General Certificate of Education Advanced Subsidiary Examination June 2012
Mathematics
MFP1
Unit Further Pure 1 Friday 18 May 2012
9.00 am to 10.30 am
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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 each question in the space provided for that question. If you require extra space, use an AQA supplementary answer book; do not use the space provided for a different question. * 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. * You do not necessarily need to use all the space provided.
P49998/Jun12/MFP1 6/6/
MFP1
2
The quadratic equation
1
5x 2 7x þ 1 ¼ 0 has roots a and b . (a)
Write down the values of a þ b and ab .
(b)
Show that
(c)
Find a quadratic equation, with integer coefficients, which has roots
(2 marks)
a b 39 þ ¼ . b a 5
(3 marks)
aþ
1 a
and b þ
1 b
(5 marks)
A curve has equation y ¼ x 4 þ x .
2 (a)
Find the gradient of the line passing through the point ð2, 14Þ and the point on the curve for which x ¼ 2 þ h . Give your answer in the form p þ qh þ rh2 þ h3 where p, q and r are integers.
(b)
(5 marks)
Show how the answer to part (a) can be used to find the gradient of the curve at the point ð2, 14Þ. State the value of this gradient. (2 marks)
It is given that z ¼ x þ iy , where x and y are real numbers.
3 (a)
Find, in terms of x and y , the real and imaginary parts of iðz þ 7Þ þ 3ðz* iÞ
(b)
Hence find the complex number z such that iðz þ 7Þ þ 3ðz* iÞ ¼ 0
4
Find the general solution, in degrees, of the equation sin 70 2 x ¼ cos 20 3
(02)
(3 marks)
(3 marks)
(6 marks)
P49998/Jun12/MFP1
3
The curve C has equation y ¼
5
x . ðx þ 1Þðx 2Þ 1
The line L has equation y ¼ 2 . (a)
Write down the equations of the asymptotes of C.
(b)
The line L intersects the curve C at two points. Find the x-coordinates of these two points. (2 marks)
(c)
Sketch C and L on the same axes. (You are given that the curve C has no stationary points.)
(3 marks)
(3 marks)
Solve the inequality
(d)
x 41 2 ðx þ 1Þðx 2Þ
(3 marks)
Using surd forms, find the matrix of a rotation about the origin through 135 anticlockwise. (2 marks)
6 (a)
" The matrix M is defined by M ¼
(b)
(i)
1
1
1
1
# .
Given that M represents an enlargement followed by a rotation, find the scale factor of the enlargement and the angle of the rotation. (3 marks)
(ii) The matrix M2 also represents an enlargement followed by a rotation. State the scale
factor of the enlargement and the angle of the rotation.
(2 marks)
(iii) Show that M4 ¼ kI , where k is an integer and I is the 2 2 identity matrix.
(2 marks) (iv) Deduce that M2012 ¼ 2n I for some positive integer n.
Turn over
s
(03)
(2 marks)
P49998/Jun12/MFP1
4
The equation
7
24x 3 þ 36x 2 þ 18x 5 ¼ 0 has one real root, a . (a)
Show that a lies in the interval 0:1 < x < 0:2 .
(b)
Starting from the interval 0:1 < x < 0:2 , use interval bisection twice to obtain an interval of width 0.025 within which a must lie. (3 marks)
(c)
Taking x1 ¼ 0:2 as a first approximation to a , use the Newton–Raphson method to find a second approximation, x2 , to a . Give your answer to four decimal places. (4 marks)
(04)
(2 marks)
P49998/Jun12/MFP1
5
The diagram shows the ellipse E with equation
8
x2 y2 þ ¼1 5 4 and the straight line L with equation y¼xþ4 y
x
(a)
(b)
(c)
Write down the coordinates of the points where the ellipse E intersects the coordinate axes. (2 marks) p The ellipse E is translated by the vector , where p is a constant. Write down 0 the equation of the translated ellipse. (2 marks) Show that, if the translated ellipse intersects the line L, the x-coordinates of the points of intersection must satisfy the equation 9x 2 ð8p 40Þx þ ð4p 2 þ 60Þ ¼ 0
(d)
(3 marks)
Given that the line L is a tangent to the translated ellipse, find the coordinates of the two possible points of contact. (No credit will be given for solutions based on differentiation.)
(8 marks)
Copyright ª 2012 AQA and its licensors. All rights reserved.
(05)
P49998/Jun12/MFP1
Key to mark scheme abbreviations M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA c sf dp
mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach candidate significant figure(s) decimal place(s)
No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.
General Certificate of Education MFP1 June 2012 Q
Solution
1(a)
(b)
Marks
7 1.4 5
Total
B1
1 0.2 5
B1
2 2
2
M1 2
7 1 2 ( ) 2 2 5 5 = = 1 5 49 1 49 2 39 2 25 5 25 5 25 39 = 1 1 1 5 5 5 5
(c) (Sum=)
7 5
1
1
=
m1
A1
3
1 7( ) 1 1 OE eg = 5 scores M1 m1 in terms of Correct expression for either (+) and or with numerical substitution of correct/c’s values from (a) CSO AG must see some intermediate evaluation, must see one of the first three expressions A0 if has wrong sign
Writing
1
1
in a correct suitable form or with numerical values
M1
7 5 1 5
(Product =) =
Comments Accept correct equivalent decimals in place of some/all fractions in the scheme
1
Correct expression for product into which substitution of numbers attempted for all terms, at least one either correct/correct ft
M1
1 39 5 5 5
OE Both
42 Sum = , Product = 13 5
SC If B0 for
A1
S=
x Sx P 0
M1
Equation is 5x2 − 42x + 65 = 0
A1
2
Total
5 10
7 in (a), and (c) 5
42 oe, P = 13 award this A1 5
Using correct general form of LHS of equation with ft substitution of c’s S and P values. PI. M0 if either S = or P = values CSO Integer coefficients and ‘= 0’ needed. Dependent on B1B1 in (a) and previous 4 marks in (c) scored
Q
Solution 2(a)
Marks
(2 h) 4 (2 h)
B1
= h 4 8h 3 24 h 2 31h 14 Gradient = =
A1F
y 2 y1 x 2 x1
h 4 8h 3 24h 2 31h 14 (14) 2 h (2)
M1
h 4 8h 3 24h 2 31h = h h 3 8h 2 24 h 31
A1
As h→ 0, gradient of line in (a) → gradient of curve at point (−2, 14)}
E1
{Gradient of curve at point (−2, 14) is} −31
E1
=
Total
(b)
(2 h) 4 ( 2 h) PI Correct expansion of (−2 + h)4 as h 4 8h 3 24 h 2 32 h 16 PI Seen separately or as part of the gradient expression. Ft one incorrect term in expansion of (−2+h)4
M1
= h 4 8h 3 24 h 2 32 h 16 −2 + h
3(a)
Comments
y x4 x
y (2 h)
(b)
Total
i z 7 3 z * i
Use of correct formula for gradient PI
5
2
The four correct terms in any order A0 if incorrect (constant/h) term ignored due printed form of answer Lim [c’s(p+qh+rh2+h3)] OE h→0 NB ‘h=0’ instead of ‘h→ 0’ gets E0 Dependent on previous E1 and printed form of answer in (a) obtained convincingly but then ft on c’s p value
7
i x i y 7 3 x iy i ix y 7i 3 x 3iy 3i
M1
M1 for use of z* = x − iy
M1
3x y i x 3 y 4
A1
M1 for i2 y = − y If the five terms correct but not grouped into Real and Imaginary parts, allow A1 retrospectively provided the correct two expressions used in the M1 line in (b)
3 x y 0,
M1
x 3y 4 0
x 9x 4 0
(or eg y 9 y 12 0 )
Solving to give z
1 3 i 2 2
A1 A1
Total
3
3 6
Attempting to equate all Real parts to zero and all Imaginary parts to zero A correct equation in either x or y PI by correct final answer 1 3 Allow x , y 2 2
Q
Solution
4
Marks
2 sin 70 x cos 20 sin 70 3 2 sin 70 x sin 110 3 2 70 x 360n "70 " 3 2 70 x 360n "110 " 3 3 x 70 70 360n 2
x
Total
B1 B1
OE; Use of a correct angle, in degrees, in other relevant quadrant PI
M1
OE; Either one, showing a correct use of 360n in forming a general solution. Condone 2n in place of 360n
3 70 110 360n 2
Comments Watch out for the many correct different forms of the general solutions OE cos20 = sin70; or cos20 = sin110 etc PI
Rearrangement of 70
x 540 n ; x 540 n 60
A2,1,0
Total
3 360n 70 OE, 2 where is from c’s sin = cos20 Condone 2n in place of 360n OE eg 540nº, 540nº−60º. Condone 0 ± 540n for ± 540n. If not A2, award (i) A1 for either correct unsimplified full general solution or (ii) A1F for correct ft full general solution, ft c’s wrong angle(s) after award of B0, may be left in unsimplified form(s) or (iii) A1 for ‘correct’ simplified full general solution but with radians present A0 for only a partial correct solution OE to x
m1
6
6
2 x 360n 3
Q
Solution
Marks
5(a) Asymptotes x=–1 x=2 y=0 (b)
B1 B1 B1
1 x 2 x 2 x 2 2 x 2 x x2
M1
x 2 x 2 0 x 1, x 2
A1
(c)
0
(2)
x
1 2
3
x = −1 OE x = 2 OE y=0
2
2 x 1 1 x 2 2 x 1 , 1 x 2
B1 B1 B1 Total
Correctly removing brackets and fractions to reach x 2 x 2 2 x OE Correct two values for x-coordinates. NMS 2 or 0 marks Three branches shown on sketch of C with either middle branch or outer two branches correct in shape All three branches, correct shape and positions and approaching correct asymptotes in a correct manner. If middle branch does clearly not go through the origin, then A0
A1
B1 (d)
Comments
M1
y
(–1)
Total
3
3 11
Correct sketch of line (L), y = − 0.5 identified Condone < for ≤ or vice versa Condone < for ≤ or vice versa All complete and correct
Q
Solution 6(a)
1 2 1 2
Marks
1 2 1 2
1 1 (b)(i) M= = 1 1
1 2 = 1 2
M1 A1
1 2 2 1 2
1 2 1 2
2
Comments
If A1 not scored, award M1A0 for all correct entries expressed in trig form eg cos 135 sin 135 sin 135 cos 135
2 0 2 0 Scale factor of enlargement is 2 1 2 1 2
Or better PI by cand. having both a correct scale factor of enlargement and a correct corresponding angle of rotation
M1
A1
Angle of rotation is 135 (degrees anticlockwise)
A1
(b)(ii) For M2, SF of enlargement = 2
B1F
Angle of rotation is 270 (degrees anticlockwise)
B1F
(iii)
Total
3
2
M1
4
1 0 M 4 4 4I 0 1
A1
OE If incorrect, ft on [c’s SF in (b)(i)]2 OE, eg − 90(degrees), eg 90 (degrees) clockwise If incorrect, ft on 2×c’s angle in (b)(i) (neither B1F B1 nor B1 B1F is possible) For complete method (matrix calculation or geometrical reasoning) Matrix for M2 could be seen earlier (M0 if >1 independent error in matrix multiplication) Geometrically SF = 4, rotation angle= 540 OE scores M1 and completion scores A1
1 1 1 1 0 2 M2 1 1 1 1 2 0 0 2 0 2 4 0 M 2 0 2 0 0 4
SF = 2 OE surd form Angle = 135 OE eg −225 If M0 give B1 for SF= 2 OE surd and B1 for angle = 135 OE
2
Either of these two forms convincingly shown
(iv) M2012 = (M4) 503 = (−4I)503 =
− (22)503I = − 21006I
E1
M2012 = − 21006I
B1
2
(Geometrically: M2012 represents an enlargement with SF 21006 followed by a rotation of angle 2012×135º ie 754.5 revolutions, being equivalent to rotation of 180º ie matrix is −I so M2012 = − 21006 I) Total
11
OE Fully explained, algebraically from (−4I)503 , or geometrically M2012 = − 21006 I (n = 1006) (B0 if FIW)
Q
Solution
Marks
Total
Comments
7(a) Let f(x) = 24 x 3 36 x 2 18 x 5
f(0.1) = −2.816, f(0.2) = 0.232
M1
Change of sign so lies between 0.1 and 0.2
A1
(b) f(0.15) = −1.409 (< 0 so root > 0.15)
2
M1
f(0.175) ≈ −0.619 (< 0 so root > 0.175)
A1
lies between 0.175 and 0.2
A1
(c) f' (x) = 72x2 + 72x + 18 (x2 =) 24(0.2) 3 36(0.2) 2 18(0.2) 5 0. 2 72(0.2) 2 72(0.2) 18
= 0.1934 (to 4dp)
f(0.15) considered first f(0.15) then f(0.175) both evaluated correctly to at least 1sf OE fractions Dependent on both previous marks gained and no other additional evaluations other than at 0.15 and 0.175
B1
PI
