AQA AS Level Mathematics Paper 2, June 2025: Question 8

Mark scheme

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Q Marking instructions AO Marks Typical solution

8 Differentiates with at least one 1.1a M1

term correct dy 2

Obtains correct derivative 1.1b A1 = 3x −10x −8

dx

dy 2.4 E1 dy

Explains that = 0 For stationary point = 0

dx dx

for stationary point OE 3x2 − 10x − 8 = 0

Sets their derivative = 0 and 2

x = − or x = 4

solves to find at least one real 1.1a M1 3

value for x d2 y

PI by both correct values of x 2 = 6x −10

dx

d2 y

2 1.1b A1 2 = −14

Obtains x = − and x = 4 When x = − , 2 ,

33 dx

maximum

d2 y

Substitutes at least one of their 1.1a M1 When x = 4 , 2 =14 , minimum

values of x to obtain a y value dx

When x = 4 , y = −46

PI by –46 or Minimum is (4,−46)

Must have obtained M1 M1

Differentiates a second time 3.1a M1

dy

using their and tests at least

dx

one of the x coordinates of their

turning points

Or

Evaluates the gradient either

side of at least one of their x

value

Or

Justifies fully from shape of

cubic with reference to a sketch

or using the nature of a positive

cubic graph

Completes a reasoned

argument to justify the minimum 2.1 R1

point at (4, –46)

Can obtain R1 without the E1

Must see derivative = 0 for R1

Must test both x values for R1

CSO

Question 8 Total 8

How to answer it

Stationary Points: Finding & Classifying a Local Minimum

📌 What this question tests

This 8-mark question assesses your ability to carry out full calculus workflows on a cubic polynomial function:

First-order differentiation: Finding dy/dx for polynomial powers using d(xⁿ)/dx = n xⁿ⁻¹.
Stationary point condition: Stating explicitly and applying dy/dx = 0 to form a quadratic equation.
Quadratic solving: Finding both stationary x-values by factorisation or the quadratic formula.
Coordinates: Substituting x back into the original curve equation y = f(x) to determine the corresponding y-value.
Rigorous justification: Using the second derivative test (d²y/dx²) to definitively prove which stationary point is the local minimum.

Question 8 (Full 8 Marks)

Curve: y = x³ − 5x² − 8x + 2

📐 Step-by-Step Solution

1 Differentiate the function with respect to x:

y = x³ − 5x² − 8x + 2
dy/dx = 3x² − 10x − 8

2 Set the derivative equal to zero:

For stationary points, dy/dx = 0 .
3x² − 10x − 8 = 0

3 Solve the quadratic equation for x:

Factorising: (3x + 2)(x − 4) = 0
Therefore: x = −2/3 or x = 4

4 Find the second derivative to classify the turning points:

d²y/dx² = d/dx (3x² − 10x − 8) = 6x − 10

5 Test both stationary values to determine their nature:

• When x = −2/3 : d²y/dx² = 6(−2/3) − 10 = −4 − 10 = −14 < 0 → Local Maximum
• When x = 4 : d²y/dx² = 6(4) − 10 = 24 − 10 = +14 > 0 → Local Minimum

6 Calculate the corresponding y-coordinate for the minimum:

Substitute x = 4 back into the original cubic equation y = x³ − 5x² − 8x + 2 :
y = (4)³ − 5(4)² − 8(4) + 2
y = 64 − 5(16) − 32 + 2
y = 64 − 80 − 32 + 2 = −46

✅ Final Answer

Coordinates of the minimum point: (4, −46)

Fully justified by: d²y/dx² = 14 > 0 at x = 4, and d²y/dx² = −14 < 0 at x = −2/3.

🧠 Mark Scheme Breakdown (8 Marks)

M1 : Differentiates with at least one non-zero term correct.
A1 : Correct derivative: 3x² − 10x − 8 .
E1 : Explains that dy/dx = 0 for stationary points.
M1 : Sets derivative to 0 and solves quadratic to find at least one real root.
A1 : Both correct roots: x = −2/3 and x = 4 .
M1 : Substitutes at least one x-value into original cubic to find y.
M1 : Differentiates a second time and evaluates at a turning point.
R1 : Fully reasoned argument classifying both points with correct conclusion (4, −46) .

💡 Key Knowledge

Condition for Stationary Points: The gradient is zero ( dy/dx = 0 ). You must write this relationship down explicitly to secure the explanation mark (E1).
Second Derivative Test:
• If d²y/dx² > 0 , the curve is concave up (smile), giving a minimum.
• If d²y/dx² < 0 , the curve is concave down (frown), giving a maximum.
Coordinates require (x, y): Always substitute your x-value back into the original function y = f(x) , NEVER into the gradient function dy/dx .

❌ Common Student Errors

Forgetting to state dy/dx = 0: Jumping straight from finding the derivative to writing 3x² − 10x − 8 = 0 forfeits the E1 explanation mark. Always write: "At stationary points, dy/dx = 0".
Testing only one point: To earn the final reasoning mark ( R1 ), examiners require a fully reasoned argument showing why x = 4 is the minimum compared to x = −2/3 (CSO - Correct Solution Only).
Plugging x into dy/dx for y: Substituting x = 4 back into 3x² − 10x − 8 just gives 0 (confirming it's stationary), not the y-coordinate on the curve!
Sign confusion in the second derivative: Mistakenly thinking a positive second derivative means "maximum" and negative means "minimum". Remember: positive = concave up = minimum point.

Examiner Insight: To achieve full marks on "fully justify your answer" questions, you must treat your solution like a proof. State your criteria clearly ( dy/dx = 0 ), evaluate both roots, demonstrate the sign of d²y/dx² for each, and present the final answer clearly as an ordered pair (x, y) .

Topics

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