NCERT Solutions for Class 11 Maths Chapter 13: Limits and Derivatives - Exercise 13.2

Exercise 13.2

1. Find the derivative of \[{x^2} - 2\]at $x = 10$

Ans: Let $f\left( x \right) = {x^2} - 2$

Accordingly,

$f'\left( 10 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 10+h \right)-f\left( 10 \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\left( {10 + h} \right)}^2} - 2} \right] - \left( {{{10}^2} - 2} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{{{10}^2} + \left( {2 \times 10 \times h} \right) + {h^2} - 2 - {{10}^2} + 2}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{20h + {h^2}}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {20 + h} \right) = 20 + 0 = 20$

Thus, the derivative of ${x^2} - 2$at $x = 10$is $20$

2. Find the derivative of $99x$at $x = 100$

Ans: Let $f\left( x \right) = 99x$

Accordingly,

     $f'\left( 100 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 100+h \right)-f\left( 100 \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{99\left( {100 + h} \right) - \left( {99 \times 100} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {99 \times 100} \right) + 99h - \left( {99 \times 100} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{99h}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {99} \right) = 99$

Thus, the derivative of $99x$at $x = 100$is $99$

3. Find the derivative of $x$at $x = 1$

Ans: Let $f\left( x \right) = x$

Accordingly,

     $f'\left( 1 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 + h} \right) - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( 1 \right) = 1$

Thus, the derivative of $x$at $x = 1$is $1$

4. Find the derivative of the following functions using the first principle.

(i). ${x^3} - 27$

Ans: Let $f\left( x \right) = {x^3} - 27$

Accordingly, from the first principle,

     $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\left( {x + h} \right)}^3} - 27} \right] - \left( {{x^3} - 27} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + {h^3} + 3{x^2}h + 3x{h^2} - {x^3}}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {{h^2} + 3{x^2} + 3xh} \right)$

$ \Rightarrow 0 + 3{x^2} + 0 = 3{x^2}$

(ii). $\left( {x - 1} \right)\left( {x - 2} \right)$

Ans: Let $f\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)$

Accordingly, from the first principle,

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h - 1} \right)\left( {x + h - 2} \right) - \left( {x - 1} \right)\left( {x - 2} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^2} + hx - 2x + hx + {h^2} - 2h - x - h + 2} \right) - \left( {{x^2} - 2x - x + 2} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {hx + hx + {h^2} - 2h - h} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{2hx + {h^2} - 3h}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {2x + h - 3} \right) = 2x - 3$

(iii). $\frac{1}{{{x^2}}}$

Ans: Let $f\left( x \right) = \frac{1}{{{x^2}}}$

Accordingly, from the first principle,

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{\left( {x + h} \right)}^2}}} - \frac{1}{{{x^2}}}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{{x^2} - {{\left( {x + h} \right)}^2}}}{{{x^2}{{\left( {x + h} \right)}^2}}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{{x^2} - {x^2} - 2hx - {h^2}}}{{{x^2}{{\left( {x + h} \right)}^2}}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - {h^2} - 2hx}}{{{x^2}{{\left( {x + h} \right)}^2}}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - {h^2} - 2x}}{{{x^2}{{\left( {x + h} \right)}^2}}}} \right] = \frac{{0 - 2x}}{{{x^2}{{\left( {x + 0} \right)}^2}}}$

$ \Rightarrow \frac{{ - 2}}{{{x^3}}}$

(iv). $\frac{{x + 1}}{{x - 1}}$

Ans: Let $f\left( x \right) = \frac{{x + 1}}{{x - 1}}$

Accordingly, from the first principle,

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{x + h + 1}}{{x + h - 1}} - \frac{{x + 1}}{{x - 1}}} \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\left( {x - 1} \right)\left( {x + h + 1} \right) - \left( {x + 1} \right)\left( {x + h - 1} \right)}}{{\left( {x - 1} \right)\left( {x + h - 1} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\left( {{x^2} + hx + x - x - h - 1} \right) - \left( {{x^2} + hx - x + x + h - 1} \right)}}{{\left( {x - 1} \right)\left( {x + h - 1} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2h}}{{\left( {x - 1} \right)\left( {x + h - 1} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - 2}}{{\left( {x - 1} \right)\left( {x + h - 1} \right)}}} \right] = \frac{{ - 2}}{{\left( {x - 1} \right)\left( {x - 1} \right)}}$

