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Question 16 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by Teachoo

f (x) = xx has a stationary point at

(A) x = e                      (B) x = 1/e

(C) x = 1                      (D) x = √e

Transcript

Question 16 f (x) = xx has a stationary point at (A) x = e (B) x = 1/๐‘’ (C) x = 1 (D) x = โˆš๐‘’ A stationary point of a function is a point where ๐’‡โ€ฒ(๐’™) = 0 For differentiating f (๐‘ฅ), we use logarithmic differentiation f (๐‘ฅ) = ๐‘ฅ^๐‘ฅ Taking log on both sides log f (๐’™) = ๐’™ log ๐’™ Differentiating w.r.t. x 1/๐‘“(๐‘ฅ) ๐‘“โ€ฒ(๐‘ฅ) = ๐‘ฅ . 1/๐‘ฅ + 1. log ๐‘ฅ 1/๐‘ฅ^๐‘ฅ ๐‘“โ€ฒ(๐‘ฅ) = 1 + log ๐‘ฅ ๐’‡โ€ฒ(๐’™) = ๐’™^๐’™(1 + log ๐’™) Putting ๐’‡โ€™(x) = 0 ๐‘ฅ^๐‘ฅ ("1 + log " ๐‘ฅ" " )=๐ŸŽ Either ๐’™^๐’™ = 0 Since, ๐‘ฅ^๐‘ฅ is exponential function it can never be zero. Or 1 + log ๐’™ = 0 log ๐‘ฅ = โˆ’1 Taking exponential on both sides ๐’†^๐’๐’๐’ˆโก๐’™ = ๐’†^(โˆ’๐Ÿ) ๐‘ฅ = ๐‘’^(โˆ’1) ๐’™ = ๐Ÿ/๐’† Hence, Stationary point is ๐’™ = ๐Ÿ/๐’† So, the correct answer is (B)
  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

    3. NCERT Exemplar - MCQs

    Tangents and Normals (using Differentiation) →
Tangents and Normals (using Differentiation) →
Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo