Given function
$f(x)=\left[\begin{array}{cc} \frac{\sin \left(x^2\right)}{x} & , x \neq 0 \\ 0 & , \text { if } x=0\end{array}\right.$
then $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin x^2}{x}=\lim _{x \rightarrow 0}$
$x \frac{\sin x^2}{x^2}=0=f(0)$
Hence, $f(x)$ is continuous at $x=0$
Now, for differentiability
$RHD ($ at $x=0)$
$=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{\sin h^2}{h^2}=1$
and $LHD$ (at $x=0)$
$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)-\lim _{h \rightarrow 0} \sin h^2}{-h}=1$
So, $f(x)$ is differentiable at $x=0$
$\therefore \quad f^{\prime}(x)=\left[\begin{array}{cc}2 \cos \left(x^2\right)-\frac{\sin x^2}{x^2} & , \text { if } x \neq 0 \\ 1 & , \text { if } x=0\end{array}\right]$ $\because \lim _{x \rightarrow 0} f^{\prime}(x)=\lim _{x \rightarrow 0}\left[2 \cos \left(x^2\right)-\frac{\sin x^2}{x^2}\right]$ $=2-1=1$ $\therefore \lim _{x \rightarrow 0} f^{\prime}(x)=f^{\prime}(0)$ So,f $f(x)$ is differentiable and the
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ The curve $y=f(x)$ passes through the point $(1,2)$
$(B)$ The curve $y=f(x)$ passes through the point $(2,-1)$
$(C)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-2}{4}$
$(D)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-1}{4}$
| $X$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
| $P(X)$ | $a$ | $6 a$ | $6 a$ | $4 a$ | $8 a$ | $8 a$ | $6 a$ | $9 a$ |