$ \Rightarrow \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}$

5. For the function $f\left( x \right) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + .... + \frac{{{x^2}}}{2} + x + 1$prove that      $f'\left( 1 \right)=100f'\left( 0 \right)$

Ans: The given function is,

$f\left( x \right) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + .... + \frac{{{x^2}}}{2} + x + 1$

$\frac{d}{{dx}}f\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + .... + \frac{{{x^2}}}{2} + x + 1} \right]$

$\frac{d}{{dx}}f\left( x \right) = \frac{d}{{dx}}\left( {\frac{{{x^{100}}}}{{100}}} \right) + \frac{d}{{dx}}\left( {\frac{{{x^{99}}}}{{99}}} \right) + ... + \frac{d}{{dx}}\left( {\frac{{{x^2}}}{2}} \right) + \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( 1 \right)$

On using theorem $\frac{d}{{dx}}\left( {n{x^{n - 1}}} \right) = n{x^{n - 1}}$, we obtain

$\frac{d}{{dx}}f\left( x \right) = \frac{{100{x^{99}}}}{{100}} + \frac{{99{x^{98}}}}{{99}} + .... + \frac{{2x}}{2} + 1 + 0$

$\Rightarrow {x^{99}} + {x^{98}} + .... + x + 1$

$\therefore \text{ }f'\left( x \right)={{x}^{99}}+{{x}^{98}}+.....x+1$

At $x = 0,$

$f'\left( 0 \right)=1$

At $x = 1,$

$f'\left( 1 \right)={{1}^{99}}+{{1}^{98}}+....+1+1={{\left[ 1+1+....+1+1 \right]}_{100terms}}$

$\Rightarrow 1 \times 100 = 100$

Thus, $f'\left( 1 \right)=100f'\left( 0 \right)$

6. Find the derivative of ${x^n} + a{x^{n - 1}} + {a^2}{x^{n - 2}} + ..... + {a^{n - 1}}x + {a^n}$ , where$a$ is some fixed real number.

Ans: Let$f\left( x \right) = {x^n} + a{x^{n - 1}} + {a^2}{x^{n - 2}} + ..... + {a^{n - 1}}x + {a^n}$

$\frac{d}{{dx}}f\left( x \right) = \frac{d}{{dx}}\left( {{x^n} + a{x^{n - 1}} + {a^2}{x^{n - 2}} + ..... + {a^{n - 1}}x + {a^n}} \right)$

$\Rightarrow \frac{{d\left( {{x^n}} \right)}}{{dx}} + a\frac{{d\left( {{x^{n - 1}}} \right)}}{{dx}} + {a^2}\frac{{d\left( {{x^{n - 2}}} \right)}}{{dx}} + .... + {a^{n - 1}}\frac{{d\left( x \right)}}{{dx}} + {a^n}\frac{{d\left( 1 \right)}}{{dx}}$

On using theorem $\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

So, we obtain

\[f'\left( x \right)=n{{x}^{n-1}}+a\left( n-1 \right){{x}^{n-2}}+{{a}^{2}}\left( n-2 \right){{x}^{n-3}}+....+{{a}^{n-1}}+{{a}^{n}}\left( 0 \right)\]

$\therefore \text{ }f'\left( x \right)=n{{x}^{n-1}}+a\left( n-1 \right){{x}^{n-2}}+{{a}^{2}}\left( n-2 \right){{x}^{n-3}}+...+{{a}^{n-1}}$

7. For some constants $a$and $b$, find the derivative of the following functions.

(a). $\left( {x - a} \right)\left( {x - b} \right)$

Ans: Let $f\left( x \right) = \left( {x - a} \right)\left( {x - b} \right)$

$\Rightarrow f\left( x \right) = {x^2} - \left( {a - b} \right)x + ab$

$\therefore \text{ }f'\left( x \right)=\frac{d}{dx}\left( {{x}^{2}}-\left( a+b \right)x+ab \right)$

