Let $f(x) = |x - \alpha| + |x - \beta|$, where $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 2 = 0$. Then the number of points in $[\alpha, \beta]$ at which $f$ is not differentiable 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:

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

Let $f:R \to R$ be a function defined by $f(x) = \max \,(x, x^3).$ The set of all points where $f(x)$ is not differentiable is

Identify the correct statement about the function $f(x) = \max(x^2 - 1, 7 - x^2, 5)$.

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) = $