PRACTICE MAT SOLUTIONS

These solutions are for the practice test on the Pearson VUE system, created ahead of MAT 2024.
The questions are all past MAT questions from 2007–2022. The answers are collated below.

1. The centre is at (3, 4) and one corner is at (1, 5), which is a displacement of (−2, 1). There’s
another corner opposite, at (3, 4) − (−2, 1) = (5, 3), but this is not one of the options. To find
the other corners, we need to find a vector with equal magnitude, but at right-angles to (−2, 1).
Rotating by 90◦ gives (1, 2). The other corners are at (3, 4) ± (1, 2), which includes (2, 2).
The answer is (2,2).
Z 1
2. Using the difference of two squares, this is e2x − x2 dx. Notice that, since the derivative of
0
e2x is 2e2x , this integral is
1
1 2x x3 e2 1 1 3e2 − 5

e − = − − =
2 3 0 2 3 2 6
3e2 − 5
The answer is .
6
3. Pair the terms to get −3 − 7 − · · · − 199. This is the sum of an arithmetic progression and is
equal to (−3 − 199) × 50/2 = −5050
The answer is −5050.
4. Connect each point to the centre of the circle to split the shape into 12 isosceles triangles, each
with angle 30◦ at the centre. Then the area of each triangle is 21 × 1 × 1 × sin 30◦ = 14 , and
there are 12 triangles, for a total area of 3.
The answer is 3.
5. The integral is
a a a
√
Z Z 
2 1/2 2 1 2 1
x + x dx = x 2
+ x dx = x3/2 + x3 = a3/2 + a3 .
0 0 3 3 0 3 3

So we have 2a3/2 + a3 = 15. This factorises as (a3/2 − 3)(a3/2 + 5) = 0. Since a > 0 we want
a3/2 > 0, so it’s 3, so a = 32/3 .
The answer is a = 32/3 .
6. The gradient at p is ep and so the tangent is y = ep (x − p) + ep . This crosses the x-axis when
ep (a−p)+ep = 0 which happens if a = p−1. Similarly q = 1−b so p−a = q−b (they’re both 1).

(q, eq )

(p, ep )

a=p−1 b=q−1
 The answer is p − a = q − b.
7. The intersection point is at ex = 1 − ex which happens when 2ex = 1, that is x = − ln 2 < 0.

The area has reflectional symmetry in the line y = 12 so we want

Z 0   
x 1 h
x x i0 1 ln 2
2 e − dx = 2 e − = 2 (1) − + = 1 − ln 2
− ln 2 2 2 − ln 2 2 2
The answer is 1 − ln 2.
x ≥
0 and x2 + 1 = 3x or x < 0 and −x2 + 1 = −3x. The first equation√has
solutions
8. Either √
1
2
3 ± 5 which are both positive. The second equation has solutions 21 3 ± 13 and only
one of these is actually negative. So there are three solutions in total.
The answer is 3.
9. The relationship between circle Cn of radius rn and circle Cn+1 of radius rn+1 is shown below.

rn+1 1

rn
1

The tangent is at right-angles to the radius, so there’s a right-angled triangle with hypotenuse
2
rn+1 and other sides rn and 1. Pythagoras gives rn+1 = rn2 + 1. Since r12 = 1, we have r22 = 2
and r32 = 3 and so on, up to r100
2
= 100, so the radius of C100 is 10.
The answer is 10.
10. Complete the square for x and for y (x − 2k)2 − 4k 2 + (y − 2)2 − 4 + 8 = k 3 − k This is the
equation of a circle with centre (2k, 2) and r2 = k 3 + 4k 2 − k − 4 provided that expression is
positive. Factorising the cubic as (k + 4)(k − 1)(k + 1) reveals that this happens if and only if
either −4 < k < −1 or if k > 1.
The answer is “if and only if either −4 < k < −1 or k > 1”.
11. 3 cos2 x + 2 sin x + 1 = 3(1 − sin2 x) + 2 sin x + 1 = −3(sin x − 1/3)2 + 1/3 + 4. This takes its
maximum value when sin x = 1/3, and that value is 4 + 1/3 = 13/3.
13
The answer is .
3
12. The tangent is y = 2ax − a2 , which crosses the x-axis at (a/2, 0).

x
Q P

Z a
The area is x2 dx − (area of APQ) where A is (a, a2 ), P is (a, 0) and Q is (a/2, 0). This is
0
a3 a3 a3
− =
3 4 12
a3
The answer is .
12
13. Factorise the numbers from 1 to 10; this gives

log10 (2 × 5 × 32 × 23 × 7 × 2 × 3 × 5 × 22 × 3 × 2) = log10 (28 34 52 7).

Use log10 (22 52 ) = 2. Use 24 34 = 64 . This gives 2+log10 (22 64 7) = 2+2 log10 2+4 log10 6+log10 7.
The answer is 2 + 2 log10 2 + 4 log10 6 + log10 7.
dy
14. The turning points have dx = 0 at x = 1 and at x = 3, so 3 + 2a + b = 0 and 27 + 6a + b = 0.
Solve these for a = −6, b = 9. Now y = 2 at x = 1 so c = −2. Then y = −2 at x = 3 gives
d = −2.
The answer is −2.
       
