d)}}{\text{.}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cx + d)}} \hfill \\ {\text{ = }}\;{\text{ - sin(cx + d)}}{\text{.c}} \hfill \\ {\text{ = }}\;{\text{ - csin(cx + d)}} \hfill \\ \end{align}

$\therefore \;\dfrac{{{\text{cos(cx + d)}}{\text{.acos(ax + b) + sin(ax + b)}}{\text{.csin(cx + d)}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(cx + d)}}}}$

\begin{align} {\text{ = }}\;\dfrac{{{\text{acos(ax + b)cos(cx + d)}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(cx + d)}}}}{\text{ + }}\dfrac{{{\text{csin(cx + d)sin(ax + b)}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(cx + d)}}}} \hfill \\ {\text{ = }}\;\dfrac{{{\text{acos(ax + b)}}}}{{{\text{cos(cx + d)}}}}{\text{ + }}\dfrac{{{\text{csin(cx + d)sin(ax + b)}}}}{{{\text{cos(cx + d)}}}}{\text{ \times cos(cx + d)}}\; \hfill \\ {\text{ = }}\;{\text{acos(ax + b) \times sec(cx + d) + csin(ax + b) \times tan(cx + d) \times sec(cx + d)}} \hfill \\ \end{align}

6. ${\text{cos}}\;{{\text{x}}^{\text{3}}}{\text{.si}}{{\text{n}}^2}{{\text{x}}^{\text{5}}}$

उत्तर: मान लीजिए ${\text{y}}\;{\text{ = }}\;$${\text{cos}}\;{{\text{x}}^{\text{3}}}{\text{.si}}{{\text{n}}^2}{{\text{x}}^{\text{5}}}$

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{cos}}{{\text{x}}^{\text{3}}}{\text{ \times si}}{{\text{n}}^{\text{2}}}{\text{(}}{{\text{x}}^5}{\text{)}}} \right)$

भागफल नियम द्वारा

${\text{ = }}\;{\text{cos}}{{\text{x}}^{\text{3}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{si}}{{\text{n}}^{\text{2}}}{\text{(}}{{\text{x}}^{\text{5}}}{\text{) + si}}{{\text{n}}^{\text{2}}}{\text{(}}{{\text{x}}^{\text{5}}}{\text{)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cos}}{{\text{x}}^{\text{3}}}\;$

शृंखला नियम द्वारा

\begin{align} {\text{ = }}\;{\text{cos}}{{\text{x}}^{\text{3}}}{\text{(2sin(}}{{\text{x}}^{\text{5}}}{\text{)) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sin(}}{{\text{x}}^{\text{5}}}{\text{) + si}}{{\text{n}}^{\text{2}}}{\text{(}}{{\text{x}}^{\text{5}}}{\text{)}}\left( {{\text{ - sin}}{{\text{x}}^{\text{3}}}} \right) \times \dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{3}}}\; \hfill \\ {\text{ = }}\;{\text{cos}}{{\text{x}}^{\text{3}}}{\text{(2sin(}}{{\text{x}}^{\text{5}}}{\text{)) \times cos(}}{{\text{x}}^{\text{5}}}{\text{)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{5}}}{\text{ + si}}{{\text{n}}^{\text{2}}}{\text{(}}{{\text{x}}^{\text{5}}}{\text{)}}\left( {{\text{ - sin}}{{\text{x}}^{\text{3}}}} \right){\text{3}}{{\text{x}}^{\text{2}}}\; \hfill \\ {\text{ = }}\;{\text{cos}}{{\text{x}}^{\text{3}}}{\text{2sin(}}{{\text{x}}^{\text{5}}}{\text{)cos}}{{\text{x}}^{\text{5}}}{\text{ \times 5}}{{\text{x}}^{\text{4}}}{\text{ + si}}{{\text{n}}^{\text{2}}}{\text{(}}{{\text{x}}^{\text{5}}}{\text{)}}\left( {{\text{ - sin}}{{\text{x}}^{\text{3}}}} \right){\text{3}}{{\text{x}}^{\text{2}}}\, \hfill \\ {\text{ = }}\;{\text{10}}{{\text{x}}^{\text{4}}}{\text{sin}}{{\text{x}}^{\text{5}}}{\text{cos}}{{\text{x}}^{\text{5}}}{\text{cos}}{{\text{x}}^{\text{3}}}{\text{ - sin}}{{\text{x}}^{\text{3}}}{\text{si}}{{\text{n}}^{\text{2}}}{{\text{x}}^{\text{5}}} \hfill \\ \end{align}

7. ${\text{2}}\sqrt {{\text{cot(}}{{\text{x}}^{\text{2}}}{\text{)}}}$

उत्तर: मान लीजिए ${\text{y}}\;{\text{ = }}\;$${\text{2}}\sqrt {{\text{cot(}}{{\text{x}}^{\text{2}}}{\text{)}}}$ , ${\text{u}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}$ , ${\text{cot}}\;{\text{u}}\;{\text{ = }}\;{\text{t}}$

फिर ${\text{y}}\;{\text{ = }}\;\sqrt {\text{t}}$

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{dy}}}}{{{\text{dt}}}} \times \dfrac{{{\text{dt}}}}{{{\text{du}}}} \times \dfrac{{{\text{du}}}}{{{\text{dx}}}} \hfill \\ \;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{2}}\sqrt {\text{t}} \times \dfrac{{\text{d}}}{{{\text{du}}}}{\text{cotu \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{2}}} \hfill \\ \end{align}

\begin{align} {\text{ = }}\;\dfrac{{\text{2}}}{{{\text{2}}\sqrt {\text{t}} }} \times \left( {{\text{ - cose}}{{\text{c}}^{\text{2}}}{\text{u}}} \right){\text{ \times 2x}} \hfill \\ \;{\text{ = }}\;\dfrac{{\text{1}}}{{\sqrt {\text{t}} }} \times \left( {{\text{ - cose}}{{\text{c}}^{\text{2}}}{\text{u}}} \right){\text{ \times 2x}}\; \hfill \\ \end{align}

${\text{ = }}\;\dfrac{{\text{1}}}{{\sqrt {{\text{cot}}{{\text{x}}^{\text{2}}} \times \left( {{\text{ - cose}}{{\text{c}}^{\text{2}}}{{\text{x}}^{\text{2}}}} \right){\text{ \times 2x}}} }}\;$

${\text{ = }}\;{\text{ - 2xcose}}{{\text{c}}^{\text{2}}}\left( {{{\text{x}}^{\text{2}}}} \right){\text{/}}\sqrt {{\text{cot}}} \left( {{{\text{x}}^{\text{2}}}} \right)$

8. ${\text{cos(}}\sqrt {\text{x}} {\text{)}}$

उत्तर: मान लीजिए ${\text{y}}\;{\text{ = }}\;$${\text{cos(}}\sqrt {\text{x}} {\text{)}}$ , $\sqrt {\text{x}} \;{\text{ = }}\;{\text{u}}$

फिर ${\text{y}}\;{\text{ = }}\;{\text{cos}}\;{\text{u}}$

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{dy}}}}{{{\text{du}}}} \times \dfrac{{{\text{du}}}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;\dfrac{{\text{d}}}{{{\text{du}}}}{\text{(cos}}\;{\text{u) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}\sqrt {\text{x}} \hfill \\ {\text{ = }}\;{\text{ - sin}}\;{\text{u \times }}\dfrac{{\text{1}}}{{{\text{2}}\sqrt {\text{x}} }} \hfill \\ {\text{ = }}\;{\text{ - }}\dfrac{{{\text{sin}}\sqrt {\text{x}} }}{{{\text{2}}\sqrt {\text{x}} }} \hfill \\ \end{align}

9. सिद्ध कीजिए कि फलन  ${\text{f(x)}}\;{\text{ = }}\;\left| {{\text{x - 1}}} \right|{\text{,}}\;{\text{x}} \in {\text{R,}}\;{\text{x}}\;{\text{ = }}\;{\text{1}}$ पर अवकलित नहीं है।

उत्तर: कोई भी फलन अवकलित नहीं होगा जब बाए पक्ष कि सीमा तथा दाए पक्ष कि सीमा बराबर नहीं होगी।

\begin{align} {\text{f(x)}}\;{\text{ = }}\;\left| {{\text{x - 1}}} \right|{\text{,}}\;{\text{x}} \in \mathbb{R} \hfill \\ {\text{f(x)}}\;{\text{ = }}\;{\text{x - 1,}}\;{\text{x - 1 > 0}} \hfill \\ {\text{f(x)}}\;{\text{ = }}\;{\text{ - (x - 1),}}\;{\text{x - 1 < 0}} \hfill \\ {\text{x}}\;{\text{ = }}\;{\text{1 ;}} \hfill \\ {\text{f(1)}}\;{\text{ = }}\;{\text{1 - 1}}\;{\text{ = }}\;{\text{0}} \hfill \\ \end{align}

LHL: $\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{{{\text{f(1 - h) - f(1)}}}}{{{\text{ - h}}}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{{1 - (1 - {\text{h) - 0}}}}{{ - {\text{h}}}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{{\text{h}}}{{{\text{ - h}}}} \hfill \\ = \; - 1 \hfill \\ \end{align}

RHL: $\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{{{\text{f(1 + h) - f(1)}}}}{{\text{h}}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{{(1 + {\text{h) - 1 - 0}}}}{{\text{h}}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{{\text{h}}}{{\text{h}}} \hfill \\ = \;1 \hfill \\ \end{align}

बाए पक्ष कि सीमा तथा दाए पक्ष कि सीमा बराबर नहीं है। अतः ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{1}}$ पर अवकलित नहीं है।

10. सिद्ध कीजिए कि महतम पूर्णांक फलन ${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right],\;0{\text{ < x < 3,}}\;{\text{x}}\;{\text{ = }}\;{\text{1}}$ तथा ${\text{x}}\;{\text{ = }}\;2$ पर अवकलित नहीं है।

उत्तर: कोई भी फलन अवकलित नहीं होगा जब बाए पक्ष कि सीमा तथा दाए पक्ष कि सीमा बराबर नहीं होगी।

${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right]{\text{,}}\;{\text{0 < x < 3}}$

(i) ${\text{x}}\;{\text{ = }}\;{\text{1}}$

LHL: $\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{{{\text{f(1 - h) - f(1)}}}}{{{\text{ - h}}}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{{(1 - {\text{h) - }}\left[ 1 \right]}}{{{\text{ - h}}}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{{{\text{0 - 1}}}}{{{\text{ - h}}}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{1}{{\text{h}}} \hfill \\ = \;\infty \hfill \\ \end{align}RHL: $\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{{{\text{f(1 + h) - f(1)}}}}{{\text{h}}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{{(1 + {\text{h) - }}\left[ 1 \right]}}{{\text{h}}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{{{\text{1 - 1}}}}{{\text{h}}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{0}{{\text{h}}} \hfill \\ = \;0 \hfill \\ \end{align}

बाए पक्ष कि सीमा तथा दाए पक्ष कि सीमा बराबर नहीं है। अतः ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;{\text{1}}$ पर अवकलित नहीं है।

(ii) ${\text{x}}\;{\text{ = }}\;2$

LHL: $\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{{{\text{f(2 - h) - f(2)}}}}{{{\text{ - h}}}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{{\left[ {2 - {\text{h}}} \right] - \left[ 2 \right]}}{{{\text{ - h}}}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0 - } \dfrac{{1 - 2}}{{{\text{ - h}}}} \hfill \\ = \;\infty \hfill \\ \end{align}

RHL: $\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{{{\text{f(2 + h) - f(2)}}}}{{\text{h}}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{{\left[ {2 + {\text{h}}} \right]{\text{ - }}\left[ 2 \right]}}{{\text{h}}}$

\begin{align} = \;\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{{2 - 2}}{{\text{h}}} \hfill \\ = \;\mathop {\lim }\limits_{{\text{h}} \to 0 + } \dfrac{0}{{\text{h}}} \hfill \\ = \;0 \hfill \\ \end{align}

बाए पक्ष कि सीमा तथा दाए पक्ष कि सीमा बराबर नहीं है। अतः ${\text{f(x),}}\;{\text{x}}\;{\text{ = }}\;2$ पर अवकलित नहीं है।

### प्रश्नावली 5.3

निम्नलिखित प्रश्नों मे $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ ज्ञात कीजिए:

1. ${\text{2x + 3y}}\;{\text{ = }}\;{\text{sin}}\;{\text{x}}$

उत्तर: दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \Rightarrow \;{\text{2}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x) + 3}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(y)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sinx)}} \hfill \\ \Rightarrow \;{\text{2 \times 1 + 3}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{cosx}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{cosx - 2}}}}{{\text{3}}} \hfill \\ \end{align}

2. ${\text{2x + 3y}}\;{\text{ = }}\;{\text{sin}}\;{\text{y}}$

उत्तर: दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \Rightarrow \;{\text{2}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x) + 3}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(y)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(siny)}} \hfill \\ \Rightarrow \;{\text{2 \times 1 + 3}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{cos}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}} \hfill \\ \Rightarrow \;{\text{2}}\;{\text{ = }}\;\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(cosy - 3)}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{2}}}{{{\text{cosy - 3}}}} \hfill \\ \end{align}

3. ${\text{ax + b}}{{\text{y}}^{\text{2}}}\;{\text{ = }}\;{\text{cos}}\;{\text{y}}$

उत्तर: दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \Rightarrow \;{\text{a}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x) + b}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(}}{{\text{y}}^{\text{2}}}{\text{)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosy)}} \hfill \\ \Rightarrow \;{\text{a \times 1 + 2by}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{ - siny}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}} \hfill \\ \Rightarrow \;{\text{a + }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(2\;by + siny)}}\;{\text{ = }}\;{\text{0}} \hfill \\ \end{align}

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{ - a}}}}{{{\text{2by + sin}}\;{\text{y}}}}$

4. ${\text{xy + }}{{\text{y}}^{\text{2}}}\;{\text{ = }}\;{\text{tanx + y}}$

उत्तर: दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \Rightarrow \;{\text{x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(y) + y}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x) + }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(}}{{\text{y}}^{\text{2}}}{\text{)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(tanx) + }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(y)}} \hfill \\ \Rightarrow \;{\text{x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + y + 2y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{se}}{{\text{c}}^{\text{2}}}{\text{x + }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}} \hfill \\ \Rightarrow \;\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(x + 2y - 1)}}\;{\text{ = }}\;{\text{se}}{{\text{c}}^{\text{2}}}{\text{x - y}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{se}}{{\text{c}}^{\text{2}}}{\text{x - y}}}}{{{\text{x + 2y - 1}}}} \hfill \\ \end{align}

5. ${{\text{x}}^{\text{2}}}{\text{ + xy + }}{{\text{y}}^{\text{2}}}\;{\text{ = }}\;{\text{100}}$

उत्तर: दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \Rightarrow \;\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^{\text{2}}}} \right){\text{ + }}\left\{ {{\text{x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + y}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x)}}} \right\}{\text{ + }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(}}{{\text{y}}^{\text{2}}}{\text{)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(1}}00) \hfill \\ \Rightarrow \;{\text{2x + x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + y \times 1 + 2y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\; = \;0 \hfill \\ \Rightarrow \;{\text{2x + x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + y + 2y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;0 \hfill \\ \Rightarrow \;{\text{x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + 2y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{ - 2x - }}1 \hfill \\ \Rightarrow \;\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(x + 2y)}}\;{\text{ = }}\;{\text{ - (2x + y)}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{ - }}\dfrac{{{\text{2x + y}}}}{{{\text{x + 2y}}}} \hfill \\ \end{align}

6. ${{\text{x}}^{\text{3}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{y + x}}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{3}}}\;{\text{ = }}\;{\text{81}}$

उत्तर: दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^{\text{3}}}} \right){\text{ + }}\left\{ {{{\text{x}}^{\text{2}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + y}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(}}{{\text{x}}^{\text{2}}}{\text{)}}} \right\}{\text{ + }}\left\{ {{\text{x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\left( {{{\text{y}}^{\text{2}}}} \right){\text{ + }}{{\text{y}}^{\text{2}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x)}}} \right\}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{y}}^{\text{3}}}} \right)\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(81)}} \hfill \\ \Rightarrow \;{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{x}}^{\text{2}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + y \times 2x + x}}{\text{.2y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ \times 1 + 3}}{{\text{y}}^{\text{2}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\; = \;0 \hfill \\ \Rightarrow \;{{\text{x}}^{\text{2}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + x}}{\text{.2y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + 3}}{{\text{y}}^{\text{2}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{ - }}\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 2xy + }}{{\text{y}}^{\text{2}}}} \right) \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{ - }}\dfrac{{\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 2xy + }}{{\text{y}}^{\text{2}}}} \right)}}{{{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 2xy + 3}}{{\text{y}}^{\text{2}}}}} \hfill \\ \end{align}

7. ${\text{si}}{{\text{n}}^{\text{2}}}{\text{y + cos}}\;{\text{xy}}\;{\text{ = }}\;{\text{k}}$

उत्तर: दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

$\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{y}}} \right){\text{ + }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosxy)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(k)}}$

\begin{align} \Rightarrow \;{\text{2sinycosy}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + ( - sinxy)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(xy)}}\;{\text{ = }}\;{\text{0}} \hfill \\ \Rightarrow \;{\text{2sinycosy}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ - sinxy}}\left[ {{\text{x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + y}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x)}}} \right]\;{\text{ = }}\;{\text{0}} \hfill \\ \Rightarrow \;{\text{2sinycosy}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ - xsinxy}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ - ysinxy}}\; = \;0 \hfill \\ \Rightarrow \;{\text{sin2y - xsinxy}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ - ysinxy}}\;{\text{ = }}\;0 \hfill \\ \Rightarrow \;\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{(sin2y - xsinxy)}}\;{\text{ = }}\;{\text{ysinxy}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{ysinxy}}}}{{{\text{(sin2y - xsinxy)}}}} \hfill \\ \end{align}

8. ${\text{si}}{{\text{n}}^{\text{2}}}{\text{x + co}}{{\text{s}}^{\text{2}}}{\text{y}}\;{\text{ = }}\;{\text{1}}$

उत्तर: दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}} \right){\text{ + }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{co}}{{\text{s}}^{\text{2}}}{\text{y}}} \right)\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(1)}} \hfill \\ \Rightarrow {\text{2sinx}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sinx) + 2cosy}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosy)}}\;{\text{ = }}\;0 \hfill \\ \Rightarrow {\text{2sinxcosx - 2cosy( - siny)}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{0}} \hfill \\ \Rightarrow {\text{2sinxcosx + 2 cosy siny}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\,{\text{ = }}\;0 \hfill \\ \Rightarrow {\text{sin2x - sin2y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{0}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{sin2x}}}}{{{\text{(sin2y)}}}} \hfill \\ \end{align}

9. ${\text{y}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(}}\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{)}}$

उत्तर: ${\text{x}}\;{\text{ = }}\;{\text{tan\theta ,}}\;{\text{\theta }}\;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

\begin{align} {\text{y}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2tan\theta }}}}{{{\text{1 + ta}}{{\text{n}}^{\text{2}}}{\text{\theta }}}}} \right) \hfill \\ {\text{y}}\,\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(sin2\theta )}}\;{\text{ = }}\;{\text{2\theta }}\,{\text{ = }}\,{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \end{align}

दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\,{\text{ = }}\;{\text{2}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right) \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{2}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}} \hfill \\ \end{align}

10. ${\text{y}}\;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(}}\dfrac{{{\text{3x - }}{{\text{x}}^{\text{3}}}}}{{{\text{1 - 3}}{{\text{x}}^{\text{2}}}}}{\text{),}}\;{\text{ - }}\dfrac{{\text{1}}}{{\sqrt {\text{3}} }}{\text{ < x < }}\dfrac{{\text{1}}}{{\sqrt {\text{3}} }}$

उत्तर: ${\text{x}}\;{\text{ = }}\;{\text{tan\theta ,}}\;{\text{\theta }}\;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

${\text{y}}\;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3tan\theta - ta}}{{\text{n}}^{\text{2}}}{\text{\theta }}}}{{{\text{1 - 3ta}}{{\text{n}}^{\text{2}}}{\text{\theta }}}}} \right)\;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{tan3\theta }}\;{\text{ = }}\;{\text{3\theta }}\;{\text{ = }}\;{\text{3ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{3}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right) \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{3}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}} \hfill \\ \end{align}

11. ${\text{y}}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{(}}\dfrac{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{),}}\;{\text{0 < x < 1}}$

उत्तर: ${\text{x}}\;{\text{ = }}\;{\text{tan\theta ,}}\;{\text{\theta }}\;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

${\text{y}}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1 - ta}}{{\text{n}}^{\text{2}}}{\text{\theta }}}}{{{\text{1 + ta}}{{\text{n}}^{\text{2}}}{\text{\theta }}}}} \right)\;{\text{ = }}\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{cos2\theta }}\;{\text{ = }}\;{\text{2\theta }}\;{\text{ = }}\;{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{2}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right) \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{2}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{y}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} \right) \hfill \\ \end{align}

12. ${\text{y}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(}}\dfrac{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{),}}\;{\text{0 < x < 1}}$

उत्तर: ${\text{x}}\;{\text{ = }}\;{\text{tan\theta ,}}\;{\text{\theta }}\;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

\begin{align} {\text{y}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{1 - ta}}{{\text{n}}^{\text{2}}}{\text{\theta }}}}{{{\text{1 + ta}}{{\text{n}}^{\text{2}}}{\text{\theta }}}}} \right)\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{sin}}\left( {\dfrac{{\text{\pi }}}{{\text{2}}}{\text{ - 2\theta }}} \right)\;{\text{ = }}\;\dfrac{{\text{\pi }}}{{\text{2}}}{\text{ - 2\theta }}\;{\text{ = }}\;\dfrac{{\text{\pi }}}{{\text{2}}}{\text{ - 2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ {\text{y}}\;{\text{ = }}\;\dfrac{{\text{\pi }}}{{\text{2}}}{\text{ - 2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \end{align}

दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\; = \;\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{\pi }{2}} \right) - 2\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right) \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\; = \;0 - \dfrac{2}{{1 + {x^2}}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\; = \; - \dfrac{2}{{1 + {x^2}}} \hfill \\ \end{align}

13. ${\text{y}}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{(}}\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{),}}\;{\text{ - 1 < x < 1}}$

उत्तर: ${\text{x}}\;{\text{ = }}\;{\text{tan\theta ,}}\;{\text{\theta }}\;{\text{ = }}\;{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

\begin{align} {\text{y}}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2tan\theta }}}}{{{\text{1 + ta}}{{\text{n}}^{\text{2}}}{\text{\theta }}}}} \right)\; = \;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{sin(2\theta ) = co}}{{\text{s}}^{{\text{ - 1}}}}{\text{cos}}\left( {\dfrac{{\text{\pi }}}{{\text{2}}}{\text{ - 2\theta }}} \right)\; = \;\dfrac{{\text{\pi }}}{{\text{2}}} - 2\theta \; = \;\dfrac{\pi }{2} - 2{\tan ^{ - 1}}x \hfill \\ {\text{y}}\;{\text{ = }}\;\dfrac{{\text{\pi }}}{{\text{2}}}{\text{ - 2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \end{align}

दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{\text{\pi }}}{{\text{2}}}} \right){\text{ - 2}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right) \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{0 - }}\dfrac{{\text{2}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{ - }}\dfrac{{\text{2}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}} \hfill \\ \end{align}

14. ${\text{y}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(2x}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{),}}\;{\text{ - }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}{\text{ < x < }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}$

उत्तर: ${\text{x}}\;{\text{ = }}\;{\text{sin\theta ,}}\;{\text{\theta }}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

\begin{align} {\text{y}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{\text{2sin\theta }}\sqrt {{\text{1 - si}}{{\text{n}}^{\text{2}}}{\text{\theta }}} } \right)\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(2sin\theta cos\theta )}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(sin2\theta )}}\;{\text{ = }}\;{\text{2\theta }}\;{\text{ = }}\;{\text{2si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ {\text{y}}\;{\text{ = }}\;{\text{2si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \end{align}

दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{2si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)\;{\text{ = }}\;\dfrac{{\text{2}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}\;\;\;\;\;(\because \;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}{\text{)}}$

15. ${\text{y}}\;{\text{ = }}\;{\text{se}}{{\text{c}}^{{\text{ - 1}}}}{\text{(}}\dfrac{{\text{1}}}{{{\text{2}}{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{\text{),}}\;{\text{0 < x < }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}$

उत्तर: ${\text{x}}\;{\text{ = }}\;{\text{cos\theta ,}}\;{\text{\theta }}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}$

\begin{align} {\text{y}}\;{\text{ = }}\;{\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{{\text{2co}}{{\text{s}}^{\text{2}}}{\text{\theta - 1}}}}} \right)\;{\text{ = }}\;{\text{se}}{{\text{c}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{{\text{cos2\theta }}}}} \right)\;{\text{ = }}\;{\text{se}}{{\text{c}}^{{\text{ - 1}}}}{\text{(sec2\theta )}}\;{\text{ = }}\;{\text{2\theta }}\;{\text{ = }}\;{\text{2co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ {\text{y}}\;{\text{ = }}\;{\text{2co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \end{align}

दोनों पक्षों का ${\text{x}}$ के सापेक्ष अवकलन करने पर,

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{2co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right)\;{\text{ = }}\;\dfrac{{{\text{ - 2}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}$

### प्रश्नावली 5.4

निम्नलिखित ${\text{x}}$ के सापेक्ष अवकलन कीजिए:

1. $\dfrac{{{{\text{e}}^{\text{x}}}}}{{{\text{sin}}\;{\text{x}}}}$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$$\dfrac{{{{\text{e}}^{\text{x}}}}}{{{\text{sin}}\;{\text{x}}}}$ है। अब श्रंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{{\text{e}}^{\text{x}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sinx - sinx \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{e}}^{\text{x}}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}}} \hfill \\ {\text{ = }}\;\dfrac{{{{\text{e}}^{\text{x}}}{\text{ \times cosx - sinx \times }}{{\text{e}}^{\text{x}}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}}} \hfill \\ {\text{ = }}\;\dfrac{{{{\text{e}}^{\text{x}}}{\text{(cosx - sinx)}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}}} \hfill \\ \end{align}

2. ${{\text{e}}^{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$${{\text{e}}^{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}$ है। अब श्रंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{e}}^{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ {\text{ = }}\;{{\text{e}}^{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}} \times \dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }} \hfill \\ {\text{ = }}\;\dfrac{{{{\text{e}}^{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }} \hfill \\ \end{align}

3. ${{\text{e}}^{{{\text{x}}^{\text{3}}}}}$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$${{\text{e}}^{{{\text{x}}^{\text{3}}}}}$ है। अब श्रंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{e}}^{{{\text{x}}^{\text{3}}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{3}}} \hfill \\ {\text{ = }}\;{{\text{e}}^{{{\text{x}}^{\text{3}}}}}{\text{ \times 3}}{{\text{x}}^{\text{2}}} \hfill \\ {\text{ = }}\;{\text{3}}{{\text{x}}^{\text{2}}}{{\text{e}}^{{{\text{x}}^{\text{3}}}}} \hfill \\ \end{align}

4. ${\text{sin}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{e}}^{{\text{ - x}}}}} \right)$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$${\text{sin}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{e}}^{{\text{ - x}}}}} \right)$ है। अब श्रंखला नियम द्वारा

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{cos}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{e}}^{{\text{ - x}}}}} \right) \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{e}}^{{\text{ - x}}}}$

${\text{ = }}\;{\text{cos}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{e}}^{{\text{ - x}}}}} \right) \times \dfrac{{\text{1}}}{{{\text{1 + }}{{\left( {{{\text{e}}^{{\text{ - x}}}}} \right)}^{\text{2}}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{e}}^{{\text{ - x}}}}$

\begin{align} {\text{ = }}\;{\text{cos}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{e}}^{{\text{ - x}}}}} \right) \times \dfrac{{\text{1}}}{{{\text{1 + }}{{\text{e}}^{{\text{ - 2x}}}}}} \times \left( {{\text{ - }}{{\text{e}}^{{\text{ - x}}}}} \right) \hfill \\ {\text{ = }}\;{\text{ - }}{{\text{e}}^{{\text{ - x}}}}{\text{cos}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{e}}^{{\text{ - 1}}}}} \right)\dfrac{1}{{\left( {{\text{1 + }}{{\text{e}}^{{\text{ - 2x}}}}} \right)}} \hfill \\ \end{align}

5. ${\text{log}}\left( {{\text{cos}}\;{{\text{e}}^{\text{x}}}} \right)$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$${\text{log}}\left( {{\text{cos}}\;{{\text{e}}^{\text{x}}}} \right)$ है। अब श्रंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{cos}}{{\text{e}}^{\text{x}}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cos}}{{\text{e}}^{\text{x}}} \hfill \\ {\text{ = }}\;\dfrac{{\dfrac{{\text{1}}}{{{\text{cos}}{{\text{e}}^{\text{x}}}}}\left( {{\text{ - sin}}{{\text{e}}^{\text{x}}}} \right){\text{d}}}}{{{\text{dx}}}}{{\text{e}}^{\text{x}}} \hfill \\ {\text{ = }}\;{\text{ - tan}}{{\text{e}}^{\text{x}}} \times {{\text{e}}^{\text{x}}} \hfill \\ \end{align}

6. ${{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}{\text{ + }}...{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{5}}}}}$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$${{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}{\text{ + }}...{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{5}}}}}$ है। अब श्रंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\;{{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{2}}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{3}}}}} \times \dfrac{{\text{d}}}{{{\text{dy}}}}{{\text{x}}^{\text{3}}}{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{4}}}}} \times \dfrac{{\text{d}}}{{{\text{dy}}}}{{\text{e}}^{\text{4}}}{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{5}}}}} \times \dfrac{{\text{d}}}{{{\text{dy}}}}{{\text{x}}^{\text{5}}} \hfill \\ {\text{ = }}\;{{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}{\text{ \times 2x + }}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}{\text{ \times 3}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{4}}}}}{\text{ \times 4}}{{\text{x}}^{\text{3}}}{\text{ + }}{{\text{e}}^{{{\text{x}}^{\text{5}}}}}{\text{ \times 5}}{{\text{x}}^{\text{4}}} \hfill \\ {\text{ = }}\;{{\text{e}}^{\text{x}}}{\text{ + 2x}}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}{\text{ + 3}}{{\text{x}}^{\text{2}}}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}{\text{ + 4}}{{\text{x}}^{\text{3}}}{{\text{e}}^{{{\text{x}}^{\text{4}}}}}{\text{ + 5}}{{\text{x}}^4}{{\text{e}}^{{x^5}}} \hfill \\ \end{align}

7. $\sqrt {{{\text{e}}^{\sqrt {\text{x}} }}} {\text{,}}\;{\text{x > 0}}$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$$\sqrt {{{\text{e}}^{\sqrt {\text{x}} }}} {\text{,}}\;{\text{x > 0}}$ है। अब श्रंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{2}}\sqrt {{{\text{e}}^{\sqrt {\text{x}} }}} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{e}}^{\sqrt {\text{x}} }} \hfill \\ {\text{ = }}\;\dfrac{{\text{1}}}{{{\text{2}}\sqrt {{{\text{e}}^{\sqrt {\text{x}} }}} }}{{\text{e}}^{\sqrt {\text{x}} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\sqrt {\text{x}} \hfill \\ {\text{ = }}\;\dfrac{{\text{1}}}{{{\text{2}}\sqrt {{{\text{e}}^{\sqrt {\text{x}} }}} }}{{\text{e}}^{\sqrt {\text{x}} }} \times \dfrac{{\text{1}}}{{{\text{2}}\sqrt {\text{x}} }} \hfill \\ {\text{ = }}\;\dfrac{{\sqrt {{{\text{e}}^{\sqrt {\text{x}} }}} }}{{{\text{4}}\sqrt {\text{x}} }} \hfill \\ \end{align}

8. ${\text{log(log}}\;{\text{x),}}\;{\text{x > 1}}$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$${\text{log(log}}\;{\text{x),}}\;{\text{x > 1}}$ है। अब श्रंखला नियम द्वारा

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{logx}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{logx}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{logx}}}} \times \dfrac{{\text{1}}}{{\text{x}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{xlogx}}}}$

9. $\dfrac{{{\text{cos}}\;{\text{x}}}}{{{\text{log}}\;{\text{x}}}}{\text{,}}\;{\text{x > 0}}$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$$\dfrac{{{\text{cos}}\;{\text{x}}}}{{{\text{log}}\;{\text{x}}}}{\text{,}}\;{\text{x > 0}}$ है। अब श्रंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{logx \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cosx - cosx \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{logx}}}}{{{{{\text{(logx)}}}^{\text{2}}}}} \hfill \\ {\text{ = }}\;\dfrac{{{\text{logx \times ( - sinx) - cosx \times }}\dfrac{{\text{1}}}{{\text{x}}}}}{{{{{\text{(logx)}}}^{\text{2}}}}} \hfill \\ {\text{ = }}\;\dfrac{{{\text{ - (xsinxlogx + cosx)}}}}{{{\text{x(logx}}{{\text{)}}^{\text{2}}}}} \hfill \\ \end{align}

10. ${\text{cos}}\left( {{\text{log}}\;{\text{x + }}{{\text{e}}^{\text{x}}}} \right)$

उत्तर: मान लीजिए कि ${\text{y}}\;{\text{ = }}\;$${\text{cos}}\left( {{\text{log}}\;{\text{x + }}{{\text{e}}^{\text{x}}}} \right)$ है। अब श्रंखला नियम द्वारा

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{ - sin}}\left( {{\text{logx + }}{{\text{e}}^{\text{x}}}} \right) \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{logx + }}{{\text{e}}^{\text{x}}}} \right) \hfill \\ {\text{ = }}\;{\text{ - sin}}\left( {{\text{logx + }}{{\text{e}}^{\text{x}}}} \right) \times \left( {\dfrac{{\text{1}}}{{\text{x}}}{\text{ + }}{{\text{e}}^{\text{x}}}} \right) \hfill \\ \end{align}

### प्रश्नावली 5.5

1 से 11 तक के प्रश्नों मे प्रदत फलनों का ${\text{x}}$ के सापेक्ष अवकलन कीजिए:

1. ${\text{cosx \times cos2x \times cos3x}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;{\text{cosx \times cos2x \times cos3x}}$

दोनों और ${\text{log}}$ लगाने पर,

\begin{align} {\text{logy}}\;{\text{ = }}\;{\text{logcosx + logcos2x + logcos3x}} \hfill \\ \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{cosx}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cosx + }}\dfrac{{\text{1}}}{{{\text{cos2x}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cos2x + }}\dfrac{{\text{1}}}{{{\text{cos3x}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cos3x}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{y}}\left[ {\dfrac{{\text{1}}}{{{\text{cosx}}}}{\text{( - sinx) + }}\dfrac{{\text{1}}}{{{\text{cos2x}}}}{\text{( - sin2x) \times 2}}} \right.\left. {{\text{ + }}\dfrac{{\text{1}}}{{{\text{cos3x}}}}{\text{( - sin3x) \times 3}}} \right] \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{cosx \times cos2x \times cos3x[ - tanx - 2tan2x - 3tan3x]}} \hfill \\ \end{align}

2. $\sqrt {\dfrac{{{\text{(x - 1)(x - 2)}}}}{{{\text{(x - 3)(x - 4)(x - 5)}}}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$$\sqrt {\dfrac{{{\text{(x - 1)(x - 2)}}}}{{{\text{(x - 3)(x - 4)(x - 5)}}}}}$

दोनों और ${\text{log}}$ लगाने पर,

\begin{align} {\text{logy}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}{\text{[log(x - 1) + log(x - 2) - log(x - 3) - log(x - 4) - log(x - 5)]}} \hfill \\ \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}\left[ {\dfrac{{\text{1}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{(x - 2)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{(x - 3)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{(x - 4)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{(x - 5)}}}}} \right] \hfill \\ \end{align}

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}\sqrt {\dfrac{{{\text{(x - 1)(x - 2)}}}}{{{\text{(x - 3)(x - 4)(x - 5)}}}}} \left[ {\dfrac{{\text{1}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{(x - 2)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{(x - 3)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{(x - 4)}}}}{\text{ - }}} \right.\left. {\dfrac{{\text{1}}}{{{\text{(x - 5)}}}}} \right]$

3. ${{\text{(logx)}}^{{\text{cosx}}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${{\text{(logx)}}^{{\text{cosx}}}}$

दोनों और ${\text{log}}$ लगाने पर,

\begin{align} {\text{logy}}\;{\text{ = }}\;{\text{log(logx}}{{\text{)}}^{{\text{cosx}}}}\;{\text{ = }}\;{\text{cosxloglogx}} \hfill \\ \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{cosx \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{loglogx + loglogx \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cosx}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{y}}\left[ {{\text{cosx}}\dfrac{{\text{1}}}{{{\text{logx}}}}\dfrac{{\text{1}}}{{\text{x}}}{\text{ + loglogx( - sinx)}}} \right] \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{(logx)}}^{{\text{cosx}}}}\left[ {\dfrac{{{\text{cosx - sinxloglogx}}}}{{{\text{xlogx}}}}} \right] \hfill \\ \end{align}

4. ${{\text{x}}^{\text{x}}}{\text{ - }}{{\text{2}}^{{\text{sinx}}}}$

उत्तर: ${\text{u}}\;{\text{ = }}\;{{\text{x}}^{\text{x}}}{\text{,}}\;{\text{v}}\;{\text{ = }}\;{{\text{2}}^{{\text{sinx}}}}$

${\text{y}}\;{\text{ = }}\;$ ${\text{u - v}}$

दोनों और ${\text{x}}$ लगाने पर,

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ - }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}$ ……………..(i)

${\text{u}}\;{\text{ = }}\;{{\text{x}}^{\text{x}}}$

दोनों और ${\text{log}}$ लगाने पर,

${\text{logu}}\;{\text{ = }}\;{\text{xlogx}}$

\begin{align} \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{logx + logx}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{x}}\;{\text{ = }}\;{\text{x \times }}\dfrac{{\text{1}}}{{\text{x}}}{\text{ + logx \times 1}}\,{\text{ = }}\;{\text{1 + logx}} \hfill \\ \dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = u}}\;{\text{[1 + logx] = }}{{\text{x}}^{\text{x}}}{\text{[1 + logx]}} \hfill \\ {\text{v}}\;{\text{ = }}\;{{\text{2}}^{{\text{sinx}}}} \hfill \\ {\text{logv}}\;{\text{ = }}\;{\text{sinxlog2}} \hfill \\ \dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{log2 \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sinx = log2 \times cosx}} \hfill \\ \dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{v[cosxlog2]}}\;{\text{ = }}\;{{\text{2}}^{{\text{sinx}}}}{\text{[cosxlog2]}} \hfill \\ \therefore \;\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{x}}^{\text{x}}}{\text{[1 + logx] - }}{{\text{2}}^{{\text{sinx}}}}{\text{[cosxlog2]}} \hfill \\ \end{align}

5. ${{\text{(x + 3)}}^{\text{2}}}{\text{ \times (x + 4}}{{\text{)}}^{\text{3}}}{\text{ \times (x + 5}}{{\text{)}}^{\text{4}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${{\text{(x + 3)}}^{\text{2}}}{\text{ \times (x + 4}}{{\text{)}}^{\text{3}}}{\text{ \times (x + 5}}{{\text{)}}^{\text{4}}}$

दोनों और ${\text{log}}$ लगाने पर,

\begin{align} {\text{logy}}\;{\text{ = }}\;{\text{2log(x + 3) + 3log(x + 4) + 4log(x + 5)}} \hfill \\ \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{2 \times }}\dfrac{{\text{1}}}{{{\text{log(x + 3)}}}}{\text{ + 3}}\dfrac{{\text{1}}}{{{\text{log(x + 4)}}}}{\text{ + 4}}\dfrac{{\text{1}}}{{{\text{log(x + 5)}}}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{y}}\left[ {\dfrac{{{\text{2(x + 4)(x + 5) + 3(x + 3)(x + 5) + 4(x + 3)(x + 4)}}}}{{{\text{(x + 3)(x + 4)(x + 5)}}}}} \right] \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{(x + 3)(x + 4}}{{\text{)}}^{\text{2}}}{{\text{(x + 5)}}^{\text{3}}}\left( {{\text{9}}{{\text{x}}^{\text{2}}}{\text{ + 70x + 133}}} \right) \hfill \\ \end{align}

6. ${\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)^{\text{x}}}{\text{ + }}{{\text{x}}^{\left( {{\text{1 + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)^{\text{x}}}{\text{ + }}{{\text{x}}^{\left( {{\text{1 + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)}}$

${\text{u}}\;{\text{ = }}\;{\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)^{\text{x}}}{\text{,}}\;{\text{v}}\;{\text{ = }}\;{{\text{x}}^{\left( {{\text{1 + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)}}$

