Let f x be a function satisfying f x + f π - x = π 2 ,   ∀ x ∈ ℝ . Then…

Let

$I={\int }_0^{\mathrm{\pi }}f\left(x\right)\sin xdx ...\left(1\right)$

Now using the property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ we get,

$\Rightarrow I={\int }_0^{\mathrm{\pi }}f\left(\mathrm{\pi }-x\right)\sin \left(\mathrm{\pi }-\mathrm{x}\right)dx$

$\Rightarrow I={\int }_0^{\mathrm{\pi }}f\left(\mathrm{\pi }-x\right)\mathrm{sinx}dx ...\left(2\right)$

Adding $\left(1\right) \& \left(2\right)$, we get

$2I={\int }_0^{\mathrm{\pi }}\left[f\left(x\right)+f\left(\pi -x\right)\right]\mathrm{sinx}dx$

$\Rightarrow 2I={\mathrm{\pi }}^2{\int }_0^{\mathrm{\pi }}\mathrm{sinx}dx$ {as given $f\left(x\right)+f\left(\mathrm{\pi }-x\right)={\mathrm{\pi }}^2$}

$\Rightarrow 2I={\mathrm{\pi }}^2{\left[-\mathrm{cosx}\right]}_0^{\mathrm{\pi }}$

$\Rightarrow 2I=2{\mathrm{\pi }}^2$

$\Rightarrow I={\mathrm{\pi }}^2$

Hence this is the correct option.