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WMA11/01 Jun 2021, Question 9: Sketching Cubics and Reciprocal Curves

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 7 marks

The question, in short

Using the tan x graph over −2π to 2π, find the rightmost asymptote's equation, sketch y = 1/x + 1, and count solutions of tan x = 1/x + 1 in given intervals.

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

Part (a) · 1 mark

Give the equation of the rightmost vertical asymptote l of y = tan x shown in Figure 4.

  1. 1Asymptotes of tan x are at x = π/2 + nπ
  2. 2The one furthest right in −2π ≤ x ≤ 2π is x = 3π/2

Answerx = 3π/2

Part (bi) · 2 marks

On Diagram 1, draw y = 1/x + 1 for −2π ≤ x ≤ 2π and state its horizontal asymptote.

  1. 1y = 1/x + 1 is y = 1/x moved up 1 unit
  2. 2Horizontal asymptote y = 1, vertical asymptote x = 0
  3. 3Right branch: above y = 1 for x > 0, decreasing towards y = 1
  4. 4Left branch: below y = 1 for x < 0, crossing the x-axis at x = −1

AnswerTwo-branch curve y = 1/x + 1: horizontal asymptote y = 1, vertical asymptote x = 0 (the y-axis); the right branch lies in the first quadrant above y = 1, falling towards it; the left branch lies below y = 1, crossing the x-axis at (−1, 0) and going down to −∞ as x → 0⁻

Part (bii) · 2 marks

Using your sketch, state how many times tan x = 1/x + 1 holds for −2π ≤ x ≤ 2π, with a reason.

  1. 1Solutions of tan x = 1/x + 1 are where the two graphs cross
  2. 2x < 0: one crossing on each of the branches in −2π < x < −3π/2, −3π/2 < x < −π/2 and −π/2 < x < 0, so 3
  3. 3x > 0: one crossing for 0 < x < π/2 and one for π/2 < x < 3π/2; none for 3π/2 < x ≤ 2π, where tan x < 0 < 1/x + 1, so 2
  4. 4So there are 3 + 2 = 5 solutions

Answer5 solutions, because the curves y = tan x and y = 1/x + 1 intersect 5 times

Part (ci) · 1 mark

State how many solutions tan x = 1/x + 1 has for 0 ≤ x ≤ 40π.

  1. 1For x > 0 each tan branch between consecutive asymptotes meets y = 1/x + 1 once
  2. 20 ≤ x ≤ 40π contains 40 such branches (one per interval of length π)
  3. 340 solutions

Answer40

Part (cii) · 1 mark

State how many solutions tan x = 1/x + 1 has for −10π ≤ x ≤ (5/2)π.

  1. 1For x > 0 up to (5/2)π: 3 intersections (on the branches starting at 0, π/2 and 3π/2)
  2. 2For −2π ≤ x < 0 there are 3 intersections; each further interval of length π to the left adds 1
  3. 3−10π ≤ x < 0: 3 + 8 = 11 intersections
  4. 4Total = 3 + 11 = 14

Answer14

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