${\text{y}}\;{\text{ = }}\;{\text{u + v}}$

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ {\text{u}}\;{\text{ = }}\;{\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)^{\text{x}}} \hfill \\ {\text{logu}}\;{\text{ = }}\;{\text{log}}{\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)^{\text{x}}} \hfill \\ {\text{logu}}\;{\text{ = }}\;{\text{xlog}}\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right) \hfill \\ \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;\left. {\;{\text{ = x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{log}}\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right){\text{ + log}}\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)} \right) \times \dfrac{{{\text{d(x)}}}}{{{\text{dx}}}} \hfill \\ \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{x}}}{{\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)}}\left[ {{\text{1 - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}} \right]{\text{ + log}}\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right) \hfill \\ \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{x}}}{{\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{\text{x}}}} \right)}}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{{{{\text{x}}^{\text{2}}}}}} \right]{\text{ + log}}\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right) \hfill \\ \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} \right){\text{ + log}}\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right) \hfill \\ \dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)^{\text{x}}}\left[ {\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} \right){\text{ + log}}\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)} \right] \hfill \\ \end{align}

\begin{align} \therefore \;{\text{V}}\;{\text{ = }}\;{{\text{x}}^{\left( {{\text{1 + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)}} \hfill \\ {\text{log(v)}}\;{\text{ = }}\;{\text{log}}\left[ {{{\text{x}}^{\left( {{\text{1 + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)}}} \right] \hfill \\ {\text{ = }}\;\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right){\text{log(x)}} \hfill \\ \dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{log(x) + log(x)}} \times \dfrac{{{\text{d}}\left( {\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)} \right)}}{{{\text{dx}}}} \hfill \\ \end{align}

$\dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)\left( {\dfrac{{\text{1}}}{{\text{x}}}} \right){\text{ + log(x)}}\left[ {{\text{1 - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}} \right]$

$\dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;\left( {{\text{1 + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}} \right){\text{ + log(x)}}\left[ {{\text{1 - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}} \right]$

\begin{align} \dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{2}}}}}{\text{ + log(x)}}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{{{{\text{x}}^{\text{2}}}}}} \right] \hfill \\ \dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{(1 + log(x)) + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}{\text{(1 - log(x))}} \hfill \\ \dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{x}}^{\left( {{\text{1 + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)}}\left[ {{\text{(1 + log(x)) + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}{\text{(1 - }}} \right.{\text{log(x))]}} \hfill \\ \therefore \;\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ \end{align}

${\text{ = }}\;{\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)^{\text{x}}}\left[ {\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} \right){\text{ + log}}\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)} \right]{\text{ + }}{{\text{x}}^{\left( {{\text{1 + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)}}\left[ {{\text{(1 + log(x)) + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}{\text{(1 - log(x))}}} \right]$

7. ${\text{log(x}}{{\text{)}}^{\text{x}}}{\text{ + }}{{\text{x}}^{{\text{log(x)}}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$ ${\text{log(x}}{{\text{)}}^{\text{x}}}{\text{ + }}{{\text{x}}^{{\text{log(x)}}}}$

\begin{align} {\text{u}}\;{\text{ = }}\;{\text{log(}}{{\text{x}}^{\text{x}}}{\text{),}}\;{\text{v}}\;{\text{ = }}\;{{\text{x}}^{{\text{log(x)}}}} \hfill \\ {\text{y}}\;{\text{ = }}\;{\text{u + v}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ \end{align}

\begin{align} {\text{u}}\;{\text{ = }}\;{\text{log(x}}{{\text{)}}^{\text{x}}} \hfill \\ {\text{log(u)}}\;{\text{ = }}\;{\text{log}}\left( {{\text{log(x}}{{\text{)}}^{\text{x}}}} \right) \hfill \\ {\text{ = xlog(log(x))}} \hfill \\ \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{log(log(x)) + log(log(x))}} \times \dfrac{{{\text{d(x)}}}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;\dfrac{{\text{x}}}{{{\text{log(x)}}}} \times \dfrac{{\text{1}}}{{\text{x}}}{\text{ + log(log(x))}} \hfill \\ {\text{ = }}\;\dfrac{{\text{1}}}{{{\text{log(x)}}}}{\text{ + log(log(x))}} \hfill \\ \end{align}

\begin{align} \dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{log(x}}{{\text{)}}^{\text{x}}}\left[ {\dfrac{{\text{1}}}{{{\text{log(x)}}}}{\text{ + log(log(x)}}} \right.{\text{)]}} \hfill \\ {\text{v}}\;{\text{ = }}\;{{\text{x}}^{{\text{log(x)}}}} \hfill \\ {\text{log(v)}}\;{\text{ = }}\;{\text{log}}\left[ {{{\text{x}}^{{\text{log(x)}}}}} \right] \hfill \\ \end{align} \begin{align} {\text{ = }}\;{\text{log(x)[log(x)] = [log(x)}}{{\text{]}}^{\text{2}}} \hfill \\ \dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{2}} \times {\text{log(x) \times }}\dfrac{{\text{1}}}{{\text{x}}} \hfill \\ \end{align}

${\text{ = }}\;\dfrac{{\text{2}}}{{\text{x}}}{\text{log(x)}}\;{\text{ = }}\;{\text{v}}\left[ {\dfrac{{\text{2}}}{{\text{x}}}{\text{log(x)}}} \right]$

$\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{x}}^{{\text{log(x)}}}}\left[ {\dfrac{{\text{2}}}{{\text{x}}}{\text{log(x)}}} \right]$

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;{\text{log(x}}{{\text{)}}^{\text{x}}}\left[ {\dfrac{{\text{1}}}{{{\text{log(x)}}}}{\text{ + log(}}} \right.{\text{log(x)) + }}{{\text{x}}^{{\text{log(x)}}}}\left[ {\dfrac{{\text{2}}}{{\text{x}}}{\text{log(x)}}} \right]{\text{]}} \hfill \\ \end{align}

8. ${{\text{(sinx)}}^{\text{x}}}{\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${{\text{(sinx)}}^{\text{x}}}{\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}}$

\begin{align} {\text{u}}\;{\text{ = }}\;{{\text{(sinx)}}^{\text{x}}}{\text{,}}\;{\text{v}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} \hfill \\ {\text{u}}\;{\text{ = }}\;{{\text{(sinx)}}^{\text{x}}} \hfill \\ {\text{log(u)}}\;{\text{ = }}\;{\text{x}}\;{\text{log[sin(x)]}} \hfill \\ \end{align}

\begin{align} \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(log[sin(x)]) + log(sin(x))}} \times \dfrac{{{\text{d(x)}}}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;\dfrac{{\text{x}}}{{{\text{sin(x)}}}}{\text{xcos(x) + log(sin(x))}} \hfill \\ {\text{ = }}\;\dfrac{{{\text{xcos(x)}}}}{{{\text{sin(x)}}}}{\text{ + log(sin(x))}} \hfill \\ \dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{(sinx)}}^{\text{x}}}\left[ {\dfrac{{{\text{xcos(x)}}}}{{{\text{sin(x)}}}}{\text{ + log(sin(x))}}} \right.{\text{]}} \hfill \\ {\text{v}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} \hfill \\ \dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\sqrt {\text{x}} }^{\text{2}}}} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(}}\sqrt {\text{x}} {\text{)}} \hfill \\ {\text{ = }}\;\dfrac{{\text{1}}}{{\sqrt {{\text{1 - x}}} }} \times \dfrac{{\text{1}}}{{{\text{2}}\sqrt {\text{x}} }} \hfill \\ \end{align}

\begin{align} {\text{ = }}\;\dfrac{{\text{1}}}{{{\text{2}}\sqrt {{\text{x - }}{{\text{x}}^{\text{2}}}} }} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;{{\text{(sinx)}}^{\text{x}}}\left[ {\dfrac{{{\text{xcos(x)}}}}{{{\text{sin(x)}}}}{\text{ + log(sin(x))}}} \right]{\text{ + }}\dfrac{{\text{1}}}{{{\text{2}}\sqrt {{\text{x - }}{{\text{x}}^{\text{2}}}} }} \hfill \\ \end{align}

9. ${{\text{x}}^{{\text{sin(x)}}}}{\text{ + si}}{{\text{n}}^{{\text{cos(x)}}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${{\text{x}}^{{\text{sin(x)}}}}{\text{ + si}}{{\text{n}}^{{\text{cos(x)}}}}$

\begin{align} {\text{u}}\;{\text{ = }}\;{{\text{x}}^{{\text{sin(x)}}}}{\text{,}}\;{\text{v}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{{\text{cos(x)}}}} \hfill \\ {\text{y}}\;{\text{ = }}\;{\text{u + v}} \hfill \\ \end{align}

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ {\text{u}}\;{\text{ = }}\;{{\text{x}}^{{\text{sin(x)}}}} \hfill \\ {\text{log(u)}}\;{\text{ = }}\;{\text{sin(x)log(x)}} \hfill \\ \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{sin(x)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(log(x))}} \times {\text{log(x)}}\dfrac{{{\text{d(sin(x))}}}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;\dfrac{{{\text{sin(x)}}}}{{\text{x}}}{\text{ + log(x)(cos(x))}} \hfill \\ {\text{ = }}\;{{\text{x}}^{{\text{sin(x)}}}}\left[ {\dfrac{{{\text{sin(x)}}}}{{\text{x}}}{\text{ + log(x)(cos(x))}}} \right] \hfill \\ \end{align}

\begin{align} {\text{v}}\;{\text{ = }}\;{\text{sin(x}}{{\text{)}}^{{\text{cos(x)}}}} \hfill \\ {\text{log(v)}}\;{\text{ = }}\;{\text{log}}\left[ {{\text{sin(x}}{{\text{)}}^{{\text{cos(x)}}}}} \right] \hfill \\ {\text{ = }}\;{\text{cos(x)[log(sin(x)]}} \hfill \\ \dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{cos(x)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(log(sin(x)) + log(sin(x))}}\dfrac{{{\text{d(cos(x))}}}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;\dfrac{{{\text{cos(x)}}}}{{{\text{sin(x)}}}} \times {\text{cos(x) + log(x)( - sin(x))}} \hfill \\ {\text{ = }}\;{\text{sin(x}}{{\text{)}}^{{\text{cos(x)}}}}{\text{[cot(x)cos(x) - log(x)(sin(x))]}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{x}}^{{\text{sin(x)}}}}\left[ {\dfrac{{{\text{sin(x)}}}}{{\text{x}}}{\text{ + log(x)(cos(x))}}} \right]{\text{ + sin(x}}{{\text{)}}^{{\text{cos(x)}}}}{\text{[cot(x)cos(x) - log(x)(sin(x))]}} \hfill \\ \end{align}

10. ${{\text{x}}^{{\text{xcos(x)}}}}{\text{ + }}\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${{\text{x}}^{{\text{xcos(x)}}}}{\text{ + }}\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}$

\begin{align} {\text{u}}\;{\text{ = }}\;{{\text{x}}^{{\text{x}}\;{\text{cos(x)}}}}{\text{,}}\;{\text{v}}\;{\text{ = }}\;\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}} \hfill \\ {\text{y}}\;{\text{ = }}\;{\text{u + v}} \hfill \\ \end{align}

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ {\text{u}}\;{\text{ = }}\;{{\text{x}}^{{\text{xcos(x)}}}} \hfill \\ {\text{log(u)}}\;{\text{ = }}\;{\text{xcos(x)log(x)}} \hfill \\ \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{xcos(x)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(log(x)) + log(x)}}\dfrac{{{\text{d(xcos(x))}}}}{{{\text{dx}}}} \hfill \\ \end{align}

$\begin{array}{*{20}{l}} {{\text{ = }}\;\dfrac{{{\text{xcos(x)}}}}{{\text{x}}}{\text{ + log(x)(x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cos(x) + cos(x) \times }}} \end{array}{\text{1)}}$

\begin{align} \begin{array}{*{20}{l}} {{\text{ = }}\;\dfrac{{{\text{xcos(x)}}}}{{\text{x}}}{\text{ + log(x)(x( - sin(x)) + cos(x)}}} \end{array}{\text{)}} \hfill \\ {\text{ = }}\;{\text{[cos(x) - xsin(x) + cos(x) + cos(x)log(x)]}} \hfill \\ {\text{ = }}\;{{\text{x}}^{{\text{xcos(x)}}}}{\text{[cos(x) - xsin(x) + cos(x)log(x)]}} \hfill \\ {\text{v}}\;{\text{ = }}\;\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}} \hfill \\ \end{align}

\begin{align} {\text{log(v)}}\;{\text{ = }}\;{\text{log}}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}} \right) \hfill \\ {\text{ = }}\;{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ - log}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right) \hfill \\ \dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ - }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right)} \right. \hfill \\ {\text{ = }}\;\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right) \hfill \\ {\text{ = }}\;\left[ {\dfrac{{{\text{2x}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{\text{ - }}\dfrac{{{\text{2x}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}} \right] \hfill \\ {\text{ = }}\;\dfrac{{{\text{ - 4x}}}}{{{{\text{x}}^{\text{4}}}{\text{ - 1}}}} \hfill \\ {\text{ = }}\;\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}\left[ {\dfrac{{{\text{ - 4x}}}}{{{{\text{x}}^{\text{4}}}{\text{ - 1}}}}} \right] \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{x}}^{{\text{xcos(x)}}}}{\text{[cos(x) - xsin(x) + cos(x)log(x)] + }}\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{\text{[}}\dfrac{{{\text{ - 4x}}}}{{{{\text{x}}^{\text{4}}}{\text{ - 1}}}}{\text{]}} \hfill \\ \end{align}

11. ${{\text{[xcos(x)]}}^{\text{x}}}{\text{ + xsin(x}}{{\text{)}}^{\dfrac{{\text{1}}}{{\text{x}}}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${{\text{[xcos(x)]}}^{\text{x}}}{\text{ + xsin(x}}{{\text{)}}^{\dfrac{{\text{1}}}{{\text{x}}}}}$

\begin{align} {\text{u}}\;{\text{ = }}\;{\text{xcos(x}}{{\text{)}}^{\text{x}}}{\text{,}}\;{\text{v}}\;{\text{ = }}\;{\text{xsin(x}}{{\text{)}}^{\dfrac{{\text{1}}}{{\text{x}}}}} \hfill \\ {\text{y}}\;{\text{ = }}\;{\text{u + v}} \hfill \\ {\text{u}}\;{\text{ = }}\;{\text{xcos(x}}{{\text{)}}^{\text{x}}} \hfill \\ \end{align}

\begin{align} {\text{log(u)}}\;{\text{ = }}\;{\text{log}}\left( {{\text{xcos(x}}{{\text{)}}^{\text{x}}}} \right) \hfill \\ {\text{ = }}\;{\text{xlog(xcos(x))}} \hfill \\ \end{align}

$\dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(log(xcos(x)) + log(xcos(x))}}\dfrac{{{\text{d(x)}}}}{{{\text{dx}}}}$

${\text{ = }}\;{\text{log(xcos(x)) + x \times }}\dfrac{{\text{1}}}{{{\text{xcos(x)}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(xcos(x))}}$

\begin{align} {\text{ = }}\;{\text{log(xcos(x)) + }}\dfrac{{\text{1}}}{{{\text{cos(x)}}}}\left[ {{{\text{x}}^{\text{*}}}{\text{ - }}} \right.{\text{sin(x) + cos(x)]}} \hfill \\ {\text{ = }}\;{\text{[ - xtan(x) + 1] + log(xcos(x))}} \hfill \\ \end{align}

\begin{align} \dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{x}}^{{\text{xcos(x)}}}}{\text{[1 - xtan(x) + log(xcos(x))]}} \hfill \\ {\text{v}}\;{\text{ = }}\;{\text{xsin(x}}{{\text{)}}^{\dfrac{{\text{1}}}{{\text{x}}}}} \hfill \\ {\text{log(v)}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{x}}}{\text{log(xsin(x))}} \hfill \\ \dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{x}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logxsin(x)) + log(xsin(x))}}\dfrac{{{\text{d(1/x)}}}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;\dfrac{{\text{1}}}{{\text{x}}} \times \dfrac{{\text{1}}}{{{\text{xsin(x)}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(xsin(x)) + log[xsin(x)]}}\left( {\dfrac{{{\text{ - 1}}}}{{{{\text{x}}^{\text{2}}}}}} \right) \hfill \\ {\text{ = }}\;\dfrac{{\text{1}}}{{\text{x}}} \times \dfrac{{\text{1}}}{{{\text{xsin(x)}}}}{\text{[xcos(x) + sin(x)] + log[xsin(x)]}}\left( {\dfrac{{{\text{ - 1}}}}{{{{\text{x}}^{\text{2}}}}}} \right) \hfill \\ {\text{ = }}\;\dfrac{{{\text{xcos(x)}}}}{{{{\text{x}}^{\text{2}}}{\text{sin(x)}}}}{\text{ + }}\dfrac{{{\text{sin(x)}}}}{{{{\text{x}}^{\text{2}}}{\text{sin(x)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}{\text{log(xsin(x))}} \hfill \\ {\text{ = }}\;{\text{xsin(x}}{{\text{)}}^{\dfrac{{\text{1}}}{{\text{x}}}}}\left[ {\dfrac{{{\text{xcos(x)}}}}{{{{\text{x}}^{\text{2}}}{\text{sin(x)}}}}{\text{ + }}} \right.\left. {\dfrac{{{\text{sin(x)}}}}{{{{\text{x}}^{\text{2}}}{\text{sin(x)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}{\text{log(xsin(x))}}} \right] \hfill \\ \end{align}

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{{\text{x}}^{{\text{xcos(x)}}}}\left[ {{\text{1 - xtan(x) + log(xcos(x)) + xsin(x}}{{\text{)}}^{\dfrac{{\text{1}}}{{\text{x}}}}}{\text{[}}} \right.\left. {\dfrac{{{\text{xcos(x)}}}}{{{{\text{x}}^{\text{2}}}{\text{sin(x)}}}}{\text{ + }}\dfrac{{{\text{sin(x)}}}}{{{{\text{x}}^{\text{2}}}{\text{sin(x)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}{\text{log(xsin(x))}}} \right] \hfill \\ \end{align}

12 से 15 तक प्रदत फलनों के लिए $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ ज्ञात कीजिए:

12. ${{\text{x}}^{\text{x}}}{\text{ + }}{{\text{y}}^{\text{x}}}\;{\text{ = }}\;{\text{1}}$

उत्तर: मान लीजिए ${\text{u}}\;{\text{ = }}\;{{\text{x}}^{\text{y}}}{\text{,}}\;{\text{v}}\;{\text{ = }}\;{{\text{y}}^{\text{x}}}$

तथा ${\text{u + v}}\;{\text{ = }}\;{\text{1}}$

दोनों तरफ का लघुगुणक लेने पर,

$\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{0}}$

अब ${\text{u}}\;{\text{ = }}\;{{\text{x}}^{\text{y}}}$

दोनों तरफ का लघुगुणक लेने पर,

${\text{log(u)}}\;{\text{ = }}\;{\text{y}}\;{\text{log(x)}}$

${\text{x}}$ के सापेक्ष अवकलन करने पर,

$\dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{y \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logx) + log(x) \times }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$

\begin{align} {\text{ = }}\;\dfrac{{\text{y}}}{{\text{x}}}{\text{ + log(x) \times }}\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right) \hfill \\ {\text{ = }}\;{\text{u \times }}\left[ {\dfrac{{\text{y}}}{{\text{x}}}{\text{ + log(x) \times }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right] \hfill \\ {\text{ = }}\;{{\text{x}}^{\text{y}}} \times \left[ {\dfrac{{\text{y}}}{{\text{x}}}{\text{ + log(x) \times }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right] \hfill \\ {\text{v}}\;{\text{ = }}\;{{\text{y}}^{\text{x}}} \hfill \\ {\text{log(v)}}\;{\text{ = }}\;{\text{xlog(y)}} \hfill \\ \dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logy) + log(y)}}\dfrac{{{\text{d(x)}}}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;\dfrac{{\text{x}}}{{\text{y}}} \times \left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right){\text{ + log(y)}} \hfill \\ {\text{ = }}\;{{\text{y}}^{\text{x}}}\left[ {\dfrac{{\text{x}}}{{\text{y}}} \times \left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right){\text{ + log(y)}}} \right] \hfill \\ {\text{ = }}\;{{\text{y}}^{\text{x}}}\left[ {\dfrac{{\text{x}}}{{\text{y}}} \times \left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right){\text{ + log(y)}}} \right]{\text{ + }}{{\text{x}}^{\text{y}}} \times \left[ {\dfrac{{\text{y}}}{{\text{x}}}{\text{ + log(x) \times }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right]\;{\text{ = }}\;{\text{0}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\left( {{{\text{x}}^{\text{y}}}{\text{ \times log(x) + }}\dfrac{{\text{x}}}{{\text{y}}} \times {{\text{y}}^{\text{x}}}} \right)\;{\text{ = }}\;{\text{ - }}{{\text{y}}^{\text{x}}}{\text{log(y) - }}{{\text{x}}^{\text{y}}} \times \dfrac{{\text{y}}}{{\text{x}}} \hfill \\ \end{align}

