Question:medium

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

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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.
Updated On: Jun 7, 2026
  • \( \{0\} \)
  • \( \{2\} \)
  • \( \{0, 2\} \)
  • \( \emptyset \)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Identify the corner points.
The function $f(x)=|x-2|\,(3^{4|x|}-1)$ has two modulus parts: $|x-2|$ gives a possible corner at $x=2$, and $|x|$ in the exponent gives a possible corner at $x=0$.
Step 2: Recall the smoothing rule.
A corner from $|x-c|$ disappears only if the factor multiplying it becomes zero at $x=c$. Otherwise the sharp point stays.
Step 3: Test $x=2$.
At $x=2$ the multiplier is $3^{4|2|}-1=3^8-1\neq0$. Since it does not vanish, the corner of $|x-2|$ remains, so $f$ is not differentiable at $x=2$.
Step 4: Test $x=0$.
At $x=0$ the term $3^{4|x|}$ has a corner because $|x|$ does. The other factor is $|0-2|=2\neq0$, so it cannot smooth the corner.
Step 5: Confirm with one-sided slopes at $0$.
The right derivative of $3^{4|x|}$ at $0$ is $4\ln3$ and the left derivative is $-4\ln3$; these differ, so $f$ is not differentiable at $x=0$ either.
Step 6: Collect the points.
Both $x=0$ and $x=2$ are non-differentiable points. \[ \boxed{\{0,\ 2\}} \]
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