Question

If f(x) = ∫x0 t(sin x-sin t)dt then

• A. $f '''(x)+f ''(x)=\sin x$
• B. $f '''(x)+f''(x)-f (x)=\cos x$
• C. $f'''(x)+f '(x)=\cos x-2x \sin x$
• D. $f '''(x)-f ''(x)=\cos x-2x \sin x$

Answer 1

The Correct Option is C

\(f\left(x\right) = \int^{^x}_{_0}t\left(sin\,x - sin\,t\right)dt\)
\(f\left(x\right) = sinx \int^{^x}_{_0}t\,dt - \int^{^x}_{_0} t \,sin\,tdt\)
\(f'\left(x\right) = \left(sinx\right) x +cosx \int^{^x}_{_0} t \,dt-x\,sinx\)
\(f'\left(x\right) = cosx \int^{^x}_{_0} tdt\) = \(x\cos x\)
\(f''\left(x\right) = \left(cosx\right) x - \left(sinx\right) \int^{^x}_{_0} tdt\)
\(f'''\left(x\right) = x\left(-sinx\right) + cosx-\left(sinx\right)x-\left(cosx\right) \int^{^x}_{_0} tdt\)
\(f'''\left(x\right)+f'\left(x\right) = cosx-2x\,sinx\)