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Unformatted text preview: 1. (a) (b) (i) (ii) (iii) N97/5 10/H(2) Let =‘p3 +p2 5p—2, then = 23 +22 10—2 = 0 (M1)(C1)
Therefore p = 2 is a solution of f ( p) = 0. (C1)
Now f(p)=(p—2)(p2 +3p+1)=0
(Students may use other approaches, use your discretion)
The rest of the solutions can be found by using the quadratic formula
onp2+3p+1=0 (MI)
Hence these solutions are p = _3 :‘5 (A1)
Arithmetic sequence: 1, 1+ p, 1+ 2p, 1+ 3p (A1)
Geometric sequence: 1, p, p2 , p3 (A1)
(1+2p)+(1+3p)=p2+p3:>p3+pZ—5p—2=0 (C2)
Therefore, from part (i), p = 2, p = 3 (C1)
(a) The sum to inﬁnity of a geometric series exists iff lpl <1, (C1)
Hence, p = 3 is the only such number. MD
(b) The sum to inﬁnity of the geometric series is
_“_ =___1__ (c2)
— r 1_ J3 — 3
2
. 2 l 1
Which converges to 5 _ J5 = '2 + 1043 (A1)
(0) The sum of the ﬁrst 20 terms of the arithmetic series can be
found by applying the sum formula
S20 =10(2a+19d)=10(2+19p) (C1)
So, 520 = t0[2 +19[‘/—§§'—3D = 265 + 95J§ . (A1) N97/5 10/H(2) 3 2
(i) (a) The vectors 71 = —2 , and 72 = 1 are parallel to L and L2
4 —1
re ectivel .
5P y (CI) Also, if we let :7 = —?.z7 +11] + 7.1;, then u'll=6—22+28=0andu12=4+ll—7=0 (CI) Therefore the required vector is perpendicular to both lines. ( C1) (b) L and L2 have the following parametric equations: L,:x =3t+2, y=1—2t, z=4t; and L,:x=2u2, y=a+1, z=2—u (CI) For an intersection to happen, the following system must be consistent
3t+2=2u—2 (1) (R1)
1  21‘ = u + l (2)
4t = 2—u (3) “311 ﬁ'om (2) and (3) we have t = 1. Hence u = 2. However these values (A1) do not satisfy (1) since 5 ¢ 6. Therefore the lines cannot meet. (R1)
 (0) Since :7 = —2? +1 1f+ 71?, is perpendicular to both L1 and L2, and since the plane contains one of the lines and is parallel to the other, therefore 17 is normal to the plane. The equation of this plane is of the form (R1)
—2x+11y+7z= d. The point (2, l, 0) lies on L, and on P, therefore (R1)
—4+11+0=d,henced=7andtheequationis (A1)
—2x+11y+7z= 7. (ii) (a) The required system is
—l k +1 —1 x O 1 1 k 2 y = k + 2
2 2 'k z kzZk—S (C2) ('1 f°r
each error) (b) det(M)=02>2(k—2)—k—(k+l)(k—2(k2))0=0 =k2—2k—8=0 (M1)
:>k=4ork=—2 (A1)
(A1) a ‘v«v a ~ — 2:Jt‘"Tlf'sﬁfv‘vugﬁﬁvqum warms; , \‘ 1‘. er" N97/5 l 0/H(2) (c) (i) Using Gaussian elimination on the following matrix will lead us to
the solution: —1 k+l 1 0
1 1 k—2 k+2
2 2 k kzZkS «1)
‘ with 2 elementary row operations, the system will be
1 k +l l O
0 k + 2 k — 3 k + 2
2 —
0 0 —k+4 k 4k 12 (CI)
Therefore the value of 2 can be read ﬁom the last row
z_E—M—u MD
(6 4—k
(ii) When k = 4, the value of z is undefined, and hence the system has
no solution. (R1)
When k = —2, the system is a homogenous system, and hence it in
; admits at least one solution. It is reduced to x + y = 0 (R1) z = 0 1%
Hence it has an inﬁnite number of solutions. (A1)
For all other values of k, the system has a unique solution. (A1) (9 ,. ‘ F VWV—vwmr v—g—nzgvgn—vt “hang. ",Vr; ~~ m v_L,1<r~_~;»4r,v~ v  » ‘A.‘Y\:." {73 “ 'JV—vwwu ~r '.".v.,\ n u“ . .\ w Us», n » arrva'ut 'r' .rwvzm n.~.~'~f\«~ 4. ~y w, ..~=‘~]~3an'_’“ N97/5 lO/H(2) 4. (i) (a) y = 2xsinx + cost fay —25ian+4xeost25in2x=4xc052x ' WZXAG)
. 1:/2 2 x/2 2
00 Required volume V = 1: Jo cos x) dx — nj; (x cos x) dx (CI)
1:]2
= ifs/2111+ cost) dx = E] (x + xcost) dx
2 o 2 0 (CI)
«l2
and now using integration by parts, or otherwise, on Jox COS 2x dx will
yield
1:/2
V = g(%x2 +éxsin2x + 1cost]
4 ° (MINA!)
2
=£[“__l_l]=1(nz _4)
2 8 4 4 16 (A1)(AG')
(ii) (a) g(x)=x+l—e‘=>g’(x)=l—e‘:Owhenx=0, also, g’(x) < 0 when x > 0 and g’(x) > 0 when x < 0; (MIX/ﬂ) g(x) will increase for x < 0 and decrease for xi> 0, hence the
function has a maximum value at x = 0, g(0) = 0.
.‘.g(x)SOVxeR,=>x+l—e‘$0,=>x+13e‘ (10%“)
(R1) Cb)(i) From the previous result, we have 6‘ 1 2 x, and since it is given
that x > O, and we know that ex > 0 dividing both sides of this inequality by xez" will yield
cx _ 1 1 (R1) ,x > 2: , that is f(x)>e'2".
xe e CI)
6“ 1 e“ l—e“ (
(‘9 f(x)=T':= x
xe e xe (R1) . (iii) also from (ii)(a) and since g(x) _<_ 0 for all x ER, and since
replacing x by —x will result in reﬂecting the function about the
y—axis, therefore its maximum will not change, and hence g(—x) S 0.
(R2) g(x) = —x + l  e“ S 0 => 1  e" s x, and dividing both sides of this inequality by the positive expression JceJr will yield f (x) < e“ .
. (M) l
(iv) The two results above give e’“ s f (x) s e" , and since
lim e‘z" =1, and lim e“ =1
Xeo’ 390’ f
then as x approaches zero from the right 1 g f (x) s l , therefore (C1)
lim f (x) = l x—>O* "(Room ..e«.~.~.~ (mg—v u; vn vA~“." —‘  " I' ' ‘I r ‘ 3 1 humm.” A w;~;~~tetramfg:vaﬁ'sjlharsrreﬁ‘lr"q'wj§wW_ N97/510/H(2) 5. (i) (a) To prove that a set is a group under an operation, we need to prove
that: (i) The set is closed under multiplication: b
consider any two elements x = [a d] and y = [ e f) in S.
0 (C1) Now x.y=(ae+bg af+bh} ce + dg cf + dh Also (CI)(A1)
(ae + bg)(cf + dh) — (ce+ dg)(af + bh) = aecf + aea'h + bgq“ + bgdh  aecf  bceh — adgf — bgdh = ad(eh—gf)+ bc(g’—eh) = (eh —g‘)(ad —bc) = 1. (M1)
Therefore xy ES. (CI)
4% (ii) The operation is associative. This is true for multiplication of
matrices, hence true for this case. (R1) (CI)
1 0
(iii) There is an identity element. Since 0 1 is such that
ad — bc = 1, hence it belongs to S.
(R1)(C1)
(iv) Every element has an inverse. Since for every x E 5,
ad  bc = det(x) ¢ 0, therefore it is invertible. (R1) ( CI)
(b)(i) To prove that G is a subgroup of S, it is enough to prove that it is
nonempty, closed, and that it contains the inverse of each of its
elements. (R1)
1 0 . . . . Clearly 0 1 e G, hence the ﬁrst condition 13 satisﬁed.
" v b (CI)
The elements of G are of the form [:2 a) , also,
[ ‘1: 3: f]:( 221); af+gfj eG, and the second
— a — e  — a ae —
_ _1 (R2)
a b a —b
condition is satisﬁed. Also _b a = b a 5 G: (R I” C 1)
Hence G is a subgroup of S.
