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

Last updated at Dec. 16, 2024 by Teachoo

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

(A) 0                   (B) 12

(C) 16                 (D) 32

Transcript

Question 15 Maximum slope of the curve y = –x3 + 3x2 + 9𝑥 – 27 is: 0 (B) 12 (C) 16 (D) 32 Given y = − 𝑥^3 + 3𝑥^2 + 9𝑥− 27 Now, Slope of the curve =𝒅𝒚/𝒅𝒙 = −3𝑥^2 + 6𝑥+ 9 We need to find maximum slope Let’s assume 𝒈(𝒙) = Slope Thus, we need to maximize 𝑔(𝑥) Maximizing 𝒈(𝒙) 𝑔(𝑥) = −3𝑥^2 + 6𝑥+ 9 Finding 𝒈’(𝒙) 𝒈^′ (𝒙)=−6𝑥+6 =−𝟔(𝒙−𝟏) Putting 𝒈^′ (𝒙)=𝟎 −6 (𝑥−1) = 0 𝑥−1 = 0 𝒙 = 1 Finding sign of g”(𝒙) at x = 1 g”(𝑥) = −6 < 0 Since g’’(x) < 0 at x = 1 ∴ g is maximum at x = 1 Finding Maximum Value of g(x) g(1) = "−3 " 〖(1)〗^2 "+ 6 (1) + 9" = −3 + 6 + 9 = −3 + 15 = 12 Hence, Maximum slope = Maximum value of g(x) = 12 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