Limit Continuity And Differentiability Ques 53
- Let $f(x)$ be a function satisfying the condition $f(-x)=f(x), \forall x$. If $f^{\prime}(0)$ exists, find its value.
(1987, 2M)
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Answer:
Correct Answer: 53.$(f^{\prime}(0)=0)$
Solution: Since, $f(-x)=f(x)$
$\therefore \quad f(x)$ is an even function.
$ \begin{aligned} \therefore \quad f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \quad \quad[\because f(-h)=f(h)] \end{aligned} $
Since, $f^{\prime}(0)$ exists.
$ \therefore \quad R f^{\prime}(0)=L f^{\prime}(0) $
$ \begin{aligned} & \therefore \quad \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{-h} \\ & \Rightarrow \quad 2 \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=0 \\ & \Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=0 \\ & \therefore \quad f^{\prime}(0)=L f^{\prime}(0) \\ & \Rightarrow \quad f^{\prime}(0)=0 \end{aligned} $
BETA
