Last Updated: 22 nd October 2014

Index
Click to go to the corresponding section.
Comparing Values
Sequences
Trapezium Rule
Number Theory
Area/Perimeter
Remainder
Theorem
Logarithms
Circles
Graph Sketching
Calculus
Reasoning about
Solutions
Trigonometry

General Points
1.
The Oxford MAT paper is the admissions test used for applicants applying to Oxford for
Mathematics and/or Computer Science, or to Mathematics at Imperial.
2.
It consists of two sections. The first is multiple choice, consisting of 10 questions each worth 4% each (for a total of 40%). The second consists of 4 longer questions, each worth 15% (for a total of 60%). We deal with the first section here.
3.
The paper is non-calculator.
4.
You need roughly 50% to be invited for interviews. However, successful maths applicants have an average of around 75%.
5.
The questions only test knowledge from C1 and C2. You must ensure you know the content of these two modules inside out. You should also keep in mind that the MAT
won’t test you on theory you wouldn’t have covered, so should think in the context
of what you can do.
6.
The multiple choice questions become progressively easier (and quicker) the more you practice. So practice these papers regularly. Even redoing a paper you’ve done before has value.
7.
I’ve grouped some of the questions from these papers here by topic, to help you spot some of the common strategies you can use.
8.
You can find a huge collection of past papers (many with mark schemes) here: http://www.mathshelper.co.uk/oxb.htm

Preliminary Tips
• You often have to compare logs. If one gives you a rational number and the other not, then form an inequality and rearrange to see if one is bigger than the other. E.g.:
Which is bigger: 𝐥𝐨𝐠
𝟗
> log
5
𝟐𝟕 or 𝐥𝐨𝐠
𝟓 log
9
𝟕 ?
27 = log
3 log
3
27
9
=
3
2
Suppose
3
2
5
3
2 > 8
8 . Taking 5 to the power of each side:
• Make approximations where appropriate. E.g. log 𝜋
10 is roughly 2.
• Similarly, form inequalities by considering approximations. E.g. log
4
15
15 < 2 is approximately 2, but slightly less than 2, i.e. we know 1 < log
4
• Remember that log 𝑎 𝑏 = 1/ log 𝑏 𝑎
• Remember that log 𝑎 < 0 when 𝑎 < 1 .

Comparing values

Key Points
1. Make sure you know your sin/cos/tan of 30/45/60 degrees.
2. Be comfortable with making estimates. 3 is somewhere between 1.5 and 2. Similarly log
2
15 is just less than 4.
3. Know your laws of logs like the back of your hand!

Comparing values

Key Points
1. Changing bases occasionally allows us to evaluate less obvious logs.
For example log
4
8 =
2. We can see that log
3 log
2 log
2
8
4
=
3
2
2 < 1 < log
2
3
3. Forming inequalities and manipulating them often helps us compare
logs. If we’re trying to find which of log
4
If log
5
10 < log
4
8 ,  log
5
10 <
3
2
 10 < 5
3
2
8 and log
5
10
(by taking 5 to the power of each side) are bigger, then:
 100 < 5 3  100 < 125 , so we were right.
We can use the same technique to show that log
2
3 is bigger.

Comparing values

Key Points
1. Again, realise that 𝐥𝐨𝐠 𝒂 𝒃 is less than 1 when 𝒂 is larger than 𝒃 .
2. The root inside outside and outside the log function has different
effects. In (d), we can put
1
2 can’t do this.
on front of the log instead. With (b), we
3. Realise that squaring a number between 0 and 1 makes it smaller.
Alternatively: log
10 𝜋 is approximately
1
2
, because log approximate (b), (c) and (d).
9
3 =
1
2
. We can use this value to

Comparing values

Preliminary Tips
• Know your formulae for the following and be able to apply them quickly:
• Sum of an arithmetic series: 𝑆 𝑛 𝑛
= 2𝑎 + 𝑛 − 1 𝑑
2 𝑎 1−𝑟 𝑛
•
•
Sum of a geometric series: 𝑆 𝑛
=
Infinite sum of a geometric series: 𝑆
1−𝑟
∞ 𝑎
=
1−𝑟
• Two series might be interleaved, e.g. 1, 3, 2, 6, 4, 12, 8, … Find the sum of each in turn.
• In some MAT questions (as well as Senior Maths Challenge) it sometimes helps to consider the ‘running total’ when you have a series that oscillates between negative and positive values. You’ll see an example of this.

Sequences
2011

Key Points
1. With sequences, remember that for the (n+1) th term, we can just replace
n with n+1. So just substitute these expressions into the inequality.
2. We can sometimes exploit the fact the questions are multiple choice.
Once we simplify to −3𝑛 2 + 15𝑛 + 8 > 0 , subbing the four different values in tells us (a) must be the answer.
3. Although of course, we could use a ‘C1’ approach to solve a quadratic inequality.