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{ - }}\dfrac{{{{\text{y}}^{\text{x}}}{\text{log(y) + }}{{\text{x}}^{\text{y}}}\dfrac{{\text{y}}}{{\text{x}}}}}{{{{\text{x}}^{\text{y}}}{\text{log(x) + }}\dfrac{{\text{x}}}{{\text{y}}}{{\text{y}}^{\text{x}}}}}$

13. ${{\text{x}}^{\text{y}}}{\text{ = }}{{\text{y}}^{\text{x}}}$

उत्तर: दोनों तरफ का लघुगुणक लेने पर,

${\text{y}}\;{\text{log(x)}}\;{\text{ = }}\;{\text{x}}\;{\text{log(y)}}$

${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} {\text{y \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logx) + log(x) \times }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logy) + log(y)}}\dfrac{{{\text{d(x)}}}}{{{\text{dx}}}} \hfill \\ \dfrac{{\text{y}}}{{\text{x}}}{\text{ + log(x) \times }}\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)\;{\text{ = }}\;\dfrac{{\text{x}}}{{\text{y}}} \times \left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right){\text{ + log(y)}} \hfill \\ \dfrac{{\text{x}}}{{\text{y}}} \times \left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right){\text{ - log(x) \times }}\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)\;{\text{ = }}\;\dfrac{{\text{y}}}{{\text{x}}}{\text{ - log(y)}} \hfill \\ \end{align}

$\left[ {\dfrac{{\text{x}}}{{\text{y}}}{\text{ - log(x)}}} \right] \times \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{y}}}{{\text{x}}}{\text{ - log(y)}}$

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\dfrac{{\text{y}}}{{\text{x}}}{\text{ - log(y)}}}}{{\dfrac{{\text{x}}}{{\text{y}}}{\text{ - log(x)}}}}\;{\text{ = }}\;\dfrac{{{\text{y[y - xlog(y)]}}}}{{{\text{x[x - ylog(x)]}}}}$

14. ${{\text{(cosx)}}^{\text{y}}}{\text{ = (cosy}}{{\text{)}}^{\text{x}}}$

उत्तर: दोनों तरफ का लघुगुणक लेने पर,

${\text{y}}\;{\text{log(cos(x))}}\;{\text{ = }}\;{\text{x}}\;{\text{log(cos(y))}}$

${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} {\text{y \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logcos(y)) + log(cos(x)) \times }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logcos(y)) + log(cos(y))}}\dfrac{{{\text{d(x)}}}}{{{\text{dx}}}} \hfill \\ \dfrac{{\text{y}}}{{{\text{cosx}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cos(x)) + log[cos(x)] \times }}\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)\;{\text{ = }}\;\dfrac{{\text{x}}}{{{\text{cos(y)}}}} \times \dfrac{{{\text{d(cosy)}}}}{{{\text{dx}}}}{\text{ + log(cosy)}} \hfill \\ \end{align}

\begin{align} \dfrac{{\text{y}}}{{{\text{cosx}}}}{\text{( - sin(x)) + log[cos(x)] \times }}\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)\;{\text{ = }}\;\dfrac{{\text{x}}}{{{\text{cos(y)}}}}{\text{ \times ( - sin(y)) \times }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + log(cosy)}} \hfill \\ {\text{[log(cosx) + xtan(y)] \times }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{log(cos(y)) + y \times tan(x)}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{log(cos(y)) + y \times tan(x)}}}}{{{\text{log(cosx) + xtan(y)}}}} \hfill \\ \end{align}

15. ${\text{x}}{\text{.y = }}{{\text{e}}^{{\text{x - y}}}}$

उत्तर: दोनों तरफ का लघुगुणक लेने पर,

\begin{align} {\text{log(xy)}}\;{\text{ = }}\;{\text{(x - y)log(e)}} \hfill \\ {\text{log(x) + log(y)}}\;{\text{ = }}\;{\text{x - y}} \hfill \\ \end{align}

${\text{x}}$ के सापेक्ष अवकलन करने पर,

$\dfrac{{\text{1}}}{{\text{x}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{1 - }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$

\begin{align} \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{1 - }}\dfrac{{\text{1}}}{{\text{x}}}\left( {\dfrac{{\text{1}}}{{\text{y}}}{\text{ + 1}}} \right) \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{1 - }}\dfrac{{\text{1}}}{{\text{x}}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{1 - }}\dfrac{{\text{1}}}{{\text{x}}}}}{{\dfrac{{\text{1}}}{{\text{y}}}{\text{ + 1}}}}\;{\text{ = }}\;\dfrac{{{\text{x - 1}}}}{{{\text{y + 1}}}} \hfill \\ \end{align}

16. ${\text{f(x)}}\;{\text{ = }}\;{\text{(1 + x)(1 + }}{{\text{x}}^{\text{2}}}{\text{)(1 + }}{{\text{x}}^{\text{3}}}{\text{)(1 + }}{{\text{x}}^{\text{8}}}{\text{)}}$ द्वारा प्रदत फलन का अवकलन ज्ञात कीजिए और इस प्रकार $\text{f}(1)$ ज्ञात कीजिए। दोनों तरफ का लघुगणक लेकर सूत्र का प्रयोग करके अवकलन कीजिए।

उत्तर: ${\text{f(x)}}\;{\text{ = }}\;{\text{(1 + x)(1 + }}{{\text{x}}^{\text{2}}}{\text{)(1 + }}{{\text{x}}^{\text{3}}}{\text{)(1 + }}{{\text{x}}^{\text{8}}}{\text{)}}$

दोनों तरफ का लघुगुणक लेने पर,

\begin{align} {\text{log(f(x))}}\;{\text{ = }}\;{\text{log((1 + x)(1 + }}{{\text{x}}^{\text{2}}}{\text{)(1 + }}{{\text{x}}^{\text{3}}}{\text{)}}\left( {{\text{1 + }}} \right.\left. {\left. {{{\text{x}}^{\text{4}}}} \right)} \right) \hfill \\ {\text{ = }}\;{\text{log(1 + x) + log}}\left( {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right){\text{ + log}}\left( {{\text{1 + }}{{\text{x}}^{\text{4}}}} \right){\text{ + log}}\left( {{\text{1 + }}{{\text{x}}^{\text{8}}}} \right) \hfill \\ \end{align}

${\text{x}}$ के सापेक्ष अवकलन करने पर,

$\begin{array}{l} =\frac{1}{(1+x)}+\frac{2 x}{\left(1+x^{2}\right)}+\frac{4 x^{3}}{\left(1+x^{4}\right)}+\frac{8 x^{7}}{\left(1+x^{8}\right)} \\ f^{\prime}(x)=f(x)\left[\frac{1}{(1+x)}+\frac{2 x}{\left(1+x^{2}\right)}+\frac{4 x^{3}}{\left(1+x^{4}\right)}+\frac{8 x^{7}}{\left(1+x^{8}\right)}\right] \\ f^{\prime}(1)=f(1)\left[\frac{1}{(1+1)}+\frac{2.1}{\left(1+(1)^{2}\right)}+\frac{4.1}{\left(1+(1)^{4}\right)}+\frac{8(1)^{7}}{\left(1+(1)^{8}\right)}\right] \\ =f(1)\left[\frac{1}{(2)}+\frac{2}{(2)}+\frac{4}{(2)}+\frac{8}{(2)}\right] \end{array}$

$\begin{array}{l} =(1)(1)(1)(1) \times \frac{15}{2} \\ =\frac{15}{2} \end{array}$

17. ${\text{(}}{{\text{x}}^{\text{2}}}{\text{ - 5x + 8)(}}{{\text{x}}^{\text{3}}}{\text{ + 7x + 9)}}$ का अवकलन निम्नलिखित तीन प्रकार से कीजिए।

(i) गुणनफल नियम का प्रयोग करके

उत्तर: ${\text{y}}\;{\text{ = }}\;$ ${\text{(}}{{\text{x}}^{\text{2}}}{\text{ - 5x + 8)(}}{{\text{x}}^{\text{3}}}{\text{ + 7x + 9)}}$

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\left( {{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}} \right)\dfrac{{\left. {{\text{d}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)} \right)}}{{{\text{dx}}}}{\text{ + }}\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)\dfrac{{{\text{d}}\left( {{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}} \right)}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;\left( {{{\text{x}}^{\text{2}}}{\text{ + 7x + 9}}} \right){\text{(2x - 5) + }}\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 7}}} \right) \hfill \\ {\text{ = }}\;{\text{3}}{{\text{x}}^{\text{4}}}{\text{ + 15(x}}{{\text{)}}^{\text{3}}}{\text{ + 24(x}}{{\text{)}}^{\text{2}}}{\text{ + 7}}{{\text{x}}^{\text{2}}}{\text{ - 35x + 56 + 2}}{{\text{x}}^{\text{4}}}{\text{ + 14}}{{\text{x}}^{\text{2}}}{\text{ + 16x - 5}}{{\text{x}}^{\text{3}}}{\text{ - 35x - 45}} \hfill \\ {\text{ = }}\;{\text{5}}{{\text{x}}^{\text{4}}}{\text{ - 20}}{{\text{x}}^{\text{3}}}{\text{ + 45}}{{\text{x}}^{\text{2}}}{\text{ - 52x + 11}} \hfill \\ \end{align}

(ii) गुणनफल के विस्तारण द्वारा एक एकल बहुपद प्राप्त करके

उत्तर: ${\text{y}}\;{\text{ = }}\;$ ${\text{(}}{{\text{x}}^{\text{2}}}{\text{ - 5x + 8)(}}{{\text{x}}^{\text{3}}}{\text{ + 7x + 9)}}$

${\text{y}}\;{\text{ = }}\;{{\text{x}}^{\text{5}}}{\text{ - 5}}{{\text{x}}^{\text{4}}}{\text{ + 15}}{{\text{x}}^{\text{3}}}{\text{ - 26}}{{\text{x}}^{\text{2}}}{\text{ + 11x + 72}}$

$\dfrac{{{\text{d(y)}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^{\text{5}}}{\text{ - 5}}{{\text{x}}^{\text{4}}}{\text{ + 15}}{{\text{x}}^{\text{3}}}{\text{ - 26}}{{\text{x}}^{\text{2}}}{\text{ + 11x + 72}}} \right)$

${\text{ = }}\;{\text{5}}{{\text{x}}^{\text{4}}}{\text{ - 20}}{{\text{x}}^{\text{3}}}{\text{ + 45}}{{\text{x}}^{\text{2}}}{\text{ - 52x + 11}}$

(iii) लघुगणकीय अवकलन द्वारा, यह भी सत्यापित कीजिए कि इस प्रकार प्राप्त तीनों उत्तर समान है।

उत्तर: समी (i) के दोनों तरफ का लघुगणक लेने पर,

\begin{align} {\text{log(y)}}\;{\text{ = }}\;{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)\left( {{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}} \right) \hfill \\ {\text{ = }}\;{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right){\text{ + log}}\left( {{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}} \right) \hfill \\ \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)}}\dfrac{{{\text{d}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)}}{{{\text{dy}}}}{\text{ + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}}}\dfrac{{{\text{d}}\left( {{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}} \right)}}{{{\text{dx}}}} \hfill \\ {\text{ = }}\;\dfrac{{{\text{2x - 5}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)}}{\text{ + }}\dfrac{{\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 7}}} \right)}}{{{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}}} \hfill \\ {\text{ = }}\;{\text{y}}\left[ {\dfrac{{{\text{2x - 5}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)}}{\text{ + }}\dfrac{{\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 7}}} \right)}}{{{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}}}} \right] \hfill \\ {\text{ = }}\;\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)\left( {{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}} \right)\left[ {\dfrac{{{\text{2x - 5}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)}}{\text{ + }}\dfrac{{\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 7}}} \right)}}{{{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}}}} \right] \hfill \\ {\text{ = }}\;\left( {{{\text{x}}^{\text{3}}}{\text{ + 7x + 9}}} \right){\text{[2x - 5] + }}\left( {{{\text{x}}^{\text{2}}}{\text{ - 5x + 8}}} \right)\left[ {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 7}}} \right] \hfill \\ \end{align}

${\text{5}}{{\text{x}}^{\text{4}}}{\text{ - 20}}{{\text{x}}^{\text{3}}}{\text{ + 45}}{{\text{x}}^{\text{2}}}{\text{ - 52x + 11}}$

18. यदि ${\text{u,}}\;{\text{v,}}\;{\text{w,}}\,{\text{x }}$ के फलन है, तो दो विधियों अर्थात प्रथम गुणनफल कि पुनरावर्ती द्वारा, द्वितीय लघुगणकीय अवकलन द्वारा दर्शाइए कि $\dfrac{{{\text{d(u \times v \times w)}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{v}}{\text{.w}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + u}}{\text{.w}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + u}}{\text{.v}}\dfrac{{{\text{dw}}}}{{{\text{dx}}}}$

उत्तर: गुणनफल नियम के प्रयोग से, $\dfrac{{{\text{d(u \times v \times w)}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{u}}{\text{.v}}\dfrac{{{\text{dw}}}}{{{\text{dx}}}}{\text{ + w}}\dfrac{{{\text{d(u \times v)}}}}{{{\text{dx}}}}$ (${\text{uv}}$ को एक फलन तथा ${\text{w}}$ को दूसरे फलन माना गया है)

\begin{align} {\text{ = }}\;{\text{u}}{\text{.v}}\dfrac{{{\text{dw}}}}{{{\text{dx}}}}{\text{ + w}}\left[ {{\text{v \times }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + u}}\dfrac{{{\text{d(v)}}}}{{{\text{dx}}}}} \right] \hfill \\ {\text{ = }}\;{\text{u}}{\text{.v \times }}\dfrac{{{\text{dw}}}}{{{\text{dx}}}}{\text{ + w}}{\text{.v}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + w \times u}}\dfrac{{{\text{d(v)}}}}{{{\text{dx}}}} \hfill \\ \end{align}

मान लीजिए ${\text{y}}\;{\text{ = }}\;{\text{f(x)}}\;{\text{ = }}\;{\text{uvw}}$ दोनों तरफ का लघुगणक लेने पर,

\begin{align} {\text{log(y)}}\;{\text{ = }}\;{\text{log( u}}{\text{.v}}{\text{.w)}} \hfill \\ {\text{ = }}\;{\text{log(u) + log(v) + log(w)}} \hfill \\ \end{align}

${\text{x}}$ के सापेक्ष अवकलन करने पर,

\begin{align} \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{w}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{y}}\left[ {\dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{w}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}} \right] \hfill \\ \end{align}

\begin{align} {\text{ = }}\;{\text{uvw}}\left[ {\dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{v}}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{w}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}} \right] \hfill \\ {\text{ = }}\;{\text{u \times v \times }}\dfrac{{{\text{dw}}}}{{{\text{dx}}}}{\text{ + w \times v}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + w}}{\text{.u}}\dfrac{{{\text{d(v)}}}}{{{\text{dx}}}} \hfill \\ \end{align}

स्पष्ट है कि परिणाम समान है।

### प्रश्नावली 5.6

यदि प्रश्न संख्या 1 से 10 तक मे ${\text{x,y}}$ दिए समीकरणों द्वारा, एक दूसरे से प्रचलिक रूप मे संबंधित हो, तो प्राचलों का विलोपन किए बिना, $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ ज्ञात कीजिए।

1. ${\text{x}}\;{\text{ = }}\;{\text{2a}}{{\text{t}}^{\text{2}}}{\text{,}}\;{\text{y}}\;{\text{ = }}\;{\text{a}}{{\text{t}}^{\text{4}}}$

उत्तर: $\dfrac{{{\text{dx}}}}{{{\text{dt}}}}\;{\text{ = }}\;{\text{2a(2t),}}\;\dfrac{{{\text{dy}}}}{{{\text{dt}}}}\;{\text{ = }}\;{\text{a}}\left( {{\text{4}}{{\text{t}}^{\text{3}}}} \right)$

${\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/dt)/(dx/dt)}}\;{\text{ = }}\;{\text{4a}}{{\text{t}}^{\text{3}}}{\text{/4at}}\;{\text{ = }}\;{{\text{t}}^{\text{2}}}$

2. ${\text{x}}\;{\text{ = }}\;{\text{acos\theta ,}}\;{\text{y}}\;{\text{ = }}\;{\text{bcos\theta }}$

उत्तर: ${\text{dx/d\theta }}\;{\text{ = }}\;{\text{a( - sin\theta ),}}\;{\text{dy/d\theta }}\;{\text{ = }}\;{\text{b( - sin\theta )}}$

${\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/d\theta )/(dx/d\theta )}}\;{\text{ = }}\;{\text{ - bsin\theta / - asin\theta }}\;{\text{ = }}\;{\text{b/a}}$

3. ${\text{x}}\;{\text{ = }}\;{\text{sint,}}\;{\text{y}}\;{\text{ = }}\;{\text{cos2t}}$

उत्तर: ${\text{dx/dt}}\;{\text{ = }}\;{\text{cost,}}\;{\text{dy/dt}}\;{\text{ = }}\;{\text{ - sin2t}}{\text{.2}}$

\begin{align} {\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/dt)/(dx/dt)}} \hfill \\ {\text{ = }}\;{\text{ - 2sin2t/cot}}\;{\text{t}} \hfill \\ {\text{ = }}\;{\text{ - 4sint}} \hfill \\ \end{align}

4. ${\text{x}}\;{\text{ = }}\;{\text{4t,}}\;{\text{y}}\;{\text{ = }}\;\dfrac{4}{{\text{t}}}$

उत्तर: ${\text{dx/dt}}\;{\text{ = }}\;{\text{4,}}\;{\text{dy/dt}}\;{\text{ = }}\;{\text{ - 4/}}{{\text{t}}^{\text{2}}}$

\begin{align} {\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/dt)/(dx/dt)}} \hfill \\ {\text{ = }}\;\left( {{\text{ - 4/}}{{\text{t}}^{\text{2}}}} \right){\text{/4}} \hfill \\ {\text{ = }}\;{\text{ - 1/}}{{\text{t}}^{\text{2}}} \hfill \\ \end{align}

5. ${\text{x}}\;{\text{ = }}\;{\text{cos\theta - cos2\theta ,}}\,{\text{y}}\;{\text{ = }}\;{\text{sin\theta - sin2\theta }}$

उत्तर: ${\text{dx/d\theta }}\;{\text{ = }}\;{\text{ - sin\theta + 2sin2\theta ,}}\;{\text{dy/d\theta }}\;{\text{ = }}\;{\text{cos\theta + 2cos2\theta }}$

\begin{align} {\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/d\theta )/(dx/d\theta )}} \hfill \\ {\text{ = }}\;{\text{(cos\theta + 2cos2\theta )/( - sin\theta + 2sin2\theta )}} \hfill \\ \end{align}

6. ${\text{x}}\;{\text{ = }}\;{\text{a(\theta - sin\theta ),}}\;{\text{y}}\;{\text{ = }}\;{\text{a(1 + cos\theta )}}$

उत्तर: ${\text{dx/d\theta }}\;{\text{ = }}\;{\text{a(1 - cos\theta ),}}\;{\text{dy/d\theta }}\;{\text{ = }}\;{\text{a(0 - sin\theta )}}$

\begin{align} {\text{dy/dx}}\,{\text{ = }}\;{\text{(dy/d\theta )/(dx/d\theta )}} \hfill \\ {\text{ = }}\;{\text{a(0 - sin\theta )/a(1 - cos\theta )}} \hfill \\ {\text{ = }}\;{\text{ - cot(\theta /2)}} \hfill \\ \end{align}

7. ${\text{x}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{\text{3}}}{\text{t/}}\sqrt {{\text{cos}}} {\text{2t,}}\;{\text{y}}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{\text{3}}}{\text{t/}}\sqrt {{\text{cos}}} {\text{2t}}$

उत्तर: ${\text{dx/dt}}\;{\text{ = }}\;\left( {{\text{si}}{{\text{n}}^{\text{3}}}{\text{t(d/dt) \times cos2t - }}\sqrt {{\text{cos2t}}} {\text{(d/dt)si}}{{\text{n}}^{\text{3}}}{\text{t}}} \right){\text{/(}}\sqrt {{\text{cos2t}}} {{\text{)}}^{\text{2}}}$

${\text{ = }}\left( {{\text{ - si}}{{\text{n}}^{\text{3}}}{\text{t \times sin2t + 3cos2t \times co}}{{\text{s}}^{\text{2}}}{\text{ \times tsint}}} \right){\text{/(cos2t}}\sqrt {{\text{cos2t}}} {\text{)}}$

\begin{align} {\text{dy/dt}}\;{\text{ = }}\;\left( {{\text{co}}{{\text{s}}^{\text{3}}}{\text{t(d/dt)}}\sqrt {{\text{cos2t}}} {\text{ - }}\sqrt {{\text{cos2t}}} {\text{(d/dt)co}}{{\text{s}}^{\text{3}}}{\text{t}}} \right){\text{/(}}\sqrt {{\text{cos2t}}} {{\text{)}}^{\text{2}}} \hfill \\ {\text{ = }}\;\left( {{\text{ - co}}{{\text{s}}^{\text{3}}}{\text{t \times sin2t + 3cos2t \times co}}{{\text{s}}^{\text{2}}}{\text{ \times tsint}}} \right){\text{/(cos2t}}\sqrt {{\text{cos2t}}} {\text{)}} \hfill \\ {\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/dt)/(dx/dt)}}\; \hfill \\ {\text{ = }}\left[ {\left( {{\text{ - co}}{{\text{s}}^{\text{3}}}{\text{t \times sin2t + 3cos2t \times co}}{{\text{s}}^{\text{2}}}{\text{ \times tsin}}} \right.} \right.{\text{t) /(cos2t}}\sqrt {{\text{cos2t}}} {\text{)]/}}\left[ {\left( {\left( {{\text{ - si}}{{\text{n}}^{\text{3}}}{\text{t \times sin2t + 3cos2t \times co}}{{\text{s}}^{\text{2}}}{\text{ \times tsin}}} \right.} \right.} \right.{\text{t) /(cos2t}}\sqrt {{\text{cos2t}}} {\text{)]}} \hfill \\ {\text{ = }}\;\left[ {{\text{cost}}\left( {{\text{ - 2co}}{{\text{s}}^{\text{2}}}{\text{t + 6cos2t - 3}}} \right)} \right]{\text{/}}\left[ {{\text{si}}{{\text{n}}^3}{\text{t}}\left( {{\text{ - 2si}}{{\text{n}}^{\text{2}}}{\text{t - 3 + 6sin2t}}} \right)} \right] \hfill \\ {\text{ = }}\;{\text{ - cos3t/sin3t}} \hfill \\ {\text{ = }}\;{\text{ - cot3t}} \hfill \\ \end{align}

8. ${\text{x}}\;{\text{ = }}\;{\text{a(cost + log}}\,{\text{tan(t/2)),}}\;{\text{y}}\;{\text{ = }}\;{\text{asint}}$

उत्तर: ${\text{dx/dt}}\;{\text{ = }}\;{\text{a}}\left[ {{\text{ - sint + }}\left( {{\text{\{ 1/tant/2\} \times }}\left\{ {{\text{se}}{{\text{c}}^{\text{2}}}{\text{t/2}}{\text{.1/2}}} \right\}} \right)} \right]$

\begin{align} {\text{ = }}\;{\text{a( - sint + \{ 1/2 \times sint/2 \times cost/2\} )}} \hfill \\ {\text{ = }}\;{\text{a}}\left( {\left\{ {{\text{ - si}}{{\text{n}}^{\text{2}}}{\text{t + 1}}} \right\}{\text{/sint}}} \right) \hfill \\ {\text{ = }}\;{\text{a}}\left( {{\text{co}}{{\text{s}}^{\text{2}}}{\text{t/sint}}} \right) \hfill \\ {\text{dy/dt}}\;{\text{ = }}\;{\text{acost}} \hfill \\ \end{align}

\begin{align} {\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/dt)/(dx/dt)}} \hfill \\ {\text{ = }}\;{\text{(acost)/[a(a(co}}{{\text{s}}^{\text{2}}}{\text{t/sint))]}} \hfill \\ {\text{ = }}\;{\text{sint/cost}} \hfill \\ {\text{ = }}\;{\text{tan}}\;{\text{t}} \hfill \\ \end{align}

9. ${\text{x}}\;{\text{ = }}\;{\text{asec\theta ,}}\;{\text{y}}\;{\text{ = }}\;{\text{btan\theta }}$

उत्तर: ${\text{dx/d\theta }}\;{\text{ = }}\;{\text{asec\theta tan\theta ,}}\;{\text{dy/d\theta }}\;{\text{ = }}\;{\text{bse}}{{\text{c}}^{\text{2}}}{\text{\theta }}$

\begin{align} {\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/d\theta )/(dx/d\theta )}}\;{\text{ = }}\;{\text{bse}}{{\text{c}}^{\text{2}}}{\text{\theta /asec\theta tan\theta }} \hfill \\ {\text{ = }}\;{\text{bsec\theta /atan\theta }} \hfill \\ {\text{ = }}\;({\text{b/a)cosec\theta }} \hfill \\ \end{align}

10. ${\text{x}}\;{\text{ = }}\;{\text{a(cos\theta + \theta sin\theta ),}}\;{\text{y}}\;{\text{ = }}\;{\text{a(sin\theta - \theta cos\theta )}}$

उत्तर: ${\text{dx/d\theta }}\;{\text{ = }}\;{\text{a[ - sin\theta + (\theta cos\theta + sin\theta )]}}\;{\text{ = }}\,{\text{a\theta cos\theta }}$

\begin{align} {\text{dy/d\theta }}\;{\text{ = }}\;{\text{a[cos\theta - ( - sin\theta + cos\theta )]}} \hfill \\ {\text{ = }}\;{\text{a\theta sin\theta }} \hfill \\ {\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/d\theta )/(dx/d\theta )}} \hfill \\ {\text{ = }}\;{\text{a\theta sin\theta /a\theta cos\theta }} \hfill \\ {\text{ = }}\;{\text{tan\theta }} \hfill \\ \end{align}

11. यदि $\left. {{\text{x}}\;{\text{ = }}\;\sqrt {\text{(}} {{\text{a}}^{{\text{si}}{{\text{n}}^{ - 1}}{\text{t}}}}} \right){\text{,}}\;{\text{y}}\;{\text{ = }}\;\sqrt {\left( {{{\text{a}}^{{\text{co}}{{\text{s}}^{ - 1}}{\text{t}}}}} \right)}$ , तो दर्शाइए की $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{y}}}{{\text{x}}}$

उत्तर: ${\text{dx/dt}}\;{\text{ = }}\;{\text{xloga/}}\sqrt {{\text{1 - }}{{\text{t}}^2}}$

\begin{align} {\text{dy/dt}}\;{\text{ = }}\;{\text{ - yloga/}}\sqrt {{\text{1 - }}{{\text{t}}^2}} \hfill \\ {\text{dy/dx}}\;{\text{ = }}\;{\text{(dy/dt)/(dx/dt)}} \hfill \\ {\text{ = }}\;\left( {{\text{ - yloga/}}\sqrt {{\text{1 - }}{{\text{t}}^2}} } \right){\text{/}}\left( {{\text{xloga/}}\sqrt {{\text{1 - }}{{\text{t}}^2}} } \right) \hfill \\ {\text{ = }}\;{\text{ - y/x}} \hfill \\ \end{align}

### प्रश्नावली 5.7

प्रश्न संख्या 1 से 10 तक दिए गए फलनों के द्वितीय कोटी के अवकलज ज्ञात कीजिए:

1. ${{\text{x}}^{\text{2}}}{\text{ + 3x + 2}}$

उत्तर: ${\text{ = }}\;{\text{2x + 3}}\;{\text{ = }}\;{\text{2}}$

2. ${{\text{x}}^{{\text{20}}}}$

उत्तर: ${\text{ = }}\;{\text{20}}{{\text{x}}^{{\text{19}}}}\,{\text{ = }}\;{\text{380}}{{\text{x}}^{{\text{18}}}}$

3. ${\text{x \times cosx}}$

उत्तर: ${\text{ = }}\;{\text{x( - sinx) + cos}}\;{\text{x}}$

\begin{align} {\text{ = }}\;{\text{( - sinx) - xcosx - sinx}} \hfill \\ {\text{ = }}\;{\text{ - (xcosx + 2sinx)}} \hfill \\ \end{align}

4. ${\text{logx}}$

उत्तर: ${\text{ = }}\;\dfrac{{\text{1}}}{{\text{x}}}\;{\text{ = }}\;{\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}$

5. ${{\text{x}}^{\text{3}}}{\text{logx}}$

उत्तर: ${\text{ = }}\;{\text{3}}{{\text{x}}^{\text{2}}}{\text{logx + }}{{\text{x}}^{\text{3}}}{\text{/x}}$

${\text{ = }}\;{\text{3}}{{\text{x}}^{\text{2}}}{\text{logx + }}{{\text{x}}^{\text{2}}}$

${\text{ = }}\;{\text{6x \times logx + 3}}{{\text{x}}^{\text{2}}}{\text{/x + 2x}}$

${\text{ = }}\;{\text{6x \times logx + 5x}}$

6. ${{\text{e}}^{\text{x}}}{\text{sin5x}}$

उत्तर: ${\text{ = }}{{\text{e}}^{\text{x}}}{\text{sin5x + }}{{\text{e}}^{\text{x}}}{\text{cos5x}}$

\begin{align} {\text{ = }}{{\text{e}}^{\text{x}}}{\text{sin5x + 2}}{{\text{e}}^{\text{x}}}{\text{cos5x - }}{{\text{e}}^{\text{x}}}{\text{sin5x}} \hfill \\ {\text{ = 2}}{{\text{e}}^{\text{x}}}{\text{cos5x}} \hfill \\ \end{align}

7. ${{\text{e}}^{{\text{6x}}}}{\text{cos3x}}$

उत्तर: ${\text{ = 6}}{{\text{e}}^{{\text{6x}}}}{\text{ \times cos3x - sin3x \times }}{{\text{e}}^{{\text{6x}}}}$

${\text{ = 36}}{{\text{e}}^{{\text{6x}}}}{\text{ \times cos3x - sin3x \times 6}}{{\text{e}}^{{\text{6x}}}}{\text{ - }}\left( {{\text{cos3x \times }}{{\text{e}}^{{\text{6x}}}}{\text{ + 6}}{{\text{e}}^{{\text{6x}}}}{\text{ \times sin3x}}} \right)$

8. ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}$

उत्तर: ${\text{ = }}\dfrac{1}{{\sqrt {1 - {{\text{x}}^{\text{2}}}} }}$

${\text{ = - }}\dfrac{{2{\text{x}}}}{{2\sqrt {1 - {{\text{x}}^{\text{2}}}} }}$

9. ${\text{log(logx)}}$

उत्तर: ${\text{ = }}\dfrac{1}{{{\text{log}}\;{\text{x}}}}$

${\text{ = }}\;{\text{x}}$

10. ${\text{sin(logx)}}$

उत्तर: ${\text{ = cos(logx)}}$

${\text{ = cos}}\left( {{{\text{x}}^{{\text{ - 1}}}}} \right)$

11. यदि ${\text{y}}\;{\text{ = }}\;{\text{5cosx - 3sinx}}$ है तो सिद्ध कीजिए की $\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{{\text{d}}^{\text{2}}}{\text{x}}}}{\text{ + y}}\;{\text{ = }}\;{\text{0}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;{\text{5cosx - 3sinx}}$

${\text{dy/dx = - 5sinx - 3cosx}}$

${{\text{d}}^{\text{2}}}{\text{y/d}}{{\text{x}}^{\text{2}}}{\text{ = - 5cosx + 3sinx}}$

$\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{{\text{d}}^{\text{2}}}{\text{x}}}}{\text{ + y}}\;{\text{ = }}\;{\text{0}}$

\begin{align} {\text{5cosx - 3sinx - 5cosx + 3sinx}}\;{\text{ = 0}} \hfill \\ {\text{0 = 0}} \hfill \\ \end{align}

12. यदि ${\text{y }}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}$ है तो $\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{{\text{d}}^{\text{2}}}{\text{x}}}}$ को केवल ${\text{y}}$ के पदों मे ज्ञात कीजिए।

उत्तर: ${\text{y }}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}$

$\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{{\text{d}}^{\text{2}}}{\text{x}}}}$ को केवल ${\text{y}}$ के पदों मे

\begin{align} {\text{y = - }}\dfrac{1}{{\sqrt {{\text{1 - }}{{\text{x}}^2}} }} \hfill \\ {\text{ y}} \times \left( {\sqrt {{\text{1 - }}{{\text{x}}^2}} } \right){\text{ = - 1}} \hfill \\ \end{align}

${{\text{y}}^{\text{2}}} \times \left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right){\text{ = 1}}$

${\text{2y \times }}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right){\text{ = 1}}$

${\text{y = }}\dfrac{1}{{{\text{2}}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)}}$

13. यदि ${\text{y}}\;{\text{ = }}\;{\text{3cos(logx) + 4sin(logx)}}$ है तो दर्शाइए की ${{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}{\text{ + x}}{{\text{y}}_{\text{1}}}{\text{ + y}}\;{\text{ = }}\;{\text{0}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;{\text{3cos(logx) + 4sin(logx)}}$

\begin{align} {{\text{y}}_1}{\text{ = 3sin(logx) \times }}\dfrac{1}{{\text{x}}}{\text{ - 4cos(logx) \times }}\dfrac{1}{{\text{x}}} \hfill \\ {{\text{y}}_1} \times {\text{x = 3sin(logx) - 4cos(logx)}} \hfill \\ \end{align}

\begin{align} {\text{(}}{{\text{y}}_2} \times {\text{x) + }}{{\text{y}}_1}{\text{ = ( - 3cos(logx) - 4sin(logx))}}\dfrac{{\text{1}}}{{\text{x}}} \hfill \\ {{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}{\text{ + x}}{{\text{y}}_{\text{1}}}\;{\text{ = }}\;{\text{ - 3cos(logx) - 4sin(logx)}} \hfill \\ {{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}{\text{ + x}}{{\text{y}}_{\text{1}}} + {{\text{y}}_{\text{1}}}\;{\text{ = }}\;{\text{0}} \hfill \\ {\text{ - 3cos(logx) - 4sin(logx) + 3cos(logx) + 4sin(logx)}}\; = \;0 \hfill \\ 0\; = \;0 \hfill \\ \end{align}

14. यदि ${\text{y}}\;{\text{ = }}\;{\text{A}}{{\text{e}}^{{\text{mx}}}}{\text{ + B}}{{\text{e}}^{{\text{nx}}}}$  है तो दर्शाइए की ${{\text{d}}^{\text{2}}}{\text{y/d}}{{\text{x}}^{\text{2}}}{\text{ - (m + n)dy/dx + mny}}\;{\text{ = }}\;{\text{0}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;{\text{A}}{{\text{e}}^{{\text{mx}}}}{\text{ + B}}{{\text{e}}^{{\text{nx}}}}$

\begin{align} {\text{dy/dx = mA}}{{\text{e}}^{{\text{mx}}}}{\text{ + nB}}{{\text{e}}^{{\text{nx}}}} \hfill \\ {\text{(m + n)(dy/dx) = }}{{\text{m}}^{\text{2}}}{\text{A}}{{\text{e}}^{{\text{mx}}}}{\text{ + mnA}}{{\text{e}}^{{\text{mx}}}}{\text{ + }}{{\text{n}}^{\text{2}}}{\text{B}}{{\text{e}}^{{\text{nx}}}}{\text{ + mnB}}{{\text{e}}^{{\text{nx}}}} \hfill \\ {{\text{d}}^{\text{2}}}{\text{y/d}}{{\text{x}}^{\text{2}}}{\text{ = }}{{\text{m}}^{\text{2}}}{\text{A}}{{\text{e}}^{{\text{mx}}}}{\text{ + }}{{\text{n}}^{\text{2}}}{\text{B}}{{\text{e}}^{{\text{nx}}}} \hfill \\ {{\text{d}}^{\text{2}}}{\text{y/d}}{{\text{x}}^{\text{2}}}{\text{ - (m + n)dy/dx + mny = 0}} \hfill \\ {{\text{m}}^{\text{2}}}{\text{A}}{{\text{e}}^{{\text{mx}}}}{\text{ + }}{{\text{n}}^{\text{2}}}{\text{B}}{{\text{e}}^{{\text{nx}}}}{\text{ - }}{{\text{m}}^{\text{2}}}{\text{A}}{{\text{e}}^{{\text{mx}}}}{\text{ - mnA}}{{\text{e}}^{{\text{mx}}}}{\text{ - }}{{\text{n}}^{\text{2}}}{\text{B}}{{\text{e}}^{{\text{nx}}}}{\text{ - mnB}}{{\text{e}}^{{\text{nx}}}}{\text{ + mnA}}{{\text{e}}^{{\text{mx}}}}{\text{ + mnB}}{{\text{e}}^{{\text{nx}}}}\;{\text{ = = }}\;{\text{0}} \hfill \\ {\text{0}}\;{\text{ = }}\;{\text{0}} \hfill \\ \end{align}

15. यदि ${\text{y}}\;{\text{ = }}\;{\text{500}}{{\text{e}}^{{\text{7x}}}}{\text{ + 600}}{{\text{e}}^{{\text{ - 7x}}}}$ है तो दर्शाइए की ${{\text{d}}^{\text{2}}}{\text{y/d}}{{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{49y}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;{\text{500}}{{\text{e}}^{{\text{7x}}}}{\text{ + 600}}{{\text{e}}^{{\text{ - 7x}}}}$

\begin{align} {\text{dy/dx = 7}}\left( {{\text{500}}{{\text{e}}^{{\text{7x}}}}{\text{ - 600}}{{\text{e}}^{{\text{ - 7x}}}}} \right) \hfill \\ {{\text{d}}^{\text{2}}}{\text{y/d}}{{\text{x}}^{\text{2}}}{\text{ = 49}}\left( {{\text{500}}{{\text{e}}^{{\text{7x}}}}{\text{ + 600}}{{\text{e}}^{{\text{ - 7x}}}}} \right) \hfill \\ {{\text{d}}^{\text{2}}}{\text{y/d}}{{\text{x}}^{\text{2}}}{\text{ = 49y}} \hfill \\ {\text{49}}\left( {{\text{500}}{{\text{e}}^{{\text{7x}}}}{\text{ + 600}}{{\text{e}}^{{\text{ - 7x}}}}} \right)\; = \;49{\text{y}} \hfill \\ 49{\text{y}}\; = \;49{\text{y}} \hfill \\ \end{align}

16. यदि ${{\text{e}}^{\text{x}}}{\text{(x + 1)}}\;{\text{ = }}\;{\text{1}}$ है तो दर्शाइए की ${{\text{d}}^{\text{2}}}{\text{y/d}}{{\text{x}}^{\text{2}}}\;{\text{ = }}\,{{\text{(dy/dx)}}^{\text{2}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;{{\text{e}}^{\text{x}}}{\text{(x + 1)}}$

\begin{align} {{\text{e}}^{\text{x}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{(x + 1)}}}} \hfill \\ {\text{log}}\;{{\text{e}}^{\text{x}}}\;{\text{ = }}\;{\text{log}}\dfrac{{\text{1}}}{{{\text{(x + 1)}}}} \hfill \\ {\text{y}}\;{\text{loge}}\;{\text{ = }}\;{\text{log1 - log(x + 1)}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{ - }}\dfrac{{\text{1}}}{{{\text{x + 1}}}} \hfill \\ \dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}\;}}\;{\text{ = }}\;{\text{ - }}\dfrac{{{\text{( - 1)}}\;}}{{{{{\text{(x + 1)}}}^{\text{2}}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{{{\text{(x + 1)}}}^{\text{2}}}}} \hfill \\ {{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}^{\text{2}}}\;{\text{ = }}\;{{\text{(}}\dfrac{{{\text{ - 1}}}}{{{\text{x + 1}}}}{\text{)}}^{\text{2}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{{{\text{(x + 1)}}}^{\text{2}}}}} \hfill \\ \end{align}

17. यदि ${\text{y}}\;{\text{ = }}\;{\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}$ है तो दर्शाइए की ${\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)^{\text{2}}}{{\text{y}}_{\text{2}}}{\text{ + 2x}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){{\text{y}}_{\text{1}}}\;{\text{ = }}\;{\text{2}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;{\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}$

\begin{align} {{\text{y}}_{\text{1}}}\;{\text{ = }}\;{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}} \hfill \\ {\text{(}}{{\text{x}}^{\text{2}}}{\text{ + 1)}}{{\text{y}}_{\text{1}}}\;{\text{ = }}\;{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ {\text{(}}{{\text{x}}^{\text{2}}}{\text{ + 1)}}{{\text{y}}_{\text{2}}}{\text{ + 2x}}{{\text{y}}_{\text{1}}}\;{\text{ = }}\;\dfrac{{\text{2}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}} \hfill \\ {{\text{(}}{{\text{x}}^{\text{2}}}{\text{ + 1)}}^{\text{2}}}{{\text{y}}_{\text{2}}}{\text{ + 2x(}}{{\text{x}}^{\text{2}}}{\text{ + 1)}}{{\text{y}}_{\text{1}}}\;{\text{ = }}\;{\text{2}} \hfill \\ \end{align}

### प्रश्नावली 5.8

1. फलन  ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ + 2x - 8,}}\;{\text{x}} \in {\text{[ - 4,2]}}$  के लिए रोले के प्रमेय को सतयापित कीजिए

उत्तर: फलन  ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ + 2x - 8}}$ , अंतराल $[ - 4,2]$ मे सतंत तथा अंतराल $( - 4,2)$ मे अवकलनीय है।

\begin{align} {\text{f( - 4) = - }}{{\text{4}}^{\text{2}}}{\text{ + 2( - 4) - 8}} \hfill \\ {\text{f( - 4) = 16 - 8 - 8 = 0}} \hfill \\ {\text{f(2) = }}{{\text{2}}^{\text{2}}}{\text{ + 2(2) - 8}} \hfill \\ {\text{f(2) = 4 + 4 - 8 = 0}} \hfill \\ \end{align}

इसलिए ${\text{f( - 4)}}\;{\text{ = }}\;{\text{f(2)}}\;{\text{ = }}\;{\text{0}}$ है अतएव ${\text{f(x)}}$ का मान ${\text{ - 4,}}\;{\text{2 }}$ पर समान है रोले के प्रेमय के अनुसार एक बिन्दु ${\text{c}} \in {\text{( - 4,2) }}$ का अस्तित्व होगा, जहा ${{\text{f}}^{\text{'}}}{\text{(c)}}\;{\text{ = }}\;{\text{0}}$ है।

चूंकि ${{\text{f}}^{\text{'}}}{\text{(x)}}\;{\text{ = }}\;{\text{2x + 2}}$ है इसलिए ${\text{2c + 2 }}\;{\text{ = }}\;{\text{0 }} \Rightarrow \;{\text{c }}\;{\text{ = }}\;{\text{ - 1 }}$ पर  ${{\text{f}}^{\text{'}}}{\text{(c)}}\;{\text{ = }}\;{\text{0}}$ और ${\text{c}}\;{\text{ = }}\;{\text{ - 1 }} \in {\text{( - 4,2)}}$ चूंकि रोले के प्रेमय की तीनों शर्ते इस फलन मे संतुष्ट है, इसलिए रोले के प्रेमय सत्यापित है।

2. जाँच कीजिये कि क्या रोले का प्रमेय निम्लिखित फलनों में से किन किन पर लागू होता हैं। इन उदाहरणों से क्या आप रोले के प्रमेय के विलोम के बारे में कुछ कह सकते हैं?

(i) ${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right]{\text{,}}\;{\text{x}} \in {\text{[5,9]}}$

उत्तर: फलन ${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right]$ , अंतराल $[5,9]$ मे संतत नहीं है नाही अंतराल $( - 4,2)$ मे अवकलनीय है, इसलिए यह फलन रोले के प्रेमय को सत्यापित नहीं करेगा।

(ii) ${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right]{\text{,}}\;{\text{x}} \in {\text{[ - 2,2]}}$

उत्तर: फलन ${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right]$ , अंतराल ${\text{[ - 2,2]}}$ मे संतत नहीं है नाही अंतराल ${\text{( - 4,2)}}$ मे अवकलनीय है इसलिए यह फलन रोले के प्रेमय को सत्यापित नहीं करेगा।

(iii) ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^2} - 1,\;{\text{x}} \in {\text{[1,2]}}$

उत्तर: फलन ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - 1}}$ अंतराल ${\text{[1,2]}}$ मे संतत तथा अंतराल ${\text{(1,2)}}$ मे अवकलनीय है।

\begin{align} {\text{f(1)}}\;{\text{ = }}\;{{\text{1}}^{\text{2}}}{\text{ - 1}}\;{\text{ = }}\;{\text{0}} \hfill \\ {\text{f(2)}}\;{\text{ = }}\;{{\text{2}}^{\text{2}}}{\text{ - 1}}\;{\text{ = }}\;{\text{2 - 1}}\;{\text{ = }}\;{\text{1}} \hfill \\ \end{align}

इसलिए ${\text{f(1)}} \ne {\text{f(2)}}$ है अतएव ${\text{f(x)}}$ का मान ${\text{1,}}\;{\text{2}}$ पर समान है इसलिए यह फलन रोले के प्रेमय को सत्यापित नहीं करेगा इन उदाहरण से हम यह कह सकते है की रोले के प्रेमय का विलोम सत्यापित नहीं है।

3. यदि ${\text{f}}\;:\;[ - 5,5]\; \to \;\mathbb{R}$ एक सतंत फलन हैं और यदि ${{\text{f}}^{\text{'}}}{\text{(x)}}$  किसी भी बिंदु पर शुन्य नहीं होता हैं तो सिद्ध कीजिए कि ${\text{f( - 5)}} \ne {\text{f(5)}}$

उत्तर: ${\text{f}}\;:\;[ - 5,5]\; \to \;\mathbb{R}$ एक सतंत फलन हैं

माध्यमान प्रेमय का प्रयोग करके हम कह सकते है की $\dfrac{{{\text{f(b) - f(a)}}}}{{{\text{b - a}}}}\;{\text{ = }}\;{{\text{f}}^{\text{'}}}{\text{(c)}}$

परंतु ${{\text{f}}^{\text{'}}}{\text{(x)}}\; \ne \;0$

इसलिए $\dfrac{{{\text{f(b) - f(a)}}}}{{{\text{b - a}}}}\; \ne \;0$

${\text{a}}\;{\text{ = }}\;{\text{ - 5,}}\;{\text{b}}\;{\text{ = }}\;{\text{5}}$

इसलिए $\dfrac{{{\text{f(5) - f( - 5)}}}}{{{\text{5 - ( - 5)}}}}\; \ne \;0$

${\text{f( - 5)}} \ne {\text{f(5)}}$ सत्यापित हुआ

4. माध्यमान प्रमेय सत्यापित कीजिये यदि अंतराल ${\text{[a,b]}}$  में, ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - 4x - 3}}$ जहाँ${\text{a}}\;{\text{ = }}\;{\text{1,}}\;{\text{b}}\;{\text{ = }}\;{\text{4}}$ हैं।

उत्तर: फलन ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - 4x - 3}}$ अंतराल $[1,4]$ मे संतत तथा अंतराल $(1,4)$ मे अवकलनीय है, क्योंकि इसका अवकलन ${{\text{f}}^{\text{'}}}{\text{(x)}}\;{\text{ = }}\;{\text{2x - 4}}$ अंतराल ${\text{(1,4)}}$ मे परिभाषित है

अब ${\text{f(1)}}\;{\text{ = }}\;{{\text{1}}^{\text{2}}}{\text{4(1) - 3}}\;{\text{ = }}\;{\text{ - 8}}$

और ${\text{f(4)}}\;{\text{ = }}\;{{\text{4}}^{\text{2}}}{\text{ - 4(4) - 3}}\;{\text{ = }}\;{\text{ - 3}}$ है इसलिए

$\dfrac{{{\text{f(4) - f(1)}}}}{{{\text{4 - 1}}}}\;{\text{ = }}\;\dfrac{{{\text{ - 3 - ( - 8)}}}}{{\text{3}}}\;{\text{ = }}\;\dfrac{{\text{5}}}{{\text{3}}}$

माध्यमान प्रेमय के अनुसार एक बिन्दु ${\text{c}} \in {\text{(1,4)}}$ ऐसा होना चाहिए ताकि ${{\text{f}}^{\text{'}}}{\text{(c)}}\;{\text{ = }}\;\dfrac{5}{3}$ हो।

यह ${{\text{f}}^{\text{'}}}{\text{(x)}}\;{\text{ = }}\;{\text{2x - 4}}$

इसलिए ${\text{2x - 4}}\;{\text{ = }}\;\dfrac{5}{3}$

${\text{x}}\;{\text{ = }}\;\dfrac{{{\text{17}}}}{{\text{6}}}$

अतः ${\text{c}}\;{\text{ = }}\;\dfrac{{{\text{17}}}}{{\text{6}}}{\text{,}}\;{{\text{f}}^{\text{'}}}{\text{(c)}}\;{\text{ = }}\;\dfrac{{\text{5}}}{{\text{3}}}$ है

तथा मध्यमान प्रेमय सत्यापित है

5. माध्यमान प्रमेय सत्यापित कीजिये यदि अंतराल ${\text{[a,b]}}$ मे ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{3}}}{\text{ - 5}}{{\text{x}}^2}{\text{ - 3x}}$  जहाँ ${\text{a}}\;{\text{ = }}\;{\text{1,}}\;{\text{b}}\;{\text{ = }}\;{\text{3}}$ हैं। ${{\text{f}}^{\text{'}}}{\text{(c)}}\; = \;0$ के लिए ${\text{c}} \in (1,3)$ को ज्ञात कीजिए।

उत्तर: फलन ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{3}}}{\text{ - 5}}{{\text{x}}^2}{\text{ - 3x}}$ अंतराल ${\text{[1,3]}}$ मे संतत तथा अंतराल ${\text{(1,3)}}$ मे अवकलनीय है क्योंकि इसका अवकलन ${{\text{f}}^{\text{'}}}{\text{(x)}}\;{\text{ = }}\;{\text{2}}{{\text{x}}^{\text{2}}}{\text{ - 10x - 3}}$ अंतराल ${\text{(1,3)}}$ मे परिभाषित है

अब ${\text{f(1)}}\;{\text{ = }}\;{{\text{1}}^3} - 5({1^2}) - 3(1)\; = \; - 7$

और ${\text{f(3)}}\;{\text{ = }}\;{3^3} - 5({3^2}) - 3()\; = \; - 27$

इसलिए $\dfrac{{{\text{f(3) - f(1)}}}}{{{\text{3 - 1}}}}\;{\text{ = }}\;\dfrac{{{\text{ - 27 - ( - 7)}}}}{{\text{2}}}\;{\text{ = }}\;{\text{ - 10}}$

मध्यमान प्रेमय के अनुसार एक बिन्दु ${\text{c}} \in {\text{(1,3)}}$ ऐसा होना चाहिए ताकि ${{\text{f}}^{\text{'}}}{\text{(c)}}\; = \; - 10$ हो

यह ${{\text{f}}^{\text{'}}}{\text{(x)}}\;{\text{ = }}\;{\text{3}}{{\text{x}}^{\text{2}}}{\text{ - 10x - 3}}$

इसलिए ${\text{3}}{{\text{x}}^{\text{2}}}{\text{ - 10x - 3}}\;{\text{ = }}\;{\text{ - 10}}$

${\text{x}}\;{\text{ = }}\;{\text{1,}}\;{\text{x}}\;{\text{ = }}\;\dfrac{7}{3}$

अतः ${\text{c}}\;{\text{ = }}\;\dfrac{{\text{7}}}{{\text{3}}}{\text{,}}\;{\text{c}}\;{\text{ = }}\;{\text{1,}}\;{{\text{f}}^{\text{'}}}{\text{(c)}}\;{\text{ = }}\;{\text{10}}$

तथा मध्यमान प्रेमय सत्यापित है

अब ${{\text{f}}^{\text{'}}}{\text{(c)}}\;{\text{ = }}\;{\text{0,}}\;{\text{c}} \in (1,3)$

\begin{align} {{\text{f}}^{\text{'}}}{\text{(c)}}\;{\text{ = }}\;{\text{3}}{{\text{c}}^{\text{2}}}{\text{ - 10c - 3}}\;{\text{ = }}\;{\text{0}} \hfill \\ {\text{c}}\;{\text{ = }}\;\dfrac{{{\text{5 \pm }}\sqrt {{\text{34}}} }}{{\text{3}}} \hfill \\ \end{align}

अतः ${\text{c}}\;{\text{ = }}\;\dfrac{{{\text{5 \pm }}\sqrt {{\text{34}}} }}{{\text{3}}},\;{{\text{f}}^{\text{'}}}{\text{(c}})\; = \;0$

6. प्रशन संख्या 2 में उपरोक्त दिए तीनो फलनों के लिए माध्यमान प्रमेय की अनुपयोगिता कि जाँच कीजिए।

उत्तर: (i) ${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right]{\text{,}}\;{\text{x}} \in [5,9]$

फलन ${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right]$ अंतराल $[5,9]$ मे संतत नहीं है नाही अंतराल $( - 4,2)$ मे अवकलनीय है, इसलिए यह फलन मध्यमान प्रेमय को सत्यापित नहीं करेगा।

(ii) ${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right]{\text{,}}\;{\text{x}} \in [ - 2,2]$

फलन ${\text{f(x)}}\;{\text{ = }}\;\left[ {\text{x}} \right]$ अंतराल $[ - 2,2]$ मे संतत नहीं है नाही अंतराल $( - 4,2)$ मे अवकलनीय है, इसलिए यह फलन मध्यमान प्रेमय को सत्यापित नहीं करेगा।

(iii) ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - 1,}}\;{\text{x}} \in [1,2]$

फलन ${\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - 1}}$ अंतराल मे संतत तथा अंतराल $(1,2)$ मे अवकलनीय है क्योंकि इसका अवकलन ${{\text{f}}^{\text{'}}}{\text{(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - 1}}$ अंतराल $(1,2)$ मे परिभाषित है।

\begin{align} {\text{f(1)}}\;{\text{ = }}\;{{\text{1}}^{\text{2}}}{\text{ - 1}}\;{\text{ = }}\;{\text{0}} \hfill \\ {\text{f(2)}}\;{\text{ = }}\;{{\text{2}}^{\text{2}}}{\text{ - 1}}\;{\text{ = }}\;{\text{2 - 1}}\;{\text{ = }}\;{\text{1}} \hfill \\ \dfrac{{{\text{f(2) - f(1)}}}}{{{\text{2 - 1}}}}\;{\text{ = }}\;\dfrac{{{\text{1 - 0}}}}{{\text{1}}}\;{\text{ = }}\;{\text{1}} \hfill \\ \end{align}

मध्यमान प्रेमय के अनुसार एक बिन्दु ${\text{c}} \in (1,2)$ ऐसा होना चाहिए ताकि ${{\text{f}}^{\text{'}}}{\text{(c)}}\,{\text{ = }}\;{\text{1}}$ हो

यह ${{\text{f}}^{\text{'}}}{\text{(x)}}\,{\text{ = }}\;2{\text{x}}$ है

इसलिए ${\text{2x}}\;{\text{ = }}\;{\text{1}}\; \Rightarrow \;{\text{x}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}$

अतः ${\text{c}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}{\text{,}}\;{{\text{f}}^{\text{'}}}{\text{(c)}}\;{\text{ = }}\;{\text{1}}$

तथा मध्यमान प्रेमय सत्यापित है।

### प्रश्नावली A5

प्रश्न संख्या 1 से 11 तक प्रदत फलनों के सापेक्ष अवकलन कीजिए:

1. ${\left( {{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 9x + 5}}} \right)^{\text{9}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$ ${\left( {{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 9x + 5}}} \right)^{\text{9}}}$

सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\left( {{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 9x + 5}}} \right)^{\text{9}}} \hfill \\ {\text{ = 9}}{\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ - 9x + 5}}} \right)^{\text{8}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 9x + 5}}} \right) \hfill \\ {\text{ = 9}}{\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ - 9x + 5}}} \right)^{\text{8}}}{\text{ \times (6x - 9)}} \hfill \\ {\text{ = 9}}{\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ - 9x + 5}}} \right)^{\text{8}}}{\text{ \times 3(2x - 9)}} \hfill \\ {\text{ = 27}}{\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ - 9x + 5}}} \right)^{\text{8}}}{\text{ \times (2x - 9)}} \hfill \\ \end{align}

2. ${\text{si}}{{\text{n}}^{\text{3}}}{\text{x + co}}{{\text{s}}^{\text{6}}}{\text{x}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$ ${\text{si}}{{\text{n}}^{\text{3}}}{\text{x + co}}{{\text{s}}^{\text{6}}}{\text{x}}$

सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{si}}{{\text{n}}^{\text{3}}}{\text{x}}} \right){\text{ + }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{co}}{{\text{s}}^{\text{6}}}{\text{x}}} \right) \hfill \\ {\text{ = 3si}}{{\text{n}}^{\text{2}}}{\text{x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sinx) + 6co}}{{\text{s}}^{\text{5}}}{\text{x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosx)}} \hfill \\ {\text{ = 3si}}{{\text{n}}^{\text{2}}}{\text{xcosx + 6co}}{{\text{s}}^{\text{5}}}{\text{x( - sinx)}} \hfill \\ {\text{ = 3sinxcosx}}\left( {{\text{sinx - 2co}}{{\text{s}}^{\text{4}}}{\text{x}}} \right) \hfill \\ \end{align}

3. ${{\text{(5x)}}^{{\text{3cosx2x}}}}$

उत्तर: दोनों पक्षों मे लघुगणक लेने पर

${\text{logy = 2 cos2x log5x}}$

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 3}}\left[ {{\text{log5x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{cos2x + cos2x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(log5x)}}} \right]} \right. \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 3y}}\left[ {{\text{log5x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cos2x) + cos2x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(log5x)}}} \right] \hfill \\ \end{align}

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 3y}}\left[ {{\text{log5x( - sin2x) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(2x) + cos2x \times }}\dfrac{{\text{1}}}{{{\text{5x}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(5x)}}} \right]$

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 3y}}\left[ {{\text{ - 2sin2xlog5x + }}\dfrac{{{\text{cos2x}}}}{{\text{x}}}} \right]$

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 3y}}\left[ {\dfrac{{{\text{3cos2x}}}}{{\text{x}}}{\text{ - 6sin2xlog5x}}} \right] \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = (5x}}{{\text{)}}^{{\text{3cos2x}}}}\left[ {\dfrac{{{\text{3cos2x}}}}{{\text{x}}}{\text{ - 6sin2xlog5x}}} \right] \hfill \\ \end{align}

4. ${\text{4si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x}}\sqrt {\text{x}} {\text{),}}\;{\text{o}} \leqslant {\text{x}} \leqslant {\text{1}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$ ${\text{4si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x}}\sqrt {\text{x}} {\text{)}}$

सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x}}\sqrt {\text{x}} {\text{)}} \hfill \\ {\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}\left( {{\text{x}}{{\sqrt {{\text{x)}}} }^{\text{2}}}} \right.} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{ \times (x}}\sqrt {\text{x}} {\text{)}} \hfill \\ {\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - (x}}{{\text{)}}^{\text{3}}}} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}} \times \left( {{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}}}} \right) \hfill \\ {\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - (x}}{{\text{)}}^{\text{3}}}} }} \times \dfrac{{\text{3}}}{{\text{2}}} \times \left( {{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}}}} \right) \hfill \\ {\text{ = }}\dfrac{{{\text{3}}\sqrt {\text{x}} }}{{{\text{2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{3}}}} }} \hfill \\ {\text{ = }}\dfrac{{\text{3}}}{{\text{3}}}\sqrt {\dfrac{{\text{x}}}{{{\text{1 - }}{{\text{x}}^{\text{3}}}}}} \hfill \\ \end{align}

5. $\dfrac{{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{2}}}}}{{\sqrt {{\text{2x + 7}}} }}{\text{,}}\;{\text{ - 2 < x < 2}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$ $\dfrac{{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{2}}}}}{{\sqrt {{\text{2x + 7}}} }}$

