
Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
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) You are here
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
Ex 6.5,29 (MCQ)
Last updated at Aug. 19, 2021 by Teachoo
Ex 6.5, 4 Prove that the following functions do not have maxima or minima: (iii) β(π₯)= π₯^3+π₯^2+π₯+1Given h (x) = x3 + x2 + x + 1 Finding maxima or minima ββ (π₯) = 3π₯^2+2π₯+1 Putting ββ (π₯)= 0 3π₯^2+2π₯+1=0 For ax2 + bx + c = 0 x = (βπ Β± β(π^2 β 4ππ))/2π Here π = 3, b = 2, & c = 1 x = (β 2 Β± β(4 β 4(3)(1)))/6 x = (β 2 Β± β(4 β 12))/6 x = (β2 Β± β(β 8))/6 x = (β 2 Β± 2β(β 2))/6 x = (βπ Β± β(β π))/π Since root has minus sign, x has no real value β΄ h (x) does not have a maxima of minima