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 You are here
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 June 12, 2023 by Teachoo
Ex 6.3, 10 Find the maximum value of 2đĽ3 â 24đĽ + 107 in the interval [1, 3]. Find the maximum value of the same function in [â3, â1]. Let f(đĽ)=2đĽ^3â24đĽ+107 Finding fâ(đĽ) fâ(đĽ)=đ(2đĽ^3 â 24đĽ + 107)/đđĽ = 2 Ă 3đĽ^2â24 = 6đĽ^2â24 = 6 (đĽ^2â4" " ) Putting f(đĽ)=0 6 (đĽ^2â4" " )=0 đĽ^2â4" = 0 " đĽ^2=4 đĽ=Âąâ4 đĽ=Âą2 Thus, đĽ=2 , â 2 Since x is in interval [1 , 3] đĽ = 2 is only Critical point Also, since given the interval đĽ= â [1 , 3] We calculate f(x) at đĽ= 1 , 2 & 3 Hence maximum value of f(đ)=đđ at đ = 3 in the interval [1 , 3] For the interval [âđ , âđ] đĽ = â2 is only Critical Point Also, since given the interval đĽ= â [â3,â1] We calculate f(x) at đĽ= â1 , â2 & â3 Hence maximum value of f(đ)=đđđ at đ = â2 in the interval [â3,â1]