Question:medium

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.
Updated On: Jun 17, 2026
  • \(\{0,1,2\}\)
  • \(\{-1,0,2\}\)
  • \(\{-2,0,2\}\)
  • \(\{-2,1,2\}\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Know the rule for modulus functions.
Any modulus is continuous everywhere, but it has a sharp corner (no derivative) wherever the inside becomes zero, and wherever an inner modulus has a corner.
Step 2: Look at the inner modulus.
In $f(x)=|2-|x||$ the inside contains $|x|$, which has a corner at $x=0$.
Step 3: Find where the outer inside is zero.
The outer modulus wraps $2-|x|$. This is zero when $|x|=2$, that is $x=2$ or $x=-2$.
Step 4: Confirm continuity.
Modulus functions never jump, so $f$ is continuous at every real number, including these points.
Step 5: Collect the corner points.
Corners appear at $x=0$ (from $|x|$) and at $x=\pm2$ (where the outer modulus folds). At each, the left and right slopes differ, so no derivative exists there.
Step 6: Write the set.
The points that are continuous but not differentiable are \[ \boxed{\{-2,0,2\}}. \]
Was this answer helpful?
0