General Certificate of Education Advanced Level Examination June 2010
Mathematics
MFP4
Unit Further Pure 4 Tuesday 15 June 2010
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 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.
P28066/Jun10/MFP4 6/6/6/
MFP4
2
The position vectors of the points P, Q and R are, respectively, 2 3 2 3 2 3 3 1 1 4 5 4 5 4 p¼ 4 , q¼ 2 and r ¼ 4 5 1 2 1
1
(a)
Show that p, q and r are linearly dependent.
(2 marks)
(b)
Determine the area of triangle PQR.
(4 marks)
1 x 1 1 4 4x 8 Let A ¼ , B¼ and C ¼ . 2 3 2 2 8x 4 4
2
(a)
Find AB in terms of x.
(2 marks)
(b)
Show that BTAT ¼ C for some value of x.
(5 marks)
The plane P1 is perpendicular to the vector 9i 8j þ 72k and passes through the point Að2, 10, 1Þ .
3
(a)
Find, in the form r . n ¼ d , a vector equation for P1 .
(b)
Determine the exact value of the cosine of the acute angle between P1 and the (4 marks) plane P2 with equation r . ði þ j þ kÞ ¼ 11 .
(3 marks)
The fixed points A and B and the variable point C have position vectors 2 3 2 3 2 3 3 2 2t a ¼4 4 5, b ¼ 4 1 5 and c ¼ 4 t 5 1 3 5
4
respectively, relative to the origin O, where t is a scalar parameter. (a)
Find an equation of the line AB in the form ðr uÞ v ¼ 0 .
(3 marks)
(b)
Determine b c in terms of t.
(4 marks)
(c) (i)
Show that a . ðb cÞ is constant for all values of t, and state the value of this constant. (2 marks)
(ii) Write down a geometrical conclusion that can be deduced from the answer to (1 mark) part (c)(i).
P28066/Jun10/MFP4
3
x Factorise fully the determinant x 2 yz
5
y y2 zx
z z 2 . xy
(8 marks)
The line L and the plane P have vector equations 2 3 2 3 2 3 2 3 2 3 7 6 2 5 1 4 5 4 5 4 5 4 5 4 r¼ 8 þt 2 and r ¼ 0 þ l 3 þ m 65 50 9 25 4 2
6
respectively. (a) (i)
Find direction cosines for L.
(2 marks)
(ii) Show that L is perpendicular to P .
(3 marks)
For the system of equations
(b)
6p þ 5q þ r ¼ 9 2p þ 3q þ 6r ¼ 8 9p þ 4q þ 2r ¼ 75 form a pair of equations in p and q only, and hence find the unique solution of this system of equations. (5 marks) It is given that L meets P at the point P.
(c) (i)
Demonstrate how the coordinates of P may be obtained from the system of equations (2 marks) in part (b).
(ii) Hence determine the coordinates of P.
(2 marks)
The transformation T is represented by the matrix M with diagonalised form
7
M ¼ U D U1
4 where U ¼ 1 (a) (i)
1 3
27 0 and D ¼ . 0 1
State the eigenvalues, and corresponding eigenvectors, of M.
(ii) Find a cartesian equation for the line of invariant points of T.
(4 marks) (2 marks)
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Turn over
P28066/Jun10/MFP4
4 (b)
Write down U1 , and hence find the matrix M in the form a b c d where a, b, c and d are integers.
(c)
(5 marks)
By finding the element in the first row, first column position of M n , prove that 4 33nþ1 þ 1 is a multiple of 13 for all positive integers n.
(5 marks)
8
12 16 The matrix represents the transformation which is the composition, in 9 36 either order, of the two plane transformations E: an enlargement, centre O and scale factor k ðk > 0Þ and S: a shear parallel to the line l which passes through O Show that k ¼ 24 and find a cartesian equation for l.
(7 marks)
END OF QUESTIONS
Copyright Ó 2010 AQA and its licensors. All rights reserved.
