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Higher Maths Past Papers
Mr JD MacBeath
March 2019
1 Introduction
It is hoped that this document will make it easier to find past paper questions, both for teachers
and students. I will do my best to keep this document up to date and include new past paper
questions as they become available. If you spot any mistakes, or want to suggest any improve-
ments, send me an email at john.macbeath@wickhigh.org.uk. I am more than happy to send you
the Tex file used to produce the document so that you can modify it as you wish.
Sincere thanks are given to Mr. C. Davie at Glenrothes High School for his extensive assistance
in producing this document.
2 How to Use
The table on the next page contains links to questions sorted by topic and year. Clicking on a
link will take you to the page of that question. The marking instructions follow directly after
each question or section of questions where more than one question is on a page. To return to the
table click on Back to Table at the top or bottom of any page. Trying to navigate the document
without doing this is tedious.
Remember - Paper 1 questions are non-calculator, while Paper 2 questions are calculator questions.
Note also that Specimen Papers (SP) are put together from previous past paper questions and so
it is possible you may see the same question twice.
Before starting any past paper questions I recommend that you print a paper copy of the Higher
Maths Sheet to avoid wasting time. A copy of the Formula Sheet is available on page 3.
i
Topic
18 P
18 P
17 P
17 P
16 P
16 P
15 P
15 P
SP P
SP P
Circles
5c, 12
12b
Differentiation& Optimisation
8, 15c
6b, 6c, 9
Exp’s & Logs
Functions &
Graphs
1,6,15a
Further Calculus
10,11b
7a
Integration
10, 15b
3b, 9
Polynomials &
Quadratics
4, 7a, 10
2, 3a
6a
Recurrence Relations
7b, 7c
Straight Lines
5a, 5b
Trig. Formulae & Equations
8b, 11a
7b
Vectors
4, 12a
Wave Function
8a
page 03
Attempt ALL questions MARKS
Total marks — 60
1. PQR is a triangle with vertices P (−2, 4), Q (4, 0) and R (3, 6).
R (3, 6)
P (−2, 4)
Q (4, 0)
Find the equation of the median through R.
2. A function g x ( ) is defined on , the set of real numbers, by
g x ( ) = x −
Find the inverse function, g ( ) x
− 1 .
3. Given h x ( ) = 3 cos 2 x , find the value of
h
[Turn over
page 05
Detailed marking instructions for each question
Question Generic scheme Illustrative scheme
Max
mark
1. ^1 find mid-point of PQ
^2 find gradient of median
3 determine equation of median
^1 1,2
^2
^3 y 2 x
Notes:
- ^2 is only available to candidates who use a midpoint to find a gradient.
-
3 is only available as a consequence of using the mid-point and the point R, or any other point
which lies on the median, eg (^) 2,4.
- At
3 accept any arrangement of a candidate’s equation where constant terms have been simplified.
- ^3 is not available as a consequence of using a perpendicular gradient.
Commonly Observed Responses:
Candidate A – Perpendicular Bisector of PQ
MPQ (^) 1,2
PQ
m m ^2
2 y 3 x 1
3 2
For other perpendicular bisectors award 0/
Candidate B – Altitude through R
PQ
m ^1 ^
m ^2
2 y 3 x 3 ^3 2
Candidate C – Median through P
MQR (^) 3 5,3
PM
m ^2 1
11 y 2 x 40 ^3 2
Candidate D – Median through Q
MPR (^) 0 5,5
QM
m ^2 1
7 y 10 x 40 ^3 2
page 07
Question Generic scheme Illustrative scheme
Max mark
3. (^) ^1 start to differentiate
^2 complete differentiation
^3 evaluate derivative
^1 3 sin 2 x stated or implied by ^2
^2 2
3
Notes:
1. Ignore the appearance of c at any stage.
- ^3 is available for evaluating an attempt at finding the derivative at 6
3. For cos
h ^3
^
award 0/3.
Commonly Observed Responses:
Candidate A
3 sin 2 x ^1
^2
3 1
Candidate B
3 sin 2 x ^1
3 3 ^3 1
Candidate C
3 sin 2 x
1
1
2
^2
^3 1
Candidate D
6 cos 2 x ^1
2
3 ^3 1
Candidate E
3 cos 2 x ^1
2 ^2 1
3 1
Candidate F
6 sin 2 x ^1
2
3 3
3 1
page 04
MARKS
4. The point K (8, −5) lies on the circle with equation x + y − x − y − =
2 2 12 6 23 0.
K (8, −5)
x + y − x − y − =
2 2 12 6 23 0
Find the equation of the tangent to the circle at K.
5. A (−3, 4, −7), B (5, t , 5) and C (7, 9, 8) are collinear.
(a) State the ratio in which B divides AC.
(b) State the value of t.
6. Find the value of log 5 − log 5
page 09
Question Generic scheme Illustrative scheme
Max mark
5. (a) ^1 state ratio explicitly ^1 4 1: 1
Notes:
- The only acceptable variations for
1 must be related explicitly to AB and BC.
For
BC 1
AB 4
AB 4
BC 1
or BC AB: 1 4: award 1/1.
- For
BC AB
award 0/1.
Commonly Observed Responses:
(b) ^2 state value of t ^2 8 1
Notes:
- The answer to part (b) must be consistent with a ratio stated in part (a) unless a valid strategy
which does not require the use of their ratio from part (a) is used.
Commonly Observed Responses:
Candidate A
1
t 8
2
Candidate B
1
t 5
2 1
page 10
Question Generic scheme^ Illustrative scheme^
Max mark
1
apply log 5 log 5
m
m x x
2
applylog 5 log 5 log 5
x
x y
y
^3 evaluate log
^1 log
1 3
2
log 5
3
^3
Notes:
- Each line of working must be equivalent to the line above within a valid strategy, however see
Candidate B.
- Do not penalise the omission of the base of the logarithm at
1 or
2 .
- For ‘3’ with no working award 0/3.
Commonly Observed Responses:
Candidate A
log 5 log 5
^1
log 5
8 3
2 1
log 5
^3 2
Candidate B
log 5
log 5
log
1 3 5
Award 1/3 1 ^
^1 is awarded for the
final two lines of working
page 11
Question Generic scheme Illustrative scheme
Max mark
7. (a) (^) ^1 state coordinates of P (^) ^1 0,5 1
Notes:
1. Accept ‘ x 0 , y 5 ’.
2. ‘ y 5 ’ alone or ‘5’ does not gain
1 .
Commonly Observed Responses:
(b) ^2 differentiate
^3 calculate gradient
^4 state equation of tangent
2 2
3 x 6 x 2
^3
^4 y 2 x 5
Notes:
3. At ^4 accept y 2 x 5 , 2 x y 5 0 , y 5 2 x or any other rearrangement of the equation
where the constant terms have been simplified.
- ^4 is only available if an attempt has been made to find the gradient from differentiation.
Commonly Observed Responses:
page 12
Question Generic scheme Illustrative scheme
Max mark
7. (c) (^) ^5 set
y line y curve and arrange in
standard form
6 factorise
7
state x - coordinate of Q
8
calculate y - coordinate of Q
5 3 2
x 3 x 0
^6 x^2 x (^3)
^7
^8
Notes:
5. ^5 is only available if ‘ 0 ’ appears at either ^5 or ^6 stage.
7 and
8 are only available as a consequence of solving a cubic equation and a linear equation simultaneously.
- For an answer of (^) 3,11with no working award 0/4.
- For an answer of (^) 3,11verified in both equations award 3/4.
- For an answer of (^) 3,11verified in both equations along with a statement such as ‘same point
on both line and curve so Q is (^) 3,11’ award 4/4.
- For candidates who work with a derivative, no further marks are available.
11. x 3 must be supported by valid working for ^7 and ^8 to be awarded.
Commonly Observed Responses:
Candidate A 3 2
x 3 x 0
5
x 3 0 ^6
x 3 ^7
y 11 ^8
Dividing by
2
x is valid since x 0 at ^6
page 06
MARKS
9. The diagram shows a triangular prism ABC,DEF.
AB = t , AC = u and AD = v.
B
D F
u C
A
t M
v
E
(a) Express BC in terms of u and t.
M is the midpoint of BC.
(b) Express MD in terms of t, u and v.
10. Given that
dy
x x
dx
2
6 3 4 , and
- y = 14 when x = 2,
express y in terms of x.
page 14
Question Generic scheme^ Illustrative scheme^
Max mark
9. (a) ^1 identify pathway ^1 t u 1
Notes:
Commonly Observed Responses:
(b) ^2 state an appropriate pathway
3
express pathway in terms of t, u
and v
^2 eg
BC CA AD
stated or
implied by ^3
^3 +
t u v
Notes:
- There is no need to simplify the expression at
3
. Eg
t u u v.
2. ^3 is only available for using a valid pathway.
- The expression at ^3 must be consistent with the candidate’s expression at ^1.
4. If the pathway in ^1 is given in terms of a single vector t , u or v , then ^3 is not available.
Commonly Observed Responses:
Candidate A
1 MD 2
t v u ^2 ^ ^3
page 07
MARKS
11. The diagram shows the curve with equation y = log 3 x.
y
O 1 x
y = log 3 x
(a) On the diagram in your answer booklet, sketch the curve with equation
y = 1 − log 3 x.
(b) Determine the exact value of the x -coordinate of the point of intersection of the
two curves.
12. Vectors a and b are such that a = 4 i − 2 j + 2 k and b = − 2 i + j + p k.
(a) Express 2 a + b in component form.
(b) Hence find the values of p for which 2 a + b = 7.
[Turn over for next question
page 16
Question Generic scheme Illustrative scheme
Max mark
11. (a) ^1 curve reflected in x - axis and
translated 1 unit vertically
^2 accurate sketch
1 a generally decreasing curve
above the x - axis for 1 x 3
^2 asymptote at x 0 and passing
through 3,0 and continuing to
decrease for x 3
Notes:
1. For any attempt which involves a horizontal translation or reflection in the y - axis award 0/2.
- For a single transformation award 0/2.
3. For any attempt involving a reflection in the line y x award 0/
Commonly Observed Responses:
Candidate A
Award 1/
(b) ^3 set ‘ y y ’
4 start to solve
5
state x coordinate
^3 log 3 x 1 log 3 x
^4 log 3
x orlog
2
3 x^ ^1
5 3 or
1
Notes:
4. ^3 may be implied bylog 3
x from symmetry of the curves.
- Do not penalise the omission of the base of the logarithm at ^3 or ^4.
6. For a solution which equates a to log 3 a , the final mark is not available.
7. If a candidate considers and then does not discard 3 in their final answer, ^5 is not available.
Commonly Observed Responses:
