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
- A 3 , when \(\alpha < 0\)
- B 5 , when \(\alpha>0\)
- C 4 , when \(\alpha>0\)
- D 5 , when \(\alpha < 0\)
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…
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