NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

Edited By Komal Miglani | Updated on Mar 29, 2025 12:40 AM IST

Did you ever wonder how scientists determine the velocity of a car at a specific moment or how economists determine how rapidly financial patterns are changing? The concept of Limits and Derivatives is central to the understanding of such real-life applications. Limits assist us in determining what value a function is getting nearer to as we approach a particular number. Limits come in handy when we are unable to use the number in an equation directly. Derivatives indicate at what rate something is changing, for instance, the velocity of a traveling vehicle. Derivatives assist in determining the slope of a curve at any given point. In simple terms, limits indicate where a function is heading, and derivatives indicate at what rate it is changing. This chapter explains the fundamental concepts of calculus so that students can understand how things are changing extremely rapidly.

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  1. NCERT Exemplar Class 11 Maths Solutions Chapter 13
  2. Important topics in Class 11 Maths NCERT Exemplar Solutions Chapter 13
  3. NCERT Exemplar Class 11 Mathematics Chapters
  4. Importance of solving NCERT Exemplar Class 11 Maths Questions
  5. NCERT solutions of class 11 - Subject-wise
  6. NCERT Notes of class 11 - Subject Wise
  7. NCERT Books and NCERT Syllabus
  8. NCERT Exemplar Class 11 Solutions

To grasp this chapter well, students must concentrate on the fundamental concepts of limits, continuity, and differentiation. Solving the problems of Limits and Derivatives from the NCERT Exemplar exercises will create a strong concept. Solving more problems from NCERT Class 11 Maths Solutions Chapter Limits and Derivatives will make students proficient in solving problems. Regular practice, solving sample papers, and previous years' questions will make them precise and confident in solving calculus problems.

NCERT Exemplar Class 11 Maths Solutions Chapter 13

Class 11 Maths Chapter 13 exemplar solutions Exercise: 13.3
Page number: 239-245
Total questions: 80

Question:1

Evaluate $\mathop{\lim}\limits_{x \rightarrow 3}\frac{x^{2}-9}{x-3}$

Answer:

Given $\mathop{\lim }\limits_{x \rightarrow 3}\frac{x^{2}-9}{x-3}=\mathop{\lim }\limits_{x \rightarrow 3}\frac{ \left( x-3 \right) \left( x+3 \right) }{x-3}=\mathop{\lim }\limits_{x \rightarrow 3}x+3=6$

Question:2

Evaluate $\mathop{\lim }\limits\limits_{x \rightarrow 1/2}\frac{4x^{2}-1}{2x-1}$

Answer:

Given that $\mathop{\lim }\limits_{x \rightarrow 1/2}\frac{4x^{2}-1}{2x-1}=\mathop{\lim }\limits_{x \rightarrow 1/2}\frac{ \left( 2x-1 \right) \left( 2x+1 \right) }{2x-1}=\mathop{\lim }\limits_{x \rightarrow 1/2}2x+1=2$

Question:3

Evaluate $\mathop{\lim }\limits_{h \rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$

Answer:

$\mathop{\lim }\limits_{h \rightarrow 0}\frac{\sqrt {x+h}-\sqrt {x}}{h}$

$\\=\mathop{\lim }\limits_{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h[\sqrt{x+h}+\sqrt{x}]} \times( \sqrt{x+h}+\sqrt{x}) \ $
[Rationalizing the denominator]
$\mathop{\lim }\limits_{h \rightarrow 0}\frac{x+h-x}{h[\sqrt{x+h}+\sqrt{x}]}$ $\\\\=\mathop{\lim }\limits_{h \rightarrow 0} \frac{1}{ \sqrt{x+h}+\sqrt{x}}\\\\$
Put limit
$\\= \frac{1}{ \sqrt{x}+\sqrt{x}}\\\\ =\frac{1}{2\sqrt{x}}$

Question:4

Evaluate: $\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( x+2 \right) ^{\frac{1}{3}}-2^{\frac{1}{3}}}{x}$

Answer:

Given $\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( x+2 \right) ^{\frac{1}{3}}-2^{\frac{1}{3}}}{x}~~ \\\\ ~$
Now put x=x-2 limits change from 0 to 2
$\\\\ =\mathop{\lim }\limits_{y \rightarrow 2}\frac{y^{\frac{1}{3}}-2^{\frac{1}{3}}}{y-2}=\frac{1}{3} \left( 2 \right) ^{\frac{1}{3}-1}=\frac{1}{3}2^{-\frac{2}{3}}$ $\left[u \sin g \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n \cdot a^{n-1}\right]$

Question:5

Evaluate : $\mathop{\lim }\limits_{x \rightarrow 1}\frac{ \left( 1+x \right) ^{6}-1}{ \left( 1+x \right) ^{2}-1}$

Answer:

$\text{Given that }\mathop{\lim }\limits_{x \rightarrow 1}\frac{ \left( 1+x \right) ^{6}-1}{ \left( 1+x \right) ^{2}-1}=\mathop{\lim }\limits_{x \rightarrow 1}\frac{ \left( \left( 1+x \right) ^{2} \right) ^{3}-1}{ \left( 1+x \right) ^{2}-1}$
$\\ =\mathop{\lim }\limits_{x \rightarrow 1}\frac{ \left( \left( 1+x \right) ^{2}-1 \right) \left[ \left( 1+x \right) ^{4}+ \left( 1+x \right) ^{2}+1 \right] }{ \left( 1+x \right) ^{2}-1} \\$

$\\ =\mathop{\lim }\limits_{x \rightarrow 1} \left( 1+x \right) ^{4}+ \left( 1+x \right) ^{2}+1 \\\\ =2^{4}+2^{2}+1=21$

Question:6

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow a}\frac{ \left( 2+x \right) ^{\frac{5}{2}}- \left( 2+a \right) ^{\frac{5}{2}}}{x-a}$

Answer:
Given that $\mathop{\lim }\limits_{x \rightarrow a}\frac{ \left( 2+x \right) ^{\frac{5}{2}}- \left( 2+a \right) ^{\frac{5}{2}}}{x-a}$

$=\mathop{\lim }\limits_{y \rightarrow a+2}\frac{ \left( y \right) ^{\frac{5}{2}}- \left( 2+a \right) ^{\frac{5}{2}}}{y- \left( a+2 \right) }=\frac{5}{2} \left( a+2 \right) ^{\frac{5}{2}-1}=\frac{5}{2} \left( a+2 \right) ^{\frac{3}{2}}$ $\left [\text{using} \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n \cdot a^{n-1}\right]$

Question:7

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 1}\frac{x^{4}-\sqrt {x}}{\sqrt {x}-1}$

Answer:

$Given~\mathop{\lim }\limits_{x \rightarrow 1}\frac{x^{4}-\sqrt {x}}{\sqrt {x}-1}=\mathop{\lim }\limits_{x \rightarrow 1}\frac{ \left( x^{4}-\sqrt {x} \right) \left( \sqrt {x}+1 \right) }{ \left( \sqrt {x}-1 \right) \left( \sqrt {x}+1 \right) }$

$\\ =\mathop{\lim }\limits_{x \rightarrow 1}\frac{x^{4}\sqrt {x}+x^{4}-x-\sqrt {x}}{x-1} \\ \\ =\mathop{\lim }\limits_{x \rightarrow 1}\frac{\sqrt {x} \left( x^{4}-1 \right) +x \left( x^{3}-1 \right) }{x-1} \\ $

$\\ =\mathop{\lim }\limits_{x \rightarrow 1}\frac{ \left( x-1 \right) \left( \sqrt {x} \left( x^{3}+x^{2}+x+1 \right) +x \left( x^{2}+x+1 \right) \right) }{x-1} \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 1}\sqrt {x} \left( x^{3}+x^{2}+x+1 \right) +x \left( x^{2}+x+1 \right) \\ \\ =1 *4+1 *3=7 \\ \\$

Question:8

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 2}\frac{x^{2}-4}{\sqrt {3x-2}-\sqrt {x+2} }$

Answer:

Given $\mathop{\lim }\limits_{x \rightarrow 2}\frac{x^{2}-4}{\sqrt {3x-2}-\sqrt {x+2}}=\mathop{\lim }\limits_{x \rightarrow 2}\frac{x^{2}-4}{ \left( \sqrt {3x-2}-\sqrt {x+2} \right) } \times\frac{\sqrt {3x-2}+\sqrt {x+2}}{\sqrt {3x-2}+\sqrt {x+2}} \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 2}\frac{x^{2}-4}{ \left( 3x-2-x-2 \right) } \times \left( \sqrt {3x-2}+\sqrt {x+2} \right) \\ $

$\\ =\mathop{\lim }\limits_{x \rightarrow 2}\frac{ \left( x-2 \right) \left( x+2 \right) }{2 \left( x-2 \right) } \times \left( \sqrt {3x-2}+\sqrt {x+2} \right) \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 2}\frac{ \left( x+2 \right) }{2} \times \left( \sqrt {3x-2}+\sqrt {x+2} \right) =\frac{4}{2} \times \left( \sqrt {3 \times2-2}+\sqrt {2+2} \right) =8 \\ \\$

Question:9

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow \sqrt {2}}\frac{x^{2}-4}{x^{2}+3\sqrt {2}x-8} \\$

Answer:

$\begin{aligned} &=\lim _{x \rightarrow \sqrt{2}} \frac{\left(x^{2}-2\right)\left(x^{2}+2\right)}{x^{2}+4 \sqrt{2} x-\sqrt{2} x-8}\\ &=\lim _{x \rightarrow \sqrt{2}} \frac{(x+\sqrt{2})(x-\sqrt{2})\left(x^{2}+2\right)}{x(x+4 \sqrt{2})-\sqrt{2}(x+4 \sqrt{2})}\\ &=\lim _{x \rightarrow \sqrt{2}} \frac{(x+\sqrt{2})(x-\sqrt{2})\left(x^{2}+2\right)}{(x+4 \sqrt{2})(x-\sqrt{2})}\\&=\lim _{x \rightarrow \sqrt{2}} \frac{(x+\sqrt{2})\left(x^{2}+2\right)}{x+4 \sqrt{2}}\\ &\text { Put limit }\\ &=\frac{(\sqrt{2}+\sqrt{2})(2+2)}{\sqrt{2}+4 \sqrt{2}}\\&=\frac{2 \sqrt{2} \times 4}{5 \sqrt{2}}=\frac{8}{5} \end{aligned}$

