

Now learn Economics at Teachoo for Class 12
Ex 6.5
Ex 6.5, 1 (ii)
Ex 6.5, 1 (iii) Important
Ex 6.5, 1 (iv)
Ex 6.5, 2 (i)
Ex 6.5, 2 (ii) Important
Ex 6.5, 2 (iii)
Ex 6.5, 2 (iv) Important
Ex 6.5, 2 (v) Important
Ex 6.5, 3 (i)
Ex 6.5, 3 (ii)
Ex 6.5, 3 (iii)
Ex 6.5, 3 (iv) Important
Ex 6.5, 3 (v)
Ex 6.5, 3 (vi)
Ex 6.5, 3 (vii) Important
Ex 6.5, 3 (viii)
Ex 6.5, 4 (i)
Ex 6.5, 4 (ii) Important
Ex 6.5, 4 (iii)
Ex 6.5, 5 (i)
Ex 6.5, 5 (ii)
Ex 6.5, 5 (iii) Important
Ex 6.5, 5 (iv)
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important
Ex 6.5,15 Important
Ex 6.5,16
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
Ex 6.5,23 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
Ex 6.5, 27 (MCQ)
Ex 6.5,28 (MCQ) Important You are here
Ex 6.5,29 (MCQ)
Last updated at Aug. 9, 2021 by Teachoo
Ex 6.5, 28 For all real values of x, the minimum value of (1 β π₯ + π₯2)/(1 + π₯ + π₯2) is (A) 0 (B) 1 (C) 3 (D) 1/3Let π(π₯)=(1 β π₯ + π₯2)/(1 + π₯ + π₯2) Finding πβ²(π) π(π₯)=(1 β π₯ + π₯2)/(1 + π₯ + π₯2) π^β²(π₯) =((1 β π₯ + π₯^2 )^β² (1 + π₯ + π₯^2 ) β (1 β π₯ + π₯^2 ) (1 + π₯ + π₯^2 )^β²)/(1 + π₯ + π₯^2 )^2 Using quotient Rule As (π’/π£)^β²=(π’^β² π£βπ£^β² π’)/π£^2 πβ²(π₯)=(β1 β π₯ β π₯^2 + 2π₯ + 2π₯^2+ 2π₯^3 β (1 β π₯ + π₯^2+ 2π₯ β 2π₯^2 + 2π₯^3 ))/(1 + π₯ + π₯^2 )^2 π(π₯)=(β1 + π₯ + π₯^2 + 2π₯^3 β (1 + π₯ β π₯^2 + 2π₯^3 ))/(1 + π₯ + π₯^2 )^2 πβ²(π₯)=(β2 + 2π₯^2)/(1 + π₯ + π₯^2 )^2 Putting π^β² (π)=π (β2 + 2π₯^2)/(1 + π₯ + π₯^2 )^2 =0 2π₯^2β2=0 2π₯^2=2 π₯^2=1 π₯=Β±1 Hence, x = 1 or x = β1 are the critical points Finding value of π(π) at critical points Hence, minimum value of f(x) is 1/3. So, (D) is the correct answer