P28066/Jun10/MFP4
MFP4 - AQA GCE Mark Scheme 2010 June series
Key to mark scheme and abbreviations used in marking 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
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
MC MR RA FW ISW FIW BOD WR FB NOS G c sf dp
mis-copy mis-read required accuracy further work ignore subsequent work from incorrect work given benefit of doubt work replaced by candidate formulae book not on scheme graph 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. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. 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.
3
MFP4 - AQA GCE Mark Scheme 2010 June series
MFP4 Q
Solution 1(a)
Marks
3 4 −1 −1 2 2 = 6 + 8 + 4 + 2 – 24 + 4 1 4 1
or 3(2 – 8) – 4(–1 – 2) – 1(–4 – 2) etc or 3(2 – 8) + 1(4 + 4) + 1(8 + 2) etc Correctly shown = 0 Or 3p + 4q = 5r
= =
1 2
1 2
i j k 4 2 −3 = 2 0 −2 2
2
4 +2 +4
M1
1 2
M1 A1
(b)
A1
⎡ 1 2⎤ ⎡1 2⎤ BTAT = (AB)T Or ⎢ ⎥ ⎥⎢ ⎣ −1 2 ⎦ ⎣ x 3 ⎦ ⎡2 x + 1 8⎤ = ⎢ ⎥ ⎣ 2 x − 1 4⎦ 2x + 1 = 4 – 4x Or 2x – 1 = 8x – 4 x = 12 1 2
in other eqn.
⎡9⎤ Clearly identifying n = ⎢⎢ −8⎥⎥ ⎢⎣ 72 ⎥⎦ ⎡9 ⎤ ⎡2⎤ d = ⎢⎢ −8⎥⎥ • ⎢⎢10 ⎥⎥ = 10 ⎢⎣ 72 ⎥⎦ ⎢⎣ 1 ⎥⎦
Use of
4 6
Method for finding magnitude of their relevant vector CSO
Good attempt (at least one entry in R19)
M1
⎡ 2 x + 1 2 x − 1⎤ AB = ⎢ 4 ⎥⎦ ⎣ 8
2
All four correct
M1 A1
Ft their (a) Or CAO
M1 A1
Ft previous answers CAO
B1
Total 3(a)
Formula used with attempt at a vector product of any 2 of the above (ignore missing 12 for now)
±(4i − 2 j + 4k )
2
Checking/noting x =
2
M1
Total
(b)
Good attempt at det M0 for | .. | = 0 and no working
M1
=3 2(a)
Comments
A1 (M1) (A1)
JJJG JJJG JJJG (b) For attempt at 2 of (±) PQ , PR , QR JJJG JJJG Area ΔPQR = 12 QP × QR e.g.
Total
5
Visibly
7
B1 M1 A1
Sc.prod. of normals prod. of their moduli
Nr. = 73 Dr. = 73 3 or 15987 1 cosθ = 3
3
M1
Must be (9i – 8j + 72k), (i + j + k) or their n from (a)
B1 B1
Ft their n from (a) only
A1 Total
4 7
4
CAO Allow unsimplified exact forms
MFP4 - AQA GCE Mark Scheme 2010 June series
MFP4 (cont) Q 4(a)
Solution ⎡1⎤ (v =) ± (a – b) = ± ⎢⎢ −5⎥⎥ ⎣⎢ 4 ⎦⎥
Marks
i j k ⎡ 3t + 5 ⎤ 2 1 −3 = ⎢⎢3t − 16 ⎥⎥ 2 − t t 5 ⎢⎣ 3t − 2 ⎥⎦ ⎡ 3 ⎤ ⎡ 3t + 5 ⎤ (c) (i) a • b × c = ⎢ −4⎥ • ⎢3t − 16 ⎥ = 77 ⎥ ⎢ ⎥ ⎢ ⎢⎣ 1 ⎥⎦ ⎢⎣ 3t − 2 ⎥⎦
(b) b × c =
(ii) C never lies in plane of O, A, B (or is a fixed distance from it) or Vol. //ppd. OABC always = 77 or Vol. tetrhdrn. OABC always = or O is never in plane of A, B, C JJJG JJJG JJJG or OA , OB , OC never co-planar
B1
3
M1 A3,2,1
4
M1
y
z
Δ = x2
y2
z2
yz
zx
xy
x = x
2
yz
y−x 2
y −x
2
CAO
B1
1
Any suitable geometrical comment
Vectors 9; points 8
y( x − z) x
1
1
y+x
z+x
yz
−z
−y
x
1
0
y+x
z−y
−z
z−y
= ( y − x)( z − x) x
= ( y − x)( z − x) x
2
2
yz
x = ( y − x)( z − x)( z − y ) x
2
yz
10
M1 M1
z 2 − x2
z( x − y)
1
By C2′ = C2 – C1 (eg) C3′ = C3 – C1 (eg)
A1 A1
0
y+x 1 −z
3 −4 1 2 1 −3 2−t t 5
77 6
z−x 2
Or starting again:
A1
Total x
Comments
JJJG M1 A0 if ± AB found but not stated/shown this is v
M1 A1
⎡3⎤ ⎡2⎤ ⎢ ⎥ u = ⎢ −4⎥ or ⎢⎢ 1 ⎥⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ −3⎥⎦
5
Total
First two factors extracted (what’s left has to be correct also)
M1
By C3′ = C3 – C2 (e.g.)
A1
3rd factor extracted
M1
Further R/C ops or expansion of remaining det (almost a dM1) CAO up to equivalents due to repositioning of the signs
1
= ( x − y )( y − z )( z − x)( xy + yz + zx)
A1 Alternatives using Cyclic Symmetry and the Factor Theorem are fine Total
8
8
5
MFP4 - AQA GCE Mark Scheme 2010 June series
MFP4 (cont) Q Solution 6(a)(i) • = 6 2 + 2 2 + 9 2 attempted and ⎛ 6 2 −9 ⎞ ±⎜ , , ⎟ ⎝• • • ⎠ • = 11 and all correct (ii)
Marks
A1
⎡ 6 ⎤ ⎡5⎤ ⎡ 6 ⎤ ⎡1 ⎤ ⎢ ⎥ ⎢ ⎥ Or ⎢ 2 ⎥ • ⎢ 3 ⎥ = 0 and ⎢⎢ 2 ⎥⎥ • ⎢⎢ 6 ⎥⎥ = 0 ⎢⎣ −9 ⎥⎦ ⎢⎣ 4 ⎥⎦ ⎢⎣ −9 ⎥⎦ ⎢⎣ 2 ⎥⎦ Explaining that d.v. of L is ⊥r to 2 (non-//) vectors in Π ⇒ L ⊥r Π
2×① – ③:
–57 = 21p + 6q
M1 A1 B1 (M1) (A1) (B1)
3
7 + 6t = −2 + 5λ + μ 0 + 3λ + 6μ (i) 8 + 2t = 50 − 9t = −25 + 4λ + 2μ
A1
Not just stating Eliminating r from any pair of eqns. Any 2 correct eqns (1 mark each)
M1
p = –5, q = 8, r = –1
± (0.545, 0.182, – 0.818) ok
Correct vector product only here
A1 A1
2×④ + 9×⑤: 605 = – 121p
→
2
M1
46 = 34p + 27q
② – 3×③: –217 = 29p – 9q
(c)
Comments
M1
⎡5 ⎤ ⎡1 ⎤ ⎡6⎤ ⎢ ⎥ ⎢ ⎥ Either ⎢ 3 ⎥ × ⎢ 6 ⎥ = −3 ⎢⎢ 2 ⎥⎥ ⎢⎣ 4 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ −9 ⎥⎦ Explaining that d.v. of L is in dirn. of Π ’s nml. ⇒ L ⊥r Π
(b) E.g. 6×① – ②:
Total
5
M1
Solving a 2×2 system (any means) in order to get values for p, q, r All 3 9 CAO
Including re-arrangement attempt
9 = −6t + 5λ + μ 8 = −2t + 3λ + 6μ 75 = 9t + 4λ + 2μ i.e. the above system with p = – t, q = λ and r = μ
