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Unformatted text preview: Milli!) y = 6x sin x MEBKS d .
5% = ex(smx + cosx)
d3 .
ﬁ=ex(smx+cosx+cosx—sinx)=2excosx C1
3
ngg = 2ex(cosx —sin x)
141 X .  x .
dx4=2e (cosx—smx—smxcosx)=4e smx=—4y C]
(b) Hence
a};  «14 dix d4 dix 2
dx8  cix4 ( dx“) —'dx4 (—4)!) _ _4dx4  (4) y  16y Cl
and the conjecture P(n) is that
d4“! _ n
dx4“  (—4) y. C2
4
(c) The statement P(n) : gig = (—4)ny is true for n = 1 from above. C1
41:
Assume P(k) is true, k 6 IN, i. e. 313% = (—4)ky. Then di‘iﬂ’xndi (1‘ka _ 24. k _ ktﬁx k
MM, — dx4 ( wk) — M r (4) y) — <4) dx4  <4) (—4)!) d4(k+l) {(+1
and so d~2x4(k+l) = (—4) y. Hence P(k) : P(k+l) and so by induction P(n) is true for all n E IN. R4 ..——_ ___—_—_——— ............................................................ (ii) The tank has dimensions L (length), w (width) and h (height) 0?) (a) If tan 9 < {1; the volume of water spilled is L X Area upper m’angle 2
=LX%w.wtan9=l—l§Eilg R3(AG) (b) If tan 9 > £ the volume of water spilled is L X Area trapezium =L>< {wh—.17h.hcot6}=L71={2w —hcot9]. R3,c2 2. L1:X2_l=YT—§=z—I:L2:3x =2 “3:2“ was Id (21) L13 X=l+2l L2 x=3_4p
V=3+3X y=§+§u
Z=l+2k z:..l+2u
andso
__) 1 2 .._) 3 ——4
l't=( 3]+7t( 3) r2=(3/2)+u[ 3/2) C1,C1
l 2 1 2 (b) For [I and L2 to intersect we must have ZA+4tL=21 BK git: —%; 2k—2u=—Z. Adding the first and the third equations gives l =% and hence it = % which do not satisfy the second equation. Hence the lines do not intersect. M1, R1 (AG)
The lines are not parallel as the dirctiOn vectors are not in the same direction. Cl
(c) A plane that is perpendicular to L3 has the equation 4x+%y +Zz=d and d can be chosen so that the plane contains (1, 3. 1). For example d = g and the planeisthen
8x—3y—42+5=0i MLAI 2 4 . 7’ I? 3 1
(d) 3]>< [3/2]: = {—12 =3£~4 MLAI 2 2 2 3 2 15 . s —4 3/2 2
—«) —9 —) —> (e) The distance is l(r1 — r2 ). nl where n isaunitvecror in the direction (1, — 4. 5). Hence it is —2 l
——————————l 3/2 . [—4 =—2 = L42 . M2,A1
2 2 5
(ii) Taking a coordinate system with origin at the beacon then the vecmr representing the path of the aeroplane is shown in the diagram below.
A vector in the direction of the ﬂight path is V: (0—6)i+(8—0)j+(5—5)k=—6i+8j R1.C1
. . . . . IPX Vi . ,
The distance from the origin to the line rs—l—V—r— where P is any pom:
ion the line. Taking P as 6i + 5k then
i j k
PX V: 6 0 5 =—40i——30j+48k MLAI
—6 8 O .andso IPX vi: «1402+ 302 + 482 = J4804. C2 Since IV I = V36 + 64 =10 then MAﬂS I P X V I 4804
t V l = 10 = 6.931.
Hence the closes: the plane comes to the beacon is 6931meues. . M1, A1 I p(X)= 1+x2. OSxSk.
0, x>k.
To be a probability density function ,
m k
jp(x)dx= lfx2=1=mk=1=k=m1=ua R1,C1
.00
Then,
:2lex 1 m1
p=d[l+xz dx : 7 loge (1 +x2) o  0.616 MLAI
and
, m1 1
2 x2 2  m 2
6 == 1+x2dx—.I.  (x—arctanx)0 —;,L = 0.17841. Thus 0' = 0.422. ' M2, A2 (ii) (a) The two component heater will operate if one or both of its components MARKS
work when the heater is switched on. Thus
P ( two component heater works )
= (l —q)3+ 2q(l ~q)
=lq3=(l—q)(l+q). Rl(AG)
The feur component heater will operate if two, three or four of its components
work when the heater is switched on. Thus
P ( four component heater works )
=(1— m4 + 4q(1q)3 + 6q2(1q)2
=(1—q)2t1—2q+q2+4q4q2+6q21
=(l—q)2{1+2q + 3q2}
=t—4q3 +3q4 R1,Cl
(b) The heaters are equally likely to operate when
l—q2 = l—4q3 + 3q4
or when
lq2 — l—4q3 + 3c;4 = 0,
i e
(3Q2 — 4q + 1)qz = 0. R2, Cl
But the heaters are equally likely to operate if q = 1 and so (1 — q) must be a
factor of the above polynomial and so it can be written as
(q — 1) (3q  m2 = 0.
Thusthe heaters are equally likely to work if q = 0% or 1. C2
(c) For% < q < l. l — q2 > 1 — 4q3 + 3q4 and so the two component heater i
is more reliable. For the remaining values, 0 < q < ~1 , the four component heater 3
is more reliable. 4 (i) (a). The curves of)! = v7: , and y = 6 — x intersect when
«[2 = 6—x =>x =4orx = 9.
The curves of y = v7: , and y = c interseCt when x = c2 and the curves of y = 6 —x and y = c intersect atx = 6 — c. Hence the region is C3. one mark
for each boundary (b) The required area is then 4
3th —c ldx + % (2c)(6—c~4)
C
4
3 3
; 3t n—cx 2 + %(2—c)2
c
2 . l
= ng—c’)—c(4c2)+ E tl—c)2
22 c2 c3
=_T6C+'2'+'3— M3,A3 (c) When 0 = 2 the line y = c goes thrOugh the point of intersection of the other
two curves and so the area is zero. Setting c = 2 in the above expression gives 7 .. _
246X2+%+§=44 72+12+16 =72 72_0 R3 3 3 6 6 ‘
(ii)(a) The area of sheet metal required for one can is equal to the surface area of
the can and the surface area is
A = area of ends of can plus area of rectangle used to form cylinder
= 2M3 + 27trh. C1 (b) Now the volume of the can is equal to area of base times the height and this is
500cm3. Thus rtrzh = 500 = h = §0_<2)
1tr
and substituting for h into A = 2m (r + h) gives
A = 2m{r + 599i = 2m2 + 1000. M2, A2
7cr2, r Now r must be is positive and S is a continuous function of r. It is seen that as
r—>0,A —>°o, andasr—aoo, A >°°. Hence A has no maximum value, any stationary point of A will be a minimum. R]
(c) Differentiate A with respect to r to give
9& = 41tr — 1000
dr 1'2
and setting this to zero gives
4an = 1000 => r= 31° . M2,A1
at:
This is the only Stationary point and so it musr be a minimum“. 7/3 R]
Then, since h = 50—3 , it follows that h = 500 (4102/3 = 3153— = 39 . C2 itr lOO‘lt 3
n \l 47: ** Alternatively, <23.  a i 2000
dr2 r3
2 X 41:
and when r: 10 this has the value 4n +—O—(i%O—O—— = 121: 3
V41: and since this is positive, the point is a minimum. 5(a) Given M be the set of2 X 2 matrices [1, A. B. C, D E} where I is the MM
identity matrix and the others are A=(?$)~B=(?J),C=("6‘Dv0=(‘iVD,E=(ii then by matrix multiplication the operation table is E
E
D
C
B C4, less 1 per
A error.
I
To be a group under matrix multiplication the operation has to be associative,
given, there has to be an identity element and each element must have an inverse. C3
Clearly I is the identity element and from the operation table the elements I, A, D,
C, B and E are the inverses of I, A, B, C, D and E respectively. R2
From the table it is clear that AB = C but BA = E and as C at E the group is not
abelian. R3
(b)There are six (3!) permutations of the numbers 1,2 and 3 and so S3 has six
elements. These are the given ones
1 2 3 1 2 3
P1= [123 P2: [231]
and the others are
Pa=[ii‘i P4=[i§3] Ps=[§§i PﬁHii] C4
The group table is pi p1 p3 p4 p5 'p5
P1 Pl P2 P3 P4 P5 P6
P2 P2 P3 P1 P5 P6 P4
P3 P3 P1 P2 P6 P4 P5
P4 P4 P6 P5 P1 P3 P2 (34.1555 1 per
P5 P5 P4 P6 P2 P1 P3 311'“
P6 P6 P5 P4 P3 P2 P1
(c) Two groups G, and G; are isomorphic if there is a one to one correspondence
f ; G1 —> (32. satisfying f(a + b) = f(a) 9 f(b) for all a, b 601 where = and 9
are the operations associated with 01 and 02 respectively. C3
From the two group tables above an isomorphism between the group in (a) and the
group in (b) is
p1—)l. p2—)B. p3—)D, p49A, p5——)E, p5—>C. C3 (Markers : See note at end of solution on next page) (d) Denoting the operation of composition by J and considering the two elements .  l 2 3 _ l 2 3
5“ [132...]a“d52‘ [321...]
