Check sibling questions

The point(s) on the curve y = x 3 – 11x + 5 at which the tangent isy  = x – 11 is/are:

(a) (−2, 19)       (b) (2, −9)

(c) (±2, 19)       (d) (−2, 19) and (2, −9)

 

This question is inspired from Ex 6.3,9 - Chapter 6 Class 12 - Application of Derivatives

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Question 39 The point(s) on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11 is/are: (a) (−2, 19) (b) (2, −9) (c) (±2, 19) (d) (−2, 19) and (2, −9) 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) But (–2, 19) does not satisfy line y = x – 11 As 19 ≠ –2 – 11 ∴ Only point is (2, –9) So, the correct answer is (B)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.