भागफल करने पर,

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\sqrt {{\text{2x + 7}}} \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{ - }}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{2}}}} \right)\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(}}\sqrt {{\text{2x - 7}}} {\text{)}}}}{{{{{\text{(}}\sqrt {{\text{2x + 7}}} {\text{)}}}^{\text{2}}}}} \hfill \\ {\text{ = }}\dfrac{{\sqrt {{\text{2x + 7}}} [\dfrac{{ - 1}}{{\left. {\sqrt {{\text{1 - }}{{\left( {\dfrac{{\text{x}}}{{\text{2}}}} \right)}^{\text{2}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{\text{x}}}{{\text{2}}}} \right)} \right]{\text{ - }}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{2}}}} \right)\dfrac{{\text{1}}}{{{\text{2}}\sqrt {{\text{2x + 7}}} {\text{dx}}}}{\text{(2x + 7)}}}}]}}{{{\text{2x + 7}}}} \hfill \\ {\text{ = }}\dfrac{{\sqrt {{\text{2x + 7}}} \dfrac{{{\text{ - 1}}}}{{\sqrt {{\text{4 - }}{{\text{x}}^{\text{2}}}} }}{\text{ - }}\left( {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{2}}}} \right)\dfrac{{\text{1}}}{{{\text{2}}\sqrt {{\text{2x + 7}}} }}}}{{{\text{2x + 7}}}} \hfill \\ {\text{ = }}\dfrac{{{\text{ - }}\sqrt {{\text{2x + 7}}} }}{{\sqrt {{\text{4 - }}{{\text{x}}^{\text{2}}}} {\text{(2x + 7)}}}}{\text{ - }}\dfrac{{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{2}}\sqrt {{\text{2x + 7}}} {\text{(2x + 7)}}}} \hfill \\ {\text{ = }}\left[ {\dfrac{{\text{1}}}{{\sqrt {{\text{4 - }}{{\text{x}}^{\text{2}}}} \sqrt {{\text{2x + 7}}} }}{\text{ + }}\dfrac{{{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{{{\text{(2x + 7)}}}^{\dfrac{{\text{3}}}{{\text{2}}}}}}}} \right] \hfill \\ \end{align}

6. ${\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left[ {\dfrac{{\sqrt {{\text{1 + sinx}}} {\text{ + }}\sqrt {{\text{1 - sinx}}} }}{{\sqrt {{\text{1 + sinx}}} {\text{ - }}\sqrt {{\text{1 - sinx}}} }}} \right]{\text{,}}\;{\text{0 < x < }}\dfrac{{\text{\pi }}}{{\text{2}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${\text{co}}{{\text{t}}^{{\text{ - 1}}}}\left[ {\dfrac{{\sqrt {{\text{1 + sinx}}} {\text{ + }}\sqrt {{\text{1 - sinx}}} }}{{\sqrt {{\text{1 + sinx}}} {\text{ - }}\sqrt {{\text{1 - sinx}}} }}} \right]$

\begin{align} {\text{ = }}\dfrac{{\sqrt {{\text{1 + sinx}}} {\text{ + }}\sqrt {{\text{1 - sinx}}} }}{{\sqrt {{\text{1 + sinx}}} {\text{ - }}\sqrt {{\text{1 - sinx}}} }} \hfill \\ {\text{ = }}\dfrac{{{{{\text{(}}\sqrt {{\text{1 + sinx}}} {\text{ + }}\sqrt {{\text{1 - sinx}}} {\text{)}}}^{\text{2}}}}}{{{\text{(}}\sqrt {{\text{1 + sinx}}} {\text{ - }}\sqrt {{\text{1 - sinx}}} {\text{)(}}\sqrt {{\text{1 + sinx}}} {\text{ + }}\sqrt {{\text{1 - sinx}}} {\text{)}}}} \hfill \\ {\text{ = }}\dfrac{{{\text{(1 + sinx) + (1 - sinx) + 2(}}\sqrt {{\text{(1 - sinx)(1 + sinx)}}} {\text{)}}}}{{{\text{(1 + sinx) - (1 - sinx)}}}} \hfill \\ {\text{ = }}\dfrac{{{\text{2 + 2}}\sqrt {{\text{1 - si}}{{\text{n}}^{\text{2}}}{\text{x}}} }}{{{\text{2sinx}}}} \hfill \\ \end{align}

\begin{align} {\text{ = }}\dfrac{{{\text{1 + cosx}}}}{{{\text{sinx}}}} \hfill \\ {\text{ = }}\dfrac{{{\text{2co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{2sin}}\dfrac{{\text{x}}}{{\text{2}}}{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}}} \hfill \\ \end{align}

\begin{align} {\text{ = cot}}\dfrac{{\text{x}}}{{\text{2}}} \hfill \\ \therefore {\text{ y = co}}{{\text{t}}^{{\text{ - 1}}}}\left( {{\text{cot}}\dfrac{{\text{x}}}{{\text{2}}}} \right) \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x)}} \hfill \\ \Rightarrow \;\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}} \hfill \\ \end{align}

7. ${{\text{(logx)}}^{{\text{logx}}}}{\text{,}}\;{\text{x > 1}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$ ${{\text{(logx)}}^{{\text{logx}}}}$

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[logx \times log(logx)]}} \hfill \\ {\text{ = log(logx)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logx) + logx \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[log(logx)]}} \hfill \\ {\text{ = y}}\left[ {{\text{log(logx) \times }}\dfrac{{\text{1}}}{{\text{x}}}{\text{ + logx \times }}\dfrac{{\text{1}}}{{{\text{logx}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logx)}}} \right] \hfill \\ {\text{ = y}}\left[ {\dfrac{{\text{1}}}{{\text{x}}}{\text{log(logx) + }}\dfrac{{\text{1}}}{{\text{x}}}} \right] \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = (logx}}{{\text{)}}^{{\text{logx}}}}\left[ {\dfrac{{\text{1}}}{{\text{x}}}{\text{ + }}\dfrac{{{\text{log(logx)}}}}{{\text{x}}}} \right] \hfill \\ \end{align}

8. ${\text{cos(acosx + bsinx)}}$ किन्ही अचर ${\text{a,}}\;{\text{b}}$ के लिए।

उत्तर: ${\text{y}}\;{\text{ = }}\;$ ${\text{cos(acosx + bsinx)}}$

सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cos(acosx + bsinx)}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = - sin(acosx + bsinx) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{acosx + bsinx}}} \right) \hfill \\ {\text{ = - sin(acosx + bsinx) \times [a( - sinx) + bsinx]}} \hfill \\ {\text{ = (asinx - bcosx) \times sin(acosx + bsinx)}} \hfill \\ \end{align}

9. ${{\text{(sinx - cosx)}}^{{\text{sinx - cosx}}}}{\text{,}}\;\dfrac{{\text{\pi }}}{{\text{4}}}{\text{ < x < }}\dfrac{{{\text{3\pi }}}}{{\text{4}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${{\text{(sinx - cosx)}}^{{\text{sinx - cosx}}}}$

दोनों पक्षों मे लघुगणक लेने पर,

\begin{align} {\text{logy = log}}\left[ {{{{\text{(sinx - cosx)}}}^{{\text{(sinx - cosx)}}}}} \right] \hfill \\ {\text{logy = (sinx - cosx) \times log(sinx - cosx)}} \hfill \\ \end{align}

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{\text{1}}}{{\text{y}}} \times \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[(sinx - cosx)log(sinx - cosx)]}} \hfill \\ \dfrac{{\text{1}}}{{\text{y}}} \times \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = log(sinx - cosx) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sinx - cosx) + (sinx - cosx) \times }}\dfrac{{\text{1}}}{{{\text{(sinx - cosx)}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sinx - cosx)}} \hfill \\ \dfrac{{\text{1}}}{{\text{y}}} \times \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = log(sinx - cosx) \times (cosx + sinx) + (sinx - cosx) \times }}\dfrac{{\text{1}}}{{{\text{(sinx - cosx)}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sinx - cosx)}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = (sinx - cosx}}{{\text{)}}^{{\text{(sinx - cosx)}}}}{\text{[(cosx + sinx) \times log(sinx - cosx) + (cosx + sinx)]}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = (sinx - cosx}}{{\text{)}}^{{\text{(sinx - cosx)}}}}{\text{(cosx + sinx)[1 + log(sinx - cosx)]}} \hfill \\ \end{align}

10. ${{\text{x}}^{\text{x}}}{\text{ + }}{{\text{x}}^{\text{a}}}{\text{ + }}{{\text{a}}^{\text{x}}}{\text{ + }}{{\text{a}}^{\text{a}}}$ किसी नियत ${\text{a > 0,}}\;{\text{x > 0 }}$

उत्तर: माना ${\text{y = }}\;$${{\text{x}}^{\text{x}}}{\text{ + }}{{\text{x}}^{\text{a}}}{\text{ + }}{{\text{a}}^{\text{x}}}{\text{ + }}{{\text{a}}^{\text{a}}}$

माना ${{\text{x}}^{\text{3}}}{\text{ = u, }}{{\text{x}}^{\text{a}}}{\text{ = v, }}{{\text{a}}^{\text{x}}}{\text{ = w, }}{{\text{a}}^{\text{a}}}{\text{ = s}}$

${\text{y = u + v + w + s}}$

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dw}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{ds}}}}{{{\text{dx}}}}$ ..........(1)

\begin{align} {\text{u = }}{{\text{x}}^{\text{x}}} \hfill \\ {\text{logu = log}}{{\text{x}}^{\text{x}}} \hfill \\ {\text{logu = xlogx}} \hfill \\ \end{align}

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ = logx}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x) + x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logx)}} \hfill \\ \dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ = u}}\left[ {{\text{logx \times 1 + x}}\dfrac{{\text{1}}}{{\text{x}}}} \right] \hfill \\ \end{align}

$\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ = }}{{\text{x}}^{\text{x}}}\left[ {{\text{logx \times 1 + x}}\dfrac{{\text{1}}}{{\text{x}}}} \right]\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{(1 + logx)}}$ ..........(2)

\begin{align} {\text{v = }}{{\text{x}}^{\text{a}}} \hfill \\ \dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{(x)}}^{\text{a}}} \hfill \\ \end{align}

$\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ = a}}{{\text{x}}^{{\text{a - 1}}}}$ ........(3)

\begin{align} {\text{w = }}{{\text{a}}^{\text{x}}} \hfill \\ {\text{logw = log}}{{\text{a}}^{\text{x}}} \hfill \\ \end{align}

${\text{logw = x(log}}\,{\text{a)}}$

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{\text{1}}}{{\text{w}}}\dfrac{{{\text{dw}}}}{{{\text{dx}}}}{\text{ = (log}}\;{\text{a) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x)}} \hfill \\ \dfrac{{{\text{dw}}}}{{{\text{dx}}}}{\text{ = w(log a)}} \hfill \\ \end{align}

$\dfrac{{{\text{dw}}}}{{{\text{dx}}}}{\text{ = }}{{\text{a}}^{\text{x}}}{\text{(log a)}}$ ........(4)

चूंकि A स्थिर है, अतः ${{\text{a}}^{\text{a}}}$ भी एक स्थिर है।

$\dfrac{{{\text{ds}}}}{{{\text{dx}}}}{\text{ = 0}}$ ..........(5)

(1), (2), (3), (4), (5) समीकरण करने पर

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}{{\text{x}}^{\text{x}}}{\text{(1 + logx) + a}}{{\text{x}}^{{\text{a - 1}}}}{\text{ + }}{{\text{a}}^{\text{x}}}{\text{(log a) + 0}} \hfill \\ {\text{ = }}{{\text{x}}^{\text{x}}}{\text{(1 + logx) + a}}{{\text{x}}^{{\text{a - 1}}}}{\text{ + }}{{\text{a}}^{\text{x}}}{\text{(log a)}} \hfill \\ \end{align}

11. ${{\text{x}}^{{{\text{x}}^{\text{2}}}{\text{ - 3}}}}{\text{ + (x - 3}}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}{\text{,}}\;{\text{x > 3}}$

उत्तर: ${\text{y = }}{{\text{x}}^{{{\text{x}}^{\text{2}}}{\text{ - 3}}}}{\text{ + (x - 3}}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}$

\begin{align} {\text{u = }}{{\text{x}}^{{{\text{x}}^{\text{2}}}{\text{ - 3}}}}{\text{, v = (x - 3}}{{\text{)}}^{{{\text{x}}^{\text{2}}}}} \hfill \\ \therefore {\text{ y = u + v}} \hfill \\ \end{align}

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}$ ..........(1)

\begin{align} {\text{u = }}\;{{\text{x}}^{{{\text{x}}^{\text{2}}}{\text{ - 3}}}} \hfill \\ {\text{logu}}\;{\text{ = }}\;{\text{log}}{{\text{x}}^{{{\text{x}}^{\text{2}}}{\text{ - 3}}}} \hfill \\ \end{align}

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ = logx \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 3}}} \right){\text{ + }}\left( {{{\text{x}}^{\text{2}}}{\text{ - 3}}} \right) \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(logx)}} \hfill \\ \dfrac{{\text{1}}}{{\text{u}}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ = logx \times 2x + }}\left( {{{\text{x}}^{\text{2}}}{\text{ - 3}}} \right) \times \dfrac{{\text{1}}}{{\text{x}}} \hfill \\ \end{align}

$\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ = }}{{\text{x}}^{{{\text{x}}^{\text{2}}}{\text{ - 3}}}}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ - 3}}}}{{\text{x}}}{\text{ + 2xlogx}}} \right]$

${\text{v = (x - 3}}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}$

\begin{align} {\text{logv = log(x - 3}}{{\text{)}}^{{{\text{x}}^{\text{2}}}}} \hfill \\ {\text{logy = }}{{\text{x}}^{\text{2}}}{\text{log(x - 3)}} \hfill \\ \end{align}

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{\text{1}}}{{\text{v}}} \times \dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ = log(x - 3) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{x}}^{\text{2}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[log(x - 3)]}} \hfill \\ \dfrac{{\text{1}}}{{\text{v}}} \times \dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ = log(x - 3) \times 2x + }}{{\text{x}}^{\text{2}}} \times \dfrac{{\text{1}}}{{{\text{x - 3}}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x - 3)}} \hfill \\ \dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ = v}}\left[ {{\text{2xlog(x - 3) + }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{x - 3}}}}{\text{ \times 1}}} \right] \hfill \\ \dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ = (x - 3}}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{x - 3}}}}{\text{ + 2xlog(x - 3)}}} \right] \hfill \\ \end{align}

$\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{,}}\;\dfrac{{{\text{dv}}}}{{{\text{dx}}}}$ का मान (1) मे रखने पर,

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}{{\text{x}}^{{{\text{x}}^{\text{2}}}{\text{ - 3}}}}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ - 3}}}}{{\text{x}}}{\text{ + 2xlogx}}} \right]{\text{ + (x - 3}}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{x - 3}}}}{\text{ + 2xlog(x - }}} \right.{\text{3)]}}$

12. यदि ${\text{y }}{\text{ = 12(1 - cost), x = 10(t - sint), - }}\dfrac{{\text{\pi }}}{{\text{2}}}{\text{ < t < }}\dfrac{{\text{\pi }}}{{\text{2}}}$ तो $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ ज्ञात कीजिए।

उत्तर: ${\text{y }}{\text{ = 12(1 - cost), x = 10(t - sint)}}$

\begin{align} \dfrac{{{\text{dx}}}}{{{\text{dt}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{[10(t - sint)] = 10 \times }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{(t - sint) = 10(1 - cost)}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dt}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{[12(1 - cost)]}}\;{\text{ = 12 \times }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{(1 - cost) = 12}} \times {\text{[0 - ( - sint)] = 12sint}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\dfrac{{{\text{dy}}}}{{{\text{dt}}}}}}{{\dfrac{{{\text{dx}}}}{{{\text{dt}}}}}}{\text{ = }}\dfrac{{{\text{12sint}}}}{{{\text{10(1 - cost)}}}}{\text{ = }}\dfrac{{{\text{12}} \times {\text{2sin}}\dfrac{{\text{t}}}{{\text{2}}}{\text{cos}}\dfrac{{\text{t}}}{{\text{2}}}}}{{{\text{10}} \times {\text{2si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{t}}}{{\text{2}}}}}{\text{ = }}\dfrac{{\text{6}}}{{\text{5}}}{\text{cot}}\dfrac{{\text{t}}}{{\text{2}}} \hfill \\ \end{align}

13. यदि ${\text{y = si}}{{\text{n}}^{{\text{ - 1}}}}{\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{, 0 < x < 1}}$ तो $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ ज्ञात कीजिए।

उत्तर: ${\text{y = si}}{{\text{n}}^{{\text{ - 1}}}}{\text{ + si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}}$

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right] \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ + }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right) \hfill \\ \end{align}

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}{\text{ + }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}\left( {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right)} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right) \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}{\text{ + }}\dfrac{{\text{1}}}{{\text{x}}}\dfrac{{\text{1}}}{{{\text{2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right) \hfill \\ \end{align}

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}{\text{ + }}\dfrac{{\text{1}}}{{{\text{2x}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}{\text{( - 2x)}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}{\text{ - }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 0}} \hfill \\ \end{align}

14. यदि ${\text{ - 1 < x < 1 }}$ के लिए ${\text{x}}\sqrt {{\text{1 + y}}} {\text{ + y}}\sqrt {{\text{1 + x}}} \;{\text{ = }}\;{\text{0}}$ तो सिद्ध कीजिए $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{{{\text{(1 + x)}}}^{\text{2}}}}}$

उत्तर: ${\text{y}}\;{\text{ = }}\;$${\text{x}}\sqrt {{\text{1 + y}}} {\text{ + y}}\sqrt {{\text{1 + x}}} \;{\text{ = }}\;{\text{0}}$

\begin{align} {{\text{x}}^{\text{2}}}{\text{(1 + y) = }}{{\text{y}}^{\text{2}}}{\text{(1 + x)}} \hfill \\ {{\text{x}}^{\text{2}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{y = }}{{\text{y}}^{\text{2}}}{\text{ + x}}{{\text{y}}^{\text{2}}} \hfill \\ {{\text{x}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{y - x}}{{\text{y}}^{\text{2}}} \hfill \\ {{\text{x}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}{\text{ = xy(y - x)}} \hfill \\ {\text{(x + y)(x - y) = xy(y - x)}} \hfill \\ {\text{x + y = - xy}} \hfill \\ {\text{(1 + x)y = - x}} \hfill \\ {\text{y = }}\dfrac{{{\text{ - x}}}}{{{\text{(1 + x)}}}} \hfill \\ \end{align}

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} {\text{y = }}\dfrac{{{\text{ - x}}}}{{{\text{(1 + x)}}}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{\text{(1 + x)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x) - x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(1 + x)}}}}{{{{{\text{(1 + x)}}}^{\text{2}}}}} \hfill \\ \end{align}

\begin{align} {\text{ = }}\dfrac{{{\text{(1 + x) - x}}}}{{{{{\text{(1 + x)}}}^{\text{2}}}}} \hfill \\ {\text{ = - }}\dfrac{{\text{1}}}{{{{{\text{(1 + x)}}}^{\text{2}}}}} \hfill \\ \end{align}

15. यदि ${\text{c > 0 }}$ के लिए ${{\text{(x - a)}}^{\text{2}}}{\text{ + (y - b}}{{\text{)}}^{\text{2}}}{\text{ = }}{{\text{c}}^{\text{2}}}$ तो सिद्ध कीजिए की $\dfrac{{{{\left[ {{\text{1 + }}{{\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)}^{\text{2}}}} \right]}^{\dfrac{{\text{3}}}{{\text{2}}}}}}}{{\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}}}$ , ${\text{a,}}\;{\text{b}}$ स्वतंत्र एक स्थिर राशि है।

उत्तर: ${\text{y = }}\;$${{\text{(x - a)}}^{\text{2}}}{\text{ + (y - b}}{{\text{)}}^{\text{2}}}{\text{ = }}{{\text{c}}^{\text{2}}}$

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} \dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {{{{\text{(x - a)}}}^{\text{2}}}} \right]{\text{ + }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {{{{\text{(y - b)}}}^{\text{2}}}} \right]{\text{ = }}\dfrac{{\text{d}}}{{{\text{dc}}}}\left( {{{\text{c}}^{\text{2}}}} \right) \hfill \\ {\text{2(x - a) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x - a) + 2(y - b) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(y - b) = 0}} \hfill \\ {\text{2(x - a) \times 1 + 2(y - b) \times }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 0}} \hfill \\ \end{align}

