# Ex 6.3, 3 (viii) - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

Ex 6.3

Ex 6.3, 1 (i)
Important

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) You are here

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

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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

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Ex 6.3,22 Important

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Ex 6.3,24 Important

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Ex 6.3, 26 Important

Ex 6.3, 27 (MCQ)

Ex 6.3,28 (MCQ) Important

Ex 6.3,29 (MCQ)

Last updated at April 16, 2024 by Teachoo

Ex 6.3, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (viii) f(đĽ) = đĽâ(1âđĽ), đĽ > 0= â(1âđĽ) + 1/(2â(1 â đĽ)) (0 â1) . đĽ = â(1âđĽ) â đĽ/(2â(1 â đĽ)) = (2(â(1 â đĽ) )^2â đĽ)/(2â(1 â đĽ)) = (2(1 â đĽ) â đĽ)/(2â(1 â đĽ)) = (2 â 2đĽ â đĽ)/(2â(1 â đĽ)) = (2 â 3đĽ)/(2â(1 â đĽ)) Putting fâ(đ)=đ (2 â 3đĽ)/(2â(1 â đĽ))=0 2 â 3đĽ = 0 Ă 2â(1âđĽ) 2 â 3đĽ=0 â 3đĽ=â2 đĽ =2/3 Finding fââ(đ) fâ(đĽ)=(2 â 3đĽ)/(2â(1 â đĽ)) fââ(đĽ)=đ/đđĽ ((2 â 3đĽ)/(2â(1 â đĽ))) = 1/2 [(đ(2 â 3đĽ)/đđĽ . â(1 â đĽ) â đ(â(1 â đĽ))/đđĽ . (2 â 3đĽ))/(â(1 â đĽ))^2 ] = 1/2 [((0 â 3) â(1 â đĽ) â 1/(2â(1 â đĽ)) . đ(1 â đĽ)/đđĽ . (2 â 3đĽ))/((1 â đĽ) )] = 1/2 [(â3â(1 â đĽ) â 1/(2â(1 â đĽ)) (0 â 1) . (2 â 3đĽ))/((1 â đĽ) )] = 1/2 [(â3â(1 â đĽ) + (2 â 3đĽ)/(2â(1 â đĽ)) )/(1 â đĽ)] = 1/2 [((â3â(1 â đĽ)) (2â(1 â đĽ)) + 2 â 3đĽ )/(2(1 â đĽ) â(1 â đĽ))] = 1/2 [(â6(1 â đĽ) + 2 â 3đĽ )/(2(1 â đĽ) â(1 â đĽ))] = 1/2 [(â6 + 6đĽ + 2 â 3đĽ )/(2(1 â đĽ) â(1 â đĽ))] = 1/4 [(â4 + 3đĽ )/(1 + đĽ)^(3/2) ] Hence, fââ(đĽ)=1/4 [(â4 + 3đĽ )/(1 + đĽ)^(3/2) ] Putting đĽ=2/3 fââ(2/3)=1/4 [(â4 + 3(2/3))/(1 + 2/3)^(3/2) ] =1/4 [(â4 + 2)/(5/3)^(3/2) ] =1/4 [(â2)/(5/3)^(3/2) ] = (â 1)/2 (3/5)^(3/2) < 0 Since fââ(đĽ)<0 when đĽ = 2/3 Hence, đĽ=2/3 is the maxima Finding Maximum value of f(đ)=đâ(đâđ) Putting đĽ=2/3 f(2/3)=2/3 â(1â2/3) =2/3 â((3 â 2)/3) Finding Maximum value of f(đ)=đâ(đâđ) Putting đĽ=2/3 f(đĽ) = đĽâ(1âđĽ) f(2/3)=2/3 â(1â2/3) =2/3 â((3 â 2)/3) =2/3 â(1/3) =2/(3â3) =2/(3â3) Ă â3/â3 =(2â3)/9 Maximum value of f(đ) is (đâđ)/đ at x = đ/đ