Let f(x)= 1, & x -1 |x|, & -1 <x <1 0,&x 1 then, f is: - Vidyadip

MCQ

Let $\text{f(x)}=\begin{cases}1, & \text{x}\leq-1\\|\text{x}|, & -1 <\text{x} <1\\0,&\text{x}\geq1\end{cases}$ then, $f$ is:

A
Continuous at $x = -1$
✓
Differentible at $x = -1$
C
Everywhere continuous.
D
Everywhere diffrentiable.

✓

Answer

Correct option: B.

Differentible at $x = -1$

$\text{f(x)}=\begin{cases}1, & \text{x}\leq-1\\|\text{x}|, & -1 <\text{x} <1\\0,&\text{x}\geq1\end{cases}$
$\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{-\text{x}+1}{\text{x}+1}=0$
Similarly,
$\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{x}+1}{\text{x}+1}=0$
Function is diffrentiable at $x = -1.$

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