In this article, I cover more on the TMUA mathematics admissions test. Specifically I will cover further question types to revise. If you have not already read part 1, please feel free to give it a read first:
In short, I will cover three question types: differentiation and optimal points, polynomials and what I term “difficult dimensions” problems.
Differentiation and Optimal Points
Relevant questions from the 2021 TMUA paper 1: 4, 6, 11, 12.
These questions can involve finding maximum or minimum points for unusual functions. In this case you could use knowledge of the nature of the function or you could use differentiation.
Alternatively you could be finding whether a function is increasing or decreasing.
Often a substitution can help. It can turn a more complicated problem into a simpler one.
Common mistakes include incorrect use of unusual functions (exponents, logs),
Example
Here is an unusual example involving the volume of a cone:
The volume of a cone \(V\)is \(V=\frac{1}{3}\pi r^2h\), where \(r\) is the radius of the circular base and \(h\) is the perpendicular height from the centre of the base to the peak of the cone. The length of the side of the cone, from the circumference of the base to the peak of the cone, is 9 centimetres. Find the greatest possible volume of the cone.
Method
To summarise the problem, we are maximising the volume. But the problem is that the volume is a function of two variables \(r\) and \(h\). In order to apply simple differentiation to the problem to find the maximum or minimum, we need to convert the problem into a one-variable optimisation.
Conveniently, we also have a constraint that enables us to do this here. Namely the perpendicular height, the radius and the length of the side of the cone must follow Pythagoras’ Law:
\[r^2+h^2=9^2=81\]
Then we can rearrange for \(r^2\) and substitute this into the volume expression:
\[\frac{1}{3}\pi (81-h^2)h\]
\[=\frac{1}{3}\pi(81h-h^3)\]
Next, to find the maximum volume, we can differentiate with respect to \(h\) and find the stationary point.
\[\frac{dV}{dh}= \frac{1}{3}\pi (81-3h^2) =0\]
\[h^2-27=0\]
\[h=\sqrt{27}=3\sqrt{3}\]
Note \(h>0\) must hold, as heights must be positive.
The question asks for the maximum volume. To find this, we can substitute this value for \(h\) back into the volume expression we derived in terms of \(h\) only. This means the volume is:
\[V= \frac{1}{3}\pi (81h-h^3) = \frac{1}{3}\pi (81 (3\sqrt{3})-27(3\sqrt{3})) \]
\[= 54 \pi \sqrt {3} \]
which is the final answer.
Moreover this technique is important. It is similar in style to several mathematical problems in economics, for example how the consumer should decide to maximise their utility by purchasing two goods, while keeping to a budget. For an example of this kind of problem in utility maximisation, which is beyond what is required for the TMUA, see the links here and here.
Graphs – Intersections and Equations of Lines
Questions in 2021 papers: paper 1 – questions 1, 8, 18, 20 ; paper 2 – questions 2, 7, 15.
These questions usually involve finding intersections of two or more lines on a graph. This can include finding tangents, intersections and so on. Alternatively you could need to derive an equation for a particular line.
So for these types of questions, make sure you are familiar with the formulae for circles, straight lines, tangents and so on. Also you may need to be aware of circle theorems in a few cases.
A common issue includes not being able to visualise the question. So I recommend a quick 10-second sketch of a diagram if you are unsure.
Example
How many times does the line \[y=\frac{1}{5}x^5-x+1\] intersect the x-axis?
Answer
It is not so easy to solve the equation [\frac{1}{5}x^5-x+1=0\]
So instead we will take a different approach.
Let’s find the stationary points of this function.
\[\frac{dy}{dx}=x^4-1=0\]
\[(x^2+1)(x^2-1)=0\]
\[(x^2+1)(x+1)(x-1)=0\]
Thus there are two roots for x values -1 and 1.
Consider the first stationary point in more detail. Does it lie above or below the x-axis and is it a maximum or minimum?
\[x=1, y=\frac{1}{5}\]
\[\frac{d^2y}{dx^2}=4>0\]
So this point lies above the x-axis and is a local minimum.
Repeat this for the other stationary point
\[x=-1, y=\frac{9}{5}\]
\[\frac{d^2y}{dx^2}=-4<0\]
So this point lies above the x-axis and is a local maximum.
