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Ex 6.3,27 (MCQ) - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at Aug. 9, 2021 by Teachoo
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Ex 6.3,27 (MCQ) You are here
Transcript
Ex 6.3, 27 The line π¦=π₯+1 is a tangent to the curve π¦2=4π₯ at the point (A) (1, 2) (B) (2, 1) (C) (1, β 2) (D) (β 1, 2)Given Curve is π¦^2=4π₯ Differentiating w.r.t. π₯ π(π¦^2 )/ππ₯=π(4π₯)/ππ₯ π(π¦^2 )/ππ¦ Γ ππ¦/ππ₯=4 2π¦ Γ ππ¦/ππ₯=4 ππ¦/ππ₯=4/2π¦ ππ¦/ππ₯=2/π¦ Given line is π¦=π₯+1 The Above line is of the form π¦=ππ₯+π when m is slope of line Slope of line π¦=π₯+1 is 1 Now Slope of tangent = Slope of line ππ¦/ππ₯=1 2/π¦=1 2=π¦ π¦=2 Finding x when π¦=2 π¦^2=4π₯ (2)^2=4π₯ 4=4π₯ 4/4=π₯ π₯=1 Hence the Required point is (x, y) = (1 , 2) Correct Answer is (A)
