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NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

Edited By Ravindra Pindel | Updated on Sep 12, 2022 01:56 PM IST

NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives is drafted by experts teachers of mathematics. The solutions are prescribed by NCERT and using the easiest or rather the most comprehensible methods. To provide reliable and authentic NCERT exemplar Class 11 Maths solutions chapter 13, we aligned to the guidelines of CBSE for the students. The lesson provides for practical application and usage of Limits and Derivatives in various functions of trigonometry, polynomials, and rational numbers.
Also, check - NCERT Class 11 Maths Solutions.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

Suggested: JEE Main: high scoring chapters | Past 10 year's papers

This Story also Contains
  1. NCERT Class 11 Exemplar Maths Solutions Chapter 13 - Question-Wise Solution
  2. More About NCERT Exemplar Solutions for Class 11 Maths Chapter 13
  3. NCERT Exemplar Class 11 Maths Solutions Chapter-Wise
  4. Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 13

NCERT Class 11 Exemplar Maths Solutions Chapter 13 - Question-Wise Solution

Question:1

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

Answer:

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

Question:2

Evaluate \mathop{\lim }_{x \rightarrow 1/2}\frac{4x^{2}-1}{2x-1}

Answer:

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

Question:3

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

Answer:

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

\\=\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h[\sqrt{x+h}+\sqrt{x}]} \times( \sqrt{x+h}+\sqrt{x}) \ $ [Rationalizing the denominator] $ \\\\=\lim _{h \rightarrow 0} \frac{x+h-x}{h[\sqrt{x+h}+\sqrt{x}]}$ $\\\\=\lim _{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 }_{x \rightarrow 0}\frac{ \left( x+2 \right) ^{\frac{1}{3}}-2^{\frac{1}{3}}}{x}

Answer:

Given~\mathop{\lim }_{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 }_{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 }_{x \rightarrow 1}\frac{ \left( 1+x \right) ^{6}-1}{ \left( 1+x \right) ^{2}-1}

Answer:

\text{~Given that }\mathop{\lim }_{x \rightarrow 1}\frac{ \left( 1+x \right) ^{6}-1}{ \left( 1+x \right) ^{2}-1}=\mathop{\lim }_{x \rightarrow 1}\frac{ \left( \left( 1+x \right) ^{2} \right) ^{3}-1}{ \left( 1+x \right) ^{2}-1}
\\ =\mathop{\lim }_{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 }_{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 }_{x \rightarrow a}\frac{ \left( 2+x \right) ^{\frac{5}{2}}- \left( 2+a \right) ^{\frac{5}{2}}}{x-a}

Answer:

\ \ \ \text{~~~~~~~~~~ Given that }\mathop{\lim }_{x \rightarrow a}\frac{ \left( 2+x \right) ^{\frac{5}{2}}- \left( 2+a \right) ^{\frac{5}{2}}}{x-a}

=\mathop{\lim }_{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[u \sin g \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n \cdot a^{n-1}\right]

Question:7

Evaluate:
\mathop{\lim }_{x \rightarrow 1}\frac{x^{4}-\sqrt {x}}{\sqrt {x}-1}
Answer:

Given~\mathop{\lim }_{x \rightarrow 1}\frac{x^{4}-\sqrt {x}}{\sqrt {x}-1}=\mathop{\lim }_{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 }_{x \rightarrow 1}\frac{x^{4}\sqrt {x}+x^{4}-x-\sqrt {x}}{x-1} \\ \\ =\mathop{\lim }_{x \rightarrow 1}\frac{\sqrt {x} \left( x^{4}-1 \right) +x \left( x^{3}-1 \right) }{x-1} \\ \\ =\mathop{\lim }_{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 }_{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 }_{x \rightarrow 2}\frac{x^{2}-4}{\sqrt {3x-2}-\sqrt {x+2} }

Answer:

\\ Given~\mathop{\lim }_{x \rightarrow 2}\frac{x^{2}-4}{\sqrt {3x-2}-\sqrt {x+2}}=\mathop{\lim }_{x \rightarrow 2}\frac{x^{2}-4}{ \left( \sqrt {3x-2}-\sqrt {x+2} \right) } *\frac{\sqrt {3x-2}+\sqrt {x+2}}{\sqrt {3x-2}+\sqrt {x+2}} \\ \\ =\mathop{\lim }_{x \rightarrow 2}\frac{x^{2}-4}{ \left( 3x-2-x-2 \right) } * \left( \sqrt {3x-2}+\sqrt {x+2} \right) \\ \\ =\mathop{\lim }_{x \rightarrow 2}\frac{ \left( x-2 \right) \left( x+2 \right) }{2 \left( x-2 \right) } * \left( \sqrt {3x-2}+\sqrt {x+2} \right) \\ \\ =\mathop{\lim }_{x \rightarrow 2}\frac{ \left( x+2 \right) }{2} * \left( \sqrt {3x-2}+\sqrt {x+2} \right) =\frac{4}{2} * \left( \sqrt {3 *2-2}+\sqrt {2+2} \right) =8 \\ \\

Question:9

Evaluate:
\mathop{\lim }_{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 }_{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'srule to this question


\\ ~~\mathop{\lim }_{x \rightarrow 1} \left( \frac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2} \right) \\ \\ \mathop{\lim }_{x \rightarrow a} \left( \frac{f \left( x \right) }{g \left( x \right) } \right) =\mathop{\lim }_{x \rightarrow a} \left( \frac{f^{'} \left( x \right) }{g^{'} \left( x \right) } \right) \\ \\ \mathop{\lim }_{x \rightarrow 1} \left( \frac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2} \right) =\mathop{\lim }_{x \rightarrow 1} \left( \frac{7x^{6}-10x^{4}}{3x^{2}-6x} \right) =\frac{7-10}{3-6}=1 \\ \\

Question:11

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

Answer:

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

Question:12

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

Answer:


Given~\mathop{\lim }_{x \rightarrow \frac{1}{2}} \left( \frac{8x-3}{2x-1}-\frac{4x^{2}+1}{4x^{2}-1} \right) \\ \\
\\=\mathop{\lim }_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) * \left( \frac{ \left( 8x-3 \right) \left( 2x+1 \right) -4x^{2}-1}{2x-1} \right) \right) \\ \\ =\mathop{\lim }_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) * \left( \frac{12x^{2}+2x-4}{2x-1} \right) \right) \\ \\


\\ =\mathop{\lim }_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) * \left( \frac{12x^{2}+8x-6x-4}{2x-1} \right) \right) \\ \\ =\mathop{\lim }_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) * \left( \frac{ \left( 3x+2 \right) \left( 4x-2 \right) }{2x-1} \right) ~ \right) \\ \\ =\mathop{\lim }_{x \rightarrow \frac{1}{2}} \left( \left( \frac{1}{2x+1} \right) *2 * \left( 3x+2 \right) ~ \right) =\frac{1}{2 *\frac{1}{2}+1} *2 * \left( 3 *\frac{1}{2}+2 \right) =\frac{7}{2} \\ \\

Question:14

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

Answer:

\\ \text{We know that }\mathop{\lim }_{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 \\ \\ 5 * \left( 2 \right) ^{5-1}=5 *16=80 \\ \\

Question:15

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

Answer:

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

Question:16

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

Answer:

Given~\mathop{\lim }_{x \rightarrow 0} \left( \frac{\sin ^{2}2x}{\sin ^{2}4x} \right) \\ \\ \\ =\mathop{\lim }_{x \rightarrow 0} \left( \frac{ \left( \frac{\sin 2x}{2x} *2x \right) ^{2}}{ \left( \frac{\sin 4x}{4x} *4x \right) ^{2}} \right) =\mathop{\lim }_{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 }_{x \rightarrow 0} \left( \frac{1-\cos 2x}{x^{2}} \right)

Answer:

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

Question:18

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

Answer:

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

Answer:

\\ \text{Given that }\mathop{\lim }_{x \rightarrow 0} \left( \frac{1-\cos mx}{1-\cos nx} \right) =\mathop{\lim }_{x \rightarrow 0} \left( \frac{\sin ^{2}\frac{mx}{2}}{\sin ^{2}\frac{nx}{2}} \right) \\ \\ =\mathop{\lim }_{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 }_{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 }_{x \rightarrow \frac{ \pi }{3}}\frac{\sqrt {1-\cos 6x}}{\sqrt {2} \left( \frac{ \pi }{3}-x \right)}

Answer:

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

Question:21

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

Answer:


\\ \text{Given that }\mathop{\lim }_{x \rightarrow \frac{ \pi }{4}}\frac{\sin x-\cos x}{x-\frac{ \pi }{4}} \\ \\ =\mathop{\lim }_{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 }_{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 }_{x \rightarrow \frac{ \pi }{6}}\frac{\sqrt {3}\sin x-\cos x}{x-\frac{ \pi }{6}} \\

Answer:

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

Question:23

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

Answer:


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


\\ =\mathop{\lim }_{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 }_{x \rightarrow a}\frac{\sin x-\sin a}{\sqrt {x}-\sqrt {a}} \\

Answer:

\\ \text{Given that }\mathop{\lim }_{x \rightarrow a}\frac{\sin x-\sin a}{\sqrt {x}-\sqrt {a}} \\ \\ =\mathop{\lim }_{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 }_{x \rightarrow a}\frac{ \left( \sin x-\sin a \right) \left( \sqrt {x}+\sqrt {a} \right) }{ \left( x-a \right) } \\ \\
\\=\mathop{\lim }_{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 }_{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 }_{x \rightarrow \frac{ \pi }{6}}\frac{\cot ^{2}x-3}{cosec x-2} \\

Answer:

\\ \text{Given that }\mathop{\lim }_{x \rightarrow \frac{ \pi }{6}}\frac{\cot ^{2}x-3}{cosec x-2} \\ \\ =\mathop{\lim }_{x \rightarrow \frac{ \pi }{6}}\frac{cosec^{2}x-1-3}{cosec x-2} \\ \\ =\mathop{\lim }_{x \rightarrow \frac{ \pi }{6}}\frac{cosec^{2}x-1-3}{cosec x-2} \\ \\ =\mathop{\lim }_{x \rightarrow \frac{ \pi }{6}}\frac{ \left( cosecx-2 \right) \left( cosecx+2 \right) }{cosec x-2}

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

Question:26

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

Answer:

\\ \text{Given that }\mathop{\lim }_{x \rightarrow 0}\frac{ \left( \sqrt {2}-\sqrt {1+\cos x} \right) }{\sin ^{2}x} \\ \\ =\mathop{\lim }_{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 }_{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 }_{x \rightarrow 0}\frac{1-\cos x}{\sin ^{2}x \left( \sqrt {2}+\sqrt {1+\cos x} \right) } \\ \\
\\=\mathop{\lim }_{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 }_{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 }_{x \rightarrow 0} \left( \frac{\sin x-2\sin 3x+\sin 5x}{x} \right) \\

Answer:

. \\ \text{Given that }\mathop{\lim }_{x \rightarrow 0} \left( \frac{\sin x-2\sin 3x+\sin 5x}{x} \right) \\ \\ =\mathop{\lim }_{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 }_{x \rightarrow 1} \left( \frac{x^{4}-1}{x-1} \right) = \mathop{\lim }_{x \rightarrow k} \left( \frac{x^{3}-k^{3}}{x^{2}-k^{2}} \right)

Answer:

\\ \text{Given that }\mathop{\lim }_{x \rightarrow 1} \left( \frac{x^{4}-1}{x-1} \right) =4 \left( 1 \right) ^{4-1}=4 \\ \\ \mathop{\lim }_{x \rightarrow k} \left( \frac{x^{3}-k^{3}}{x^{2}-k^{2}} \right) =\mathop{\lim }_{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 }_{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 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 thaty= \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 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\textsuperscript{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 \textsuperscript{3}x \cos \textsuperscript{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 }_{ \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 }_{ \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 }_{ \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 }_{ \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 }_{ \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 }_{ \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 }_{ \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 }_{ \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 }_{ \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 }_{ \Delta x \rightarrow 0}\frac{ \left( x+ \Delta x \right) ^{\frac{2}{3}}-x^{\frac{2}{3}}}{ \Delta x} \\ \\ =\mathop{\lim }_{ \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 }_{ \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 }_{ \Delta x \rightarrow 0}\frac{x^{\frac{2}{3}} \left( \frac{2}{3}\frac{ \Delta x}{x} \right) }{ \Delta x}=\mathop{\lim }_{ \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 }_{ \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 }_{ \Delta x \rightarrow 0} \left( \frac{x\cos \left( x+ \Delta x \right) -x\cos x}{ \Delta x} \right) +\mathop{\lim }_{ \Delta x \rightarrow 0} \left( \frac{ \Delta x\cos \left( x+ \Delta x \right) }{ \Delta x} \right) \\ \\ =\mathop{\lim }_{ \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 }_{ \Delta x \rightarrow 0}\cos \left( x+ \Delta x \right) \\ \\ =\mathop{\lim }_{ \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 }_{ \Delta x \rightarrow 0}\cos \left( x+ \Delta x \right) \\ \\


