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Unformatted text preview: 26 pages INTERNATIONAL BACCALAUREATE
BACCALAUREAT INTERNATIONAL 
BACHILLERATO INTERNACIONAL MARKSCHEME May 1999 MATHEMATICS
Higher Level Paper 2 M99/510/H(2)M — 6  M99/510/H(2)M (3) Since the coordinates of the points P, Q and R are (4,],—l), (3,3,5) and (1,0, 25'), —> —;
respectively, the vectors QR and PR are given by QR = —23 —3} H204); (MIXAI)
ER = —3E — j + (2c+ 1):: (MINA!)
QAR is perpendicular to RR if and only if QqR RR = 0 La. 6 + 3 + (2c — 5) (2c +1)= 0 (M1) => 4028c+4=0 => (61)2 =0 (M1) => c =1 (A1) (b) PR = —3F —}+3i£, PS: —3F+ 31? (MI)(M1)
f I? PSx PR = —3 0 3 (M1)
—3 —1 3 = 35+ 3:? (A1) (c) The parametric equation of a line I which passes through the point (3,3,5) and is _,
parallel to the vector PR is given by F = (3i+3} +5E)+r(—3i —] +311) (MI)(M1)
=3(1—t)F+(3—z)j+(5+3t)12 (—oo<:<oo) (A1) Note: If —oo < t < no is not mentioned, do not penalise. Also note that some candidates may give the parametric equation of the line in the form
x=3(1—t),y=3—t,z=(5+3t),—oo<t<oo continued... — 7 — M99/S 10/H(2)M Question 1 continued (d) Let P] and P2 be points on the line 1 corresponding to (=0 and I =1, respectively.
Hence, for F = xi +y:1z +211,
x=3(l—t), y=(3—t) and z=5+3t.
Putting 1:0 and (=1, we get the coordinates of points P1 and P2 as (3,3,5) and
(O, 2, 8) , respectively. Vectors SP1 and SP2 are given by SP! = 2? +2} + 3:? and $32 = F + f +61? . (M1)
2" f I?
A vector perpendicular to both SP1 and SP2 is SP, ><SP2 = 2 2 3 (M1)
—1 1 6
=9f—15}+4E (AI)
Let T(x,y, 2) be any point ofthe plane 7:. Since S =(1,1,2),
§r=(x—1)F+(y1)}+(z—2)E is avector in 2:. Hence TSﬁ =0
tie. 9(x—1)—15(y—1)+4(z—2) =0 => 9x—15y+4z—2 :0 (A1)
OR
6s: (1 —3)f +(1—3)} +(2—5)E (M1)
= —2i —2}—3IE (M1)
The equation of the plane containing the line I and passing through the point S is
determined by I and the vector SQ. Hence, the equation is:
3 —3 2
F = 3 +x‘L —1 +11 ——2
5 3 ‘3 (M11041)
—1 9
__, 2 —15
d PQ Fr 6 4
(e) Shortest istance =——.—_.—._
151 J322 (MIMMI)
— 1—5 (MINA!
322 )
OR
The distance of P from It is:
9(4)— 15(1)+4(—1)—2
J92 +152 +42 (M1)(A1)
_ 15
322 (MINA!) Note: Accept 0.836 (3 s.f.) (i) (ii) (iii) — 8 —— M99/510/H(2)M
Let the arithmetic sequence be written as a , a + a’ , a + 2d , . ..
Then a + 4d  i
a+11d _ 13 (M1)
Sol3a+52d=6a+66d=>7a=l4d (M1)
=> 0 = 2d. (A1) Since each term is positive, both a and d are positive. We are given a(a+2d) = 32,
setting a = 2d, we get 2d(2d +2d) = M2 = 32.
=> d = :2. Hence, d = 2 and a = 4 and sum to 100 terms of this sequence is 13—0{(2)(4)+(100—1)2}_ =10300 (a) Since a) is a complex number which satisﬁes (03 — 1 = 0, a) at 1. Hence, (b) (wx+a)2y)(a)2x+coy) =w3x2 +codyx+w2xy+wsy2.