Sequences
2010

Key Points
1. If you have two interweaved sequences, find formulae for them separately.
This means we just want the sum of the first n terms from each of the two.
2. Know your formulae for the sum/infinite sum of a geometric series like
the back of your hand.

Sequences
2009

Key Points
1. Sometimes it helps to think about the ‘running total’ as we progress along
the sum. Our cumulative totals here are 1, -1, 2, -2, 3, -3, …
2. Alternatively, try to spot when you can pair off terms such that things either cancel or become the same. In this case, 1-2 = -1, 3-4=-1, and so on.
Although this makes it harder to spot exactly when we hit 100 in this case.
3. Think carefully about what happens at the end. Looking at the running totals, if the 1 st is 1, the 3 rd is 2, the 5 th 3, then the (2n-1) th gives us n. So when our running total was 100, 2n-1 = 199.

Preliminary Tips
• The Trapezium Rule is:
𝐴 = 𝑤
2 𝑦
1
+ 2 𝑦 where w is the width of each strip.
2
+ ⋯ + 𝑦 𝑛−1
+ 𝑦 𝑛
• Know when the trapezium rule overestimates or underestimates area.
Overestimates when line curves upwards.
Underestimates when line curves downwards.

Trapezium Rule
2010

Key Points
1. Remember that that the trapezium rule uses equal length
intervals. This suggests that the boundaries of the strips have to coincide with the dots in the diagram, otherwise our trapeziums wouldn’t exactly match the function.
2. (a) we can eliminate because multiples of
1
4 won’t include a
1
3
.
We can eliminate (b) and (c) in a similar way.
3. In general, it’ll be the LCM of the
denominators.

Trapezium Rule
2009

Key Points
1. This is all about being proficient with the sum of a geometric series. The question is more a test of your dexterity, not your problem solving skills.
1
2. By the trapezium rule, we get
2𝑁
1
2
𝑁
1
(1 + 2 2
𝑁
2
+ 2
𝑁
+ ⋯ + 2
𝑁−1
𝑁
1 and 𝑛 = 𝑁 − 1 , for the summation in the middle we get
2 𝑁
1
1−2 𝑁 𝑁−1
1
1−2 𝑁
1
+ 2) . So using 𝑎 = 2
𝑁 and 𝑟 =
. Simplifying, we end up with (b).
3. Occasionally we can absorb terms into the geometric series to make our life easier. Since the first length inside the outer brackets was 2 0 , it makes much more sense to make 𝑛 = 𝑁 and 𝑎 = 1 .
4. Be careful in considering the number of points you use in the trapezium rule. If you have 𝑁 regions, then you actually have 𝑁 + 1 points, 𝑁 − 1 of which are duplicated.

Trapezium Rule
2008

Key Points
1. Thinking about the question visually helps. A function which curves
upwards will give an overestimate, and a function which curves downwards gives an underestimate.
2. (d) is the only transform which changes the shape of the curve, giving us a reflection on the y-axis (in the line y=1). A curve for example curving up will now curve down, giving us an underestimate.

Preliminary Tips
• For many problems it helps to find the prime factorisation of a number:
• A number is square if and only if the powers in its prime factorisation are even.
• Similarly, a number is a cube if and only if its powers are divisible by 3, and son on.
• We can get the number of factors a number has by adding 1 to each power. E.g. 60 = 2 2 × 3 1 × 5 1 , so it has 3 × 2 × 2 = 12 factors.
• Diophantine Equations are equations where you’re trying to identify integer solutions. There will never be anything too difficult on this front relative to the
Maths Olympiad for example, but it may be worth seeing the RZC Number
Theory slides on this topic.
• Know your divisibility rules (also on the RZC Number Theory slides): A number is divisible by 3 if its digits add up to a multiple of 3, divisible by 4 if its last two digits are divisible by 4, divisible by 6 if it’s both divisible by 2 and 3, and so on.

Number Theory
2008

Key Points
1. These kinds of questions are quite common in Maths Challenge papers. The key is to systematically consider how many times each digit appears for
the units digit, and then the tens digit (rather than considering each full number in turn).
2. For the units digit, 1 to 9 is each seen 10 times (0 is seen 9 times, but this doesn’t matter because it doesn’t contribute to the sum). And 10 ×
1 + 2 + 3 + ⋯ + 9 = 450 .
3. Each tens digit, 1 to 9, occurs 10 times. This again gives 450, so our total is
900.

Number Theory
2008

Key Points
1. From the RZC lecture notes, remember that a number is square if all the
powers in the prime factorisation are even.
2. The prime factorisation is 2 𝑘 × 2 2𝑚 × 2 3𝑛 = 2 𝑘+2𝑚+3𝑛
.
3. Be adept with recognising odd/evenness when considering sums and
products. If 𝑘 + 2𝑚 + 3𝑛 is even, then consider which of the four statements supports this. E.g. If 𝑘 + 2𝑚 + 3𝑛 is even, 𝑘 + 3𝑛 is even
(because 𝑒𝑣𝑒𝑛 + 𝑒𝑣𝑒𝑛 = 𝑒𝑣𝑒𝑛 ), and since 𝑘 + 3𝑛 = 𝑘 + 𝑛 + 2𝑛 , 𝑘 + 𝑛 is even.

