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


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
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 bymnjuly1970 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 bygsk2684 last updated on 10/Jul/21

	$w e l c o m e$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum