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

The function $f(x) = x^2 \sin \frac{1}{x}$ for $x \ne 0$ and $f(0) = 0$ at $x = 0$ is:

Let $f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3$,where $a_0, a_1, a_2, a_3$ are real constants. Then $f(x)$ is differentiable at $x=0$ if and only if:

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:

Let $f(x) = \begin{cases} x^2 \left| \cos \frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x=0 \end{cases}$,$x \in \mathbb{R}$,then $f$ is

If $f(x) = \begin{cases} x + 1, & x < 2 \\ 2x - 1, & x \ge 2 \end{cases}$,then $f'(2)$ equals