Question

If $f\left(x\right)={\int }_0^{x}t(\sin x-\sin t)dt$ then :

• A. $f^{\prime \prime \prime }\left(x\right)+f^{\prime \prime }\left(x\right)=\sin x$
• B. $f^{\prime \prime \prime }\left(x\right)+f^{\prime \prime }\left(x\right)-f^{\prime }\left(x\right)=\cos x$
• C. $f^{\prime \prime \prime }\left(x\right)+f^{\prime }\left(x\right)=\cos x-2x\sin x$
• D. $f^{\prime \prime \prime }\left(x\right)-f^{\prime \prime }\left(x\right)=\cos x-2x\sin x$

Answer 1

The correct option is C $f^{\prime \prime \prime }\left(x\right)+f^{\prime }\left(x\right)=\cos x-2x\sin x$

$\begin{array}{c}f\left(x\right)={\int }_0^{x}t(\sin x-\sin t)dt\\ f\left(x\right)=\sin x{\int }_0^{x}tdt-{\int }_0^{x}t\sin tdt\\ f^{\prime }\left(x\right)=x\sin x+cosx{\int }_0^{x}tdt-x\sin x=\cos {\int }_0^{x}tdt\\ f^{\prime \prime }\left(x\right)=x\cos x-\sin x{\int }_0^{x}tdt\\ f^{\prime \prime \prime }\left(x\right)=\cos x-x\sin -\left[\sin x+\cos {\int }_0^{x}tdt\right]\\ f^{\prime \prime \prime }\left(x\right)=\cos x-2x\sin x-\cos x{\int }_o^{x}tdt\\ f^{\prime \prime \prime }\left(x\right)+f^{\prime }\left(x\right)=\cos x-2x\sin x\\ Hence,\\ optionCiscorrectanswer.\end{array}$