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 $
