Edexcel Pure Paper 1 2024 Question 3

Edexcel Pure Paper 1 2024 Question 3 – Newton Raphson Method

❓ The Question

You’re given a function, asked to show a root exists in an interval, and then improve an estimate using Newton–Raphson.

🧠 First Thoughts

This is a standard Newton–Raphson question, but it’s easy to lose marks on small things.

There are really three parts here:

checking a root exists
running the iteration
then saying what’s actually happening

Nothing complicated, but it only works if each step is clear.

One thing to watch — don’t rush into the formula straight away. The first part is just about showing a sign change, and that’s usually a separate mark.

✏️ Working

Part (a)

Function:

$f(x) = x^3 + 4x^2 - 10$

Check the endpoints.

At $x = 1$ :

$f(1) = 1 + 4 - 10 = -5$

At $x = 2$ :

$f(2) = 8 + 16 - 10 = 14$

So one is negative, the other positive.

That’s enough.

There’s a root between them.

Part (b)

Now the iteration.

Formula:

$x_{n+1} = x_n – \frac{f(x_n)}{f'(x_n)}$

Differentiate first:

$f'(x) = 3x^2 + 8x$

Start value:

$x_0 = 1.5$

Work out the function value:

$f(1.5) = 1.5^3 + 4(1.5)^2 - 10$

$= 3.375 + 9 - 10 = 2.375$

Now the derivative:

$f'(1.5) = 3(1.5)^2 + 8(1.5)$

$= 3(2.25) + 12 = 6.75 + 12 = 18.75$

Put it into the formula:

$x_1 = 1.5 – \frac{2.375}{18.75}$

That fraction is about $0.1267$

So:

$x_1 \approx 1.3733$

Go again.

You don’t need to re-derive anything — just substitute.

$f(1.3733) \approx 0.018$

$f'(1.3733) \approx 16.6$

So:

$x_2 = 1.3733 – \frac{0.018}{16.6}$

That correction is tiny.

So:

$x_2 \approx 1.3722$

At this point it’s clearly settling down.

Final value (to 3 d.p.):

$x \approx 1.372$

Part (c)

Look at the numbers:

$1.5 \rightarrow 1.3733 \rightarrow 1.3722$

They’re getting closer each time.

So the method is converging to a root.

🎯 How marks are awarded

Marks are pretty method-driven here.

substituting correctly into $f(1)$ and $f(2)$
actually stating there’s a root
writing the Newton–Raphson formula
substituting values properly
carrying out the iteration
giving the final answer to the right accuracy
recognising what the iteration is doing

You don’t need perfect numbers at every stage, but the structure has to be clear.

⚠️ What went wrong

Quite a few small things rather than one big issue:

students checked values but didn’t say a root exists
mistakes in the derivative (usually rushed)
rounding too early, which throws off later values
calculator slips in the fraction

And for part (c), a lot of answers were just vague.

Saying “it gets closer” isn’t quite enough — it needs to be linked to a root.

🚀 If this felt a bit off

If the process felt messy, it’s usually not the idea — just keeping everything organised.

Working through a few more with an online A level maths tutor can help with that structure.

If the whole method still feels unclear, a maths revision course for A level can make the steps feel more predictable.

🔗 Next Steps

← Question 2
→ Question 4

👨‍🏫Author Bio

S. Mahandru is an experienced A Level Maths teacher and founder of Exam.Tips, specialising in exam-focused revision techniques and helping students achieve top grades.

❓ Frequently Asked Questions

📌 Do I need both iterations?

Only if the question asks — but usually yes.

📌 Why not round early?

Because small changes affect the next step.

📌 What does convergence mean here?

The values are settling towards a single solution.

📌 Where are marks usually lost?

In the setup and in missing statements, not the final answer.

Edexcel Pure Paper 1 2024 Question 3