Using 603 =1 and 1:04 =0), we get,
(cox +m2y)(co2x+ my) = (x2 +y2)+(co2 +co)xy
=3:2 +y2 —xy, (Since l+cu+(1)2 =0) Let she) be the statement; 22" — 3n — 1 is divisible by 9.
Since 22 —3—1 = 0, 5(1) is true. Assume as the induction hypothesis S(k) tie. 22" —3k —1 is divisible by 9. We shall show that S(k +1) is true.
so: +1) = 220‘“) —3(k + 1) —1
=4(22")—3k—4
=49?" —3k—1)+9k By the induction hypothesis 2“  3k 1 is divisible by 9. Since 9k is also divisible by
9, S(k +1) is true.
Thus, by mathematical induction S(n) is true, 1': = 1, 2, (M1) (A1) (M1) (A1) (MIX/11) (M1)
(M1)
(M1) (A1) (C1)
(C1)
(M1) (M1)
(A1) (M1)
(R1) — 9 — M99f5 10/11(2)M (i) Let D be the event that the patient has the disease and S be the event that the new blood
test shows that the patient has the disease. Let D' be the complement of D, tie. the
patient does not have the disease. Now the given probabilities can be written as p(s D) = 0.99, p(D) = 0.0001,p(s D’) = 0.05. (A1)(AI)(A1) Since the blood test shows that the patient has the disease, we are required to ﬁnd p[D S
By Bayes’ theorem, p(S D)p(D)
p D S = ————,—, (M1)
( ' ) pol 0mm + an 0 )p(0)
(0.99)(0.0001)
= m (M1)
(0.99)(0.0001) +(0.05)(1— 0.0001)
=0.001976...= 0.00198 (3 s.f.) (A1)
OR 0.99 3
0.0001 D
0.01 S.
0.05 5
0.9999 0’
0.95 S.
(A3)
Note: Award (A1) for 0.99, (A1) for 0.0001, (A1) for 0.05
Therefore p(S) = 0.0001 x 0.99 + 0.9999 x 0.05
= 0.0500939 (A1)
0.000] X 0.99
D S = ——— MI
p( I ) 0.0500939 ( )
=0.00198 (3 s.f.) (A1)
(ii) Let n = number of chips: 1000,
p = the probability that a randomly chosen chip is defective = 0.02.
Hence, the mean np = (1000)(0.02) = 20 and the
variance = np(1— p) = (1000)(0.02) (0.98) = 19.6. (A1 )(A1)
Suppose X is the normal random variable that approximates the binomial distribution.
The X = N(20, 19.6). (MI)(M1)
19.5 —20 305—20
Thus l9.5$XS30.5 = 523 M1 M1
= p(—0.11$ Z S 2.37)
= 0.5349 (A 1} Note: Line before last should bep(F0.113 5 Z S 2.37) = 0.536. Accept 0.535 or 0.536 If student’s work is not shown but there is evidence that hefshe used the calculator to ﬁnd the
answer, accept the answer. — IO  M99/510/H(2)M (a) Notes: Award (A1) for the correct intercepts
(A1) for graphing over the correct interval
(A1) for the correct x  coordinate of the maximum point. (b) Required area: [010:— x2)dx— f0:  x2)dx (MI) x2 x31 x2 x3 "
{3‘7} {77] W) 0 1 4i k_’ 11 195 6 2 3 6 3 3 2 (M1)
= ﬁg + 21:3 — 318) (AU
0R
‘ k
Requlred Area: I! 05—38)de (M1)
x2 x3 k
= ____ (MIN/11)
2 3]
3 2
=k?_’fz_+% (A1) continued ._11... Question 4 continued (ii) (a) (b) (c) In! 2—lnt
t=—. S 't = . J7 2:3” Hence, g’(r)>0 when 2>lnt or int<2 or r <e2. Since the domain of g(t) is {tzt > 0}, g’(t) > 0 when 0 < t < e2. _ , 2—lnt ,, 2J?—3J52—1:
Since g (t): 2:312 ,g (t)=—# __JF[s—3in:]
4:3 Hence g”(t) > 0 when 8—31nt< 0 ie. :> e“.
Similarly, g"(t) < 0 when 0 < t < em. g”(t)=0 when t=0 or 8=3ln!.
Since, the domain of g is {tzt > 0} , g”(t) = 0 whent = em.
Since g”(t) > 0 when r > e“ and g”(t) < 0 when t< em, 3 8 . . . . . .
[em , —e"‘”3] IS the pomt of 1nﬂex10n. The required value oft 15 em. Note: Award (A1) for evaluating t as ea” . (d) (e) g’(t)=0 when lnt=2 ort=e2. 2 2
Also g”(e2) = —J—e¥ = via—5 < 0. Hence 1" = e2 At (t‘ , g(t')) the tangent is horizontal. So the normal at the point (t‘ , g(t")) is the line I = t". Thus, it meets the I axis at the point t = t‘ = e2 and hence the point is (e2 , 0). M99/510/H(2)M (MIN/41) (M1) (A1) (M2) (A1) (M1)
(A1) (M1)
(M1) (AI) (M 1)
(M1)
(A1)
(M1) (M1)
(AI) — 12 — M99/510/H(2)M (i) (a) S is the group of permutations of {1, 2, 3} under the composition of permutation. (M1)
Since 31:6,order of S=6. (RI)
1 2 3 1 2 3 1 2 3
b M b f8 = = =
a e... [I 2 3],. [2 3 I}... [3 I 2), _ l 2 3 _ l 2 3 _ l 2 3
193—] 3 22194—2 1 32175—3 2 1 (A2) Note: Award (A2) for 3 correct permutations;
(A1) for 2 correct permutations;
(A0) for 1 correct permutation. , 123123123
Slncep3°p4=132213=312 d _123123_"123 I
anp“p3"213132_231’ (M) .03 ° A: at A ° 12; (RI) Note: There are other possibilities to show that the group is not Abelian.
() 2_123123_123_
C“"‘231231—312'pi 3_123123_123_ M1
p"312231_123_p°' () (Note that p0 is the identity of the group S.) Hence {po , p, , p2} form a cyclic group of order 3 under the binary operation of
composition of permutations. (R1) Note: Some candidates may write {p0 , p1, p2} is a subgroup of order 3, (award (A1 D, and write
the following table, (award (RD): continued...  l3  M99/510/H(2)M Question 5 continued b a (ii) (a) Lam“: }a,beR,az+b2¢0}. a b c d
Let [ ] and [ ] be two elements of A. Then —b a a’ c a b c d _ ac—bd ad+bc b a —d c “ —bc—ad —bd+ac (M1)
0! l3 . . = _ﬂ a ,lfwewrltea=ac—bd,B=ad+bc. (R1) Since a, [3 are real numbers/I is closed under matrix multiplication. A550ciativity follows from the fact that matrix multiplication is an associative
binary operation on the collection of all 2x2 matrices and the fact that A is closed with respect to the binary operation of multiplication of matrices. (R1)
10Ad_th_dl_ab10_lOab_ab M1
0 I E an rs etentltysmce “b a 0 I — O 1 _b a — _b a ( )(MI)
a b
forany element[ b (1)0134. (RI)
, a b 2 2
The determinant b =a +b #0. (M1)
—— a
H th t' abh ‘ h'h' 1 611, £1.41
ence emanx _b a asanrnverse,w1c lS a2+b2 b a ( )( )
, a b 1 a —b l 0 and
Since =
—b a c12+.b2 b a 0 I
1 a —b a b 1 (12+!)2 0 1 0
2 2 = 2 2 2 2 = (R1)
a +1: b a —b a a +1; 0 a +1: 0 1 _ a l 0 a 1)
Note: Some candidates may use b _
_ — a Award marks as appropriate. continued... — 14 — M991510/H(2)M Question 5 (ii) continued ®)W%383G3G3l l 0
Mcontains the identity [0 1]. (A1) The following results show that the set M is closed with respect to matrix
multiplication as the binary operation. (RI) 10—10_—1o
0101'0—1’
10
—1 l Gihlﬁﬂ
U3ﬁ$ﬁﬂ
Gﬂhﬂﬁil
GﬂijH] m Since the product of each element with itself is the identity, every element of M
is its own inverse. Hence M is a group. (R1) (c) The order of the group M is 4. If it had a subgroup of order 3, then by the
corollary of the Lagrange’s theorem, 3 must be a divisor of 4. (M1)
Since it is not true, M can not have a subgroup of order 3. (R1) (iii) (a) Let (G,o) and (H ,0) be two groups. They are said to be isomorphic if there
exists a one—toone transformation f : G ——> H which is surjective (onto) with the (C1)
property that for all x, y e G, f(xoy) = f(x)of(y). (CI) Note: Some candidates may say that the groups (G, o) and (H, o) are isomorphic if they have the
same Cayley table (or group table). In that case award (CI). (b) Since f:G—>H,f(x)t—:H forsomexeG. Since e’ is the identity element in H, (M1)
B"f(x)=f(x)=f(x°€)=f(€)'f(x) (MIX/41)
By the right cancellation law, 3’ = f (e). (RI) continued... #15. Question 5 (in) continued (C) (‘0 Suppose G =< a >, the cyclic group generated by a, i. e. n is the smallest positive integer such that a” = e, the identity in G.
Let f : G “a H be an isomorphism. Let f(a) = b e H. f(a2) =f(a°a)=f(a)°f(a) = man? In general f(a’") = (f(a))’" , 1 Sm Sn. By (iii) (b) (f (a))" = e’, the identity in H. Hence b” = e’ and consequently H is
a cyclic group of order n with generator 17. Z4 = {[0],[1], [2],[3]}. Let $4 be the binary operation.
[11934 [1] = [21,[1194 [2] =[3], [11594 [3] = {0]
Hence [1] is a generator of the group (Z4 , 634) and 24 is cyclic. ([3] is also a generator)
Since each element of M is of order 2, M is not cyclic.
Hence Z4 is not isomorphic to M. M995 10/H(2)M (C1) (M1)
(A1) (R1)
(M1)
(MIXAI) (M1)
(RI) _ 15 (i) (a) If G is a directed graph with n vertices, then an adjacency matrix of G is an n xn
matrix whose (i, j) entry is the number of directed edges from the Vertex vi to
the vertex vi, or zero if there is no directed edge connecting v, and v j.
(b) The sum of the entries in row i' of the adjacency matrix of the directed graph G is
equal to the outdegree of the vertex v, of G, ale. the number of directed edges
from vertex vi.
(c) The sum of the entries in column j equals the indegree of the vertex j tie. the
number of directed edges to vertex vj.
(ii) (a)
“5‘
Graph H
Note: Award (A3) for a correct graph, (A2) for a graph with 1 error;
(A1) for a graph with 2 errors, (AU) for a graph with 3 or more errors
(b) Since the element in the second row and the fourth column of A2 is 1, there is
only one path of length 2 from v2 to v4.
(iii) (a) (i) A complete graph K5 is a graph on 5 vertices where every vertex is connected to every other vertex. Zi> Example: M99/510/H(2)M (C2) (C2) (C2) (A3) (MINRI) (A2) (A1) continued... — 17 — M99/510/H(2)M Question 6 (iii) continued (ii) In K“ vertices are divided into two disjoint sets A and B, each having three vertices such that no two vertices in A are adjacent and no two vertices in B
are adjacent, but every vertex of A is adjacent to every vertex of B. (A1) Example: v4 v5 v6 (A1) (b) To ﬁnd a Hamiltonian circuit in K13, we show that there is a circuit which goes through each vertex exactly once. Such a circuit is shown below: (M1)
“1 V2 v3
v4 vs vs (MINAI) Please check the circuit drawn by the candidate and award (M2)(A1) for a correct answer.
