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

Transcript

Ex 6.3, 16 Show that the tangents to the curve ๐‘ฆ=7๐‘ฅ3+11 at the points where ๐‘ฅ=2 and ๐‘ฅ =โˆ’2 are parallel.We know that 2 lines are parallel y Slope of 1st line = Slope of 2nd line ๐‘š1=๐‘š2 We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Given Curve is ๐‘ฆ=7๐‘ฅ^3+11 Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(7๐‘ฅ3 + 11)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=21๐‘ฅ^2 We need to show that tangent at ๐‘ฅ=2 & tangent at ๐‘ฅ=โˆ’2 are parallel i.e. we need to show (Slope of tangent at ๐‘ฅ=2) = (Slope of tangent at ๐‘ฅ=โˆ’2) Now, Slope of tangent = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=21๐‘ฅ^2 Slope of tangent at ๐‘ฅ=2 ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_(๐‘ฅ = 2)=21(2)^2=21 ร—4=84 & Slope of tangent at ๐‘ฅ=โˆ’2 ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_(๐‘ฅ =โˆ’ 2)=21(โˆ’2)^2=21 ร—4=84 Since, (Slope of tangent at ๐‘ฅ=2) = (Slope of tangent at ๐‘ฅ=โˆ’2) Thus, tangent at ๐‘ฅ=2 & tangent at ๐‘ฅ are parallel Hence Proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.