if g(x)=((x^( 2) −x)/(2x−1)) , D_g = [1 , ∞) , lim_(x→∞) ((g^( −1) (x))/(ax + b)) = b−a (a <0 ) then find the value of Max (b ) D


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Question Number 146063 by mnjuly1970 last updated on 10/Jul/21

$i f g (x) = \frac{x^{2} - x}{2 x - 1}, D_{g} = [1, \infty)$ $, {l i m}_{x \to \infty} \frac{g^{- 1} (x)}{a x + b} = b - a (a < 0)$ $t h e n f i n d t h e v a l u e o f M a x (b)$ $D_{g} = D o m a i n$
	Answered by gsk2684 last updated on 10/Jul/21

	$g^{- 1} (x) = y \Rightarrow x = g (y) = \frac{y^{2} - y}{2 y - 1}$ $\Rightarrow y = \frac{2 x + 1 + \sqrt{4 x^{2} + 1}}{2}$ $t h e n \lim_{x \to \infty} \frac{2 x + 1 + \sqrt{4 x^{2} + 1}}{2 (a x + b)} = b - a$ $\lim_{x \to \infty} \frac{2 x + 1 + x \sqrt{4 + \frac{1}{x^{2}}}}{2 (a x + b)} = b - a$ $\lim_{\frac{1}{x} \to 0} \frac{2 + \frac{1}{x} + \sqrt{4 + \frac{1}{x^{2}}}}{2 (a + \frac{b}{x})} = b - a$ $\frac{2 + \sqrt{4}}{2 a} = b - a$ $b = a + \frac{2}{a} ⩽ - 2 \sqrt{a . \frac{2}{a}} (∵ a < 0)$ $b ⩽ - 2 \sqrt{2}$ $m a x o f b i s - 2 \sqrt{2}$
	Commented by mnjuly1970 last updated on 10/Jul/21

	$v e r y n i c e m r g . s . k t h a n k y o u . .$
	Commented by gsk2684 last updated on 10/Jul/21

	$w e l c o m e$
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