Let $f : (-1, 1) \to \mathbb{R}$ be a function defined by $f(x) = \min\{-|x|, -\sqrt{1 - x^2}\}$. If $K$ is the set of all points at which $f$ is not differentiable,then $K$ has exactly

  • A
    five elements
  • B
    one element
  • C
    three elements
  • D
    two elements

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Consider the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=e^{-\left|\log _e x\right|}$. If $m$ and $n$ are respectively the number of points at which $f$ is not continuous and $f$ is not differentiable,then $m+n$ is

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

$f(x) = \begin{cases} [\cos \pi x]; & x \leqslant 1 \\ 2\{x\} - 1; & x > 1 \end{cases}$ Comment on the derivability at $x = 1$,where $[\cdot]$ denotes the greatest integer function and $\{\cdot\}$ denotes the fractional part function.

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 $f(x) = |x| + |x - 1|$ is

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