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

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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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 3 𝑥﷮2﷯−11=1 3 𝑥﷮2﷯=1+11 3 𝑥﷮2﷯=12 𝑥﷮2﷯= 12﷮3﷯ 𝑥﷮2﷯=4 𝑥=±2 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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