# Ex 6.5,29 - Chapter 6 Class 12 Application of Derivatives

Last updated at Sept. 25, 2018 by Teachoo

Last updated at Sept. 25, 2018 by Teachoo

Transcript

Ex 6.5,29 The maximum value of [𝑥 𝑥−1+1] 13 0 ≤ x ≤ 1 is (A) 13 13 (B) 12 (C) 1 (D) 0 Let f 𝑥= 𝑥 𝑥−1+1 13 Step 1: Finding f’ 𝑥 𝑓 𝑥= 𝑥 𝑥−1+1 13 𝑓 𝑥= 𝑥2−𝑥+1 13 𝑓′ 𝑥= 𝑑 𝑥2 − 𝑥 + 1 13𝑑𝑥 𝑓′ 𝑥= 13 𝑥2−𝑥+1 13 − 1 . 𝑑 𝑥2 − 𝑥 + 1𝑑𝑥 𝑓′ 𝑥= 13 𝑥2−𝑥+1 −23 2𝑥−1 𝑓′ 𝑥= 13 𝑥2 − 𝑥 + 1 23 . 2𝑥−1 𝑓′ 𝑥= 2𝑥 − 13 𝑥2 − 𝑥 + 1 23 Step 2: Putting f’ 𝑥=0 2𝑥−13 𝑥2 − 𝑥 + 1 23=0 2𝑥−1=0 2𝑥=1 𝑥= 12 Since, 0 ≤ x ≤ 1 Hence, critical points are 𝑥=0 , 12 , & 1 Hence Maximum value of is 1 at 𝑥=0 , 1 ⇒ Hence , correct answer is C

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

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.