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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


Transcript

Ex 6.3, 12 Find the equations of all lines having slope 0 which are tangent to the curve 𝑦 = 1/(π‘₯2 βˆ’2π‘₯ + 3) Given Curve is 𝑦 = 1/(π‘₯2 βˆ’2π‘₯ + 3) Slope of tangent is 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑(1/(π‘₯2 βˆ’ 2π‘₯ + 3))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=(𝑑(π‘₯2 βˆ’ 2π‘₯ + 3)^(βˆ’1))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=βˆ’1(π‘₯2βˆ’2π‘₯+3)^(βˆ’2) . 𝑑(π‘₯^2βˆ’ 2π‘₯ + 3)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=βˆ’(π‘₯2βˆ’2π‘₯+3)^(βˆ’2) (2π‘₯βˆ’2) 𝑑𝑦/𝑑π‘₯=βˆ’2(π‘₯2βˆ’2π‘₯+3)^(βˆ’2) (π‘₯βˆ’1) 𝑑𝑦/𝑑π‘₯=(βˆ’2(π‘₯ βˆ’ 1))/(π‘₯2 βˆ’ 2π‘₯ + 3)^2 Hence Slope of tangent is (βˆ’2(π‘₯ βˆ’ 1))/(π‘₯2 βˆ’ 2π‘₯ + 3)^2 Given Slope of tangent is 0 β‡’ 𝑑𝑦/𝑑π‘₯=0 (βˆ’2(π‘₯ βˆ’ 1))/(π‘₯2 βˆ’ 2π‘₯ + 3)^2 =0 βˆ’2(π‘₯ βˆ’ 1)=0 Γ—(π‘₯2 βˆ’ 2π‘₯ + 3)^2 βˆ’2(π‘₯ βˆ’ 1)=0 (π‘₯βˆ’1)=0 π‘₯=1 Finding y when π‘₯=1 𝑦=1/(π‘₯2 βˆ’ 2π‘₯ + 3) 𝑦=1/((1)^2 βˆ’ 2(1) + 3) 𝑦=1/(1 βˆ’ 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 βˆ’1/2)=0(π‘₯βˆ’1) 𝑦 βˆ’1/2=0 𝑦=1/2 Hence Equation of tangent is π’š=𝟏/𝟐 We know that Equation of time passing through (π‘₯ , 𝑦) & having Slope m is (𝑦 βˆ’π‘¦1)=π‘š(π‘₯ βˆ’π‘₯2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.