Question 26 (MCQ) - Finding point when tangent is parallel/ perpendicular - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Finding point when tangent is parallel/ perpendicular
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 19 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Question 16 Deleted for CBSE Board 2024 Exams
Question 26 (MCQ) Important Deleted for CBSE Board 2024 Exams You are here
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 25 Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 23 Important Deleted for CBSE Board 2024 Exams
Question 17 Deleted for CBSE Board 2024 Exams
Question 18 Important Deleted for CBSE Board 2024 Exams
Finding point when tangent is parallel/ perpendicular
Last updated at April 16, 2024 by Teachoo
Question 26 The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is (A) 3 (B) 1/3 (C) β 3 (D) β 1/3Slope of tangent is ππ¦/ππ₯ π¦=2π₯^2+3 sinβ‘π₯ Differentiating w.r.t. π₯ ππ¦/ππ₯=π(2π₯^2 +3 sinβ‘π₯ )/ππ₯ ππ¦/ππ₯=4π₯+3 cosβ‘π₯ We know that Slope of tangent Γ Slope of Normal =β1 (4π₯+3 cosβ‘π₯ ) Γ Slope of Normal =β1 Slope of Normal = (β1)/(4π₯ + 3 cosβ‘π₯ ) We need to find Slope of Normal at π₯=0 At x = 0 Slope of Normal =(β 1 )/(4(0) + 3 cosβ‘γ0Β°γ ) =(β1)/(0 + 3(1) )=(β1)/( 3) Hence, Correct Answer is D