If $f(\pi)=2$ and $\int\limits_{0}^{\pi}(f(x)+f''(x))\sin x\,dx=5$ then f(0) is equal to : (It is given that f(x) is continuous in [0, π])
Answer & explanation
Correct answer: option 2
$I=\int\limits_{0}^{\pi}f(x).\sin x\,dx+\int\limits_{0}^{\pi}f''(x)\sin x\,dx$
$=|f(x)(-\cos x)|_{0}^{\pi}-\int\limits_{0}^{\pi}f'(x)(-\cos x)dx+|\sin x(f'(x))|_{0}^{\pi}-\int\limits_{0}^{\pi}\cos x.f'(x)dx=5$
$=|f'(x)\sin x-f(x)\cos x|_{0}^{\pi}=5$
$⇒f(\pi)+f(0)=5$
$⇒f(0)=3$