If the function $g(x)=\left\{\begin{array}{cl}K \sqrt{x+1} & , 0 \leq x \leq 3 \\ m x+2 & , 3 < x \leq 5\end{array}\right.$ is differentiable, then $K+m=$
Consider the following statements
Assertion (A) For $x \in R-\{1\}$;
$$ \frac{d}{d x}\left(\tan ^{-1}\left(\frac{1+x}{1-x}\right)\right)=\frac{d}{d x}\left(\tan ^{-1} x\right) $$
Reason (R) For $x<1, \tan ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{\pi}{4}+\tan ^{-1} x$, for
$$ x>1, \tan ^{-1}\left(\frac{1+x}{1-x}\right)=-\frac{3 \pi}{4}+\tan ^{-1} x $$
The correct answer is
If $\frac{d}{d x}\left\{\left(\frac{x-1}{x-\sqrt{x}}\right) e^{2 x+1}\right\}=\frac{x-1}{x-\sqrt{x}} e^{2 x+1} f(x)$, then $f(4)=$
If $y=\left(\sin ^{-1} x\right)^2$, then $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=$
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