If $f:R \rightarrow R$ is an even function which is twice differentiable on $R$ and $f^{''}(\pi) = 1,\$then $f^{''}( - \pi$) is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
Let the even function be
$f(x) = \cos x$
$\Rightarrow \ f^{'}(x) = - \sin x\ \ \ \$
$\Rightarrow \ f^{''}(x) = \  - \cos x$
At $x = \pi,\ \ {\ f}^{''}(\pi) = \  - \cos\pi\$=1
$	herefore\ \ Our\ assumption\ is\ true.$
$	herefore At\ \ x = - \pi$
$\ f^{''}( - \pi) = \  - \cos{( - \pi}) = 1$
Alternate
Since the function is twice differentiable
$	herefore f^{''}(x) = const.\ \forall xf^{''}( - \pi) = - f^{''}(\pi) = 1\$

Q. If $f:R \rightarrow R$ is an even function which is twice differentiable on $R$ and $f^{''}(\pi) = 1,\$then $f^{''}( - \pi$) is equal to

Explanation

Reviewed by Dr. Abubacker

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