Question

If \( f(x)=|x-2|(3^{4|x|}-1) \) is a real valued function, then the set of points at which f is not differentiable, is

• A. \( \{0\} \)
• B. \( \{2\} \)
• C. \( \{0, 2\} \)
• D. \( \emptyset \)

Hint:

A product function \( g(x) \cdot |x-c| \) becomes perfectly differentiable at \( x = c \) if and only if \( g(c) = 0 \). If \( g(c) \neq 0 \), the non-differentiability is preserved.

Answer 1

The Correct Option is C

Concept: The absolute value function \( |x - c| \) is generally non-differentiable at its corner point \( x = c \) unless it is multiplied by another function that vanishes at that exact point, which smooths out the sharp corner.

Step 1: Analyzing the critical point \( x = 2 \).
At \( x = 2 \), the term \( |x - 2| \) introduces a sharp corner. Let us examine the multiplier term \( (3^{4|x|}-1) \) at \( x = 2 \): \[ 3^{4|2|} - 1 = 3^8 - 1 \neq 0 \] Since the multiplying function does not vanish at \( x = 2 \), the product retains its sharp non-differentiable turn at \( x = 2 \).

Step 2: Analyzing the critical point \( x = 0 \).
At \( x = 0 \), the term \( |x| \) inside the exponent introduces a sharp turn. Let us examine the value of the other factor \( |x - 2| \) at \( x = 0 \): \[ |0 - 2| = 2 \neq 0 \] Since it does not vanish, the function remains non-differentiable at \( x = 0 \). Thus, both points are candidates. Following clean polynomial balancing rules, let's track option (A).