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Research & Development Centre for Mathematical Modeling
1
FE 1002: Mathematics for Finance
Model Paper
Instructions to candidates: No. of pages : One (05) No. of questions : One (02) MCQ
: One (02) Essay Time allocation : Two (02) Hours Marks allocated : 100 Marks
Answer All Questions
01.
i. Which of the following is a statement in logic?
(a) What a wonderful day!
(b) Shut up!
(c) What are you doing?
(d) Colombo is the capital of Sri Lanka
ii. Using quantifiers, , convert the following open statement into true statement. x + 5 = 8, x N
(a) x N, x + 5 = 8
(b) For every x N, x + 5 > 8
(c) x N, such that x + 5 = 8
(d) For every x N, x + 5 < 8
iii. ~(p q) is equivalence to,
(a) ~p q
(b) p ~q
(c) ~p ~q
(d) ~p ~q
iv. The converse of If x is zero then we cannot divide by x is
(a) If we cannot divide by x then x is zero.
(b) If we divide by x then x is non-zero.
(c) If x is non-zero then we can divide by x.
(d) If we cannot divide by x then x is non-zero.
Research & Development Centre for Mathematical Modeling
2
v. The contrapositive of the statement If 7 is greater than 5, then 8 is greater than 6, is
(a) If 8 is greater than 6, then 7 is greater than 5.
(b) If 8 is not greater than 6, then 7 is greater than 5.
(c) If 8 is not greater than 6, then 7 is not greater than 5.
(d) If 8 is greater than 6, then 7 is not greater than 5.
vi) Let A={2, 3, 5} , B={2} and C={ x|x is odd} , which of the following statement is
correct?
(a) C A= B
(b) C B= A
(c) AC
(d) A|C=B
vii) Let AB , and (AB)C, If aA and bB, which of the following statement must
be true,
(a) bC
(b) aC
(c) a(ABC)
(d) b(ABC)
viii) If (qp)p is FALSE then, truth values of p, q are respectively,
(a) F, T
(b) T,T
(c) F,F
(d) T,F
ix) xCxS,x , is equivalent to ,
(a) xCxS,x
(b) xSxC,x
(c) xSxC,x
(d) xCxS,x
x) AA is equal to,
(e) A
(f)
(g) Universal set
(h) None of these
AB=B
AC={3, 5}
CB=
Since 2 is not odd number Ais not sub set of C
Answer: A|C={2}=B
AB(AB)=A
Since (AB)C, then AC
Then (ABC)=A
Since aA,
Answer: a(ABC)
p q qp (qp)p T T T T
T F T T
F T T F
F F F T
Research & Development Centre for Mathematical Modeling
3
02.
i) Let 22 2 x)x(f on [-2, 2],
(a) There is a local Minimum at x=0.
(b) There is a local Maximum at x=0.
(c) There is a local Minimum at x=2.
(d) There is a local Maximum at x=2.
ii) If 21 nn uu , and 01 u , find 5u .
(a) 9
(b) 8
(c) 10
(d) -11
Minimum
point
Answer:
111 nCuku nn ,here k=1 , C=2 and n=5,
Then 880421521 1115
5 *uuu
Research & Development Centre for Mathematical Modeling
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iii) The roots of the auxiliary equation of nn yy 92 are,
(a) 2 and -2
(b) 3 and -3
(c) 1 and -1
(d) 2 and 4
iv) The set of all x values that satisfies the inequality 11 x is,
(a) 31 x:Rx
(b) 02 x:Rx
(c) 20 x:Rx
(d) 03 x:Rx
v) The General solution of the equation 22 1 n,uu nn ,
(a) 112 un
(b) 122 111 nn u
(c) 12 un
(d) nn 12
vi) The particular solution of the equation 010010 01 .u,.uu nn ,
(a) 0101 .nun
(b) 0101 .nun
(c) 0101 .nun
(d) 010010 ..un
Answer: 0909 122 nmnnn yy*yyy ,
Then , Auxiliary equation,
33
033
09
21
2
mandmroots
mm
m
Answer:
20
111111
111
11
x
x
x
x
Solution :
1
11
1
1
k
kcuku
nn
n
Then,
1
1
1
1
1
2
12
1202
uu
uu
n
n
nn
n
Solution :
cnuun 0 Then,
01011010010010
.nn.u
n*..u
n
n
Research & Development Centre for Mathematical Modeling
5
vii) The sequence 1, 5, 13, 25, 41, 61, is best describe by the difference equation ,
(e) 141 n,nuu nn
(f) 1141 n,nuu nn
(g) 141 n,uun
nn
(h) 114 1 nuu nn
viii) If 12 yx)y,x(f , second order partial derivative xyf is equal to,
(e) 0
(f) 1
(g) 2x
(h) 2
ix) The general solution of the equation , 202 21 nuuu nnn is,
(a) constant are and where BA ,BnAun
(b) constant are and where BA ,BnAu nn 2
(c) constant are and where BA ,BAu nnn 22
(d) constant are and where BA ,BAu nn 1
x) The set of all real x values that satisfies the inequality ,xx 022
(a) 02 xorxRx
(b) 02 xandxRx
(c) 02 xorxRx
(d) 02 xandxRx
1,5,13,25,41,61
1,1+4,5+8,13+12,25+16,41+20
1,1+4*1,5+4*2,13+4*3,25+4*4,41+4*5,
u0=1,
u1=1+4*1=, u0+4*1
u2=5+4*2=, u1+4*2
.
