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 β y = 0
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This question is similar to Ex 6.3, 14 (i) - Chapter 6 Class 12 - Application of Derivatives
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Last updated at Dec. 4, 2021 by Teachoo
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This question is similar to Ex 6.3, 14 (i) - Chapter 6 Class 12 - Application of Derivatives
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 β y = 0 π¦=sinβ‘π₯ Since Slope of normal =(β1)/(ππ¦/ππ₯) Differentiating π¦ w.r.t. π₯ ππ¦/ππ₯=ππ¨π¬β‘π Since given point is (0, 0) Putting π=π in (1) π π/π π =cosβ‘0 ππ¦/ππ₯=π Hence, Slope of normal =(β1)/(ππ¦/ππ₯) =(β1)/1 =βπ Finding equation of normal Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) β΄ Equation of Normal at (0, 0) & Slope -1 is (πβπ)=βπ(πβπ) π¦=β1(π₯) π¦=βπ₯ π+π=π Hence, equation of normal is π+π=π So, the correct answer is (C)