in which each number after 3 is unchanged, we obtain sx=sz= [Hi :i:] with all other elemnts unchanged but
2 3 ...
t 2 . .. ] Hence st '~‘ 53 at 53 c 51 and so the group is not Abelian. t; In particular S 3 is not abelian and as S3 is isomorphic to the set M in (a) under
multiplication. that group is not abelian either. (c) From the group table in (a) it is seen that
AE=DandD—1 = B =EA = 13—1 A—l. This suggests that if x and y are elements of a group G then
(xyrl = fl K‘l
Consider the operation
0000"1 X‘1 ) = X( y y“) X*1
by associativity,
=xe x‘1 = x x‘1= and so the suggested result is true for all groups. Note
It is possible that some candidates will have a group table that is the mirror image of the one given, mirrored about the main diagonal, as a result of taking the
compositions in the reverse order. For example, in the above table p2p4 is 134 ﬁrst then p2, to give p6 . In the other order the result would be p5. Award the marks
provided the same approach is consistently applied. MARK§ C1 C1 R2 R2 C2 R2 R4, split at
discretion 6 (i) Starting at E. though any junction could be the starting point. use the edge M.__ARK§
EF, the one of minimum length. Then take the edge of minimum length from E or ' F. other than EF. This is EH (or it could be FC). Then take HI, the edge of
minimum length from E, F or H. In this way we obtain the minimum spanning tree EF, EH, HI, FC, HG, CB, BD, DA M5, C5 split
with a total length of 1+ 3 + 2 + 3 + 4 + 4 + 3 + 2 = 22. at discretion C1
0
} C1
0
There are thus two ways of getting from v1 to V4 using two edges. R1
Squaring again
gives
2 0 0 2 2 0 O 2 8 O 0 8
~[333] [333] [3333] ex
2 O 0 2 2 0 0 2 8 0 0 8
and there are thus eight ways of getting from v1 to V4 using four edges. R1
HenCe the tatal is ten ways. C1
(b) If the adjacency matrix A is raised to the power k then the number afj) in the
matrix Ak is the number of ways of getting from the vertex v, to the vertex v,
using exactly k edges. This result can be used to find the shortest path from the
vertex v; to the vertex Vj by successively calculating the matrices A, A2, ' R4, split at A3, ..... , until there is a non zero number in the (i, j)th place in the mam‘x. discretion. (iii) Since the matrix is 5 X 6. there are 5 vertices and 6 edges. The element bi, MARKS
in B is 1 if the edge ej is incident with the vertex v; and zero if nor. Hence the
graph is of the form C4 C6. Split at
discretion (iv) There is an Eulerian path since there are exactly two vertices with odd degree. namely those with the totals 5 and i. The rest are even. C3 There is no Hamiltonian circuit since there is only one edge from V3, so a circuit cannot continue beyond there. C3 The sum of the column totals is 42, and this is twice the number of edges. Hence the number is 21. _ R2 Then, there are l0 columns corresponding to 10 vertices and from Eulers formula, if the graph is planar, V—E +F=23 lO21+F=2=>F=13. R2 7(a) Model : X1 is Poisson with mean 5. LARKS (i) Prtx. =0) = 5‘5 =0.0()67 (:2
(ii) Pr(X1 > 10): Pr(X12 ll)= 0.0l37 C2
The total weekly amount in sales is 31X] where a] is 200. .
The mean is alE[X1] = 200x 5:1000 c2
7
The variance is a'l' Var [X1]: (200)? X 5 = 200,000 C2
For both models the mean is at E[X‘] + a2 Elle = (200 X 5) + (250 X 3)
= 1750 C2
7 .
and the variance is a? Var [xx] + a5 Var [x2]: 200,000 + 187,500
= 387.500 C1
giving a standard deviation of 622.49. All results in dollars. Cl
(b) Model : X A is NQLA , 6A2}
.. 6A2
2» XA isN(uA,—— ). C3
"A
Now SEA = 57.5 is a reading of YA and 942 is unknown estimate using
sA = 3.7. Then
t ___ xA ” “A
SA / \l “A
is a reading of t with nA — I = 24 degrees of freedom. C3 The 95% conﬁdence interval for HA is SA X t tongs (14 d.f.) A
V "A and for the given values this is 3.7 57.5 i 2.06 X —~
‘1 25
= 57.5 i 1.52 and so the conﬁdence interval is (55.98, 59.02). C4
Since the 11A value of 60 is not included in this 95% interval the assumption of a
mean service time of 60 minutes is not accepted. C2
Note to markers: Candidates may have been taught to use the unbiased estimat for the variance, namely (3.7)2 X g . If this is used the interval becomes (55.94, 59.06) and full marks should be awarded. (c) Assume that 2A is NmA , A )and that is is NmB . —B: ) C2
n
i A B
Assume further a common variance 62.
