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

Transcript

Ex 6.3, 26 The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is (A) 3 (B) 1/3 (C) – 3 (D) – 1/3 Slope of tangent is 𝑑𝑦/𝑑π‘₯ 𝑦=2π‘₯^2+3 sin⁑π‘₯ Differentiating w.r.t. π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑(2π‘₯^2 +3 sin⁑π‘₯ )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=4π‘₯+3 cos⁑π‘₯ We know that Slope of tangent Γ— Slope of Normal =βˆ’1 (4π‘₯+3 cos⁑π‘₯ ) Γ— Slope of Normal =βˆ’1 Slope of Normal = (βˆ’1)/(4π‘₯ + 3 cos⁑π‘₯ ) We need to find Slope of Normal at π‘₯=0 At x = 0 Slope of Normal =(βˆ’ 1 )/(4(0) + 3 cos⁑〖0Β°γ€— ) =(βˆ’1)/(0 + 3(1) )=(βˆ’1)/( 3) Hence, Correct Answer is D

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.