M02/540/S(2)M++
INTERNATIONAL BACCALAUREATE
BACCALAURÉAT INTERNATIONAL
BACHILLERATO INTERNACIONAL
c
MARKSCHEME
May 2002
FURTHER MATHEMATICS
Standard Level
Paper 2
12 pages
 3  M02/540/S(2)M+
Paper 2 Markscheme
Instructions to Examiners
1 Method of marking
(a) All marking must be done using a red pen.
(b) Marks should be noted on candidates scripts as in the markscheme:
show the breakdown of individual marks using the abbreviations (M1), (A2) etc.
write down each part mark total, indicated on the markscheme (for example, [3 marks] )  it
is suggested that this be written at the end of each part, and underlined;
write down and circle the total for each question at the end of the question.
2 Abbreviations
The markscheme may make use of the following abbreviations:
M Marks awarded for Method
A Marks awarded for an Answer or for Accuracy
G Marks awarded for correct solutions, generally obtained from a Graphic Display Calculator,
irrespective of working shown
C Marks awarded for Correct statements
R Marks awarded for clear Reasoning
AG Answer Given in the question and consequently marks are not awarded
3 Follow Through (ft) Marks
Questions in this paper were constructed to enable a candidate to:
show, step by step, what he or she knows and is able to do;
use an answer obtained in one part of a question to obtain answers in the later parts of a question.
Thus errors made at any step of the solution can affect all working that follows. Furthermore, errors
made early in the solution can affect more steps or parts of the solution than similar errors made later.
To limit the severity of the penalty for errors made at any step of a solution, follow through (ft)
marks should be awarded. The procedures for awarding these marks require that all examiners:
(i) penalise an error when it first occurs;
(ii) accept the incorrect answer as the appropriate value or quantity to be used in all subsequent
parts of the question;
 4  M02/540/S(2)M+
(iii) award M marks for a correct method, and A(ft) marks if the subsequent working contains no
further errors.
Follow through procedures may be applied repeatedly throughout the same problem.
The errors made by a candidate may be: arithmetical errors; errors in algebraic manipulation; errors in
geometrical representation; use of an incorrect formula; errors in conceptual understanding.
The following illustrates a use of the follow through procedure:
Markscheme Candidate s Script Marking
$ 600 × 1.02 M1 Amount earned = $ 600 × 1.02 M1
= $ 612 A1 = $602
× A0
$ (306 × 1.02) + (306 × 1.04) M1 Amount = 301 × 1.02 + 301 × 1.04
M1
= $ 630.36 A1 = $ 620.06
A1(ft)
Note that the candidate made an arithmetical error at line 2; the candidate used a correct method at
lines 3, 4; the candidate s working at lines 3, 4 is correct.
However, if a question is transformed by an error into a different, much simpler question then:
(i) fewer marks should be awarded at the discretion of the Examiner;
(ii) marks awarded should be followed by  (d) (to indicate that these marks have been awarded at
the discretion of the Examiner);
(iii) a brief note should be written on the script explaining how these marks have been awarded.
4 Using the Markscheme
(a) This markscheme presents a particular way in which each question may be worked and how it
should be marked. Alternative methods have not always been included. Thus, if an answer is
wrong then the working must be carefully analysed in order that marks are awarded for a
different method in a manner which is consistent with the markscheme.
In this case:
(i) a mark should be awarded followed by  (d) (to indicate that these marks have
been awarded at the discretion of the Examiner);
(ii) a brief note should be written on the script explaining how these marks have been
awarded.
Alternative solutions are indicated by OR. Where these are accompanied by G marks, they
usually signify that the answer is acceptable from a graphic display calculator without showing
working. For example:
Mean = 7906/134 (M1)
= 59 (A1)
OR
Mean = 59 (G2)
 5  M02/540/S(2)M+
sin¸
(b) Unless the question specifies otherwise, accept equivalent forms. For example: for tan¸
cos¸
These equivalent numerical or algebraic forms may be written in brackets after the required answer.
(c) As this is an international examination, all alternative forms of notation should be accepted.
For example: 1.7 , 1Å" 7 , 1,7 ; different forms of vector notation such as , , u ; tan-1 x for
u
u
arctan x.