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{\text{ - (x - a)}}}}{{{\text{y - b}}}}$ ........(1)

\begin{align} \dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {\dfrac{{{\text{ - (x - a)}}}}{{{\text{y - b}}}}} \right] \hfill \\ {\text{ = }}\left[ {\dfrac{{{\text{(y - b) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x - a) - (x - a) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(y - b)}}}}{{{{{\text{(y - b)}}}^{\text{2}}}}}} \right] \hfill \\ {\text{ = }}\left[ {\dfrac{{{\text{(y - b) - (x - a)}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}}{{{{{\text{(y - b)}}}^{\text{2}}}}}} \right] \hfill \\ \end{align}

${\text{ = }}\left[ {\dfrac{{{\text{(y - b) - (x - a) \times }}\left\{ {\dfrac{{{\text{ - (x - a)}}}}{{{\text{y - b}}}}} \right\}}}{{{{{\text{(y - b)}}}^{\text{2}}}}}} \right]$                [(1) का उपयोग करते हुए]

${\text{ = }}\left[ {\dfrac{{{{{\text{(y - b)}}}^{\text{2}}}{\text{ + (x - a}}{{\text{)}}^{\text{2}}}}}{{{{{\text{(y - b)}}}^{\text{3}}}}}} \right]$

${[\dfrac{{{\text{1 + }}{{\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)}^{\text{2}}}}}{{\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}}}{\text{]}}^{3/2}}{\text{ = }}\dfrac{{{{\left[ {{\text{1 + }}\dfrac{{{{{\text{(x - a)}}}^{\text{2}}}}}{{{{{\text{(y - b)}}}^{\text{2}}}}}} \right]}^{\dfrac{{\text{3}}}{{\text{2}}}}}}}{{{\text{ - }}\left[ {\dfrac{{{{{\text{(y - b)}}}^{\text{2}}}{\text{ + (x - a}}{{\text{)}}^{\text{2}}}}}{{{{{\text{(y - b)}}}^{\text{3}}}}}} \right]}}{\text{ = }}\dfrac{{{{\left[ {\dfrac{{{{{\text{(y - b)}}}^{\text{2}}}{\text{ + (x - a}}{{\text{)}}^{\text{2}}}}}{{{{{\text{(y - b)}}}^{\text{2}}}}}} \right]}^{\dfrac{{\text{3}}}{{\text{2}}}}}}}{{{\text{ - }}\left[ {\dfrac{{{{{\text{(y - b)}}}^{\text{2}}}{\text{ + (x - a}}{{\text{)}}^{\text{2}}}}}{{{{{\text{(y - b)}}}^{\text{3}}}}}} \right]}}$

\begin{align} {\text{ = }}\dfrac{{{{\left[ {\dfrac{{{{\text{c}}^{\text{2}}}}}{{{{{\text{(y - b)}}}^{\text{2}}}}}} \right]}^{\dfrac{{\text{3}}}{{\text{2}}}}}}}{{{\text{ - }}\dfrac{{{{\text{c}}^{\text{2}}}}}{{{{{\text{(y - b)}}}^{\text{3}}}}}}}{\text{ = }}\dfrac{{\dfrac{{{{\text{c}}^{\text{3}}}}}{{{{{\text{(y - b)}}}^{\text{3}}}}}}}{{{\text{ - }}\dfrac{{{{\text{c}}^{\text{2}}}}}{{{{{\text{(y - b)}}}^{\text{3}}}}}}} \hfill \\ {\text{ = - c}} \hfill \\ \end{align}

अतः ${\text{ - c}}$ स्थिर और स्वतंत्र राशि है।

16. यदि ${\text{cosy = xcos(a + y), cosa}} \ne {\text{ \pm 1}}$ तो सिद्ध कीजिए की $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(a + y)}}}}{{{\text{sina}}}}$

उत्तर: ${\text{y = }}\;$${\text{cosy = xcos(a + y)}}$

\begin{align} \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[cosy] = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[xcos(a + y)]}} \hfill \\ {\text{ - siny}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = cos(a + y) \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x) + x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[cos(a + y)]}} \hfill \\ {\text{ - siny}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = cos(a + y) + x \times [sin(a + y)]}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}} \hfill \\ \end{align}

${\text{[xsin(a + y) - siny]}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = cos(a + y)}}$ ......(1)

${\text{cosy = xcos(a + y), x = }}\dfrac{{{\text{cosy}}}}{{{\text{cos(a + y)}}}}$

जब (1) समीकरण कम हो

\begin{align} \left[ {\dfrac{{{\text{cosy}}}}{{{\text{cos(a + y)}}}}{\text{ \times sin(a + y) - siny}}} \right]\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = cos(a + y)}} \hfill \\ {\text{[cosy \times sin(a + y) - siny \times cos(a + y)]}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = co}}{{\text{s}}^{\text{1}}}{\text{(a + y)}} \hfill \\ {\text{sin(a + y - y)}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = co}}{{\text{s}}^{\text{2}}}{\text{(a + b)}} \hfill \\ \Rightarrow \;\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(a + b)}}}}{{{\text{sina}}}} \hfill \\ \end{align}

17. यदि ${\text{x = a(cost + tsint), y = a(sint - tcost)}}$ तो $\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}$ ज्ञात कीजिए।

उत्तर: दिया है ${\text{x = a(cost + tsint), y = a(sint - tcost)}}$

$\dfrac{{{\text{dx}}}}{{{\text{dt}}}}{\text{ = a}}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{(cost + tsint)}}$

${\text{ = a}}\left[ {{\text{ - sint + sint \times }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{(t) + t \times }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{(sint)}}} \right]$

\begin{align} {\text{ = a[ - sint + sint + tcost]}} \hfill \\ {\text{ = atcost}} \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dt}}}}{\text{ = a \times }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{(sint}}{\text{.tcost)}} \hfill \\ {\text{ = a}}\left[ {{\text{cost - }}\left\{ {{\text{cost \times }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{(t) + t \times }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{(cost)}}} \right\}} \right] \hfill \\ {\text{ = a[cost - \{ cost - tsint\} ]}} \hfill \\ {\text{ = atsint}} \hfill \\ \end{align}

\begin{align} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\left( {\dfrac{{{\text{dy}}}}{{{\text{dt}}}}} \right)}}{{\left( {\dfrac{{{\text{dx}}}}{{{\text{dt}}}}} \right)}}{\text{ = }}\dfrac{{{\text{atsint}}}}{{{\text{atcost}}}}{\text{ = tant}} \hfill \\ \dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right){\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(tant) = se}}{{\text{c}}^{\text{2}}}{\text{t \times }}\dfrac{{{\text{dt}}}}{{{\text{dx}}}} \hfill \\ {\text{ = se}}{{\text{c}}^{\text{2}}}{\text{t \times }}\dfrac{{\text{1}}}{{{\text{atcost}}}}\;\;\;\;\;\;\;\left[ {\because \dfrac{{{\text{dx}}}}{{{\text{dt}}}}{\text{ = atcost}}\; \Rightarrow \;\dfrac{{{\text{dt}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{atcost}}}}} \right] \hfill \\ {\text{ = }}\dfrac{{{\text{se}}{{\text{c}}^3}{\text{t}}}}{{{\text{at}}}}{\text{, 0 < t < }}\dfrac{{\text{\pi }}}{{\text{2}}} \hfill \\ \end{align}

18. यदि ${\text{f(x)}}\;{\text{ = }}\;{\left| {\text{x}} \right|^{\text{2}}}$ तो प्रमाणित कीजिए की ${{\text{f}}^{''}}{\text{(x)}}$ का अस्तित्व है और इसे ज्ञात भी कीजिए।

उत्तर: $\left| {\text{x}} \right|\;{\text{ = }}\;{\text{\{ }}\begin{array}{*{20}{c}} {{\text{x,}}}&{x \geqslant 0} \\ {{\text{ - x,}}}&{{\text{x < 0}}} \end{array}$दिया है ${\text{f(x) = |x}}{{\text{|}}^{\text{3}}}{\text{ = }}{{\text{x}}^{\text{3}}}$

जब ${\text{x}} \geq0,\;f'(x)=3x^2$

अतः ${{\text{f}}^{''}}{\text{(x) = 6x}}$

जब ${\text{x < 0, f(x) = |x}}{{\text{|}}^{\text{3}}}{\text{ = ( - x}}{{\text{)}}^{\text{3}}}{\text{ = - }}{{\text{x}}^{\text{3}}}$

अतः $f'(x)=6x$

${{\text{f}}^{''}}{\text{(x) = - 6x}}$

जो प्रमाणित है की ${\text{f(x)}}\;{\text{ = }}\;{\left| {\text{x}} \right|^{\text{3}}}{\text{,}}\;{{\text{f}}^{{\text{''(x)}}}}$ का अस्तित्व है ${{\text{f}}^{{\text{''}}}}{\text{(x)}}\;{\text{ = }}\;{\text{\{ }}\begin{array}{*{20}{c}} {{\text{6x,}}}&{x \geqslant 0} \\ {{\text{ - 6x,}}}&{{\text{x < 0}}} \end{array}$

19. गणितीय आगमन के सिद्धांत के प्रत्येक द्वारा, सिद्ध कीजिए की सभी धन पूर्णांक ${\text{n}}$ के लिए $\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{n}}}\;{\text{ = }}\;{\text{n}}{{\text{x}}^{{\text{n - 1}}}}$ है।

उत्तर: सभी सकारात्मक पूर्णांक के लिए $\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{n}}}\;{\text{ = }}\;{\text{n}}{{\text{x}}^{{\text{n - 1}}}}$

${\text{n = 1}}$

${\text{P(1):}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x) = 1 = 1}}{\text{.}}{{\text{x}}^{{\text{1 - 1}}}}$

${\text{P(n)}}$ सत्य है ${\text{n = 1}}$

सकारात्मक पूर्णांक के लिए ${\text{P(k)}}$ सत्य है

${\text{P(k):}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^{\text{k}}}} \right){\text{ = k}}{{\text{x}}^{{\text{k - 1}}}}$

${\text{P(k + 1)}}$ सत्य सिद्ध करने के लिए माना

\begin{align} \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^{{\text{k + 1}}}}} \right){\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{x \times }}{{\text{x}}^{\text{k}}}} \right) \hfill \\ {\text{ = }}{{\text{x}}^{\text{k}}} \times \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x) + x \times }}\dfrac{{\text{d}}}{{{\text{dx}}}} \times \left( {{{\text{x}}^{\text{k}}}} \right) \hfill \\ {\text{ = }}{{\text{x}}^{\text{k}}}{\text{ \times 1 + x \times k \times }}{{\text{x}}^{{\text{k - 1}}}} \hfill \\ {\text{ = }}{{\text{x}}^{\text{k}}}{\text{ + k \times }}{{\text{x}}^{\text{k}}} \hfill \\ \end{align}

\begin{align} {\text{ = (k + 1) \times }}{{\text{x}}^{\text{k}}} \hfill \\ {\text{ = (k + 1) \times }}{{\text{x}}^{{\text{(k + 1) - 1}}}} \hfill \\ \end{align}

अतः ${\text{P(k + 1)}}$ सत्य है जब ${\text{P(k)}}$ सत्य है।

20. ${\text{sin(A + B) = sinAcosB + cosAsinB}}$ का प्रयोग करते हुए अवकलन द्वारा cosines के लिए योग सूत्र ज्ञात कीजिए।

उत्तर: ${\text{sin(A + B) = sinAcosB + cosAsinB}}$

${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

$\begin{array}{l} \frac{\mathrm{d}}{\mathrm{dx}}[\sin (\mathrm{A}+\mathrm{B})]=\frac{\mathrm{d}}{\mathrm{dx}}\left((\sin \mathrm{A} \cos \mathrm{B})+\frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{A} \sin \mathrm{B})\right) \\ \cos (\mathrm{A}+\mathrm{B}) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{A}+\mathrm{B})=\cos \mathrm{B} \times \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{A})+\sin \mathrm{A} \times \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{B})+\sin \mathrm{b} \times \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{A})+\cos \mathrm{A} \times \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{B}) \\ \cos (\mathrm{A}+\mathrm{B}) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{A}+\mathrm{B})=\cos \mathrm{B} \times \cos \mathrm{A} \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}+\sin \mathrm{A}(-\sin \mathrm{B}) \frac{\mathrm{d} \mathrm{B}}{\mathrm{dx}}+\sin \mathrm{B}(-\sin \mathrm{A}) \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}+\cos \mathrm{A} \cos \mathrm{B} \frac{\mathrm{d} \mathrm{B}}{\mathrm{dx}} \\ \cos (\mathrm{A}+\mathrm{B})\left[\frac{\mathrm{d} \mathrm{A}}{\mathrm{d} \mathrm{x}}+\frac{\mathrm{d} \mathrm{B}}{\mathrm{d} \mathrm{x}}\right]=\left(\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B} \times\left[\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}+\frac{\mathrm{d} \mathrm{B}}{\mathrm{dx}}\right]\right. \\ \cos (\mathrm{A}+\mathrm{B})=\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B} \end{array}$

21. क्या एक ऐसे फलन का अस्तित्व है जो प्रत्येक बिन्दु पर संतत हो किन्तु केवल दो बिन्दुओ पर अवकलन न हो। अपने उत्तर का ओचित्य भी बतलाइए।

उत्तर: फलन ${\text{f(x)}}\;{\text{ = }}\;\left| {{\text{x - 1}}} \right|{\text{ + }}\left| {{\text{x - 3}}} \right|$ का वास्तविक बिन्दु के लिए निरंतर है, लेकिन दो बिन्दु ${\text{(x }}\;{\text{ = }}\;{\text{1,3)}}$ पर भिन्न नहीं है।

22. यदि ${\text{y = }}\left| {\begin{array}{*{20}{c}} {{\text{f(x)}}}&{{\text{g(x)}}}&{{\text{h(x)}}} \\ {\text{l}}&{\text{m}}&{\text{n}} \\ {\text{a}}&{\text{b}}&{\text{c}} \end{array}} \right|$ हो तो ज्ञात कीजिए की \$\frac{dy}{dx}=\begin{vmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{vmatrix}$

उत्तर: यदि ${\text{y = }}\left| {\begin{array}{*{20}{c}} {{\text{f(x)}}}&{{\text{g(x)}}}&{{\text{h(x)}}} \\ {\text{l}}&{\text{m}}&{\text{n}} \\ {\text{a}}&{\text{b}}&{\text{c}} \end{array}} \right|$

$y=\text{(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x))}$

$\frac{dy}{dx}=\frac{d}{dx}\text{((mc-nb)f(x)})-\frac{d}{dx}\text{((lc-na)g(x)})+\frac{d}{dx}\text{((lb-ma)h(x)})$

$=\text{(mc-nb)f'(x)}-\text{(lc-na)g(x)}+\text{(lb-ma)h(x)}$

$=\begin{vmatrix} f'(x) & g'(x) &h'(x) \\ l& m &n \\ a &b & c \end{vmatrix}$

$\frac{dy}{dx}=\begin{vmatrix} f'(x) & g'(x) &h'(x) \\ l& m &n \\ a &b & c \end{vmatrix}$

23. यदि ${\text{y = }}{{\text{e}}^{{\text{aco}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}}{\text{, - 1}} \leqslant {\text{x}} \leqslant {\text{1}}$ तो दर्शाइए की ${\text{(1 - }}{{\text{x}}^{\text{2}}})\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ - }}{{\text{a}}^{\text{2}}}{\text{y = 0}}$

उत्तर: यदि ${\text{y = }}{{\text{e}}^{{\text{aco}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}}$

दोनों पक्षों मव लघुगणक लेने पर

\begin{align} {\text{logy = aco}}{{\text{s}}^{{\text{ - 1}}}}{\text{xloge}} \hfill \\ {\text{logy = aco}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}} \hfill \\ \end{align} ${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर \begin{align} \dfrac{{\text{1}}}{{\text{y}}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = a}}\left[ {\dfrac{{{\text{ - 1}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} \right] \hfill \\ \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{\text{ - ay}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }} \hfill \\ \end{align}

दोनों पक्षों का वर्ग करने पर

\begin{align} {\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)^{\text{2}}}{\text{ = }}\dfrac{{{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}}}{{{\text{1 - }}{{\text{x}}^{\text{2}}}}} \hfill \\ \left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right){\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}} \hfill \\ \end{align}

पुन: ${\text{x}}$ के संबंध मे दोनों पक्षों के सापेक्ष अवकलन करने पर

\begin{align} {\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)^{\text{2}}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right){\text{ + }}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right) \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {{{\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)}^{\text{2}}}} \right]{\text{ = }}{{\text{a}}^{\text{2}}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{y}}^{\text{2}}}} \right) \hfill \\ {\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)^{\text{2}}}{\text{( - 2x) + }}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right) \times \text{2} \dfrac{{{\text{dy}}}}{{{\text{dx}}}}\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}{{\text{a}}^{\text{2}}}{\text{2y}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}} \hfill \\ {\text{ - x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + }}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}{{\text{a}}^{\text{2}}}{\text{y }}\left[ {\because \;\dfrac{{{\text{dy}}}}{{{\text{dx}}}} \ne {\text{0}}} \right] \hfill \\ \left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ - }}{{\text{a}}^{\text{2}}}{\text{y = 0}} \hfill \\ \end{align}

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability in Hindi

Chapter-wise NCERT Solutions are provided everywhere on the internet with an aim to help the students to gain a comprehensive understanding. Class 12 Maths Chapter 5 solution Hindi medium are created by our in-house experts keeping the understanding ability of all types of candidates in mind. NCERT textbooks and solutions are built to give a strong foundation to every concept. These NCERT Solutions for Class 12 Maths Chapter 5 in Hindi ensure a smooth understanding of all the concepts including the advanced concepts covered in the textbook.

NCERT Solutions in Hindi medium have been created keeping those students in mind who are studying in a Hindi medium school. These NCERT Solutions for Class 12 Maths Continuity and Differentiability in Hindi medium pdf download have innumerable benefits as these are created in simple and easy-to-understand language. The best feature of these solutions is a free download option. Students of Class 12 can download these solutions at any time as per their convenience for self-study purpose.

These solutions are nothing but a compilation of all the answers to the questions of the textbook exercises. The answers/solutions are given in a stepwise format and very well researched by the subject matter experts who have relevant experience in this field. Relevant diagrams, graphs, illustrations are provided along with the answers wherever required. In nutshell, NCERT Solutions for Class 12 Maths in Hindi come really handy in exam preparation and quick revision as well prior to the final examinations.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 5 - In Hindi

1. What are the topics covered in Chapter 5 of Class 12 Maths?

The chapter covers the fundamental concepts and theorems related to complex numbers and quadratic equations. There are four exercises in the chapter that deal with prominent concepts of Chapter 5. The first exercise deals with the clarity of the concept of determining the multiplicative inverse, the second exercise explains the concept of modulus and argument of a given set of numbers, and the third exercise deals with the methods of quadratic equations. The fundamental concept of this chapter is to understand complex numbers and solving equations.

2. Why should I choose Vedantu’s NCERT Solutions for Chapter 5 of Class 12 Maths?

NCERT Solutions explain complex and new chapters such as the complex number and quadratic equations in a simple yet detailed manner. The answers and explanations are explained step by step for a complete understanding of the concept. The NCERT solutions are curated by a panel of experienced subject experts that are in the education system for a long time and are also aware of the latest updates in the curriculum. These NCERT solutions also abide by the CBSE guidelines. These solutions are available on the Vedantu website and the app.

3. How many questions are there in Chapter 5 of Class 12 Maths?

Chapter 5- Complex Numbers And Quadratic Equations consist of four exercises in total that deal with the prominent topic of the chapter. Questions combined of all the four exercises sum up to 51 questions. NCERT solution provides you with the solution to all the 51 questions in these four exercises in English as well as the Hindi language for the convenience of the students. NCERT solutions are available to students for free in PDF format on the official website of Vedantu.

4. How can I score good marks in Chapter 5 of Class 12 Maths?

To secure good marks in Chapter 5, the students must be well versed with all the theorems, rules, and most importantly the theorems related to the complex numbers and quadratic equations. The next essential step is to practice as many questions of quadratic equations. You solve practice papers, examples, and extra questions provided by the NCERT solutions. With these simple rules, you can achieve desired scores in Chapter 5 which is a scoring chapter.

5. What are practical applications of complex numbers and quadratic equations?

Students often mug up concepts for the sake of examinations but when students understand the application of those concepts in practical worlds they often develop an interest in the topic or subject. Complex numbers and quadratic equations is not an irrelevant topic in your syllabus. It is used in various modern-day concepts such as in the calculation of voltage, current, or resistance in an AC circuit, in quantum mechanics, in Fourier Transform under signal processing, etc.