Question

If $f\left(\pi \right)=2$ and ${\int }_0^{\pi }(f\left(x\right)+f"\left(x\right))\sin xdx=5$, then $f\left(0\right)$ is equal to (it given that $f\left(x\right)$ is continuous in $[0,\pi ])$

• A. $7$
• B. $3$
• C. $5$
• D. $1$

Answer 1

The correct option is B $3$

$f\left(\pi \right)=2$ $\&$ ${\int }_0^{\pi }\{f\left(x\right)+f^{{}^{\prime \prime }}\left(x\right)\}\sin xdx=5$

$\Rightarrow {\int }_0^{\pi }f\left(x\right)\sin xdx+{\int }_0^{\pi }{f}^{\prime \prime }\left(x\right)\sin xdx=5$

$\Rightarrow {[-f\left(x\right)\cos x+\int f^{{}^{\prime }}\left(x\right)\cos xdx]}_0^{\pi }+{\int }_0^{\pi }{f}^{\prime \prime }\left(x\right)\sin xdx=5$

$\Rightarrow {[-f\left(x\right)\cos x+f^{\prime }\left(x\right)\sin x-\int f^{{}^{\prime \prime }}\left(x\right)\sin xdx]}_0^{\pi }+{\int }_0^{\pi }{f}^{\prime \prime }\left(x\right)\sin xdx=5$

$\Rightarrow [-f\left(\pi \right)\cos \pi +f\left(0\right)\cos \left(0\right)]+[f^{{}^{\prime }}\left(x\right)\sin \pi -f^{{}^{\prime }}\left(0\right)\sin 0]-{\int }_0^{\pi }{f}^{\prime \prime }\left(x\right)\sin xdx+{\int }_0^{\pi }{f}^{\prime \prime }\left(x\right)\sin xdx=5$

$\Rightarrow -2\times (-1)+f\left(0\right)+0-0=5$

$\Rightarrow f\left(0\right)=3.$

Hence, the answer is $3.$