Edexcel Pure Paper 1 2024 Question 6

Edexcel Pure Paper 1 2024 Question 6 – Modulus Graph, Equations and Range of k

❓ The Question

🧠 First Thoughts

Nothing here is especially complicated on its own.

But it builds a bit. If part (b) goes wrong, part (c) usually follows.

So it’s worth slowing down slightly — especially when the modulus appears.

✏️ Working

Part (a)

This one is quick if you recognise the form.

$y = a|x - h| + b$

The turning point is always at $(h, b)$ .

So here it’s just:

$(2, 5)$

Part (b)

Now solve:

$16 - 4x = 3|x - 2| + 5$

First step — just tidy it:

$11 - 4x = 3|x - 2|$

At this point, you have to split it. No way around it.

If $x \ge 2$ , then:

$|x - 2| = x - 2$

So:

$11 - 4x = 3(x - 2)$

$11 - 4x = 3x - 6$

$17 = 7x$

$x = \frac{17}{7}$

That one works.

Now the other side.

If $x < 2$ :

$|x - 2| = 2 - x$

So:

$11 - 4x = 3(2 - x)$

$11 - 4x = 6 - 3x$

$5 = x$

But this doesn’t fit $x < 2$ , so it gets rejected.

That check is important — easy to forget.

Part (c)

This is where it usually gets a bit less comfortable.

You’ve got:

$y = kx + 4$

and you want it to cut the graph twice.

Rather than jumping into algebra, it’s easier to just think about the shape.

The graph is a V. Vertex at latex[/latex].

Two intersections means the line crosses both sides.

So rewrite the modulus as two lines.

Right side:

$y = 3x - 1$

Left side:

$y = -3x + 11$

Now compare slopes.

If the line is too steep, it only hits once. Same if it’s too flat.

So the gradient has to sit between the two:

$-3 < k < 3$

That’s the range.

✅ Final Answers

(a) $(2, 5)$

(b) $x = \frac{17}{7}$

🎯 How marks are awarded

This question is quite structured, so most of the marks come from doing the steps in the right order.

Part (a)

B1 for identifying the vertex correctly

This is just recognition. If you know the form $y = a|x - h| + b$ , this is a quick mark.

Part (b)

M1 for splitting the modulus into two cases
A1 for solving one case correctly
A1 for identifying and rejecting the invalid solution

The key thing here is the case split. If that isn’t shown, it’s very difficult to pick up full marks.

Even if one side goes wrong, you can still recover marks if the method is clear.

Part (c)

M1 for rewriting the modulus as two linear functions
A1 for reasoning about gradients correctly
A1 for giving the correct range of $k$

This part is more about interpretation than algebra.

You’re not expected to solve simultaneous equations — you’re expected to understand how the graph behaves.

Overall

Most of the marks are method marks (M).

That means:

showing structure matters
writing steps clearly matters
even if the final answer isn’t perfect, you can still score well

💡 Small but important point

A lot of students lose marks not because they don’t know what to do, but because they skip writing it down.

Especially in part (b), if the two cases aren’t clearly shown, the examiner has nothing to award the method marks to.

⚠️ What Actually Went Wrong

A few things came up quite a lot:

not splitting the modulus properly
keeping both answers in (b), even when one doesn’t fit
and in (c), trying to force algebra instead of thinking about the graph

Part (c) especially — people overcomplicated it.

💡 One Thing That Helps

As soon as you see a modulus, think of two straight lines.

Write them down early.

It makes everything after that much clearer.

🚀 If This Didn’t Feel Great

If part (b) felt awkward, it’s usually just getting used to splitting cases and checking them properly.

Working through a few more with maths tutoring online helps with that quite quickly.

If part (c) was the sticking point, that’s more about seeing the graph than doing algebra. An A level maths revision programme can help make that link clearer.

🔗 Next Steps

← Question 5
→ Question 7

👨‍🏫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 always need two cases for modulus?

Yes — unless you’re working directly from a graph.

📌 Why reject one solution?

Because it doesn’t match the condition you assumed.

📌 How should I approach part (c)?

Think about the shape first. Don’t jump straight into equations.

📌 What’s the main idea here?

A modulus graph is really just two straight lines.

Edexcel Pure Paper 1 2024 Question 6