.. _ _ 62 63
Then XA — X315 NmA uB,n— + ) C2
A “3
Estimate the unknown 02 by the pooled van'ance escimate MARKS
2 2
52: (nA — USA + (nB — 1)sB C"
n + n — 2 A B
The 95% conﬁdence interval for Li  “B is i». — RE i to_025(nA + rh2)Xs and for the given data this becomes 57.5 —‘61.71 20st \H + ‘16 7 = —4.2: 20st 1‘36
where a 1 52 = 24 X (3.7)‘ 4; 19 X (3.1)* = 11.887
and so 3 = 3.45. M3, A2
Hence the interval is ‘ 54.2 i 2.02 X 3.45 x % = 4252.09 or (—6.29, —2.11). C2 The postulated value of “A  “B = 0 does not lie in this 95% interval and so the hypothesis that [a = “B is rejected at the 5% signiﬁcance level. C2 0° S(i)(a) Given the series Zn“ . let {(1) be such that f(n) = Lil for all n Elli .
n=l If ﬁx ) is positive. continuous and decreasing for x 2 1 then the series converges
00 if [fan dx is ﬁnite and diverges otherwise. Taking the function to be
x f ::
(I) x2+l then clearly it is positive forx 2 1, it is continuous and 0° 0° I
x 1 2x 1 limit 2x
{[12+1dx _ ilfoildx it—wo 12+1dx
l
1 limit I 1 limit 1
= i [goo logc(t2+ l) l =  m logt  510%.? The natural logarithm tends to inﬁnity asx tends to inﬁnity and so the mt teml
does not exist. Hence the series diverges. (b) The series 2 LP converges if p > 1 and diverges if 0 < p S 1.
n=1 n 00 Setting p =% yields the series 2 ‘71.; which diverges. on 00
B com arison Z < Z —1
y p a 7+ \f n=1" n=l n
. 1
Since O< _<— nelN,
. 2 + *l n ‘5
W
1 . . .
but even though 2 — diverges this tells us nothing about the series.
n=l ‘5
OO
2' I _. However setting p=1. and comparing 2 _ with
n=12+~In n_122+~Jn
co
1 . . 1 1
 t tht2+ <n=> >forn>4.
“:1, n 1 IS seen a ﬁ 2 + \Iﬁ n
00 W
1
Hence, 2 _ > Zoo ~
“:1 2 + ‘jn “:1 n
. 1 . l
and as the series  diverges, so does ..
“:1 n “=1 2 + \In Alternatively, the comparison test with p = 1/2 can be used by considering
WT: < 2/(2+\/Ti)forn>4. MARKS C2 M2,A1 C1 M2,C2 R4, split at
discretion (ii)(a) The first two terms ofthe Taylor series for f(x) about a point .tn are MARKS
fix) = f(In) + ix “ In) F0511) C2
and if we denote this by p(:c) we obtain
PC!) = ﬁxn) + (X " In) f“(/tn) = 0 If xn+1 is a real number such that p(1n+t) = 0 then
ﬁx“) + (X n+l “ In) f’(Xn) = O and solving for x n+1 yields ﬂirt). t'txn)‘ Given a value x0. the above expression generates the sequence x0, x1, x2. ..... . This is the Newmn Raphson iterative method for approximating a root of ﬁx) . R2, C2 xn+l=xn ‘ (b) Applying the above method to the equation x2 — c = 0 gives X  C 1 C
_ ._“__.__.   __
xn+l xn " an " 2 {In + xn} . C3
Then, adding or subtracting NF: from each side gives
1 c 1 2
Xn+i iv; = §{Xn + xn}i‘lE =‘2_x;{xni'2‘]EXn +0}
= 7;; (xn i m2 C3
Taking the ratio of the above expressions gives
Xn+!—‘J‘E=Xn‘jzz C2
Xn+1 + \IC Xn + '\/E
and similarly
—  2
“Ln—Li; = _ﬂ;L___‘j_‘i C2
xn + \lc xn1+ \lc ‘
and so putting these together yields
  2 2
x9” — xii: xni  \lg C2
an + \10 xn_1 + ~1c
Repeating this process eventually gives
Xn+1 — ‘12 =VX“ "" "16 2n+1 c3
Xn+1 + \I c xo + «I c ‘
Then, ifxo > 0, the ratio Fifi—E < 1 and so as n increases the right hand
x0 6
side of the above tends to zero. Hence the left hand side tends to zero and so
R3 1n+l "’ J5. ...
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 Winter '10
 Chang
 Math, Calculus

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