AQA A-Level Mathematics Paper 1, June 2025: Question 15

12 marks · Medium difficulty · Multi-step Problem

Use differentiation or normal line properties to form a cubic equation for the point on a parabola closest to a given point, derive the Newton–Raphson iteration formula, and calculate the minimum distance.

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Question

Question 15 features the parabola y = x^2 and a point Q with coordinates (3, 2.5). A point P on the curve is closest to point Q. Part (a) asks to show that the x-coordinate of P satisfies 2x^3 - 4x - 3 = 0 (4 marks). Part (b) asks to show that the Newton–Raphson method applied to this equation generates the recurrence formula x_{n+1} = (4x_n^3 + 3) / (6x_n^2 - 4) (4 marks). Part (c) asks to calculate x_3 starting from x_0 = 3 to three decimal places (2 marks). Part (d) asks to find the distance PQ to two decimal places (2 marks).
Question text

15 A curve has equation

y = x2

The point Q has coordinates (3, 2.5)

The point P on the curve which is closest to the point Q is shown on the diagram below.

y

P

Q (3, 2.5)

O x

15 (a) Show that the x‑coordinate of P satisfies the equation

2x3 – 4x – 3 = 0

[4 marks]

… 21

(20)

15 (b) The Newton–Raphson method is to be used to find an approximate solution to

the equation

2x3 – 4x – 3 = 0

Show that the Newton–Raphson method generates the iterative formula

4x3 + 3

n

xn+1 = 2

6xn – 4

[4 marks]

… U

(21)

15 (c) Starting with x0 = 3, use the iterative formula given in part (b) to find the value of x3

Give your answer to three decimal places.

[2 marks]

(22)

15 (d) Hence find the distance PQ

Give your answer to two decimal places.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 15: Part (a) awards marks for setting up the normal gradient equation -1/(2x) = (y - 2.5)/(x - 3) or differentiating the squared distance (x - 3)^2 + (x^2 - 2.5)^2, leading to 2x^3 - 4x - 3 = 0 (4 marks). Part (b) differentiates f(x) to 6x^2 - 4, applies Newton–Raphson formula x_n - f(x_n)/f'(x_n), and simplifies to the required form (4 marks). Part (c) shows intermediate values x_1 = 2.22, x_2 = 1.8288... and final value x_3 = 1.709 (2 marks). Part (d) calculates distance using sqrt((1.709 - 3)^2 + (1.709^2 - 2.5)^2) = 1.36 (2 marks).

Q Marking instructions AO Marks Typical solution

15(a) 1 y − 2 5.

Obtains = or dy

2x x −3 = 2x

y −2 5. dx

2x =

x −3 1 y − 2 5.

− =

OE eg in the form of the 2x x − 3

equation of the straight line. 3.1a M1

1 x2 − 2 5.

Or − =

Uses distance formula (may be 2x x − 3

in terms of x and y) − ( x − 3) = 2x3 − 5x

( 2 ) 2 2

d = (x −3) +( y − 2.5) 2x3 − 4x −3 = 0

Obtains

1 y − 2 5.

− = OE

2x x −3

y x2 3.1a A1

May have replaced with

Or

22 2

Obtains (x −3) +(x − 2.5)

Obtains

1 x2 − 2 5.

= OE

2x x − 3

Or

x2 − 2 5.

2x = OE

x −3

3.1a M1

Or

Differentiates their distance or

distance2 expression

d (d 2 )

= 2x − 6 + 2( x2 − 2.5)× 2x

dx

= 4x3 − 8x − 6

Completes reasoned argument

to show

2x3 − 4x − 3 = 0 2.1 R1

Must see at least one line of – A-LEVEL MATHEMATICS – 7357/1 –

correct intermediate working.

Subtotal 4

15(b) Obtains 6x2 − 4

May be explicit or seen as f ( x) = 2x3 − 4x − 3 = 0

denominator in their 1.1b B1 2

3 f (x) = 6x − 4

2xn − 4xn − 3

2 2x 3 − 4x − 3

6xn − 4 x = x − n n

n+1 n 2

3 6xn − 4

2xn − 4xn −3

Forms xn − 2 ( 2 ) 3

Their 6"xn − 4" xn 6xn − 4 2x − 4x −3

= − n n

20 Condone missing or inconsistent 6x2 − 4 6x2 − 4

1.1a M1 n n

subscripts 3 3

( ) 2 6xn − 4xn − 2xn + 4xn + 3

FT their f x = ax − 4 if stated =

6x2 − 4

explicitly n

Obtains two correct fractions eg 4x3 + 3

= n

x (6x2 − 4) 3 2

n n 2xn − 4xn − 3 6xn − 4

2 − 2

6xn − 4 6xn − 4

Or 3.1a A1

obtains a single correct fraction

with five terms in the numerator

Condone missing or inconsistent

subscripts

Completes reasoned argument

with a single correct fraction

4x3 + 3

before obtaining x = n

n+1 x2 −

6 n 4

Condone missing or inconsistent 2.1 R1

4x3 + 3

subscripts but x = n

n+1 x2 −

6 n 4

must be stated. – A-LEVEL MATHEMATICS – –

AG

Subtotal 4

15(c) Obtains 2.22 or AWRT 1.829 or x1 = 2.22

AWRT 1.709

For 2.22 accept 2.220 or correct x2 = 1.828840...