Number Theory
2009

Key Points
1. This clearly looks like a binomial expansion! As per the RZC algebra slides, you should always try to spot potential factorisations in number theory
problems.
Factorising gives 𝑥 + 2𝑦
2. Then 𝑥 + 2𝑦 = 2 10
3 = 2 so 𝑥 = 2 10
30
− 2𝑦 = 2(2 9 − 𝑦) . Since 𝑥 and 𝑦 must be positive (which does not include 0), 𝑦 can have any value between 1 and
2 9 − 1 so that 𝑥 remains at least 1.

Number Theory
2007

Number Theory
2010

Preliminary Tips
• Remember the ‘ 2 ’ trick: The diagonal of square is 2 times longer than the side. Similarly, the side is 2 times shorter than the diagonal.
• For circles, add key radii at strategic places.
• Split up the shape into manageable chunks (e.g. right-angled triangles). You’re likely to be able to use either simple trigonometry or Pythagoras.
• You MUST memorise sin 30/cos 30/sin 60/cos 60. Remember also that sin 𝑥 = sin(180 − 𝑥) and cos 𝑥 = cos(360 − 𝑥) , which helps if you’re trying to work out sin 150 without a calculator.
• Don’t forget your circle theorems (although they don’t really feature very prominently in MAT).
• Sometimes you can come up with two expressions for the same length. Example:
1 1
2 r
The radius of the big circle is 1. What is the radius of the small circle?
2 = 1 + 𝑟 + 2𝑟
2−1
So 𝑟 =
2+1
?

Area/Perimeter
2012

Key Points
10
2𝜋
120°
10
2𝜋
• As per usual, draw in the radius of the circle at
strategic places (in this case, where the triangle touches the circle).
• This allows us to divide up the triangle into manageable chunks.
• The area of one third of the triangle is
1 1 100 75 3
2 𝑎𝑏 sin 𝜃 =
2 4𝜋
2 sin 120 =
4𝜋
2
• By splitting each isosceles triangle into two right angle ones, then each half of the side of the
10 triangle is
2𝜋 sin 60 =
10 3
. There’s 6 of these
4𝜋 half lengths.
• Using these A and P, we find (a) is true.

Area/Perimeter
2012

Key Points
1. I initially tried to think how the orientation of the ‘spotlight’ affected the area covered. I realised that pointing it symmetrically at the opposite end maximised the area.
2. My approach was to draw radii from Q and R to the centre (where angle QOR = 𝜃 using
2 circle theorems) and P to the centre. This gave me a sector and two triangles, and a bit of simple geometry led me to (b).
3. However, we could again exploit the fact we have multiple choice to try a specific case. If 𝜋 we were to choose 𝜃 = , we have a semicircle and a right-angled triangle, which quickly
2 gives us 𝜋
2
+ 1 . This gives us (b).

Area/Perimeter
2011

Key Points
1. A sensible first step is to form equations for the perimeter and area. Say 𝑃 =
2𝑥 + 2𝑦 and 𝐴 = 𝑥𝑦 .
2. Often we can introduce an inequality when equations are involved by using the
discriminant.
3. Using substitution: 𝑃 = 2𝑥 + 2
Then using the discriminant: 𝑃
2
𝐴
→ Px = 2x
2 𝑥
− 16𝐴 ≥ 0
+ 2A → 2x
2
− Px + 2A = 0 .
Alternatively, we could have exploited the fact that the question is multiple choice.
Choosing a few possible widths and heights will eliminate the incorrect possibilities.

Area/Perimeter
2006

Key Points
1. If the sides of the triangle are 𝑎 , 𝑏 and 𝑐 , start with 𝑎 2 + 𝑏 2 = 𝑐 2 by
Pythagoras.
2. We could always find the areas of the triangles, but it’s quicker to realise that the area of each triangle is proportional to the area of each implied square around the triangle, say 𝑘𝑐 2
, so 𝐴 + 𝐵 = 𝐶 .
𝐴 = 𝑘𝑎
2
, 𝐵 = 𝑘𝑏
2 and 𝐶 = 𝑘𝑐
2
. Then 𝑘𝑎
2
+ 𝑘𝑏
2
=

Preliminary Tips
• Nothing much to say here!
• The remainder when 𝑓(𝑥) is divided by (𝑎𝑥 + 𝑏) is 𝑓 − 𝑏
. If (𝑎𝑥 + 𝑏) is a 𝑎 𝑏 factor then clearly 𝑓 − = 0 because there’s no remainder.
𝑎
• You might have to factorise the factor first! E.g. If 𝑓(𝑥) is divisible by 𝑥 2 + 𝑥 = x(x + 1) , then 𝑓 0 = 0 and 𝑓 −1 = 0 .

Remainder Theorem
2008

Remainder Theorem
2006

Remainder Theorem
2009

Key Points
• If 𝑥
2
− 1 is a factor then both 𝑥 + 1 and 𝑥 − 1 is a factor.
• By the remainder theorem, 𝑓 −1 = 0 and 𝑓 1 = 0 . For the latter, we get 𝑛 = 15 or 𝑛 = 15 .
• However, for the former, the middle term will be positive or negative depending on whether 𝑛 is odd or even. 𝑛 can only be 10 or -15.
• 𝑛 = 10 is the only case when both are satisfied.

Preliminary Tips
• If you see numbers like 2 and 8 together for example, you should be able to spot that 2 3 = 8 and somehow use that to simplify. The ‘related groups’ you’ll see are usually powers of 2: {2, 4, 8, 16, 32, … } and powers of 3:
{3, 9, 27, 81, … } .
• log 𝑎 𝑏 = log
1 𝑏 𝑎
• You need to know how to change the base: log 𝑎 𝑏 = log 𝑐 log 𝑐 𝑏 𝑎
• log 𝑎
• 𝑎 𝑥 and 𝑎 log 𝑎 𝑥 𝑥 are the inverse of each other. This means for example that:
= 𝑥
• log 𝑎 𝑎 𝑥 = 𝑥

Logarithms
2011

Key Points
1. Some log tips that frequently crop up: 𝑎 log 𝑎 𝑥 = 𝑥 and log 𝑎 𝑎 𝑥 = 𝑥 . This is because the exponential and logarithm functions are inverses of each other, so applying one then the other to 𝑥 gets us back to 𝑥 .
2. Change bases of the exponential to be consistent with the log, so we can cancel in this way.
3. Whenever you have cubics (in this case 𝑥 3 − 2𝑥 2 − 𝑥 + 2 = 0 ), MAT papers are generous, usually allowing you to factorise the first two terms and the last two terms. i.e. 𝑥 2 𝑥 − 2 − 1 𝑥 − 2 = 𝑥 2 − 1 𝑥 − 2 = (𝑥 + 1)(𝑥 − 1)(𝑥 − 2) .

Logarithms
2007

Key Points
1. The principle that squared terms are always > 0 comes up a LOT in MAT papers (as well as BMO/SMC), so it’s worth building up a sixth sense for spotting when we can apply this principle.
2. Note that the squaring is outside the log function, so we can’t move the 2s to the coefficients.
3. To make a maximum, we need to make b minimum. When 𝑏 = 1 then its log will be
0. Then solving, we end up with (c). Note that choosing 𝑏 < 1 would make this term larger, as the log of a number in the range 0 < 𝑏 < 1 is negative, but this would be squared.

Preliminary Tips
• As per C2, if you have a mixture of 𝑥 2
, 𝑥 , 𝑦 2 and 𝑦 , complete the square.
• When considering ‘nearest points’ on a circle’s circumference, draw a line from the centre of the circle to the point of interest/centre of another
circle (see below).
• Consider the conditions for which the equation of a circle are valid: The radius 𝑟 must clearly be positive, and similarly 𝑟 2 must be positive.
The nearest point to this dot on the circumference of the circle can be found by drawing this straight line from the centre.
Nearest points to each other.

Circles
2012

Key Points
1. Drawing a quick sketch will often massively help (and immediately eliminates a number of possibilities).
2. (a) goes through the top-most and right-most point of the circle.
3. For (c), the x value occurs before the rightmost point where x=2.
4. Similarly, (d) the y-intercept of 2 occurs before the top-most point where y=2.
5. This leaves (b). We needn’t do any geometry, but it wouldn’t be too difficult to do so if we wanted to verify it.

Circles
2011

Key Points
1. A general principle is to reflect on the valid range of values for different standard formulae.
1. With a quadratic for example, the discriminant must be > 0 if it has a solution.
2. Similarly here, for 𝑥 − 𝑎
2
+ 𝑦 − 𝑏
2
= 𝑟
2
, then 𝑟
2 must clearly be positive, since a squared number is always positive.
2. As you would usually do in a C2 exam, complete the square. This gives us (𝑥 +

Circles
2009

Key Points
(-3, -4)
1. As per usual, complete the square, and
then a sketch may help.
2. By inspection, we can see the nearest point to the origin must be on the line that goes from the centre and through the origin.
3. Since the radius is 10 and the distance from the centre to the origin is 5 (by Pythagoras), then the answer must be 5.

Circles
2007

Key Points
1. Again, draw a diagram!
2. Drawing a line between the centres of the circles often helps for questions like these. We can see visually that the nearest point must lie on this line.

