## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.4) Exercise 5.4

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## Download PDF of NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.4) Exercise 5.4

A. $f(x) = {x^{\dfrac{1}{3}}}$

B. $f(x) = \dfrac{{|x|}}{x}$

C. $f(x) = {e^{ - x}}$

D. $f(x) = \tan x$

{ - 2;if{\text{ }}x \leqslant - 1} \\

{2x;if{\text{ }} - 1 < x \leqslant 1} \\

{2;if{\text{ }}x > 1}

\end{array}} \right\}\].

Every continuous function is differentiable.

{x^2} - x + 2{\text{ }}if{\text{ }}x \leqslant 1 \\

- {x^2} + x{\text{ }}if{\text{ }}x > 1 \\

\right.$.

A) \[\mathop {\lim }\limits_{x \to {0^ + }} f\left( {\dfrac{1}{x}} \right) = 1\]

B) \[\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {\dfrac{1}{x}} \right) = 2\]

C) \[\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'\left( x \right) = 0\]

D) \[\mathop {\lim }\limits_{x \to 0} f'\left( {\dfrac{1}{x}} \right) = 1\]

Function whose jump (non-negative difference of \[LHL\]and \[RHL\]) of discontinuity is greater than or equal to one, is are-

A. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}

{\dfrac{{\left( {{e^{\dfrac{1}{x}}} + 1} \right)}}{{\left( {{e^{\dfrac{1}{x}}} - 1} \right)}},\,\,\,\,\,x < 0} \\

{\dfrac{{\left( {1 - \cos x} \right)}}{x},\,\,\,\,\,x < 0}

\end{array}} \right.\]

B. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}

{\dfrac{{\left( {{x^{\dfrac{1}{3}}} + 1} \right)}}{{\left( {{x^{\dfrac{1}{2}}} - 1} \right)}},\,\,\,\,\,x > 1} \\

{\dfrac{{\ln x}}{{\left( {x - 1} \right)}},\,\,\,\,\,\dfrac{1}{2} < x < 1}

\end{array}} \right.\]

C. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}

{\dfrac{{{{\sin }^{ - 1}}2x}}{{{{\tan }^{ - 1}}3x}},\,\,\,\,\,x \in \left( {0,\left. {\dfrac{1}{2}} \right]} \right.} \\

{\dfrac{{\left| {\sin x} \right|}}{x},\,\,\,\,\,x < 0}

\end{array}} \right.\]

D. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}

{{{\log }_3}\left( {x + 2} \right),\,\,\,\,\,x > 2} \\

{{{\log }_{\dfrac{1}{2}}}\left( {{x^2} + 5} \right),\,\,\,\,\,x < 2}

\end{array}} \right.\]

{\dfrac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \left( {\dfrac{x}{p}} \right)\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}};{\text{ }}x \ne 0} \\

{{\text{ }}12{{\left( {\log 4} \right)}^3}{\text{ }};{\text{ }}x = 0}

\end{array}} \right.\] is continuous at \[x = 0\], is

(a) 4

(b) 2

(c) 3

(d) 1

## Access NCERT Solutions for Class 12 Maths Chapter-5 Continuity and Differentiability

Exercise 5.4

1. Differentiating the following w.r.t. \[x:\frac{{{e^x}}}{{\sin x}}\]

Ans:

Let $y = \frac{{{e^x}}}{{\sin x}}$

Now differentiating w.r.t. $x$ , we get

$\frac{{dy}}{{dx}} = \frac{{\sin x \times \tfrac{d}{{dx}}({e^x}) - {e^x} \times \tfrac{d}{{dx}}(\sin x)}}{{{{\sin }^2}x}}$

\[ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\sin x \times {e^x} - {e^x} \times \cos x}}{{{{\sin }^2}x}}\]

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^x}(\sin x - \cos x)}}{{{{\sin }^2}x}},x \ne n\pi ,n \in \mathbb{Z}$

