AQA AS Level Mathematics Paper 2, June 2025: Question 7

Mark scheme

Show the mark scheme

Q Marking instructions AO Marks Typical solution

7(a) Obtains a = 3 1.1b B1

3 ( x2 + x) − 8

1 1.1b B1

Obtains b = 2

21 35

3 x + −

35 1.1b B1 2 4

Obtains c = −

Subtotal 3

7(b)(i) 35 2.2a B1F

States −

4 35

Or −

FT their value of c from (a)

Subtotal 1

7(b)(ii) Equates sinθ to their –b, 3.1a M1 1

– ≤ b ≤ sinθ = −

OE provided 1 1 2

PI by correct answer θ = −

θ = 180 − −(30)

Obtains 210 1.1b A1 θ = 210°

ACF

Subtotal 2

Question 7 Total 6

How to answer it

Completing the Square & Trigonometric Minima

AQA AS Mathematics • Pure Mathematics • Algebra & Trigonometry

📌 What This Question Tests

Completing the square when the leading coefficient a > 1 , handling fractional terms correctly.
Identifying turning points: reading the minimum value of a quadratic function directly from its completed-square form.
Function substitution & trigonometric equations: linking f(sin θ) to quadratic behaviour and solving basic equations of the form sin θ = k to find the smallest positive angle.

Part (a)

Express f(x) = 3x² + 3x − 8 in the form a(x + b)² + c [3 marks]

📐 Step-by-Step Calculation

Factor out the coefficient of x² from the variable terms:
3(x² + x) − 8
Complete the square inside the bracket:
Half of the coefficient of x (which is 1) is 1/2 .
x² + x = (x + 1/2)² − (1/2)² = (x + 1/2)² − 1/4
Expand through by 3:
3[(x + 1/2)² − 1/4] − 8 = 3(x + 1/2)² − 3/4 − 8
Combine the constant terms:
−3/4 − 32/4 = −35/4
Result: 3(x + 1/2)² − 35/4

✅ Correct Answer & Marks

Values of constants:

a = 3
b = 1/2 (or 0.5)
c = −35/4 (or −8.75)

Form: 3(x + 1/2)² − 35/4

Mark Scheme Breakdown:
• B1 for obtaining a = 3
• B1 for obtaining b = 1/2
• B1 for obtaining c = −35/4

💡 Key Knowledge

For any quadratic Ax² + Bx + C :

Always factor A out of the first two terms first: A[x² + (B/A)x] + C .
Remember that x² + px = (x + p/2)² − (p/2)² .
When expanding the square brackets back out, don't forget to multiply the subtracted squared term by A !

❌ Common Errors

Forgetting to multiply by 3: Writing c = −1/4 − 8 = −33/4 instead of multiplying −1/4 by 3 first.
Sign confusion inside the bracket: Writing (x − 1/2)² despite the linear coefficient being positive (+3x).
Premature rounding: Converting fractions to rounded decimals rather than keeping exact fractions.

Part (b)(i)

State the minimum value of 3 sin²θ + 3 sin θ − 8 [1 mark]

✅ Correct Answer

Minimum value = −35/4 (or −8.75 )

Mark Scheme Breakdown:
• B1F: Correct value, or follow-through (FT) their value of constant c from part (a).

🧠 Exam Technique & Logic

Notice the structure: letting x = sin θ transforms f(x) into 3 sin²θ + 3 sin θ − 8 = 3(sin θ + 1/2)² − 35/4 .
Since (sin θ + 1/2)² ≥ 0 for all real values, the minimum occurs when the squared bracket equals 0.
Since sin θ = −1/2 is fully valid (as −1 ≤ sin θ ≤ 1 ), the true minimum value of the expression is simply c = −35/4 .

Part (b)(ii)

Find the smallest positive value of θ for which this minimum value occurs [2 marks]

📐 Step-by-Step Calculation

Set the squared term to zero:
The expression is minimized when sin θ + 1/2 = 0 .
sin θ = −1/2
Find the principal angle:
sin⁻¹(−1/2) = −30° (or reference angle is 30° in the 3rd and 4th quadrants).
Determine positive solutions:
Quadrant 3: θ = 180° − (−30°) = 210° (or 180° + 30° = 210° )
Quadrant 4: θ = 360° + (−30°) = 330°
Select the smallest positive value:
θ = 210° (or 7π/6 radians)

✅ Correct Answer & Marks

Smallest positive value: 210° (or 7π/6 )

Mark Scheme Breakdown:
• M1: Equates sin θ to their −b (where −1 ≤ b ≤ 1 ). Implied by the correct answer.
• A1: Obtains 210 (or equivalent in radians: 7π/6 ).

🧠 Examiner Commentary

Check the condition: Always check that the value of −b falls within the range [−1, 1] . If b had been greater than 1, sin θ could never equal −b , and the minimum would instead occur at an endpoint ( sin θ = ±1 ).
Watch the word "positive": A principal angle of −30° is not positive! You must add to find values in the range θ > 0 .

❌ Common Errors

Giving −30° as final answer: The question specifically asked for the smallest positive angle.
Missing the 3rd quadrant: Jumping directly from −30° to 330° ( 360° − 30° ) and missing that 210° is smaller.
Solving the wrong equation: Setting sin θ = 1/2 instead of −1/2 (forgetting that sin θ + b = 0 ⇒ sin θ = −b ).

Topics

Pure Mathematics · B: Algebra and functions · E: Trigonometry