MCQ
Let $f(x)$ be a function satisfying $f(x)+f(\pi-x)=$ $\pi^2, \forall x \in R$. Then $\int \limits_0^\pi f(x) \sin x d x$ is equal to $...........$.
- A$\frac{\pi^2}{4}$
- B$\frac{\pi^2}{2}$
- C$2 \pi^2$
- ✓$\pi^2$
$I=\int \limits_0^\pi f(x) \sin x d x$
Applying King's Rule
$I=\int \limits_0^\pi f(\pi-x) \cdot \sin (\pi-x) d x$
$2 I=\int \limits_0^\pi[f(x)+f(\pi-x)] \sin x d x$
$2 I=\int \limits_0^\pi \pi^2 \sin x d x$
$2 I=\pi^2 \cdot \int \limits_0^\pi \sin x d x$
$2 I=\pi^2 \times 2$
$I=\pi^2$
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$f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0 < x < \frac{\pi}{2} \\ a-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{b}{a}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.$
Where $a, b \in Z$. If $f$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to ..........