1.1a M1

111 x3 = 1.709

fraction

Obtains AWRT 1.709 as their

1.1b A1

final answer.

Subtotal 2

15(d) Uses expression for distance 2 2 2

(1.709 − 3) + (1.709 − 2.5)

with their x3 or a better 3.1a M1

approximation. = 1.3578...

Deduces AWRT 1.36 1.1b A1 = 1.36

Subtotal 2

Question 15 Total 12

How to answer it

Closest Point on a Curve & Newton–Raphson Method

📌 What this question tests

This 12-mark multi-topic question connects geometry, calculus, and numerical methods:

  • Geometric Calculus / Optimisation: Expressing the condition for the shortest distance between a point and a curve (either via the normal line perpendicular to the tangent, or by differentiating the distance-squared function).
  • Algebraic Proof: Setting up and simplifying coordinate geometry relationships to obtain a target cubic equation.
  • Newton–Raphson Derivation: Differentiating a polynomial and combining algebraic fractions over a common denominator to deduce a recurrence relation.
  • Iterative Computation & Coordinate Evaluation: Running sequential iterations accurately on a calculator and using the resulting coordinates to evaluate Cartesian distance.

Question 15 (a) — Establishing the Governing Equation

Show that the x-coordinate of P satisfies 2x³ − 4x − 3 = 0 [4 marks]

📐 Calculations (Two Valid Methods)

Method 1: Normal to Curve is Perpendicular to Tangent

1. Find curve tangent gradient: For y = x², dy/dx = 2x.
The gradient of the normal to the curve at P(x, x²) is −1 / (2x).
2. For P to be closest to Q(3, 2.5), the line PQ must be normal to the curve:
Gradient of line PQ = (y − 2.5) / (x − 3) = (x² − 2.5) / (x − 3).
3. Equate the two gradients:
−1 / (2x) = (x² − 2.5) / (x − 3)
4. Cross-multiply and rearrange:
−(x − 3) = 2x(x² − 2.5)
−x + 3 = 2x³ − 5x
2x³ − 4x − 3 = 0 (as required)

Method 2: Distance Minimisation

Let d² = (x − 3)² + (y − 2.5)² = (x − 3)² + (x² − 2.5)².
Differentiate with respect to x using the chain rule:
d(d²)/dx = 2(x − 3) + 2(x² − 2.5)(2x) = 2x − 6 + 4x³ − 10x = 4x³ − 8x − 6.
Set d(d²)/dx = 0: 4x³ − 8x − 6 = 0 ⇒ divide by 2: 2x³ − 4x − 3 = 0.

🧠 Exam Technique & Mark Breakdown

  • M1 (AO3.1a) : Valid start. Either write down normal gradient equated to chord gradient ±1/(2x) = (y − 2.5)/(x − 3) OR set up the distance formula d² = (x − 3)² + (y − 2.5)² .
  • A1 (AO3.1a) : Fully correct unsimplified gradient relation with negative sign correct, or full expression in x for d²: (x − 3)² + (x² − 2.5)² .
  • M1 (AO3.1a) : Substitute y = x² into the gradient equation, OR correctly differentiate the distance expression to get 4x³ − 8x − 6 .
  • R1 (AO2.1) : Clear reasoned progression to reach 2x³ − 4x − 3 = 0 without any missing algebraic steps.

💡 Key Knowledge

The shortest distance from an external point Q to a smooth curve always lies along the normal line at point P. This avoids squaring binomials and differentiating quartic polynomials!

❌ Common Errors

  • Forgetting the negative reciprocal for the normal: mistakenly writing 2x = (y − 2.5)/(x − 3) .
  • Sign slip when expanding brackets: −(x − 3) = −x − 3 instead of −x + 3 .
  • In "show that" questions, jumping directly from unsimplified forms to the target without showing cross-multiplication loses the final R1 mark.

