Q. If $f\left(x\right) = \int^\limits{x}_{0} t\left(\sin x-\sin t\right)dt$ then

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A

$f '''(x)+f ''(x)=\sin x$

16%

B

$f '''(x)+f''(x)-f (x)=\cos x$

23%

C

$f'''(x)+f '(x)=\cos x-2x \sin x$

42%

D

$f '''(x)-f ''(x)=\cos x-2x \sin x$

19%

Solution:

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

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