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 $f(x) = \begin{cases} |4x^2 - 8x + 5|, & \text{if } 8x^2 - 6x + 1 \geq 0 \\ [4x^2 - 8x + 5], & \text{if } 8x^2 - 6x + 1 < 0 \end{cases}$,where $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$. Then the number of points in $\mathbb{R}$ where $f$ is not differentiable is $.......$

The function given by $y = ||x| - 1|$ is differentiable for all real numbers except the points

The function $f(x) = \begin{cases} e^x + ax, & x < 0 \\ b(x - 1)^2, & x \geq 0 \end{cases}$ is differentiable at $x = 0$. Then

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:

Suppose $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \\ 2\{x\} - 1, & x > 1 \end{cases}$,where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer function and the fractional part of $x$ respectively,then at $x = 1$: