Chapter 13: Limits and Derivatives

Ans: Given:

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

We know that,Differentiating the \[f\left( x \right)\] w.r.t  \[x\]

Then, putting \[x = 1\] in the solved equation

Then,

\[{\text{          }}f\left( x \right) = 1 + x + \dfrac{{{x^2}}}{2} + ... + \dfrac{{{x^{100}}}}{{100}}\]

 Differentiating the equation w.r.t  \[x\]

\[{\text{         }}f\left( x \right)' = 0 + 1 + 2 \times \dfrac{x}{2} + ... + \dfrac{{{x^{100}}}}{{100}}\]

\[\therefore {\text{      }}f\left( x \right)' = 1 + x + {x^2} + ... + {x^{99}}\]

Now, putting \[x = 1\]in the solved equation

\[{\text{Now, }}f\left( x \right)' = 1 + 1 + 1 + ... + 1\left( {100{\text{ times}}} \right)\]

\[{\text{                  }} = 100\]

Correct Option :  B

Ans: Given:

\[{\text{ }}f\left( x \right) = \dfrac{{{x^n} - {a^n}}}{{x - a}}\]

We know that, Differentiating the given equation by quotient rule

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

Putting  \[a\] in the solved equation to get the final answer

Then,

\[{\text{ }}f\left( x \right) = \dfrac{{{x^n} - {a^n}}}{{x - a}}\]

Differentiating the given equation w.r.t  \[x\]

$\therefore {\text{ }}f'(x) = \dfrac{{(x - a)n{x^{n - 1}} - \left( {{x^n} - {a^n}} \right)(1)}}{{{{(x - a)}^2}}}{\text{ }}\left[ {{\text{by quotient rule}}} \right]$

$\Rightarrow f'(x) = \dfrac{{n{x^{n - 1}}(x - a) - {x^n} + {a^n}}}{{{{(x - a)}^2}}}$

${\text{Now, }}f'(a) = \dfrac{{n{a^{n - 1}}(0) - {a^n} + {a^n}}}{{{{(x - a)}^2}}}$

$\Rightarrow {\text{      }}{f^\prime }(a) = \dfrac{0}{0}$

So, \[f'\left( a \right)\] does not exist,

Since, \[f\left( x \right)\] is not defined at \[x = a\].

Hence, \[f'\left( x \right)\] at \[x = a\] does not exist.

Correct Option :  D

Ans: Given:

\[{\text{ }}f\left( x \right) = {x^{100}} + {x^{99}} + ... + x + 1\]

We know that,Differentiating $f\left( x \right)$ w.r.t $x$

Then using this formula to find sum

${S_n} = \dfrac{n}{2}\{ 2a + (n - 1)d\}$

Then, $f(x) = {x^{100}} + {x^{99}} +  \ldots  + x + 1$

Differentiating w.r.t $x$

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

$= 100{x^{99}} + 99{x^{98}} +  \ldots  + 1$

${\text{Now, }}f'(1) = 100 + 99 +  \ldots  + 1$

${\text{                 }} = \dfrac{{100}}{2}[2 \times 100 + (100 - 1)( - 1)]\,\left[ {\because {S_n} = \dfrac{n}{2}\{ 2a + (n - 1)d\} } \right]$

${\text{                 }} = 50[200 - 99]\quad$

${\text{                 }} = 50 \times 101$

${\text{                 }} = 5050$

Correct Option:  A

Ans: Given:

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

We know that, Differentiating $f\left( x \right)$ w.r.t $x$

Then using this formula to find sum

${S_n} = \dfrac{n}{2}\{ 2a + (n - 1)d\}$

Then,

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

Differentiating above equation w.r.t $x$

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

$=  - 1 + 2x - 3{x^2} +  \ldots  - 99{x^{98}} + 100{x^{99}}{f^\prime }(1)$

$=  - 1 + 2 - 3 +  \ldots  - 99 + 100$

$\therefore f'\left( x \right) = ( - 1 - 3 - 5 -  \ldots  - 99) + (2 + 4 +  \ldots  + 100)$

