If f:[1, ∞)→[1,∞) is defined by f(x)=2x(x−1) then f−1(x)=

Let f−1(x)=y⇒ x=f(y)⇒ x=2y(y−1)⇒ log2x=y2−y⇒ y2−y−log2x=0∴ y=1±1+4log2x2∵ y>0⇒ y=1+1+4log2x2

If f:[1, ∞)→[1,∞) is defined by f(x)=2x(x−1) then f−1(x)=

Let f−1(x)=y⇒ x=f(y)⇒ x=2y(y−1)⇒ log2x=y2−y⇒ y2−y−log2x=0∴ y=1±1+4log2x2∵ y>0⇒ y=1+1+4log2x2

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

$I f f : [1, \infty) \to [1, \infty) i s d e f i n e d b y f (x) = 2^{x (x - 1)} t h e n f^{- 1} (x) =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{x (x - 1)}{2}$

$\frac{(1 + \sqrt{1 + 4 \log_{2} x})}{2}$

$\frac{(1 - \sqrt{1 + 4 \log_{2} x})}{2}$

$2 x (x + 1)$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} L e t f^{- 1} (x) = y \Rightarrow x = f (y) \\ \Rightarrow x = 2^{y (y - 1)} \\ \Rightarrow \log_{2}^{x} = y^{2} - y \\ \Rightarrow y^{2} - y - \log_{2}^{x} = 0 \\ ∴ y = \frac{1 \pm \sqrt{1 + 4 \log_{2}^{x}}}{2} \\ ∵ y > 0 \Rightarrow y = \frac{1 + \sqrt{1 + 4 \log_{2}^{x}}}{2} \end{array}$