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

Answer:

\\ \text{Given\: \: that\: \: }\mathop{\lim }_{y \rightarrow 0}\frac{ \left( x+y \right) \sec \left( x+y \right) -x\sec x}{y}=\mathop{\lim }_{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 }_{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 }_{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 }_{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 }_{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 }_{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 }_{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 }_{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 }_{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 }_{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 }_{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 }_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\tan ^{3}x-\tan x}{\cos \left( x+\frac{ \pi }{4} \right) } \right)
Answer:


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

Question:50

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

Answer:

\\ \mathop{\lim }_{x \rightarrow 4}\frac{ \vert x-4 \vert }{x-4} \\ \\ LHL=\mathop{\lim }_{x \rightarrow \mathop{4}^{-}}\frac{-x+4}{x-4}=-1 \\ \\ RHL=\mathop{\lim }_{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 }_{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 }_{h \rightarrow \mathop{0}^{-}} \left( \frac{k\cos \left( \frac{ \pi }{2}-h \right) }{2h} \right) \\ \\ =\mathop{\lim }_{h \rightarrow \mathop{0}^{-}} \left( \frac{k\sin h}{2h} \right) =\frac{k}{2} \\ \\ RHL=\mathop{\lim }_{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 }_{h \rightarrow \mathop{0}^{+}} \left( \frac{k\cos \left( \frac{ \pi }{2}+h \right) }{-2h} \right) \\ \\
\\=\mathop{\lim }_{h \rightarrow \mathop{0}^{-}} \left( \frac{-k\sin h}{-2h} \right) =\frac{k}{2} \\ \\ \mathop{\lim }_{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 }_{x \rightarrow \mathop{-1}^{-}} \left( x+2 \right) =-1+2=1 \\ \\ RHL=\mathop{\lim }_{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 }_{x \rightarrow \pi }\frac{\sin x}{x- \pi } is
A. 1
B. 2
C. –1
D. –2

Answer:

\\ \mathop{\lim }_{x \rightarrow \pi }\frac{\sin x}{x- \pi }=\mathop{\lim }_{x \rightarrow \pi }\frac{\sin \left( \pi -x \right) }{- \left( \pi -x \right) }=-1~ \\ \\ \mathop{\lim }_{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 }_{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 }_{x \rightarrow 0}\frac{x^{2}\cos x}{1-\cos x}=\mathop{\lim }_{x \rightarrow 0}\frac{x^{2}\cos x}{2\sin ^{2}\frac{x}{2}} \\ \\ =\mathop{\lim }_{x \rightarrow 0}\frac{x^{2}\cos x}{\frac{2x^{2}}{4}\frac{\sin ^{2}\frac{x}{2}}{\frac{x^{2}}{4}}}=\mathop{\lim }_{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 }_{x \rightarrow 0}\frac{ \left( 1+x \right) ^{n}-1}{x}~ is
A. n
B. 1
C. –n
D. 0
Answer:

