If $f(x) = \begin{cases} x + 2, & -1 < x < 3 \\ 5, & x = 3 \\ 8 - x, & x > 3 \end{cases}$,then at $x = 3$,$f'(x) = $

The set of all those points,where the function $f(x) = \frac{x}{1 + |x|}$ is differentiable,is

$f(x) = \begin{cases} 4, & -\infty < x < -\sqrt{5} \\ x^2-1, & -\sqrt{5} \leq x \leq \sqrt{5} \\ 4, & \sqrt{5} < x < \infty \end{cases}$
If $k$ is the number of points where $f(x)$ is not differentiable,then $k-2=$

Let $[x]$ denote the greatest integer less than or equal to $x$. Then the number of points where the function $y = [x] + |1 - x|$ for $-1 \leq x \leq 3$ is not differentiable,is

If $f(x + y) = f(x) + f(y) + |x|y + xy^2$,$\forall x, y \in R$ and $f'(0) = 0$,then

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} x^{5} \sin \left(\frac{1}{x}\right) + 5x^{2} & , x < 0 \\ 0 & , x = 0 \\ x^{5} \cos \left(\frac{1}{x}\right) + \lambda x^{2} & , x > 0 \end{cases}$. The value of $\lambda$ for which $f''(0)$ exists is: