Edexcel Pure Paper 2 2024 Question 9 Solution

Edexcel Pure Paper 2 2024 Question 9 – Quadratic Modelling

❓ The Question

🧠 Before you start

This one is more about modelling than anything else.

You’re given points and a maximum, and the question expects you to build a quadratic from that. There are a couple of ways you could approach it, but the safest is just to stick with:

$H = ax^2 + bx + c$

Nothing fancy — just use the information carefully.

✏️ Working

Part (a)

We assume:

$H = ax^2 + bx + c$

Step 1: Use point A

From the diagram:

$(0, 2)$

Substitute $x = 0$ , $H = 2$ :

$2 = c$

So:

$c = 2$

Step 2: Use point B

Point B is:

$(20, 0.8)$

Substitute into the model:

$0.8 = 400a + 20b + 2$

Rearrange:

$400a + 20b = -1.2 \quad (1)$

Step 3: Use the maximum point

We’re told the maximum occurs at $x = 9$ .

At a maximum:

$\frac{dH}{dx} = 0$

Differentiate:

$\frac{dH}{dx} = 2ax + b$

Substitute $x = 9$ :

$18a + b = 0 \quad (2)$

Step 4: Solve simultaneous equations

From (2):

$b = -18a$

Substitute into (1):

$400a + 20(-18a) = -1.2$

$400a - 360a = -1.2$

$40a = -1.2$

$a = -0.03$

Now find $b$ :

$b = -18(-0.03) = 0.54$

Final equation:

$H = -0.03x^2 + 0.54x + 2$

Part (b)

We need a limitation of the model.

A valid answer:

The model assumes the path is perfectly quadratic, but in reality the ball’s path may not be exactly parabolic.

Other acceptable ideas (from examiner report) include:

wind affecting the path
air resistance not considered
ball not travelling exactly in a vertical plane

The key is linking it to the model itself.

Part (c)

Now we test whether Chandra can catch the ball.

She is at:

$x = 16$

Substitute into the model:

$H = -0.03(16)^2 + 0.54(16) + 2$

Step 1: Calculate

$16^2 = 256$

$-0.03 \times 256 = -7.68$

$0.54 \times 16 = 8.64$

Step 2: Add everything

$H = -7.68 + 8.64 + 2$

$H = 2.96$

Step 3: Compare

Chandra can catch the ball if:

$H < 2.5$

But:

$2.96 > 2.5$

Final conclusion:

Chandra cannot catch the ball.

🎯 Where the Marks Are

Most marks sit in part (a):

forming equations correctly
using the turning point condition
solving cleanly

Part (b) is just one clear statement.

Part (c) is substitution + interpretation.

⚠️ What Went Wrong

From the examiner report, most students did reasonably well — but a few things stood out.

Some didn’t recognise that $c = 2$ straight away, which made everything longer than it needed to be.

Others formed only one equation and missed the derivative condition completely.

There were also a few answers that stopped after finding $a$ and $b$ , without writing the full equation for $H$ .

In part (b), vague answers like “air resistance” didn’t always get credit unless they were linked properly to the model.

💡 One Small Tip

If you’re given a maximum, always think derivative.

That’s usually the quickest way to get your second equation.

🚀 If This Felt Difficult

If this felt like a lot, it’s probably because modelling questions bring several ideas together at once.

You’re not just using algebra — you’re interpreting a situation as well.

Working through more of these helps build that link, especially if you’re following a complete maths support programme where these ideas are connected properly.

And once that starts to click, it becomes much easier to improve exam performance in maths, because you’re not guessing how to set things up.

🔗 Next Steps

👨‍🏫Author Bio

S Mahandru focuses on helping students understand how exam questions are built, not just how to answer them. The goal is to make methods feel familiar so they can be applied confidently under pressure.

❓ Frequently Asked Questions

📌Why is c = 2?

Because when x = 0, H = 2 — so c must be 2.

📌Why use differentiation?

Because the maximum point means the gradient is zero.

📌Do I need a full sentence in part (c)?

Yes — you must clearly state whether Chandra can catch the ball.

📌Is this a common exam question?

Yes, quadratic modelling appears regularly.