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:

The function $f(x) = a \sin |x| + b e^{|x|}$ is differentiable at $x = 0$ when

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

The domain of the derivative of the function $f(x) = \frac{x}{1+|x|}$ is

Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.
Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.
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: