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

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

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The question, in short

For the curve y = (2x − 1)/(x + 4), x ≠ −4, find asymptotes, axis intersections, sketch the curve, and use calculus to locate points where tangents have gradient 1, then find k₁ and k₂.

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: the one parallel to the x-axis and the one parallel to the y-axis.

  1. 1As x → ±∞, y → 2/1 = 2
  2. 2The denominator is zero at x = −4

Answer(i) y = 2 (ii) x = −4

Part (b) · 2 marks

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

  1. 1y = 0: 2x − 1 = 0, x = 1/2
  2. 2x = 0: y = −1/4

Answer(1/2, 0) and (0, −1/4)

Part (c) · 3 marks

Sketch C on the given axes, clearly showing both asymptotes and the axis-intercept coordinates.

  1. 1Draw the asymptotes x = −4 and y = 2
  2. 2Right branch: passes through (0, −1/4) and (1/2, 0), approaches y = 2 from below
  3. 3Left branch: above y = 2, approaching both asymptotes

AnswerRectangular-hyperbola shape: one branch to the right of x = −4 and below y = 2, passing through (0, −1/4) and (1/2, 0); the other branch to the left of x = −4 and above y = 2; asymptotes x = −4 and y = 2 drawn and labelled

Part (d) · 8 marks

Using calculus, find the coordinates of P and Q, where the tangents to C have gradient 1 and x_P > x_Q.

  1. 1dy/dx = (2(x + 4) − (2x − 1))/(x + 4)² = 9/(x + 4)²
  2. 2The tangent y = x + k has gradient 1
  3. 39/(x + 4)² = 1, so (x + 4)² = 9
  4. 4x + 4 = ±3, x = −1 or x = −7
  5. 5x = −1: y = −3/3 = −1; x = −7: y = −15/(−3) = 5
  6. 6P has the larger x: P(−1, −1), Q(−7, 5)

Answer(i) P(−1, −1) (ii) Q(−7, 5)

Part (e) · 3 marks

Using your coordinates for P and Q, find the values of k₁ and k₂ in the tangent equations y = x + k₁ and y = x + k₂.

  1. 1At P: −1 = −1 + k₁, so k₁ = 0
  2. 2At Q: 5 = −7 + k₂, so k₂ = 12

Answer(i) k₁ = 0 (ii) k₂ = 12

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