$=  - \dfrac{{50}}{2}[2 \times 1 + (50 - 1)2] + \dfrac{{50}}{2}[2 \times 2 + (50 - 1)2]\,{\text{ }}\left[ {{S_n} = \dfrac{n}{2}\{ 2a + (n - 1)d\} } \right]$

$=  - 25[2 + 49 \times 2] + 25[4 + 49 \times 2]$

$=  - 25(2 + 98) + 25(4 + 98)$

$=  - 25 \times 100 + 25 \times 102$

$=  - 2500 + 2550$

$= 50$

Correct Option :  A

Fill in the blanks in Exercises 77 to 80: 

Ans: Given:

\[f\left( x \right) = \dfrac{{\tan x}}{{x - \pi }}\]

We know that,

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

Then,

\[f\left( x \right) = \dfrac{{\tan x}}{{x - \pi }}\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi } f\left( x \right) = \mathop {\lim }\limits_{x \to \pi } \dfrac{{\tan x}}{{x - \pi }}\]

\[{\text{                  }} = \mathop {\lim }\limits_{\pi  - x \to 0} \dfrac{{ - \tan \left( {\pi  - x} \right)}}{{x - \pi }}\]$\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1{\text{ }}} \right]$

\[{\text{                  }} = 1\]

Ans: Given:

\[\mathop {\lim }\limits_{x \to 0} \left( {\sin mx{\text{cot}}\dfrac{x}{{\sqrt 3 }}} \right) = 2\]

We know that,

\[\because {\text{cot}}\dfrac{x}{{\sqrt 3 }} = \dfrac{1}{{\tan \dfrac{x}{{\sqrt 3 }}}}\]

Then,

\[\mathop {\lim }\limits_{x \to 0} \left( {\sin mx{\text{cot}}\dfrac{x}{{\sqrt 3 }}} \right) = 2\] (given)

Multiplying $\dfrac{{mx}}{{mx}}$ in L.H.S  $\left[ {\because \dfrac{{mx}}{{mx}} = 1} \right]$

$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin mx}}{{mx}} \cdot mx \cdot \dfrac{1}{{\tan \dfrac{x}{{\sqrt 3 }}}} = 2$\[\left[ {\because {\text{cot}}\dfrac{x}{{\sqrt 3 }} = \dfrac{1}{{\tan \dfrac{x}{{\sqrt 3 }}}}} \right]\]

Multiplying and dividing \[\dfrac{x}{{\sqrt 3 }}\]in numerator and denominator in L.H.S

$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin mx}}{{mx}} \cdot mx \cdot \dfrac{{\dfrac{x}{{\sqrt 3 }}}}{{\tan \dfrac{x}{{\sqrt 3 }}}} \cdot \dfrac{1}{{\dfrac{x}{{\sqrt 3 }}}} = 2$

$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin mx}}{{mx}} \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{x}{{\sqrt 3 }}}}{{\tan \dfrac{x}{{\sqrt 3 }}}} \cdot \mathop {\lim }\limits_{x \to 0} \dfrac{{mx}}{{\dfrac{x}{{\sqrt 3 }}}} = 2$

$\Rightarrow \sqrt 3 x = 2$

$\therefore m = \dfrac{{2\sqrt 3 }}{3}$

Ans: Given:

$y = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ...$

Then,

$y = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} +  \ldots$ (given)

$\therefore \dfrac{{dy}}{{dx}} = 0 + 1 + \dfrac{{2x}}{2} + \dfrac{{3{x^2}}}{6} + \dfrac{{4{x^3}}}{{4!}}$

${\text{       }} = 1 + x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{6} +  \ldots$

${\text{       }} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} +  \ldots$

${\text{        }} = y$

Ans: $\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{x}{{\left[ x \right]}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {3 + h} \right)}}{{\left[ {3 + h} \right]}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {3 + h} \right)}}{{\left[ 3 \right]}}$

$= 1$ 

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