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(C) For $x \in [-2, 2]$,$f(x) = \min \{|x|, 2-x^2\}$.Solving $|x| = 2-x^2$,we get $x^2 + |x| - 2 = 0$,which gives $(|x|+2)(|x|-1) = 0$. Since $|x| \ge

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$5$

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(C) For $x \in [-2, 2]$,$f(x) = \min \{|x|, 2-x^2\}$.Solving $|x| = 2-x^2$,we get $x^2 + |x| - 2 = 0$,which gives $(|x|+2)(|x|-1) = 0$. Since $|x| \geq 0$,we have $|x| = 1$,so $x = \pm 1$.Thus,$f(x) = |x|$ for $x \in [-1, 1]$ and $f(x) = 2-x^2$ for $x \in [-2, -1) \cup (1, 2]$.Points of non-differentiability in $(-2, 2)$ are $x = -1, 0, 1$ (due to $|x|$ and the intersection of curves).For $2 For $x \in (2, 3]$,$f(x) = [x] = 2$ for $x \in (2, 3)$ and $f(3) = 3$. This is discontinuous at $x = 3$ (not in $(-3, 3)$) and $x = 2$ (jump discontinuity).For $x \in [-3, -2)$,$f(x) = [|x|] = [|x|]$. For $x \in (-3, -2)$,$f(x) = 2$. At $x = -2$,$f(-2) = 2$. At $x = -3$,$f(-3) = 3$.Checking points: $x = -2, -1, 0, 1, 2$. At these $5$ points,the function is either discontinuous or has a sharp corner.Therefore,the number of points of non-differentiability is $5$.

$A$ function $f$ is defined on $[-3,3]$ as
$f(x) = \begin{cases} \min \{|x|, 2-x^{2}\} & , -2 \leq x \leq 2 \\ [|x|] & , 2 < |x| \leq 3 \end{cases}$
where $[x]$ denotes the greatest integer $\leq x$. The number of points,where $f$ is not differentiable in $(-3,3)$ is

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$A$ function $f$ is defined on $[-3,3]$ as$f(x) = \begin{cases} \min \{|x|, 2-x^{2}\} & , -2 \leq x \leq 2 \\ [|x|] & , 2 < |x| \leq 3 \end{cases}$where $[x]$ denotes the greatest integer $\leq x$. The number of points,where $f$ is not differentiable in $(-3,3)$ is