Answer :- (C)
Definition of Derivative of function 'f' at x0 says :-
$f'(x_{0}) = \lim_{x\rightarrow x_{0}} \frac{f(x) - f(x_{0})}{x-x_{0}}$
So,
1) Left Hand Derivative(LHD) = $f'(x_{0}^{-}) = \lim_{x\rightarrow x_{0}^{-}} \frac{f(x) - f(x_{0})}{x-x_{0}}$
(or) we can write it as :-
$f'(x_{0}^{-}) = \lim_{h\rightarrow 0} \frac{f(x_{0} - h) - f(x_{0})}{x_{0}- h -x_{0}} = \lim_{h\rightarrow 0} \frac{f(x_{0} - h) - f(x_{0})}{-h}$
2) Right Hand Derivative(RHD) = $f'(x_{0}^{+}) = \lim_{x\rightarrow x_{0}^{+}} \frac{f(x) - f(x_{0})}{x-x_{0}}$
(or) we can write it as :-
$f'(x_{0}^{+}) = \lim_{h\rightarrow 0} \frac{f(x_{0} + h) - f(x_{0})}{x_{0}+ h -x_{0}} = \lim_{h\rightarrow 0} \frac{f(x_{0} + h) - f(x_{0})}{h}$
Since , here , $x_{0}$ = 0
So, LHD = $\lim_{h\rightarrow 0} \frac{f(0- h) - f(0)}{-h}$ = 5
$\Rightarrow$ $\lim_{h\rightarrow 0} \frac{f(- h) - f(0)}{-h}$ = 5
$\Rightarrow$ $\lim_{h\rightarrow 0} \frac{f(- h) - f(0)}{h}$ = -5
Since, $f(x)$ is an even function . So, $f(-h) = f(h)$
So, $\lim_{h\rightarrow 0} \frac{f(h) - f(0)}{h}$ = -5
So, RHD = -5
Since, at x = 0 , LHD $\neq$ RHD . So , $f(x) $is not differentiable at x = 0
Now , To check whether RHD = $\lim_{h\rightarrow 0} \frac{f(h) - f(0)}{h}$ exists or not , we have to check whether Left Hand Limit(LHL) = Right Hand Limit(RHL) or not .
LHL = $\lim_{h\rightarrow 0^{-}} \frac{f(h) - f(0)}{h}$ = $\lim_{z\rightarrow 0} \frac{f(0-z) - f(0)}{0-z}$ = $\lim_{z\rightarrow 0} \frac{f(-z) - f(0)}{-z}$ = $\lim_{z\rightarrow 0} \frac{-f'(-z)}{-1}$ = $\lim_{z\rightarrow 0} f'(-z)$ = $f'(0)$
RHL = $\lim_{h\rightarrow 0^{+}} \frac{f(h) - f(0)}{h}$ = $\lim_{z\rightarrow 0} \frac{f(0+z) - f(0)}{0+z}$ = $\lim_{z\rightarrow 0} \frac{f'(z)}{1}$ = $\lim_{z\rightarrow 0} f'(z)$ = $f'(0)$
Since , LHL = RHL , So, $\lim_{h\rightarrow 0} \frac{f(h) - f(0)}{h}$ exists.
So, we can say that RHD exists but it is not equal to LHD.
So, $ f(x)$ is not differentiable at $x = 0$