If $f(x)=\left\{\begin{array}{cc}\frac{a \sin x-b x+c x^2+x^3}{2 \log (1+x)-2 x^3+x^4} & , x \neq 0 \\ 0 & , x=0\end{array}\right.$

is continuous at $x=0$, then

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 $f(4)=$