Question

If \( \alpha \in \mathbb{R} \setminus \{-1\} \) and \[ f(x) = |x| + \alpha |x|(|x| - 1), \] then the number of points at which \( f(x) \) is not differentiable is:

• A. 3, when \( \alpha<0 \)
• B. 5, when \( \alpha>0 \)
• C. 4, when \( \alpha>0 \)
• D. 5, when \( \alpha<0 \)

Hint:

Check piecewise absolute value expressions for corner points causing non-differentiability.

Answer 1

The Correct Option is D

Break the function based on \( x \geq 0 \) and \( x<0 \) Key points where non-differentiability occurs: - At \( x = 0 \): due to \( |x| \) - At \( x = \pm1 \): due to piece \( |x| - 1 \) - Additional critical points due to behavior of \( \alpha \): test derivative left and right Total: 5 points of non-differentiability when \( \alpha<0 \)