AQA AS Level Mathematics Paper 1, June 2025: Question 10

Mark scheme

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Q Marking instructions AO Marks Typical solution

10 1 1

Obtains gradient of tangent = 1.1b B1 Gradient of tangent =

= 3e3x 1.2 B1 3x

Recalls gradient function Gradient of curve = 3e

Equates their gradient function 3.1a M1 1

3e3x =

to their numerical tangent

gradient 9

Uses natural log correctly with 3x 1

e =

their equation of the form 27

3x 1.1a M1

ke = m to obtain an equation 1

without e 3x = ln

Obtains x = ln 1

3 x = ln

1.1b A1 3

ACF

ISW

Accept AWRT –1.1

Question 10 Total 5

How to answer it

Tangents to Exponential Curves & Exact Logarithms

What this question tests

This problem tests your ability to connect coordinate geometry with calculus and logarithms:

Rearranging linear equations into the form y = mx + c to find the gradient of a parallel line.
Differentiating exponential functions of the form y = e^kx using the chain rule.
Setting up an equation relating a derivative to a numerical gradient.
Solving an exponential equation using natural logarithms ( ln ) and applying log laws to simplify to an exact value.

Question 10 Breakdown (5 Marks)

Finding the exact x-coordinate of point P on the curve y = e^3x

📐 Step-by-Step Solution

Step 1: Find the gradient of the parallel line
Rearrange x − 9y = 23 into y = mx + c :
9y = x − 23 ⇒ y = (1/9)x − 23/9
Therefore, gradient of line = 1/9 .
Since parallel lines share the same gradient, gradient of tangent = 1/9.

[B1] for finding the gradient of the tangent is 1/9

Step 2: Differentiate the exponential curve
Given y = e^3x :
dy/dx = 3e^3x

[B1] for correctly recalling/finding the derivative 3e^3x

Step 3: Equate derivative to the tangent gradient
3e^3x = 1/9

[M1] for equating their gradient function to their numerical gradient

Step 4: Solve for x using natural logarithms
Divide by 3: e^3x = 1/27
Take natural log of both sides:
3x = ln(1/27)

[M1] for correctly applying natural logs to an equation of the form ke^3x = m to eliminate e

Step 5: Simplify to final answer
Notice that 1/27 = (1/3)³ = 3⁻³ :
3x = ln((1/3)³) = 3 ln(1/3)
Divide by 3:
x = ln(1/3) (or −ln 3 )

[A1] for obtaining x = ln(1/3), x = −ln 3, or awrt −1.1

✅ Acceptable Final Forms

x = ln(1/3) (Exact form)
x = −ln 3 (Exact equivalent)
x = (1/3)ln(1/27) (Exact unsimplified)
x ≈ −1.10 (Accepts AWRT −1.1)

Note: The mark scheme allows "Any Correct Form" (ACF) and "Ignore Subsequent Working" (ISW). However, leaving answers in exact logarithmic form is standard best practice at A-Level.

💡 Key Knowledge

Derivative of e^kx: d/dx [e^kx] = k·e^kx (Do NOT drop the power by 1; it is not a polynomial!).
Log Power Law: ln(aⁿ) = n·ln(a) .
Reciprocal Log Law: ln(1/a) = ln(a⁻¹) = −ln(a) .
Parallel Lines: Tangent gradient matches the line gradient exactly ( m₁ = m₂ ).

🧠 Exam Technique & Strategy

Check the Diagram: Point P lies on the curve to the left of the y-axis ( x < 0 ). Since x = ln(1/3) ≈ −1.10 is negative, this provides an instant sanity check!
Isolate e^3x First: Always divide by the coefficient 3 to get e^3x = 1/27 before taking logarithms. Trying to take ln(3e^3x) often leads to algebraic errors like writing 3·3x .

❌ Common Errors to Avoid

Sign error in line gradient: Incorrectly rearranging x − 9y = 23 to get m = −1/9 or m = 9 .
Confusing tangents with normals: Taking the negative reciprocal ( −1/m ). The question explicitly says parallel, which means equal gradient.
Log distribution trap: Writing ln(3e^3x) = 3x·ln(3) instead of the correct identity ln(3) + 3x . Avoid this by dividing by 3 first!
Forgetting chain rule: Differentiating e^3x to get just e^3x or 3xe^3x−1 .

Topics

Pure Mathematics · C: Coordinate geometry in the (x, y) plane · F: Exponentials and logarithms · G: Differentiation