5 Accuracy of Answers
There are two types of accuracy errors, incorrect level of accuracy, and rounding errors. Unless the
level of accuracy is specified in the question, candidates should be penalized once only IN THE
PAPER for any accuracy error (AP). This could be an incorrect level of accuracy, or a rounding
error. Hence, on the first occasion in the paper when a correct answer is given to the wrong degree of
accuracy, or rounded incorrectly, maximum marks are not awarded, but on all subsequent occasions
when accuracy errors occur, then maximum marks are awarded.
There are also situations (particularly in some of the options) where giving an answer to more
than 3 significant figures is acceptable. This will be noted in the markscheme.
(a) Level of accuracy
(i) In the case when the accuracy of the answer is specified in the question (for example:
 find the size of angle A to the nearest degree ) the maximum mark is awarded only if
the correct answer is given to the accuracy required.
(ii) When the accuracy is not specified in the question, then the general rule applies:
Unless otherwise stated in the question, all numerical answers must
be given exactly or to three significant figures.
(b) Rounding errors
Rounding errors should only be penalized at the final answer stage. This does not apply to
intermediate answers, only those asked for as part of a question. Premature rounding which
leads to incorrect answers should only be penalized at the answer stage.
Incorrect answers are wrong, and should not be considered under (a) or (b).
Examples
A question leads to the answer 4.6789& .
4.68 is the correct 3 s.f. answer.
4.7, 4.679 are to the wrong level of accuracy, and should be penalised the first time this type of
error occurs.
4.67 is incorrectly rounded  penalise on the first occurrence.
Note: All these  incorrect answers may be assumed to come from 4.6789..., even if that value is not
seen, but previous correct working is shown. However, 4.60 is wrong, as is 4.5, 4.8, and these should
be penalised as being incorrect answers, not as examples of accuracy errors.
 6  M02/540/S(2)M+
Markscheme Candidate s Script (A) Marking
M1 M1
(a) a = 2.31 × 3.43 (a) a = 2.31 × 3.43
= 7.9233 = 7.92 (3 s.f.) A1 = 7.9233 = 7.92 A1
(b) 2a = 2 × 7.9233 (b) 2a = 2 × 7.29 = 14.58
A1 A0
= 15.8466 = 15.8 (3 s.f.) = 14.5
A1 A0(AP)
Total 2 marks
Notes: Award A1 for either the exact answer 7.9233 or the 3 s.f. answer 7.92.
In line 3, Candidate A has incorrectly transcribed the answer for part (a), but then
performs the calculation correctly, and would normally gain the follow through marks.
However, the final answer is incorrectly rounded, and the AP applies.
Candidate s Script (B) Marking Candidate s Script (C) Marking
M1 M1
(a) a = 2.31 × 3.43 = 7.9233 (a) a = 2.31 × 3.43 = 7.9233
= 7.92 A1 = 7.93 A0(AP)
(b) 2a = 2 × 7.9233 (b) 2a = 2 × 7.93
A1 A1(ft)
= 15.8466 = 15.85 = 15.86 = 15.8
A0(AP) A1(ft)
Total 3 marks Total 3 marks
Notes: Candidate B has given the answer to part (b) to the wrong level of accuracy, AP applies.
Candidate C has incorrectly rounded the answers to both parts (a) and (b), is penalised
(AP) on the first occurrence (line 2), and awarded follow through marks for part (b).
Candidate s Script (D) Marking Candidate s Script (E) Marking
M1 M1
(a) a = 2.31 × 3.43 (a) a = 2.31 × 3.43 = 7.923
= 7.923 = 7.9 A0(AP) = 7.93 A0(AP)
(b) 2a = 2 × 7.923 (b) 2a = 2 × 7.93
A1(ft) A1(ft)
= 19.446 = 19.5 = 15.86
A0 A1(ft)
Total 2 marks Total 3 marks
Notes: Candidate D has given the answer to part (a) to the wrong level of accuracy, and therefore
loses 1 mark (AP). The answer to part (b) is wrong.
Candidate E has incorrectly rounded the answer to part (a), therefore loses 1 mark
(AP), is awarded follow through marks for part (b), and does not lose a mark for the
wrong level of accuracy.
6 Graphic Display Calculators
Many candidates will be obtaining solutions directly from their calculators, often without showing
any working. They have been advised that they must use mathematical notation, not calculator
commands when explaining what they are doing. Incorrect answers without working will receive no
marks. However, if there is written evidence of using a graphic display calculator correctly, method
marks may be awarded. Where possible, examples will be provided to guide examiners in awarding
these method marks.
