If R is the set of all real numbers and if f: R−{2}→R is defined by f(x)=2+x2−x for x∈R−{2}, then the range of f is

f(x)=2+x2−x=y⇒ 2+x=2y−xy⇒ x+xy=2y−2⇒ x=2y−2y+1 is definedif y+1≠0 y≠−1∴ range = R−{−1}

If R is the set of all real numbers and if f: R−{2}→R is defined by f(x)=2+x2−x for x∈R−{2}, then the range of f is

f(x)=2+x2−x=y⇒ 2+x=2y−xy⇒ x+xy=2y−2⇒ x=2y−2y+1 is definedif y+1≠0 y≠−1∴ range = R−{−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

$I f R i s t h e s e t o f a l l r e a l n u m b e r s a n d i f f : R - {2} \to R i s d e f i n e d b y f (x) = \frac{2 + x}{2 - x} f o r x \in R - {2}, t h e n t h e r a n g e o f f i s$

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

$R - {- 2}$

$R$

$R - {1}$

$R - {- 1}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = \frac{2 + x}{2 - x} = y \\ \Rightarrow 2 + x = 2 y - x y \\ \Rightarrow x + x y = 2 y - 2 \\ \Rightarrow x = \frac{2 y - 2}{y + 1} i s d e f i n e d \\ i f y + 1 \neq 0 \\ y \neq - 1 \\ ∴ r a n g e = R - {- 1} \end{array}$