The solution of the equation $$\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$$ is

The differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\sqrt{1-y^2}}{y}$$ determines a family of circles with

A square plate is contracting at the uniform rate $$4 \mathrm{~cm}^2 / \mathrm{sec}$$, then the rate at which the perimeter is decreasing, when side of the square is $$20 \mathrm{~cm}$$, is

If the function $$\mathrm{f}(x)$$ is continuous in $$0 \leq x \leq \pi$$, then the value of $$2 a+3 b$$ is where

$$f(x)= \begin{cases}x+a \sqrt{2} \sin x & \text { if } 0 \leq x < \frac{\pi}{4} \\ 2 x \cot x+b & \text { if } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ \operatorname{acos} 2 x-b \sin x & \text { if } \frac{\pi}{2} < x \leq \pi\end{cases}$$