The maximum value of [x (x - 1) + 1 ] (1/3) , 0 ≤ ๐‘ฅ ≤ 1 is:

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

 

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

Slide111.JPG

Advertisement

Slide112.JPG

Advertisement

Slide113.JPG

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

Transcript

Question 43 The maximum value of ["๐‘ฅ (๐‘ฅ โˆ’ 1) + 1" ]^(1/3) , 0 โ‰ค ๐‘ฅ โ‰ค 1 is: (a) 0 (b) 1/2 (c) 1 (d) โˆ›(1/3) Let f(๐‘ฅ)=[๐‘ฅ(๐‘ฅโˆ’1)+1]^(1/3) Finding fโ€™(๐’™) ๐‘“(๐‘ฅ)=[๐‘ฅ[๐‘ฅโˆ’1]+1]^(1/3) ๐‘“(๐‘ฅ)=[๐‘ฅ^2โˆ’๐‘ฅ+1]^(1/3) ๐‘“^โ€ฒ (๐‘ฅ)=(๐‘‘(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)^(1/3))/๐‘‘๐‘ฅ ๐‘“^โ€ฒ (๐‘ฅ)=1/3 (๐‘ฅ^2โˆ’๐‘ฅ+1)^(1/3 โˆ’ 1) . ๐‘‘(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)/๐‘‘๐‘ฅ ๐‘“^โ€ฒ (๐‘ฅ)=1/3 (๐‘ฅ^2โˆ’๐‘ฅ+1)^((โˆ’2)/3) (2๐‘ฅโˆ’1) ๐‘“^โ€ฒ (๐‘ฅ)=1/(3(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)^(2/3) ) .(2๐‘ฅโˆ’1) ๐‘“^โ€ฒ (๐‘ฅ)=(2๐‘ฅ โˆ’ 1)/(3(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)^(2/3) ) Putting fโ€™(๐’™)=๐ŸŽ (2๐‘ฅโˆ’1)/(3(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)^(2/3) )=0 2๐‘ฅโˆ’1=0 2๐‘ฅ=1 ๐’™=๐Ÿ/๐Ÿ Since, 0 โ‰ค x โ‰ค 1 Hence, critical points are ๐’™=๐ŸŽ ,๐Ÿ/๐Ÿ , & 1 Hence, Maximum value is 1 at ๐‘ฅ=0 , 1 So, the correct answer is (C)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.