If $\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8}$, then $x^2+1=$

If $p \equiv$ The switch $S_1$ is closed, $q \equiv$ The switch $\mathrm{S}_2$ is closed, $\mathrm{r} \equiv$ switch $\mathrm{S}_3$ is closed, then symbolic form of following switching circuit is equivalent to

Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable function having $\mathrm{f}(3)=3, \mathrm{f}^{\prime}(3)=\frac{1}{27}$ and $\mathrm{g}(x)= \begin{cases}\int_3^{\mathrm{f}(x)} \frac{3 \mathrm{t}^2}{x-3} \mathrm{dt}, & \text { if } x \neq 3 \\ \mathrm{~K}, & \text { if } x=3\end{cases}$ is continuous at $x=3$, then $\mathrm{K}=$

The area bounded by the curve $y=4 x-x^2$ and X - axis in square units, is $\qquad$