AQA A-Level Mathematics Paper 2, June 2025: Question 7

9 marks · Medium difficulty · Multi-step Problem

Find unknown coefficients of a cubic curve given a stationary point and determine the coordinates of the second stationary point.

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Question

Question text

7 The point A lies on the curve with equation

y = x3 + px2 + qx + 12

7 (a) Given that A has coordinates (–5, 37), show that

5p – q = 30

[2 marks]

7 (b) Given that A is a stationary point, show that

10p – q = 75

[3 marks]

7 (c) Hence find the value of p and the value of q

[1 mark]

7 (d) The curve with equation

(10)

y = x3 + px2 + qx + 12

has a second stationary point B

Find the coordinates of B

Fully justify your answer.

[3 marks]

Mark scheme

Show the mark scheme

Q Marking instructions AO Marks Typical solution

7(a) Substitutes x = –5 and y = 37 3.1a M1 ( )3 ( )2 ( )

37 = −5 + −5 p + −5 q +12

y = x3 + px2 + qx +12

into

37 = −125 + 25 p − 5q +12

PI 37 = −125 + 25p −5q +12

150 = 25 p − 5q

Completes reasoned argument 2.1 R1

to show 5p −q = 30 5 p − q = 30

There must be at least one

intermediate step after (-5)3 and

(-5)2 evaluated

Subtotal 2

7(b) Differentiates x3 and 12 correctly 1.1b B1 dy

= 3x2 + 2px + q

Differentiates px2 correctly or 3.1a M1 dx

differentiates qx correctly 2

0 = 3 × (–5) – 10p + q

and

dy 10p – q = 75

equates their to zero

Completes reasoned argument 2.1 R1

by substituting x = –5 into a

correct to obtain

10p – q = 75 – A-LEVEL MATHEMATICS – 7357/2 –

Subtotal 3

7(c) Obtains p = 9, q =15 1.1b B1 5 p − q = 30

10 p − q = 75

p = 9, q =15

Subtotal 1

7(d) Solves 3x2 + 2px + q (= 0) for 3.1a M1 dy

Stationary points occur when = 0

their p and q dx

Condone x2 + 2px + q (= 0) dy

= 3x2 + 18x + 15 = 0

Obtains x = –1 from 1.1a A1 dx

3x2 + 18x + 15 (= 0) OE x = –5 or –1

Completes reasoned argument 2.1 R1 At B, x = –1

to obtain (–1, 5) Therefore, stationary point is (–1, 5)

Must have obtained x = –5 and

x = –1

when solving = 0

and

explained that stationary points 9

occur when = 0 OE

Subtotal 3

Question 7 Total 9

How to answer it

Stationary Points & Unknown Coefficients on a Cubic Curve

📋 What this question tests

This question assesses your fluency with polynomial curves and coordinate geometry at A-Level:

Substituting coordinates of a known point into a curve's Cartesian equation.
Differentiating polynomial terms ( xⁿ → n xⁿ⁻¹ ) treating unknown constants ( p, q ) correctly.
Applying the stationary point condition ( dy/dx = 0 ).
Setting up and solving simultaneous linear equations to deduce parameter values.
Finding second turning points by solving quadratic derivatives and providing full, clear mathematical justification for concluding coordinates.

Part 7 (a) • 2 Marks

Show that 5p − q = 30 given point A(−5, 37) lies on the curve

Curve: y = x³ + px² + qx + 12

📐 Step-by-Step Calculation

Substitute: Replace x = −5 and y = 37 :
37 = (−5)³ + p(−5)² + q(−5) + 12
Evaluate powers:
37 = −125 + 25p − 5q + 12
Collect constant terms:
37 = −113 + 25p − 5q
150 = 25p − 5q
Divide by 5:
30 = 5p − q ⟹ 5p − q = 30 (as required).

🧠 Exam Technique & "Show That" Proofs

Because the final equation is given in the question, you must not skip intermediate steps.
The mark scheme requires at least one clear line of working after powers (−5)³ and (−5)² are evaluated (e.g. showing 150 = 25p − 5q ).
Always state powers of negative numbers using brackets to avoid sign slips.

❌ Common Errors

Writing −5² = −25 instead of (−5)² = 25 .
Jumping straight from substitution to 5p − q = 30 without showing simplification, losing the reasoned argument mark (R1).