Also note that some candidates may not show the Hamiltonian circuit but mention that each 6 . . . . . .
vertex of it“ has degree 3 = Hence there IS a Hamiltonian circuit. In this case award (M2)(AI). continued... — 13 — M99/510/H(2)M Question 6 continued (iv) Front door
1"! Let the front door, rear door and each room of the house be labelled as shown. We take each of these as a vertex, and draw edges when they are connected by a door. The (M1)
graph of the floor plan is:
(MIXAI)
The degree sequence for this graph v1, v2 ,..., vH is 1,2, 4, 2, 3, 4, 3, 2, 4,2,1. (MI)(AI)
The degree of more than two vertices is odd and hence there is no Eulerian path. (R1)
Hence it is not possible to enter the house through the front door and exit at the rear
(R1) door, travelling through the house going through each internal doorway exactly once. continued... — i9 — M99/510/H(2)M Question 6 continued (v) (a) The degree sequence yields the number of edges at each vertex and an
isomorphism preserves adjacency of vertices. (M1)
Hence, if two groups are isomorphic and the degree sequences are not the same then the isomorphism can not preserve the adjacency of vertices contradicting our
observation. (M1) Thus an isomorphism must preserve degree sequences. (R1) (b) The degree sequences ofthe graphs G and Hare 2,3, 3, 3, 3,4 and 2, 3, 3, 3, 5,2 (MIXMI)
respectively.
Since they are not the same, the two graphs are not isomorphic by (a). (R1) (vi) We start with point A and write S as the set of vertices and T as the set of edges. The
weights on each edge will be used in applying Prim’s algorithm.
Initially, S = {A}, T = o. In each subsequent stage, we shall update S and T. Step 1: Add edge It: So S = {A, D}, T: {/1} Step 2: Add edge e: So S: {A,D, E} T: {h,e} Step 3: Add edge d: Then S={A,D,E,F} T={h,e,d} Step4: Add edgea: Then S={A,D,E,F,B} T={h,e,d,a} Step5: Addedgei: Then S={A,D,E,F,B,G} T={h,e,d,a,i} Step6: Addedgeg: Then S={A,D,E,F,B,G,C} T={h,e,d,a,i,g} (M4)(A3) Notes: Award (M4)(A3) for all 6 correct, (M4)(A2) for 5 correct;
(M3)(A2) for 4 correct, (M3)(AI) for 3 correct;
(M1)(AI) for 2 correct, MIX/10) for 1 correct OR (M2) for the correct deﬁnition of Prim’s algorithm,
(M2) for the correct application of Prim’s algorithm,
(A3) for the correct answers at the last three stages. Now S has all the vertices and the minimal spanning tree is obtained.
The weight ofthe edges in Tis 5+3+5+7+5+6
= 31 (A1) — 20 — M99!510/H(2)M 7. Note: Throughout the whole question, students may be using their graphic diSplay calculators
and should not be penalized if they do not show as much work as the marking scheme. (i) (a) Let X denote the number of flaws in one metre of the wire. Then E (X) = 2.3 (MIXMI)
(2.3): 2!
