Ex 6.5, 29 - The max value of [x(x - 1) + 1]1/3 is - Ex 6.5

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Slide68.JPG

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.5,29 The maximum value of [𝑥 𝑥−1﷯+1]﷮ 1﷮3﷯﷯ 0 ≤ x ≤ 1 is (A) 1﷮3﷯﷯﷮ 1﷮3﷯﷯ (B) 1﷮2﷯ (C) 1 (D) 0 Let f 𝑥﷯= 𝑥 𝑥−1﷯+1﷯﷮ 1﷮3﷯﷯ Step 1: Finding f’ 𝑥﷯ 𝑓 𝑥﷯= 𝑥 𝑥−1﷯+1﷯﷮ 1﷮3﷯﷯ 𝑓 𝑥﷯= 𝑥﷮2﷯−𝑥+1﷯﷮ 1﷮3﷯﷯ 𝑓﷮′﷯ 𝑥﷯= 𝑑 𝑥﷮2﷯ − 𝑥 + 1﷯﷮ 1﷮3﷯﷯﷮𝑑𝑥﷯ 𝑓﷮′﷯ 𝑥﷯= 1﷮3﷯ 𝑥﷮2﷯−𝑥+1﷯﷮ 1﷮3﷯ − 1﷯ . 𝑑 𝑥﷮2﷯ − 𝑥 + 1﷯﷮𝑑𝑥﷯ 𝑓﷮′﷯ 𝑥﷯= 1﷮3﷯ 𝑥﷮2﷯−𝑥+1﷯﷮ −2﷮3﷯﷯ 2𝑥−1﷯ 𝑓﷮′﷯ 𝑥﷯= 1﷮3 𝑥﷮2﷯ − 𝑥 + 1﷯﷮ 2﷮3﷯﷯﷯ . 2𝑥−1﷯ 𝑓﷮′﷯ 𝑥﷯= 2𝑥 − 1﷮3 𝑥﷮2﷯ − 𝑥 + 1﷯﷮ 2﷮3﷯﷯﷯ Step 2: Putting f’ 𝑥﷯=0 2𝑥−1﷮3 𝑥﷮2﷯ − 𝑥 + 1﷯﷮ 2﷮3﷯﷯﷯=0 2𝑥−1=0 2𝑥=1 𝑥= 1﷮2﷯ Since, 0 ≤ x ≤ 1 Hence, critical points are 𝑥=0 , 1﷮2﷯ , & 1 Hence Maximum value of is 1 at 𝑥=0 , 1 ⇒ Hence , correct answer is C

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.