Question:10

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2} \right) \\ \\$

Answer:

Let us apply LH rule i.e. L.Hospita's rule to this question

$\\ ~~\mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2} \right) \\ \\ \mathop{\lim }\limits_{x \rightarrow a} \left( \frac{f \left( x \right) }{g \left( x \right) } \right) =\mathop{\lim }\limits_{x \rightarrow a} \left( \frac{f^{'} \left( x \right) }{g^{'} \left( x \right) } \right) \\$

$ \\ \mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2} \right) =\mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{7x^{6}-10x^{4}}{3x^{2}-6x} \right) =\frac{7-10}{3-6}=1 \\ \\$

Question:11

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sqrt {1+x^{3}}-\sqrt {1-x^{3}}}{x^{2}} \right) \\ \\$

Answer:

$\text{Given that }\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sqrt {1+x^{3}}-\sqrt {1-x^{3}}}{x^{2}} \right) \\ \\$
$\\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sqrt {1+x^{3}}-\sqrt {1-x^{3}}}{x^{2}} *\frac{\sqrt {1+x^{3}}+\sqrt {1-x^{3}}}{\sqrt {1+x^{3}}+\sqrt {1-x^{3}}} \right) \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{1+x^{3}-1+x^{3}}{x^{2}} *\frac{1}{\sqrt {1+x^{3}}+\sqrt {1-x^{3}}} \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{2x}{\sqrt {1+x^{3}}+\sqrt {1-x^{3}}} \right) =0 \\ \\$

Question:12

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow -3} \left( \frac{x^{3}+27}{x^{5}+243} \right)$

Answer:

$\begin{aligned} &=\lim _{x \rightarrow -3} \frac{\frac{x^{3}+(3)^{3}}{x+3}}{\frac{x^{3}+(3)^{3}}{x+3}} \text { [Dividing the numerator and denominator by } \left.x+3\right]\\ &=\frac{\lim _{x \rightarrow -3}\left(\frac{x^{3}+(3)^{3}}{x+3}\right)}{\lim _{x \rightarrow -3}\left(\frac{x^{3}+(3)^{5}}{x+3}\right)}\left[\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}\right]\\ &=\frac{\lim _{x \rightarrow -3}\left(x^2-3x+9\right)}{\lim _{x \rightarrow -3}\left( x^4 - 3x^3 + 9x^2 - 27x + 81\right)} \\ &=\frac{9+9+9}{81+81+81+81+81}=\frac{3 \times 9}{5 \times 81 }=\frac{1}{5 \times 3}=\frac{1}{15} \end{aligned}$

Question:13

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow \frac{1}{2}} \left( \frac{8x-3}{2x-1}-\frac{4x^{2}+1}{4x^{2}-1} \right) \\$

Answer:

Given $\mathop{\lim }\limits_{x \rightarrow \frac{1}{2}} \left( \frac{8x-3}{2x-1}-\frac{4x^{2}+1}{4x^{2}-1} \right) \\ \\$
$\\=\mathop{\lim }\limits_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) \times \left( \frac{ \left( 8x-3 \right) \left( 2x+1 \right) -4x^{2}-1}{2x-1} \right) \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) \times \left( \frac{12x^{2}+2x-4}{2x-1} \right) \right) \\ \\$


$\\ =\mathop{\lim }\limits_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) \times \left( \frac{12x^{2}+8x-6x-4}{2x-1} \right) \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) \times \left( \frac{ \left( 3x+2 \right) \left( 4x-2 \right) }{2x-1} \right) ~ \right) \\$

$\\ =\mathop{\lim }\limits_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) \times2 \left( 3x+2 \right) ~ \right) =\frac{1}{2 \times\frac{1}{2}+1} \times2 \times \left( 3 \times\frac{1}{2}+2 \right) =\frac{7}{2} \\ \\$

Question:14

Evaluate: Find ‘n’, if
$\mathop{\lim }\limits_{x \rightarrow 2} \left( \frac{x^{n}-2^{n}}{x-2} \right) =80, n \in N \\ \\$

Answer:

$\\ \text{We know that }\mathop{\lim }\limits_{x \rightarrow 2} \left( \frac{x^{n}-2^{n}}{x-2} \right) =n \left( 2 \right) ^{n-1} \\ $

$\\ \\n \left( 2 \right) ^{n-1}=80 \\ $

$\\ n= 5 \times \left( 2 \right) ^{5-1}=5 \times16=80 \\ \\$

Question:15

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow a} \left( \frac{\sin 3x}{\sin 7x} \right) \\$

Answer:

Given $\mathop{\lim }\limits_{x \rightarrow a} \left( \frac{\sin 3x}{\sin 7x} \right) =\frac{\sin 3a}{\sin 7a} \\ \\$

Question:16

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin ^{2}2x}{\sin ^{2}4x} \right) \\$

Answer:

$Given~\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin ^{2}2x}{\sin ^{2}4x} \right) \\ \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{ \left( \frac{\sin 2x}{2x} *2x \right) ^{2}}{ \left( \frac{\sin 4x}{4x} *4x \right) ^{2}} \right) =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{4}{16} *\frac{ \left( \frac{\sin 2x}{2x} \right) ^{2}}{ \left( \frac{\sin 4x}{4x} \right) ^{2}} \right) =\frac{4}{16}=\frac{1}{4} \\ \\$

Question:17

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{1-\cos 2x}{x^{2}} \right)$

Answer:

Given that
$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{1-\cos 2x}{x^{2}} \right) =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{2\sin ^{2}x}{x^{2}} \right) =\mathop{\lim }\limits_{x \rightarrow 0} \left( 2 * \left( \frac{\sin x}{x} \right) ^{2} \right) =2 \\ \\$

Question:18

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{2\sin x-\sin 2x}{x^{3}} \right) \\$

Answer:

$\\Given~\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{2\sin x-\sin 2x}{x^{3}} \right) =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{2\sin x \left( 1-\cos x \right) }{x^{3}} \right) =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{2\sin x \left( 2\sin ^{2}\frac{x}{2} \right) }{x^{3}} \right) \\ $

$\\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( 2 *\frac{\sin x}{x} *2 * \left( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right) ^{2} *\frac{1}{4} \right) =1 \\ \\$

Question:19

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{1-\cos mx}{1-\cos nx} \right) \\$

Answer:

$\\ \text{Given that }\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{1-\cos mx}{1-\cos nx} \right) =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin ^{2}\frac{mx}{2}}{\sin ^{2}\frac{nx}{2}} \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\frac{\sin ^{2}\frac{mx}{2}}{ \left( \frac{mx}{2} \right) ^{2}} * \left( \frac{mx}{2} \right) ^{2}}{\frac{\sin ^{2}\frac{nx}{2}}{ \left( \frac{nx}{2} \right) ^{2}} * \left( \frac{nx}{2} \right) ^{2}} \right) \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\frac{\sin ^{2}\frac{mx}{2}}{ \left( \frac{mx}{2} \right) ^{2}} *m^{2}}{\frac{\sin ^{2}\frac{nx}{2}}{ \left( \frac{nx}{2} \right) ^{2}} *n^{2}} \right) =\frac{m^{2}}{n^{2}} \\ \\$

Question:20

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{3}}\frac{\sqrt {1-\cos 6x}}{\sqrt {2} \left( \frac{ \pi }{3}-x \right)}$

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{3}}\frac{\sqrt {1-\cos 6x}}{\sqrt {2} \left( \frac{ \pi }{3}-x \right) }~~ \\ \\ ~Here\cos 6x= 1-2\sin ^{2}3x \\ \\$


$\\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{3}}\frac{\sqrt {2\sin ^{2}3x}}{\sqrt {2} \left( \frac{ \pi }{3}-x \right) }=\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{3}}\frac{ \vert \sin 3x \vert }{ \left( \frac{ \pi -3x}{3} \right) }=\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{3}}\frac{ \vert \sin \left( \pi -3x \right) \vert }{ \left( \frac{ \pi -3x}{3} \right) } \\ \\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{3}}3 *\frac{ \vert \sin \left( \pi -3x \right) \vert }{ \pi -3x}=3 *1=3 \\$

Question:21

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}}\frac{\sin x-\cos x}{x-\frac{ \pi }{4} }\\$

Answer:


$\\ \text{Given that }\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}}\frac{\sin x-\cos x}{x-\frac{ \pi }{4}} \\ \\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}}\frac{\sqrt {2} \left( \sin x\cos \frac{ \pi }{4}-\cos x\sin \frac{ \pi }{4} \right) }{x-\frac{ \pi }{4}}=\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}}\frac{\sqrt {2}\sin \left( x-\frac{ \pi }{4} \right) }{x-\frac{ \pi }{4}}= \sqrt {2} \\ \\$

Question:22

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{\sqrt {3}\sin x-\cos x}{x-\frac{ \pi }{6}} \\$

Answer:

$\\\text{Given that }\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{\sqrt {3}\sin x-\cos x}{x-\frac{ \pi }{6}}=\\\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{2 \left( \frac{\sqrt {3}}{2}\sin x-\frac{1}{2}\cos x \right) }{x-\frac{ \pi }{6}}=\\ \mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{2 \left( \cos \frac{ \pi }{6}\sin x-\sin \frac{ \pi }{6}\cos x \right) }{x-\frac{ \pi }{6}} \\ \\$

$=\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{2\sin \left( x-\frac{ \pi }{6} \right) }{x-\frac{ \pi }{6}}=2 \\ \\$

Question:23

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 0}\frac{\sin 2x+3x}{2x+\tan 3x} \\$

Answer:


$\\ \text{Given that }\mathop{\lim }\limits_{x \rightarrow 0}\frac{\sin 2x+3x}{2x+\tan 3x} \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{2x *\frac{\sin 2x}{2x}+3x}{2x+3x *\frac{\tan 3x}{3x}} \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{x \left( 2 *\frac{\sin 2x}{2x}+3 \right) }{x \left( 2+3 *\frac{\tan 3x}{3x} \right) } \\ \\$