. 0 —l —l O l 0
(i1) k=[1 o],andsincek2=[ o 1J,andk“=[o I),
therefore the order of k is 4. (M2)
(A1) 10 ...—‘,..._V.«_..._.!...V“,I _ ‘,3_,$‘_.‘._,‘.§...“eta. ,7 . ._. .7. .,. .,,. . a" .l y. l .a . 0., .mwr. .. _...'. ‘ NV... .. _. ., .  .t we a.“ . T. ~_..—. 7. yr.  « gv— « "."vge.;v‘_ ...1 (iii) The elements of H are k, k2, and k‘ which were found in part (b) 0 1
above. The fourth element is k3 =[ 1 0). (iv) The multiplication table for H is (c) (i) By setting up a multiplication table, we can verify that the set is a
group. The set is closed under the operation since the product of any two
elements is another element of the set. Also there is an identity
element 1, and every element has an inverse since 1 appears in
every row once. The operation is associative. Since multiplication of complex numbers is a commutative
operation, the group is Abelian. (ii) Two groups G and H are isomorphic if there is a one to one
correspondence ﬁ'orn G to H which preServe the group operation. The correspondence, 1<—>k4, —1<—>k2,i<—>k3,i <> kis sucha 11 correspondence.
Hence H and K are isomorphic. Since the two groups are isomorphic, and K is Abelian, then H has
to be Abelian. (ii) A cyclic group is a group G with at least one element a, such that
every element b e G can be expressed as b = ai for some i < n,
and a" = e where n is the order of the group. Clearly a° =1, and a generates the group elements. Also an“ _— e2"i =1, hence the order is n + 1. (iii) For the group to be Abelian, ab = ba for all a and b in G.
NOW, (ab)2 = (ab)(ab), and (ab)2 = azbz, therefore, abab : aabb, multiplying on the left by a“ and on the right by b“ will yield
ab = ba. Hence, the group is Abelian. 11 N97/510/H(2) (C2) (C2)
(1 for each
error) (R3) (C1) (C1) (C3) (R1) . (C2)
(R2) (CI) (M2)
(R1) . ~v~g~.cvwrw 5‘99???“ at: 6. (i) (a) Any layout is acceptable. 1%;
4/4: (b) An isomorphism is a 1—l correspondence so that images of any two (0) adjacent vertices are also adjacent and conversely. By looking at the
adjacency matrix of G, we observe that F has an out degree of l and so does X in H, we also observe that B has an out degree of 4 and so does Z. Therefore these should be images of each other in the required
isomorphism. We also see that C has an out degree of 3 and so does U
in H. D has a 2 out and 3 in degrees, so does W. By similar reasoning,
we have A and Vcorrespond as well as E and Y. Hence the required
isomorphism is: A<—>V,B<—>Z,C<)U,D<—>W,E<—>Y, andFHX. For a simple connected graph G, with at least three edges, to be planar,
then according to Euler’s formula e s 3v — 6 but 14 >18 — 6, therefore G cannot be planar. (ii) (a) Prim’s algorithm requires that we start‘at any vertex at random and consider it as a tree, then look for the shortest path that joins a vertex
on this tree to any of the remaining vertices and add it, then repeat the last step until all the vertices are on the tree.
Starting at F, (other forms can be used) 1. The shortest path ﬁ'om F to ABCDEGH is to E, add FE => weight
added = 90 2. The shortest path from FE to ABCDGH is to G, add F G with
weight 120, 3. The shortest path ﬁom GFE to ABCDH is to H, add GH with
weight 220, 4. The shortest path ﬁom GFEH to ABCD is to C, add F C with
weight 230, 5. (ABD) Add CB with weight 240,
6. (AD) Add BA with weight 180,
7. (D) Add AD with weight 200. Therefore a minimal tree has a weight of 1280. 12 N97/5 1 0/H(2) (C4) (1 for each
error) (C1) (R1) (R2) T (C1) (R1)(C1)
(R1)(CI) (R5)
(1 for each
error) fl"! a. ',1\"<("V'1“q" v»v~,~. mq_—.,.., N97/5 10/H(2) 6R, A
(b) To ﬁnd the minimum path ﬁ'om A to H using Dijksu'a’s algorithm we (A1)
assign a weight of zero to the ﬁrst vertex and a weight of co to all
vertices that are not in the path. Then we start adding vertices adjacent
to the present ones that have a minimum weight. At each iteration, the
length of the path to a given vertex is evaluated and the minimum is
assi ed.
g“ (R2)
A (0) AB = 180, => B (130)
AD = 200, z: D (200)
AC = 270, => 0 (270) ADF = 480 is the minimum path to F, a F (480)
ADE = 480 is the minimum path to E, => E (480) (R5) (1 for ACG = 620 is the minimum path to G, => G (620) eaCh
error)
ADFG = 600 is the'minimum to G, = New G (600) @
ADEH = 840 is the minimum to H, :9 H (840) i '
ADFGH = 820 is the minimum to H, => NEWH (820).
