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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
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5 , when $\alpha < 0$
Since, $\alpha \in R-\{-1\}$
$\begin{aligned} & \text { and } \mathrm{f}(\mathrm{x})=|(|x|+\alpha)(|x|-1)| \\ & |x|+\alpha=0 \Rightarrow|x|=-\alpha \Rightarrow \alpha < 0 \\ & |x|-1=0 \Rightarrow x= \pm 1\end{aligned}$
Now, graph of $f(x)$ is
It is clear from graph that there are 5 points at which $\mathrm{f}(\mathrm{x})$ is not differentiable and $\alpha < 0$
$\begin{aligned} & \text { and } \mathrm{f}(\mathrm{x})=|(|x|+\alpha)(|x|-1)| \\ & |x|+\alpha=0 \Rightarrow|x|=-\alpha \Rightarrow \alpha < 0 \\ & |x|-1=0 \Rightarrow x= \pm 1\end{aligned}$
Now, graph of $f(x)$ is
It is clear from graph that there are 5 points at which $\mathrm{f}(\mathrm{x})$ is not differentiable and $\alpha < 0$
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