If $\mathrm{f}(x)=\frac{1+\cos \pi x}{\pi(1-x)^2}$, for $x \neq 1$ is continuous at $x=1$, then $\mathrm{f}(1)$ is equal to

Let K be the set of all real values of $x$, where the function $\mathrm{f}(x)=\sin |x|-|x|+2(x-\pi) \cos |x|$ is not differentiable. Then the set K is

The function $\mathrm{f}$ defined on $$\left(-\frac{1}{3}, \frac{1}{3}\right)$$ by $$\mathrm{f}(x)=\left\{\begin{array}{cc} \frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right) & , \quad x \neq 0 \\ \mathrm{k} & , \quad x=0 \end{array}\right.$$ is continuous at $$x=0$$, then $$\mathrm{k}$$ is

If $$f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$$, then as $$x$$ approaches a, $$\frac{\mathrm{g}(x) \mathrm{f}(\mathrm{a})-\mathrm{g}(\mathrm{a}) \mathrm{f}(x)}{(x-\mathrm{a})}$$ approaches