If alpha in R - {-1} and f(x) = |(|x| + alpha | vedclass

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(D) Given $f(x) = |(|x| + \alpha)(|x| - 1)|$. Let $g(x) = (|x| + \alpha)(|x| - 1)$. The function $f(x) = |g(x)|$ is not differentiable at the points w

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$5$,when $\alpha < 0$

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(D) Given $f(x) = |(|x| + \alpha)(|x| - 1)|$. Let $g(x) = (|x| + \alpha)(|x| - 1)$. The function $f(x) = |g(x)|$ is not differentiable at the points where $g(x) = 0$ (provided the roots are simple) and at $x = 0$ due to the $|x|$ term. Case $1$: If $\alpha > 0$,then $|x| + \alpha = 0$ has no real solution. The roots of $g(x) = 0$ are $|x| = 1$,i.e.,$x = 1, -1$. At $x = 1, -1$,the function $f(x)$ has sharp corners. Also,at $x = 0$,$f(x) = |\alpha(-1)| = |-\alpha| = \alpha$. Since $f'(0^+)$ and $f'(0^-)$ will differ due to the $|x|$ term,$f(x)$ is not differentiable at $x = 0$. Thus,for $\alpha > 0$,there are $3$ points of non-differentiability $(x = -1, 0, 1)$. Case $2$: If $\alpha  0$ and $k 
eq 1$. Then $g(x) = (|x| - k)(|x| - 1)$. The roots are $|x| = k$ and $|x| = 1$,which gives $x = \pm k$ and $x = \pm 1$. These are $4$ distinct points where $f(x) = 0$. Additionally,$f(x)$ is not differentiable at $x = 0$ because of the $|x|$ term. Thus,for $\alpha Comparing with the options,$D$ is correct.

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

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