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Ex 6.3, 12 - Find equations of all lines having slope 0, tangent

Ex 6.3,12 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,12 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,12 - Chapter 6 Class 12 Application of Derivatives - Part 4

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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)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.