10 1 3 10
15. In order to make the vector we would need a +b = where a is the number
  8 1 2  8
1 3
of times we pick and b is the number of times we pick .
1 2
Since we have six vectors, a + b = 6. Solving the simultaneous equations a + 3b = 10 and
a + 2b = 8, we get a = 4 and b = 2, and we can  check that a + b = 6 for this solution! So
3
we want exactly two of the six vectors to be . There are 6 C2 = 15 ways that this could
2
1 15
happen, each with probability , so the answer is .
64 64
15
The answer is .
64
16. The tangent at a is y = (3a2 − 3)(x − a) + (a3 − 3a), which passes through (2, 0) if and only if
0 = (3a2 − 3)(2 − a) + (a3 − 3a). This simplifies to 2a3 − 6a2 + 6 = 0. The left-hand side is a
cubic in a and we’d like to know how many roots it has.

The turning points of 2a3 − 6a2 + 6 are at a = 0 and a = 2, where the value of the cubic is
6 and −2 respectively. So this cubic starts negative, rises to a positive local maximum, then
decreases to a negative local minimum before rising again. There are therefore three roots for
this cubic, so three values of a for which the tangent to the original cubic passes through the
point (2, 0).
The answer is “three values of a”.

17. We can use the fact that sin2 (90◦ − n) = cos2 (n) for any n, and sin2 x + cos2 x = 1. So we have
sin2 1◦ + sin2 89◦ = 1 and sin2 2◦ + sin2 88◦ = 1 and so on up to sin2 44◦ + sin2 46◦ = 1. We also
have sin2 45◦ = 12 and sin2 90◦ = 1 for a total of 45 12 .
The answer is 45 21 .

18. The function inside the brackets is 6 sin2 x − 8 sin x + 3 which is a quadratic for sin x. We could
therefore consider the quadratic 6u2 − 8u + 3 for −1 ≤ u ≤ 1. Complete the square to write
2
this as 6 u − 32 + 13 . This reaches a minimum value when u = 32 . For u = sin x in the range

0 ≤ x
≤ 360◦ , this happens for two values of x both in 0 < x < 180◦ . The value there is
log2 31 < 0. Only one of the graphs reaches a negative minimum value twice in that range.
The answer is this graph;

x
360◦

4
19. a1 = 8 × (3)4 and a2 = 8(8 × 34 )4 = 8 × 84 × (34 )4 and a3 = 8 × (8 × 84 × (34 )4 ) . In a similar
way,
n−1 n
an = 8 × 84 × 816 × · · · × 84 × 3(4 )
The exponent on the 8s is 1+4+16+· · ·+4n−1 = 31 (4n − 1). Then 81/3 = 2. Also use 4n = 22n .
2n
22n −1 (22n ) 62
an = 2 3 =
2
20
6(2 )
So for n = 10 we get .
20
2
6(2 )
The answer is .
2
20. We’ll calculate the square of (x + 1 + x−1 ) first;

x2 + 2x + 3 + 2x−1 + x−2

(I drew a square grid to keep track of all the cross-terms, and there’s a nice pattern which
helps). Now if we were to square this expression, the constant term independent of x would be

2(x2 )(x−2 ) + 2(2x)(2x−1 ) + 32

Most of the terms have a factor of 2 because they occur in either order. This sum is 2+8+9 = 19.
The answer is 19.

21. One of the options looks like a quadratic, three of the options look like cubics, and one of the
options looks like a quartic (a polynomial of degree 4), so we might expect that one of the
cubics is the derivative of the quartic, and the quadratic is the derivative of another of the
cubics, leaving one remaining cubic that could be h(x). Checking the sign and the locations of
the zeros of the possible derivatives leaves one function over as a possible candidate for h(x);

x
−2 2

−5

We should check that it’s definitely not the derivative of any of the other options, and that its
derivative is not any of the other options. Note that it’s only got one zero, and it’s negative
for large x. This is enough to see that it’s not the derivative of any of the other functions.
It’s also not the case that the derivative of this graph is any of the other graphs; such a graph
would have two zeros for the two turning points, but only of the functions has two zeros, and
the sign of that graph is wrong (positive where the gradient of this graph is negative). So we
conclude that the graph above is the “odd one out”.
The answer is the graph above.
1
22. The geometric sum converges if |tan x|
< 1, and converges to

1
1 1 1 tan x 1
+ 2
+ 3
+ ··· = 1 = .
tan x tan x tan x 1 − tan x tan x − 1

Let u = tan√ x, then this sum is equal to tan x if and only if u2 − u = 1, so if √

and only if
1 1
u = 2 (1 ± 5). But the condition | tan x| > 1 means that we must take u = 2 (1 + 5). There
is one value of x in the range −90◦ < x < 90◦ with this value of tan x.
The answer is 1.
 √
23. For r < 1, A = 0 and B = 4 − πr2 . For r > 2, A = πr2 − 4 and B = 0, which eliminates two
of the options. Now two of the remaining options have A + B = 0 at some value of r, which
would require A = 0 and B = 0, that is; no area inside the square that’s not also inside the
circle and vice versa. This is impossible. There is only one remaining option.
The answer is this graph