\\ \mathop{\lim }_{x \rightarrow 0}\frac{ \left( 1+x \right) ^{n}-1}{x}=\mathop{\lim }_{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 }_{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 }_{x \rightarrow 1}\frac{x^{m}-1}{x^{n}-1} \\ \\ =\mathop{\lim }_{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 }_{ \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 }_{ \theta \rightarrow 0} \left( \frac{1-\cos 4 \theta }{1-\cos 6 \theta } \right) \\ \\ =\mathop{\lim }_{ \theta \rightarrow 0} \left( \frac{2\sin ^{2}2 \theta }{2\sin ^{2}3 \theta } \right) \\ \\ =\mathop{\lim }_{ \theta \rightarrow 0} \left( \frac{\sin 2 \theta }{\sin 3 \theta } \right) ^{2}=\mathop{\lim }_{ \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 }_{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 }_{x \rightarrow 0} \left( \frac{cosec x-\cot x}{x} \right) \\ \\ =\mathop{\lim }_{x \rightarrow 0} \left( \frac{1-\cos x}{x\sin x} \right) \\ \\ =\mathop{\lim }_{x \rightarrow 0} \left( \frac{2\sin ^{2}x/2}{2x\sin x/2\cos x/2} \right) \\ \\ =\mathop{\lim }_{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 }_{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 }_{x \rightarrow 0} \left( \frac{\sin x}{\sqrt {x+1}-\sqrt {1-x}} \right) \\ \\ =\mathop{\lim }_{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 }_{x \rightarrow 0} \left( \frac{\sin x \left( \sqrt {x+1}+\sqrt {1-x} \right) }{ \left( x+1-1+x \right) } \right) \\ \\ =\mathop{\lim }_{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 }_{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 }_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\sec ^{2}x-2}{\tan x-1} \right)\\ =\mathop{\lim }_{x \rightarrow \frac{ \pi }{4}} \left( \frac{1+\tan ^{2}x-2}{\tan x-1} \right) \\=\mathop{\lim }_{x \rightarrow \frac{ \pi }{4}} \left( \frac{\tan ^{2}x-1}{\tan x-1} \right)\\ =\mathop{\lim }_{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 }_{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 }_{x \rightarrow 1} \left( \frac{ \left( \sqrt {x}-1 \right) \left( 2x-3 \right) }{2x^{2}+x-3} \right) \\ \\ =\mathop{\lim }_{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 }_{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 }_{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 }_{x \rightarrow \mathop{0}^{-}} \left( \frac{\sin \left[ x \right] }{ \left[ x \right] } \right) =\mathop{\lim }_{h \rightarrow \mathop{0}^{-}} \left( \frac{\sin \left[ 0-h \right] }{ \left[ 0-h \right] } \right) =\mathop{\lim }_{h \rightarrow \mathop{0}^{-}} \left( \frac{\sin \left[ -h \right] }{ \left[ -h \right] } \right) \\=\mathop{\lim }_{h \rightarrow \mathop{0}^{-}} \left( \frac{\sin \left( -1 \right) }{-1} \right) =\sin 1 \\ \\
RHL=\mathop{\lim }_{x \rightarrow \mathop{0}^{+}} \left( \frac{\sin \left[ x \right] }{ \left[ x \right] } \right) =\mathop{\lim }_{h \rightarrow \mathop{0}^{+}} \left( \frac{\sin \left[ 0+h \right] }{ \left[ 0+h \right] } \right) =\mathop{\lim }_{h \rightarrow \mathop{0}^{+}} \left( \frac{\sin \left[ h \right] }{ \left[ h \right] } \right) =\mathop{\lim }_{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 }_{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 }_{x \rightarrow \mathop{0}^{-}} \left( \frac{ \vert \sin x \vert }{x} \right) =\mathop{\lim }_{h \rightarrow \mathop{0}^{-}} \left( \frac{ \vert \sin \left( 0-h \right) \vert }{ \left( 0-h \right) } \right) =\mathop{\lim }_{h \rightarrow \mathop{0}^{-}} \left( \frac{-\sin \left( -h \right) }{-h} \right) \\=-1 \\ \\ RHL=\mathop{\lim }_{x \rightarrow \mathop{0}^{+}} \left( \frac{ \vert \sin x \vert }{x} \right) =\mathop{\lim }_{h \rightarrow \mathop{0}^{+}} \left( \frac{ \vert \sin \left( 0+h \right) \vert }{ \left( 0+h \right) } \right) =\mathop{\lim }_{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\textsuperscript{2} - 6x + 9 = 0\\ B. x\textsuperscript{2} -7x + 8 = 0\\ C. x\textsuperscript{2} - 14x + 49 = 0\\ D. x\textsuperscript{2} - 10x + 21 = 0 \\

