# Ex 6.3,26 (MCQ) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Aug. 9, 2021 by Teachoo

Last updated at Aug. 9, 2021 by Teachoo

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Ex 6.3, 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

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