2
r
1 2

24. Let’s call the product of the first n terms bn . Then we have bn = an bn−1 (that’s how the product
works). We also have the definition of an to interpret; it’s one more than the previous product,
so an = bn−1 + 1. We can use this to eliminate bn and bn−1 from the previous equation, to get
an+1 − 1 = an (an − 1). Adjust the subscripts and rearrange for an = an−1 (an−1 − 1) + 1.
The answer is an = an−1 (an−1 − 1) + 1.
p p
25. We must have |AB| = |BC| so (b − a)2 + (c − b)2 = (c − b)2 + (d − c)2 .
p p
We must also have |BC| = |CD| so (c − b)2 + (d − c)2 = (d − c)2 + (a − d)2 .
These conditions are equivalent to (b − a)2 = (d − c)2 and (c − b)2 = (a − d)2 respectively.
Using the difference of two squares, the first is equivalent to (a − b + c − d)(a − b − c + d) = 0
and the second is equivalent to (a − b + c − d)(a + b − c − d) = 0.
In each case, we can’t have both brackets equal to zero because c ̸= d and b ̸= c because the
numbers are distinct. So either a − b + c − d = 0 or both of a − b − c + d = 0 and a + b − c − d = 0.
That second case would imply that a − c = 0, but the numbers are distinct so that’s impossible.
p just the case that a −p
So we’re left with b + c − d = 0. We can also check that CD = DA in this
case, because (d − c)2 + (a − d)2 = (a − d)2 + (b − a)2 rearranges to (d − c)2 = (b − a)2
which is one of the equations we already had.
The answer is “if and only if a − b + c − d = 0”.
 26

(i) Bob’s number is 1. Alice then realises her number is 2, because it can’t be 0.
If Bob had any other number n, then Alice’s number could be either n+1 or n-1, so Alice
wouldn’t know her number.

(ii) Write A for Alice’s number, and B for Bob’s number.

After Charlie’s initial statement, the possibilities for (A,B) are
(2,1) (2,3)
(3,2) (3,4)
(5,4) (5,6)
(7,6) (7,8)
If Alice could see 1, 2, 3 or 8, she would deduce her number, so those possibilities are ruled out.
(3,4)
(5,4) (5,6)
(7,6)
If Bob could then see 3 or 7, he would deduce his number was 4 or 6, respectively, so those
possibilities are ruled out.
(5,4) (5,6)
Hence Alice’s number is 5.

(iii) Alice’s statement implies Bob’s number is not 1 or 10. Charlie’s second statement implies that
precisely one out of B-1 and B+1 is a square.
The possibilities are:
B=2 and A is 1 or 3
B=3 and A is 2 or 4
B=5 and A is 4 or 6
B=8 and A is 7 or 9
If Bob could see 1, 3, 2, 6, 7 or 9, he would know his number. Hence Alice’s number is 4.
 27

(i) Let’s consider dates of the form d1 d2 / m1 m2 / 2 0 y3 y4.

Clearly m1 is 1 to avoid repetition of 0.
But then m2 is 0 or 1 or 2, each of which would be a repetition.
Hence there are no such dates.

(ii) As there are no such dates this century, let’s try dates of the form d1 d2 / m1 m2 / 1 9 y3 y4.
The last possible year is 1987; then the last possible month is 06; and then the last possible
day is 25.
This gives the date 25/06/1987.

(iii) As there are no such dates this century, let’s try dates of the form d1 d2 / m1 m2 / 2 1 y3 y4.
Now m1 is 0 to avoid repetitions.
Then d1 is 3 to avoid repetitions.
But that leaves no possible value for d1.
Clearly there is no such date of the form d1 d2 / m1 m2 / 2 2 y3 y4.
For dates of the form d1 d2 / m1 m2 / 2 3 y3 y4, if m1 is 1 then m2 is 0 and there is no
possibility left for d1. So m1 is 0 and d1 is 1.
Of the dates 1 d2 / 0 m2 / 2 3 y3 y4, the earliest possible year is 2345, then the earliest possible
month is 06, and then the earliest possible day is 17.
This gives the date 17/06/2345.

(iv) Let’s consider dates of the form d1 d2 / m1 m2 / 1 9 y3 y4.

Clearly m1 is 0.
If d1 is 3, then d2 is 0 or 1, either of which would be a repetition. Hence d1 is 2.
We therefore have dates of the form 2 d2 / 0 m2 / 1 9 y3 y4.
The remaining spaces can be filled with arbitrary distinct values from 3, 4, 5, 6, 7, 8, giving
6x5x4x3=360 possibilities. Each such possibility is a valid date.