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WMA11/01 Jun 2022, Question 10: Tangent and Normal Problems

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 12 marks

The question, in short

For the cubic curve y = (2/7)x³ + (1/7)x² − (5/2)x + k, differentiate it, use the normal at A (x = −7/2) that is tangent at B to derive and solve a quadratic for B's x coordinate, then find k given l's y intercept is −1.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 2 marks

Differentiate the given cubic equation to find dy/dx.

  1. 1Differentiate each term: (2/7)x³ → (6/7)x², (1/7)x² → (2/7)x
  2. 2−(5/2)x → −5/2 and k → 0, so dy/dx = (6/7)x² + (2/7)x − 5/2

Answerdy/dx = (6/7)x² + (2/7)x − 5/2

Part (b) · 4 marks

Using the normal at A with x = −7/2 also tangent at B, show B's x coordinate satisfies 12x² + 4x − 33 = 0.

  1. 1At x = −7/2: dy/dx = (6/7)(49/4) − 1 − 5/2 = 7
  2. 2The normal at A has gradient −1/7, so the tangent at B has gradient −1/7
  3. 3(6/7)x² + (2/7)x − 5/2 = −1/7
  4. 4Multiply by 14: 12x² + 4x − 35 = −2, so 12x² + 4x − 33 = 0

AnswerShown: gradient at A is 7, so l has gradient −1/7; (6/7)x² + (2/7)x − 5/2 = −1/7 ⇒ 12x² + 4x − 35 = −2 ⇒ 12x² + 4x − 33 = 0

Part (c) · 2 marks

Solve the quadratic to find the x coordinate of B, justifying which root is valid.

  1. 112x² + 4x − 33 = (2x − 3)(6x + 11) = 0
  2. 2x = 3/2 or x = −11/6
  3. 3The figure shows B has a positive x coordinate, so x = 3/2

Answerx = 3/2 (the other root −11/6 is negative, but B has a positive x coordinate)

Part (d) · 4 marks

Given the y intercept of line l is −1, determine the constant k.

  1. 1Equation of l: y = −(1/7)x − 1
  2. 2At A, x = −7/2: y = 1/2 − 1 = −1/2
  3. 3Substitute (−7/2, −1/2) into C: (2/7)(−343/8) + (1/7)(49/4) + 35/4 + k = −1/2
  4. 4−49/4 + 7/4 + 35/4 + k = −1/2, so −7/4 + k = −1/2
  5. 5k = 5/4

Answerk = 5/4

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