Question

Let $f(x) = \begin{cases} - 1, - 2 \le x < 0 \\ x^2 - 1 , 0 \le x \le 2 \end{cases} $ and $g(x) = |f (x) + f (| x |)|$ . Then, in the interval $(-2, 2), g$ is :

• A. differentiable at all points
• B. not differentiable at two points
• C. not continuous
• D. not differentiable at one point

Answer 1

Answer: (D)

$|f(x) | = \begin{cases} 1 &, - 2\le x < 0 \\ 1 - x^2 & , 0 \le x < 1 \\ x^2 - 1 &, 1 \le x \le 2 \end{cases} $
and $f(|x|) = x^2 - 1 , x \in [-2 , 2 ]$
Hence $g(x) = \begin{cases} x^2 &, x \in [-2, 0)\\ 0 & , x \in [0,1) \\ 2(x^2 - 1) &, x \in [1, 2 ] \end{cases}$
It is not differentiable at $x = 1$