Let f(x) = x|x|,g(x) = sin x and h(x) = (g ci | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = x|x|$ and $g(x) = \sin x$.$h(x) = g(f(x)) = \sin(x|x|)$.Since $x|x| = x^2$ for $x \ge 0$ and $-x^2$ for $x < 0$,we have $h(x) = \beg

Vedclass · Accepted Answer

$h'(x)$ is continuous at $x = 0$ but it is not differentiable at $x = 0$.

Vedclass · Answer

(C) Given $f(x) = x|x|$ and $g(x) = \sin x$.$h(x) = g(f(x)) = \sin(x|x|)$.Since $x|x| = x^2$ for $x \ge 0$ and $-x^2$ for $x Now,$h'(x) = \begin{cases} 2x \cos(x^2) & x \ge 0 \ -2x \cos(x^2) & x At $x = 0$,$LHL = \lim_{x 	o 0^-} (-2x \cos(x^2)) = 0$ and $RHL = \lim_{x 	o 0^+} (2x \cos(x^2)) = 0$. Since $h'(0) = 0$,$h'(x)$ is continuous at $x = 0$.Now,check differentiability of $h'(x)$ at $x = 0$ by finding $h''(x)$:$h''(x) = \begin{cases} 2 \cos(x^2) - 4x^2 \sin(x^2) & x > 0 \ -2 \cos(x^2) + 4x^2 \sin(x^2) & x $LHD = \lim_{x 	o 0^-} (-2 \cos(x^2) + 4x^2 \sin(x^2)) = -2$.$RHD = \lim_{x 	o 0^+} (2 \cos(x^2) - 4x^2 \sin(x^2)) = 2$.Since $LHD 
eq RHD$,$h'(x)$ is not differentiable at $x = 0$.

List-$I$	List-$II$
$a$. If $y=\|x\|+\|x-2\|$,then at $x=2, \frac{dy}{dx}=$	$i$. $2$
$b$. If $f(x)=\|\cos 2x\|$,then $f^{\prime}(\frac{\pi}{4}+)=$	$ii$. $0$
$c$. If $f(x)=\sin(\pi[x])$,where $[\cdot]$ denotes the greatest integer function,then $f^{\prime}(1-)=$	$iii$. $-2$
$d$. If $f(x)=\log\|x-1\|, x \neq 1$,then $f^{\prime}(\frac{1}{2})=$	$iv$. Does not exist

Let $f(x) = x|x|$,$g(x) = \sin x$ and $h(x) = (g \circ f)(x)$. Then

Similar Questions

The function $f(x) = \begin{cases} |x - 3| & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} & x < 1 \end{cases}$ is :

If $f(x) = \begin{cases} \int_{0}^{x} (5 + |1-t|) \, dt, & x > 2 \\ 5x + 1, & x \leq 2 \end{cases}$,then:

The number of points,where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$,$x \in R$ cuts the $x$-axis,is equal to

The first derivative of the function $f(x) = \cos^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right) + x^x$ with respect to $x$ at $x=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module