Let f:(-2,2) rightarrow mathbb{R} be defined | vedclass

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(D) The function is defined as $f(x) = \begin{cases} x[x] & , -2 < x < 0 \ (x-1)[x] & , 0 \leq x < 2 \end{cases}$.For $-2 < x < -1$,$[x] = -2$,so $f(

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$4$

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(D) The function is defined as $f(x) = \begin{cases} x[x] & , -2 For $-2 For $-1 \leq x For $0 \leq x For $1 \leq x Now consider $g(x) = |f(x)|$:$g(x) = \begin{cases} |-2x| = -2x & , -2 Actually,for $1 \leq x Checking continuity:At $x = -1$: $LHL = \lim_{x 	o -1^-} (-2x) = 2$,$RHL = \lim_{x 	o -1^+} (-x) = 1$. Discontinuous at $x = -1$.At $x = 0$: $LHL = \lim_{x 	o 0^-} (-x) = 0$,$RHL = f(0) = 0$. Continuous at $x = 0$.At $x = 1$: $LHL = \lim_{x 	o 1^-} (0) = 0$,$RHL = f(1) = 1-1 = 0$. Continuous at $x = 1$.Thus,$m = 1$ (point of discontinuity is $x = -1$).Checking differentiability:At $x = -1$: Not continuous,so not differentiable.At $x = 0$: $LHD = \frac{d}{dx}(-x) = -1$,$RHD = \frac{d}{dx}(0) = 0$. Not differentiable.At $x = 1$: $LHD = \frac{d}{dx}(0) = 0$,$RHD = \frac{d}{dx}(x-1) = 1$. Not differentiable.Thus,$n = 3$ (points of non-differentiability are $x = -1, 0, 1$).Therefore,$m + n = 1 + 3 = 4$.

Column $I$	Column $II$
$A$. $x\|x\|$	$I$. Strictly increasing and continuous in $(-1,1)$
$B$. $\sqrt{\|x\|}$	$II$. Continuous but not differentiable in $(-1,1)$
$C$. $x+[x]$	$III$. Differentiable in $(-1,1)$
$D$. $\|x-1\|+\|x+1\|+\|x\|$	$IV$. Differentiable in $(-1,0) \cup (0,1)$
	$V$. Strictly increasing and not differentiable in $(-1,1)$

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