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Ex 6.5

Ex 6.5, 1 (i)
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Ex 6.5, 27 (MCQ)

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Ex 6.5,29 (MCQ) You are here

Last updated at Aug. 9, 2021 by Teachoo

Ex 6.5, 29 The maximum value of γ[π₯(π₯β1)+1]γ^(1/3) 0 β€ x β€ 1 is (A) (1/3)^(1/3) (B) 1/2 (C) 1 (D) 0 Let f(π₯)=[π₯(π₯β1)+1]^(1/3) Finding fβ(π) π(π₯)=[π₯[π₯β1]+1]^(1/3) π(π₯)=[π₯^2βπ₯+1]^(1/3) π^β² (π₯)=(π(π₯^2 β π₯ + 1)^(1/3))/ππ₯ π^β² (π₯)=1/3 (π₯^2βπ₯+1)^(1/3 β 1) . π(π₯^2 β π₯ + 1)/ππ₯ π^β² (π₯)=1/3 (π₯^2βπ₯+1)^((β2)/3) (2π₯β1) π^β² (π₯)=1/(3(π₯^2 β π₯ + 1)^(2/3) ) .(2π₯β1) π^β² (π₯)=(2π₯ β 1)/(3(π₯^2 β π₯ + 1)^(2/3) ) Putting fβ(π)=π (2π₯β1)/(3(π₯^2 β π₯ + 1)^(2/3) )=0 2π₯β1=0 2π₯=1 π₯=1/2 Since, 0 β€ x β€ 1 Hence, critical points are π₯=0 ,1/2 , & 1 Hence, Maximum value is 1 at π₯=0 , 1 β΄ Hence, correct answer is C