The minimum is above the x-axis and so is the maximum. The maximum comes before the minimum. We know the function is continuous. We also know that the line will start below the x-axis for large negative x values and be above the x-axis for large positive x values.
Hence there is only one intersection of the x-axis. The graph looks something like this:
Example 2
This combines the two question types above:
Suppose \(y=x^3-3x^2+2xb\), where \(b\) is a constant. For what range of values of b are there both two stationary points and at most one real root?
Method
Firstly, at stationary points, the first derivative of \(y\) with respect to \(x\) is zero. We can use this fact to identify a condition that must hold at any stationary point:
\[\frac{dy}{dx}=3x^2-6x+2b=0\]
The number of unique solutions to this equation will determine the number of stationary points. Consequently, to have two stationary points, the discriminant of the quadratic equation above must be strictly greater than zero. In other words:
\[(-6)^2-4(3)(2b)> 0\]
\[36-24b>0\]
\[b<\frac{3}{2}\]
Furthermore, the second condition required on \(b\) is that there must be at most one real root. To find real roots, we set \(y=0\):
\[y=x(x^2-3x+2b)=0\]
Hence this factorisation gives the first root \(x=0\). Now, for there to be at most one root, the second component \(x^2-3x+2b\) must have no roots itself. As this is again another quadratic equation, we should consider the discriminant of this quadratic too. We require the discriminant to be below zero so that there are no further roots:
\[(-3)^2-4(1)(2b)<0\]
\[9-8b<0\]
\[b>\frac{9}{8}\]
Then, combining these conditions on \(b\) we get the final answer:
\[\frac{9}{8}<b<\frac{3}{2}\]
Difficult Dimensions
This is an extra question type. Note it has appeared rarely in recent papers but is possible to come up. Consider this question type an extension task.
Finally, the question type “difficult dimensions” refers often to questions asking for the area or volume of non-standard shapes.
In these types of questions, try to break down the difficult shapes into simpler shapes, whose areas you can calculate separately. You should also be familiar with the formulae for areas of circles, sectors, segments, other shapes as well as circle theorems.
To demonstrate, let me run through following example:
Example
Consider the following shape as below. The circle has four tangents of equal length \(AB=CD=DE=AF\). Let angle BAF = angle CDE = \(x\) radians. Let the radius of the circle be \(r\) Calculate the area of the shaded regions divided by the area of the circle, in terms of \(x\) and \(r\) only.
Method – First Shape Breakdown
To summarise, the technique for these types of questions is to break down the shapes into manageable sections. Consider the following shape:
Note O is the centre of the circle. We know ABO is a right angle from the circle theorem that the tangent meets the radius of the circle at a right angle. We know OB has length \(r\) and the inner angle OAB is \(0.5x\). So the area of the triangle OAB is
\[0.5r (AB)\]
Note \(tan(0.5x)=\frac{r}{AB}\), so we can rewrite the area of OAB in terms of \(r\) only:
\[\frac{0.5r^2}{tan(0.5x)}\]
Second Shape Breakdown
Next, consider the following shape:
In addition to the diagram above, I have now added to the diagram a quarter of the circle. To work out one quarter of the shaded area, we need to subtract the relevant sector of the circle away from the area of the triangle.
First of all, the area of the quarter of the circle is \(0.25\pi r^2\).
Then for sector OBG, note OB and OG are both radii of the circle and so have length \(r\). Also the inner angle BOG is equal to 90 minus the inner angle AOB, in other words \(90-(180-90-(0.5x))=0.5x\)
Following this, the area of the sector OBG is then
\[\frac{0.5x\pi r^2}{2\pi}=0.25xr^2\]
After this, subtract the area of sector OBG from the area of the quarter circle: to give the area of the relevant sector:
\[ 0.25\pi r^2 – 0.25xr^2 =0.25(\pi-x)r^2\]
Next, to find one quarter of the shaded area, subtract the relevant sector of the circle away from the triangle’s area:
\[\frac{0.5r^2}{tan(0.5x)}- 0.25(\pi-x)r^2 \]
As a result, the total shaded area is then four times this:
\[\frac{2r^2}{tan(0.5x)}-(\pi-x)r^2 \]
which is the final answer.
Soon, more resources should follow on the TMUA. For more resources on the TMUA, personal statements and Oxbridge interviews (particularly for economics), see the link below: