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Ex 6.3

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Ex 6.3, 9 - Find point on y = x3 - 11x + 5 at which tangent

Ex 6.3,9 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,9 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Transcript

Ex 6.3, 9 Find the point on the curve 𝑦=𝑥^3−11𝑥+5 at which the tangent is 𝑦=𝑥 −11.Equation of Curve is 𝑦=𝑥^3−11𝑥+5 We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑(𝑥^3 − 11𝑥 + 5)/𝑑𝑥 𝑑𝑦/𝑑𝑥=〖3𝑥〗^2−11 Also, Given tangent is 𝑦=𝑥−12 Comparing with 𝑦=𝑚𝑥+𝑐 , when m is the Slope Slope of tangent =1 From (1) and (2) 𝑑𝑦/𝑑𝑥=1 3𝑥^2−11=1 3𝑥^2=1+11 3𝑥^2=12 𝑥^2=12/3 𝑥^2=4 𝑥=±2 When 𝒙=𝟐 𝑦=(2)^3−11(2)+5 𝑦=8−22+5 𝑦=− 9 So, Point is (2 , −9) When 𝒙=−𝟐 𝑦=(−2)^3−11(−2)+5 𝑦=− 8+22+5 𝑦=19 So, Point is (−2 , 19) Hence , Points (2 , −9) & (−2 , 19) But (–2, 19) does not satisfy line y = x – 11 As 19 ≠ –2 – 11 ∴ Only point is (2, –9)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.