AQA A-Level Mathematics Paper 3, June 2025: Question 9

✅ Model Solution & Accepted Sequences

Any of the following fully described pairs in the correct order:

• Method 1: A stretch in the y-direction, scale factor 3 followed by a translation by vector 4
  0 . (Or vice versa: translation first, then vertical stretch).
• Method 2: A stretch in the x-direction, scale factor 1/3 followed by a translation by vector 4
  0 .
• Method 3: A translation by vector 4/3
  0 followed by a stretch in the x-direction, scale factor 1/3.

💡 Key Knowledge

• Vertical Stretch: y = a·f(x) stretches f(x) vertically by scale factor a (here a = 3).
• Horizontal Translation: y = f(x − c) translates f(x) by vector c
  0 (here c = 4).
• Since the operations act on different variables (one in y, one in x), order between vertical stretch and horizontal translation is interchangeable!

🧠 Exam Technique

• The mark scheme specifically states: "Only accept vectors for translations". Writing "shift right 4" or "move 4 units along x-axis" will lose marks. Always use column vector notation: 4
  0 .
• For stretches, explicitly specify: (1) stretch, (2) direction ("y-direction" or "parallel to y-axis"), and (3) scale factor ("scale factor 3").

❌ Common Errors

• Writing coordinates (4, 0) instead of a column vector 4
  0 .
• Sign flip error: translating by -4
  0 instead of 4
  0 .
• Combining horizontal stretch and horizontal translation in the wrong order without adjusting the vector.

✅ Model Solution

x = 4

💡 Key Knowledge

A rational function of the form y = k / (x − c) is undefined when its denominator equals zero. The vertical line that the curve approaches as x → c is the vertical asymptote, given by the full linear equation x = c .

❌ Common Errors

• Writing just the number 4 instead of an equation. Asymptotes are straight lines and must be given as equations (e.g. x = 4 ).
• Confusing x and y: writing y = 4 or y = 0 . (Note: y = 0 is the horizontal asymptote).

✅ Model Solution

Reason 1:

The function is discontinuous on the interval [3, 5] because there is a vertical asymptote at x = 4 (the points x = 3 and x = 5 lie on opposite sides of the asymptote).

Reason 2:

There could be an even number of roots (e.g. more than one solution / two solutions) between x = 3 and x = 5, which would result in no overall sign change between the endpoints.

💡 Key Knowledge: Sign Change Conditions

For the Intermediate Value Theorem / Sign Change rule to guarantee roots:

• Continuity is mandatory: A sign change only guarantees a root if f(x) is continuous on [a, b]. A sign change across an asymptote does not indicate a root, and a discontinuity can mask existing roots.
• No sign change ≠ No roots: If f(a) and f(b) have the same sign, the curve could still cross the x-axis multiple times (any even number of times: 2, 4, etc.) or touch the axis at a turning point.

📐 Mathematical Reality Check

Solving the actual equation algebraically:

1. 3 / (x − 4) = x
2. 3 = x(x − 4) ⇒ x² − 4x − 3 = 0
3. Using the quadratic formula: x = [4 ± √(16 − 4(1)(−3))] / 2 = [4 ± √28] / 2 = 2 ± √7
4. Roots: x ≈ −0.65 and x ≈ 4.65

Notice that x ≈ 4.65 lies inside [3, 5]! A root genuinely exists in the interval despite f(3) < 0 and f(5) < 0, proving the student's deduction completely false.

🧠 Exam Technique & Pitfalls

• Link directly to context: Mentioning the specific asymptote at x = 4 makes your answer robust and unambiguously awards the continuity mark.
• Avoid vague answers such as "the student did not check enough numbers" or "the interval is too wide". Examiners require precise mathematical conditions: discontinuity / asymptote and possibility of multiple roots.