A1
2
M1 A1
2
(ii) Substg. t = 5 into L’s eqn. Or λ = 8 and μ = – 1 into ∏ ’s eqn. P = (37, 18, 5)
Total
14
6
CAO
MFP4 - AQA GCE Mark Scheme 2010 June series
MFP4 (cont) Q Solution 7(a)(i) Evals λ = 27 1 4 ⎡ ⎤ ⎡ −1⎤ Evecs (α) ⎢ ⎥ and (β) ⎢ ⎥ ⎣1 ⎦ ⎣3⎦ (ii)
Marks B1
B1 B1 B1
y = – 3x
B1
⎡ 3 1⎤ ⎢ −1 4 ⎥ ⎣ ⎦ M = U D U–1 ⎡ 4 −1⎤ ⎡ 27 0 ⎤ ⎡ 3 1 ⎤ = 131 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 1 3 ⎦ ⎣ 0 1 ⎦ ⎣ −1 4 ⎦ ⎡ 4 −1⎤ ⎡ 81 27 ⎤ = 131 ⎢ ⎥⎢ ⎥ ⎣ 1 3 ⎦ ⎣ −1 4 ⎦ 1 13
or
1 13
⎡108 −1⎤ ⎡ 3 1 ⎤ ⎢ 27 3 ⎥ ⎢ −1 4 ⎥ ⎣ ⎦ ⎣ ⎦
⎡ 25 8⎤ = ⎢ ⎥ ⎣ 6 3⎦
Correctly matched up with evals (look out for λ1, v1 notations) Ft 4y = x if evecs mis-matched
2
Must say why they have chosen this one
1 13
3n
Det; mtx
M1
Including attempt to multiply (at least U D …)
A1
Ft incorrect/missing U – 1 for one product; ignore missing 131 until the end 5
(12 × 27
n
)
n
0⎤ ⎥ 1⎦
3n+1
B1 A1
+1
So 4 × 3 × 3 + 1 = 4 × 3
CAO Including attempt to multiply
M1
⎡ 27 Dn = ⎢ ⎣ 0
Mn (1,1) =
4
B1 B1
A1
(c) Mn = U Dn U – 1
Comments
Both
B1
from λ = 1 –1 (b) U =
Total
+ 1 div. by 13
Since M has all integer elements, each element of Mn is an integer also Total
Legitimately so from their working, from fact that the element is an integer
E1 E1
5 16
7
Explaining why it must be an integer
MFP4 - AQA GCE Mark Scheme 2010 June series
MFP4 (cont) Q Solution 8 det W = 12.36 + 9.16 = 576 = k2 ⇒ k = 24
Marks M1 A1
⎡ 1 2⎤ W = ⎢ 2 3 33 ⎥ ⎣− 8 2 ⎦ ⎡ 12 32 ⎤ ⎡ x ⎤ ⎡ 12 x + 23 y ⎤ ⎢ 3 3⎥⎢ ⎥ = ⎢3 3 ⎥ ⎣− 8 2 ⎦ ⎣ y ⎦ ⎣ 2 y − 8 x ⎦ 1 24
Total
Comments Attempt at det. = k2
2
B1 M1 A1
⎡ x⎤ Equating this to ⎢ ⎥ = ⎣ y⎦
M1
y=
3 4
x
A1
CAO
ALT. 1
⎡ 1 2⎤ W = ⎢ 2 3 33 ⎥ ⎣− 8 2 ⎦ ⎡ 12 23 ⎤ ⎡ x ⎤ ⎡( 12 + 23 m ) x ⎤ ⎥ ⎢ 3 3⎥⎢ ⎥ = ⎢ 3 3 ⎣ − 8 2 ⎦ ⎣ mx ⎦ ⎣ ( 2 m − 8 ) x ⎦ Setting y′ = mx′ and solving for m y = 34 x 1 24
(B1) (M1) (A1) (M1) (A1)
Get (4m – 3)2 = 0 CAO
(M1) (A1)
This may simply be stated or assumed
ALT. 2
λ2 – 2λ + 1 = 0
⇒ λ = 1 (twice) − 12 x + 23 y = 0 λ=1 ⇒ 3 − 8 x + 12 y = 0 y=
3 4
x
(M1) (A1) (A1)
Total TOTAL
8
5 7 75
AQA – Further pure 4 – Jun 2010 – Answers Question 1:
Exam report
a) The vectors p, q and r are linearly dependent if and only if p.(q × r ) = 0. 3 4 −1 −1 2 −1 2 2 2 p.(q × r ) = −1 2 2= 3 −4 −1 4 1 1 1 1 4 1 4 1 = 3 × −6 − 4 × −3 − 1× −6 = −6 + 12 − 6 = 0. p, q and r are linearly dependent.
1 a × b but then thought that a and b were 2
1 b) Area of PQR = PQ × PR 2 PQ = −4i − 2 j + 3k and PR = −2i + 2k i j k 2 PQ × PR = −4 2 3 = 0 −2 0 2 1 Area of PQR = PQ × = PR 2
This was generally done very well. The only common mistake arose when candidates quoted the area formula for a triangle as
p and q (for instance) rather than
PQ and PR (say).
−4 3 −4 2 3 i− j+ k= 4i − 2 j + 4k −2 2 −2 0 2 1 2 1 4 + (−2) 2 += 42 = 36 3 2 2
Question 2:
Exam report
1 x 1 −1 1 + 2 x a) A × = B = × 2 3 2 2 8 1 + 2 x 8 T b) BT AT= ( AB ) = = C= 2 x − 1 4
2 x − 1 4
4 − 4 x 8 8 x − 4 4 if and only if 1 + 2 x = 4 − 4 x and 2 x − 1 = 8 x − 4 6 x 3= and 6 x 3 = 1 x= 2
This was another straightforward question and was usually well-received and successfully attempted. In part (b), candidates seemed evenly split between those who recognised T T T T T that B A = (AB) and those who simply worked out B A directly. However, the very last mark on the question was often not gained, as many failed to show visibly that x =
1 2
worked in both of the equations that arose when comparing elements of the two matrices.
Question 3:
Exam report
2 9 a ) a =10 n = −8 a.n =18 − 80 + 72 =10 1 72 An equation of the plan ∏1 is r.n = 10 b) A vector perpendicular to ∏ 2 is u = i +j+k Cos θ =
n.u = n u
Cos= θ
1 = 3
9 − 8 + 72 73 = = 81 + 64 + 5184 × 1 + 1 + 1 73 3 3 3
1 3
A similar fault was all too common here in part (a), even amongst those who presumably knew what they were doing. A lot of candidates correctly found d = 10 from the use of the vector 9i – 8j + 72k, yet still lost a mark by nowhere identifying this as the “n” of the question … not even to the extent of writing down the answer in the form requested. This is just carelessness. Apart from a small minority who failed to identify n correctly, part (b) was found to be very straightforward indeed.