Mark Scheme Breakdown:
• M1 (3.1a): Correct substitution of x = −5 and y = 37 into curve equation.
• R1 (2.1): Fully reasoned argument leading to 5p − q = 30 with at least one intermediate step evaluated.

Part 7 (b) • 3 Marks

Show that 10p − q = 75 given A is a stationary point

Condition: dy/dx = 0 at stationary points

📐 Step-by-Step Calculation

Differentiate y with respect to x:
dy/dx = 3x² + 2px + q
Apply stationary point condition:
At stationary point A , dy/dx = 0 when x = −5 :
0 = 3(−5)² + 2p(−5) + q
Simplify and rearrange:
0 = 3(25) − 10p + q
0 = 75 − 10p + q
10p − q = 75 (as required).

💡 Key Knowledge

Differentiating polynomial terms with constants:
• d/dx (x³) = 3x²
• d/dx (px²) = 2px
• d/dx (qx) = q
• d/dx (12) = 0
The phrase "stationary point" mathematically defines the gradient: dy/dx = 0 .

Mark Scheme Breakdown:
• B1 (1.1b): Correct differentiation of x³ and 12 (giving 3x² and 0 ).
• M1 (3.1a): Differentiates px² or qx correctly AND sets their dy/dx = 0 .
• R1 (2.1): Completes reasoned argument substituting x = −5 into correct derivative to arrive at 10p − q = 75 .

Part 7 (c) • 1 Mark

Hence find the values of p and q

📐 Solving the Simultaneous Equations

We have two simultaneous linear equations from parts (a) and (b):

(1) 5p − q = 30
(2) 10p − q = 75

Subtract equation (1) from (2):
(10p − q) − (5p − q) = 75 − 30
5p = 45 ⟹ p = 9

Substitute p = 9 back into (1):
5(9) − q = 30
45 − q = 30 ⟹ q = 15

✅ Correct Answer

p = 9
q = 15

Check in equation (2): 10(9) − 15 = 90 − 15 = 75 (Consistent).

Mark Scheme Breakdown:
• B1 (1.1b): Both correct values obtained ( p = 9 and q = 15 ).

Part 7 (d) • 3 Marks

Find the coordinates of the second stationary point B

Requirement: "Fully justify your answer"

📐 Step-by-Step Calculation

Update derivative with p and q:
dy/dx = 3x² + 2(9)x + 15 = 3x² + 18x + 15
State condition and solve for x:
Stationary points occur when dy/dx = 0 :
3x² + 18x + 15 = 0
Divide through by 3:
x² + 6x + 5 = 0
(x + 5)(x + 1) = 0
Roots: x = −5 or x = −1
Select root for point B:
Since x = −5 corresponds to point A , stationary point B has x = −1 .
Find y-coordinate of B:
Substitute x = −1 , p = 9 , q = 15 into curve equation:
y = (−1)³ + 9(−1)² + 15(−1) + 12
y = −1 + 9 − 15 + 12 = 5
Final Coordinates: (−1, 5)

🧠 Full Justification Breakdown

To secure the final reasoning mark (R1), examiners look for specific criteria:

Explicitly stating that stationary points occur when dy/dx = 0 .
Showing both roots ( x = −5 and x = −1 ) from the quadratic equation.
Explaining or clearly identifying that x = −1 corresponds to point B because x = −5 belongs to A .
Calculating the y-coordinate using the original curve equation, not dy/dx .

❌ Common Errors to Avoid

Forgetting to find y: Stopping after finding x = −1 . The question asks for coordinates (an ordered pair (x, y) ).
Substituting into dy/dx: Substituting x = −1 into the derivative gives 0, not the y-coordinate. Always substitute into the curve equation for y .
Missing statement: Omitting dy/dx = 0 loses the justification mark.

✅ Correct Coordinates

B(−1, 5)

Mark Scheme Breakdown:
• M1 (3.1a): Sets 3x² + 2px + q = 0 with their p and q and attempts to solve.
• A1 (1.1a): Obtains x = −1 correctly from 3x² + 18x + 15 = 0 .
• R1 (2.1): Complete reasoned argument to achieve coordinates (−1, 5) , including stating dy/dx = 0 and displaying both roots x = −5 and x = −1 .

Topics

Pure Mathematics · B: Algebra and functions · G: Differentiation

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.