# Ex 6.3,16 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 14, 2021 by Teachoo

Last updated at April 14, 2021 by Teachoo

Transcript

Ex 6.3, 16 Show that the tangents to the curve ๐ฆ=7๐ฅ3+11 at the points where ๐ฅ=2 and ๐ฅ =โ2 are parallel.We know that 2 lines are parallel y Slope of 1st line = Slope of 2nd line ๐1=๐2 We know that Slope of tangent is ๐๐ฆ/๐๐ฅ Given Curve is ๐ฆ=7๐ฅ^3+11 Differentiating w.r.t.๐ฅ ๐๐ฆ/๐๐ฅ=๐(7๐ฅ3 + 11)/๐๐ฅ ๐๐ฆ/๐๐ฅ=21๐ฅ^2 We need to show that tangent at ๐ฅ=2 & tangent at ๐ฅ=โ2 are parallel i.e. we need to show (Slope of tangent at ๐ฅ=2) = (Slope of tangent at ๐ฅ=โ2) Now, Slope of tangent = ๐๐ฆ/๐๐ฅ=21๐ฅ^2 Slope of tangent at ๐ฅ=2 ใ๐๐ฆ/๐๐ฅโใ_(๐ฅ = 2)=21(2)^2=21 ร4=84 & Slope of tangent at ๐ฅ=โ2 ใ๐๐ฆ/๐๐ฅโใ_(๐ฅ =โ 2)=21(โ2)^2=21 ร4=84 Since, (Slope of tangent at ๐ฅ=2) = (Slope of tangent at ๐ฅ=โ2) Thus, tangent at ๐ฅ=2 & tangent at ๐ฅ 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.