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(B) $f(x) = x|x|$. This is continuous and differentiable everywhere,including $x=0$. Since $f'(x) = 2|x| \ge 0$,it is strictly increasing. Thus,$(A) \

Vedclass · Accepted Answer

$A-(p, q, r), B-(p, s), C-(r, s), D-(p, q)$

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(B) $f(x) = x|x|$. This is continuous and differentiable everywhere,including $x=0$. Since $f'(x) = 2|x| \ge 0$,it is strictly increasing. Thus,$(A) 	o (p, q, r)$.$(B)$ $f(x) = \sqrt{|x|}$. This is continuous everywhere. At $x=0$,it has a cusp,so it is not differentiable at $x=0$. Thus,$(B) 	o (p, s)$.$(C)$ $f(x) = x + [x]$. This function is discontinuous at every integer point. In $(-1, 1)$,it is discontinuous at $x=0$. It is strictly increasing on $(-1, 0)$ and $[0, 1)$. It is not differentiable at $x=0$. Thus,$(C) 	o (r, s)$.$(D)$ $f(x) = |x - 1| + |x + 1|$. In $(-1, 1)$,$f(x) = (1 - x) + (x + 1) = 2$. This is a constant function,which is continuous and differentiable everywhere. Thus,$(D) 	o (p, q)$.

Column $I$	Column $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

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