Circles
2006

Preliminary Tips
• The RZC Graph Theory slides go through a lot of detail on this!
• Since this is multiple choice, you should look out for the following features:
• Is the y-intercept correct?
• What happens as 𝑥 → ±∞ ?
• Do the turning points look right?
• Do the roots (i.e. x-intercepts) look right?
• Think about whether a transformation is inside some other function or
outside. Compare sin 𝑥 2 and sin 𝑥 for example: in the first, as 𝑥 increases, the input we use for sin increases more rapidly, so we accelerate across the sin graph more rapidly. The result is that is gradually oscillates more quickly.
• Squaring a function always results in a positive value.
• In polynomial equations:
• A single factor of (𝑥 − 𝑎) means the curve crosses at 𝑥 = 𝑎 .
• A repeated factor of 𝑥 − 𝑎 2 means the curve touches the x-axis at 𝑥 = 𝑎 .

Graph Sketching
2011

Key Points
1. Consider the y-intercept (by using x = 0)
2. Consider the roots (by using y = 0)
3. Cubics given in MAT exams are often conveniently factorisable. In this case, we can see we can factorise it to 𝑥 2 𝑥 − 1 − 1 𝑥 − 1 = 𝑥 2 − 1 𝑥 − 1 = 𝑥 + 1 𝑥 − 1 2
.
4. But we didn’t even need to do this in this case, because we can eliminate (a) because it’s in the wrong direction, eliminate (b) because the y-intercept is wrong, and eliminate (d) because 𝑓 1 ≠ 0 .

Graph Sketching
2008

Key Points
1. We could again immediately eliminate (a) and (b) by considering the y-intercept.
2. We could eliminate (d) by considering a small value of x just above 0, and seeing if the value increases or decreases relative to the y-intercept of −
1
5
. E.g. If 𝑥 = 0.1
, then 𝑦 = −
1
4.61
. This is more negative, which eliminates (d) where y increases.
3. But this is an inelegant method which shouldn’t be used on anything other than multiple choice, but we can’t guarantee our ‘small value’ occurs before the turning point. You could instead complete
1 the square to get − 𝑥−2 2 +1
. We can then see that the turning point occurs when 𝑥 = 2 , more easily eliminating (d).
4. In general, completing the square (or differentiation) allows us to find turning points. The fact we’ve taken the reciprocal doesn’t change the 𝑥 at which they occur.

Graph Sketching
2007

Key Points
1. It’s quite common in MAT (and in interviews) to consider the effects both inside and
outside of trigonometric functions.
2. Inside the sin, 𝑥 2 increases more rapidly than 𝑥 would, so our periodicity decreases.
3. Remember that a squared value is always positive!
4. Consider the shape of 2 −𝑥
. This gradually decreases (although is always positive). Since we’re multiplying this by our trigonometric function, the peaks will gradually get
smaller.
5. This evidence so far gives us (a) or (d). But sin 0 = 0 , which eliminates (d).

Graph Sketching
2010

Key Points
1. This is similar to the previous question!
2. Inside the sin, 𝑥 increases more slowly than 𝑥 . So we’re going across a sin graph increasingly slowly, reducing the periodicity. This eliminates (a) and (d).
3. We’re squaring the value, so it must be positive (although we’ve already eliminated (a)).
4. Now consider what happens to the peaks. The peaks will all still be 1, since squaring the value or changing the periodicity won’t change this maximum. That eliminates (c).

Graph Sketching
2012

Key Points
1. This question tests whether you’re comfortable with recognising that repeated factors lead to the x-axis being touched (straight from C1).
2. It can’t be (b) because the origin is not a root in the graph.
3. It can’t be (a) because the curve should touch at -3 and +3 but cross at +1, which is not consistent with the graph.
4. Equation (d) factorises to 𝒚 = 𝒙 − 𝟏 𝟐 𝒙 + 𝟏 𝟐 (𝟑 − 𝒙) . So both (c) and (d) initially seem feasible. But considering the y-intercept, we see that only (d) is positive.

Preliminary Tips
• Remember that definite integration represents the area under the graph. For some questions, you needn’t actually perform the integration, you just need to consider graphically if the area is positive or negative.
• It may sometimes help to sketch the graph (particularly in light of the above).
• MAT only expects you to know how to differentiate and integrate polynomials.
They can’t expect you to know how to deal with trigonometric functions or exponential functions – in such instances you need to more generally reason.
• Whenever you see 𝒅𝑨 𝒅𝑩
, you should think to yourself “how does A change as B
changes?”. Thinking this statement may help you which conceptually more difficult questions, or where you wouldn’t actually be expected to work out what 𝑑𝐴 is (e.g. because you don’t know how to differentiate such a function).
𝑑𝐵
• If you want to integrate a function where the gradient suddenly changes, e.g.
1
−1
|𝑥| 𝑑𝑥 , then you have to split it up into pieces based on where the
1
‘breaks’ occur and integrate each. e.g.
−1
|𝑥|𝑑𝑥 = 2
0
1 𝑥 𝑑𝑥
• Whenever you see the words ‘minimum’ or ‘maximum’, think differentiation!

