If f(x)=\ array cl x x (1+x^ 2 ) & , x ≠ 0 0 & , x=0 array . then f(x) is

Correct answer: A. continuous as well as differentiable at x=0 We have, aligned Lf^ (0) &= _ h arrow 0 f(0-h)-f(0) -h = _ h arrow 0 -h -h (1+h^ 2 ) =& _ h arrow 0 (1+h^ 2 ) ((0)/(0) form ) aligned aligned &= _ h arrow 0 - h 2 h /(1+h^ 2 ) =-1 / 2 Rf^ (0) &= _ h arrow 0 f(0+h)-f(0) h = _ h arrow 0 h h (1+h^ 2 ) &= _ h arrow 0 h (1+h^ 2 ) ((0)/(0) form ) &= _ h arrow 0 - h 2 h /(1+h^ 2 ) =(-1)/(2) aligned Since Lf^ (0)=Rf^ (0), therefore f(x) is dif- ferentiable at x=0 Since differentiability continuity, there-fore f(x) is continuous at x=0 .

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Mathematics
Chapter: Continuity and Differentiability

BITSAT 2016 · Mathematics · Continuity and Differentiability

If f(x)=\ array cl x x (1+x^ 2 ) & , x ≠ 0 0 & , x=0 array . then...

f(x)=\left\{

\begin{array}{cl}\frac{x \log \cos x}{\log \left(1+x^{2}\right)} & , x \neq 0 \\ 0 & , x=0\end{array}

\right.

then

f(x)

Options

Acontinuous as well as differentiable at $\mathrm{x}=0$
Bcontinuous but not differentiable at $x=0$
Cdifferentiable but not continuous at $\mathrm{x}=0$
Dneither continuous nor differentiable at $\mathrm{x}=0$

Correct answer

A. continuous as well as differentiable at $\mathrm{x}=0$

Step-by-step solution

We have,

\begin{aligned}\mathrm{Lf}^{\prime}(0) &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h} \log \cosh }{-\mathrm{h} \log \left(1+\mathrm{h}^{2}\right)}\\ =& \lim _{\mathrm{h} \rightarrow 0} \frac{\log \cosh }{\log \left(1+\mathrm{h}^{2}\right)}\left(\frac{0}{0} \text{ form }\right) \end{aligned}

\begin{aligned} &=\lim _{\mathrm{h} \rightarrow 0} \frac{-\tan \mathrm{h}}{2 \mathrm{ h} /\left(1+\mathrm{h}^{2}\right)}=-1 / 2 \\ \mathrm{Rf}^{\prime}(0) &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \log \cosh }{\mathrm{h} \log \left(1+\mathrm{h}^{2}\right)} \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{\log \cos \mathrm{h}}{\log \left(1+\mathrm{h}^{2}\right)}\left(\frac{0}{0} \text{ form }\right) \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{-\tan \mathrm{h}}{2 \mathrm{ h} /\left(1+\mathrm{h}^{2}\right)}=\frac{-1}{2} \end{aligned}

Since

\mathrm{Lf}^{\prime}(0)=\operatorname{Rf}^{\prime}(0),

therefore

\mathrm{f}(\mathrm{x})

is dif- ferentiable at

x=0

Since differentiability \Rightarrow continuity, there-fore

\mathrm{f}(\mathrm{x})

is continuous at

\mathrm{x}=0$.

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