$\\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( 2 *\frac{\sin 2x}{2x}+3 \right) }{ \left( 2+3 *\frac{\tan 3x}{3x} \right) } \\ \\ =\frac{2+3}{2+3}=\frac{5}{5}=1 \\ \\$

Question:24

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow a}\frac{\sin x-\sin a}{\sqrt {x}-\sqrt {a}} \\$

Answer:

$\\ \text{Given that }\mathop{\lim }\limits_{x \rightarrow a}\frac{\sin x-\sin a}{\sqrt {x}-\sqrt {a}} \\ \\ =\mathop{\lim }\limits_{x \rightarrow a}\frac{ \left( \sin x-\sin a \right) \left( \sqrt {x}+\sqrt {a} \right) }{ \left( \sqrt {x}-\sqrt {a} \right) \left( \sqrt {x}+\sqrt {a} \right) } \\ \\ =\mathop{\lim }\limits_{x \rightarrow a}\frac{ \left( \sin x-\sin a \right) \left( \sqrt {x}+\sqrt {a} \right) }{ \left( x-a \right) } \\ \\$
$\\=\mathop{\lim }\limits_{x \rightarrow a}\frac{ \left( 2\cos \left( \frac{x+a}{2} \right) \sin \left( \frac{x-a}{2} \right) \right) \left( \sqrt {x}+\sqrt {a} \right) }{ \left( x-a \right) } \\ $

$\\ =\mathop{\lim }\limits_{x \rightarrow a}\cos \left( \frac{x+a}{2} \right) \frac{\sin \left( \frac{x-a}{2} \right) }{\frac{x-a}{2}} \left( \sqrt {x}+\sqrt {a} \right) \\ \\ =\cos \left( a \right) *1 * \left( 2\sqrt {a} \right) =2\sqrt {a}\cos a \\ \\$

Question:25

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{\cot ^{2}x-3}{cosec x-2} \\$

Answer:

$\\ \text{Given that }\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{\cot ^{2}x-3}{cosec x-2} \\ \\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{cosec^{2}x-1-3}{cosec x-2} \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{cosec^{2}x-1-3}{cosec x-2} \\ \\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}}\frac{ \left( cosecx-2 \right) \left( cosecx+2 \right) }{cosec x-2}$

$\\ \\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{6}} \left( cosecx+2 \right) =2+2=4 \\ \\$

Question:26

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( \sqrt {2}-\sqrt {1+\cos x} \right) }{\sin ^{2}x} \\$

Answer:

$\\ \text{Given that }\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( \sqrt {2}-\sqrt {1+\cos x} \right) }{\sin ^{2}x} \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( \sqrt {2}-\sqrt {1+\cos x} \right) \left( \sqrt {2}+\sqrt {1+\cos x} \right) }{\sin ^{2}x \left( \sqrt {2}+\sqrt {1+\cos x} \right) } \\ $

$\\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( 2- \left( 1+\cos x \right) \right) }{\sin ^{2}x \left( \sqrt {2}+\sqrt {1+\cos x} \right) } \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{1-\cos x}{\sin ^{2}x \left( \sqrt {2}+\sqrt {1+\cos x} \right) } \\ \\$
$\\=\mathop{\lim }\limits_{x \rightarrow 0}\frac{1-\cos x}{ \left( 1-\cos x \right) \left( 1+\cos x \right) \left( \sqrt {2}+\sqrt {1+\cos x} \right) } \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{1}{ \left( 1+\cos x \right) \left( \sqrt {2}+\sqrt {1+\cos x} \right) } \\ \\ =\frac{1}{ \left( 1+1 \right) * \left( 2\sqrt {2} \right) }=\frac{1}{4\sqrt {2}} \\ \\$

Question:27

Evaluate:
$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin x-2\sin 3x+\sin 5x}{x} \right) \\$

Answer:

$. \\ \text{Given that }\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin x-2\sin 3x+\sin 5x}{x} \right) \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin x}{x}-\frac{2\sin 3x}{3x} *3+\frac{\sin 5x}{5x} *5 \right) =1-2 *3+5=0 \\ \\$

Question:28

Evaluate: If
$\\ \mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{x^{4}-1}{x-1} \right) = \mathop{\lim }\limits_{x \rightarrow k} \left( \frac{x^{3}-k^{3}}{x^{2}-k^{2}} \right)$

Answer:

$\\ \text{Given that }$

$\mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{x^{4}-1}{x-1} \right) =4 \left( 1 \right) ^{4-1}=4 \\ \\ \mathop{\lim }\limits_{x \rightarrow k} \left( \frac{x^{3}-k^{3}}{x^{2}-k^{2}} \right)$

$ =\mathop{\lim }\limits_{x \rightarrow k} \left( \frac{ \left( x-k \right) \left( x^{2}+k^{2}+xk \right) }{ \left( x-k \right) \left( x+k \right) } \right) =\mathop{\lim }\limits_{x \rightarrow k} \left( \frac{x^{2}+k^{2}+xk}{x+k} \right) \\=\frac{3k}{2} \\ \\$
$\frac{3k}{2}=4 \\ \\ k=8/3 \\ \\$

Question:29

Differentiate each of the functions w.r.t x in
$\frac{x^{4}+x^{3}+x^{2}+1}{x}$

Answer:

$\\\text{ Let y}=\frac{x^{4}+x^{3}+x^{2}+1}{x} \\ \\ \frac{dy}{dx}=\frac{d \left( x^{3} \right) }{dx}+\frac{d \left( x^{2} \right) }{dx}+\frac{d \left( x \right) }{dx}+\frac{d \left( x^{-1} \right) }{dx} \\$

$ \\ =3x^{2}+2x+1-\frac{1}{x^{2}} \\ \\ =\frac{3x^{4}+2x^{3}+x^{2}-1}{x^{2}} \\ \\$

Question:30

Differentiate each of the functions w.r. to x in
$\left(x+\frac{1}{x}\right)^{3}$

Answer:

Let y= $\left( x+\frac{1}{x} \right) ^{3} \\ \\$

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

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

$ \\ =3 \left( x^{2}+2+\frac{1}{x^{2}}-1-\frac{2}{x^{2}}-\frac{1}{x^{4}} \right) \\$

$ \\ =3x^{2}+3-\frac{3}{x^{2}}-\frac{3}{x^{4}} \\ \\$

Question:31

Differentiate each of the functions w.r. to x in
$(3x + 5) (1 + tan x)$

Answer:

Given that $y= \left( 3x+5 \right) \left( 1+\tan x \right) ~ \\ \\$

Applying product rule of differentiation we get

$\\ \frac{dy}{dx}= \left( 1+\tan x \right) \frac{d}{dx} \left( 3x+5 \right) + \left( 3x+5 \right) \frac{d}{dx} \left( 1+\tan x \right) \\ $

$\\ =3 \left( 1+\tan x \right) + \left( 3x+5 \right) \sec ^{2}x \\ \\$

Question:32

Differentiate each of the functions w.r. to x in
$(sec x - 1) (sec x + 1)$

Answer:

y= $\left( \sec x-1 \right) \left( \sec x+1 \right) =\sec ^{2}x-1=\tan ^{2}x~ \\ \\$

Now applying the concept of chain rule


$\frac{dy}{dx}=\frac{d \left( \tan ^{2}x \right) }{dx}=2\tan x\sec ^{2}x \\ \\$

Question:33

Differentiate each of the functions w.r. to x in
$\frac{3x+4}{5x^{2}-7x+9 }\\$

Answer:

Given that $y=\frac{3x+4}{5x^{2}-7x+9} \\ \\$

Applying division rule of differentiation that is

$\\ \frac{dy}{dx}=\frac{d}{dx} \left( \frac{3x+4}{5x^{2}-7x+9} \right) \\$

$ \\ =\frac{ \left( 5x^{2}-7x+9 \right) \frac{d}{dx} \left( 3x+4 \right) - \left( 3x+4 \right) \frac{d}{dx} \left( 5x^{2}-7x+9 \right) }{ \left( 5x^{2}-7x+9 \right) ^{2}} \\ $

$\\ =\frac{ \left( 5x^{2}-7x+9 \right) \left( 3 \right) - \left( 3x+4 \right) \left( 10x-7 \right) }{ \left( 5x^{2}-7x+9 \right) ^{2}} \\$

$ \\ =-\frac{5 \left( 3x^{2}+8x-11 \right) }{ \left( 5x^{2}-7x+9 \right) ^{2}} \\$
$\\ =\frac{5 \left( 3x+11 \right) \left( 1-x \right) }{ \left( 5x^{2}-7x+9 \right) ^{2}} \\ \\$

Question:34

Differentiate each of the functions w.r. to x in
$\frac{x^{5}-\cos x}{\sin x} \\$

Answer:

Given that $y=\frac{x^{5}-\cos x}{\sin x}=\frac{x^{5}}{\sin x}-\frac{\cos x}{\sin x} \\ \\$

Applying division rule of differentiation that is

$\\ \frac{dy}{dx}=\frac{\sin x\frac{d}{dx} \left( x^{5} \right) -x^{5}\frac{d}{dx} \left( sinx \right) }{\sin ^{2}x}-\frac{d}{dx} \left( \cot x \right) \\ \\ =\frac{5x^{4}\sin x-x^{5}\cos x}{\sin ^{2}x}+\mathrm{cosec} ^{2}x \\ \\$

Question:35

Differentiate each of the functions w.r. to x in
$\frac{x^{2}\cos \frac{ \pi }{4}}{\sin x }$

Answer:

$y=\frac{x^{2}\cos \frac{ \pi }{4}}{\sin x} \\ \\$

Applying division rule of differentiation that is

$\\ \frac{dy}{dx}=\frac{\cos \frac{ \pi }{4} \left( \sin x\frac{d}{dx} \left( x^{2} \right) -x^{2}\frac{d}{dx} \left( \sin x \right) \right) }{\sin ^{2}x} \\ \\ =\frac{1}{\sqrt {2}} \left( \frac{2x\sin x-x^{2}\cos x}{\sin ^{2}x} \right) \\ \\$

Question:36
Differentiate each of the functions w.r. to x in

(ax\textsuperscript{2} + cotx) (p + q cosx)}} \\

Answer:

Given that $y= \left( ax^{2}+\cot x \right) \left( p+q\cos x \right) ~~~ \\ \\$

Applying the division rule of differentiation that is


$\frac{dy}{dx}= \left( p+q\cos x \right) \frac{d}{dx} \left( ax^{2}+\cot x \right) + \left( ax^{2}+\cot x \right) \frac{d}{dx} \left( p+q\cos x \right) \\ \\$
$= \left( p+q\cos x \right) \left( 2ax-cosec^{2}x \right) + \left( ax^{2}+\cot x \right) \left( -q\sin x \right) \\ \\$

Question:37

Differentiate each of the functions w.r. to x in
$\left( \frac{a+b\sin x}{c+d\cos x} \right)$

Answer:

Given that$y= \left( \frac{a+b\sin x}{c+d\cos x} \right) \\ \\$

Applying division rule of differentiation that is
$\\ \frac{dy}{dt}=\frac{ \left( c+d\cos x \right) \frac{d}{dx} \left( a+b\sin x \right) - \left( a+b\sin x \right) \frac{d}{dx} \left( c+d\cos x \right) }{ \left( c+d\cos x \right) ^{2}} \\ \\ =\frac{b\cos x \left( c+d\cos x \right) - \left( -d\sin x \right) \left( a+b\sin x \right) }{ \left( c+d\cos x \right) ^{2}} \\ \\$

$=\frac{bc\cos x+bd\cos ^{2}x+ad\sin x+bd\sin ^{2}x}{ \left( c+d\cos x \right) ^{2}}=\frac{bd+bc\cos x+ad\sin x}{ \left( c+d\cos x \right) ^{2}} \\ \\$

Question:38

Differentiate each of the functions w.r. to x in

(sin x + cosx)^2

Answer:

Given that
$\\y= \left( \sin x+\cos x \right) ^{2}=\sin ^{2}x+\cos ^{2}x+2\sin x\cos x=1+2\sin x\cos x\\=1+\sin 2x~ \\ \\$
Applying the concept of chain rule
$\frac{dy}{dx}=\frac{d}{dx} \left( 1+\sin 2x \right) =0+2 *\cos 2x=2\cos 2x=2 \left( \cos ^{2}x-\sin ^{2}x \right) \\ \\$

Question:39

Differentiate each of the functions w.r. to x in

(2x - 7)^{2} (3x + 5)^{3}

Answer:

This question will involve the concept of both chain rule and product rule

Given that $y= \left( 2x-7 \right) ^{2} \left( 3x+5 \right) ^{3}~ \\ \\$

Applying the product rule of differentiation
$\\ \frac{dy}{dx}= \left( 3x+5 \right) ^{3}\frac{d}{dx} \left( 2x-7 \right) ^{2}+ \left( 2x-7 \right) ^{2}\frac{d}{dx} \left( 3x+5 \right) ^{3} \\ $

$\\ = \left( 3x+5 \right) ^{3} *2 * \left( 2x-7 \right) *2+ \left( 2x-7 \right) ^{2} *3 * \left( 3x+5 \right) ^{2} *3 \\$

$ \\ = \left( 2x-7 \right) \left( 3x+5 \right) ^{2} \left[ 4 \left( 3x+5 \right) +9 \left( 2x-7 \right) \right] \\ \\ = \left( 2x-7 \right) \left( 3x+5 \right) ^{2} \left( 30x-43 \right) \\ \\$

Question:40

Differentiate each of the functions w.r. to x in
$x^{2} sinx + cos2x$

Answer:

This question will involve the concept of both chain rule and product rule

$\\Given \: \: that\: \: y=x^{2}\sin x+\cos 2x \\$

$ \\ \frac{dy}{dx}=\frac{d}{dx} \left( x^{2}\sin x \right) +\frac{d}{dx} \left( \cos 2x \right) \\ $

$\\ =\sin x\frac{d}{dx} \left( x^{2} \right) +x^{2}\frac{d}{dx} \left( \sin x \right) +\frac{d}{dx} \left( \cos 2x \right) \\ \\ =2x\sin x+x^{2}\cos x-2\sin 2x \\ \\$

Question:41

Differentiate each of the functions w.r.to x in
$sin ^{3}x cos ^{3}x $

Answer:

The question involves the concept of chain rule

$\\Given\: \: that \: \: y=\sin ^{3}x\cos ^{3}x \\ \\ y=\frac{1}{8} \left( 2\sin x\cos x \right) ^{3}$

$=\frac{1}{8}\sin ^{3}2x \\ \\ \frac{dy}{dx}=\frac{3}{8}\sin ^{2}2x \left( 2\cos 2x \right) =\frac{3}{4}\sin ^{2}2x\cos 2x \\ \\$

Question:42

Differentiate each of the functions w.r. to x in
$\frac{1}{ax^{2}+bx+c}$

Answer:

The question involves the concept of chain rule

$\\ Given\: \: that\: \: y=\frac{1}{ax^{2}+bx+c}= \left( ax^{2}+bx+c \right) ^{-1} \\ $

$\\ \frac{dy}{dx}=-1 \left( ax^{2}+bx+c \right) ^{-2} \left( 2ax+b \right) =-\frac{2ax+b}{ax^{2}+bx+c} \\ \\$

Question:43

Differentiate each of the functions with respect to ‘x’Differentiate using first principle

\cos (x^{2} + 1)

Answer:

$\\ \text{Let f} \left( x \right) =\cos \left( x^{2}+1 \right) \ldots \ldots .. \left( i \right) \\$

$ \\ f \left( x+ \Delta x \right) =\cos \left( \left( x+ \Delta x \right) ^{2}+1 \right) \ldots \ldots \left( ii \right) \\$

$ \\ \text{ Subtracting equation } \left( i \right) from \: \: equation \left( ii \right) \\$

$ \\ \frac{f \left( x+ \Delta x \right) -f \left( x \right) }{ \Delta x}=\frac{\cos \left( \left( x+ \Delta x \right) ^{2}+1 \right) -\cos \left( x^{2}+1 \right) }{ \Delta x} \\ \\$
$\\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{\cos \left( \left( x+ \Delta x \right) ^{2}+1 \right) -\cos \left( x^{2}+1 \right) }{ \Delta x} \\ $

$\\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{-2\sin \left[ \frac{ \left( x+ \Delta x \right) ^{2}+1-x^{2}-1}{2} \right] \sin \left[ \frac{ \left( x+ \Delta x \right) ^{2}+1+x^{2}+1}{2} \right] }{ \Delta x} \\ \\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{-2\sin \left[ \frac{2x \Delta x+ \left( \Delta x \right) ^{2}}{2} \right] \sin \left[ \frac{2x^{2}+2x \Delta x+2}{2} \right] }{ \Delta x} \\$

$ \\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0} \left( \frac{2x+ \Delta x}{2} \right) \left( ~\frac{-2\sin \left[ \frac{2x \Delta x+ \left( \Delta x \right) ^{2}}{2} \right] \sin \left[ \frac{2x^{2}+2x \Delta x+2}{2} \right] }{\frac{ \Delta x \left( 2x+ \Delta x \right) }{2}} \right) \\ \\$
$\\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0} \left( \frac{2x+ \Delta x}{2} \right) \left( -2\sin \left[ \frac{2x^{2}+2x \Delta x+2}{2} \right] \right) \left( ~\frac{\sin \left[ \frac{2x \Delta x+ \left( \Delta x \right) ^{2}}{2} \right] }{\frac{ \Delta x \left( 2x+ \Delta x \right) }{2}} \right) \\$

$ \\ =- \left( \frac{2x+0}{2} \right) \left( 2\sin \left( x^{2}+0+1 \right) \right) \left( 1 \right) =-2x\sin \left( x^{2}+1 \right) \text{is the required answer} \\ \\$

Question:44

Differentiate each of the functions with respect to ‘x’
Differentiate using first principle $\frac{ax+b}{cx+d}$

Answer:

$\\ \text{Let f} \left( x \right) =\frac{ax+b}{cx+d} \ldots .. \left( i \right) \\$

$ \\ f \left( x+ \Delta x \right) =\frac{a \left( x+ \Delta x \right) +b}{c \left( x+ \Delta x \right) +d} \ldots \ldots \left( ii \right) \\$

$ \\ \text{Subtracting equation } \left( i \right) \text{from equation } \left( ii \right) \\ $

$\\ \frac{f \left( x+ \Delta x \right) -f \left( x \right) }{ \Delta x}=\frac{\frac{a \left( x+ \Delta x \right) +b}{c \left( x+ \Delta x \right) +d}-\frac{ax+b}{cx+d}}{ \Delta x}~ \\ \\$
$\\ \text{Taking the limit} \\ \\ f^{'} \left( x \right) =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{\frac{a \left( x+ \Delta x \right) +b}{c \left( x+ \Delta x \right) +d}-\frac{ax+b}{cx+d}}{ \Delta x} \\ $

$\\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{\frac{ \left( cx+d \right) \left( ax+a \Delta x+b \right) - \left( ax+b \right) \left( cx+c \Delta x+d \right) }{ \left( cx+c \Delta x+d \right) \left( cx+d \right) }}{ \Delta x} \\$

$ \\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{ \left( ad-bc \right) \Delta x}{ \left( cx+c \Delta x+d \right) \left( cx+d \right) \Delta x} \\ \\$

$\\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{ \left( ad-bc \right) }{ \left( cx+c \Delta x+d \right) \left( cx+d \right) }= \frac{ad-bc}{ \left( cx+d \right) ^{2}}\text{ is the required answer} \\ \\$

Question:45

Differentiate each of the functions with respect to ‘x’
Differentiate using first principle $X^{\frac{2}{3}}$

Answer:

$\\ f^{'} \left( x \right) =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{ \left( x+ \Delta x \right) ^{\frac{2}{3}}-x^{\frac{2}{3}}}{ \Delta x} \\ \\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{x^{\frac{2}{3}} \left[ \left( 1+\frac{ \Delta x}{x} \right) ^{\frac{2}{3}}-1 \right] }{ \Delta x} \\ \\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{x^{\frac{2}{3}} \left[ \left( 1+\frac{2}{3}\frac{ \Delta x}{x}+ \ldots \right) -1 \right] }{ \Delta x} \\ \\$

Expanding by binomial theorem and rejecting the higher powers of $\Delta x \: \: as\: \: \Delta x \rightarrow 0 \\ \\$


$\\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{x^{\frac{2}{3}} \left( \frac{2}{3}\frac{ \Delta x}{x} \right) }{ \Delta x}=\mathop{\lim }\limits_{ \Delta x \rightarrow 0} \left( \frac{2}{3}\frac{x^{\frac{2}{3}}}{x} \right) =\frac{2}{3}x^{-\frac{1}{3}}$

Question:46

Differentiate each of the functions with respect to ‘x’
Differentiate using first principle x cos x.

Answer:

$\\ Given\: \: that\: \: y=x\cos x \ldots . \left( i \right) \\$

$ \\ y+ \Delta y= \left( x+ \Delta x \right) \cos \left( x+ \Delta x \right) \ldots . \left( ii \right) ~ \\ $

$\\ \text{ Subtracting equation } \left( i \right) from equation \left( ii \right) \\ $

$\\ y^{'}=\mathop{\lim }\limits_{ \Delta x \rightarrow 0} \left( \frac{ \left( \left( x+ \Delta x \right) \cos \left( x+ \Delta x \right) -x\cos x \right) }{ \Delta x} \right) \\ \\$


$\\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0} \left( \frac{x\cos \left( x+ \Delta x \right) -x\cos x}{ \Delta x} \right) +\mathop{\lim }\limits_{ \Delta x \rightarrow 0} \left( \frac{ \Delta x\cos \left( x+ \Delta x \right) }{ \Delta x} \right) \\$

$ \\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0} \left( \frac{x \left( -2\sin \frac{x+ \Delta x-x}{2}\sin \frac{x+ \Delta x+x}{2} \right) }{ \Delta x} \right) +\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\cos \left( x+ \Delta x \right) \\$

$ \\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\frac{- \left( 2\sin \left( x+\frac{ \Delta x}{2} \right) \sin \left( \frac{ \Delta x}{2} \right) x \right) }{ \Delta x}+\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\cos \left( x+ \Delta x \right) \\ \\$


$\\ =\mathop{\lim }\limits_{ \Delta x \rightarrow 0}-x\sin \left( x+\frac{ \Delta x}{2} \right) \frac{ \left( \sin \left( \frac{ \Delta x}{2} \right) \right) }{\frac{ \Delta x}{2}}+\mathop{\lim }\limits_{ \Delta x \rightarrow 0}\cos \left( x+ \Delta x \right) \\ \\ =-x\sin x+\cos x \\ \\$

Question:47

Evaluate each of the following limits

$\mathop{\lim }\limits_{y \rightarrow 0}\frac{ \left( x+y \right) \sec \left( x+y \right) -x\sec x}{y }$

Answer:

$\\ \text{Given that}$

$\mathop{\lim }\limits_{y \rightarrow 0}\frac{ \left( x+y \right) \sec \left( x+y \right) -x\sec x}{y}=\mathop{\lim }\limits_{y \rightarrow 0} \left( \frac{x\sec \left( x+y \right) -x\sec x}{y}+\frac{y\sec \left( x+y \right) }{y} \right) \\$

$ \\ =\mathop{\lim }\limits_{y \rightarrow 0} \left( \frac{x}{y} \left[ \frac{\cos x-\cos \left( x+y \right) }{\cos x\cos \left( x+y \right) } \right] +\sec \left( x+y \right) \right) \\ $

$\\ =\mathop{\lim }\limits_{y \rightarrow 0} \left( \frac{x}{y} \left[ \frac{-2\sin \left( x+\frac{y}{2} \right) \sin \left( \frac{-y}{2} \right) }{\cos x\cos \left( x+y \right) } \right] +\sec \left( x+y \right) \right) \\ \\$
$\\ =\mathop{\lim }\limits_{y \rightarrow 0} \left( \frac{x\sin \left( x+\frac{y}{2} \right) }{\cos x\cos \left( x+y \right) } \left[ \frac{\sin \left( \frac{y}{2} \right) }{\frac{y}{2}} \right] +\sec \left( x+y \right) \right) \\ \\ =x\sec x\tan x+\sec x \\ \\ =\sec x \left( x\tan x+1 \right) \\ \\$

Question:48

Evaluate each of the following limits
$\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left[ \sin \left( \alpha + \beta \right) x+\sin \left( \alpha - \beta \right) x+\sin 2 \alpha x \right] }{\cos 2 \beta x-\cos 2 \alpha x}x \\$

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left[ \sin \left( \alpha + \beta \right) x+\sin \left( \alpha - \beta \right) x+\sin 2 \alpha x \right] }{\cos 2 \beta x-\cos 2 \alpha x}x \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left[ 2\sin \alpha x\cos \beta x+2\sin \alpha x\cos \alpha x \right] }{ \left[ 2\sin \left( \alpha + \beta \right) x~\sin \left( \alpha - \beta \right) x \right] }x \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left[ \sin \alpha x \left( \cos \beta x+\cos \alpha x \right) \right] }{ \left[ \sin \left( \alpha + \beta \right) x~\sin \left( \alpha - \beta \right) x \right] }x \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left[ \sin \alpha x \left( 2\cos \frac{ \left( \alpha + \beta \right) x}{2}\cos \frac{ \left( \alpha - \beta \right) x}{2} \right) \right] }{ \left[ \sin \left( \alpha + \beta \right) x~\sin \left( \alpha - \beta \right) x \right] }x \\$

$ \\$
$\\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left[ \sin \alpha x \right] }{ \left[ 2\sin \frac{ \left( \alpha + \beta \right) x}{2}\sin \frac{ \left( \alpha - \beta \right) x}{2} \right] }x \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left[ \frac{\sin \alpha x}{ \alpha x} \right] * \alpha x *x}{ \left[ 2 \left( \frac{\sin \frac{ \left( \alpha + \beta \right) x}{2}}{\frac{ \left( \alpha + \beta \right) x}{2}} \right) \left( \frac{\sin \frac{ \left( \alpha - \beta \right) x}{2}}{\frac{ \left( \alpha - \beta \right) x}{2}} \right) *\frac{ \left( \alpha + \beta \right) x}{2} *\frac{ \left( \alpha - \beta \right) x}{2} \right] } \\$

$ \\ =\frac{ \alpha }{2 *\frac{ \left( \alpha + \beta \right) }{2} *\frac{ \left( \alpha - \beta \right) }{2}}=\frac{2 \alpha }{ \alpha ^{2}- \beta ^{2}} \\ \\$

Question:49

Evaluate each of the following limits
$\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\tan ^{3}x-\tan x}{\cos \left( x+\frac{ \pi }{4} \right) } \right)$

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\tan ^{3}x-\tan x}{\cos \left( x+\frac{ \pi }{4} \right) } \right) =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \tan x \right) *\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\tan ^{2}x-1}{\cos \left( x+\frac{ \pi }{4} \right) } \right) \\$

$ \\ =1 *\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{- \left( 1-\tan x \right) \left( 1+\tan x \right) }{\cos \left( x+\frac{ \pi }{4} \right) } \right) \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}}- \left( 1+\tan x \right) *\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{ \left( 1-\tan x \right) }{\cos \left( x+\frac{ \pi }{4} \right) } \right) \\$

$ \\ =-2\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{ \left( \cos x-\sin x \right) }{\cos \left( x+\frac{ \pi }{4} \right) \cos x} \right) \\ \\$
$\\ =-2\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\sqrt {2} \left( \cos \frac{ \pi }{4}\cos x-\sin \frac{ \pi }{4}\sin x \right) }{\cos \left( x+\frac{ \pi }{4} \right) \cos x} \right) \\ $

$\\ =-2\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\sqrt {2} \left( \cos \left( x+\frac{ \pi }{4} \right) \right) }{\cos \left( x+\frac{ \pi }{4} \right) \cos x} \right) =-2\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\sqrt {2}}{\cos x} \right) =-4 \\ \\$

Question:50

Evaluate each of the following limits
$\mathop{\lim }\limits_{x \rightarrow \pi } \left( \frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2} \left( \cos \frac{x}{4}-\sin \frac{x}{4} \right) } \right)$

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow \pi } \left( \frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2} \left( \cos \frac{x}{4}-\sin \frac{x}{4} \right) } \right) =\mathop{\lim }\limits_{x \rightarrow \pi } \left( \frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2} \left( \cos \frac{x}{4}-\sin \frac{x}{4} \right) } \right) \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow \pi } \left( \frac{\cos ^{2}\frac{x}{4}+\sin ^{2}\frac{x}{4}-2\sin \frac{x}{4}\cos \frac{x}{4}}{ \left( \cos ^{2}\frac{x}{4}-\sin ^{2}\frac{x}{4} \right) \left( \cos \frac{x}{4}-\sin \frac{x}{4} \right) } \right) \\ \\$
$\\ =\mathop{\lim }\limits_{x \rightarrow \pi } \left( \frac{ \left( \cos \frac{x}{4}-\sin \frac{x}{4} \right) ^{2}}{ \left( \cos ^{2}\frac{x}{4}-\sin ^{2}\frac{x}{4} \right) \left( \cos \frac{x}{4}-\sin \frac{x}{4} \right) } \right) =\mathop{\lim }\limits_{x \rightarrow \pi } \left( \frac{ \left( \cos \frac{x}{4}-\sin \frac{x}{4} \right) }{ \left( \cos ^{2}\frac{x}{4}-\sin ^{2}\frac{x}{4} \right) } \right) \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow \pi } \left( \frac{1}{ \left( \cos \frac{x}{4}+\sin \frac{x}{4} \right) } \right) =\frac{1}{\frac{1}{\sqrt {2}}+\frac{1}{\sqrt {2}}}=\frac{1}{\sqrt {2}} \\ \\$

Question:51

Evaluate each of the following limits
Show that $\mathop{\lim }\limits_{x \rightarrow 4}\frac{ \vert x-4 \vert }{x-4}$
does not exist.

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 4}\frac{ \vert x-4 \vert }{x-4} \\$

$ \\ LHL=\mathop{\lim }\limits_{x \rightarrow \mathop{4}^{-}}\frac{-x+4}{x-4}=-1 \\$

$ \\ RHL=\mathop{\lim }\limits_{x \rightarrow \mathop{4}^{+}}\frac{x-4}{x-4}=1~ \\$

$ \\ LHL \neq RHL \\ \\$

Hence, the limit doesnot exist

Question:52

Let $f \left( x \right) =\frac{k\cos x}{ \pi -2x}$ when $x \neq \frac{\pi}{2}$and
$f(x)=3$ if $\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)$ find the value of k.

Answer:

$\\ LHL=\mathop{\lim }\limits_{x \rightarrow \mathop{\frac{ \pi }{2}}^{-}} \left( \frac{k\cos \left( \frac{ \pi }{2}-h \right) }{ \pi -2 \left( \frac{ \pi }{2}-h \right) } \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{-}} \left( \frac{k\cos \left( \frac{ \pi }{2}-h \right) }{2h} \right) \\ \\ =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{-}} \left( \frac{k\sin h}{2h} \right) =\frac{k}{2} \\ $

$\\ RHL=\mathop{\lim }\limits_{x \rightarrow \mathop{\frac{ \pi }{2}}^{+}} \left( \frac{k\cos \left( \frac{ \pi }{2}+h \right) }{ \pi -2 \left( \frac{ \pi }{2}+h \right) } \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{+}} \left( \frac{k\cos \left( \frac{ \pi }{2}+h \right) }{-2h} \right) \\ \\$$\\=\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{-}} \left( \frac{-k\sin h}{-2h} \right) =\frac{k}{2} \\$

$ \\ \mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{2}}f \left( x \right) =\frac{k}{2}=3 \\ \\ k=6 \\ \\$

Question:53

Evaluate each of the following limits
Let $f(x)=\begin{array}{ll} x+2 & x \leq-1 \\ c x^{2} & x>-1 \end{array}$ find ‘c’ if $\lim _{x \rightarrow-1} f(x)$ exists.

Answer:

$\\ LHL=\mathop{\lim }\limits_{x \rightarrow \mathop{-1}^{-}} \left( x+2 \right) =-1+2=1 \\ $

$\\ RHL=\mathop{\lim }\limits_{x \rightarrow \mathop{-1}^{+}}cx^{2}=c \\$

$ \\ LHL=RHL \\$

$ \\ c=1 \\ \\$

Question:54

Choose the correct answer out of 4 options given against each Question
$\mathop{\lim }\limits_{x \rightarrow \pi }\frac{\sin x}{x- \pi }$ is

A. 1
B. 2
C. –1
D. –2

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow \pi }\frac{\sin x}{x- \pi }=\mathop{\lim }\limits_{x \rightarrow \pi }\frac{\sin \left( \pi -x \right) }{- \left( \pi -x \right) }=-1~ \\ \\ \mathop{\lim }\limits_{x \rightarrow 0}\frac{\sin x}{x}=1~~~ \\ \\ \pi -x \rightarrow 0~~~x \rightarrow \pi \\ \\$

Hence, the answer is option C

Question:55

Choose the correct answer out of 4 options given against each Question
$\mathop{\lim }\limits_{x \rightarrow 0}\frac{x^{2}\cos x}{1-\cos x}~$ is

A. 2
B. $\frac{3}{2}$
C. $\frac{-3}{2}$
D. 1

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 0}\frac{x^{2}\cos x}{1-\cos x}=\mathop{\lim }\limits_{x \rightarrow 0}\frac{x^{2}\cos x}{2\sin ^{2}\frac{x}{2}} \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0}\frac{x^{2}\cos x}{\frac{2x^{2}}{4}\frac{\sin ^{2}\frac{x}{2}}{\frac{x^{2}}{4}}}=\mathop{\lim }\limits_{x \rightarrow 0}\frac{\cos x}{\frac{2}{4}\frac{\sin ^{2}\frac{x}{2}}{\frac{x^{2}}{4}}}=2 \\ \\$
Hence, the answer is option A

Question:56

Choose the correct answer out of 4 options given against each Question
$\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( 1+x \right) ^{n}-1}{x}~$ is

A. n
B. 1
C. –n
D. 0
Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( 1+x \right) ^{n}-1}{x}=\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( 1+x \right) ^{n}-1^{n}}{ \left( 1+x \right) -1}=n \left( 1 \right) ^{n-1}=n \\ \\$
Hence, the answer is option A

Question:57

Choose the correct answer out of 4 options given against each Question
$\mathop{\lim }\limits_{x \rightarrow 1}\frac{x^{m}-1}{x^{n}-1}$ is

A. 1

B. $\frac{m}{n}$
C. $-\frac{m}{n}$
D. $\frac{m^2}{n^2}$

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 1}\frac{x^{m}-1}{x^{n}-1} \\ \\ =\mathop{\lim }\limits_{x \rightarrow 1}\frac{\frac{x^{m}-1}{x-1}}{\frac{x^{n}-1}{x-1}}=\frac{m \left( 1 \right) ^{m-1}}{n \left( 1 \right) ^{n-1}}=\frac{m}{n} \\ \\$
Hence, the answer is option B

Question:58

Choose the correct answer out of 4 options given against each Question

$\mathop{\lim }\limits_{ \theta \rightarrow 0} \left( \frac{1-\cos 4 \theta }{1-\cos 6 \theta } \right)$is


A.$\frac{4}{9}$
B.$\frac{1}{2}$
C.$-\frac{1}{2}$
D.$-1$
Answer:

$\\ \mathop{\lim }\limits_{ \theta \rightarrow 0} \left( \frac{1-\cos 4 \theta }{1-\cos 6 \theta } \right) \\ \\ =\mathop{\lim }\limits_{ \theta \rightarrow 0} \left( \frac{2\sin ^{2}2 \theta }{2\sin ^{2}3 \theta } \right) \\ \\ =\mathop{\lim }\limits_{ \theta \rightarrow 0} \left( \frac{\sin 2 \theta }{\sin 3 \theta } \right) ^{2}=\mathop{\lim }\limits_{ \theta \rightarrow 0} \left( \frac{\frac{\sin 2 \theta }{2 \theta } *2 \theta }{ \left( \frac{\sin 3 \theta }{3 \theta } \right) *3 \theta } \right) ^{2}=\frac{4}{9} \\ \\$
Hence, the answer is option A

Question:59

Choose the correct answer out of 4 options given against each Question
$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{cosec x-\cot x}{x} \right)$ is

A. $-\frac{1}{2}$

B. 1
C. $\frac{1}{2}$
D. –1

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{cosec x-\cot x}{x} \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{1-\cos x}{x\sin x} \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{2\sin ^{2}x/2}{2x\sin x/2\cos x/2} \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin x/2}{x\cos x/2} \right) =\frac{1}{2} \\ \\$
Hence, the answer is option C

Question:60

Choose the correct answer out of 4 options given against each Question
$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin x}{\sqrt {x+1}-\sqrt {1-x}} \right)$ is

A. 2
B. 0
C. 1
D. –1

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin x}{\sqrt {x+1}-\sqrt {1-x}} \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin x \left( \sqrt {x+1}+\sqrt {1-x} \right) }{ \left( \sqrt {x+1}-\sqrt {1-x} \right) \left( \sqrt {x+1}+\sqrt {1-x} \right) } \right) \\ $

$\\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin x \left( \sqrt {x+1}+\sqrt {1-x} \right) }{ \left( x+1-1+x \right) } \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{\sin x \left( \sqrt {x+1}+\sqrt {1-x} \right) }{2x} \right) \\ \\ =1 \\ \\$
Hence, the answer is option C

Question:61

Choose the correct answer out of 4 options given against each Question

$\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\sec ^{2}x-2}{\tan x-1} \right)$ is
A. 3
B. 1
C. 0
D. $\sqrt{2}$

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\sec ^{2}x-2}{\tan x-1} \right)\\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{1+\tan ^{2}x-2}{\tan x-1} \right) \\=\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\tan ^{2}x-1}{\tan x-1} \right)\\ =\mathop{\lim }\limits_{x \rightarrow \frac{ \pi }{4}} \left( \tan x+1 \right)\\ =2 \\ \\$
Hence, the answer is option D

Question:62

Choose the correct answer out of 4 options given against each Question
$\mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{ \left( \sqrt {x}-1 \right) \left( 2x-3 \right) }{2x^{2}+x-3} \right)$ is

A. $\frac{1}{10}$
B. $\frac{-1}{10}$
C. 1
D. None of these

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{ \left( \sqrt {x}-1 \right) \left( 2x-3 \right) }{2x^{2}+x-3} \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{ \left( \sqrt {x}-1 \right) \left( 2x-3 \right) }{ \left( 2x+3 \right) \left( x-1 \right) } \right) \\$

$ \\ =\mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{ \left( \sqrt {x}-1 \right) \left( 2x-3 \right) }{ \left( 2x+3 \right) \left( \sqrt {x}-1 \right) \left( \sqrt {x}+1 \right) } \right) \\ \\ =\mathop{\lim }\limits_{x \rightarrow 1} \left( \frac{ \left( 2x-3 \right) }{ \left( 2x+3 \right) \left( \sqrt {x}+1 \right) } \right) =-\frac{1}{10} \\ \\$
Hence, the answer is option B

Question:63

Choose the correct answer out of 4 options given against each Question
If $\begin{aligned} f(x)=\frac{\sin [x]}{[x]}, &[x] \neq 0 \\ 0, &[x]=0 \end{aligned}$ where [.] denotes the greatest integer function, then $\lim_{x\rightarrow 0}f(x)$ is equal to

A. 1
B. 0
C. –1
D. None of these
Answer:

$LHL=\mathop{\lim }\limits_{x \rightarrow \mathop{0}^{-}} \left( \frac{\sin \left[ x \right] }{ \left[ x \right] } \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{-}} \left( \frac{\sin \left[ 0-h \right] }{ \left[ 0-h \right] } \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{-}} \left( \frac{\sin \left[ -h \right] }{ \left[ -h \right] } \right) \\=\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{-}} \left( \frac{\sin \left( -1 \right) }{-1} \right) =\sin 1 \\ \\$
$RHL=\mathop{\lim }\limits_{x \rightarrow \mathop{0}^{+}} \left( \frac{\sin \left[ x \right] }{ \left[ x \right] } \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{+}} \left( \frac{\sin \left[ 0+h \right] }{ \left[ 0+h \right] } \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{+}} \left( \frac{\sin \left[ h \right] }{ \left[ h \right] } \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{+}} \left( \frac{\sin \left( 0 \right) }{0} \right) \\ \\$
Limit doesn’t exist
Hence, the answer is option D