Question 15 (b) — Newton–Raphson Formula Derivation

Show that Newton–Raphson generates xn+1 = (4xn³ + 3) / (6xn² − 4) [4 marks]

📐 Step-by-Step Algebraic Derivation

Step 1: Identify f(x) and f'(x)
f(x) = 2x³ − 4x − 3
f'(x) = 6x² − 4
Step 2: Quote standard Newton–Raphson formula
xn+1 = xn − [ f(xn) / f'(xn) ]
xn+1 = xn − (2xn³ − 4xn − 3) / (6xn² − 4)
Step 3: Put over a common denominator
xn+1 = [ xn(6xn² − 4) ] / (6xn² − 4) − (2xn³ − 4xn − 3) / (6xn² − 4)
Step 4: Expand and subtract carefully
xn+1 = [ 6xn³ − 4xn − (2xn³ − 4xn − 3) ] / (6xn² − 4)
xn+1 = [ 6xn³ − 4xn − 2xn³ + 4xn + 3 ] / (6xn² − 4)
Step 5: Collect like terms
xn+1 = (4xn³ + 3) / (6xn² − 4) (as required)

🧠 Mark Scheme Breakdown

  • B1 (AO1.1b) : Differentiating correctly to obtain 6x² − 4 (either seen explicitly or as the denominator).
  • M1 (AO1.1a) : Correctly substituting into Newton–Raphson: xn − (2xn³ − 4xn − 3)/(their f') . Subscripts may be omitted here.
  • A1 (AO3.1a) : Showing the common denominator step with two separate fractions or one single fraction with all five terms expanded in the numerator.
  • R1 (AO2.1) : Fully convincing simplification leading to the final form with correct subscripts and xn+1 = stated.

❌ Common Sign Traps

The minus sign before the fraction applies to the entire numerator:

− (2xn³ − 4xn − 3) = − 2xn³ + 4xn + 3

Many students write − 4xn instead of + 4xn , failing to cancel out the middle linear terms.

✅ What Top Students Include

Always show the step where the numerator has all 5 terms visible before combining:

[ 6xn³ − 4xn − 2xn³ + 4xn + 3 ] / (6xn² − 4)

This guarantees full credit under the strict R1 marking rule.

Question 15 (c) — Performing the Iterations

Starting with x₀ = 3, find x₃ to three decimal places [2 marks]

📐 Step-by-Step Iterations

Set up calculator: Type 3 and press = (stores to Ans ).
Enter: (4×Ans³ + 3) / (6×Ans² − 4)
First iteration (n = 0):
x₁ = [ 4(3)³ + 3 ] / [ 6(3)² − 4 ] = 111 / 50 = 2.22
Second iteration (n = 1):
x₂ = [ 4(2.22)³ + 3 ] / [ 6(2.22)² − 4 ] ≈ 1.828840...
Third iteration (n = 2):
x₃ = [ 4(1.828840...)³ + 3 ] / [ 6(1.828840...)² − 4 ] ≈ 1.70932...

🧠 Mark Scheme & Examiner Advice

  • M1 (AO1.1a) : Obtains either x₁ = 2.22 (or 111/50), or x₂ = AWRT 1.829 , or x₃ = AWRT 1.709 .
  • A1 (AO1.1b) : Final answer strictly 1.709 (AWRT 1.709).
💡 Accuracy tip: Never round intermediate values. Keep the full precision in your calculator's Ans memory to prevent compound rounding errors.

Question 15 (d) — Finding the Minimum Distance PQ

Find the distance PQ to two decimal places [2 marks]

📐 Step-by-Step Distance Calculation

1. Find coordinates of point P:
x ≈ 1.70932...
y = x² ≈ (1.70932...)² ≈ 2.9218...
2. Use Cartesian distance formula between P(x, y) and Q(3, 2.5):
PQ = √[ (x − 3)² + (y − 2.5)² ]
PQ = √[ (1.70932... − 3)² + (2.9218... − 2.5)² ]
3. Evaluate components:
(1.70932... − 3)² ≈ (−1.29068)² ≈ 1.66585
(2.9218... − 2.5)² ≈ (0.4218)² ≈ 0.17791
PQ = √[ 1.66585 + 0.17791 ] = √[ 1.84376 ] ≈ 1.35785...
4. Round to 2 decimal places:
Distance PQ = 1.36

🧠 Mark Scheme Breakdown

  • M1 (AO3.1a) : Uses standard distance formula √[(x − 3)² + (x² − 2.5)²] with their x₃ from part (c) (or a more accurate value).
  • A1 (AO1.1b) : Deduces 1.36 (AWRT 1.36).

❌ Common Errors

  • Forgetting to square the x-coordinate to find the y-coordinate of P, mistakenly using 2.5 or ignoring y entirely.
  • Premature rounding: using 1.71 can lead to slight inaccuracies in the 2nd decimal place.
  • Leaving the final answer as d² ≈ 1.84 instead of taking the square root to find distance d .

✅ Final Answer Summary

  • x-coordinate of P: 1.709 (3 d.p.)
  • y-coordinate of P: 2.922
  • Distance PQ: 1.36 (2 d.p.)

Topics

Pure Mathematics · C: Coordinate geometry in the (x, y) plane · G: Differentiation · I: Numerical methods

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.