JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 10)
If $$f(x) = \int\limits_0^x {t\left( {\sin x - \sin t} \right)dt\,\,\,} $$ then :
f'''(x) + f''(x) = sinx
f'''(x) + f''(x) $$-$$ f'(x) = cosx
f'''(x) + f'(x) = cosx $$-$$ 2x sinx
f'''(x) $$-$$ f''(x) = cosx $$-$$ 2x sinx
Explanation
f(x) = $$\int_0^x {t(\sin x - \sin t).dt} $$
= sin x$$\int_0^x {t.dt - \int_0^x {t\sin t.dt} } $$
= $${{{x^2}} \over 2}$$ sin x +$$\left[ {t\cos t} \right]_0^x$$ + sin x
$$ \Rightarrow $$f(x) = $${{{x^2}} \over 2}$$ sinx + xcosx + sinx
f'(x) = $${{{x^2}} \over 2}$$ cosx + 2cos x
f''(x) = x cos x $$-$$ $${{{x^2}} \over 2}$$ sin x $$-$$ 2sin x
f'''(x) = cos x $$-$$ 2x sin x $$-$$ $${{{x^2}} \over 2}$$ cos x $$-$$ 2cos x
$$\therefore\,\,\,$$ f'''(x) + f'(x) = cos x $$-$$ 2x sin x
= sin x$$\int_0^x {t.dt - \int_0^x {t\sin t.dt} } $$
= $${{{x^2}} \over 2}$$ sin x +$$\left[ {t\cos t} \right]_0^x$$ + sin x
$$ \Rightarrow $$f(x) = $${{{x^2}} \over 2}$$ sinx + xcosx + sinx
f'(x) = $${{{x^2}} \over 2}$$ cosx + 2cos x
f''(x) = x cos x $$-$$ $${{{x^2}} \over 2}$$ sin x $$-$$ 2sin x
f'''(x) = cos x $$-$$ 2x sin x $$-$$ $${{{x^2}} \over 2}$$ cos x $$-$$ 2cos x
$$\therefore\,\,\,$$ f'''(x) + f'(x) = cos x $$-$$ 2x sin x
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