Calculus
2009

Key Points
1. It might be helpful to first find what I(a) is. If we expand out the bracket, we get I 𝑎 =
0
1 𝑥 4 − 2𝑎𝑥 2 + 𝑎 2 𝑑𝑥 =
1
−
2 𝑎 + 𝑎 2
5 3
2. We’re asked for the smallest value of 𝐼(𝑎) as 𝑎 varies, so differentiate w.r.t. 𝑎 and set to
0 (as we have a stationary point) and we get 𝑎 =
1
3
.
1 1 2 1 1
2
4
3. So the question is asking us for 𝐼 = − + =
3 5 3 3 3 45
4. A key thing to reflect on here is that the question is purposely trying to bamboozle you by combining integration and differentiation, as well as the fact that you’re integrating with respect to 𝒙 , but differentiating with respect to 𝒂 (i.e. different variables). But as long as you carefully consider what you’re trying to do at each step, and with respect to what variable you’re differentiating/integrating, then you’ll be OK!

Calculus
2012

Key Points
1. You don’t need to actually do any integration here (and you won’t be able to unless you’ve done C3!)
2. Looking at the multiple choice options, we only care if T is positive/negative/0.
3. Thus for each of the integrals, we only care whether the area under the graph is above
the x-axis or below the x-axis.
4. By sketching the 3 graphs, we find the first area is positive, the second negative and the third positive. Thus T is negative.

Calculus
2011

Key Points
This one is pretty tricky!
The complication obviously comes from the fact that we’re transforming the input we’re using for the function we’re integrating.
A good place to start is to note we’re integrating 𝑥 between 0 and 1, i.e. −1 ≤ 𝑥 ≤ 1 . Then
− 1 ≤ 𝑥
2
− 1 < 0 , so after 𝑥 is transformed before being used in the function, the input to the function is going to vary between only −1 and 0. The equation of the function in this range is f(x) = 𝑥 + 1 . But 𝑓 𝑥
2
− 1 = 𝑥
2
− 1 + 1 = 𝑥
2
.
1
So we have
−1 𝑥 2 𝑑𝑥 =
2
3

Calculus
2010

Key Points
1. It helps that you know that the paper can’t expect you to know how to integrate 2 𝑥
2
! It suggests we’re going to have consider a graphical method.
2. If in doubt, sketch your function.
3 𝑦 = 4 − 2
√2 𝑎 𝑥
2
3. I(a) gives us the area up to 𝑥 = 𝑎 .
4. Now consider what 𝑑𝐼/𝑑𝑎 , actually means.
It’s asking when the area doesn’t change as 𝒂
changes. Notice that as 𝑎 goes past 2 , the area starts to decrease because there’s a negative area beyond 2 . Thus the area stops increasing and is about to decrease when 𝑎 = 2 .

Calculus
2008

Calculus
2007

Key Points
1. The key part here it that divide an area up into smaller chunks, and sum the areas of these smaller chunks. For example
0
2 𝑓 𝑥 𝑑𝑥 =
0
1 𝑓 𝑥 𝑑𝑥 +
1
2 𝑓 𝑥 𝑑𝑥
2. Also note that we can take the constant factor outside the integral.
3. So if we let for example 𝑎 =
0
1 𝑓 𝑥 𝑑𝑥 and 𝑏 =
1
2 𝑓 𝑥 𝑑𝑥 , then we’re trying to find 𝑎 + 𝑏 , when we know that 3𝑎 + 2𝑏 = 7 and 𝑎 + 𝑏 + 𝑏 = 1 . These are obviously just simultaneous equations.

Calculus
Important Note: This question is sufficiently old that it was before the ‘C’ modules existed at A Level (instead of C1-4 and FP1-3, there was P1-6). This kind of content would now appear in C3, and thus a question like this would no longer appear in a MAT.
2006

Preliminary Tips
There’s three ways to consider the number of solutions:
METHOD 1: Factorise (when possible!) e.g. 𝑥 3 − 𝑥 2 − 𝑥 + 1 = 0
This cubic conveniently factorises to: 𝑥 2 𝑥 − 1 − 1 𝑥 − 1
= 𝑥 2 − 1 𝑥 − 1
= 𝑥 + 1 𝑥 − 1 2
We can see it has three solutions (two of them equal).
Look out for the difference of two
squares!!!
METHOD 3: Consider the discriminant
Remember that if there are real solutions, 𝑏 2 − 4𝑎𝑐 ≥ 0
METHOD 2: Reason about the graph
(Example from STEP)
Sketch 𝑦 = 𝑥 4 − 6𝑥 2 + 9 . Thus state the values of 𝑏 for which the equation 𝑦 = 𝑥 4 − 6𝑥 2 + 𝑏 has:
(a) 0 solutions
(b) 1 solution
(c) 2 solutions
(d) 3 solutions 𝑦 = 𝑥 4 − 6𝑥 2 + 9
− 3 + 3
As 𝑏 changes, the graph slides up and down, we can see the number of roots change. Clearly if 𝑏 >
9 then the curve won’t touch the 𝑥 -axis for example.