B1 B1
B1 for numerator in correct formula B1 for denominator in correct formula
B1 Total
3
Both attempted and at least one evaluated correctly to at least 1sf rounded or truncated OE fraction Need both evaluations correct to above degree of accuracy and ‘change of sign OE’ and relevant reference to 0.1 and 0.2
4 9
CAO Must be 0.1934 Do not apply ISW NMS scores 0/4
Q
Solution
Marks
8(a)
8(b)
( x p) 2 y 2 1 5 4
M1 A1
8(c)
( x p ) 2 ( x 4) 2 1 5 4
M1
5, 0 ,
0 , 2
B2,1
Total
2
2
m1
4x2 − 8px + 4p2 + 5x2 + 40x + 80 = 20
(d)
9x2 − (8p − 40)x + 4p2 + 60 = 0
A1
Discriminant is (8p − 40)2 – 4 (9) (4p2 + 60)
B1
If not B2, award B1 if either at least two of these 4 correct pts or if ‘ x 5 and y = ± 2’ Replacing x by either x + p or x − p and keeping y unchanged or as y ± 0 ACF Substitution into c’s (b) eqn of y = x+4 to eliminate y Denominators 5 and 4 cleared in a correct manner and at least either a correct expansion of (x ± p)2 or a correct expansion of (x + 4)2
4(x − p)2 + 5(x + 4)2 = 4 × 5 4(x2 − 2px + p2) + 5(x2 + 8x + 16) = 20
Comments
3
CSO No errors in any line of working. AG. Must see brackets correctly removed and all terms involving x, p correctly rearranged to same side before the printed answer is stated. Must have ‘= 0’ although brackets around 4p2 + 60 may be omitted OE Must be isolated, not just within the quadratic formula
For tangency (8p − 40)2 – 4 (9) (4p2 + 60) = 0
M1
p2 + 8p + 7 = 0
A1 B1
p = −7: 9x2 + 96x + 256 (= 0)
M1
{(p + 1) (p + 7) = 0 } p = −1, p = −7 (*) p = −1: 9x2 + 48x + 64 (= 0)
8 3
A1
16 3
A1
p = −1: 9x2 + 48x + 64 (= 0) x
p = −7: 9x2 + 96x + 256 (=0) x
8 4 16 4 x , y ; x , y 3 3 3 3 4 8 4 16 , , 3 3 3 3 Total TOTAL
A1
OE Equating c’s discriminant to zero before obtaining any values for p ACF with like terms collected Correct values −1, −7 for p Substitutes at least one of c’s two values for p either into the given quadratic in (c) 8 p 40 OE or into 18 8 x OE as only root from the 3 8 p 40 quadratic or from . Apply FIW if 18 (*) is B0 16 x=− OE as only root from the 3 8 p 40 . Apply FIW if quadratic or from 18 (*) is B0 8
15 75
CSO Previous 7 marks must have been awarded and coordinates of both points need to be correct and exact but accept in either format
Scaled mark unit grade boundaries - June 2012 exams A-level Max. Scaled Mark
Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E
Code
Title
MD01
MATHEMATICS UNIT MD01
75
-
60
55
50
45
40
MD02
MATHEMATICS UNIT MD02
75
68
61
53
45
38
31
MFP1
MATHEMATICS UNIT MFP1
75
-
61
54
47
41
35
MFP2
MATHEMATICS UNIT MFP2
75
68
63
56
49
42
35
MFP3
MATHEMATICS UNIT MFP3
75
70
65
57
49
41
33
MFP4
MATHEMATICS UNIT MFP4
75
61
55
48
41
34
28
MM1A
MATHEMATICS UNIT MM1A
100
-
79
69
59
49
39
MM1A/W
MATHEMATICS UNIT MM1A - WRITTEN
75
59
29
MM1A/C
MATHEMATICS UNIT MM1A - COURSEWORK
25
20
10
MM1B
MATHEMATICS UNIT MM1B
75
-
57
49
41
33
26
MM2B
MATHEMATICS UNIT MM2B
75
69
63
55
48
41
34
MM03
MATHEMATICS UNIT MM03
75
62
55
48
41
34
27
MM04
MATHEMATICS UNIT MM04
75
67
60
52
44
37
30
MM05
MATHEMATICS UNIT MM05
75
67
60
52
44
37
30
MPC1
MATHEMATICS UNIT MPC1
75
-
58
51
44
37
30
MPC2
MATHEMATICS UNIT MPC2
75
-
51
46
41
36
31
MPC3
MATHEMATICS UNIT MPC3
75
67
61
55
49
43
38
MPC4
MATHEMATICS UNIT MPC4
75
59
53
47
41
36
31
MS1A
MATHEMATICS UNIT MS1A
100
-
76
67
59
51
43