Answer:

\\ LHL=\mathop{\lim }_{x \rightarrow \mathop{2}^{-}}f \left( x \right) =\mathop{\lim }_{x \rightarrow \mathop{2}^{-}} \left( x^{2}-1 \right) =3 \\ \\ RHL=\mathop{\lim }_{x \rightarrow \mathop{2}^{+}}f \left( x \right) =\mathop{\lim }_{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 }_{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 }_{x \rightarrow 0}\frac{ \left( \tan 2x-x \right) }{3x-\sin x}=\mathop{\lim }_{x \rightarrow 0}\frac{ \left( \frac{\tan 2x}{x}-1 \right) }{ \left( 3-\frac{\sin x}{x} \right) }=\mathop{\lim }_{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 }_{h \rightarrow 0}\frac{ \left( f \left( \frac{1}{2}-h \right) -f \left( \frac{1}{2} \right) \right) }{-h} \\ \\ =\mathop{\lim }_{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 }_{h \rightarrow 0}\frac{-h}{-h}=1 \\ \\ RHD=\mathop{\lim }_{h \rightarrow 0}\frac{ \left( f \left( \frac{1}{2}+h \right) -f \left( \frac{1}{2} \right) \right) }{h} \\ \\ =\mathop{\lim }_{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 }_{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\textsuperscript{100} + x\textsuperscript{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\textsuperscript{2} - x\textsuperscript{3} $ \ldots $ -x\textsuperscript{99} + x\textsuperscript{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 }_{x \rightarrow \pi }\frac{\tan x}{x- \pi }=\mathop{\lim }_{x \rightarrow \pi }\frac{-\tan \left( \pi -x \right) }{- \left( \pi -x \right) }=\mathop{\lim }_{ \pi -x \rightarrow 0}\frac{\tan \left( \pi -x \right) }{ \left( \pi -x \right) }=1 \\

Question:78

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

Answer:

\\ \mathop{\lim }_{x \rightarrow 0}\sin mx\cot \frac{x}{\sqrt {3}}=2 \\ \\ \mathop{\lim }_{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 }_{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 }_{x \rightarrow \mathop{3}^{+}}\frac{x}{ \left[ x \right] }=...............

Answer:

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

More About NCERT Exemplar Solutions for Class 11 Maths Chapter 13

NCERT Exemplar Class 11 Maths chapter 13 solutions is a great way for students to learn and understand the concept of limits and derivatives through easier methods prescribed by the experts and develop their base for the same.

NCERT Exemplar Class 11 Maths solutions chapter 13 PDF download is useful for students to read offline. Use an online webpage to PDF tool for this. These solutions make learning more convenient and includes the detailed study of methods and guidelines of CBSE and NCERT.

The NCERT Exemplar solutions for Class 11 Maths chapter 13 will provide access to efficient and carefully drafted solutions to students to aid preparation and learning process for a better outcome.

Topics and Subtopics 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

What will the Students Learn in NCERT Exemplar Class 11 Maths Chapter 13 solutions?

  • 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 calculating the limits of any function which is important for calculus and mathematical analysis that are further used to define integrals, derivatives, and continuity of different functions.

NCERT Exemplar Class 11 Maths Solutions Chapter-Wise

Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 13

Some of the important topics for students to review are as follows:

  • 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.

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NCERT Exemplar Class 11 Solutions

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Frequently Asked Questions (FAQs)

1. Where can we download the solutions to this chapter?

NCERT exemplar Class 11 Maths solutions chapter 13 pdf download can be accessed by using the webpage to PDF tool.

2. Is the chapter important from the perspective of Board and competitive exams?

Yes, the chapter has prominent weightage in both Board and competitive exams and helps to understand the basics terms and methods important from an examination perspective.

3. Which are the important topics that students must cover in this chapter?

The students must cover limits for trigonometric functions and also for polynomials. Derivatives also make a base for future learning in this NCERT exemplar Class 11 Maths chapter 13 solutions.

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Popular Questions

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
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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

 
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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
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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
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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 .
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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
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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.
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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.
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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
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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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