N01/540/S(2)M
INTERNATIONAL BACCALAUREATE
BACCALAURÉAT INTERNATIONAL
BACHILLERATO INTERNACIONAL
MARKSCHEME
November 2001
FURTHER MATHEMATICS
Standard Level
Paper 2
13 pages
7 N01/540/S(2)M
1. (i) Let X , Y be the mass of one bag and 10 bags respectively.
X ~ N(100 g,1 g2) Y = X1 + X2 +& + X10 , so Y ~ N(1000 g,10 g2)
(M1)(A1)
P(995 < Y <1005) = 0.886 (3 s.f.) (M1)(A1)
OR
P(995 < Y <1005) = 0.886 (G2)
[4 marks]
fi
"xi
(M1)(AG)
(ii) (a) X == 2.1
100
[1 mark]
i
mx
(b) P(X = xi ) = e-m
(M1)
xi !
2.10
P(X = 0) = e-2.1 = 0.122 Ò! a =12.2 (A1)
0!
2.12
P(X = 2) = e-2.1 = 0.270 Ò! b = 27.0 c 100%(1- P(%d" 5)) = 2.1 (A1)
. Also
2!
H0 : X can be modelled by the Poisson distribution Po(2.1)
(C1)
H1 : X can not be modelled by the Poisson distribution Po(2.1) .
5
2
Ç =
(M1)(A1)
"( fo - fe)2 = 2.38 (3 s.f.)
fe
i=0
OR
2
Ç = 2.38 (G2)
degrees of freedom Å = 4 (M1)
2
so Ç(4, 5 %) = 9.488 > 2.38
(A1)
H0 (R1)
We can not reject and conclude that we do not have enough evidence to
say that the data cannot be modelled by a Poisson distribution with mean 2.1.
[9 marks]
Total [14 marks]
8 N01/540/S(2)M
2. (i) (a) (i) Since the graph can be redrawn as follows:
E
A
F
B
(A1)
C
D
And since this graph contains K3, 3 as a sub-graph, then it cannot be
planar because its sub-graph is not planar. (M1)(R1)(A1)
(ii) Second graph can be drawn in the following way:
PQ
U R W
T S
(M1)
Therefore the graph is planar. (C1)
[6 marks]
(b) In the graph the orders of the vertices are 2, 4, 4, 4, 4, and 2 (A1)
so by the theorem we can deduce that there is a Hamiltonian cycle. (M1)(C1)
One possible cycle is 1, 2, 3, 4, 5, 6,1 (A1)
Note: Vertices are numbered in an anti-clockwise direction
starting with the vertex of degree 2 at the top right corner.
[4 marks]
continued&
Question 2 continued
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9 N01/540/S(2)M
(ii) n =19x + 4 and n =11y +1 (M1)
Ò!19x -11y = -3
Since (19,11) =1, the equation has an integer solution. (M1)
Applying Euclid s algorithm we find:
19 =11×1+ 8 a = b + r1r1 = a - b
üÅ‚ (M1)
ôÅ‚
11 = 8×1+ 3 b = r1 + r2 r2 = 2b - a
ôÅ‚ (M1)
Ò!
żł
8 = 3× 2 + 2 r1 = 2r2 + r3 r3 = 3a - 5b
ôÅ‚
ôÅ‚
3 = 2×1+1 r2 = r3 + r4 r4 = -4a + 7b
þÅ‚
Since r4 =1 the particular solutions are to be found by the following:
19× (-4) -11× (-7) =1/ multiply by (-3) Ò!19×12 -11× 21 = -3
(M1)
(A1)
So x0 =12 and y0 = 21.
The general solutions are x =12 -11t , y = 21-19t , t " (M1)
Ò! n = 232 - 209t , t " . (A1)
For values of t "{1, 0, -1} , the solutions are 23, 232, and 441. (A1)
[9 marks]
Total [19 marks]
10 N01/540/S(2)M
-1
3. (i) Reflexive: AR A , because A = I AI and I is an invertible matrix. (C1)(C1)
Symmetrical: AR B , then there is an invertible matrix X such that
(C1)
-1 -1 -1 -1
B = X AX Ò! A = (X )-1BX , where X is an invertible matrix, (M1)
so B R A . (C1)
Transitive: AR B and B RC , means that there are invertible matrices (C1)
-1 -1 -1 -1
X and Y such that B = X AX , C = Y BY Ò! C = Y X AXY = (XY )-1 A(XY ) , (M1)
where XY is an invertible matrix so consequently ARC . (C1)
[8 marks]
(ii) Theorem: if a and b are two elements of a subgroup then ab-1 is also an element of
the sub group. (M1)
Let S1 and S2 be two subgroups and S1 )" S2 be the intersection.
and
a, b " S1 )" S2 Ò! a, b " S1 Ò! ab-1 " S1 a, b " S2 Ò! ab-1 " S2 (M2)
Ò! ab-1 " S1 )" S2 .
(C1)
Therefore S1 )" S2 is a subgroup of the same group.
[4 marks]
(iii) (a) Using n to represent the equivalence class for n, the elements of × will
p p
be written as (i, j) where i, j "{0,1,& , p}. (C1)
Since the order of × is 9, then the possible order of a subgroup is 1, 3,
3 3
(R1)(C1)
or 9. Obviously (0, 0) and × itself are two of the subgroups.
3 3
We need to take the subgroups of order 3.
Take the subgroup (0, 0), (0,1), (0, 2) . We can represent it as since
{ } (0,1)
it is generated by (0, 1). (C1)
The other groups are: (1, 0) , (1,1) , and (1, 2) . Each group can be
(A1)(C1)
generated by any of its non-zero elements they are cyclic.
p2
(b) For × the possible orders are again 0, and p by Lagrange s
p p
(R1)
Theorem.
So, the only groups we need to look for are the ones with order p. (C1)
Since the number of elements of × is p2 there there are p2 -1
p p
generators for the subgroups. (C1)
Also for each subgroup we have -1 generators. Therefore the number of
p
p2 -1
subgroups of order p is , and the total number of subgroups is
= p +1
p -1
(R1)(A1)
then p + 3
[11 marks]
Total [23 marks]
11 N01/540/S(2)M
4. (i) (a) (i) Since f (1) = g (1) =1, x =1 is a solution.
(M1)(A1)
(ii) To use the Newton-Raphson method we consider the equation
h(x) = f (x) - g (x) = 0 .
h(xn ) 3 - 2x - e1-x
xn+1 = xn - = xn -
(M1)(M1)(A1)
2
h (xn ) e1-x - 2
By applying four iterations of the Newton-Raphson method we get
(G1)
x =-0.256 .
(iii) h is continuous and differentiable over the set of real numbers, and
(C1)
2
h (x) = e1-x - 2 .
2
h (x) = 0 when x =1- ln 2 . So, by Rolle s theorem, h must have two
zeros one before and one after 0, which has been verified above.
2
H cannot have solutions anywhere else, otherwise h (x) will have to
have another zero which is not possible. Therefore 0.256 and 1 are the
only two zeros. (R3)
[10 marks]
(b) Since this function is differentiable to the fourth order over the interval
[ 0.256, 1], then the error of the estimate for 8 intervals satisfies the
following inequality:
(b - a)5 M 1.2565
(4)
E d"= Å" M with M = max f (x) on [-0.256,1] (R2)
.
180n4 180×84
(4)
Now f (x) = e1-x which has a max of approximately 4, when x =-0.256 .
(M1)(A1)
(G1)(AG)
This will give an error of 0.000017 < 0.00002.
[5 marks]
(ii) Maclaurin s series for sin x requires that x be expressed in radians, hence
Ä„ Ä„
3 = ×3 = (M1)
180 60
Also,
x2
2 2 2
f (x) = f (0) + xf (0) + f (0) + & + Rn Ò!
2!
x3 xn+1 (n+1)
sin x = 0 + x - + & + f (c) (M1)(A1)
3! (n +1)!
(n+1) (n+1)
Since f (x) =Ä…sin x or Ä… cos x, then f (c) =1 (C1)
n+1
Ä„
60
Thus , then by trial and error we find that the minimum
Rn d" d" 0.000005
(n +1)!
n = 3 , this yields
3
Ä„
ëÅ‚ öÅ‚
ìÅ‚ ÷Å‚
Ä„
60
íÅ‚ Å‚Å‚
sin3 H" - H" 0.05234 (M1)(A1)(A1)
60 3!
[7 marks]
Total [22 marks]
12 N01/540/S(2)M
5. (i) (a) C
P
Q
S
A R D B
AR ×SD
Area "(ARS) =
2
RB×SD
(M1)
and Area "(RBS) =
2
Area "(ARS) AR
(A1)(AG)
=
Area "(RBS) RB
[2 marks]
AR BP CQ Area "(ARS) Area "(BPS) Area "(CQS)
(b) × × = × ×
(M1)(C1)
RB PC QA Area "(RBS) Area "(PCS) Area "(QAS)
SA ×SR × sin(RSA) SB×SP × sin(BSP) SC ×SQ × sin(CSQ)
=× ×=1
(M1)(C1)(R1)
SR ×SB× sin(BSR) SP ×SC × sin(PSC) SQ ×SA × sin(QSA)
AR BP CQ
i.e. = = =1 (Ceva s theorem)
(AG)
RB PC QA
[5 marks]
dy
dy cost gradient -1
dt
= = Ò! =
(ii) (a) (M1)(A1)
dx
dx -4sin t tangent 4tan t
dt
For the line MP, the gradient m = 4tant .
Therefore y - sint = 4 tan t (x - 4cost)
(M1)
(A1)
Ò! y = 4x tant -15sint .
The diameter through the point N goes through the origin (centre of the circle)
so its equation is (M1)
y = x tant . (A1)
y = 4x tan t -15sin t
Now we have to solve the system ,
(M1)
{
y = x tan t
Ò! x = 5cost , y = 5sin t . (A1)
Which is the parametric equation of a circle with radius 5. (R1)(A1)
[10 marks]
continued&
13 N01/540/S(2)M
Question 5 (ii) continued
(b) The diameter through the point R has equation y = -x tan t . (A1)
y = 4x tan t -15sin t
Solving the system
(M1)
{
y =-x tan t
the coordinates of the point Q are found to be (3cost , - 3sin t) . (A1)
(A1)
Ä„
Å‚Å‚ îÅ‚
Since the locus is an arc of the circle,
t " 0,
śł ïÅ‚
2
ûÅ‚ ðÅ‚
the centre at the origin, and radius 3. The arc goes from the point (0, - 3) to
the point (3, 0) , excluding the endpoints.
(R1)
[5 marks]
Total [22 marks]
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