= 0.265. (A1) ﬂaws and P(X = 2): 8‘2'3 Note: Award (C3) for a correct answer from a graphic display calculator. (b) Let Y denote the number of ﬂaws in two metres of wire. Then 1’ has a Poisson distribution with mean E (Y) = 2 x 2.3 = 4.6 ﬂaws for 2 metres. (M1)
Hence, P(Y21)=1P(Y=0)=l—e""6 (M1)
= 0.990 (3 s.f.) (AI) Note: Accept l—e'”. (ii) Let E be the sample mean, JLi be the pepulation mean and o‘be the p0pulation standard
deviation. Hence, 0}, the standard error is given by (If =—= where n is the sample size (M1)(A1) which is required. The 95 % confidence interval of the population mean ,u is given by (3?  p)i1.960';. (MINA!) /% s\ 1.96 p 1.96
Hence, we need to ﬁnd n so that (090.96) S 0.25 (MIXRI)
=> [%](1.96) S 025
=> S 43:56) => 52 48:56) = 31.36 (MI)(M1)
Hence, n 2 983.44 (A1)
The sample size required is 984. (R1) continued — 21 * M99/510/H(2)M Question 7 continued (iii) Let Jul and ,uz be the weekly mean wage of science and humanities students
respectively. The hypotheses to be tested are H0411 ‘ruz :0 (AI)
Him. nu: #0 (AI) It is a twotailed test. The test statistic is 9?] —:?z. The two random variables are independent tie. probability of selection of a science student is not affected by the
selection of a humanity student. of a"; —__+_.._ “ire = 0’ Gane z
"i "2 where 0'1, 02 are the population standard deviations of science and humanity students
reSpectively. Since the samples are large one can approximate 0', , 0'2 by s] ,52 ,
reSpectively. The resulting estimated standard error is given by 2 2
s. 4 = 5145—2 (M2)
1 2 2 2
_ 4—+ (4'5) (A1)
100 200
= 051... (A1)
0.01 0.01
2.33 2.33
From the normal distribution table we obtain the decision rule that if the computed
value of z > 2.33 or z < —2.33 then we reject H0. (MI)(AI)
f —.'i'
B t = ' 2
“ 2 5H: (M1)
120.5 — 1 15
=§—_ (M1)
051
=10.784 > 233 (A1)
So we reject H0. (R1) Thus, we conclude the wages of science and humanity students are not the same. Notes: An alternative approach by calculating critical values for a rejection region is possible. Please
check the working and award marks accordingly.
Award full marks for answers obtained using a graphic display calculator with a pvalue of 5.4 X 10‘” (exponent is —27), a zvalue of 10.761, and a correct conclusion. continued... — 22 — M9915 I 0/H(2)M Question 7 continued (iv) H0: the cost of the automobile is independent of the number of complaints.
H]: the cost of an automobile is not independent of the number of complaints. (A1) The observed frequencies are ﬁrst totalled, and then the expected frequencies under H0
are calculated from the formula: (row total) (column total) grand total
mu
m
In.
Innmu Table of expected frequencies expected frequency = (A4) Note: Award (A4) for 9 correct bold entries
(A3) for 7 or 8 correct bold entries
(AZ) for 5 or 6 correct bold entries
(A1) for 4 correct bold entries (A0) for 3 or less bold entries (M1) 2
We shall calculate x2 from a table where we shall list f0, fc ,(fn  fa)2 ,giffJ, where f,, is the observed number of complaints and f3 is the expected number of
complaints. continued — 23 — M9915 I 0/H(2)M Question 7 (iv) continued (A3) Calculation of x2
v = number ofdegrees of freedom = (3 — 1) (3 — 1) = 4 (M1)
At the 5 % level of signiﬁcance, 153105 with four degrees of freedom, x305 = 9.49 (/11) Decision rule: If x2 > 9.49, reject HG. If x2 S 9.49, then accept H0. Our computed x2 = 79.43 > 9.49 and hence according to our decision
rule, we reject H0. Conclusion: The cost of the automobile is not independent of the number of
c0mplaints. (R1) — 24 — M995 1 O/H(2)M (i) (a) Let f(x) = x3 —3x—5. Hence f(1)= #7 < 0 and f(3) = 13> 0. (M1)
Hence, by the intermediate value theorem there is one zero of f (x) = x3 — 3x — 5 (R1)
in the interval 15 x S 3. (b) f(x)=x3—3x—5 f’(x) = 3x2 — 3
_ foe.) _ xS—sx.—5
xnn xn fun ) — xn 3x: _ 3 (MINA!)
n
n 2.333333 0.333 333
. 2.333333 2.280556 0.052 777
2.280556 2.279 020 0.001536
2.279020 2.279019
2.279019 2.279019 (A3) Note: Award (A3) for any indication that the NewtonRaphson method has been correctly used. Since we need accuracy 10‘5 , we need Ixn h x“, [< 5x104.
So x = 2.279 02 is the solution off(x) = 0. (RI)
(ii) (a) Formula for the trapezium rule with n = 6 gives,
ﬂandx = le[g(xn)+?g(x1)+Zg(xz)+2g(x3) + 2g1x4)+ 2.306) + 21x0] (M2)
= 0.745 (A1)
Formula for Simpson’s rule gives,
I;g(x)dx = ﬁkm) +4g(X1) +Zg(x2)+4g(x3) + 21304) + 43065) + 30%)] (M2)
= 0.747 (A1)
(b) lError by the Simpson’s rulel S g(4)(x) (MIX/II) continued 25_ Question 8 (it) continued (0) (iii) (80 (iv) (b) (a) 1
Let' Sn( g) be the approximation to Lg(x)dx by Simpson’s rule over n intervals
of equal length. 6 So a
(180)n S [Jamsag) _ 1
30114 ’ Since g<4)(x)se for 05x51. To be correct to 5 d.p., we need to ﬁnd :1 so that 4.5x10‘6
30H
105 4 10" 106
i.e. 30n‘2—orn 2—0rn2‘——=9.036
5 150 150 Hence, we need 10 intervals. If g(x) is continuous for a s x S b and differentiable for a < x < b , then there
exists a .5 in a < 5 <5 so that g(b)g(a)=(b~a)g’(¢),a<€<b By the mean value theorem h(x) — 11(0) = h'(c)(x  0) , for 0 < c < 7.
Therefore h(x)~h(0)ls 10x, for 0s xs 7, since lh’(x)S 10.
Thus —70+h(0) s hm s h(0)+ 70 => —70—4 Sh(x)S—4+70 => ~74 Sh(x)$66 Hence h(x) 2 —74, for 05 x S 7. Letuk=k3:1.
2 k
Then limuk+‘=lim(kl:l)x 3 =l<1.
H uk H» 3 (k+1) 3 and hence the series converges by the ratio test. M99/510/H(2)M (M1) (A1) (M1) (M1) (M1) {A1} (A 1)
(A 1) (M1)
(A1) (M1) (A1) (M2) (R1) continued — 26 — M99/5 I O/H(2)M Question 8 (iv) continued (b) (c) 1
Jr(lnx)3 Let f(x) = Since xan)3 is an increasing function of x for x>1, f(x) is a positive decreasing function of x for x > 1. (M1)
Further f (k)  l
’ k(lnk)3 °’ 1
Hence, by the integral test, 2 3 converges or diverges according to the kg: k(ln 1:)
co e nce or diver ence of Jun ldx lirn R dx nv r e =
g g 2 sr(lnx)3 Rm 2 x(1nx)3 (M1)
(1 )'2 R 1 1 1
= lim "" = lim — , + 2 = 2 (A1)
aw —2 2 1H» 2(1n R)" 2(ln2) 2(1n2)
So the series converges, by the integral test. (R1)
°° 1+1 k .  .
2(—1) 2 IS an alternating series.
M k +1
When f(x) x f’(x)——1—'i‘2—<0 I (M1)(A1)
x2+l’ (x2+l)2— , k . .
when x 21, the sequence k2 I IS a decreasmg sequence. + . k . 1
Also 11m 2 = 11m :0. (M1)(AI)
15—»: k + 1 k—Iw k + i
1: Hence the series converges by the altemating series test. (R1) ...
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 Winter '10
 Chang
 Math, Calculus

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