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4PM1/01 Jun 2021, Question 3: The Product and Quotient Rules

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 10 marks

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The question, in short

Curve C: y = (ax − 3)/(x + 5), x ≠ −5, gradient at x = 2 is 18/49. Show a = 3, then find asymptotes, axis crossings and sketch C.

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

Part (a) · 3 marks

Differentiate the quotient and use the gradient 18/49 at x = 2 to show that a = 3.

  1. 1dy/dx = (a(x + 5) − (ax − 3))/(x + 5)²
  2. 2At x = 2: (7a − 2a + 3)/49 = 18/49
  3. 35a + 3 = 18
  4. 4a = 3

Answera = 3 shown

Part (b) · 2 marks

With a = 3, state the horizontal asymptote (i) and the vertical asymptote (ii) of C.

  1. 1y = (3x − 3)/(x + 5) → 3 as x → ±∞, so y = 3
  2. 2Denominator is zero at x = −5, so x = −5

Answer(i) y = 3; (ii) x = −5

Part (c) · 2 marks

Find where C meets the x-axis (i) and the y-axis (ii), giving coordinates.

  1. 1y = 0: 3x − 3 = 0, so x = 1, point (1, 0)
  2. 2x = 0: y = −3/5, point (0, −3/5)

Answer(i) (1, 0); (ii) (0, −3/5)

Part (d) · 3 marks

Sketch C, clearly marking both asymptotes and the coordinates of the axis crossing points.

  1. 1Draw the asymptotes x = −5 and y = 3
  2. 2For x > −5 the curve rises from −∞ towards y = 3, crossing the y-axis at (0, −3/5) and the x-axis at (1, 0)
  3. 3For x < −5 the curve lies above y = 3, rising to +∞ as x → −5

AnswerTwo-branch reciprocal-type curve with asymptotes x = −5 and y = 3 drawn and labelled; one branch top-left of the asymptotes, the other bottom-right passing through (0, −3/5) and (1, 0)

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