If $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5}, & \text{for } x \neq 1 \\ -\frac{1}{3}, & \text{for } x=1 \end{cases}$,then $f^{\prime}(1)$ is equal to:

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:

The function $f(x) = |\cos x|$ is

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

If $f(x)=\begin{cases} \frac{2 x e^{\frac{1}{2 x}}-3 x e^{\frac{-1}{2 x}}}{e^{\frac{1}{2 x}}+4 e^{\frac{-1}{2 x}}} & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$ is a real valued function,then:

If $x + |y| = 2y,$ then $y$ as a function of $x,$ at $x = 0$ is