અહી f: R rightarrow R એ મુજબ વ્યાખ્યાયિત છે&n | vedclass

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Question

(d)

Given function

$f(x)=\left[\begin{array}{cc} \frac{\sin \left(x^2ight)}{x} & , x 
eq 0 \ 0 & , 	ext { if } x=0\end{array}ight

Vedclass · Accepted Answer

વિકલનીય છે અને વિકલન સતત છે.

Vedclass · Answer

(d)

Given function

$f(x)=\left[\begin{array}{cc} \frac{\sin \left(x^2ight)}{x} & , x 
eq 0 \ 0 & , 	ext { if } x=0\end{array}ight.$

then $\lim _{x ightarrow 0} f(x)=\lim _{x ightarrow 0} \frac{\sin x^2}{x}=\lim _{x ightarrow 0}$

$x \frac{\sin x^2}{x^2}=0=f(0)$

Hence, $f(x)$ is continuous at $x=0$

Now, for differentiability

$RHD ($ at $x=0)$

$=\lim _{h ightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h ightarrow 0} \frac{\sin h^2}{h^2}=1$

and $LHD$ (at $x=0)$

$=\lim _{h ightarrow 0} \frac{f(0-h)-f(0)-\lim _{h ightarrow 0} \sin h^2}{-h}=1$

So, $f(x)$ is differentiable at $x=0$

$	herefore \quad f^{\prime}(x)=\left[\begin{array}{cc}2 \cos \left(x^2ight)-\frac{\sin x^2}{x^2} & , 	ext { if } x 
eq 0 \ 1 & , 	ext { if } x=0\end{array}ight]$ $\because \lim _{x ightarrow 0} f^{\prime}(x)=\lim _{x ightarrow 0}\left[2 \cos \left(x^2ight)-\frac{\sin x^2}{x^2}ight]$ $=2-1=1$ $	herefore \lim _{x ightarrow 0} f^{\prime}(x)=f^{\prime}(0)$ So,f $f(x)$ is differentiable and the

અહી $f: R \rightarrow R$ એ મુજબ વ્યાખ્યાયિત છે $f(x)=\left\{\begin{array}{l}\frac{\sin \left(x^2\right)}{x} \text { if } x \neq 0 \\ 0 \text { if } x=0\end{array}\right\}$ હોય તો $x=0$ આગળ $f$ એ . . .

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