This is a minimum path ﬁ'om A to H.
(A1) (iii) (a)
A » B C D (b) There is no such walk since the graph has 4 vertices with odd degree. Therefore there is no Euler’s cycle nor a path. (RI)
' (C1) If the door between A and B is locked then there will be a Euler path through the whole graph. There will be only two vertices, C and F with (121) an odd degree.
13 .w ,_ . an. . .. h... N, ., a”: . x FM”: "A. ~. .w i.“ ....< . .w . w. , I~iq'\'7§5\')‘31"¢"¥w wvwwﬂf? N97/5 l Oil1(2) Therefore our walk should start at one end at the other. Such a walk (RI)
could be the following:
FBFEFDBEDCAFACF C. T (A2) (G) Since the degree of each vertex is larger than 11 = 3 , there is a . . . . . . 2 (R1)
Harmltoman c1rcmt. Such a c1rcu1t could be FABEDCF. (A1) 14 ‘ ~VJ~MKMW...¥._V,W...\.Vuh,v.07,” . ... ,. r . _. .. «  w r M w. .. V’ r. .. 7 , . _ .. ..ﬁ ».:«,y..;i.3.\.‘$.¢wnmﬂl 7. (a) The conﬁdence interval gives a range of values which has a 95%
probability of containing the mean of the population. 0’ 0‘
That is the interval 1? “196—, I? — 196—] if calculated r eatedl ,
( J? J; ep y
then 95% of such intervals will contain the population mean. 1.96 is the number of standard deviations away from the mean and is read from the
standard normal table. According to the scheme above, the conﬁdence intervals are: Male: 17411.96=(172.824, 175.176) Female: 173 i 1.96 = (170.844, 175.156) Not enough evidence of any difference between the performance of males
and females since each of the confidence intervals contain the mean of the
opposite sample. (b) Finding the conﬁdence interval for the difference between the means is similar to the process above, however the standard error is different. The
conﬁdence interval is 2 2
(ﬂu‘ﬂr)=EM‘iFi—196 i+i£ nM n;
2 2
=174—173~_~196.’1—2—+l = (1.456, 3.456)
400 100 The result here conﬁrms our previous ﬁndings, since this interval contains
zero. Which indicates that the situation where the two means are equal is
also a member of the conﬁdence interval. 1. Form the null hypothesis: How” ~,u, = 0
The alternate hypothesis: H a : y M — M; > O 2. The test statistic is
_ 5M —EF —(pM —y,) = 174—173—0 z,_ =o.79s
5;, s; (144+121
__+_. —" —
M 400 100 "M 3. Critical value for Zc for a onetail test is 1.645.
Since Z. = 0.798 < 26 = 1.645  it is nonrejection region  we do
not have enough evidence that males out perform females on
mathematics exams. 15 971f‘, "rawaw [gnu \f‘ﬂf"ﬂ"?‘f7"""“?1av<ﬁ~ ._...y— 1...... .i , 4 "a. e .4.V......,~.4.‘.».\... v.,...»..,i...., .7 iv N97/5 10/H(2) (R2) ( C1)(A1) (CD041) (R2) (RI)(C1) (C1)(A1) (R1)( C1) (C2) (C2)(AI) (C1)
(R1) (A1) ' ‘ 'x”"'“5""~" 7“" "WT"W 3W???“ (d) This is a matchedpairs difference test since the samples are not (6) M. \...»...:.~...T_,.;, ‘1“. “mm”, 1 . independent. We study the differences between pairs and we test the
differences with a tdistribution since the sample size is small and we do
not know the standard deviation of the population. ' The number of degrees of freedom is 9. The differences are: l, 2,—2,—1, 0,2,—1,1,1,—2. Hence the average
difference is E = —03 and standard deviation sd = 1.494. 1 Hozd = 0
H1217 e o
2. Test statistic T = L031 = —0.635
1.494/Jﬁ 3. The critical value with 9 d.f. is £262.
Since —2.262 < T < 2.262 , there is not enough evidence of any
difference in speed between the two groups. This is a goodness of fit test. If the distribution were normal, then the
expected frequencies should not be very different from the observed ones.
Using the normal distribution with mean 174 and standard deviation 12, the probabilities corresponding to the different classes are 0.0228, 0.0989,
0.248, 0.322, 0.217 and 0.0912 respectively. Therefore the corresponding expected frequencies are 9, 40, 99, 129, 87 and 36 respectively. H 0: distribution is normal with mean 174 and standard deviation 12.