2. Differentiating the following ${e^{{{\sin }^{ - 1}}x}}$

Ans:

Let $y = {e^{{{\sin }^{ - 1}}x}}$

Now differentiating w.r.t. $x$ , we get

$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^{{{\sin }^{ - 1}}x}}} \right)$

$ \Rightarrow \frac{{dy}}{{dx}} = {e^{{{\sin }^{ - 1}}x}} \times \frac{d}{{dx}}({\sin ^{ - 1}}x)$

$ \Rightarrow \frac{{dy}}{{dx}} = {e^{{{\sin }^{ - 1}}x}} \times \frac{1}{{\sqrt {1 - {x^2}} }}$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }},x \in ( - 1,1)$

3. Differentiating the following w.r.t. $x:{e^{{x^3}}}$

Ans: Let $y = {e^{{x^3}}}$

By using the quotient rule, we get

$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right)$

$ \Rightarrow \frac{{dy}}{{dx}} = {e^{{x^3}}} \times \frac{d}{{dx}}\left( {{x^3}} \right)$

$ \Rightarrow \frac{{dy}}{{dx}} = {e^{{x^3}}} \times 3{x^2}$

$ \Rightarrow \frac{{dy}}{{dx}} = 3{x^2}{e^{{x^3}}}$

4. Differentiating the following w.r.t. $x:\sin ({\tan ^{ - 1}}{e^{ - x}})$

Ans:

Let $y = \sin ({\tan ^{ - 1}}{e^{ - x}})$

By using the chain rule we get

\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sin ({{\tan }^{ - 1}}{e^{ - x}})} \right]\]

$ \Rightarrow \frac{{dy}}{{dx}} = \cos ({\tan ^{ - 1}}{e^{ - x}}) \times \frac{d}{{dx}}({\tan ^{ - 1}}{e^{ - x}})$

$ \Rightarrow \frac{{dy}}{{dx}} = \cos ({\tan ^{ - 1}}{e^{ - x}}) \times \frac{1}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}}\frac{d}{{dx}}({e^{ - x}})$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\cos ({{\tan }^{ - 1}}{e^{ - x}})}}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}} \times {e^{ - x}} \times \frac{d}{{dx}}( - x)$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{ - x}}\cos ({{\tan }^{ - 1}}{e^{ - x}})}}{{1 + {e^{ - 2x}}}} \times ( - 1)$

$ \Rightarrow \frac{{dy}}{{dx}} = - \frac{{{e^{ - x}}\cos ({{\tan }^{ - 1}}{e^{ - x}})}}{{1 + {e^{ - 2x}}}}$

5. Differentiating the following w.r.t. \[x:\log (\cos {e^x})\]

Ans:

Let $y = \log (\cos {e^x})$

By using the chain rule, we get

\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\log (\cos {e^x})} \right]\]

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\cos {e^x}}} \times \frac{d}{{dx}}\left( {\cos {e^x}} \right)$

\[ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\cos {e^x}}} \times ( - \sin {e^x}) \times \frac{d}{{dx}}\left( {{e^x}} \right)\]

\[ \Rightarrow \frac{{dy}}{{dx}} = - \frac{{\sin {e^x}}}{{\cos {e^x}}} \times \left( {{e^x}} \right)\]

\[ \Rightarrow \frac{{dy}}{{dx}} = - {e^x}\tan {e^x},{e^x} \ne (2n + 1)\frac{\pi }{2},n \in \mathbb{N}\]

6. Differentiating the following w.r.t. $x:{e^x} + {e^{{x^2}}} + ... + {e^{{x^5}}}$

Ans:

