The function y = frac{2x - 1}{x - 2} (x neq 2 | vedclass

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(D) Given $y = \frac{2x - 1}{x - 2}$ where $x 
eq 2$. To find the inverse,we solve for $x$ in terms of $y$: $y(x - 2) = 2x - 1$ $xy - 2y = 2x - 1$ $x

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

Vedclass · Answer

(D) Given $y = \frac{2x - 1}{x - 2}$ where $x 
eq 2$. To find the inverse,we solve for $x$ in terms of $y$: $y(x - 2) = 2x - 1$ $xy - 2y = 2x - 1$ $xy - 2x = 2y - 1$ $x(y - 2) = 2y - 1$ $x = \frac{2y - 1}{y - 2}$. Since the form of the inverse function $f^{-1}(y) = \frac{2y - 1}{y - 2}$ is identical to $f(x)$,the function is its own inverse. Now,differentiating $y$ with respect to $x$: $y' = \frac{(x - 2)(2) - (2x - 1)(1)}{(x - 2)^2} = \frac{2x - 4 - 2x + 1}{(x - 2)^2} = \frac{-3}{(x - 2)^2}$. Since $(x - 2)^2 > 0$ for all $x 
eq 2$,$y' Since both $(A)$ and $(B)$ are correct,the correct option is $(D)$.

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

The function $y = \frac{2x - 1}{x - 2}$ $(x \neq 2)$:

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