For all real values x, the minimum value of 1−x+x21+x+x2 is

y=1−x+x21+x+x2 (y−1)x2+(y+1)x+(y−1)=0 x∈R ⇒ Δ ≥ 0 ⇒(y+1)2−4(y−1)2≥0 ⇒−3y2+10y−3≥0 ⇒3y2−10y+3≤0 ⇒(3y−1) (y−3) ≤0 ⇒y∈[13,3] Minimum value = 13

For all real values x, the minimum value of 1−x+x21+x+x2 is

y=1−x+x21+x+x2 (y−1)x2+(y+1)x+(y−1)=0 x∈R ⇒ Δ ≥ 0 ⇒(y+1)2−4(y−1)2≥0 ⇒−3y2+10y−3≥0 ⇒3y2−10y+3≤0 ⇒(3y−1) (y−3) ≤0 ⇒y∈[13,3] Minimum value = 13

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

For all real values x, the minimum value of $\frac{1 - x + x^{2}}{1 + x + x^{2}}$ is

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

$\frac{1}{3}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$y = \frac{1 - x + x^{2}}{1 + x + x^{2}} (y - 1) x^{2} + (y + 1) x + (y - 1) = 0 x \in R \Rightarrow Δ \geq 0 \Rightarrow {(y + 1)}^{2} - 4 {(y - 1)}^{2} \geq 0 \Rightarrow - 3 y^{2} + 10 y - 3 \geq 0 \Rightarrow 3 y^{2} - 10 y + 3 \leq 0 \Rightarrow (3 y - 1) (y - 3) \leq 0 \Rightarrow y \in [\frac{1}{3}, 3] M i n i m u m v a l u e = \frac{1}{3}$