Versatility | Oxford Maths Admissions Test Livestream

Versatility Solutions

Part of the Oxford Maths Admissions Test Livestream 2026

These are the solutions for the Versatility worksheet

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Warm-up

  • Let's see what happens if we continue the student's method. They've got to \[\frac{x-2}{\sqrt{(x-1)^2+(x-3)^2}}+\frac{x-4}{\sqrt{(x-2)^2+(x-6)^2}}=0.\]
  • So now it would probably make sense to move a term to the right-hand side, square both sides, and multiply up. But let's be careful, because squaring both sides of an equation can introduce additional solutions.
  • This gives \[ (x-2)^2\left((x-2)^2+(x-6)^2\right) = (x-4)^2\left((x-1)^2+(x-3)^2\right) \]
  • This looks like a mess, but I can see that the $x^4$ term will cancel. Multiplying everything out and bringing the remaining terms to the left reveals that in fact the $x^3$ term cancels too, and it all simplifies to the quadratic $3x^2-8x=0$. The value $x=0$ isn't the one I want (it doesn't satisfy the equation that we had before we squared), so I'll take $x=\frac{8}{3}$. Then $y=\frac{5}{3}$ and the minimum value of the original function is $\sqrt{\frac{26}{9}} + \sqrt{\frac{4}{9}+\frac{100}{9}}=\sqrt{26}$.

  • Alternatively, we could interpret the question geometrically. The question wants the point on the line $y=x-1$ that minimises the sum of the distances to $(1,2)$ and $(2,5)$. Let's write $A$ for the point $(1,2)$, $B$ for the point $(2,5)$, and $P$ for the unknown point on the line.

    A straight line y=x-1 with four points marked. The point P is on the line, points A and B are above the line, and the point A-prime is below the line. The lines from P to A and to B are shown, and a dashed line from P to A-prime is shown. In this diagram, the path from A-prime to P to B is not quite straight.
  • This problem is not easy! Let's use an idea from geometry; imagine reflecting the point $A$ in the line $y=x-1$, as if the line $y=x-1$ is a mirror, to give a new point $A'$ on the other side of the line. Then the distance $|A'P|$ will be the same as the distance $|AP|$. Then the key idea is that the sum $|A'P|+|BP|$ will be minimised when the segments $A'P$ and $BP$ form a straight line; the quickest route from $A'$ to $B$ is a straight line.
  • So we need to find the mirror image $A'$, and then we need to find the straight line through $A'$ and $B$. I find that $A'$ is at $(3,0)$ and the line is $y=15-5x$. This intersects $y=x-1$ at $x=\frac{8}{3}$, $y=\frac{5}{3}$. As above, the original function has value $\sqrt{26}$.

 

Questions

TMUA 2020 Paper 1 Question 19

  • It's tempting to just evaluate the quadratic at each of the six values and find out whether it's positive.
  • This is lots of work though;
    • $26^2-52\times 26 - 52 = -728$
    • $27^2-52\times 27 - 52 = -727$
    • $51^2-52\times 51 - 52 = -103$
    • $52^2-52\times 52 - 52 = -52$
    • $53^2-52\times 53 - 52 = 1$
    • I can stop now!
  • Instead of persisting through all that, perhaps we should rewrite the quadratic to make it easier to evaluate.
  • One way to do this is to complete the square for $(x-26)^2-26^2-52$.
  • This is a bit of a pain, because I don't know $26^2$ and I don't want to work it out.
  • Perhaps I should use the difference of two squares to simplify those first two terms. Said differently, I could have just factored the $x$ out of the first two terms of the original quadratic.
  • The expression is $x(x-52)-52$.
  • This is much easier to work with!
  • I can see that the expression will definitely be negative for $0\leq x \leq 52$ because $x(x-52)$ will be less than or equal to zero. When $x=53$, the expression will be $53\times (1) -52=1$, which is positive.
  • The answer is E.

Extension

  • It's a good thing that we didn't try to find the roots of the quadratic $x^2-52x-52=0$, because they're not very nice numbers. As a challenge, find the exact value of the larger root of the quadratic and prove that it's between $x=52$ and $x=53$.

  • Suppose that we have some polynomial of degree $n$ and we would like to evaluate it at some horrible value of $x$. Explain how you can do this in a way that involves multiplying by $x$ exactly $n$ times (so you cannot work out $x^{n}$ and then $x^{n-1}$ and so on).

    If you're interested in this, you can look up "Horner's method" online, but you've probably just invented it yourself.

 

TMUA 2021 Paper 1 Question 16

  • We're told the third term in ascending powers is $105x^2$. That looks a little odd to me, because the exponent is $2$ not $3$, but this is just because the first term in ascending powers of $x$ would be the term that doesn't involve $x$ at all (or, said differently, the term involves $x^0$). Careful not to be out-by-one when counting!
  • The third term would be $\binom{n}{2}a^{n-2}b^{2}x^2$.
  • Similarly, the fourth term would be $\binom{n}{3}a^{n-3}b^3x^3$.
  • Now we're told about the fourth term in descending powers of $x$. Before I write out an equation for this, I'm struck by the fact that it's $210x^3$ again. What's going on here?
  • The fourth term in ascending powers of $x$ has the same coefficient as the fourth term in descending powers of $x$. But, more to the point, it has the same power of $x$.
  • We must be looking at the middle term in the expansion.
  • So there are, in total, seven different powers of $x$ in this expansion. So $n=6$... careful not to be out-by-one when counting!
  • This is really helpful for my equations above. I need to know expressions for the binomial coefficients \[ \binom{n}{2}=\frac{n(n-1)}{2}\quad \text{and}\quad \binom{n}{3}=\frac{n(n-1)(n-2)}{6} \] so that I can work out $\binom{6}{2}=15$ and $\binom{6}{3}=20$.
  • I have \[ a^4b^2=7\quad \text{and}\quad a^3b^3=\frac{21}{2} \]
  • We're asked for $\left(\frac{a}{b}\right)^2$. That's a bit unusual, but we can work it out from the equations. We could try to solve for $a$ and $b$, but I don't think that we need to. Instead, let's divide the first by the second to get $\frac{a}{b}$ on the left, and then we'll be almost done. I have \[ \frac{a^4b^2}{a^3b^3}=\frac{7}{\frac{21}{2}},\quad\text{so}\quad \frac{a}{b}=\frac{2}{3} \]
  • Then the square of this is $\frac{4}{9}$.
  • The answer is B.

Extension

  • Using your solution, find the third term, in descending powers of $x$, of the expansion.
  • We have three unknowns ($a$ and $b$ and $n$) and we've been given three bits of information, so we might hope that we can solve any version of this question. Investigate this by choosing your own values of $a$ and $b$ and $n$ (not too large!), and calculating the expansion. Then select three terms from your expansion, try to forget $a$ and $b$ and $n$, and see if you can deduce them from your selected terms. Can you always do this? What might go wrong?

 

TMUA 2021 Paper 2 Question 19

  • I could do some algebra, perhaps moving one term to the other side and squaring both sides. I know that's dangerous though, because an equation like $x=1-2x$ squares to the same thing as $x=2x-1$, even though they have totally different solutions.
  • Instead, let's try to simplify first. It's odd that the question has two square-roots multiplied together. You don't see that very much. Why not? Well, because we can simplify it. If $a\geq 0$ and $b\geq 0$ then $\sqrt{a}\sqrt{b}=\sqrt{ab}$, and that's normally a nicer thing to write down.
  • So I can simplify $\sqrt{1+\sin \theta}\sqrt{1-\sin \theta}$ to $\sqrt{1-\sin^2 \theta}$. The next step is where I might make a mistake.
  • I might spot that $1-\sin^2\theta = \cos^2\theta$ and then I might want to write $\sqrt{1-\sin^2 \theta}$ as $\cos \theta$. Not so fast!
  • Remember that $\sqrt{x^2}$ is not equal to $x$ if $x\lt 0$. Instead of writing $\cos \theta$, I must write $\left| \cos \theta \right|$.
  • So, remembering this fact about square-roots, the equation simplifies to \[ \sin \theta \left| \cos\theta \right| + \cos\theta \left|\sin\theta\right| = 0 \]
  • Whenever I have absolute value signs, I like to think about separate cases depending on whether the expression(s) inside are positive or negative.
  • In this case, if $\sin \theta$ and $\cos \theta$ are both positive, then I'll have the sum of two positive terms. That can't be zero.
  • Similarly, if $\sin \theta$ and $\cos \theta$ are both negative, then I'll have the sum of two negative terms. That can't be zero.
  • I want exactly one of $\sin\theta$ and $\cos\theta$ to be positive, and the other to be negative. Or for one of them to be zero, I suppose (because then both terms will be zero).

