Versatility
Part of the Oxford Maths Admissions Test Livestream 2026
Solutions to follow, will be available here
General Advice
Sometimes an idea isn't working, and you need to decide whether to salvage or start over.
- Lay out your work in a way that makes sense to you, so that if you need to re-read it later you can follow what you were doing.
- It's helpful if you can read your own handwriting!
- Don't scribble out rough work so much that you can't read it, because there's some chance you'll need to come back to it.
- If you're in a non-calculator test and everything has got very complicated (for example, a quadratic equation with three-digit coefficients), then it's possible that a different method exists that is easier to work with.
- There's a clear link to the Fluency worksheet; if you can predict what's going to happen when you try a method, then you can decide ahead of time whether to go down that route or not.
Warm-up
A student is attempting this question:
Given $y=x-1$, find the minimum value of $\sqrt{(x-1)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-5)^2}$.
The student has performed the following calculations, which you are welcome to read if you like. You may assume that the student has not made any mistakes.
- Make the substitution $y=x-1$ to get $\sqrt{(x-1)^2+(x-3)^2}+\sqrt{(x-2)^2+(x-6)^2}$.
- Differentiate this with the chain rule and set the derivative equal to zero. This gives $$\frac{1}{2}\times\frac{2(x-1)+2(x-3)}{\sqrt{(x-1)^2+(x-3)^2}}+\frac{1}{2}\times\frac{2(x-2)+2(x-6)}{\sqrt{(x-2)^2+(x-6)^2}}=0.$$
- This simplifies to $$\frac{x-2}{\sqrt{(x-1)^2+(x-3)^2}}+\frac{x-4}{\sqrt{(x-2)^2+(x-6)^2}}=0.$$
Decide whether you would like to continue with this method, or start again with a different method.
Work on the question yourself and consider whether you've made the right decision. It's not too late to change your mind!
Questions
TMUA 2020 Paper 1 Question 19
Find the lowest positive integer for which $x^2 - 52x - 52$ is positive.
(A) 26
(B) 27
(C) 51
(D) 52
(E) 53
(F) 54
[Scroll down for a hint]
TMUA 2021 Paper 1 Question 16
Consider the expansion of \[ (a + bx)^n \]
The third term, in ascending powers of $x$, is $105x^2$
The fourth term, in ascending powers of $x$, is $210x^3$
The fourth term, in descending powers of $x$, is $210x^3$
Find the value of $\left(\dfrac{a}{b}\right)^{\!2}$
(A) $\dfrac{1}{4}$
(B) $\dfrac{4}{9}$
(C) $\dfrac{25}{36}$
(D) $\dfrac{5}{6}$
(E) $1$
[Scroll down for a hint]
TMUA 2021 Paper 2 Question 19
The angle $\theta$ can take any of the values $1^\circ, 2^\circ, 3^\circ, \ldots, 359^\circ, 360^\circ$.
For how many of these values of $\theta$ is it true that \[ \sin\theta\sqrt{1+\sin\theta}\sqrt{1-\sin\theta} + \cos\theta\sqrt{1+\cos\theta}\sqrt{1-\cos\theta} = 0 \]
(A) $0$
(B) $1$
(C) $2$
(D) $4$
(E) $93$
(F) $182$
(G) $271$
(H) $360$
[Scroll down for a hint]
TMUA 2022 Paper 1 Question 16
The solutions to $7x^4 - 6x^2 + 1 = 0$ are $\pm\cos\theta$ and $\pm\cos\beta$.
Which one of the following equations has solutions $\pm\sin\theta$ and $\pm\sin\beta$?
(A) $7x^4 - 8x^2 - 5 = 0$
(B) $7x^4 - 8x^2 + 2 = 0$
(C) $7x^4 - 6x^2 - 2 = 0$
(D) $7x^4 - 6x^2 + 1 = 0$
(E) $7x^4 + 6x^2 - 1 = 0$
(F) $7x^4 + 6x^2 + 5 = 0$
[Scroll down for a hint]
MAT 2019 Q1I
The positive real numbers $x$ and $y$ satisfy $0\lt x\lt y$ and \begin{equation*} x2^x=y2^y \end{equation*} for
(a) no pairs $x$ and $y$.
(b) exactly one pair $x$ and $y$.
(c) exactly two pairs $x$ and $y$.
(d) exactly four pairs $x$ and $y$.
(e) infinitely many pairs $x$ and $y$.
Part of an Interview
Adapted from an interview question used by James Munro for Maths interviews at Oxford. Reproduced here with permission.
Suppose that $A$, $B$, and $C$ are points on the parabola $y=x^2$, with $x$-coordinates $a$, $b$, and $c$. Suppose that $A$ and $C$ are fixed points, and that $B$ lies somewhere on the curve in between $A$ and $C$.
I would like to maximise the area of triangle $ABC$.
In terms of $a$ and $c$, where should I put the point $B$?
(If you have a guess based on something that you know to be between $a$ and $c$, that's fine, but let's try to replace the guesswork with a calculation and/or some reasoning!)
(You know lots of different formulas for the area of a triangle, perhaps including $\displaystyle\frac{1}{2}ab$ for a right-angled triangle, or $\displaystyle\frac{1}{2}bh$, or $\displaystyle\frac{1}{2}ab\sin C$, or maybe you've even seen more exotic formulas like $\sqrt{s(s-a)(s-b)(s-c)}$ or $\displaystyle \frac{abc}{4R}$.
For this question, some of these formulas are not good ideas! If you've picked something that's not working, then the interviewer could help you to pick another direction.)
Hints
TMUA 2020 Paper 1 Question 19
- If you've started evaluating that quadratic for each of the values (for example, you've worked out $26^2-52\times 26 -52=-728$), that's a lot of work. Stop and think about simplifying the quadratic first.
TMUA 2021 Paper 1 Question 16
- If you've written out equations $$\binom{n}{2}a^{n-2}b^{2} = 105, \quad \text{and}\quad \binom{n}{3}a^{n-2}b^{3}=210 \quad \text{and}\quad \binom{n}{n-3}a^{3}b^{n-3}=210$$ and you're trying to solve those simultaneously, stop and think of an easier way to get to the value of $n$.
TMUA 2021 Paper 2 Question 19
- If you've squared that whole expression, it's probably got quite messy. Can you start by simplifying any of the square-roots in the expression, before you do anything complicated like squaring?
TMUA 2022 Paper 1 Question 16
- If you've solved the equation for the four values of $x$ then well done for spotting that you could do that, but your nested square-roots are going to be a little tricky to deal with. Proceed with care, or consider whether you needed a solution for $x$ anyway.