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
NCERT Exemplar - MCQs
Question 2 Important
Question 3 Important
Question 4 Important You are here
Question 5 Important
Question 6
Question 7 Important
Question 8 Important
Question 9 Important
Question 10
Question 11 Important
Question 12 Important
Question 13
Question 14
Question 15 Important
Question 16 Important
Question 17 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams You are here
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Last updated at April 16, 2024 by Teachoo
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)