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Edexcel Pure Paper 2 2024 Question 15 Solution
Edexcel Pure Paper 2 2024 Question 15 – Implicit Differentiation and Proof
❓ The Question
🧠 First thoughts
This is very much a “finish strong” question.
Nothing in part (a) is unfamiliar, but it’s easy to lose structure. Part (b) is quick if (a) is correct. Then part (c) is where it turns into a proper proof.
So the main challenge here isn’t knowing what to do — it’s carrying it through cleanly.
✏️ Working
Part (a)
Start with:
(x + y)^3 = 3x^2 – 3y – 2
Differentiate both sides with respect to x.
Left-hand side uses the chain rule:
\frac{d}{dx}(x+y)^3 = 3(x+y)^2\left(1 + \frac{dy}{dx}\right)
That bracket matters. Missing the 1 is a very common error.
Right-hand side:
\frac{d}{dx}(3x^2 – 3y – 2) = 6x – 3\frac{dy}{dx}
So:
3(x+y)^2\left(1 + \frac{dy}{dx}\right) = 6x – 3\frac{dy}{dx}
Expand slightly:
3(x+y)^2 + 3(x+y)^2\frac{dy}{dx} = 6x – 3\frac{dy}{dx}
Now gather \frac{dy}{dx} terms:
3(x+y)^2\frac{dy}{dx} + 3\frac{dy}{dx} = 6x – 3(x+y)^2
Factor:
\frac{dy}{dx}\left(3(x+y)^2 + 3\right) = 6x – 3(x+y)^2
So:
\frac{dy}{dx} = \frac{6x – 3(x+y)^2}{3(x+y)^2 + 3}
Part (b)
Point given: latex[/latex]
Substitute into derivative:
\frac{dy}{dx} = \frac{6(1) – 3(1+0)^2}{3(1+0)^2 + 3}
Simplify:
Top:
6 – 3 = 3
Bottom:
3 + 3 = 6
So:
\frac{dy}{dx} = \frac{1}{2}
Normal gradient:
-2
Equation:
y – 0 = -2(x – 1)
y = -2x + 2
Now substitute y = -2x + 2 into the original equation:
(x + y)^3 = 3x^2 – 3y – 2
Substitute:
(x – 2x + 2)^3 = 3x^2 – 3(-2x+2) – 2
Simplify:
Left:
(2 – x)^3
Right:
3x^2 + 6x – 6 – 2 = 3x^2 + 6x – 8
So:
(2 – x)^3 = 3x^2 + 6x – 8
Expand left:
(2 – x)^3 = 8 – 12x + 6x^2 – x^3
Form equation:
8 – 12x + 6x^2 – x^3 = 3x^2 + 6x – 8
Rearrange:
-x^3 + 6x^2 – 12x + 8 – 3x^2 – 6x + 8 = 0
Simplify:
-x^3 + 3x^2 – 18x + 16 = 0
Multiply by -1:
x^3 – 3x^2 + 18x – 16 = 0
We know x = 1 is a root:
(x – 1)(x^2 – 2x + 16) = 0
Check the quadratic:
b^2 – 4ac = (-2)^2 – 4(1)(16) = 4 – 64 = -60
No real roots.
So only one real solution → the normal does not meet the curve again.
🎯 Marks Breakdown (All Parts)
Part (a) – 5 marks
- M1: chain rule applied correctly
- A1: correct RHS differentiation
- A1: correct expansion structure
- M1: collecting \frac{dy}{dx} terms
- A1: final expression
Part (b) – 2 marks
- M1: correct substitution into derivative
- A1: correct normal equation
Part (c) – 5 marks
- M1: correct substitution into curve
- A1: correct cubic formed
- dM1: factorising using x = 1
- dM1: discriminant argument
- A1: clear conclusion
⚠️ What went wrong (important)
This question exposed a few consistent issues.
A lot of students expanded (x+y)^3 straight away. That made the differentiation longer and introduced mistakes early.
Another common one — missing the (1 + \frac{dy}{dx}) from the chain rule. That breaks the whole structure.
In part (c), many reached the cubic but stopped there. Some tried to use a calculator instead of proving it.
Others said “no real roots” but didn’t link it back to the question — so they didn’t fully justify the conclusion.
💡 One small takeaway
If you see “does not meet again”, you’re aiming to show:
- one known solution
- everything else not real
That’s the whole idea.
🚀 Final thought
This is a proper end-of-paper question — it’s not testing one skill, it’s testing how well everything holds together under pressure.
If these feel inconsistent, getting maths revision support for A level can really help with structure.
And for questions like this, where accuracy matters more than speed, one-to-one maths tuition often makes the biggest difference.
🔗 Next Steps
👨🏫Author Bio
S Mahandru focuses on helping students handle high-mark exam questions with clear structure, careful algebra, and strong final reasoning.
❓ Frequently Asked Questions
📌Why do we use implicit differentiation in this question?
Because y appears inside (x+y)^3, so it cannot be separated easily from x. This means we must differentiate both sides treating y as a function of x.
📌What is the most common mistake in part (a)?
Forgetting the chain rule. When differentiating (x+y)^3, the result must include (1 + \frac{dy}{dx}). Missing this leads to an incorrect derivative.
📌Why is the gradient of the normal equal to -2 in part (b)?
The gradient of the tangent is \frac{1}{2}. The normal is perpendicular, so its gradient is the negative reciprocal, which gives -2.
📌How do we show the normal does not meet the curve again?
After substituting y=-2x+2 into the original equation, you form a cubic with one known root x=1. The remaining quadratic has no real solutions, so there are no further intersection points.