f:[-1,1] R be a function defined by f(x)= cl x^2 ( x ) & for x 0 … - Vidyadip

MCQ

$f:[-1,1] \rightarrow R$ be a function defined by $f(x)=\left\{\begin{array}{cl}x^2 \mid \cos \left(\frac{\pi}{x}\right) & \text { for } x \neq 0 \\ 0 & \text { for } x=0\end{array}\right.$

The set of points where $f$ is not differentiable is

A
$\{x \in[-1,1]: x \neq 0\}$
B
$\left\{x \in[-1,1]: x=0\right.$ or $\left.x=\frac{2}{2 n+1}, n \in Z\right\}$
✓
$\left\{x \in[-1,1]: x=\frac{2}{2 n+1}, n \in Z\right\}$
D
$[-1,1]$

✓

Answer

Correct option: C.

$\left\{x \in[-1,1]: x=\frac{2}{2 n+1}, n \in Z\right\}$

c
(c)

We have,

$f(x)=\left\{\begin{array}{cc} x^2 \mid \cos \left(\frac{\pi}{x}\right) & \text { for } x \neq 0 \\ 0 & \text { for } x=0 \end{array}\right.$

$f(x)$ is not differentiable at

$\cos \frac{\pi}{x} \mid =0$

$\cos \frac{\pi}{x} =0 \Rightarrow \frac{\pi}{x}=(2 n+1) \frac{\pi}{2}$

$\cos \frac{\pi}{x}=0 \Rightarrow \frac{\pi}{x}=(2 n+1) \frac{\pi}{2}$

$x=\frac{2}{2 n+1}$

$\because f(x)$ is not differentiable at

$x \in[-1,1]: x=\frac{2}{2 n+1}, n \in Z$

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