Question:
The set of all the points at which
\[
f(x)=|2-|x||
\]
is continuous but not differentiable is:
The set of all the points at which
\[
f(x)=|2-|x||
\]
is continuous but not differentiable is:
Show Hint
For modulus functions:
\[
|g(x)|
\]
check:
• where \(g(x)=0\),
• where \(g(x)\) is itself non-differentiable.
Those are the possible points of non-differentiability.
• where \(g(x)=0\),
• where \(g(x)\) is itself non-differentiable.
Those are the possible points of non-differentiability.
Updated On: Oct 5, 2026
- \(\{0,1,2\}\)
- \(\{-1,0,2\}\)
- \(\{-2,0,2\}\)
- \(\{-2,1,2\}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
Absolute value functions are continuous everywhere but may fail to be differentiable where the expression inside modulus changes sign.
For:
\[
|g(x)|
\]
non-differentiability occurs at points where:
\[
g(x)=0
\]
or where \(g(x)\) itself is non-differentiable.
Step 1: Analyze the given function carefully.
Given: \[ f(x)=|2-|x|| \] We first examine the inner modulus: \[ |x| \] which is not differentiable at: \[ x=0 \] Next consider: \[ 2-|x| \] This becomes zero when: \[ 2-|x|=0 \] \[ |x|=2 \] Thus, \[ x=\pm2 \] At these points, the outer modulus changes sign.
Step 2: Determine differentiability.
The function is continuous everywhere because modulus functions are continuous. However:
• At \(x=0\), \(|x|\) is non-differentiable.
• At \(x=2\) and \(x=-2\), the outer modulus introduces sharp corners.
Hence the function is not differentiable at: \[ x=-2,\;0,\;2 \] Therefore the required set is: \[ \boxed{\{-2,0,2\}} \]
Step 1: Analyze the given function carefully.
Given: \[ f(x)=|2-|x|| \] We first examine the inner modulus: \[ |x| \] which is not differentiable at: \[ x=0 \] Next consider: \[ 2-|x| \] This becomes zero when: \[ 2-|x|=0 \] \[ |x|=2 \] Thus, \[ x=\pm2 \] At these points, the outer modulus changes sign.
Step 2: Determine differentiability.
The function is continuous everywhere because modulus functions are continuous. However:
• At \(x=0\), \(|x|\) is non-differentiable.
• At \(x=2\) and \(x=-2\), the outer modulus introduces sharp corners.
Hence the function is not differentiable at: \[ x=-2,\;0,\;2 \] Therefore the required set is: \[ \boxed{\{-2,0,2\}} \]
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