fe 1002 model paper answers

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.


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


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


4

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


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


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


7

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


8

(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


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

*)*(*

(**)

fe 1002 model paper answers

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b b c b

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x c xsxc

x values

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c ac d ac

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