# Ex 6.3,27 (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

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)

Ex 6.3

Ex 6.3, 1

Ex 6.3,2

Ex 6.3,3 Important

Ex 6.3,4

Ex 6.3, 5 Important

Ex 6.3,6

Ex 6.3,7 Important

Ex 6.3,8

Ex 6.3,9 Important

Ex 6.3,10

Ex 6.3,11 Important

Ex 6.3,12

Ex 6.3,13

Ex 6.3, 14 (i)

Ex 6.3, 14 (ii) Important

Ex 6.3, 14 (iii)

Ex 6.3, 14 (iv) Important

Ex 6.3, 14 (v)

Ex 6.3,15 Important

Ex 6.3,16

Ex 6.3,17

Ex 6.3,18 Important

Ex 6.3,19

Ex 6.3,20

Ex 6.3,21 Important

Ex 6.3,22

Ex 6.3,23 Important

Ex 6.3,24 Important

Ex 6.3,25

Ex 6.3,26 (MCQ) Important

Ex 6.3,27 (MCQ) You are here

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.