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$f'''(x) + f''(x) = \,\sin \,x$
$ = \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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$\lim _{n \rightarrow 0^{+}} \int_n^{1-n} t^{-3}(1-t)^{a-1} d t$
exists. Let this limit be $g(a)$. In addition, it is given that the function $g(a)$ is differentiable on $(0,1)$.
$1.$ The value of $g\left(\frac{1}{2}\right)$ is
$(A)$ $\pi$ $(B)$ $2 \pi$ $(C)$ $\frac{\pi}{2}$ $(D)$ $\frac{\pi}{4}$
$2.$ The value of $g ^{\prime}\left(\frac{1}{2}\right)$ is
$(A)$ $\frac{\pi}{2}$ $(B)$ $\pi$ $(C)$ $-\frac{\pi}{2}$ $(D)$ $0$
Give the answer question $1$ and $2.$