Continuity and ...

If $$y={ \left( \tan ^{ -1 }{ x }  \right)  }^{ 2 }$$ and $${ \left( { x }^{ 2 }+1 \right)  }^{ 2 }\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +2x\left( { x }^{ 2 }+1 \right) \dfrac { dy }{ dx } =k$$, then the value of $$k$$ is

Which of the following  is not differentiable in the interval $$(-1,2)$$?

Let  $$f:(0,\infty ) \to R$$ be a differentiable function such that $$f'(x) = 2 - {{f(x)} \over x}$$ for all $$x \in (0,\infty )$$ and $$f(1) \ne 1$$

Let $$f(x)=\begin{cases}
(x-1) \sin \left( \large{\frac{1}{x-1}}\right) \ &\mbox{if}& \ x \ne 1\\
0, \ &\mbox{if}& \ x \ne 1
\end{cases}$$.
Then which one of the following is true?

Let $$f(x)={ x }^{ 3 }-{ x }^{ 2 }+x+1$$ and $$g\left( x \right) =\begin{cases} max\{ f\left( t \right) ;0\le t\le x\} \quad ;\quad 0\le x\le 1 \\ 3-x\quad \quad \quad \quad \quad \quad \quad \quad   ;\quad 1<x\le 2 \end{cases}$$ then

If $$x=a{ t }^{ 2 },y=2at$$, then $$\cfrac { dy }{ dx } =$$_____;$$t\neq 0$$

If the function $$f(x)=\dfrac{e^{x^{2}}-\cos x}{x^{2}}$$ for $$x \neq 0$$ continuous at $$x=0$$ then $$f(0)=$$

The function $$f : R /{0} \rightarrow R$$ given by $$f(x) =
\dfrac{1}{x} - \dfrac{2}{e^{2x} -1}$$ can be made continuous at $$x=0$$ by
defining $$f(0)$$ as 

The function $$f(x) =
\ sin^{-1} (\ cosx)$$ is 

Let $$f(x) = \begin{cases} x & x<1 \\ 2-x & 1 \leq x \leq 2 \\ -2+3x-x^2 & x>2 \end{cases}$$
 then $$f(x)$$ is