Edexcel Pure Paper 1 2024 Question 8

Edexcel Pure Paper 1 2024 Question 8 – Functions, Inverse and Composite Functions

❓ The Question

🧠 Before you start

This one isn’t difficult in terms of content, but it can feel a bit stop-start.

Each part is testing something slightly different. If you rush, it’s easy to mix things up — especially the order in composite functions.

Best approach is just to slow it down and treat each part on its own. Don’t try to “link” them too much.

✏️ Working

Part (a)

You’re looking for $fg(2)$ , so that’s $f(g(2))$ .

Start inside.

$g(2) = \frac{5}{2(2) – 9}$

$= \frac{5}{4 – 9}$

$= -1$

Now feed that into $f$ .

$f(-1) = 4 - 3(-1)^2$

$= 4 - 3$

$= 1$

That part’s just careful substitution. Nothing more than that.

Part (b)

Now the inverse.

Write it as:

$y = \frac{5}{2x – 9}$

Swap the variables:

$x = \frac{5}{2y – 9}$

At this point, just solve it like a normal equation.

Multiply through:

$x(2y - 9) = 5$

Expand:

$2xy - 9x = 5$

Rearrange:

$2xy = 5 + 9x$

Divide by $2x$ :

$y = \frac{5 + 9x}{2x}$

That’s it. Just make sure the final answer is written in terms of $x$ .

Part (c)(i)

Now you’re doing $gf(x)$ .

So take $f(x)$ and drop it straight into $g$ .

$g(f(x)) = \frac{5}{2(4 – 3x^2) – 9}$

Work through the denominator:

$= \frac{5}{8 – 6x^2 – 9}$

$= \frac{5}{-1 – 6x^2}$

It’s easier to read like this:

$= -\frac{5}{6x^2 + 1}$

That last step matters more than it looks — it makes the range part much clearer.

Part (c)(ii)

Now think about what values this can take.

The bottom is $6x^2 + 1$ .

That can’t go below 1.

So the fraction can’t go above 5.

Because of the minus sign, everything flips below the axis.

When $x = 0$ :

$y = -5$

As $|x|$ increases, the denominator grows.

So the value moves up towards 0… but never reaches it.

So the range sits between:

$-5 \le y < 0$

Part (d)

Now solve:

$f(x) = h(x)$

Write them out:

$4 - 3x^2 = 2x^2 - 6x + k$

Move everything across:

$0 = 5x^2 - 6x + (k - 4)$

This is just a quadratic.

For no real solutions, the discriminant has to be negative.

$x^2 - 4(5)(k - 4) < 0$

Work it through:

$36 - 20(k - 4) < 0$

$36 - 20k + 80 < 0$

$116 - 20k < 0$

So:

$k > \frac{29}{5}$

🎯 Where the Marks Are

Marks are picked up bit by bit here.

The early part is mostly substitution — if that’s clean, you’re fine.

The composite function and range carry more weight. That’s where accuracy matters.

Final part is method. Once you recognise discriminant, it’s straightforward.

⚠️ What Went Wrong

Quite a lot of answers were nearly right.

Main issues were small:

wrong order in composite functions
algebra slips when simplifying
not thinking carefully about the range
missing the discriminant step

Some answers stopped just short — especially in the range part.

💡 One Small Tip

If you’re stuck on range, don’t overthink it.

Just ask: what can the denominator be?

That usually tells you everything.

🚀 If This Felt Difficult

If switching between these ideas feels awkward, that’s normal.

Working through a few more with maths tuition online helps build that fluency.

If you want something more structured, an A level maths learning course can help connect these topics so they don’t feel so separate.

🔗 Next Steps

← Question 7
→ Question 9

👨‍🏫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

📌 What’s the most common mistake here?

Mixing up composite functions.

📌 How do I check an inverse?

Substitute it back into the original function.

📌 Why is the range negative?

Because the expression is always negative.

📌 What’s key in part (d)?

Using the discriminant correctly.

Edexcel Pure Paper 1 2024 Question 8