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An even function $f(x)$ has left derivative $5$ at $x=0$. Then

  1. the right derivative of $f(x)$ at $x=0$ need not exist
  2. the right derivative of $f(x)$ at $x=0$ exists and is equal to $5$
  3. the right derivative of $f(x)$ at $x=0$ exists and is equal to $-5$
  4. none of the above is necessarily true
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14 votes

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 $x = 0$

 

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4 votes

Since the function is even, hence


$\begin{align*} f(x)&=f(-x)\\ \Rightarrow f'(x)&=-f'(-x) \\ \Rightarrow \lim_{x \to 0^{-}} f'(x)&=\lim_{x \to 0^{-}} -f'(-x) \\ \Rightarrow f'(0^-)&= -f'(-0^-) \\ \therefore f'(0^-)&= -f'(0^+) ; ~~[\because -0^-=0^+] \end{align*}$

 

Here, $f'(0^-)=5 \\\Rightarrow f'(0^+)=-5$

$\therefore$ The right derivative of $f(x)$ at $x=0$ is $-5$. So the answer is C.

2 votes
2 votes

Even function $f(x)=f(-x)$

$L.H.S.=R.H.S.=5$

answer is (B)

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