JEE Main 201411 AprilOnline

Exam: JEE Main
Year: 2014
Date: 11 April
Mode: Online
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2014 · 11 April · Online · Mathematics · Continuity and Differentiability

Let f(x)=x|x|, g(x)= x and h(x)=(g f)(x) . Then

Let

f(x)=x|x|, g(x)=\sin x

and

h(x)=(g \circ f)(x)

. Then

Options

A $\mathrm{h}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$ .
B $\mathrm{h}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$ , but $\mathrm{h}^{\prime}(\mathrm{x})$ is not continuous at $x=0$
C $\mathrm{h}^{\prime}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ but it is not differentiable at $x=0$
D $\mathrm{h}^{\prime}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$

Correct answer

C. $\mathrm{h}^{\prime}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ but it is not differentiable at $x=0$

Step-by-step solution

Let

f(x)=x|x|=x|x|, g(x)=\sin x

and

h(x)=g o f(x)=g[f(x)]

\begin{aligned} &\therefore h(x)= \begin{cases}\sin x^2, & x \geq 0 \\ -\sin x^2, & x < 0\end{cases} \\ &\text{ Now, } h^{\prime}(x)=\left\{\begin{array}{cc} 2 x \cos x^2, & x \geq 0 \\ -2 x \cos x^2, & x < 0 \end{array}\right. \end{aligned}

Since, L.H.L and

R.H.L atx=0 of h^{\prime}(x)

is equal to 0 therefore

h^{\prime}(x)

is continuous at

x=0

Now, suppose

h^{\prime}(x)

is differentiable

\begin{aligned} &\therefore h^{\prime \prime}(x) \\ &= \begin{cases}2\left(\cos x^2+2 x^2\left(-\sin x^2\right),\right. & x \geq 0 \\ 2\left(-\cos x^2+2 x^2 \sin x^2\right), & x < 0\end{cases} \end{aligned}

Since, L.H.L

and R.H.L atx=0 of h^{\prime \prime}(x)

are different therefore

h^{\prime \prime}(x)

is not continuous.

\Rightarrow h^{\prime \prime}(x)

is not differentiable

\Rightarrow

our assumption is wrong Hence

h^{\prime}(x)

is not differentiable at

x=0

.

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