Ex 6.3,29 (MCQ) - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
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
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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
Ex 6.3,29 (MCQ) You are here
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 29 The maximum value of ă[đĽ(đĽâ1)+1]ă^(1/3) 0 ⤠x ⤠1 is (A) (1/3)^(1/3) (B) 1/2 (C) 1 (D) 0 đ^Ⲡ(đĽ)=1/3 (đĽ^2âđĽ+1)^((â2)/3) (2đĽâ1) đ^Ⲡ(đĽ)=1/(3(đĽ^2 â đĽ + 1)^(2/3) ) .(2đĽâ1) đ^Ⲡ(đĽ)=(2đĽ â 1)/(3(đĽ^2 â đĽ + 1)^(2/3) ) Putting fâ(đ)=đ (2đĽâ1)/(3(đĽ^2 â đĽ + 1)^(2/3) )=0 2đĽâ1=0 2đĽ=1 đĽ=1/2 Since, 0 ⤠x ⤠1 Hence, critical points are đĽ=0 ,1/2 , & 1 Hence, Maximum value is 1 at đĽ=0 , 1 â´ Hence, correct answer is C