Preliminary Tips
More on METHOD 2: Reason about the graph
• Cubics will always have at least one solution, because the y-value goes from −∞ to
∞ . In general, this is true when the greatest power is an odd number.
• Polynomials where the greatest power is even however (e.g. quadratic/quartic) have a global minimum or maximum.
• Considering the turning points often allows us to reason about solutions.
𝑚𝑎𝑥 → 𝑚𝑖𝑛 → 𝑚𝑎𝑥 → 𝑚𝑖𝑛 𝑚𝑎𝑥 → 𝑚𝑖𝑛 𝑚𝑖𝑛 → 𝑚𝑎𝑥 → 𝑚𝑖𝑛
Cubic (where coefficient of 𝑥 3 is positive)
Quartic (where coefficient of 𝑥
4 is positive)
Quintic (where coefficient of 𝑥 5 is positive)

Reasoning about Solutions
2009

Key Points
•
•
•
• Spot when you can use the difference of two squares (also useful for the SMC/BMO!)
• Make use of the discriminant.
Thus 𝑥 4 − 𝑥 − 𝑐 2 = 0 𝑥 2 + 𝑥 − 𝑐 𝑥 2 − 𝑥 + 𝑐 = 0
If 𝑥 2 + 𝑥 − 𝑐 = 0 , and using the discriminant on the first, 𝑐 ≥ −
1
4
.
1
Using the discriminant on the second 𝑐 ≤
4

Reasoning about Solutions
2009

Key Points
• By differentiating to find the turning points: 12𝑥 𝑥 3 − 4𝑥 2
3
+ 3𝑥 = 0 𝑥 𝑥 2 − 4𝑥 + 3 = 0
− 48𝑥
2
+ 36𝑥 = 0 𝑥 𝑥 − 1 𝑥 − 3 = 0
So the turning points occur at 𝑥 = 0, 1, 3 . Then considering the graph of the quartic:
If the x-axis is anywhere in the horizontal trip between the maximum and the greater of the two minimums
(whichever it is), we’ll have four solutions because the line will cross the axis 4 times. The y-values of the turning points are 𝑘, 5 + 𝑘 and 𝑘 − 27 respectively. So 5 + 𝑘 > 0 so the maximum is above the x-axis, and 𝑘 < 0 so that the greater of the two minimums occurs below the x-axis.

Reasoning about Solutions
2008

Key Points
•
• If we could form a quadratic, then when can clearly use the discriminant: 𝑏
2
− 4𝑎𝑐 ≥ 0
• This is very much in the style of a C2 log question. The only trickier thing here is realising that
3 2𝑥
9 𝑥 = 3
= 3 𝑥 2
.
2 𝑥 = 3 𝑥 2
, whereas in C2 papers, you only needed to be able to spot that
9
The quadratic we want is therefore 𝑦 2 − 3𝑦 − 𝑘 , where 𝑦 = 3 𝑥
. This gives 𝑘 ≥ −
4
• However, if 3 𝑥
=
3± 9+4𝑘
, we know 3 𝑥 must be positive (because exponential functions
2 always give positive values). The larger root will always be positive, so this is therefore not an issue.

Reasoning about Solutions
2007

Key Points
• It’s the same deal! The equation factorises to (2 𝑥 + 2)(2 𝑥 − 2)(2 𝑥 − 1) . As per usual, we were given a cubic which was nice and easy to factorise, and we could factorise the difference of two squares.
• So 2 𝑥
= −1 or 2 𝑥
= 2 or 2 𝑥 real solutions.
= 1 . The first one doesn’t give a real solution, so there’s two

Reasoning about Solutions
2006

Key Points
• When you have 𝒇 𝒙 + 𝒈(𝒙) , it’s often it’s useful to consider the graphs separately and
then add them:
Now it’s clear there’s no solutions.
But it’s even easier when we realise that because the modulus function always leaves a positive value, the sum of |𝑥| and |𝑥 − 1| must be positive. Thus to sum to zero, both have to be 0. From the first, 𝑥 = 0 . But then |𝑥 − 1| would be 1. So there can’t be any solutions.

Preliminary Tips
1. Often when considering the number of possible solutions to an equation which involves trigonometric functions, we need only consider the range of the trig
functions, as this may well constrain our solutions considerably: a) sin 𝑥 and cos 𝑥 obviously vary between -1 and 1.
b) sin
2 𝑥 and cos
2 𝑥 vary between 0 and 1.
2. In particularly, as before, remember that things that are squared are always
positive, i.e. their minimum value is 0.
3. Sometimes dividing the whole equation by 𝒔𝒊𝒏 / 𝒄𝒐𝒔 / 𝐬𝐢𝐧
𝟐 𝒙 etc. puts our equation just in terms of one trigonometric function (e.g. leading to a quadratic equation).

Trigonometry
2007

Key Points
• We know sin 𝜃 varies between -1 and 1 and thus sin
2 𝜃 between 0 and 1. The “ 10𝑥 +
11 ” has just been put there to throw you: we can still get any value to input into the sin.
• 𝑓(𝑥) therefore varies between 16 and 49. Thus the answer is (c).

Trigonometry
2008

Key Points
• Our usual trick is to consider how each expression can vary as sin/cos varies.
• We can see that 3 + cos 𝑥 since sin 8 𝑥
2 must be at least 4. Similarly, 4 − 2 sin is positive. Thus both expressions must be equal to 4.
8 𝑥 must be at most 4
• This happens when cos 𝑥 = −1 and sin 𝑥 = 0 . This gives the one solution 𝑥 = 𝜋 .

Trigonometry
2009

Key Points
• You should build up a sixth sense of thinking POSITIVE!!! whenever you see something squared.
• This means that sin 2 𝑥 and cos 2 𝑥 must be at least 0. So the minimum value of the LHS is
1 + 1 = 2 .
• Thus both have to be 0. But sin 𝑥 and cos 𝑥 aren’t both 0 at the same time, so there’s no solutions.

Trigonometry
2011

Key Points
• I personally drew a quick sketch of 𝑦 = sin 8 𝑥 and 𝑦 = cos 6 𝑥 , which are similar to sin 𝑥 and cos 𝑥 in that the peaks are still at 1 (with the same x values), but negative values are now positive, and the values (except 1) will be much closer to the 𝑥 -axis since we’re doing the power of a value < 1 . That means we’ll have solutions when sin 8 𝑥 = 1 (i.e. sin 𝑥 = ±1 ) and cos 6 𝑥 = 0 (i.e. sin 𝑥 = 0 ) or vice versa, which gives 𝑥 = 0, 𝜋
2
, 𝜋,
3𝜋
. But you can sort see from the sketch that they’ll be no solutions in
2 between, due to the lack of symmetry in the line 𝑦 =
1
2
, given that the powers of 𝑠𝑖𝑛 and 𝑐𝑜𝑠 are different.
• But here’s a better solution: The equation looks similar to sin 2 𝑥 = 1 and cos 2 𝑥 = 0 or vice versa, sin 8 𝑥 + cos sin
6
2 𝑥 + cos 𝑥 < sin 2
2 𝑥 = 1 , and thus unless 𝑥 + cos 2 𝑥 because taking the power of a number less than 1 reduces the value. Thus sin 8 𝑥 + cos 6 𝑥 < 1 except in these circumstances, and there are no other solutions. Remember you have to consider when both cos 𝑥 = 1 and cos 𝑥 = −1 , (and the same for sin) otherwise you’ll miss a solution.

Trigonometry
2010

Key Points
• The key here is dividing by cos 2 𝑥 . This gives us a quadratic in terms of tan 𝑥 . Factorising we get tan 𝑥 + 2 tan 𝑥 + 1 = 0 .
• Alternatively, you could have spotted that you can factorise the original equation as
(sin 𝑥 + 2 cos 𝑥) sin 𝑥 + cos 𝑥 = 0
• Each gives us 2 solutions.
• In general, think about what you might be able to divide by to simplify the equation.

Trigonometry
2008

Key Points
The key here is to get expressions for 𝑥 and 𝑦 in terms of 𝜃 . The only reason we might not have a solution for 𝜃 is if we had some division by 0. By suitable simplification we find there’s no division at all, so the answer must be (a).
To avoid a notational mess, it’s helpful to replace cos 𝜃 with just 𝑐 and so on, i.e. 𝑐𝑠 − 𝑠𝑦 = 2 and 𝑠𝑥 + 𝑐𝑦 = 1 . Then substitute one into the other!

Trigonometry
2011

Key Points
• The fact I saw the tangent suggested to me that I could form a right-angled triangle if
I drew in the radius again. I just let the radius be 1.
• I could then see the two triangles share a common length (say 𝑥 ). For the bottom triangle 𝑥 = 1/ sin 𝛼 . For the top triangle, using the sin 𝑥 rule: = 1/ sin 𝛽 .
sin 𝛾
• Putting these two together, we get (b).

Trigonometry
2011

Key Points
• It’s helpful to draw out the two graphs on the same axis, and then shade the appropriate regions.

Trigonometry
2011

Key Points
• Clearly if sin 𝑦 = cos 𝑥 , then 𝑥 = 𝑦 , which rules out (b) and (d), as no line goes through the original. But note that since cos/sin repeats every 2𝜋 , in general we have sin 𝑦 = cos 𝑥 + 2𝜋𝑛 for any integer 𝑛 , thus 𝑦 = 𝑥 + 2𝜋𝑛 . This gives us all the lines in (a).
• But note also that sin 𝑥 = sin 𝜋 − 𝑥 . So 𝜋 − 𝑦 = 𝑥 + 2𝜋𝑛 , i.e.
𝑦 = −𝑥 + 2𝜋 + 1 𝑛 .
This gives us the lines with gradient -1 in (c).