H1: distribution is not normal with mean 174 and standard deviation 12. 2
2 (fa—f.) 1 o 9 81 81 16
= ——_=_.+._+_.+__+.__+_,
N°W 1‘ 2 f, 9 4o 99 129 87 36 that is z} = 221. But Z3 =11.07 (5 d.f., 5% upper tail). And since the test statistic does not lie in the rejection region, we do not have enough evidence to reject the hypothesis that the distribution is normal with mean 174 and standard
deviation 12. Students using the continuity correction will get:
12 = 2.75.
Conclusion does not change. 16 N97/5 lO/H(2) (R1)
(R1)
(C1) (A2) (C1) (C1) ( C1)
(Al) (R1) (R1)
(A1) ' (RI) (C2) (A1) (R2) (i) (a) The ratio test is appropriate here. The ratio of consecutive terms in the 36 ,. 36
series is r(n) = $11 = M. = ’ and an "3641114 4 n
hence lim r(n) = i x 1 = i < 1 . Thus the series converges by the ratio test. (b) The ratio of consecutive terms in the series is 31 " !
Whig: (3n+ )“1100 n
a,l (3n)!100 (n+1)!
1 1
_ﬁxmx(3n+3)(3n+2)(3n+l)
=I3_Ox(3n+2)(sn+1) and as n > so, r(n) —> co, and so the series diverges. (ii) (a) Using the trapezium rule, the area under the curve is‘approximated by b — a
L = 2n [f(xo) + f(x,) + 2{f(xx )+.. .+f(xu1 M
1 0100+O.199+0296+0297+0.479+
' ‘[0 + 0'8“ + 2{ 0565 + 0.644 + 0.717 + 0.783 7;, = 0.450 2
(b) The volume in question is V = 1t£[ f (x)] dx , hence the function values in the table have to be squared. b—a
Vz{ 3n }[y0 +4y1+2y2+...+4yn_l+yn]x 1t 02 +4x 01002 +2 x 01992 +4 x 02962 +
lief—0 2x02972 +4xo.4792 +2x05652 +
4x 0.6442 + 2 x 0.7172 +4 x 0.7832 + 0.8412 V 2: 02681: = 0.843 cubic units. 17 .. .VJ‘HWr i7 ..:.. ,V—‘n ._..—‘_~.., q~1g':~hv ‘I— r;......~.w,——m»_ , 3..., , v. .7 .,...\r.,‘ w “A”.  w . «a»: ..V. “NW«w . ~f‘wr‘1v_".‘v“"1“ﬂ"fWlﬁ1'w’S'FW’53‘" N97/510/H(2) (C1)
(M2)(A1) (R1)(A1) ‘ (CI) J (M2) (A1) (R1)(C1) (R2) 012) (A1) (R2) (C2) (M2) (A1) N97/5 I 0/H(2) (iii) (a) The Maclaurin series for a function f (x) is x2 xn xn+l
x= 0+ ’0+— "O+...+— (“0+ ("we
f() f()xf() 2!f() nlf()(n+1)!f () (R2)
where 0 < < x . Thus,
.763 XS
sinx=0+x+0xx2———+0xx" +—cosc, O<c<x.
3! 5! (C2)
2 ’53
(b) In the equationx = smx , replace smx byx ~33 to get (R1)
3
sinx z x—% ,or x3 + 6x2 —6x = 0, yields the non zero roots as
“L... «WP—Mg (CZ)
Obviously x = —3 — Jl‘S— cannot approximate x2 = sinx , (CI)
thus x0 = —3 + Jig is the only approximate solution. (AG)
Now
lsinx —le= x §+£gCOSC—X2 (C1)
0 0 0 0
x3 x5
= x0 mfg—x3 +5—zcos
x5
= 3—00st (C1)
5!
where 0 < c < Jig — 3. Since {cosxl $1 for all x , it follows that
s
5 J3 —3 s
lmﬂzis xo <(__l<£<_1_ (x2)
5! 5! 120 200
x5
Since sinx0 —x3 = geese, where 0 <c < 5—3, and inthis interval
the cosine function is positive, it follows that Sin Io ‘xg is positive. (122)
 2
smx x . .
(c) I1=xo&;roi_2—x°o, andbysettmgx0=—3+s/l_5_weobtam (RI)
2:, =O.8767. (A1) 18 ...
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 Winter '10
 Chang
 Math, Calculus

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