Let $y = {e^x} + {e^{{x^2}}} + ... + {e^{{x^5}}}$

$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^x} + {e^{{x^2}}} + ... + {e^{{x^5}}}} \right)$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^x}} \right) + \frac{d}{{dx}}\left( {{e^{{x^2}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^4}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^5}}}} \right)$

\[ \Rightarrow \frac{{dy}}{{dx}} = {e^x} + \left[ {{e^{{x^2}}}\frac{d}{{dx}}\left( {{x^2}} \right)} \right] + \left[ {{e^{{x^3}}}\frac{d}{{dx}}\left( {{x^3}} \right)} \right] + \left[ {{e^{{x^4}}}\frac{d}{{dx}}\left( {{x^4}} \right)} \right] + \left[ {{e^{{x^5}}}\frac{d}{{dx}}\left( {{x^5}} \right)} \right]\]

\[ \Rightarrow \frac{{dy}}{{dx}} = {e^x} + \left[ {{e^{{x^2}}} \times 2x} \right] + \left[ {{e^{{x^3}}} \times 3{x^2}} \right] + \left[ {{e^{{x^4}}} \times 4{x^3}} \right] + \left[ {{e^{{x^5}}} \times 5{x^4}} \right]\]

\[ \Rightarrow \frac{{dy}}{{dx}} = {e^x} + 2x{e^{{x^2}}} + 3{x^2}{e^{{x^3}}} + 4{x^3}{e^{{x^4}}} + 5{x^4}{e^{{x^5}}}\]

7. Differentiating the following w.r.t. $x:\sqrt {{e^{\sqrt x }}} ,x > 0$

Ans:

Let $y = \sqrt {{e^{\sqrt x }}} $

Then ${y^2} = {e^{\sqrt x }}$

By differentiating the above equation with respect to $x$ , we get

${y^2} = {e^{\sqrt x }}$

$ \Rightarrow 2y\frac{{dy}}{{dx}} = {e^{\sqrt x }}\frac{d}{{dx}}\left( {\sqrt x } \right)$ (By applying the chain rule)

$ \Rightarrow 2y\frac{{dy}}{{dx}} = {e^{\sqrt x }} \times \frac{1}{2} \times \frac{1}{{\sqrt x }}$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4y\sqrt x }}$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4\sqrt {{e^{\sqrt x }}} \sqrt x }}$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4\sqrt {x{e^{\sqrt x }}} }},x > 0$

8. Differentiating the following w.r.t. $x:\log (\log x),x > 1$

Ans:

Let $y = \log (\log x)$

By using the chain rule in the above equation, we get

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\log (\log x)} \right]$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\log x}} \times \frac{d}{{dx}}(\log x)$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\log x}} \times \frac{1}{x}$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{x\log x}},x > 1$

9. Differentiating the following w.r.t. $x:\frac{{\cos x}}{{\log x}},x > 0$

Ans:

Let $y = \frac{{\cos x}}{{\log x}}$

By using the quotient rule, we obtain

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\tfrac{d}{{dx}}(\cos x) \times \log x - \cos x \times \tfrac{d}{{dx}}(\log x)}}{{{{\left( {\log x} \right)}^2}}}$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{( - \sin x)\log x - \cos x \times \frac{1}{x}}}{{{{(\log x)}^2}}}$

$ \Rightarrow \frac{{dy}}{{dx}} = - \frac{{x\log x \times \sin + \cos x}}{{x{{(\log x)}^2}}},x > 0$

10. Differentiating the following w.r.t. $x:\cos (\log x + {e^x}),x > 0$

Ans:

Let $y = \cos (\log x + {e^x})$

By using the chain rule in the above equation, we get

$y = \cos (\log x + {e^x})$

$\frac{{dy}}{{dx}} = - \sin (\log x + {e^x}) \times \frac{d}{{dx}}(\log x + {e^x})$

$ = - \sin (\log x + {e^x}) \times \left[ {\frac{d}{{dx}}(\log x) + \frac{d}{{dx}}({e^x})} \right]$

$ = - \sin (\log x + {e^x}) \times \left( {\frac{1}{x} + {e^x}} \right)$

$ = - \left( {\frac{1}{x} + {e^x}} \right)\sin (\log x + {e^x}),x > 0$

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