Match the items of List-I with those of List- | vedclass

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(A-(III), B-(V), C-(I), D-(II)) For $A$: Let $y = 	an^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}} = 	an^{-1}\sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}} = 	an

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(A-(III), B-(V), C-(I), D-(II)) For $A$: Let $y = 	an^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}} = 	an^{-1}\sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}} = 	an^{-1}(	an(x/2)) = \frac{x}{2}$.Then $\frac{dy}{dx} = \frac{1}{2}$. Thus,$A ightarrow (iii)$.For $B$: Let $y = \frac{3+|x-1|}{3x+4}$. The function $|x-1|$ is not differentiable at $x=1$. Thus,the expression is not differentiable at $x=1$. Thus,$B ightarrow (v)$.For $C$: Let $y = \sinh^{-1} x$. We know that $\sinh^{-1} x = \log(x+\sqrt{1+x^2})$. Thus,$C ightarrow (i)$.For $D$: Let $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight) = 2	an^{-1} x$.Then $\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = 2(1+x^2)^{-1}$.Then $\frac{d^2y}{dx^2} = 2(-1)(1+x^2)^{-2}(2x) = -\frac{4x}{(1+x^2)^2}$. Thus,$D ightarrow (ii)$.The correct matching is $A-(iii), B-(v), C-(i), D-(ii)$.

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+\|x-1\|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

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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