Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.3, 10 Find the equation of all lines having slope โ€“1 that are tangents to the curve ๐‘ฆ=1/(๐‘ฅ โˆ’ 1) , ๐‘ฅโ‰ 1. Equation of Curve is ๐‘ฆ=1/(๐‘ฅ โˆ’ 1) Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(1/(๐‘ฅ โˆ’ 1))/๐‘‘๐‘ฅ (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ =(๐‘‘ )/๐‘‘๐‘ฅ (๐‘ฅโˆ’1)^(โˆ’1) (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ =โˆ’1ใ€– ร— (๐‘ฅโˆ’1)ใ€—^(โˆ’1โˆ’1) (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ =โˆ’(๐‘ฅโˆ’1)^(โˆ’2) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(โˆ’ 1)/(๐‘ฅ โˆ’ 1)^2 Given that slope = โˆ’1 Hence, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’1 โˆด (โˆ’ 1)/(๐‘ฅ โˆ’ 1)^2 =โˆ’1 1/(๐‘ฅ โˆ’ 1)^2 =1 1=(๐‘ฅ โˆ’ 1)^2 (๐‘ฅ โˆ’ 1)^2=1 ๐‘ฅ โˆ’ 1=ยฑ1 x โˆ’ 1 = 1 x = 2 x โˆ’ 1 = โˆ’1 x = 0 So, x = 2 & x = 0 Finding value of y If x = 2 y = 1/(๐‘ฅ โˆ’ 1) y = 1/(2 โˆ’ 1) y = 1/1 y = 1 Thus, point is (2, 1) If x = 0 y = 1/(๐‘ฅ โˆ’ 1) y = 1/(0 โˆ’ 1) y = 1/(โˆ’1) y = โˆ’1 Thus, point is (0, โ€“1) Thus, there are 2 tangents to the curve with slope 2 and passing through points (2, 1) and (0, โˆ’1) We know that Equation of line at (๐‘ฅ1 , ๐‘ฆ1)& having Slope m is ๐‘ฆโˆ’๐‘ฆ1=๐‘š(๐‘ฅโˆ’๐‘ฅ1) Equation of tangent through (2, 1) is ๐‘ฆ โˆ’1 =โˆ’1 (๐‘ฅ โˆ’2) ๐‘ฆ โˆ’1 =โˆ’๐‘ฅ+2 ๐’š+๐’™โˆ’๐Ÿ‘ = ๐ŸŽ Equation of tangent through (0, โˆ’1) is ๐‘ฆ โˆ’(โˆ’1)=โˆ’1 (๐‘ฅ โˆ’0) ๐‘ฆ +1 =โˆ’๐‘ฅ ๐’š+๐’™ + ๐Ÿ = ๐ŸŽ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.