Question:64

Choose the correct answer out of 4 options given against each Question

$\mathop{\lim }\limits_{x \rightarrow 0} \left( \frac{ \vert \sin x \vert }{x} \right) ~$ is

A. 1
B. –1
C. does not exist
D. None of these

Answer:

$\\ LHL=\mathop{\lim }\limits_{x \rightarrow \mathop{0}^{-}} \left( \frac{ \vert \sin x \vert }{x} \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{-}} \left( \frac{ \vert \sin \left( 0-h \right) \vert }{ \left( 0-h \right) } \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{-}} \left( \frac{-\sin \left( -h \right) }{-h} \right) \\=-1 \\ $

$RHL=\mathop{\lim }\limits_{x \rightarrow \mathop{0}^{+}} \left( \frac{ \vert \sin x \vert }{x} \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{+}} \left( \frac{ \vert \sin \left( 0+h \right) \vert }{ \left( 0+h \right) } \right) =\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{+}} \left( \frac{\sin \left( h \right) }{h} \right) =1 \\ \\$
$\text{ So, limit does}n^{'}\text{t exists} \\ \\ \text{Hence, the answer is option C} \\ \\$

Question:65

Choose the correct answer out of 4 options given against each Question
Let $f(x)=\begin{array}{l} x^{2}-1,0<x<2 \\ 2 x+3,2 \leq x<3 \end{array}$ the quadratic equation whose roots are $\lim _{x \rightarrow 2} f(x) \text { and } \lim _{x \rightarrow 2} f(x)$ is

$\\A. x^{2} - 6x + 9 = 0\\ B. x^{2} -7x + 8 = 0\\ C. x^{2} - 14x + 49 = 0\\ D. x^{2} - 10x + 21 = 0 \\$

Answer:

$\\ LHL=\mathop{\lim }\limits_{x \rightarrow \mathop{2}^{-}}f \left( x \right) =\mathop{\lim }\limits_{x \rightarrow \mathop{2}^{-}} \left( x^{2}-1 \right) =3 \\ $

$\\ RHL=\mathop{\lim }\limits_{x \rightarrow \mathop{2}^{+}}f \left( x \right)$

$ =\mathop{\lim }\limits_{x \rightarrow \mathop{2}^{+}} \left( 2x+3 \right) =7 \\$

$ \\ \text{The quadratic equation whose roots are 3 and 7 are } \left( x-3 \right) \left( x-7 \right)\\ $

$=x^{2}-10x+21 \\ \\$
Hence, the answer is option D

Question:66

Choose the correct answer out of 4 options given against each Question
$\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( \tan 2x-x \right) }{3x-\sin x}~$ is

A. 2
B. $\frac{1}{2}$
C. $\frac{-1}{2}$
D. $\frac{1}{4}$

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( \tan 2x-x \right) }{3x-\sin x}=\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( \frac{\tan 2x}{x}-1 \right) }{ \left( 3-\frac{\sin x}{x} \right) }=\mathop{\lim }\limits_{x \rightarrow 0}\frac{ \left( \frac{\text{2 tan}2x}{2x}-1 \right) }{ \left( 3-\frac{\sin x}{x} \right) }=\frac{2-1}{3-1}=\frac{1}{2} \\ \\$
Hence, the answer is option B

Question:67

Choose the correct answer out of 4 options given against each Question
Let f(x) = x – [x], $\in \mathrm{R}, \text { then } \mathrm{f}^{\prime} \frac{1}{2}$ is

A. 3/2
B. 1
C. 0
D. –1

Answer:

$\\ LHD=\mathop{\lim }\limits_{h \rightarrow 0}\frac{ \left( f \left( \frac{1}{2}-h \right) -f \left( \frac{1}{2} \right) \right) }{-h} \\ \\ =\mathop{\lim }\limits_{h \rightarrow 0}\frac{ \left( \left( \frac{1}{2}-h \right) - \left[ \left( \frac{1}{2}-h \right) \right] - \left( \frac{1}{2} \right) + \left[ \frac{1}{2} \right] \right) }{-h}=\mathop{\lim }\limits_{h \rightarrow 0}\frac{-h}{-h}=1 \\ $

$\\ RHD=\mathop{\lim }\limits_{h \rightarrow 0}\frac{ \left( f \left( \frac{1}{2}+h \right) -f \left( \frac{1}{2} \right) \right) }{h} \\ \\ =\mathop{\lim }\limits_{h \rightarrow 0}\frac{ \left( \left( \frac{1}{2}+h \right) - \left[ \left( \frac{1}{2}+h \right) \right] - \left( \frac{1}{2} \right) + \left[ \frac{1}{2} \right] \right) }{h}=\mathop{\lim }\limits_{h \rightarrow 0}\frac{h}{h}=1 \\ \\$
Hence, the answer is option B

Question:68

Choose the correct answer out of 4 options given against each Question
If $If\: \: y=\sqrt {x}+\frac{1}{x}, then \frac{dy}{dx}$ at x = 1 is

A. 1
B. $\frac{1}{2}$
C. $\frac{1}{\sqrt{2}}$
D. 0

Answer:

$\\ y=\sqrt {x}+\frac{1}{x} \\ \\ \frac{dy}{dx}=\frac{1}{2\sqrt {x}}-\frac{1}{x^{2}} \\ $

$\\ \left( \frac{dy}{dx} \right) _{x=1}=\frac{1}{2}-1=-\frac{1}{2} \\ \\$

Hence, the answer is option D

Question:69

Choose the correct answer out of 4 options given against each Question
If $~f \left( x \right) =\frac{x-4}{2\sqrt {x}}$ then f’(1) is

A. $\frac{5}{4}$
B. $\frac{4}{5}$
C. 1
D. 0

Answer:

$\\ f \left( x \right) =\frac{x-4}{2\sqrt {x}} \\$

$ \\ f^{'} \left( x \right) =\frac{1}{2} \left[ \frac{\sqrt {x} *1- \left( x-4 \right) *\frac{1}{2\sqrt {x}}}{x} \right] =\frac{1}{2} \left[ \frac{x+4}{2x^{\frac{3}{2}}} \right] \\$

$ \\ f^{'} \left( 1 \right) =\frac{1}{2} \left[ \frac{1+4}{2 \left( 1 \right) } \right] =\frac{5}{4} \\ \\$
Hence, the answer is option A

Question:70

Choose the correct answer out of 4 options given against each Question

If $y=\frac{1+\frac{1}{x^{2}}}{1-\frac{1}{x^{2}}}$ then $\frac{dy}{dx}$ is

A.$\frac{-4x}{\left(x^{2}-1\right)^{2}}$
B.$\frac {-4 x}{x^{2}-1}$
C.$\frac{1-x^{2}}{4 x}$
D.$\frac{4 x}{x^{2}-1}$

Answer:

$\\y=\frac{1+\frac{1}{x^{2}}}{1-\frac{1}{x^{2}}}=\frac{x^{2}+1}{x^{2}-1} \\$

$ \\ \frac{dy}{dx}=\frac{ \left( x^{2}-1 \right) \left( 2x \right) - \left( x^{2}+1 \right) \left( 2x \right) }{ \left( x^{2}-1 \right) ^{2}}=\frac{ \left( 2x \right) \left( x^{2}-1-x^{2}-1 \right) }{ \left( x^{2}-1 \right) ^{2}}=-\frac{4x}{ \left( x^{2}-1 \right) ^{2}} \\ \\$
Hence, the answer is option A

Question:71

Choose the correct answer out of 4 options given against each Question

If $y=\frac{\sin x+\cos x}{\sin x-\cos x}$ then $\frac{dy}{dx}_{at\: \: x=0}$ is


A. –2
B. 0
C. $\frac{1}{2}$
D. does not exist
Answer:

$\\ y=\frac{\sin x+\cos x}{\sin x-\cos x} \\ \\ \frac{dy}{dx}=\frac{ \left( \sin x-\cos x \right) \left( \cos x-\sin x \right) - \left( \sin x+\cos x \right) \left( \sin x+\cos x \right) }{ \left( \sin x-\cos x \right) ^{2}} \\ $

$\\ =\frac{- \left( \sin ^{2}x+\cos ^{2}x-2\sin x\cos x \right) - \left( \sin ^{2}x+\cos ^{2}x+2\sin x\cos x \right) }{ \left( \sin x-\cos x \right) ^{2}}=-\frac{2}{ \left( \sin x-\cos x \right) ^{2}} \\ $

$\\ \left( \frac{dy}{dx} \right) _{x=0}=-\frac{2}{ \left( -1 \right) ^{2}}=-2 \\ \\$
Hence, the answer is option A

Question:72

Choose the correct answer out of 4 options given against each Question

If $y=\frac{\sin \left( x+9 \right) }{\cos x}$ then $\frac{d y}{d x} \text { at } x=0$ is

A. cos 9
B. sin 9
C. 0
D. 1

Answer:

$\\ y=\frac{\sin \left( x+9 \right) }{\cos x} \\ \\ \frac{dy}{dx}=\frac{\cos x\cos \left( x+9 \right) -\sin \left( x+9 \right) \left( -\sin x \right) }{\cos ^{2}x} \\$

$ \\ =\frac{\cos x\cos \left( x+9 \right) +\sin \left( x+9 \right) \sin x}{\cos ^{2}x} \\$

$ \\ =\frac{\cos \left( x+9-x \right) }{\cos ^{2}x}=\frac{\cos 9}{\cos ^{2}x} \\$

$ \\ \left( \frac{dy}{dx} \right) _{x=0}=\frac{\cos 9}{ \left( 1 \right) ^{2}}=\cos 9 \\ \\$


Hence, the answer is option A

Question:73

Choose the correct answer out of 4 options given against each Question

If $f \left( x \right) =1+x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+ \ldots +\frac{x^{100}}{100}$ then f’(1) is equal to

A. 1/100
B. 100
C. does not exist
D. 0

Answer:

$\\ f \left( x \right) =1+x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+ \ldots +\frac{x^{100}}{100} \\$

$ \\ f^{'} \left( x \right) =0+1+\frac{2x}{2}+\frac{3x^{2}}{3}+ \ldots +\frac{100x^{99}}{100} \\$

$ \\ f^{'} \left( x \right) =0+1+x+x^{2}+ \ldots x^{99} \\$

$ \\ f^{'} \left( 1 \right) =1+1+1+ \ldots +1~~ \left( \text{100 times} \right) =100 \\ \\$


Hence, the answer is option B

Question:74

Choose the correct answer out of 4 options given against each Question

If $f \left( x \right) =\frac{x^{n}-a^{n}}{x-a}$ for some constant ‘a’, then f’(a) is

A. 1
B. 0
C. does not exist
D. 1/2

Answer:

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

$\\ f^{'} \left( x \right) =\frac{ \left( x-a \right) \left( nx^{n-1} \right) - \left( x^{n}-a^{n} \right) \left( 1 \right) }{ \left( x-a \right) ^{2}} \\$

$ \\ f^{'} \left( a \right) =\frac{ \left( a-a \right) \left( na^{n-1} \right) - \left( a^{n}-a^{n} \right) \left( 1 \right) }{ \left( a-a \right) ^{2}}=\frac{0}{0} \\ \\$
Hence, the answer is option B

Question:75

Choose the correct answer out of 4 options given against each Question
If $f(x) = x^{100} + x^{99} + $ \ldots $ x + 1,$, then f’(1) is equal to

A. 5050
B. 5049
C. 5051
D. 50051

Answer:

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

$\\ f^{'} \left( x \right) =100x^{99}+99x^{98}+ \ldots +1+0 \\ $

$\\ f^{'} \left( 1 \right) =100+99+98+ \ldots +2+1=\frac{100 *101}{2}=5050 \\ \\$


Hence, the answer is option A

Question:76

Choose the correct answer out of 4 options given against each Question
If $f(x) = 1 - x + x^{2} - x^{3} $ \ldots $ -x^{99} + x^{100}$, then f’(1) is equal to

A. 150
B. –50
C. –150
D. 50

Answer:

$\\ f \left( x \right) =1-x+x^{2}-x^{3}+ \ldots -x^{99}+x^{100} \\ $

$\\ f^{'} \left( x \right) =0-1+2x-3x^{2}+ \ldots -99x^{98}+100x^{99} \\ $

$\\ f^{'} \left( 1 \right) $$=-1+2-3+4- \ldots -99+100 \\ $

$\\ = \left( 2-1 \right) + \left( 4-3 \right) + \left( 6-5 \right) + \ldots \left( 100-99 \right) =1+1+ \ldots +1~~ \left( \text{50 times} \right) =50 \\ \\$
Hence, the answer is option D

Question:77

Fill in the blanks
If $f(x)=\frac{\tan x}{x-\pi}, \lim _{x \rightarrow \pi} f(x)=$

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow \pi }\frac{\tan x}{x- \pi }=\mathop{\lim }\limits_{x \rightarrow \pi }\frac{-\tan \left( \pi -x \right) }{- \left( \pi -x \right) }=\mathop{\lim }\limits_{ \pi -x \rightarrow 0}\frac{\tan \left( \pi -x \right) }{ \left( \pi -x \right) }=1 \\$

Question:78

Fill in the blanks
$\mathop{\lim }\limits_{x \rightarrow 0}\sin mx\cot \frac{x}{\sqrt {3}}=2$ then m = ________

Answer:

$\\ \mathop{\lim }\limits_{x \rightarrow 0}\sin mx\cot \frac{x}{\sqrt {3}}=2 \\$

$ \\ \mathop{\lim }\limits_{x \rightarrow 0}mx * \left( \frac{\sin mx}{mx} \right) * \left( \frac{\frac{x}{\sqrt {3}}}{\tan \frac{x}{\sqrt {3}}} \right) * \left( \frac{\sqrt{3}}{x} \right) =2 \\ $

$\\ \mathop{\lim }\limits_{x \rightarrow 0}mx * \left( \frac{\sqrt {3}}{x} \right) =2 \\$

$ \\ \sqrt {3}m=2 \\ \\ m=\frac{2}{\sqrt {3}}=\frac{2\sqrt {3}}{3} \\ \\$

Question:79

Fill in the blanks
If $y=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+ \ldots$ then $\frac{dy}{dx}=$ ........

Answer:

$\\ y=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+ \ldots \\$

$ \\ \frac{dy}{dx}=0+1+\frac{2x}{2!}+\frac{3x^{2}}{3!}+\frac{4x^{3}}{4!}+ \ldots \\ $

$\\ \frac{dy}{dx}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+ \ldots \\ $

$\\ \frac{dy}{dx}=y \\$

Question:80

Fill in the blanks
$\mathop{\lim }\limits_{x \rightarrow \mathop{3}^{+}}\frac{x}{ \left[ x \right] }=$...............

Answer:

$\mathop{\lim }\limits_{x \rightarrow \mathop{3}^{+}}\frac{x}{ \left[ x \right] }=\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{+}}\frac{3+h}{ \left[ 3+h \right] }=\mathop{\lim }\limits_{h \rightarrow \mathop{0}^{+}}\frac{3+h}{3}=\frac{3}{3}=1 \\$

Important topics in Class 11 Maths NCERT Exemplar Solutions Chapter 13

  • 13.1 Introduction
  • 13.2 Intuitive Idea of Derivatives
  • 13.3 Limits
  • 13.3.1 Algebra of Limits
  • 13.3.2 Limits of polynomials and rational functions
  • 13.4 Limits of Trigonometric function
  • 13.5 Derivatives
  • 13.5.1 Algebra of Derivative of functions
  • 13.5.2 Derivatives of polynomials and trigonometric functions
  • 13.6 Miscellaneous Examples

NCERT Exemplar Class 11 Mathematics Chapters

Importance of solving NCERT Exemplar Class 11 Maths Questions

NCERT Exemplar Class 11 Maths solutions chapter 13 covers the really important topic of limits and derivatives of any function which is a very important concept for mathematics as well as physics.

  • The students will learn about the limits of different functions from NCERT exemplar solutions for Class 11 Maths chapter 13

  • The students will be able to define the limits and derivatives of different trigonometric, polynomial and rational number functions.

  • The NCERT exemplar Class 11 Maths solutions chapter 13 covers various solved examples along with solutions for better understanding and learning of different concepts.

  • The students should practice the application of different formulas provided in the chapter along with solutions and solved examples, take help from Class 11 Maths NCERT exemplar solutions chapter 13.


NCERT solutions of class 11 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 11:

NCERT Notes of class 11 - Subject Wise

Given below are the subject-wise NCERT Notes of class 11 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 11:

NCERT Exemplar Class 11 Solutions

Given below are the subject-wise exemplar solutions of class 11 NCERT:

Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Exemplar Class 11 Maths Chapter 13?

Chapter 13 of Class 11 Maths, titled "Limits and Derivatives," covers several important concepts that are foundational to understanding calculus. Key topics include limits of functions, which help in understanding the behavior of functions as they approach a specific point. The chapter also covers continuity and the conditions under which a function is continuous. Another critical concept is derivatives, where you learn how to find the rate of change of a function at a point, which is essential for understanding slopes of curves and real-world applications like speed and acceleration. Additionally, the chapter introduces you to derivatives from first principles, which is a fundamental method of finding derivatives. The chapter also includes problems on algebra of limitslimits involving infinity, and the chain rule for derivatives, which are useful tools in solving more complex calculus problems.

2. How to solve limit problems in NCERT Exemplar Class 11 Maths Chapter 13?

To solve limit problems in Chapter 13, begin by analyzing the function involved. The first step is to substitute the given value into the function and check if the limit exists. If substituting the value results in an indeterminate form like 0/0?, then you must simplify the expression using algebraic techniques, such as factoring or rationalizing. Another approach is to use standard limit laws and theorems, such as the limit of a sumproduct, or quotient of functions. If the expression is complex, you might need to apply trigonometric limits or limits involving infinity. In some cases, you can use L'Hopital's Rule, but this is more advanced and typically not covered in basic exercises. Remember to practice a variety of problems to understand the different strategies for solving limit problems effectively.

3. What are the basic concepts of derivatives in Class 11 Maths?

In Class 11 Maths, the derivative of a function represents the rate of change of the function with respect to its variable. The basic concept involves understanding the slope of the tangent to the curve of a function at any given point. This is important for describing motion, growth rates, or changes in various contexts. Derivatives are typically found using two methods: algebraic differentiation and first principles. The derivative of a function can also represent how a function behaves as it increases or decreases, helping to identify maxima, minima, and points of inflection on a graph. The rules of differentiation, such as the power ruleproduct rulequotient rule, and the chain rule, are essential for calculating derivatives efficiently and handling more complicated functions.

4. How to find the derivative of a function using first principles?

f'(x) = lim(h → 0) [f(x + h) - f(x)] / h
Here's the process:
1. Start by substituting f(x+h) and f(x) into the formula.
2. Simplify the expression in the numerator, which involves expanding and reducing terms.
3. The next step is to find the limit of the expression as h approaches zero. This step often requires factoring, expanding, or simplifying the terms in the numerator.
4. Finally, after taking the limit, you will obtain the derivative of the function. Using first principles is a more detailed method of finding derivatives and is especially useful for understanding the concept behind derivatives, even though more efficient rules like the power rule are often used in practice.

5. What is L'Hôpital's Rule, and is it included in NCERT Class 11 Maths?

L'Hôpital's Rule is a technique used to evaluate limits that result in indeterminate forms such as 0/0? or ∞/∞. The rule states that for functions f(x) and g(x) with limits of the form 0/0 or ∞/∞?, the limit of f(x)/g(x) as x→a can be found by differentiating the numerator and denominator separately and then evaluating the limit of the resulting quotient:

lim (x → a) [f(x) / g(x)] = lim (x → a) [f'(x) / g'(x)]

This process is repeated if necessary, until a determinate form is obtained. However, L'Hôpital's Rule is generally not included in the NCERT Class 11 Maths curriculum, as the focus is primarily on basic limit calculations and derivative concepts. The rule is more commonly introduced in higher-level calculus, typically in Class 12 or in more advanced studies of calculus.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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