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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  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Question 28 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
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.