Question

JEE ADVANCE :-

Que :- Let f be an even function and f'(0) exists ,then find the value of f'(0).​

Answer 1

Given

f is an even function and f'(0) exists.

To find

The value of f'(0)

Concept

• A function which satisfies f(x)= f(-x) is called even function.An even function is symmetric about y -axis.

• A function which satisfies f(x)= - f(-x) is called odd function.An odd function is symmetric about origin.
• If f is an even function, f' will be odd function and vice versa.

• For an odd function f, f(0) = 0.

Solution

Given, f is an even function. It implies that f'(x) will be an odd function .

Hence, f'(0) = 0 (f'(x) is an odd function}.

Alternative Solution

f'(0) = $\displaystyle \lim _{x \: to \: {0}^{ + }} \frac{f(h) - f(0)}{h} = \displaystyle \lim _{x \: to \: {0}^{ - }} \frac{f( - h) - f(0)}{ - h} \\ \\ f \: is \: an \: evenfunction \: f(h) = f( - h)$

$\displaystyle \lim _{x \: to \: {0}^{ + }} \frac{f(h) - f(0)}{h} = \displaystyle \lim _{x \: to \: {0}^{ - }} \frac{f( h) - f(0)}{ - h}$

$= > 2\displaystyle \lim _{x \: to \: {0}^{ + }} \frac{f(h) - f(0)}{h} = 0 \\ \\ = > 2f'(0) = 0 \\ \\ = > f'(0) = 0$

Answer 2

Answer:

f be an even function

So

f(-x) = f(x)

Differentiating both sides with respect to x

-f'(-x) = f'(x)

Putting x =0 in both sides

-f'(0) = f'(0)

2 f'(0) = 0

f'(0) = 0