Question:

The function
\(f(x) = \begin{cases} \frac{\pi}{4} + \tan^{-1}x, & |x| \le 1 \\ \frac12(|x| - 1), & |x| \gt 1 \end{cases}\)
is :

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Check the one-sided limits and one-sided derivatives at the break points \(x = 1\) and \(x = -1\).
Updated On: Nov 5, 2026
  • continuous on \(R - \{-1\}\) and differentiable on \(R - \{-1, 1\}\)
  • both continuous and differentiable on \(R - \{1\}\)
  • both continuous and differentiable on \(R - \{-1\}\)
  • continuous on \(R - \{1\}\) and differentiable on \(R - \{-1, 1\}\)
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The Correct Option is D

Solution and Explanation

At \(x = 1\): LHL \(= \frac\pi4 + \frac\pi4 = \frac\pi2\), RHL \(= \frac12(1-1) = 0\). Not equal, so f is discontinuous at \(x=1\) (hence also not differentiable there).
At \(x = -1\): LHL \(= \frac12(1-1) = 0\); RHL \(= \frac\pi4 + \tan^{-1}(-1) = 0\); \(f(-1) = 0\). So f is continuous at \(x=-1\).
Derivative at \(-1\): for \(x \lt -1\), \(f = \frac12(-x-1)\), \(f' = -\frac12\); for \(x \gt -1\), \(f' = \frac{1}{1+x^2} = \frac12\). Left and right derivatives differ, so f is not differentiable at \(-1\).
Everywhere else f is built from continuous and differentiable pieces.
Hence f is continuous on \(R - \{1\}\) and differentiable on \(R - \{-1, 1\}\).
Answer: (4) continuous on \(R - \{1\}\) and differentiable on \(R - \{-1, 1\}\)
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