Question 15 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by

Maximum slope of the curve y = –x 3 + 3x 2 + 9x – 27 is:

(A) 0                   (B) 12

(C) 16                 (D) 32

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Transcript

Question 15 Maximum slope of the curve y = –x3 + 3x2 + 9π‘₯ – 27 is: 0 (B) 12 (C) 16 (D) 32 Given y = βˆ’ π‘₯^3 + 3π‘₯^2 + 9π‘₯βˆ’ 27 Now, Slope of the curve =π’…π’š/𝒅𝒙 = βˆ’3π‘₯^2 + 6π‘₯+ 9 We need to find maximum slope Let’s assume π’ˆ(𝒙) = Slope Thus, we need to maximize 𝑔(π‘₯) Maximizing π’ˆ(𝒙) 𝑔(π‘₯) = βˆ’3π‘₯^2 + 6π‘₯+ 9 Finding π’ˆβ€™(𝒙) π’ˆ^β€² (𝒙)=βˆ’6π‘₯+6 =βˆ’πŸ”(π’™βˆ’πŸ) Putting π’ˆ^β€² (𝒙)=𝟎 βˆ’6 (π‘₯βˆ’1) = 0 π‘₯βˆ’1 = 0 𝒙 = 1 Finding sign of g”(𝒙) at x = 1 g”(π‘₯) = βˆ’6 < 0 Since g’’(x) < 0 at x = 1 ∴ g is maximum at x = 1 Finding Maximum Value of g(x) g(1) = "βˆ’3 " γ€–(1)γ€—^2 "+ 6 (1) + 9" = βˆ’3 + 6 + 9 = βˆ’3 + 15 = 12 Hence, Maximum slope = Maximum value of g(x) = 12 So, the correct answer is (B)

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.