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If $f: R \rightarrow R$ is an even function which is twice differentiable on $R$ and $f^{\prime \prime}(\pi)=1$, then $f^{\prime \prime}(-\pi)$ is equal to
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Let the even function be
$f(x)=\cos x$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=-\sin x$
Again differentiating w.r.t. $x$, we get
$f^{\prime \prime}(x)=-\cos x$
at $x=\pi$
$f^{\prime \prime}(\pi)=-\cos \pi=1$
$\therefore$ Our assumption is true.
at
$x=-\pi$
$f^{\prime \prime}(-\pi)=-\cos (-\pi)=1$
$f(x)=\cos x$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=-\sin x$
Again differentiating w.r.t. $x$, we get
$f^{\prime \prime}(x)=-\cos x$
at $x=\pi$
$f^{\prime \prime}(\pi)=-\cos \pi=1$
$\therefore$ Our assumption is true.
at
$x=-\pi$
$f^{\prime \prime}(-\pi)=-\cos (-\pi)=1$
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