# Ex 6.3,16 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.3,16 Show that the tangents to the curve 𝑦=7𝑥3+11 at the points where 𝑥=2 and 𝑥 =−2 are parallel. We know that 2 lines are parallel y Slope of 1st line = Slope of 2nd line 𝑚1=𝑚2 We know that Slope of tangent is 𝑑𝑦𝑑𝑥 Given Curve is 𝑦=7𝑥3+11 Differentiating w.r.t.𝑥 𝑑𝑦𝑑𝑥=𝑑7𝑥3 + 11𝑑𝑥 𝑑𝑦𝑑𝑥=21𝑥2 We need to show that tangent at 𝑥=2 & tangent at 𝑥=−2 are parallel i.e. we need to show (Slope of tangent at 𝑥=2) = (Slope of tangent at 𝑥=−2) Now, Slope of tangent = 𝑑𝑦𝑑𝑥=21𝑥2 Slope of tangent at 𝑥=2 𝑑𝑦𝑑𝑥│𝑥 = 2=2122=21 ×4=84 & Slope of tangent at 𝑥=−2 𝑑𝑦𝑑𝑥│𝑥 =− 2=21−22=21 ×4=84 Since, (Slope of tangent at 𝑥=2) = (Slope of tangent at 𝑥=−2) Thus , tangent at 𝑥=2 & tangent at 𝑥 are parallel Hence Proved

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Ex 6.3,16 You are here

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.