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Ex 6.3, 16 - Show that tangents to y = 7x3 + 11 at x = 2, -2

Ex 6.3,16 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,16 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Transcript

Ex 6.3, 16 Show that the tangents to the curve 𝑦=7π‘₯3+11 at the points where π‘₯=2 and π‘₯ =βˆ’2 are parallel.We know that 2 lines are parallel y Slope of 1st line = Slope of 2nd line π‘š1=π‘š2 We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ Given Curve is 𝑦=7π‘₯^3+11 Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑(7π‘₯3 + 11)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=21π‘₯^2 We need to show that tangent at π‘₯=2 & tangent at π‘₯=βˆ’2 are parallel i.e. we need to show (Slope of tangent at π‘₯=2) = (Slope of tangent at π‘₯=βˆ’2) Now, Slope of tangent = 𝑑𝑦/𝑑π‘₯=21π‘₯^2 Slope of tangent at π‘₯=2 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_(π‘₯ = 2)=21(2)^2=21 Γ—4=84 & Slope of tangent at π‘₯=βˆ’2 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_(π‘₯ =βˆ’ 2)=21(βˆ’2)^2=21 Γ—4=84 Since, (Slope of tangent at π‘₯=2) = (Slope of tangent at π‘₯=βˆ’2) Thus, tangent at π‘₯=2 & tangent at π‘₯ are parallel Hence Proved

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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.