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ARAB OPEN UNIVERSITY
FACULTY OF COMPUTER STUDIES
INFORMATION TECHNOLOGY AND COMPUTING
MST 121: USING MATHEMATICS
FALL 2005 PART-1 EXAM
EXAM PERIOD: 150 MINUTES
MIDTERM EXAM (FORM A) SOLVED
Student ID:
Student Name:
Group Number:
Tutor Name:
THIS EXAM IS IN TWO PARTS:
GRADE FOR PART I (OUT OF 36 POINTS) =
GRADE FOR PART II (OUT OF 64 POINTS) =
EXAM GRADE (OUT OF 100 POINTS) =
1
I. MULTIPLE CHOICE PART (WORTH 36 POINTS).
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Please answers as many questions as possible. Each question is worth 3
points. In this part, your grade is the total of the best twelve (12) grades
for answers to twelve (12) questions, from amongst your answers to the
available sixteen (16) questions. From the given choices, a, b, c & d,
please do box the choice that solves the question best, and write down
the letter corresponding to your choice in the space provided. Should
you think that the correct answer is not among the four choices, please
write the word: NONE, for none of the above, in the space provided.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>



Question - 1: The angle, the vector, A  6i  8 j makes with the x-axis, is:
(a)53 degrees
(c) tan( 8 / 6)
(b) – 53 degrees
(d) tan 1 (4 / 3)
The answer is: b
………………………………………………………………………………...
Question - 2: The solution to the equation, log 3 ( x)  3 , is:
(a) x =1/ 27
(b) x = 1/81
(c) x = 27
(d) x = 9
The answer is: a
………………………………………………………………………………...
a  b
Question - 3: The matrix 
 2a
(a) a =– b
(b)b = a
The answer is: b
2b 
 , is non- invertible only if :
a  b 
(c) b = a – 1
(d) b = a+1
………………………………………………………………………………………
1 2 

A  
Question - 4: The matrix
 0  3
is :
(a) diagonal (b) non- invertible (c) not square (d) not upper triangular
The answer is: none
……………………………………………………………………………….
20
Question - 5: The value of the sum
 (3  2i) :
i 1
(a)60
The answer is: c
(b) 420
(c) – 360
2
(d) 720
………………………………………………………………………………..
Question - 6: A triangle ABC has side lengths a, b and c (opposite the angles
A, B and C, respectively). If a = 4, b = 5, and c = 7, then the value of the
angle C is given by:
(a) cos1 (29 / 35)
(b) cos1 (20 / 14)
(c) cos1 (8 / 40)
(d) 90o
The answer is: c
……………………………………………………………………………………...
Question -7: A triangle ABC has side lengths a, b and c (opposite the angles
A, B and C, respectively), where the angle C is a right angle. If the sides, b =
12, and c = 13, then tan(A), is given by:
(a)5/12
(b) 5/13
(c) 12/13
(d)13/12
The answer is: a
...........................................................................................................................
1 / 2 0 0 


Question - 8: The inverse of the matrix, B   0 1 / 3 0  , if it exists, is
 0
0 0 

given by :
 2 0 0


(a)  0 3 0 
 0 0 4


 2 0 0


(b)  0 3 0 
 0 0 0


 2 1 1


(d)  1 3 1
 1 1 1


 2 0 0


(c)  0 3 0 
 0 0 1


The answer is: none
...........................................................................................................................
Question -9: The sequence an = (– 4/3) n , in the long run,
(a) goes to 0
(b) goes to ∞
(c) diverges
(d) goes to 4/3
The answer is: c
………………………………………………………………………………...

Question - 10: If a = 1– (1/3)n, n = 0,1,2,3,..., then a is equal to:

n
(a) -3/2
n0
(c) - ∞
(b) -1/2
n
(d) ∞
The answer is: d
………………………………………………………………………………...
3
Question - 11: If an = 100 (1 – 1/ n 2 ), (n = 1,2,3,...), then a10 , is equal
to:
(a)99
(b)100
(c) 10
(d) 90
The answer is: a
………………………………………………………………………………...
Question - 12: If x0 = – 10, and xn+1 = 3+ xn, (n = 0,1,2,3,...), then the
closed form of xnis given by:
(a) xn = – (10+3n) (b) xn= – (10 – 3n) (c) xn= (10 – 3n) (d) xn= (10+3n)
The answer is: b
..……………………………………………………………………………………..
Question -13: Lines parallel to the line given by the equation, 2y + 3x = 6,
all have the slope:
(a)6/2
(b) 2/3
(c) 3/2
(d) – 3/2
The answer is: d
………………………………………………………………………………..
Question - 14: The function f (x) = (x– 2) / (x2 – 5x + 6), has a removable
discontinuity at the number(s), x, equal to:
(a) 2 & 3
(b) 2
(c) 3
(d) – 2
The answer is: b
……………………………………………………………………………….
Question - 15: The distance between the two points (5, – 3) and (1, – 6) is:
(a) 5
(b) sqrt (117)
(c) (4,3)
(d) 25
The answer is: a
...........................................................................................................................
Question –16:The circle described by the equation, x 2 + 8x + y 2 – 6y = 0,
has:
(a)
radius 25.
(b)
centre (– 4, – 3) .
(c)
centre (– 4, 3), and radius 5.
(d)
centre (4, – 3), and radius 5.
The answer is: c
4
II. ESSAY QUESTIONS PART:(WORTH 64 POINTS).
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Please answers as many questions as possible. Each question has 4 subparts and each sub- part is worth 4 points, for a total of 16 points for
each question. In this part, your grade is the total of the best four (4)
grades for answers to four (4) questions, from amongst your answers to
the available six (6) questions. Please show your work, and write down
your answers neatly in the space provided directly below each sub- part.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
…………………………………………………………………………...........
Question - 1: Consider the functions, f and g , given by:
f (x) = exp (–x) , and g (x) = 3 sin (2x).
(a) Find the domains and ranges of the functions f and g.
The domain (f) = all real numbers, while the range (f) = (0, infinity).
The domain (g) = all real numbers, while the range (g) = [– 3, 3].
(b) Show that g is not one-one, and determine sub-domains on which g is
one-one. Find the inverse of g on one of these sub-domains. [ Hint: it may be
helpful to sketch the graph of g first].
Note that g (0) = g (Pi/2) = g (Pi) = 0, and hence g is not one-one. Since g
is periodic with period Pi, g is one-one for instance on the interval [– Pi/4,
Pi/4]. On this interval the inverse function is given by : y = (1/2) sin-1 (x/3).
(c) Show that f is one-one, and find the inverse function of f and its domain
and range.
For a < b, we have, f (a)= exp (-a)= 1/exp(a) > 1/exp(b) = exp (–b) = f (b).
So f is decreasing and hence f is one-one , and therefore admits an inverse.
If exp(–x) = y , then ln (exp (–x))= –x = ln (y), or x= ln (1/y).
This means that f -1(x) = ln (1/x) = – ln(x),
with domain = (0, infinity), and range = all real numbers.
(d) Consider the function h given by h(x) = g(x)f(x). Determine the long
term behaviour of h(x) and (1/h(x)) as x goes to infinity.
As x goes to infinity, f(x) goes to zero, and the limit of g(x) does not exist,
but is bounded so the long term behavior of fg is zero. However, the long
term behavior of 1/ h(x), cannot be determined since the function g
periodically changes sign (so 1/h(x) periodically goes to 1/0+ and 1/0 ,
periodically).
5
.………………………………………………………………………………
 1 0
 , and J =
 0 1
Question – 2: Consider the matrices I = 
0 1 

 .
 1 0 
(a) Show that the matrix, J 2 = – I, and for constants a and b, write the
matrix M = (aI– bJ)2 , in terms of the matrices, I and J. For what values of
a and b is the matrix, M, equal to the null matrix.
 1 0 

 = – I.
 0  1
 0 1  0 1

 
 =
 1 0  1 0
M = (aI-bJ)2 = a2I–2abIJ+ b 2J2 = a2I –2abJ – b 2 I= (a2 – b 2 )I –2abJ.
M is equal to the null matrix, when a = b = 0.
(b) Find the matrix A= I+J. Show that the matrix A is invertible. Find the
inverse, A-1, of the matrix A. Can you write the matrix, A-1 , in terms of the
matrices, I, and J.
 1
 1
A = 
1
 , then det A = 2. So A-1 = (1/2)
1
1  1
 = (1/2) (I – J).