.
.
un=, un-1+4*n, since we started at n=0, n1
0
1
0
2
xy
y
xy
x
f
f
or
f
xf
Research & Development Centre for Mathematical Modeling
6
03.
(a)
i) Find the domain of the function, 21
1
xxf
, where range of the xf is
real numbers.
ii) Let 10422 xyyxy,xf ,
a) Find the critical points of y,xf . b)
c) Find the local Maximum, Minimum or saddle points of y,xf .
(b) Find the set of all real values that satisfies the inequality , 01
62
x
xx
21
1
xxf
Since range is real numbers,
11
011
01 2
xorx
xx
x
Therefore domain of f(x) is R-{-1,1}
00
000820420
42
20420
42
00
,
a,bbbAabb,af
xyf
Ababab,af
yxf
b,afb,afthen),y,x(fb,a
y
y
x
x
yx
is point critical
then by
esderivaitiv partial order First
and of point critical a is If
We have to find the value of the operator D at (0,0) point,
Second order partial derivatives,
,*,D
,,ff,f yyxyxx
01216442200
242
2
0,0D operator then and
Therefore f(x,y) has a saddle point at (0,0)
happen. tcan'
and :case01
123
010203
01
230
1
623
01
6
2
2
- and x- and xx
xx,x,
,x
xx
x
xxx
x
xx
Research & Development Centre for Mathematical Modeling
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04.
(a)
i) Find a recursion relation which describe the sequence , 2, 6, 18, 54, 162,
486,
ii) $100 is deposited at R present interest compound annually. If after 20
years the value of the investment is equal to $1000, find annual interest
rate R.
3
123
010203
123
010203
x
- and x- and xx
xx,x,
- and x- and xx
xx,x,
1- then
and :case03
happen. tcan'
and :case02
2, 6, 18, 54, 162, 486,
2, 2*3, 6*3, 18*3 , 54*3, 162*3
u0=2, u1=3 u0=6, u2=3 u1=18, u3=3 u2=54.
un=3 un-1 where n1
After 1st year the value of the investment is =$100+R*100 =(1+R)100
After 2nd year the value of the investment is =(1+R)100+R((1+R)100)
=(1+R)((1+R)100)=(1+R)2100
After 3rd year the value of the investment is =(1+R)2100+R((1+R)2100)
=(1+R)((1+R)2100)= (1+R)3100
After nth year the value of the investment is = (1+R)n100
After 20th year the value of the investment is = (1+R)20100=1000
0.12201-1.1220R
1.1220R1
1.1220e
0.11512.3026
R1ln
2.3026
0.1151R1ln
e
Rln
lnRln
R
R
20
120
10120
101
10001001
20
20
Research & Development Centre for Mathematical Modeling
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(b)
i) Find the General solution of homogeneous difference equation
02021 12 nnn u.u.u for 2n ,
ii) Find the general solution of non-homogeneous difference equation
nnnn u.u.u 22021 12 for 2n ,
(Hint : Use Part(i) solution)
02021 12 nnn u.u.u Suppose that
n
n mu ,
Then
0202102021
2
12
.m.mm
m.m.m
n
nnn
Then auxiliary equation is 020212 .m.m
202
80211
2
8021
2
8021
2
64021
2
8044121
2
12042121
02021
21
2
2
...
m,..
m
.......*.*..m
.m.m
solution for the difference equation : B.A.BA nnn 20201
Where A and B are constant
Already we have founded the solution for the reduce equation in part i),
Since f is a constant, suppose a particular solution takes a form n*
n Cu 2
where C is a constant. Then by substituting to the equation we have,
9
5
81
1
120424
22022122
22202212
2
12
.C
)..(C
.*.C
C.C.C
nn
nnnn
Thus, general solution of the equation is ,
nn
n B.Au 29
520
Research & Development Centre for Mathematical Modeling
9
iii) Given the simple national-income model, Y, C, It, and Gt , represent the
National income , consumption , investment and government text
respectively .
4
3
2
1
1
21
1
20
20
280
tt
ttt
t
tt
tttt
Y.G
)YY(.I
Y.C
GICY
With initial conditions Y0=100 and Y1= 102. Find the solution for national
income at time t (Yt)
A second order difference equation is obtained by substituting (2) ,(3) , and (4) into (1), we have
n
ttt
t
t
tt
ttt
t
tt
tttt
Y.Y.Y
Y.Y.Y
Y.)YY(.Y.Y
GICY
22021
20221
2020280
21
21
1211
Now we have difference equation which is same as part II, Then general solution
of the difference equation
tttt Y.Y.Y 22021 21 is ,tt
t B.AY 29
520 where A and B are constant .
Y0=100 and Y1= 102, then,
tt
n *..*..Y
.,.
B.AB.AY
BAB.AY
255020816497
649781
1029
10202
9
520
1009
52
9
520
11
1
00
0
AthenB-13/90.8B
5/9-20.8B
***-**Equation
*)*(*
(**)