$\Rightarrow \frac{d}{{dx}}\left( {{x^2}} \right) - \left( {a + b} \right)\frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {ab} \right)$

On using theorem $\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, we obtain

$f'\left( x \right)=2x-\left( a+b \right)+0=2x-a-b$

(b). ${\left( {a{x^2} + b} \right)^2}$

Ans: Let, $f\left( x \right) = {\left( {a{x^2} + b} \right)^2}$

$\Rightarrow f\left( x \right) = {a^2}{x^4} + 2ab{x^2} + {b^2}$

$\therefore \text{ }f'\left( x \right)=\frac{d}{dx}\left( {{a}^{2}}{{x}^{4}}+2ab{{x}^{2}}+{{b}^{2}} \right)$

$\Rightarrow {a^2}\frac{d}{{dx}}\left( {{x^4}} \right) + 2ab\frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}{b^2}$

On using theorem $\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, we obtain

$f'\left( x \right)={{a}^{2}}\left( 4{{x}^{3}} \right)+2ab\left( 2x \right)+{{b}^{2}}\left( 0 \right)$

$\Rightarrow 4{a^2}{x^3} + 4abx = 4ax\left( {a{x^2} + b} \right)$

(c). $\frac{{x - a}}{{x - b}}$

Ans: Let, $f\left( x \right) = \frac{{x - a}}{{x - b}}$

$\Rightarrow \text{ }f'\left( x \right)=\frac{d}{dx}\left( \frac{x-a}{x-b} \right)$

By quotient rule,

$f'\left( x \right)=\frac{\left( x-b \right)\frac{d}{dx}\left( x-a \right)-\left( x-a \right)\frac{d}{dx}\left( x-b \right)}{{{\left( x-b \right)}^{2}}}$

$\Rightarrow \frac{{\left( {x - b} \right) \times 1 - \left( {x - a} \right) \times 1}}{{{{\left( {x - b} \right)}^2}}}$

$\Rightarrow \frac{{x - b - x + a}}{{{{\left( {x - b} \right)}^2}}} = \frac{{a - b}}{{{{\left( {x - b} \right)}^2}}}$

8. Find the derivative of $\frac{{{x^n} - {a^n}}}{{x - a}}$for any constant $a$

Ans: Let $f\left( x \right) = \frac{{{x^n} - {a^n}}}{{x - a}}$

$\Rightarrow \text{ }f'\left( x \right)=\frac{d}{dx}\left( \frac{{{x}^{n}}-{{a}^{n}}}{x-a} \right)$

By quotient rule,

$f'\left( x \right)=\frac{\left( x-a \right)\frac{d}{dx}\left( {{x}^{n}}-{{a}^{n}} \right)-\left( {{x}^{n}}-{{a}^{n}} \right)\frac{d}{dx}\left( x-a \right)}{{{\left( x-a \right)}^{2}}}$

$\Rightarrow \frac{{\left( {x - a} \right)\left( {n{x^{n - 1}} - 0} \right) - \left( {{x^n} - {a^n}} \right)}}{{{{\left( {x - a} \right)}^2}}}$

$\Rightarrow \frac{{n{x^n} - an{x^{n - 1}} - {x^n} + {a^n}}}{{{{\left( {x - a} \right)}^2}}}$

9. Find the derivative of the following functions

(a). $2x - \frac{3}{4}$

Ans: Let $f\left( x \right) = 2x - \frac{3}{4}$

$f'\left( x \right)=\frac{d}{dx}\left( 2x-\frac{3}{4} \right)$

$\Rightarrow 2\frac{d}{{dx}}\left( x \right) - \frac{d}{{dx}}\left( {\frac{3}{4}} \right)$

$\Rightarrow 2 - 0 = 2$

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

Ans: Let $f\left( x \right) = \left( {5{x^3} + 3x - 1} \right)\left( {x - 1} \right)$

By Leibnitz product rule,

\[f'\left( x \right)=\left( 5{{x}^{3}}+3x-1 \right)\frac{d}{dx}\left( x-1 \right)+\left( x-1 \right)\frac{d}{dx}\left( 5{{x}^{3}}+3x-1 \right)\]

$\Rightarrow \left( {\left( {5{x^3} + 3x - 1} \right) \times 1} \right) + \left( {x - 1} \right)\left( {5 \times 3{x^2} + 3 - 0} \right)$

