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) \[f'''(x)+f''(x)=\sin x\]
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 \[f'''(x)+f'(x)=\cos x-2x\sin x\]
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