Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
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Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important You are here
Ex 6.3,29 (MCQ)
Last updated at June 12, 2023 by Teachoo
Ex 6.3, 28 For all real values of x, the minimum value of (1 − 𝑥 + 𝑥2)/(1 + 𝑥 + 𝑥2) is (A) 0 (B) 1 (C) 3 (D) 1/3Let 𝑓(𝑥)=(1 − 𝑥 + 𝑥2)/(1 + 𝑥 + 𝑥2) Finding 𝒇′(𝒙) 𝑓(𝑥)=(1 − 𝑥 + 𝑥2)/(1 + 𝑥 + 𝑥2) 𝑓^′(𝑥) =((1 − 𝑥 + 𝑥^2 )^′ (1 + 𝑥 + 𝑥^2 ) − (1 − 𝑥 + 𝑥^2 ) (1 + 𝑥 + 𝑥^2 )^′)/(1 + 𝑥 + 𝑥^2 )^2 𝑓′(𝑥)=(−1 − 𝑥 − 𝑥^2 + 2𝑥 + 2𝑥^2+ 2𝑥^3 − (1 − 𝑥 + 𝑥^2+ 2𝑥 − 2𝑥^2 + 2𝑥^3 ))/(1 + 𝑥 + 𝑥^2 )^2 𝑓(𝑥)=(−1 + 𝑥 + 𝑥^2 + 2𝑥^3 − (1 + 𝑥 − 𝑥^2 + 2𝑥^3 ))/(1 + 𝑥 + 𝑥^2 )^2 𝑓′(𝑥)=(−2 + 2𝑥^2)/(1 + 𝑥 + 𝑥^2 )^2 Putting 𝒇^′ (𝒙)=𝟎 (−2 + 2𝑥^2)/(1 + 𝑥 + 𝑥^2 )^2 =0 2𝑥^2−2=0 2𝑥^2=2 𝑥^2=1 𝑥=±1 Hence, x = 1 or x = –1 are the critical points Finding value of 𝒇(𝒙) at critical points Hence, minimum value of f(x) is 1/3. So, (D) is the correct answer