Question

If $f\left(x\right)=\cos x-\underset{0}{\overset{x}{\int }}(x-t)f\left(t\right)dt$, then $f^{\prime \prime }\left(x\right)+f\left(x\right)$ is equal to

• A. $-\cos x$
• B. $-\sin x$
• C. $\underset{0}{\overset{x}{\int }}(x-t)f\left(t\right)dt$
• D. $0$

Answer 1

The correct option is A $-\cos x$

$f\left(x\right)=\cos x-\underset{0}{\overset{x}{\int }}(x-t)f\left(t\right)dt\Rightarrow f\left(x\right)=\cos x-x\underset{0}{\overset{x}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}tf\left(t\right)dt$
$\Rightarrow f^{\prime }\left(x\right)=-\sin x-xf\left(x\right)-\underset{0}{\overset{x}{\int }}f\left(t\right)dt+xf\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=-\sin x-\underset{0}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow f^{\prime \prime }\left(x\right)=-\cos x-f\left(x\right)\Rightarrow f^{\prime \prime }\left(x\right)+f\left(x\right)=-\cos x$