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4PM1/02R Jun 2022, Question 10: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 18 marks

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The question, in short

For the curve y = (7x − 2)/(2x − 3), x ≠ 3/2, find asymptotes, axis intersections, show the gradient is always negative, sketch it, and find where a given normal meets the curve again.

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

Part (a) · 2 marks

State the equations of the two asymptotes of C, one parallel to the y-axis and one parallel to the x-axis.

  1. 1Vertical asymptote where 2x − 3 = 0: x = 3/2
  2. 2Horizontal asymptote: y → 7/2 as x → ±∞

Answer(i) x = 3/2 (ii) y = 7/2

Part (b) · 2 marks

Find the coordinates where C crosses the x-axis and the y-axis.

  1. 1y = 0: 7x − 2 = 0, x = 2/7
  2. 2x = 0: y = (−2)/(−3) = 2/3

Answer(2/7, 0) and (0, 2/3)

Part (c) · 4 marks

Use differentiation to show the gradient of C is negative at every point on the curve.

  1. 1dy/dx = [7(2x − 3) − 2(7x − 2)]/(2x − 3)²
  2. 2= (14x − 21 − 14x + 4)/(2x − 3)²
  3. 3= −17/(2x − 3)²
  4. 4(2x − 3)² > 0 for x ≠ 3/2 and the numerator is negative, so the gradient is always negative

Answerdy/dx = −17/(2x − 3)², which is always negative — shown

Part (d) · 3 marks

Sketch C, labelling the asymptotes with their equations and the axis intersection coordinates.

  1. 1Draw dashed asymptotes x = 3/2 and y = 7/2
  2. 2Top-right branch: decreasing towards both asymptotes
  3. 3Bottom-left branch: decreasing, through (0, 2/3) and (2/7, 0), dropping down to x = 3/2

AnswerTwo branches: one above y = 7/2 and to the right of x = 3/2; the other below y = 7/2 and left of x = 3/2, crossing the axes at (0, 2/3) and (2/7, 0); asymptotes x = 3/2 and y = 7/2 drawn and labelled

Part (e) · 7 marks

Given normal l at A (positive x, gradient 17) meets C again at B, find the exact coordinates of B.

  1. 1Gradient of curve at A = −1/17: −17/(2x − 3)² = −1/17
  2. 2(2x − 3)² = 289, 2x − 3 = 17 (x > 0), so x = 10, y = 4: A(10, 4)
  3. 3Normal: y − 4 = 17(x − 10), so y = 17x − 166
  4. 417x − 166 = (7x − 2)/(2x − 3)
  5. 517x² − 195x + 250 = 0, (x − 10)(17x − 25) = 0
  6. 6x = 25/17, y = 17(25/17) − 166 = −141

AnswerB(25/17, −141)

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Also searched as: 4PM1 2R June 2022 · 4PM1/02R · 4pm1 jun22 2r · question 10.
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