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Question 4 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at March 22, 2023 by

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

(A)x = 0Β  Β  Β  Β  Β  Β  Β  Β  Β  (B) y = 0

(C) x + y = 0Β  Β  Β  Β  Β  Β (D) x – y = 0

Β 

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

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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)

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