Question 4:
Exam report
−1 a ) AB = b − a = 5 −4 3 −1 An equation of the line AB is r − −4 × 5 =0 1 −4 i j k −3 1 −3 2 2 1 −3 b) = 2 1= b×c i− j+ k t 5 2−t 5 2−t t 2−t t 5 = (5 + 3t )i − (16 − 3t ) j + ( 3t − 2 ) k 3 5 + 3t c) i ) a. ( b × c ) = −4 . 3t − 16 = 15 + 9t − 12t + 64 + 3t − 2 = 77 1 3t − 2 ii ) The volume of the parallelepiped form by OABC is constant. (see M.S for other options) Question 5:
x x2
y y2
yz
zx
z x 2 z = x2 xy yz
x ( x − y )( x − z ) x 2 = yz
This proved to be a good question for all but the weakest candidates, although it is still surprising how many mistakes are made over minus signs by further mathematicians, even when armed with a calculator. A little bit of care was needed in part(b) – almost invariably with the minus sign that accompanies the j-component – and it was hard to recover and answer part(c)(i) sensibly if part (b)’s answer was incorrect, as the t failed to cancel out. Part (c)(ii), where another explanation was required, received very poor answers indeed. There are so many valid possible “geometrical” conclusions that could be offered that it was very disappointing to see so few being offered. Most candidates opted to interpret the result as one of linear independence, which isn’t a geometrical statement at all. The key issue at stake here is that the answer is always 77, not just that it is 77 on some specific occasion, and the candidates’ remarks were expected to reflect this.
Exam report
x− y x2 − y 2
x−z x2 − z 2
yz − zx
yz − xy
1 x+ y −z
x ( x − y )( x − z )( y − z ) x 2 = yz
1
x ( x − y )( x − z ) x 2 x+z = yz −y 1
0
x+ y 1 −z 1
x x 1 =( x − y )( x − z )( y − z ) − + 2 yz − z x = ( x − y )( x − z )( y − z ) ( xz + yz + xy )
x+ y
1
1
0
x+ y
y−z
−z
y−z
As usual, the general treatment of determinants using row/column operations was variably received: those who could, did … and generally scored at least six of the eight marks available. Those who were less clear about the topic generally scored nothing or, at best, scrabbled around for a couple of method marks somewhere. Fortunately, the proportion of candidates in the former camp continues to grow year-by-year. A major bug-bear for the examiners is the decreasing inclination among candidates to give any hint as to what they are attempting to do with the rows and/or columns, leaving it to the markers to figure it out for themselves. In simple cases such as this, it isn’t too much of a problem, but in more difficult cases examiners may be unable to award any marks unless there is a clear method.
Question 6:
Exam report
6 a )i ) A direction vector of L is u = 2 −9 u=
36 + 4 + 81= 11
The direction cosines are
6 2 −9 , , 11 11 11
5 1 i j k 3 4 5 4 5 3 ii ) 3 × 6 = 5 3 4 = i− j+ k 1 2 1 6 4 2 1 6 2 6 2 = −18i − 6 j + 27k = −3u L is perpendicular to the plane Π. 9 6 p + 5q + r = b) 2 p + 3q + 6r = 8 −9 p + 4q + 2r = 75
l1 l2 l3
9 l1 6 p + 5q + r = 6l1 − l2 + 27 q 46 34 p= 21 p + 6q = 2l1 − l2 −57 27 q 46 68 p += 54q 92 34 p += This gives −19 −189 p − 54q = 513 7 p + 2 q = By adding the equations :
605 − 121 p = p = −5
then 7 × −5 + 2q = −19 and r =9 − 6 p − 5q =9 + 30 − 40
q =8
r = −1 c) By solving the equation simultaneously, we have 7 + 6t −2 + 5λ + µ 8 + 2t = 0 + 3λ + 6 µ 50 − 9t −25 + 4λ + 2 µ 9 −6t + 5λ + µ = 8 −2t + 3λ + 6 µ = 9t + 4λ + 2 µ = 75 This is equivalent to the system in part b) with t= − p, λ = q and µ = r so t =− p =5 7 + 6t 37 and r = 8 + 2t = 18 so P = ( 37,18, 5 ) 50 − 9t 5
There were several easy marks to be had in this question, and the rest was helpfully signposted. Part (a)(i) was routine, as was part (a)(ii) … or it would have been had not, again, some sort of explanation been required for full marks. Approaches were evenly split between using the scalar-product and the vector-product. In the former case, a lot of candidates seemed to think they only needed to show that the line’s direction vector was perpendicular to one vector parallel to the plane. In the second case, many candidates were content merely to find the vector product of the two vectors “in” (parallel to) the plane; others noted that it was a multiple of the d.v. of L rd without further comment. For the 3 mark on this part of the question, we wanted an explanation as to why their scalar- or vector-product work established the given outcome – it is not enough for candidates merely to repeat the result that we gave them. Given results always require more of candidates by way of ‘dotting the i’s and crossing the t’s’. The more discerning teachers and students of MFP4 will understand why part (b) was set up as it was, since this is the only way to force candidates not just to resort to calculatoroutput generated answers. In part (c), remarkably few got past equating the equations of L and Π; the majority of candidates had no real idea what to do and so there were lots of answers seen in which the point of intersection was taken to be (–5, 8, –1).
Question 7:
Exam report
a )i ) The eigenvalues are 27 and 1 4 −1 The eigenvectors are and 1 3
This question was handled very well indeed in its technical aspects – writing down eigenvalues and eigenvectors; finding a 2×2 inverse matrix; determining M, and the accompanying matrix multiplications. So a lot of candidates scored lots of marks on this question. However, there were very few candidates indeed who managed to score more than 12 or 13, due to the explanations that were being looked for. At its simplest level, this began in part (a)(i), where candidates were required to write down “the eigenvalues and corresponding eigenvectors …” – in other words, to say which went with which. Examiners did not accept a fortuitously ordered separate pair of each (with a 50-50 chance of a correct pairing). In part (a)(ii), we also wanted to know why they had chosen one of the two possible lines as a line of invariant points (as opposed to just an invariant line). Part (c) had been designed to require thought of the candidates, so the difficulties that arose here were not unexpected. There were two marks for explaining why the given expression was divisible by 13. The first was awarded to anyone who n noted that the 1-1 element of M is an integer; the second, subtler, one was for the explanation of why it had to be an integer. Disappointingly few candidates got both marks.
−1 ii ) The invariant line has direction vector for λ = 1 3 x−0 y−0 y = −3 x Cartesian equations : = 3 −1 b) det(U) = 12 + 1 = 13 1 3 1 13 −1 4 4 −1 27 0 108 −1 ×D U= × = 1 3 0 1 27 3 1 108 −1 3 1 1 325 104 −1 × = UDU= 13 27 3 −1 4 13 78 39 U −1 =
27 n n −1 = UD U and Dn 0 1 4 −1 27 n 0 3 = Mn × 13 1 3 0 1 −1 c) M n
25 8 M= 6 3
0 1 1 4
1 4 −1 3 × 27 n 27 n × 13 1 3 −1 4 1 1 so M n1,1 = (12 × 27 n + 1= ) ( 4 × 33n+1 + 1) 13 13 Since all the elements of M are integers, Mn =
all the elements of M n are also integers including M n1,1 this mean that 4 × 33n +1 + 1is a multiple of 13.
Question 8:
Exam report
If the enlargement has a scale factor k ,
This question was not a success for most candidates. There were a lot of blanks drawn amongst candidates, and a lot more “solutions” which consisted of little more than “hopeful manipulation” of numbers. It appeared that a large number of candidates had simply stumbled on the determinant of 576 and then pointed out that k must be its square root. Many candidates took the opportunity to play around with the (given) 24 in all sorts of ways, some of which only fortuitously resembled correct working. Full or coherent explanations were in very short supply.
the area scale factor is k 2 = det(M ) k2 =
12 16 −9 36
= 12 × 36 + 9 × 36 = 576
k = 576 = 24.
A line going through the origin is y = mx This line is invariant through the shear. 12 x + 16mx x ' x 1 24 M × = = 24 mx −9 x + 36mx mx ' 24 12x + 16mx −9 x + 36mx )= 24 24 2 12m + 16m =−9 + 36m
this gives m(
16m 2 − 24m − 9 = 0 (4m − 3) 2 = 0
Grade Mark
Max 75
m=
3 3 and l : y = x 4 4
A* 67
Grade boundaries A 60
B 52
C 44
D 36
E 29