$\Rightarrow \left( {5{x^3} + 3x - 1} \right) + \left( {x - 1} \right)\left( {15{x^2} + 3} \right)$

$\Rightarrow 5{x^3} + 3x - 1 + 15{x^3} + 3x - 15{x^2} - 3$

$\Rightarrow 20{x^3} - 15{x^2} + 6x - 4$

(c). ${x^{ - 3}}\left( {5 + 3x} \right)$

Ans: Let, $f\left( x \right) = {x^{ - 3}}\left( {5 + 3x} \right)$

By Leibnitz product rule,

$f'\left( x \right) = {x^{ - 3}}\frac{d}{{dx}}\left( {5 + 3x} \right) + \left( {5 + 3x} \right)\frac{d}{{dx}}\left( {{x^{ - 3}}} \right)$

$\Rightarrow {x^{ - 3}}\left( {0 + 3} \right) + \left( {5 + 3x} \right)\left( {3{x^{ - 3 - 1}}} \right)$

$\Rightarrow 3{x^{ - 3}} + \left( {5 + 3x} \right)\left( { - 3{x^{ - 4}}} \right) = 3{x^{ - 3}} - 15{x^{ - 4}} - 9{x^{ - 3}}$

$\Rightarrow  - 6{x^{ - 3}} - 15{x^{ - 4}} =  - 3{x^{ - 3}}\left( {2 + \frac{5}{x}} \right) = \frac{{ - 3{x^{ - 3}}}}{x}\left( {2x + 5} \right)$

$\Rightarrow \frac{{ - 3}}{{{x^4}}}\left( {5 + 2x} \right)$

(d). ${x^5}\left( {3 - 6{x^{ - 9}}} \right)$

Ans: Let, $f\left( x \right) = {x^5}\left( {3 - 6{x^{ - 9}}} \right)$

By Leibnitz product rule,

$f'\left( x \right) = {x^5}\frac{d}{{dx}}\left( {3 - 6{x^{ - 9}}} \right) + \left( {3 - 6{x^{ - 9}}} \right)\frac{d}{{dx}}{x^5}$

$\Rightarrow {x^5}\left\{ {0 - 6\left( { - 9} \right){x^{ - 9 - 1}}} \right\} + \left( {3 - 6{x^{ - 9}}} \right)\left( {5{x^4}} \right)$

$\Rightarrow {x^5}\left( {54{x^{ - 10}}} \right) + 15{x^4} - 30{x^{ - 5}} = 54{x^{ - 5}} + 15{x^4} - 30{x^{ - 5}}$

$\Rightarrow 24{x^{ - 5}} + 15{x^4} = 15{x^4} + \frac{{24}}{{{x^5}}}$

(e). ${x^{ - 4}}\left( {3 - 4{x^{ - 5}}} \right)$

Ans: Let $f\left( x \right) = {x^{ - 4}}\left( {3 - 4{x^{ - 5}}} \right)$

By Leibnitz product rule,

$f'\left( x \right) = {x^{ - 4}}\frac{d}{{dx}}\left( {3 - 4{x^{ - 5}}} \right) + \left( {3 - 4{x^{ - 5}}} \right)\frac{d}{{dx}}\left( {{x^{ - 4}}} \right)$

$\Rightarrow {x^{ - 4}}\left\{ {0 - 4\left( { - 5} \right){x^{ - 5 - 1}}} \right\} + \left( {3 - 4{x^{ - 5}}} \right)\left( { - 4} \right){x^{ - 4 - 1}}$

$\Rightarrow {x^{ - 4}}\left( {20{x^{ - 6}}} \right) + \left( {3 - 4{x^{ - 5}}} \right)\left( { - 4{x^{ - 5}}} \right)$

$\Rightarrow 20{x^{ - 10}} - 12{x^{ - 5}} + 16{x^{ - 10}} = 36{x^{ - 10}} - 12{x^{ - 5}}$

$\Rightarrow \frac{{36}}{{{x^{10}}}} - \frac{{12}}{{{x^5}}}$

(f). $\frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}$

Ans: $f\left( x \right) = \frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}$

