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AP EAMCET · Maths · Continuity and Differentiability

If \(\alpha \in \mathbb{R}-\{-1\}\) and \(\mathrm{f}(\mathrm{x})=|(|\mathrm{x}|+\alpha)(|\mathrm{x}|-1)|\), then the number of points at which \(f(x)\) is not differentiable, is

  1. A 3 , when \(\alpha < 0\)
  2. B 5 , when \(\alpha>0\)
  3. C 4 , when \(\alpha>0\)
  4. D 5 , when \(\alpha < 0\)
Verified Solution

Answer & Solution

Correct Answer

(D) 5 , when \(\alpha < 0\)

Step-by-step Solution

Detailed explanation

Since, \(\alpha \in R-\{-1\}\) \(\begin{aligned} & \text { and } \mathrm{f}(\mathrm{x})=|(|x|+\alpha)(|x|-1)| \\ & |x|+\alpha=0 \Rightarrow|x|=-\alpha \Rightarrow \alpha It is clear from graph that there are 5 points at which…