1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
  3. NCERT Exemplar - MCQs
Check sibling questions

Question 4 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by Teachoo

The equation of the normal to the curve y = sin x at (0, 0) is:

(A)x = 0                  (B) y = 0

(C) x + y = 0           (D) x – y = 0

 

This question is similar to Ex 6.3, 14 (i) - Chapter 6 Class 12 - Application of Derivatives

Transcript

Question 4 The equation of the normal to the curve y = sin x at (0, 0) is: x = 0 (B) y = 0 (C) x + y = 0 (D) x – y = 0 𝑦=sin⁑π‘₯ Since Slope of normal =(βˆ’1)/(𝑑𝑦/𝑑π‘₯) Differentiating 𝑦 w.r.t. π‘₯ 𝑑𝑦/𝑑π‘₯=πœπ¨π¬β‘π’™ Since given point is (0, 0) Putting 𝒙=𝟎 in (1) π’…π’š/𝒅𝒙 =cos⁑0 𝑑𝑦/𝑑π‘₯=𝟏 Hence, Slope of normal =(βˆ’1)/(𝑑𝑦/𝑑π‘₯) =(βˆ’1)/1 =βˆ’πŸ Finding equation of normal Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) ∴ Equation of Normal at (0, 0) & Slope -1 is (π’šβˆ’πŸŽ)=βˆ’πŸ(π’™βˆ’πŸŽ) 𝑦=βˆ’1(π‘₯) 𝑦=βˆ’π‘₯ π’š+𝒙=𝟎 Hence, equation of normal is π’š+𝒙=𝟎 So, the correct answer is (C)
  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

About the Author

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