  • Time to draw graphs of $\sin \theta$ and $\cos\theta$ to find out when this happens.

    Standard graphs of sine and cosine, from 0 degrees to 180 degrees. As a reminder, sin is positive between 0 and 180, and cosine is positive between 0 and 90 and then also between 270 and 360.
  • I want the range $90^\circ$ to $180^\circ$ inclusive, and the range $270^\circ$ to $360^\circ$ inclusive. That's 182 integer values in total.
  • The answer is F.

Extension

  • Check your understanding; replace the $+$ in the question with a $-$. Which values of $\theta$ satisfy this new equation?

 

TMUA 2022 Paper 1 Question 16

  • The quadratic equation gives \[x^2=\frac{3\pm \sqrt{2}}{7}\]
  • If $x=\cos \theta$ then we're looking at $\cos^2\theta$, so $\sin^2\theta$ will be $1$ minus this. Similarly for $\beta$ instead of $\theta$.
  • So we're looking for a new polynomial where the roots are \[x^2 = 1-\frac{3\pm\sqrt{2}}{7} = \frac{4\mp \sqrt{2}}{7}\]
  • I've used the $\mp$ symbol there to keep track of the fact that the root with the $+$ sign corresponds to the root for $x^2$ with the $-$ sign in the original polynomial, but this is not very important.
  • The key observation is that these would be the roots of some quadratic for $x^2$, and if we can work out the coefficients then we're done.
  • All the options have leading coefficient $7$, so let's work with that. The denominator for the quadratic formula is supposed to be $2a$, so I'll multiply both numerator and denominator by $2$ to get $\frac{8\mp \sqrt{8}}{14}$. Can I just take $b=-8$ and then work out $c$?
  • I'd need $b^2-4ac = 8$, with $b=-8$ and $a=7$. I can do this if I take $c=2$. So my quadratic for $x^2$ would be $7x^4-8x^2+2=0$.
  • Alternatively, finish solving the original polynomial for $x=\pm \sqrt{\frac{3+\sqrt{2}}{7}}$, or $x=\pm \sqrt{\frac{3-\sqrt{2}}{7}}$, then calculate $\sin \theta$ and $\sin \beta$ from these values of $\cos\theta$ and $\cos\beta$, and then multiply out $(x-\sin \theta)(x+\sin\theta)(x-\sin \beta)(x+\sin \beta)=0$. You'll get $x^4-\frac{8}{7}x^2+\frac{2}{7}=0$.
  • The answer is B.

Extension

  • Explain why a polynomial of degree four with roots $\pm\cos\theta$ and $\pm\cos\beta$ will never have an $x^3$ term or an $x^1$ term.

    Explain why replacing $x^2$ with $1-x^2$ in such a polynomial will always give a new polynomial with roots $\pm \sin\theta$ and $\pm\sin\beta$.

    Check that replacing $x^2$ with $1-x^2$ in $7x^4-6x^2+1=0$ gives $7x^4-8x^2+2=0$.

 

MAT 2019 Q1I

  • I might try to rearrange this for \[\frac{x}{y}=2^{y-x}.\]
  • I could even take logarithms of both sides for \[\log_2 x -\log_2 y = y-x.\]
  • But now what?
  • Instead, let's think about the function $f(x)=x2^x$. Perhaps that's not the best choice of variable, because I'd like to say that I'm evaluating that function at $x$ and at $y$, as two different inputs. If it helps, imagine rewriting the question to use $a$ and $b$ with $0\lt a\lt b$ and $a2^a = b2^b$. Then I'm looking at $f(a)=f(b)$.

  • Does this ever happen? I should probably try to sketch $f(x)$ and see if it repeats any values.

    An increasing function that grows faster and faster. It's a bit like 2 to the x but starts off a little slower (from zero instead of from 1) and then ultimately grows faster than 2 to the x.
  • Doing this, I realise that the function does not repeat any values, because $f(x)=x2^x$ is an increasing function for $x\gt 0$. It's the product of two positive increasing functions.
  • Since we're told that $x\lt y$ (no possibility of $x=y$), we conclude that $x2^x = y2^y$ simply never happens.
  • The answer is (a).

Extension

  • True or false? "If $\mathrm{f}(x)$ is increasing and $\mathrm{g}(x)$ is increasing, then $\mathrm{f}(x)\mathrm{g}(x)$ is increasing."
  • True or false? "If $\mathrm{f}(x)$ is decreasing and $\mathrm{g}(x)$ is decreasing, then $\mathrm{f}(x)\mathrm{g}(x)$ is decreasing."
  • Compare and contrast with the similar Extension question on the Graphs worksheet following TMUA 2020 Paper 2 Question 5.

 

Part of an Interview

  • It's helpful to draw a diagram.

    The parabola y equals x squared is shown, with three points marked. A is on the left of the y-axis, and B and C are on the right of the y-axis, with B closer to the y-axis.
  • It's tempting to write out an expression for the area in terms of $a$ and $b$ and $c$, perhaps with the hope of differentiating it with respect to $b$. This is really messy if you use Heron's formula, it's a bit of a pain if you use the shoelace formula, and it's more or less a non-starter if you want to work out any angles. If you want to check your answer, the area simplifies down to $\frac{1}{2}(b-a)(c-a)(c-b)$ provided $a\lt b\lt c$.
  • Instead of doing all that, let's think about how moving the point $B$ changes the area of the triangle. The points $A$ and $C$ stay fixed, so it makes sense to think of that as the "base" of the triangle, with the different positions for $B$ changing the "perpendicular height" of the triangle. It's perhaps a little odd to call $AC$ the base, because it's the top of the triangle, but the formula $\frac{1}{2}bh$ still works!
  • At this point, you might like to do some algebra to find the perpendicular distance from $B$ to the line $AC$, aiming to maximise this by differentiating. That's still a bit messy, so let's think more about where this will happen.

  • I'll draw on some lines that show the "height" of the triangle $ABC$, like a height chart. I'll draw dashed lines that are parallel to $AC$.

    The same diagram as before, but with the line segment AC drawn in, and four dashed lines, each parallel to AC, equally spaced, all below the line AC, are also shown. The point B is on the curve somewhere between the lowest two of these dashed lines.
  • I want to put $B$ as far from the line $AC$ as possible, so I should put it at a point where one of those dashed lines is tangent to the curve. So I should find the point where the gradient of the curve matches the gradient of the line $AC$.
  • This algebra is easy, just set $b$ to be the value of $x$ where $2x = \frac{c^2-a^2}{c-a}$. This simplifies to \[b=\frac{a+c}{2}\]
  • The answer turned out to be the arithmetic mean of $a$ and $c$; perhaps that's what you would have guessed!
  • Depending on how the interview is going, we might repeat the calculation with $y=x^4$ instead of $y=x^2$. The general formula for the area is horrible in this case, so you should probably use this gradient method. The answer is not the arithmetic mean of $a$ and $c$ in this case. Perhaps we could say that this second example proves that we can't just guess (sadly).
  • The approach we've used above is foreshadowing for the method of Lagrange multipliers, where you maximise a target function subject to some constraint function by choosing a point where the gradients align. This is something we teach to first-year students on the Oxford Mathematics degree, and I wouldn't expect people to have seen it before an interview.
  • The first year of the Oxford Mathematics degree also contains something called the Mean Value Theorem, which guarantees that for any two points on a smooth curve, there's a point in between where the gradient of the curve matches the gradient of the chord between the two points. (Note: "smooth" has a technical definition that is a bit stronger than I need here, but I'm hoping that you'll read it simply as a condition on the sorts of curve that we're talking about. Curves like $y=|x|$ are not covered by the Mean Value Theorem).
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