The range of fx=x21+x2 is

fx=x2x2+1Let y=x2x2+1⇒ x2y+y=x2⇒ x2y−1+y=0∀x∈R Δ≥0⇒ b2−4ac≥0⇒ 0−4y−1y≥0⇒ yy−1≤0 and y≠1∴ y∈[0, 1)

The range of fx=x21+x2 is

fx=x2x2+1Let y=x2x2+1⇒ x2y+y=x2⇒ x2y−1+y=0∀x∈R Δ≥0⇒ b2−4ac≥0⇒ 0−4y−1y≥0⇒ yy−1≤0 and y≠1∴ y∈[0, 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

$T h e r a n g e o f f (x) = \frac{x^{2}}{1 + x^{2}} i s$

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

$[0, 1)$

$(0, 1)$

$(0, \infty)$

$(0, 2)$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = \frac{x^{2}}{x^{2} + 1} \\ L e t y = \frac{x^{2}}{x^{2} + 1} \\ \Rightarrow x^{2} y + y = x^{2} \\ \Rightarrow x^{2} (y - 1) + y = 0 \\ \forall x \in R Δ \geq 0 \Rightarrow b^{2} - 4 a c \geq 0 \\ \Rightarrow 0 - 4 (y - 1) (y) \geq 0 \\ \Rightarrow y (y - 1) \leq 0 a n d y \neq 1 \\ ∴ y \in [0, 1) \end{array}$