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Class 12
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The maximum value of [x (x - 1) + 1 ] (1/3) , 0 ≀ π‘₯ ≀ 1 is:

(a) 0Β  Β  Β  Β (b) 1/2Β  Β  Β  Β  Β (c) 1Β  Β  Β (d) βˆ›(1/3)

Β 

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

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Question 43 The maximum value of ["π‘₯ (π‘₯ βˆ’ 1) + 1" ]^(1/3) , 0 ≀ π‘₯ ≀ 1 is: (a) 0 (b) 1/2 (c) 1 (d) βˆ›(1/3) Let f(π‘₯)=[π‘₯(π‘₯βˆ’1)+1]^(1/3) Finding f’(𝒙) 𝑓(π‘₯)=[π‘₯[π‘₯βˆ’1]+1]^(1/3) 𝑓(π‘₯)=[π‘₯^2βˆ’π‘₯+1]^(1/3) 𝑓^β€² (π‘₯)=(𝑑(π‘₯^2 βˆ’ π‘₯ + 1)^(1/3))/𝑑π‘₯ 𝑓^β€² (π‘₯)=1/3 (π‘₯^2βˆ’π‘₯+1)^(1/3 βˆ’ 1) . 𝑑(π‘₯^2 βˆ’ π‘₯ + 1)/𝑑π‘₯ 𝑓^β€² (π‘₯)=1/3 (π‘₯^2βˆ’π‘₯+1)^((βˆ’2)/3) (2π‘₯βˆ’1) 𝑓^β€² (π‘₯)=1/(3(π‘₯^2 βˆ’ π‘₯ + 1)^(2/3) ) .(2π‘₯βˆ’1) 𝑓^β€² (π‘₯)=(2π‘₯ βˆ’ 1)/(3(π‘₯^2 βˆ’ π‘₯ + 1)^(2/3) ) Putting f’(𝒙)=𝟎 (2π‘₯βˆ’1)/(3(π‘₯^2 βˆ’ π‘₯ + 1)^(2/3) )=0 2π‘₯βˆ’1=0 2π‘₯=1 𝒙=𝟏/𝟐 Since, 0 ≀ x ≀ 1 Hence, critical points are 𝒙=𝟎 ,𝟏/𝟐 , & 1 Hence, Maximum value is 1 at π‘₯=0 , 1 So, the correct answer is (C)

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