A Level Mathematics Practice Paper F - Pure Mathematics Mark Scheme
A Level Mathematics Practice Paper F - Pure Mathematics Mark Scheme
A Level Mathematics Practice Paper F - Pure Mathematics Mark Scheme
Pearson
Progression Step
1 Scheme Marks AOs
and Progress
descriptor
Multiplies both of the equations in an effort to equate one of the M1* 1.1b
two variables.
Finds A = 8 A1 1.1b
Find B = −2 A1 1.1b
(5 marks)
Notes
Alternative method
Uses the substitution method, having first obtained this equation: A(2x + 5) + B(5x -1) º 6x + 42
5 27
Substitutes x = - to obtain B = 27 (M1)
2 2
1 27
Substitutes x = to obtain A = 43.2 (M1)
5 5
1 B1 2.4
Recognises that this contradicts the statement that 5a + 3b = ,
5
1
as is not an integer. Therefore there do not exist integers a and
5
b such that 25a +15b = 1’
(4 marks)
Notes
(3)
5 M1 ft 1.1b
Finds the values of x and y at t
6
5 1 5 1
x cos 2 and y sin
6 2 6 2
1 1
For example, y 2 x is seen.
2 2
(5)
(8 marks)
Notes
(b) Award ft marks for a correct answer using an incorrect answer from part a.
1 A1 1.1b
States a fully correct answer ln sin3x C
3
(3 marks)
Notes
(6 marks)
Notes
Find composite
functions.
States or implies that qp( x) 5 2 x 2 M1 2.2a
25 20 x 4 x 2 5 2 x 2 or 6 x 2 20 x 20 0 is seen.
(4)
(2)
(6 marks)
Notes
(b) Alternative Method
2 2
5 65 5 65
M1: Uses the method of completing the square to show that 3 x 0 or 3 x
3 9 3 9
B1: Concludes that this equation will have no real solutions.
Consider a new number that is one greater than the product of M1 1.1b
all the existing prime numbers:
Let N = ( p1 ´ p2 ´ p3 ´ ...´ pn ) +1
(6 marks)
Notes
If N is prime, it is a new prime number separate to the finite list of prime numbers, p1, p2 , p3 ,...pn .
If N is divisible by a previously unknown prime number, that prime number is also separate to the finite list of
prime numbers.
(2 marks)
Notes
(a) Recognises that it is a geometric series with a first term a 100 M1 3.1a 6th
and common ratio r 1.05
Use geometric
Attempts to use the sum of a geometric series. For example, M1* 2.2a sequences and
S9
100 1 1.059
or
S9
100 1.059 1 is seen.
series in context.
1 1.05 1.05 1
Finds S9 £1102.66 A1 1.1b
(3)
(b)
States
100 1.05n 1 6000 or 100 1 1.05 6000
n
M1 3.1a 5th
1.05 1 1 1.05 Use arithmetic
sequences and
Begins to simplify. 1.05n 4 or 1.05n 4 M1 1.1b series in context.
Applies law of logarithms correctly M1 2.2a
n log1.05 log 4 or n log1.05 log 4
log 4 A1 1.1b
States n
log1.05
(4)
(2)
(9 marks)
Notes
M1
Award mark if attempt to calculate the amount of money after 1, 2, 3,….,8 and 9 months is seen.
cos 6x dx 2 1 cos12x dx
2 1 M1 1.1b
States that
1 1 A1 1.1b
Correctly states x sin12 x C
2 12
(5 marks)
Notes
Student does not need to state ‘+C’ to be awarded the third method mark. Must be stated in the final answer.
(a) Writes tanx and secx in terms of sinx and cosx. For example, M1 2.1 5th
sin x 1 Understand the
tan x sec x cos x cos x functions sec,
cosec and cot.
1 sin x 1 sin x
1
sin x 1 1 M1 1.1b
Manipulates the expression to find
cos x 1 sin x
1 A1 1.1b
Simplifies to find - = - sec x
cos x
(3)
(3)
(6 marks)
Notes
1 2 5 A1 1.1b
Finds y x x
64 2
(3)
(b) Deduces that the width of the arch can be found by substituting M1 3.4 8th
t 10 into x 8 t 10 Use parametric
equations in
Finds x = 0 and x = 160 and deduces the width of the arch is A1 3.2a modelling in a
160 m. variety of
contexts.
(2)
(c) Deduces that the greatest height occurs when M1 3.4 8th
dy
0 2t 0 t 0 Use parametric
dt equations in
modelling in a
Deduces that the height is 100 m. A1 3.2a variety of
contexts.
(2)
(7 marks)
Notes
138
For the final accuracy mark either D = 138 or or the
x6
remainder is 138 must be seen.
x 2 2 x 21
x 6 x3 8 x 2 9 x 12
x3 6 x 2
2 x2 9 x
2 x 2 12 x
21x 12
21x 126
138
(5 marks)
Notes
This question can be solved by first writing ( Ax 2 + Bx + C)(x + 6) + D º x 3 + 8x 2 - 9x +12 and then solving for
A, B, C and D. Award 1 mark for the setting up the problem as described. Then award 1 mark for each correct
coefficient found. For example:
Equating the coefficients of x3: A = 1
Equating the coefficients of x2: 6 + B = 8, so B = 2
Equating the coefficients of x: 12 + C = −9, so C = −21
Equating the constant terms: −126 + D = 12, so D = 138.
dV M1 3.1a 8th
Recognises the need to use the chain rule to find
dt Construct
dV dV dr dS differential
For example is seen. equations in a
dt dr dS dt range of contexts.
dV dS M1 2.2a
Finds 4r 2 and 8r
dr dr
dV A1 1.1b
Simplifies and states 6r
dt
(4 marks)
Notes
15
Recognises the need to write sin 3 x sin x sin 2 x M1 2.2a 6th
Integrate using
Selects the correct trigonometric identity to write M1 2.2a trigonometric
sin x sin 2 x sin x 1 cos 2 x . Could also write
identities.
Makes an attempt to find sin x sin x cos2 x dx M1 1.1b
1 A1 1.1b
Correctly states answer cos x cos3 x C
3
(4 marks)
Notes
(2)
h( x0 ) 0.2903...
x1 x0 x1 19.35
h( x0 ) 13.6792...
(5)
(3)
(10 marks)
Notes
(a) Minimum required is that answer states there is a sign change in the interval and that this implies a root in
the given interval.
A1 1.1b
Finds |KL| 32 02 62 45
Finds |LM | 2 2 5 2 4 2 45
(7)
(3)
(10 marks)
Notes
(b) Award ft marks for a correct answer to part a using their incorrect answer from earlier in part a.