 7  M02/540/S(2)M+
1. (i) (a)
xi
30 20 10 0 -10
pi
0.2 0.5 0.2 0.06 0.04
(A1)
5
µ = E(X ) = pi
(M1)
"xi
i=1
=17.6 cents (A1)
5
22
(b) Ã = Var(X ) = pi - µ2 = 94.24
(M1)
"xi
i=1
Ò! Ã = 9.71 (A1)
cents
For n = 700000 eggs, and by the central limit theorem the profit P has a (M1)
normal distribution, with µ1 = n × µ = $123200, Ã1 = n ×à = $ 81.24
(A1)(A1)
P(P >123000) = 0.9931
(A1)
[9 marks]
(ii) (a) X <" P(1) and Y <" P(2) . Let the variable Z denote all the accidents in the
factories and . Since X and Y are independent:
Z = X + Y (M1)
P(Z =1) = P(X = 0)× P(Y =1) + P(X =1)× P(Y = 0)
(M1)
1 2 1 2 1
= e- 2 e- + 1 e- e- = e-( +2 )(1 + 2) (A1)(AG)
OR
X + Y <" P(1 + 2 )
(M1)
1
(1 + 2 )1 e-( +2 )
Ò! P(X + Y =1) = (A1)
1!
1
= e-( +2 ) (1 + 2) (A1)(AG)
k
(b) P(Z = k) = P(X + Y = k) =
(M1)
"P(X = i)× P(Y = k - i)
i=0
kk
21 2 1
ëÅ‚
2 e- öÅ‚ ëÅ‚ 1 e- 2 -1 e- öÅ‚ ëÅ‚ 1k e- öÅ‚
1 2
= e- × + × + & + × e- ÷Å‚
(A1)
ìÅ‚÷Å‚ ìÅ‚ ÷Å‚ ìÅ‚
k!1! (k -1)! k!
íÅ‚Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
1
e-( +2 ) k (k -1)
ëÅ‚k + k2 -1 k
k
= + 122 -2 + & + 1k öÅ‚ (A1)
ìÅ‚ 2 1 ÷Å‚
k!2
íÅ‚Å‚Å‚
1
e-( +2 ) k
(M1)(A1)
= 1 + 2
( ) (by the binomial theorem).
k!
1 + 2
The variable Z has a Poisson distribution with parameter . (R1)
[9 marks]
Total [18 marks]
 8  M02/540/S(2)M+
2. (i) Closure:
x + y
(M1)
x y =
1+ xy
x + y -(1+ xy) = x -1+ y - xy (M1)
= (x -1)(1- y)< 0 (since x , y <1)
(M1)
x + y
Ò! x + y <1 + xy and <1
(C1)
1 + xy
Also, x + y +1+ xy = (x +1)(y +1) > 0
Ò! x + y> -(1+ xy)
(M1)
x + y
Ò! > -1
(C1)
1+ xy
therefore
x y " S
(C1)
Associativity: (M1)
x + y x + y + z + xyz
(x y) z = z =
1+ xy 1+ xy + xz + yz
y + z x + xyz + y + z
x ( y z) = x =
(A1)
1+ yz 1+ yz + xy + xz
Commutativity: (M1)
x + y y + x
x y = = = y x
(A1)
1+ xy 1+ yx
Identity element: (M1)
x + e
x e = x Ò! = x Ò! e = ex2 , "x " S Ò! e = 0 (A1)
1+ xe
Inverse element: (M1)
x + y
x y = 0 Ò! = 0 Ò! x + y = 0 Ò! y = -x ,
1+ xy
for all elements the inverse element is its opposite number. (A1)
(S , ) forms an Abelian group.
(AG)
[14 marks]
(ii) f (x1, y1) = f (x2 , y2) Ò! (x1 cos y1 , sin y1) = (x2 cos y2 ,sin y2)
(M1)
x1 cos y1 = x2 cos y2
Å„Å‚ Ä„
îÅ‚0, îÅ‚
y1, y2 " Ò! y1 = y2
. Since .
òÅ‚ (A1)
ïÅ‚ ïÅ‚
sin y1 = sin y2 2
ðÅ‚ ðÅ‚
ół
Hence x1 cos y1 = x2 cos y2 Ò! x1 = x2 Ò! (x1, y1) = (x2 , y2)
(A1)
The function is injective. (C1)
xcos y = a
Å„Å‚
(a, b)" B Ò! f (x, y) = (a, b) Ò!
(M1)
òÅ‚
sin y = b
ół
aa
b" 0,1 Ò! y = arcsinb Ò! x = =
Since [ [
(A1)
cos(arcsinb)
1- b2
The function is surjective. (C1)
ëÅ‚öÅ‚
ëÅ‚öÅ‚
xx
-1
ìÅ‚÷Å‚
So f (x, y) = , arcsin y or , arcsin y .
(A1)
ìÅ‚÷Å‚
ìÅ‚÷Å‚
cos(arcsin y)
íÅ‚Å‚Å‚ 1- y2
íÅ‚Å‚Å‚
[8 marks]
 9  M02/540/S(2)M+
Total [22 marks]
3. (i) (a)
3 (M1)
A B
Prim s algorithm:
1
Edge Weight Decision
4
2
B-C 1 add to the tree
F
C
B-D 2 add to the tree
B-A 3 add to the tree
3
C-E 3 add to the tree
D B-F 4 add to the tree
E
There are 10 possible minimum spanning trees. The weight is 13. (M2)(A1)
[4 marks]
(b)
We are going to take edges of the maximum
A B
weight instead of the minimum weight.
4
Start with the edge AF.
7
Edge Weight Decision
C
F
A-F 7 add to the tree
F-E 7 add to the tree
6
6
7
F-D 6 add to the tree
D-C 6 add to the tree
E
D
F-B 4 add to the tree
The weight of the tree is 30. (M4)(A1)
[5 marks]
Ò! a + 2b a" 0 mod 7
(ii) (a) The sum of the digits is divisible by 7
(M1)
abb =100a +10b + b
a"100a +11b mod 7
(M1)
a"2a +4b mod 7
(A1)
a"2(a +2b) a" 0 mod 7
(C1)
therefore abb is divisible by 7.
[4 marks]
(b) gcd (57, 13) = 1 and we are going to apply Euclid s algorithm.
57 =13× 4 + 5 a = b × 4 + r1 r1 = a - 4b
üÅ‚ üÅ‚
ôÅ‚
13 = 5× 2 + 3 b = r1 × 2 + r2 ôÅ‚ r2 = b - 2r1 = 9b - 2a
ôÅ‚ôÅ‚
Ò!Ò!
żłżł
5 = 3×1+ 2 r1 = r2 ×1+ r3 ôÅ‚ r3 = r1 - r2 = 3a -13b
(M1)
ôÅ‚
ôÅ‚þÅ‚
3 = 2×1+1 r2 = r3 ×1+ r4 ôÅ‚ r4 = r2 - r3 = 22b - 5a
(M1)
þÅ‚
Since r4 =1 we can find the particular solution.
(M1)
-5a + 22b =1Ò! -35a +154b = 7 Ò! 57 × (-35) -13× (-154) = 7
(M1)
So the particular solution is (-35, -154) .
(A1)
Now the general solutions are given by the formula
x =-35 +13p
Å„Å‚
, p "
(M1)
òÅ‚
y =-154 + 57 p
ół
For p = 3 we get the smallest positive integers that are x = 4and y =17 .
(A1)
[7 marks]
[20 marks]
 10  M02/540/S(2)M+
2 2 2
4. (i) (a) f (x) = xe-2x Ò! f (x) = (1- 2x)e-2x Ò! f (x) = 4(x -1)e-2x Ò!
(3) (4)
f (x) = 4(3 - 2x)e-2x Ò! f (x) =16(x - 2)e-2x (A1)
(b - a)h4
(4)
The error term is given by E =× M , where M e" max f (x) .
0180
ëÅ‚ - a)5 % M öÅ‚
(b
ìÅ‚or 180n4 ÷Å‚
íÅ‚Å‚Å‚
Since M = 16(0 - 2)e0 = 32 , the error term is
(M1)
2× 0.24
.
E =×32 = 0.000569 (A1)
180
[3 marks]
2 32
h = Ò! ×32 < 5×10-6
(b) (M1)
n 180n4
Ò! n > 32.7 (A1)
We have to take at least 34 strips (n is an even number). (C1)
[3 marks]
1 a b c
(ii) (a) = + +
(M1)
x(x -1)(x - 2) x x -1 x - 2
(a + b + c) x2 - (3a + 2b + c) x + 2a
=
(A1)
x(x -1)(x - 2)
a + b + c = 0
Å„Å‚
ôÅ‚3a + 2b + c = 0
òÅ‚ (M1)
ôÅ‚2a =1
ół
Ò! a = 0.5, b = -1, c = 0.5
(A1)
dx 0.5 1 0.5
ëÅ‚öÅ‚dx
= - +
(M1)
ìÅ‚÷Å‚
+"+"íÅ‚Å‚Å‚
x(x -1)(x - 2) x x -1 x - 2
x (x - 2)
= 0.5ln x - ln x -1 + 0.5ln x - 2 + c = ln + c, c " . (A2)
x -1
[7 marks]
1
(b) f (x) => 0 and decreasing for all values of x e" 4 ,
(M1)
x(x -1)(x - 2)
so we can apply the integral test. (M1)
a
îÅ‚ Å‚Å‚
a
x2 - 2x
f (x)dx = ïÅ‚ln śł (M1)
+"
4
x -1
ïÅ‚ śł4
ðÅ‚ ûÅ‚
a2 - 2a 2 2
(A1)
= ln - ln
a -13
ëÅ‚öÅ‚
a2 - 2a a2 - 2a
ìÅ‚÷Å‚
lim ln = ln lim = ln1 = 0
(M1)
a" ìÅ‚ a" ÷Å‚
a -1 a -1
íÅ‚Å‚Å‚
a
2 2
lim f (x)dx = 0 -ln < " (A1)
+"
a" 4
3
111
+ ++ &
The series converges. (C1)
2×3× 4 3× 4×5 4×5× 6
[7 marks]
Total [20 marks]
 11  M02/540/S(2)M+
5. (i) (a) If a transversal meets the lines (BC), (CA) and (AB) that carry the sides of the
BD CE AF
triangle ABC at the points D, E and F, respectively, then × × = -1
.
(C2)
DC EA FB
Converse:
If the points D, E and F on the lines (BC), (CA) and (AB) that carry the
BD CE AF
sides of the triangle ABC are such that × × = -1
then D, E and F
DC EA FB
are collinear. (C1)
[3 marks]
(b)
C
A
D
E F
P B
Ć Ć Ć
(M1)
FAP = BAP = BCP = DCP (ângles subtended by the same chord)
Ć
(M1)
PBF = PBÂ = PCÂ = PCE (ângles subtended by the same chord)
Ć
(M1)
PBD = PB
= PÂE since APBC is a cyclic quadrilateral .
We can see that one point is outside triangle so the product of the sensed
magnitudes will be negative. (M1)
Ć Ć
BD CE AF PBcosPBD PCcos PCÂ PAcosFAP
× × = × × =1 (A2)
Ć
DC EA FB
PCcosDCP PÂcosPAE PB
osPBF
By the converse theorem the points D, E and F are collinear. (R1)
OR
EAFP, BDFP, and APDC lie on circles with diameters AP, BP and CP
(R1)
respectively.
C is supplementary to EPD and to APB (R1)
(R1)
Ò! EPD = Ä™PB
Subtract from both (M1)
APD Ò! EPA = Å‚PD
But, AFE =•PA and DFB =Å‚PD (angles on the same arc) (R1)
therefore AFE = DFB (C1)
Since A, F, B are on the same line, E, F, D must be collinear. (R1)
[7 marks]
continued&
 12  M02/540/S(2)M+
Question 5 (ii) continued
2 p
dy 2 p
(ii) (a) , so the gradient of the tangent at the point (x0 , y0) is mt =
=
(M1)(A1)
y0
dx y
2 p
2
y - y0 = (x - x0) Ò! yy0 = 2 px - 2 px0 + y0
(M1)
y0
2
Since y0 = 4 px0 Ò! yy0 = 2 p(x + x0) (A1)(AG)
[4 marks]
(b)
y
D
L
Ä…
Ä…
F x
T 0
(M1)
Let DT be the tangent to the parabola at D.
From part (a):
y y0 =2 p (x + x0 )
Ò! at T, y = 0
(M1)
Ò! 2 p(xT + x0 ) = 0 Ò! xT = -x0 = 0Å„
Ò! FT = p + x0
(A1)
But DL = x0 + p, and
(C1)
DL = DF (definition of parabola) (R1)
Ò! DFT is isosceles
Ò! TDF = FTD = Ä… (A1)
Ò! DT bisects FDL
[6 marks]
Total [20 marks]