General Certificate of Education Advanced Subsidiary Examination June 2011
Mathematics
MFP1
Unit Further Pure 1 Friday 20 May 2011
1.30 pm to 3.00 pm
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For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
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Time allowed * 1 hour 30 minutes
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Instructions * Use black ink or black ball-point pen. Pencil should only be used for drawing. * Fill in the boxes at the top of this page. * Answer all questions. * Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. * You must answer the questions in the spaces provided. Do not write outside the box around each page. * Show all necessary working; otherwise marks for method may be lost. * Do all rough work in this book. Cross through any work that you do not want to be marked.
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Information * The marks for questions are shown in brackets. * The maximum mark for this paper is 75. Advice * Unless stated otherwise, you may quote formulae, without proof, from the booklet.
P39432/Jun11/MFP1 6/6/6/
MFP1
2
A curve passes through the point ð2, 3Þ and satisfies the differential equation
1
dy 1 ¼ pffiffiffiffiffiffiffiffiffiffiffi 2þx dx Starting at the point ð2, 3Þ, use a step-by-step method with a step length of 0.5 to estimate the value of y at x ¼ 3 . Give your answer to four decimal places. (5 marks)
The equation
2
4x 2 þ 6x þ 3 ¼ 0 has roots a and b . (a)
Write down the values of a þ b and ab .
(b)
Show that a 2 þ b 2 ¼ 4 .
(c)
Find an equation, with integer coefficients, which has roots
3
(2 marks) (2 marks)
3a b and 3b a
(5 marks)
It is given that z ¼ x þ iy , where x and y are real.
3 (a)
Find, in terms of x and y, the real and imaginary parts of ðz iÞðz* iÞ
(b)
(3 marks)
Given that ðz iÞðz* iÞ ¼ 24 8i find the two possible values of z.
(02)
(4 marks)
P39432/Jun11/MFP1
3
The variables x and Y , where Y ¼ log10 y , are related by the equation
4
Y ¼ mx þ c where m and c are constants. (a)
Given that y ¼ abx , express a in terms of c, and b in terms of m.
(b)
It is given that y ¼ 12 when x ¼ 1 and that y ¼ 27 when x ¼ 5 .
(3 marks)
On the diagram below, draw a linear graph relating x and Y .
(3 marks)
Use your graph to estimate, to two significant figures:
(c) (i)
the value of y when x ¼ 3 ;
(2 marks)
(ii) the value of a.
(2 marks)
Y~ 2–
1–
–
–
–
1
2
3
4
5
~
–
5 (a)
–
–
0– 0
x
Find the general solution of the equation pffiffiffi p 3 cos 3x ¼ 6 2
giving your answer in terms of p . (b)
Use your general solution to find the smallest solution of this equation which is greater than 5p . (2 marks)
Turn over
s
(03)
(5 marks)
P39432/Jun11/MFP1
4
Expand ð5 þ hÞ3 .
6 (a)
(1 mark)
A curve has equation y ¼ x 3 x 2 .
(b) (i)
Find the gradient of the line passing through the point ð5, 100Þ and the point on the curve for which x ¼ 5 þ h . Give your answer in the form p þ qh þ rh2 where p, q and r are integers.
(4 marks)
(ii) Show how the answer to part (b)(i) can be used to find the gradient of the curve at
the point ð5, 100Þ. State the value of this gradient.
(2 marks)
The matrix A is defined by
7
" A¼ (a) (i)
pffiffiffi # 1 3 pffiffiffi 3 1
Calculate the matrix A2 .
(2 marks)
(ii) Show that A3 ¼ kI , where k is an integer and I is the 2 2 identity matrix.
(2 marks) Describe the single geometrical transformation, or combination of two geometrical transformations, corresponding to each of the matrices:
(b)
(i)
A3 ;
(2 marks)
(ii) A .
(3 marks)
1 A curve has equation y ¼ 2 . x 4
8 (a) (i)
Write down the equations of the three asymptotes of the curve.
(3 marks)
(ii) Sketch the curve, showing the coordinates of any points of intersection with the
coordinate axes. (b)
(4 marks)
Hence, or otherwise, solve the inequality 1 < 2 x2 4
(04)
(3 marks)
P39432/Jun11/MFP1
5 1
The diagram shows a parabola P which has equation y ¼ 8 x 2 , and another parabola Q which is the image of P under a reflection in the line y ¼ x .
9
The parabolas P and Q intersect at the origin and again at a point A. The line L is a tangent to both P and Q. y
A
Q P
O
x
L
(a) (i)
Find the coordinates of the point A.
(2 marks)
(ii) Write down an equation for Q.
(1 mark)
(iii) Give a reason why the gradient of L must be 1 .
(1 mark)
(b) (i)
Given that the line y ¼ x þ c intersects the parabola P at two distinct points, show that c > 2
(3 marks)
(ii) Find the coordinates of the points at which the line L touches the parabolas P and Q.
(No credit will be given for solutions based on differentiation.)
(4 marks)
END OF QUESTIONS
Copyright Ó 2011 AQA and its licensors. All rights reserved.
(05)
P39432/Jun11/MFP1
Q
Solution 1 Attempt at 0.5 × y′(2) (= 0.25)
Marks M1
y(2.5) ≈ 3.25 y(3)
m1 A1F A1 B1B1
5 5 2
M1A1
2
Total 2(a) α + β = − , αβ =
3 4
(b) α 2 + β 2 = (− 3 )2 − 2( 3 ) = 2 4
3 4
(c) Sum = 2(α + β) = −3
(
B1F
)
Product = 10αβ − 3 α 2 + β 2 = x2 – Sx + P (= 0)
21 4
Eqn is 4x + 12x + 21 = 0
M1A1F A1
Total 3(a) Use of z* = x − iy (z − i)(z* − i) = (x2 + y2 − 1) − 2ix (b) Equating R and I parts − 2 x = −8 so x = 4
AG; A0 if α + β has wrong sign
ft wrong values Signs must be correct for the M1 5 9
Integer coeffs and ‘= 0’ needed
3
A1 may be earned in (b)
4
A0 if x = –4 used
M1 A1
16 + y 2 − 1 = 24 so y = ± 3 (z = 4 ± 3i) Total 4(a) Use of one law of logs or exponentials lg a = c and lg b = m So a = 10c and b = 10m
(b) Points (1, 1.08), (5, 1.43) plotted Straight line drawn through points (c)(i) Attempt at antilog of Y(3) When x = 3, Y ≈ 1.25 so y ≈ 18
m1A1
7
M1 A1 A1
3
M1A1 A1F
3
M1 A0 if one point correct ft small inaccuracy
2
OE Allow AWRT 18
M1 A1
(ii) Attempt at a as antilog of Y-intercept a ≈ 9.3 to 10 Total 3 π 5(a) cos 6 = 2
cos(− π6 ) = 23 Introduction of 2nπ Going from 3 x − π6 to x GS: x = 18π ± 18π + 23 nπ
M1 A1
π (≈ 16.755)
2 10
5
M1 A1
Total
OE; both needed
B1 B1F M1 m1 A1F
(b) n = 8 will give the required solution
... which is
M1 m1A1
PI; OE; ft c’s value for y(2.5) 4 dp needed
ft wrong value for α + β
M1
2
16 3
Comments Other variations are allowed
A1
≈ 3.25 + 0.5 y′(2.5) ≈ 3.25 + 0.2357(0) ≈ 3.4857 3 2
Total
OE AWRT OE stated or used; deg/dec penalised at 5th mark OE; ft wrong first value (or nπ) at any stage incl division of all terms by 3 ft wrong first value GS must include
2 7
2 3
nπ for this
from correct GS; π or dec approx allow 48 9
Q Solution 3 (5 + h ) = 125 + 75h + 15h2 + h3 6(a)
Marks B1
(b)(i) y(5 + h) = 100 + 65h + 14h2 + h3
B1F
Use of correct formula for gradient Gradient is 65 + 14h + h2 (ii) As h → 0 this → 65
−2 2 3 A2 = − 2 3 − 2 (ii) 8 0 A3 = 0 8
PI; ft numerical error in (a)
A2,1F
4
E2,1F
2
2
M1
(b)(i) A3 gives enlargement with SF 8 (centre the origin) (ii) Enlargement and rotation Enlargement scale factor 2 Rotation through 120° (antic’wise) Total 8(a)(i) Asymptotes x = −2, x = 2, y = 0
2
M1A1F
2
M1 A1 A1 B1 × 3 B1 B1 B1 B1
(b) y = –2 when x = ± 3.5 Sol’n − 2 < x < − 3.5 , 3.5 < x < 2
B1
B2,1
Total 9(a)(i) Elimination to give x = x A is (8, 8)
2
M1 if at least two entries correct if at least two entries correct
A1
(ii) Middle branch generally correct Other branches generally correct All branches approaching asymps Intersection at (0,− 14 ) indicated
1 8
A1 if one numerical error made; ft numerical error already penalised E1 for ‘h = 0’; ft wrong values for p, q, r
7
M1A1
……..= 8I
Comments Accept unsimplified coefficients
M1
Total 7(a)(i)
Total 1
M1 for enlargement (only); ft wrong value for k Some detail needed
3 9 3 Allow if max pt not in right place 4
Asymps must be shown correctly on diagram or elsewhere; B0 if any other intersections are shown Allow NMS
3
Condone dec approx’n for B1 if ≤ used instead of
0 Δ = 64 + 32c, so c > −2 (ii) For tangent c = −2, so x2 + 8x + 16 = 0 ... and x = −4, y = 2
Reflection in y = x x = 2, y = −4 Total TOTAL
3 .5 ;
M1 E1 A1
3
stated or implied convincingly shown (AG)
M1 A1
OE
M1 A1F
or other complete method ft wrong answer for first point; allow NMS 2/2
4 11 75
Scaled mark unit grade boundaries - June 2011 exams A-level Max. Scaled Mark
Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E
Code
Title
HBIO2 HBI3T HBI3X HBIO4 HBIO5 HBI6T HBI6X
GCE HUMAN BIOLOGY UNIT 2 GCE HUMAN BIOLOGY UNIT 3T GCE HUMAN BIOLOGY UNIT 3X GCE HUMAN BIOLOGY UNIT 4 GCE HUMAN BIOLOGY UNIT 5 GCE HUMAN BIOLOGY UNIT 6T GCE HUMAN BIOLOGY UNIT 6X
80 50 50 90 90 50 50
61 65 43 43
56 41 34 56 60 40 39
51 38 30 51 55 37 35
46 35 26 46 50 34 31
41 33 22 41 45 31 27
37 31 19 36 41 28 24
INFO1 INFO2 INFO3 INFO4
GCE INFO AND COMM TECH UNIT 1 GCE INFO AND COMM TECH UNIT 2 GCE INFO AND COMM TECH UNIT 3 GCE INFO AND COMM TECH UNIT 4
80 80 100 70
72 63
53 53 66 57
48 47 60 50
43 41 54 43
38 35 48 36
33 30 43 30
LAW01 LAW02 LAW03 LAW04
GCE LAW GCE LAW GCE LAW GCE LAW
96 94 80 85
69 73
73 70 62 66
66 61 55 59
59 52 48 53
52 43 42 47
46 35 36 41
XMCAS MD01 MFP1 MM1A MM1B MPC1
GCE MATHEMATICS UNIT XMCAS GCE MATHEMATICS UNIT D01 GCE MATHEMATICS UNIT FP1 GCE MATHEMATICS UNIT M1A GCE MATHEMATICS UNIT M1B GCE MATHEMATICS UNIT PC1
125 75 75 100 75 75
-
99 89 79 70 59 52 45 39 61 54 48 42 No candidates were entered for this unit 62 55 48 41 62 55 48 41
61 33 36
GCE MATHEMATICS UNIT S1A GCE MATHEMATICS UNIT S1A - WRITTEN GCE MATHEMATICS UNIT S1A - COURSEWORK
100 75 25
-
73 53 20
36 26 10
MS1A MS/SS1A/W MS/SS1A/C
UNIT 1 UNIT 2 UNIT 3 UNIT 4
-
63
54
45
34 34
