CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

(a) 0Β  Β  Β  Β (b) 1/2Β  Β  Β  Β  Β (c) 1Β  Β  Β (d) β(1/3)

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This question is inspired from Ex 6.5,29 (MCQ) - Chapter 6 Class 12 - Application of Derivatives

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Transcript

Question 43 The maximum value of ["π₯ (π₯ β 1) + 1" ]^(1/3) , 0 β€ π₯ β€ 1 is: (a) 0 (b) 1/2 (c) 1 (d) β(1/3) Let f(π₯)=[π₯(π₯β1)+1]^(1/3) Finding fβ(π) π(π₯)=[π₯[π₯β1]+1]^(1/3) π(π₯)=[π₯^2βπ₯+1]^(1/3) π^β² (π₯)=(π(π₯^2 β π₯ + 1)^(1/3))/ππ₯ π^β² (π₯)=1/3 (π₯^2βπ₯+1)^(1/3 β 1) . π(π₯^2 β π₯ + 1)/ππ₯ π^β² (π₯)=1/3 (π₯^2βπ₯+1)^((β2)/3) (2π₯β1) π^β² (π₯)=1/(3(π₯^2 β π₯ + 1)^(2/3) ) .(2π₯β1) π^β² (π₯)=(2π₯ β 1)/(3(π₯^2 β π₯ + 1)^(2/3) ) Putting fβ(π)=π (2π₯β1)/(3(π₯^2 β π₯ + 1)^(2/3) )=0 2π₯β1=0 2π₯=1 π=π/π Since, 0 β€ x β€ 1 Hence, critical points are π=π ,π/π , & 1 Hence, Maximum value is 1 at π₯=0 , 1 So, the correct answer is (C)