Question:
Let \(f(x)=\begin{cases}\max\{|x|,x^2\} & |x|\le2\\ 8-2|x| & 2<|x|\le4\end{cases}\)
Let \(S\) be the set of points in the interval \((-4,4)\) at which \(f\) is not differentiable. Then \(S\)
Let \(f(x)=\begin{cases}\max\{|x|,x^2\} & |x|\le2\\ 8-2|x| & 2<|x|\le4\end{cases}\)
Let \(S\) be the set of points in the interval \((-4,4)\) at which \(f\) is not differentiable. Then \(S\)
Let \(S\) be the set of points in the interval \((-4,4)\) at which \(f\) is not differentiable. Then \(S\)
Show Hint
Write \(f\) piecewise: \(|x|\) on \([-1,1]\), \(x^2\) for \(1\le|x|\le2\), \(8-2|x|\) beyond 2; compare one-sided slopes at \(0,\pm1,\pm2\).
Updated On: Nov 5, 2026
- equals \(\{-2,-1,1,2\}\)
- equals \(\{-2,-1,0,1,2\}\)
- equals \(\{-2,2\}\)
- is an empty set
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The Correct Option is B
Solution and Explanation
For \(|x|\le1\), \(|x|\ge x^2\); for \(1\le|x|\le2\), \(x^2\ge|x|\). So on \([-2,2]\): \(f=-x\) on \([-1,0]\), \(x\) on \([0,1]\), \(x^2\) on \([1,2]\) (symmetric on the left).
At \(x=0\): slopes \(-1\) and \(+1\): not differentiable.
At \(x=1\): slopes \(1\) and \(2x=2\): not differentiable (similarly \(x=-1\)).
At \(x=2\): left slope \(2x=4\), right slope \(-2\) (and \(f\) is continuous: \(4=8-4\)): not differentiable (similarly \(x=-2\)).
Elsewhere \(f\) is smooth.
\(S=\{-2,-1,0,1,2\}\).
Answer: (2) equals \(\{-2,-1,0,1,2\}\)
At \(x=0\): slopes \(-1\) and \(+1\): not differentiable.
At \(x=1\): slopes \(1\) and \(2x=2\): not differentiable (similarly \(x=-1\)).
At \(x=2\): left slope \(2x=4\), right slope \(-2\) (and \(f\) is continuous: \(4=8-4\)): not differentiable (similarly \(x=-2\)).
Elsewhere \(f\) is smooth.
\(S=\{-2,-1,0,1,2\}\).
Answer: (2) equals \(\{-2,-1,0,1,2\}\)
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