f(x) = frac{x}{ln x} and g(x) = frac{ln x}{x} | vedclass

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(A) Given $f(x) = \frac{x}{\ln x}$ and $g(x) = \frac{\ln x}{x}$.For $f(x)$,the domain is $x \in (0, 1) \cup (1, \infty)$.For $g(x)$,the domain is $x \

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$\frac{1}{g(x)}$ and $f(x)$ are identical functions.

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(A) Given $f(x) = \frac{x}{\ln x}$ and $g(x) = \frac{\ln x}{x}$.For $f(x)$,the domain is $x \in (0, 1) \cup (1, \infty)$.For $g(x)$,the domain is $x \in (0, 1) \cup (1, \infty)$.Check $(A)$: $\frac{1}{g(x)} = \frac{1}{\frac{\ln x}{x}} = \frac{x}{\ln x} = f(x)$. Since both have the same domain $x \in (0, 1) \cup (1, \infty)$,they are identical functions.Check $(B)$: $\frac{1}{f(x)} = \frac{\ln x}{x} = g(x)$. However,$f(x)$ is undefined at $x=1$,so $\frac{1}{f(x)}$ is also undefined at $x=1$,whereas $g(1) = 0$. Thus,they are not identical.Check $(C)$ and $(D)$: $f(x) \cdot g(x) = \frac{x}{\ln x} \cdot \frac{\ln x}{x} = 1$. This is valid only for $x \in (0, 1) \cup (1, \infty)$,not for all $x > 0$ (as $x=1$ is excluded). Therefore,$(A)$ is the correct statement.

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

$f(x) = \frac{x}{\ln x}$ and $g(x) = \frac{\ln x}{x}$. Identify the $CORRECT$ statement.

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