AQA AS Level Mathematics Paper 1, June 2025: Question 9

Mark scheme

Show the mark scheme

Q Marking instructions AO Marks Typical solution

9(a) Expresses at least one term in x 1.1a M1 (1− 5x)7 = 1+ 7C (−5x) + 7C (−5x)2

or x2 correctly

= 1− 35x + 525x2

May be unsimplified

Condone missing – sign

Obtains –35x or +525x2 1.1b A1

1− 35x + 525x2 1.1b A1

Obtains

Subtotal 3

9(b) Multiplies the coefficient of their 3.1a M1 Coefficient of x2:

x term by k −35×k + 525×3 =1477

Or −35k +1575 = 1477

Multiplies the coefficient of their

x2 term by 3 −35k = −98

Condone one sign error 14

Multiplies the coefficient of their 1.1a M1 k =

x term by k and multiplies the

coefficient of their x2 term by 3

and equates the sum to 1477

Condone one sign error

14 1.1b A1

Obtains k =

Subtotal 3

Question 9 Total 6

How to answer it

Binomial Expansion with Unknown Coefficients

📋 What this question tests

This question assesses your mastery of core algebraic expansion and problem-solving skills at AS Level:

Binomial Expansion formula: Expanding expressions of the form (1 + bx)ⁿ for positive integer n up to specified powers of x.
Handling negative terms and indices: Correctly squaring and cubing bracketed terms containing negative coefficients, e.g. (-5x)².
Extracting specific coefficients: Finding the coefficient of a targeted power (x²) when two algebraic polynomials are multiplied together without expanding unnecessarily.
Forming and solving linear equations: Setting up an algebraic equation involving an unknown constant k and solving it accurately.

Part (a)

Finding the first three terms of (1 − 5x)⁷ [3 Marks]

📐 Step-by-Step Calculation

State the expansion formula:
(1 + y)ⁿ = 1 + n y + [n(n − 1) / 2!] y² + ...
Here, n = 7 and y = -5x.
Write unsimplified terms:
Term 0: 1
Term 1: ⁷C₁(-5x)¹ = 7(-5x)
Term 2: ⁷C₂(-5x)² = 21(-5x)²
Evaluate coefficients carefully:
⁷C₁ = 7
⁷C₂ = (7 × 6) / 2 = 21
(-5x)² = (-5)² × x² = +25x²
Simplify each term:
1 + 7(-5x) + 21(25x²)
= 1 − 35x + 525x²

✅ Final Answer & Marks

1 − 35x + 525x²

Mark Scheme Breakdown:

[M1] (AO 1.1a): Expresses at least one term in x or x² correctly unsimplified (e.g. ⁷C₁(-5x) or ⁷C₂(-5x)²). Condones missing negative sign at this stage.
[A1] (AO 1.1b): Correctly evaluates either the x term as −35x or the x² term as +525x².
[A1] (AO 1.1b): All three terms fully simplified and correct: 1 − 35x + 525x².

💡 Key Knowledge

"First three terms in ascending powers of x" means terms with x⁰ (constant), x¹, and x².
Always put negative components in brackets before squaring: (-5x)² = +25x² , not -25x² .
Combinations on calculator: 7 nCr 1 = 7 , 7 nCr 2 = 21 .

❌ Common Errors (Examiner Warnings)

Sign Drop: Writing 7(5x) instead of 7(-5x), resulting in +35x instead of -35x.
Squaring Bracket Trap: Writing -5x² or -25x² instead of (-5x)² = +25x² . This gives -525x², losing 2 marks immediately.
Listing without '+' or '−': Writing terms separated by commas ( 1, -35x, 525x² ) instead of as an algebraic polynomial expression.

Part (b)

Finding the value of k in (3 + kx)(1 − 5x)⁷ [3 Marks]

📐 Step-by-Step Calculation

Set up the product using part (a):
(3 + kx)(1 − 35x + 525x² + ...)
Select only combinations that produce x²:
• Constant from 1st bracket × x² term from 2nd:
3 × (525x²) = 1575x²
• x term from 1st bracket × x term from 2nd:
(kx) × (−35x) = −35kx²
Form the coefficient expression:
Total x² coefficient = 1575 − 35k
Equate to the given value and solve:
1575 − 35k = 1477
−35k = 1477 − 1575
−35k = −98
k = −98 / −35 = 98 / 35 = 14/5 (or 2.8)

✅ Final Answer & Marks

k = 14/5 (or 2.8)

Mark Scheme Breakdown:

[M1] (AO 3.1a): Multiplies the coefficient of their x term by k (i.e. −35 × k) OR multiplies their x² term by 3 (i.e. 525 × 3). One sign slip condoned.
[M1] (AO 1.1a): Forms a complete equation setting the sum of both products equal to 1477: 3(525) + k(−35) = 1477 .
[A1] (AO 1.1b): Correctly solves to obtain k = 14/5 or 2.8.

🧠 Exam Technique & Strategy

Do NOT expand the full expression: You only need terms that multiply to give x². Multiplying out all brackets wastes valuable exam time.
Coefficient vs Term: The question states "coefficient of x² is 1477". A coefficient is a number, not including x². Keep x² out of your linear equation.
Follow-Through (Ecf): If you made an arithmetic slip in part (a), you can still gain method marks in part (b) by correctly applying your values.

❌ Common Errors (Examiner Warnings)

Missing a Product: Only considering 3 × 525x² and forgetting that (kx) × (−35x) also contributes to the x² term.
Sign Confusion: Writing 1575 + 35k = 1477 by dropping the negative sign from the (−35x) term.
Retaining x² in the equation: Writing 1575x² − 35kx² = 1477 without cancelling x², leading to confusion over what to solve.

Topics

Pure Mathematics · D: Sequences and series