Chapter 13: Limits and Derivatives

Ans:

Let,

$f\left( x \right) = \dfrac{{\sin \left[ x \right]}}{{\left[ x \right]}},{\text{  }}\left[ x \right] \ne 0$

$0,{\text{           }}\left[ x \right] = 0$ 

${\text{LHL}} = \mathop {\lim }\limits_{x \to {0^ - }} f(x)$

${\text{       }} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin [x]}}{{[x]}}$

${\text{       }} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin [0 - h]}}{{[0 - h]}}$

${\text{       }} = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \sin [ - h]}}{{[ - h]}} =  - 1$

${\text{RHL}} = \mathop {\lim }\limits_{x \to {0^ + }} f(x)$

${\text{        }} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin [x]}}{{[x]}}$

${\text{        }} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin [0 + h]}}{{[0 + h]}}$

${\text{        }} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin [h]}}{{[h]}} = 1$

\[{\text{LHL}} \ne {\text{RHL}}\]

Therefore, the limit does not exist.

Correct Option:  D

Ans: 

\[\therefore {\text{ }}LHL = \mathop {\lim }\limits_{x \to {0^ - }} \left( {\dfrac{{ - \sin x}}{x}} \right)\]

\[{\text{           }} =  - \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin x}}{x}\]

\[{\text{           }} =  - 1\]

\[{\text{   }}RHL = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin x}}{x}\]

\[{\text{           }} = 1\]

\[\because LHL \ne RHL\]

Thus, the limit does not exist.

Correct Option :  C

Ans: $\therefore \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 1} \right)$

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

${\text{                    }} = \mathop {\lim }\limits_{h \to 0} \left( {4 + {h^2} - 4h - 1} \right)$

${\text{                    }} = \mathop {\lim }\limits_{h \to 0} \left( {{h^2} - 4h + 3} \right)$

${\text{                    }} = 3$

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

${\text{                    }} = \mathop {\lim }\limits_{h \to 0} [2(2 + h) + 3]$

${\text{                    }} = \mathop {\lim }\limits_{h \to 0} (4 + 2h + 3)$

${\text{                    }} = 7$

Thus, quadratic equation whose roots are $3{\text{ and }}7$

${x^2} - \left( {3 + 7} \right)x + 3 \times 7 = 0$

$i.e,{\text{ }}{x^2} - 10x + 21 = 0$

Correct Option :  D

Ans: Correct Option :  B

Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 2x - x}}{{3x - \sin x}}$

We know that,

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1$

$x \to 0 \Rightarrow kx \to 0$

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 2x - x}}{{3x - \sin x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{x\left[ {\dfrac{{\tan 2x}}{x} - 1} \right]}}{{x\left[ {3 - \dfrac{{\sin x}}{x}} \right]}}$

$= \dfrac{{\mathop {\lim }\limits_{x \to 0} 2 \times \dfrac{{\tan 2x}}{{2x}} - 1}}{{3 - \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}}}$ 

${\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1}$

$\left[ {x \to 0 \Rightarrow kx \to 0} \right]$

$\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1} \right]$ 

$= \dfrac{{2 - 1}}{{3 - 1}}$

$= \dfrac{1}{2}$

Ans: $f\left( x \right) = x - \left[ x \right]$ (given)

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

${\text{         }} = 1$ 

\[\left[ {\because \dfrac{d}{{dx}}x = 1{\text{ and }}\dfrac{d}{{dx}}\left[ x \right] = 0} \right]\]

Therefore, \[f'\left( {\dfrac{1}{2}} \right) = 1\]

Correct Option:  B

Ans: Given:

\[y = \sqrt x  + \dfrac{1}{{\sqrt x }}\]

We know that,

We will differentiate the given equation using chain rule

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

Then,

$y = \sqrt x  + \dfrac{1}{{\sqrt x }}$

${\text{now, }}\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt x }} - \dfrac{1}{{2{x^{3/2}}}}$

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

${\text{                 }} = 0$

Correct Option:  D

Ans: Given:

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

We know that, Differentiate  \[f\left( x \right)\]

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

At last putting \[1\] in \[f'\left( x \right)\] to get the final answer

Then,$f\left( x \right) = \dfrac{{x - 4}}{{2\sqrt x }}$

$f'\left( x \right) = \dfrac{{2\sqrt x  - (x - 4) \cdot 2 \cdot \dfrac{1}{{2\sqrt x }}}}{{4x}}$       (by quotient rule)

${\text{         }} = \dfrac{{2x - (x - 4)}}{{4{x^{3/2}}}}$

${\text{          }} = \dfrac{{2x - x + 4}}{{4{x^{3/2}}}}$

${\text{          }} = \dfrac{{x + 4}}{{4{x^{3/2}}}}$

$f'\left( 1 \right) = \dfrac{{1 + 4}}{{4 \times {{(1)}^{3/2}}}}$

${\text{           = }}\dfrac{5}{4}$

Correct Option :  A

Ans: Given:

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

We know that, We will differentiate the given equation using quotient rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

Then,

\[y = \dfrac{{1 + \dfrac{1}{{{x^2}}}}}{{1 - \dfrac{1}{{{x^2}}}}}\]

\[{\text{  }} = \dfrac{{{x^2} + 1}}{{{x^2} - 1}}\]

Differentiating \[y\] w.r.t  \[x\]

\[\dfrac{{dy}}{{dx}} = \dfrac{{\left( {{x^2} - 1} \right)2x - \left( {{x^2} + 1} \right)(2x)}}{{{{\left( {{x^2} - 1} \right)}^2}}}{\text{ }}\left[ {{\text{By Quotient rule}}} \right]\]

\[ = \dfrac{{2x\left( {{x^2} - 1 - {x^2} - 1} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}\]

\[ = \dfrac{{2x( - 2)}}{{{{\left( {{x^2} - 1} \right)}^2}}}\]

\[ = \dfrac{{ - 4x}}{{{{\left( {{x^2} - 1} \right)}^2}}}\]

Correct Option:  A

Ans: Given, 

\[y = \dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}\]

Now, Differentiating \[y\] w.r.t  \[x\]

\[\dfrac{{dy}}{{dx}} = \dfrac{{\sin x + \cos x}}{{\sin x - \cos x}}\]

\[\dfrac{{dy}}{{dx}} = \dfrac{{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}}{{{{(\sin x - \cos x)}^2}}}{\text{ }}\left[ {{\text{by quotient rule}}} \right]\]$= \dfrac{{ - {{(\sin x - \cos x)}^2} - {{(\sin x + \cos x)}^2}}}{{{{(\sin x - \cos x)}^2}}}$

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

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

$= \dfrac{{ - 2}}{{{{(\sin x - \cos x)}^2}}}$

\[\therefore {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} =  - 2\]

Correct Option:  A

Ans: Given, $y = \dfrac{{\sin (x + 9)}}{{\cos x}}$

 Differentiating \[y\] w.r.t  \[x\]

$\dfrac{{dy}}{{dx}} = \dfrac{{\cos x\cos (x + 9) - \sin (x + 9)( - \sin x)}}{{{{(\cos x)}^2}}}\left[ {{\text{by quotient rule}}} \right]$

$= \dfrac{{\cos x\cos (x + 9) + \sin x\sin (x + 9)}}{{{{\cos }^2}x}}\left[ {\because \cos x\cos y + \sin x\sin y = \cos \left( {x - y} \right)} \right]$

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

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

$= \cos 9$

Correct Option:  A