Match the following: List-IList-IIA. frac{x}{ | vedclass

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(C) Step $1$: Analyze $A$. Let $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. Check $f(-x) = \frac{-x}{e^{-x}-1} + \frac{-x}{2} + 4 = \frac{-x e^x}{1-e^x

Vedclass · Accepted Answer

$A-II, B-III, C-IV, D-I$

Vedclass · Answer

(C) Step $1$: Analyze $A$. Let $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. Check $f(-x) = \frac{-x}{e^{-x}-1} + \frac{-x}{2} + 4 = \frac{-x e^x}{1-e^x} - \frac{x}{2} + 4 = \frac{x e^x}{e^x-1} - \frac{x}{2} + 4$. Since $\frac{e^x}{e^x-1} = 1 + \frac{1}{e^x-1}$,$f(-x) = x(1 + \frac{1}{e^x-1}) - \frac{x}{2} + 4 = x + \frac{x}{e^x-1} - \frac{x}{2} + 4 = \frac{x}{e^x-1} + \frac{x}{2} + 4 = f(x)$. Thus,$A$ is an even function $(II)$.Step $2$: Analyze $B$. Let $f(x) = 	an^{-1}(\log|x+\sqrt{x^2+1}|)$. Since $x > 0$,the domain is restricted. However,considering the nature of the expression,it is neither odd nor even $(I)$.Step $3$: Analyze $C$. For $3 Step $4$: Analyze $D$. $f(x) = \sin 2x + \sin^2 x + \cos 3x$. This is a combination of trigonometric functions and is neither odd nor even $(I)$.Wait,re-evaluating $A$: $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. This is a known even function. $C$ is identity. $B$ is neither. $D$ is neither. Checking options,$A-II$ is correct. $C-IV$ is correct. Thus,option $C$ is the correct match.

List-$I$	List-$II$
$A$. $\frac{x}{e^x-1} + \frac{x}{2} + 4; x \neq 0$	$I$. is neither odd nor even function
$B$. $\tan^{-1}(\log\|x+\sqrt{x^2+1}\|), x > 0$	$II$. is an even function
$C$. For $3 < x < 5, \|x-2\|+\|x-3\|+\|x-5\|$	$III$. is an odd function
$D$. $\sin 2x + \sin^2 x + \cos 3x, \forall x \in \mathbb{R}$	$IV$. is the identity function
	$V$. is a constant function

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