Question

${\int }_0^{\pi }\left\{f\right(x)+f^{{}^{\prime \prime }}(x\left)\right\}\sin xdx=5,f\left(\pi \right)=2.$ Then $f\left(0\right)$ equals

• A. $2$
• B. $3$
• C. $5$
• D. $4$

Answer 1

Answer: (B)

$3$

$\int f^{\prime \prime }\left(x\right)\sin xdx$

$=\sin x{f}^{\prime }\left(x\right)-\int \cos x{f}^{\prime }\left(x\right)dx$

$=\sin x{f}^{\prime }\left(x\right)-\left[\cos xf\right(x)+\int \sin xf(x\left)dx\right]$

$\Rightarrow \int \left[f\right(x)+f^{\prime \prime }(x\left)\right]\sin xdx=\sin x{f}^{\prime }\left(x\right)dx-\cos xf\left(x\right)$

$\Rightarrow {\int }_0^{\pi }\left[f\right(x)+f^{\prime \prime }(x\left)\right]\sin xdx={\left[\sin x{f}^{\prime }\right(x)dx-\cos xf(x\left)\right]}_0^{\pi }$

$\Rightarrow 5=f\left(\pi \right)+f\left(0\right)$

$\Rightarrow f\left(0\right)=5-2=3$