If f(x)=ex+e−x2 then the inverse of f(x) is

f(x)=ex+e−x2 y=ex+e−x2 ⇒ex+1ex=2y ⇒(ex)2-2yex+1=0 ⇒ex=2y±4y2-42=y±y2-1 x=log(y±y2-1)

If f(x)=ex+e−x2 then the inverse of f(x) is

f(x)=ex+e−x2 y=ex+e−x2 ⇒ex+1ex=2y ⇒(ex)2-2yex+1=0 ⇒ex=2y±4y2-42=y±y2-1 x=log(y±y2-1)

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $f (x) = \frac{e^{x} + e^{- x}}{2}$ then the inverse of $f (x)$ is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\log_{e} (x - \sqrt{x^{2} - 1})$

$\log_{e} (x + \sqrt{x^{2} - 1})$

$\log_{e} (x + \sqrt{x^{2} + 1})$

$\log_{e} \sqrt{x^{2} + 1}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = \frac{e^{x} + e^{- x}}{2} y = \frac{e^{x} + e^{- x}}{2} \Rightarrow e^{x} + \frac{1}{e^{x}} = 2 y \Rightarrow {(e^{x})}^{2} - 2 {ye}^{x} + 1 = 0 \Rightarrow e^{x} = \frac{2 y \pm \sqrt{4 y^{2} - 4}}{2} = y \pm \sqrt{y^{2} - 1} x = \log (y \pm \sqrt{y^{2} - 1})$