MCQ
If $f(x) = \int_0^x {t(\sin \,\,x\, - \sin \,\,t)\,dt} $ then ?
  • $f'''(x) + f'(x) = \cos \,x\, - 2x\,\,\sin \,x$
  • B
    $f'''(x) + f''(x) - f'(x) = \cos \,x\,$
  • C
    $f'''(x) - f''(x) = \cos \,x\,\, - \,2x\,\,\sin \,x$
  • D
    $f'''(x) + f''(x) = \,\sin \,x$

Answer

Correct option: A.
$f'''(x) + f'(x) = \cos \,x\, - 2x\,\,\sin \,x$
a
$f\left( x \right) = \int_0^x {t\left( {\sin x - \sin t} \right)} .dt$

$ = \sin x\int_0^x {t.dt}  - \int_0^x {t\sin t.dt} $

$ = \frac{{{x^2}}}{2}\sin x + \left[ {t\cos t_0^x} \right] + \sin x$

$ \Rightarrow f\left( x \right) = \frac{{{x^2}}}{2}\sin x + x\cos x + \sin x$

$f'\left( x \right) = \frac{{{x^2}}}{2}\cos x + 2\,\cos x$

$f''\left( x \right) = x\cos x - \frac{{{x^2}}}{2}\sin x - 2\sin x$

$f'''\left( x \right) = \cos x - 2x\sin x - \frac{{{x^2}}}{2}\cos x - 2\cos x$

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