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(2 marks)
States thatcos3
cot3sin3
xx
x
M1
Makes an attempt to find
cos3d
sin 3
xx
x
Writing
f 'd ln f
f
xx x
x or writing ln (sin x) constitutes an attempt.
M1
States a fully correct answer 1
ln sin33
x C A1
TOTAL: 3 marks
Begins the proof by assuming the opposite is true.
‘Assumption: there exists a rational numbera
bsuch that
a
bis the greatest positive rational
number.’
B1
Makes an attempt to consider a number that is clearly greater thana
b:
‘Consider the number 1a
b , which must be greater than
a
b’
M1
Simplifies 1a
b and concludes that this is a rational number.
1a a b a b
b b b b
By definition,a b
b
is a rational number.
M1
Makes a valid conclusion.
This contradicts the assumption that there exists a greatest positive rational number,
so we can conclude that there is not a greatest positive rational number.
B1
TOTAL: 4 marks
1
PURE MATHEMATICS
A level Practice Papers
PAPER O
MARK SCHEME
2
Makes an attempt to find
3sin 4 1 cos4 dx x x
Raising the power by 1 would constitute an attempt.
M1
States a fully correct answe r 3 41
sin 4 1 cos4 d 1 cos416
x x x x C M1
Makes an attempt to substitute the limits 4
41 11 0 1
16 2
M1 ft
Correctly states answer is 15
256
A1 ft
TOTAL: 4 marks
Makes an attempt to factor all the quadratics on the left-hand side of the identity. M1
Correctly factors each expression on the left-hand side of the identity:
(x - 6)(x + 6)
(x - 5)(x - 6)´
(5- x)(5+ x)
Ax2 + Bx + C´
(3x -1)(2x + 3)
(3x -1)(x + 6)
A1
Successfully cancels common factors:
(–1)(5+ x)(2x + 3)
Ax2 + Bx + Cº
x + 5
(-1)(x - 6)
M1
States that Ax2 + Bx + C º (2x + 3)(x - 6) M1
States or implies that A = 2, B = −9 and C = −18 A1
TOTAL: 5 marks
NOTES: Alternative method
Makes an attempt to substitute x = 0 (M1)
Finds C = −18 (A1)
Substitutes x = 1 to give A + B = −7 (M1)
Substitutes x = −1 to give A − B = 11 (M1)
Solves to get A = 2, B = −9 and C = −18 (A1)
3
4
Begins the proof by assuming the opposite is true.
‘Assumption: there exists a number n such that n is odd and n3 + 1 is also odd.’
B1
Defines an odd number: ‘Let 2k + 1 be an odd number.’ B1
Successfully calculates (2k +1)3 +1
(2k +1)3 +1º 8k3 +12k2 + 6k +1( ) +1º 8k3 +12k2 + 6k + 2
M1
Factors the expression and concludes that this number must be even.
8k3 +12k2 + 6k + 2 º 2 4k3 + 6k2 + 3k +1( )
2 4k3 + 6k2 + 3k +1( ) is even.
M1
Makes a valid conclusion.
This contradicts the assumption that there exists a number n such that n is odd and n3 + 1 is also
odd, so if n is odd, then n3 + 1 is even.
B1
TOTAL: 5 marks
NOTES: Alternative method
Assume the opposite is true: there exists a number n such that n is odd and n3 + 1 is also odd. (B1)
If n3 + 1 is odd, then n3 is even. (B1)
So 2 is a factor of n3. (M1)
This implies 2 is a factor of n. (M1)
This contradicts the statement n is odd. (B1)
Understands that for the series to be convergent 1r or states 4 1x M1
Correctly concludes that 1
4x . Accept
1 1
4 4x A1
(2 marks)
Understands to use the sum to infinity formula. For example, states 1
41 4x
M1
Makes an attempt to solve for x. For example, 3
44
x is seen. M1
States 3
16x A1
(3 marks)
TOTAL: 5 marks
5
6a
°
6b
°
States 10a b and 7 5 2a b M1
Makes an attempt to solve the pair of simultaneous equations.
Attempt could include making a substitution or multiplying the first equation by 5 or by 7.
M1
Finds a = −4 A1
Finds b = 6 A1
States −2abc = −96 M1
Finds c = −2 A1
TOTAL: 6 marks
Understands the need to complete the square, and makes an attempt to do this.
For example, 2
4x is seen.
M1
Correctly writes 2
g( ) 4 9x x A1
Demonstrates an understanding of the method for finding the inverse is to switch the x and y.
For example, 2
4 9x y is seen.
B1
Makes an attempt to rearrange to make y the subject.
Attempt must include taking the square root.
M1
Correctly states 1g ( ) 9 4x x A1
Correctly states domain is x > −9 and range is y > 4 B1
TOTAL: 6 marks
7
8
Makes an attempt to set up a long division.
For example: 9x2 + 0x -16 9x2 + 25x +16 is seen.
The ‘0x’ being seen is not necessary to award the mark.
M1
Long division completed so that a ‘1’ is seen in the quotient
and a remainder of 25x + 32 is also seen.
9x2 + 0x -16 9x2 + 25x +16
9x2 + 0x -16
25x + 32
1
M1
States B(3x + 4) + C(3x - 4) º 25x - 32 M1
Equates the various terms.
Equating the coefficients of x: 3B+3C = 25
Equating constant terms: 4B- 4C = 32
M1
Multiplies one or both of the equations in an effort to equate one of the two variables. M1
Finds 49
6B
A1
Finds 1
6C
A1
TOTAL: 7 marks
NOTES: Alternative method
Writes A+
B
3x - 4+
C
3x + 4 as
A(3x - 4)(3x + 4)
9x2 -16+
B(3x + 4)
9x2 -16+
C(3x - 4)
9x2 -16
States A(3x - 4)(3x + 4) + B(3x + 4) + C(3x - 4) º 9x2 + 25x +16
Substitutes x =
4
3 to obtain:
8B =
196
3Þ B =
49
6
Substitutes x = -
4
3 to obtain:
-8C = -
4
3Þ C =
1
6
Equating the coefficients of x2: 9A = 9 Þ A =1
9
States that sin1
BD and concludes that BD = sinq
M1
States that cos1
AD and concludes that AD = cosq
M1
States that ÐDBC =q M1
States that tansin
DC
and concludes that
2sin
cosDC
oe.
M1
States thatsin
cosBC
and concludes that BC = tanq oe.
M1
Recognises the need to use Pythagoras’ theorem. For example, AB2 + BC2 = AC2 M1
Makes substitutions and begins to manipulate the equation:
22
2 cos sin1 tan
1 cos
22 2
2 cos sin1 tan
cos
M1
Uses a clear algebraic progression to arrive at the final answer:
2
2 11 tan
cos
1+ tan2q = sec2q
A1
TOTAL: 8 marks
10
States4
sin7
xt
and
3cos
7
yt
M1
Recognises that the identity 2 2sin cos 1t t can be used to find the cartesian equation. M1
Makes the substitution to find 2 2 24 3 7x y A1
(3 marks)
States or implies that the curve is a circle with centre (−4, 3) and radius 7 M1 ft
Substitutesπ
2t to find x = −11 and y = 3 (−11, 3)
Substitutesπ
3t to find x ≈ 2.06 and y = 6.5 (2.06, 6.5)
Could also substitute t = 0 to find x = −4 and y = 10 (−4, 10)
M1 ft
A1 ft
(3 marks)
Makes an attempt to find the length of the curve by recognising that the length is part of the
circumference. Must at least attempt to find the circumference to award method mark.
2 π 7 14πC
M1 ft
Uses the fact that the arc is5
12of the circumference to write: arc length =
35π
6 A1 ft
(2 marks)
TOTAL: 8 marks
NOTES:
11b Award ft marks for correct sketch using incorrect values from part a.
11c Award ft marks for correct answer using incorrect values from part a.
11c Alternative method: use s r , with 7r and5
6
Award one mark for the attempt and one for the correct answer.
11a
11c
11b
Draws fully correct
curve.
Finds d
2sin 2d
xt
t and
dcos
d
yt
t
M1
Writes −2sin 2t = − 4sin t cos t M1
Calculates d cos 1
cosecd 4sin cos 4
y tt
x t t
A1
(3 marks)
Evaluatesd
d
y
xat
5
6t
dy
dx=
-1
4sin -5p
6
æ
èçö
ø÷
=1
2
A1 ft
Understands that the gradient of the tangent is1
2, and then the gradient of the normal is −2
M1 ft
Finds the values of x and y at5
6t
5 1
cos 26 2
x
and 5 1
sin6 2
y
M1 ft
Attempts to substitute values into 1 1( )y y m x x
For example,1 1
22 2
y x
is seen.
M1 ft
Shows logical progression to simplify algebra, arriving at:
1
22
y x or 4 2 1 0x y
A1
(5 marks)
TOTAL: 8 marks
NOTES: 12b
Award ft marks for a correct answer using an incorrect answer from part a.
12a
12b
M1
States that y =
2
x -1meets
y = ex in just one place, therefore
2
x -1= ex has just one root
g(x) = 0 has just one root
A1
(2 marks)
Makes an attempt to rearrange the equation. For example,
2
x -1- ex = 0 Þ xex - ex = 2
M1
Shows logical progression to state x = 2e-x +1
For example, x =
2 + ex
exis seen.
A1
(2 marks)
Attempts to use iterative procedure to find subsequent values. M1
Correctly finds: x1 = 1.4463 x2 = 1.4709 x3 = 1.4594 x4 = 1.4647 A1
(2 marks)
Correctly finds
¢g (x) = -2
(x -1)2- ex
A1
Finds g(1.5) = -0.4816...and ¢g (1.5) = -12.4816... M1
Attempts to find x
1: 0
1 0 1
0
g( ) 0.4816...1.5
g ( ) 12.4816...
xx x x
x
M1
Finds x
1= 1.461 A1
(4 marks)
TOTAL: 10 marks
13a
13b
13c
13d
Attempts to sketch
both y =
2
x -1and
y = ex
NOTES:
13a
Uses their graphing calculator to sketch g(x) =
2
x -1- ex (M1)
States that as g(x) only intersects the x-axis in one place, there is only one solution. (A1)
13c
Award M1 if finds at least one correct answer.
Writes:
1+ x
1- 2x as
(1+ x)(1- 2x)
-1
2 M1
Uses the binomial expansion to write:
21
2
1 3( 2 )
1 2 2(1 2 ) 1 ( 2 ) ...
2 2
x
x x
M1
Simplifies to obtain:
(1+ x)(1- 2x)-
1
2 = (1+ x) 1+ x +3
2x2æ
èçö
ø÷ M1
Writes the correct final answer: 251 2
2x x … A1 ft
(4 marks)
Either states1
2x or states
-
1
2< x <
1
2 B1
(1 mark)
Makes an attempt to substitute 1
100x into
1+ x
1- 2x
For example
1 101
100 100
981
100100
1
1 2
M1
Continues to simplify the expression:
101
100´
100
98 And states the correct final answer:
101 2
140
A1
(2 marks)
Substitutes 1
100x into 25
1 22
x x Obtains:
1+ 21
100
æ
èçö
ø÷+
5
2
1
100
æ
èçö
ø÷
2
»1.02025 M1 ft
States that 1.02025 »
101 2
140
M1 ft
Deduces that140 1.02025
2 1.41421101
A1 ft
(3 marks)
TOTAL: 10 marks
NOTES:
14a Award 3 marks if a student has used an incorrect expansion but worked out all the other steps correctly.
14d Award all three marks if a student provided an incorrect answer in part a, but accurately works out
an approximation for root 2 consistent with this incorrect answer.
14a
14d
14c
14b
Correctly substitutes x = 1.5 into 3 214
2y x x and obtains 2.2323… A1
(1 mark)
States or implies formula for the trapezium rule: 0 1 2 3 422
hA y y y y y
M1
Makes an attempt to substitute into the formula 0.5
0 2 0.12103 0.86603 2.23235 02
A M1
States correct final answer 1.610 (4 s.f.) A1
(3 marks)
Recognises the need to make a substitution.
Method 1
24u x is seen.
M1
Correctly statesd
2d
ux
x and finds new
limits 0 4x u and 2 0x u
M1
Correctly transforms the integral
23 2
0
14 d
2
x
xx x x
into
1 30
2 2
4
14 d
4
u
uu u u
M1
Correctly finds the integral
03 5
2 2
4
1 8 2
4 3 5u u
M1
Makes an attempt to substitute the limits
3 5
2 2
3 5
2 2
8 20 0
3 51
4 8 24 4
3 5
M1
Correctly finds answer32
15
A1
(6 marks)
Using more strips would improve the accuracy of the answer. B1
(1 mark)
TOTAL: 11 marks
(TOTAL: 100 MARKS)
15a
15b
15c
15d
Method 2
1
2 24u x is seen
States 2 24u x and finds d
2 2d
uu x
x and finds
new limits 0 2x u and 2 0x u
into
02 4
2
14 d
2
u
uu u u
0
3 5
2
1 4 1
2 3 5u u
3 5
3 5
4 10 0
3 51
2 4 12 2
3 5
15c Either method is acceptable.
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