# Ex 6.3,12

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 6.3,12 Find the equations of all lines having slope 0 which are tangent to the curve 𝑦 = 1𝑥2 −2𝑥 + 3 Given Curve is 𝑦 = 1𝑥2 −2𝑥 + 3 Slope of tangent is 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥=𝑑1𝑥2 − 2𝑥 + 3𝑑𝑥 𝑑𝑦𝑑𝑥=𝑑𝑥2 − 2𝑥 + 3−1𝑑𝑥 𝑑𝑦𝑑𝑥=−1𝑥2−2𝑥+3−2 . 𝑑𝑥2− 2𝑥 + 3𝑑𝑥 𝑑𝑦𝑑𝑥=−𝑥2−2𝑥+3−22𝑥−2 𝑑𝑦𝑑𝑥=−2𝑥2−2𝑥+3−2𝑥−1 𝑑𝑦𝑑𝑥=−2𝑥 − 1𝑥2 − 2𝑥 + 32 Hence Slope of tangent is −2𝑥 − 1𝑥2 − 2𝑥 + 32 Given Slope of tangent is 0 ⇒ 𝑑𝑦𝑑𝑥=0 −2𝑥 − 1𝑥2 − 2𝑥 + 32=0 −2𝑥 − 1=0 ×𝑥2 − 2𝑥 + 32 −2𝑥 − 1=0 𝑥−1=0 𝑥=1 Finding y when 𝑥=1 𝑦=1𝑥2 − 2𝑥 + 3 𝑦=112 − 21 + 3 𝑦=11 − 2 + 3 𝑦=12 Point is 𝟏 , 𝟏𝟐 Thus , tangent passes through 1 , 12 Equation of tangent at 1 , 12 & having Slope zero is 𝑦 −12=0𝑥−1 𝑦 −12=0 𝑦=12 Hence Equation of tangent is 𝒚=𝟏𝟐

Ex 6.3,1

Ex 6.3,2

Ex 6.3,3

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

Ex 6.3,10

Ex 6.3,11

Ex 6.3,12 Important You are here

Ex 6.3,13

Ex 6.3,14

Ex 6.3,15 Important

Ex 6.3,16

Ex 6.3,17

Ex 6.3,18

Ex 6.3,19

Ex 6.3,20

Ex 6.3,21

Ex 6.3,22

Ex 6.3,23

Ex 6.3,24

Ex 6.3,25

Ex 6.3,26 Important

Ex 6.3,27

Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .