


Ex 6.3
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 You are here
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)
Last updated at April 14, 2021 by Teachoo
Ex 6.3, 12 Find the equations of all lines having slope 0 which are tangent to the curve 𝑦 = 1/(𝑥2 −2𝑥 + 3) Given Curve is 𝑦 = 1/(𝑥2 −2𝑥 + 3) Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑(1/(𝑥2 − 2𝑥 + 3))/𝑑𝑥 𝑑𝑦/𝑑𝑥=(𝑑(𝑥2 − 2𝑥 + 3)^(−1))/𝑑𝑥 𝑑𝑦/𝑑𝑥=−1(𝑥2−2𝑥+3)^(−2) . 𝑑(𝑥^2− 2𝑥 + 3)/𝑑𝑥 𝑑𝑦/𝑑𝑥=−(𝑥2−2𝑥+3)^(−2) (2𝑥−2) 𝑑𝑦/𝑑𝑥=−2(𝑥2−2𝑥+3)^(−2) (𝑥−1) 𝑑𝑦/𝑑𝑥=(−2(𝑥 − 1))/(𝑥2 − 2𝑥 + 3)^2 Hence Slope of tangent is (−2(𝑥 − 1))/(𝑥2 − 2𝑥 + 3)^2 Given Slope of tangent is 0 ⇒ 𝑑𝑦/𝑑𝑥=0 (−2(𝑥 − 1))/(𝑥2 − 2𝑥 + 3)^2 =0 −2(𝑥 − 1)=0 ×(𝑥2 − 2𝑥 + 3)^2 −2(𝑥 − 1)=0 (𝑥−1)=0 𝑥=1 Finding y when 𝑥=1 𝑦=1/(𝑥2 − 2𝑥 + 3) 𝑦=1/((1)^2 − 2(1) + 3) 𝑦=1/(1 − 2 + 3) 𝑦=1/(2 ) Point is (𝟏 , 𝟏/𝟐) Thus , tangent passes through (1 , 1/2) Equation of tangent at (1 , 1/2) & having Slope zero is (𝑦 −1/2)=0(𝑥−1) 𝑦 −1/2=0 𝑦=1/2 Hence Equation of tangent is 𝒚=𝟏/𝟐 We know that Equation of time passing through (𝑥 , 𝑦) & having Slope m is (𝑦 −𝑦1)=𝑚(𝑥 −𝑥2)