Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be defined by \( f(x) = \dfrac{|x| + |x|}{1 + x^2} \), then the set of points where \( f(x) \) is not differentiable is

• A. \( \{0, 1, -1\} \)
• B. \( \{-1\} \)
• C. \( \{0\} \)
• D. \( \{1\} \)

Hint:

Absolute value function \( |x| \) is non-differentiable at \( x = 0 \).

Answer 1

The Correct Option is C

Given the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \dfrac{|x| + |x|}{1 + x^2} \), we analyze where it is not differentiable.

First, note that \( |x| + |x| = 2|x| \). So, \( f(x) = \dfrac{2|x|}{1 + x^2} \).

To determine differentiability, we need to examine points where the absolute value function may not be differentiable:

• The function \( |x| \) is not differentiable at \( x = 0 \).

Therefore, suspect \( x=0 \) as a non-differentiable point – the function has the form \( 2|x| \) which is not differentiable at \( x = 0 \).

Check other potential points such as \( x = 1 \) and \( x = -1 \):

• At these, \( |x| \) is not causing non-differentiability since it only breaks at zero.

Thus, \( f(x) \) is only non-differentiable at \( x=0 \).

The set of points where \( f(x) \) is not differentiable is \(\{0\}\).