$f'\left( x \right) = \frac{d}{{dx}}\left( {\frac{2}{{x + 1}}} \right) - \frac{d}{{dx}}\left( {\frac{{{x^2}}}{{3x - 1}}} \right)$

By quotient rule,

$f'\left( x \right) = \left[ {\frac{{\left( {x + 1} \right)\frac{d}{{dx}}\left( 2 \right) - 2\frac{d}{{dx}}\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}} \right] - \left[ {\frac{{\left( {3x - 1} \right)\frac{d}{{dx}}\left( {{x^2}} \right) - {x^2}\frac{d}{{dx}}\left( {3x - 1} \right)}}{{{{\left( {3x - 1} \right)}^2}}}} \right]$

$\Rightarrow \left[ {\frac{{\left( {x + 1} \right)\left( 0 \right) - 2\left( 0 \right)}}{{{{\left( {x + 1} \right)}^2}}}} \right] - \left[ {\frac{{\left( {3x - 1} \right)\left( {2x} \right) - {x^2}\left( 3 \right)}}{{{{\left( {3x - 1} \right)}^2}}}} \right]$

$\Rightarrow \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} - \left[ {\frac{{6{x^2} - 2x - 3{x^2}}}{{{{\left( {3x - 1} \right)}^2}}}} \right] = \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} - \left[ {\frac{{3{x^2} - 2x}}{{{{\left( {3x - 1} \right)}^2}}}} \right]$

$\Rightarrow \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} - \frac{{x\left( {3x - 2} \right)}}{{{{\left( {3x - 1} \right)}^2}}}$

10. Find the derivative of $\cos x$from the first principle

Ans: Let $f\left( x \right) = \cos x$

Accordingly from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos \left( {x + h} \right) - \cos \left( x \right)}}{h}} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos x\cosh  - \sin x\sinh  - \cos x}}{h}} \right] = \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cosh } \right)}}{h} - \frac{{\sin x\sinh }}{h}} \right]$

$\Rightarrow  - \cos x\left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{1 - \cosh }}{h}} \right)} \right] - \sin x\left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{\sinh }}{h}} \right)} \right]$

We know, $\mathop {\lim }\limits_{h \to 0} \frac{{1 - \cosh }}{h} = 0$ and $\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h} = 1$

$\Rightarrow f'\left( x \right) =  - \cos x \times 0 - \sin x \times 1$

$\therefore f'\left( x \right) =  - \sin x$

11. Find the derivative of the functions below:

(i). $\sin x\cos x$

Ans: Let $f\left( x \right) = \sin x\cos x$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {x + h} \right)\cos \left( {x + h} \right) - \sin x\cos x}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\sin \left( {x + h} \right)\cos \left( {x + h} \right) - 2\sin x\cos x} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {\sin 2\left( {x + h} \right) - \sin 2x} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\cos \frac{{2x + 2h + 2x}}{2}.\sin \frac{{2x + 2h - 2x}}{2}} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\cos \frac{{4x + 2h}}{2}.\sin \frac{{2h}}{2}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {\cos \left( {2x + h} \right)\sinh } \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \cos \left( {2x + h} \right).\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}$

$\Rightarrow \cos \left( {2x + h} \right) \times 1 = \cos 2x$

(ii). $\sec x$

Ans: Let $f\left( x \right) = \sec x$

Accordingly, from the first principle

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\sec \left( {x + h} \right) - \sec x}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos \left( {x + h} \right)}} - \frac{1}{{\cos x}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right]$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right]$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\frac{{\left[ { - 2\sin \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right]}}{{\cos \left( {x + h} \right)}}$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}$

$\Rightarrow \frac{1}{{\cos x}} \times 1 \times \frac{{\sin x}}{{\cos x}} = \sec x\tan x$

(iii). $5\sec x + 4\cos x$

Ans: Let, 

$f\left( x \right) = 5\sec x + 4\cos x$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \frac{{\lim }}{{h \to 0}}\frac{{5\sec \left( {x + h} \right) + 4\cos \left( {x + h} \right) - \left[ {5\sec x + 4\cos x} \right]}}{h}$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sec \left( {x + h} \right) - \sec x} \right]}}{h} + 4\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\cos \left( {x + h} \right) - \cos x} \right]}}{h}$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos \left( {x + h} \right)}} - \frac{1}{{\cos x}}} \right] + 4\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \left( {x + h} \right) - \cos x} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right] + 4\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos x\cosh  - \sin x\sinh  - \cos x} \right]$

$ \Rightarrow \frac{5}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right] + 4\left[ { - \cos x\mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 - \cos x} \right)}}{h} - \sin x\mathop {\lim \frac{{\sinh }}{h}}\limits_{h \to 0} } \right]$

$ \Rightarrow \frac{5}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sin \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right]}}{{\cos \left( {x + h} \right)}} + 4\left[ { - \cos x\left( 0 \right) - \sin x\left( 1 \right)} \right]$

$ \Rightarrow \frac{5}{{\cos x}}\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right] - 4\sin x$

$ \Rightarrow \frac{5}{{\cos x}} \times \frac{{\sin x}}{{\cos x}} \times 1 - 4\sin x = 5\sec x\tan x - 4\sin x$

(iv). $\cos ecx$

Ans: Let $f\left( x \right) = \cos ecx$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos ec\left( {x + h} \right) - \cos ecx} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\sin \left( {x + h} \right)}} - \frac{1}{{\sin x}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x - \sin \left( {x + h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left[ { - \cos \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right]}}{{\sin x\sin \left( {x + h} \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - \cos \left( {\frac{{2x + h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right)\mathop {\lim }\limits_{\frac{h}{2} \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}$

$ \Rightarrow \left( {\frac{{ - \cos x}}{{\sin x\sin x}}} \right) \times 1 =  - \cos ecx\cot x$

(v). $3\cot x + 5\cos ecx$

Ans: Let $f\left( x \right) = 3\cot x + 5\cos ecx$

Accordingly, from the first principle

 $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {3\cot \left( {x + h} \right) + 5\cos ec\left( {x + h} \right) - 3\cot x - 5\cos ecx} \right]$

\[ \Rightarrow 3\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cot \left( {x + h} \right) - \cot x} \right] + 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos ec\left( {x + h} \right) - \cos ecx} \right]\,\,\,\,\,\,\,\,\,\,....\left( 1 \right)\]

Now, $\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cot \left( {x + h} \right) - \cot x} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos \left( {x + h} \right)}}{{\sin \left( {x + h} \right)}} - \frac{{\cos x}}{{\sin x}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos \left( {x + h} \right)\sin x - \cos x\sin \left( {x + h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin \left( {x - x - h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin \left( { - h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}.\mathop {\lim }\limits_{h \to 0} \left[ {\frac{1}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow  - 1 \times \frac{1}{{\sin x\sin \left( {x + h} \right)}} = \frac{{ - 1}}{{{{\sin }^2}x}} =  - \cos e{c^2}x\,\,\,\,\,\,......(2)$

$\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos ec\left( {x + h} \right) - \cos ecx} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\sin \left( {x + h} \right)}} - \frac{1}{{\sin x}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x - \sin \left( {x + h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{ - \cos \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}}}{{\sin x\sin \left( {x + h} \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - \cos \left( {\frac{{2x + h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right)\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}} = \left( {\frac{{ - \cos x}}{{\sin x\sin x}}} \right).1$

$ \Rightarrow  - \cos ecx\cot x\,\,\,\,\,\,......(3)$

From (1), (2), and (3), we obtain

$f'\left( x \right) =  - 3\cos e{c^2}x - 5\cos ecx\cot x$

(vi). $5\sin x - 6\cos x + 7$

Ans: Let $f\left( x \right) = 5\sin x - 6\cos x + 7$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {5\sin \left( {x + h} \right) - 6\cos \left( {x + h} \right) + 7 - 5\sin x + 6\cos x - 7} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\sin \left( {x + h} \right) - \sin x} \right] - 6\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \left( {x + h} \right) - \cos x} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\cos \left( {\frac{{x + h + x}}{2}} \right).\sin \left( {\frac{{x + h - x}}{2}} \right)} \right] - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos x\cosh  - \sin x\sinh  - \cos x}}{h}} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\cos \left( {\frac{{2x + h}}{2}} \right).\sin \left( {\frac{h}{2}} \right)} \right] - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cosh } \right) - \sin x\sinh }}{h}} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right] - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cosh } \right)}}{h} - \frac{{\sin x\sinh }}{h}} \right]$

$ \Rightarrow 5\left[ {\mathop {\lim }\limits_{h \to 0} \cos \left( {\frac{{2x + h}}{2}} \right)} \right]\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right] - 6\left[ { - \cos x\left( {\mathop {\lim }\limits_{h \to 0} \frac{{1 - \cosh }}{h}} \right) - \sin x\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}} \right)} \right]$

$ \Rightarrow 5\cos x.1 - 6\left[ {\left( { - \cos x} \right).\left( 0 \right) - \sin x.1} \right]$

$ \Rightarrow 5\cos x + 6\sin x$

(vii). $2\tan x - 7\sec x$

Ans: Let $f\left( x \right) = 2\tan x - 7\sec x$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\tan \left( {x + h} \right) - 7\sec \left( {x + h} \right) - 2\tan x + 7\sec x}\right]$

  $ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\tan \left( {x + h} \right) - \tan x} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\sec \left( {x + h} \right) - \sec x} \right]$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin \left( {x + h} \right)}}{{\cos \left( {x + h} \right)}} - \frac{{\sin x}}{{\cos x}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos ec\left( {x + h} \right)}} - \frac{1}{{\cos ecx}}} \right]$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x\sin \left( {x + h} \right) - \sin x\cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x + h - x}}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$ \Rightarrow 2\left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{\sinh }}{h}} \right)\frac{1}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$ \Rightarrow 2\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}} \right)\left[ {\mathop {\lim }\limits_{h \to 0} \frac{1}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}} \right)\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right)$

$ \Rightarrow 2 \times 1 \times 1 \times \frac{1}{{\cos x\cos x}} - 7\left( {1 \times \frac{{\sin x}}{{\cos x\cos x}}} \right) = 2{\sec ^2}x - 7\sec x\tan x$

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 13 All Exercises

All Topics and Important Formulae Under NCERT Class 11 Maths for Exercise 13.2

The topics and related formulae covered under exercise 13.2 Maths Class 11 NCERT are given below.

Section 13.5: Derivatives

1. If f is the real-valued function and x is the point exist in the domain of f then derivative of f i.e. f’ can be find as:

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$

Section 13.5.1: Derivatives of Functions Algebra

1. \[\frac{d}{dx}\](f(x) + g(x)) = \[\frac{d}{dx}\]f(x) + \[\frac{d}{dx}\]g(x)

2. \[\frac{d}{dx}\](f(x) - g(x)) = \[\frac{d}{dx}\]f(x) - \[\frac{d}{dx}\]g(x)

3. \[\frac{d}{dx}\](f(x) . g(x)) = \[\frac{d}{dx}\]f(x).g(x) - f(x).\[\frac{d}{dx}\]g(x)

4. \[\frac{d}{dx}(\frac{f(x)}{g(x)})\] = (\[\frac{ \frac{d}{dx}f(x).g(x) - f(x).\frac{d}{dx}g(x)}{g(x)^2} \])

Section 13.5.2: To find the derivatives of polynomials and trigonometric functions

1. If the polynomial function f(x) is given as  

f (x) = anxn + an-1xn-1 +  an-2xn-2  + an-3xn-3 …… +  a2x2 + a1x1 + a0x0  

Then the derivative of the function f (x) w.r.t. X is  given as

f’ (x) = nan-1xn-1 + (n - 1) an-2xn-2  + an-3xn-3 …… +  2a2x + a1

2. Derivatives of some trigonometric functions are given as

• \[\frac{d}{dx}\](sin x) = cos x

• \[\frac{d}{dx}\](cos x) = – sin x

• \[\frac{d}{dx}\](tan x) = sec2 x

• \[\frac{d}{dx}\](cot x) = – cosec2 x

• \[\frac{d}{dx}\](sec x) = tan x . sec x

• \[\frac{d}{dx}\](cosec x) = – cot x . cosec x

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.2

Opting for the NCERT solutions for Ex 13.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 13 Exercise 13.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 13 Exercise 13.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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