Question

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

• A. {0, 1, −1}
• B. $\pm$1
• C. 1
• D. −1

Answer 1

Answer: (A)

{0, 1, −1}

Answer 2

Given function is y = | |x| − 1 |

Or $y = \left{ \begin{matrix}

• |x| + 1if & |x| < 1 \ |x| - 1if & |x| \geq 1 \end{matrix} \right.\ $

$= \left{ \begin{matrix}

• |x| + 1if & - 1 < x < 1 \ |x| - 1if & x \leq - 1orx \geq 1 \end{matrix} \right.\ $=$\left{ \begin{matrix}
• x - 1 \ x + 1 \
• x + 1 \ x - 1 \end{matrix} \right.\ \begin{matrix} if \ if \ if \ if \end{matrix}\begin{matrix} x \leq - 1 \
• 1 < x < 0 \ 0 \leq x < 1 \ x \geq 1 \end{matrix}$

Here Ly’ (−1) = −1 and Ry’ (−1) = 1

Ly’ (0) = 1 and Ry’ (0) = −1 and Ly’(1) = −1 and Ry’(0) = 1

⇒ y is not differentiable at x = −1, 0, 1. ∴ (1) is the correct option

ALTERNATIVE :

Graph of y = | |x| − 1| is as follows:

Which has sharp turnings at x = −1, 0 and 1 and hence not differentiable at x = −1, 0, 1.