
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) You are here
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
Ex 6.3,19 Important
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
Ex 6.3,29 (MCQ)
Last updated at May 29, 2023 by Teachoo
Ex 6.3, 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