JEE Main On 16 April 2018 Question 19

Question: If $ f(x)=\int_0^{x}{t(\sin x-\sin t)}dt $ then? [JEE Main 16-4-2018]

Options:

A) $ 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 $

Show Answer

Answer:

Correct Answer: A

Solution:

$ f(x)=\int_0^{2}{t(\sin x-\sin t)}.dt $

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

$ =\frac{x^{2}}{2}\sin x+t\cos t\int_0^{x}{+\sin x} $

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

$ f’(x)=\frac{x^{2}}{2}\cos x+2\cos x $

$ f’’’(x)=-x\sin x+\cos x-x\sin x-\frac{x^{2}}{2}\cos x-2\cos x $

Add $ f’’’(x) $ and $ f’(x), $ we get $ f’’’(x)+f’(x)=\cos x-2x\sin x $