If $f(x)=\frac{x\left(a^x-1\right)}{1-\cos x}$ and $g(x)=\frac{x\left(1-a^x\right)}{a^x\left(\sqrt{1-x^2}-\sqrt{1+x^2}\right)}$, then $\lim _{x \rightarrow 0}(f(x)-g(x))=$
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
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