1 1
.
(c) Solve the two by two equations system S = {x + y =10, –x + y = 4} by
the elimination method [i.e. by eliminating one variable at a time].
Adding the two equations of the system, we get, 2y = 14, and hence y = 7.
Subtracting the second equation from the first , we get 2x = 6, and hence
x = 3. So the solution set is (x = 3, y = 7).
d) Solve the system, S, by matrix methods.
 x
10 
 

Consider the vectors v =   , and w =   , then we can write the system S
y
4

as Av = w, and hence to find the values of the entries of the vector v, we
compute:
 x
1  1 10 
6
   = (1/2)   . So x = 3, y = 7.
v =   = A-1 w = (1/2) 
 y
14 
1 1   4 
6
……………………………………………………………………………..

 



Question – 3: Consider the vectors a  4i  2 j , b  3i  5 j .



(a) Find the coordinates, and magnitude of the vector c  2a  3b .



c  2a  3b = (–17, –11),
Hence the vector’s magnitude = sqrt(410)= 2.25.




(b)Find the angle between the vectors a , and b .
The angle between the given vectors a , and b is given by:
tan 1 (2 /( 4))  tan 1 (5 / 3) = – 85.6 degrees.
(c) Find the area A of
the parallelogram the two sides of which are made by


the vectors a , and b .

A = a b sin( 85.6) | 0.997 20 34  26
(d) Use the law of cosines to find the length S of the side joining the heads

(tips) of the vectors a , and b .
If we use the law of cosines, we get :
2 2
 
S 2  a  b  2 a b cos( 85.6) =20+34 -2sqrt(20)sqrt(34)cos(–85.6)=54–4=50,
So, S = 7.07
…………………………………………………………………………….…
7
Question – 4: Consider the points A (7, 2) and B (1, –6) in the plane.
(a) Find the length d of the segment AB.
d = sqrt (36 + 64) = sqrt (100) = 10.
(b) Find the equation of the line, L1, that contains the points A and B.
The equation of the line, L1 , is: y = (4/3) x + (–22/3).
(c) Find the equation of the line L2 perpendicular to the line L1, and passing
through the midpoint, M, of the segment AB.
The midpoint M of the segment AB has coordinates: (4, – 2).
A perpendicular line L2 to the line L1 going through M, has a slope: –3/4,
and hence we have: y = (–3/4) x + b, and substituting the coordinates of M,
we get: y = (–3/4) x + 1.
(d) Find the equation of the circle C, with center A and radius, r = 5. Are
the points B and M, on, inside or outside the circle C. Justify your answer.
2
2
The circle C is then described by the equation: ( x7 ) ( y2 ) 25
The point, M=(4, –2), is on the circle since its coordinates satisfy the circle
equation. The point B is outside the circle C because substituting the
coordinates in the LHS of the equation we get 100 which is larger than 25.
This should be obvious from (or at least consistent with) part (a).
……………………………………………………………………………..…
8
Question – 5 An industrial cylindrical pool in an oil refinery designed to
contain some oil products has depth h = 24 m, and radius r =10 m.
(a) The pool is to be covered with a special light metal cover which costs
450 $ per square meter. Find the cost C, of the pool surface cover.
A= Surface Cover Area = r2 Pi = 100 Pi = 314.16 m2;
Cover Cost = 141372 $.
(b) Find the volume, V, of the pool.
V = Pool Volume = hr2 Pi =24*100 Pi = 7539.8.
Approximately, V = 7540 m3
(c) If the inside wall surface and the bottom of the pool need to be covered
by a coat of special paint, find the surface area, S, to be painted.
S=2hrPi + r2Pi = 24*10*2*Pi + 100 Pi = (480+ 100) Pi = 1822.12 m2.
(d) Find the volume, B, of the largest spherical ball that can be placed inside
the pool, and the volume of oil that can still be stored in the pool with the
ball inside the pool.
The largest sphere will have radius =10m , and a maximum volume :
B = 4/3 R3 Pi= (4/3) (1000) Pi= 4188.8 m 3 = 4189 approximately.
The Volume left for oil in the pool is given by the difference of the pool
volume and the sphere volume: V- B = 7540 – 4189 = 3351 m 3.
9
.........................................................................................................................
Question– 6: Consider the sequences xn  (1 / 9) n , and yn  2n  1, n = 0,1,2...
(a) Find the first four terms, and the long term behaviour of the sequence xn .
The sequence xn is a geometric sequence with ratio r = -1/9 , and |r| <1,
hence , x0  1 , x1  1/9, x2  1/81, x3  1/729, and in the long term xn
goes to zero .

(b) Find, if possible, the value of the sum
x .
i 0
Since |1/9| < 1,
i

 x =1/(1+1/9)= 9/10 = 0.9.
i 0
i
99
(c) Find the value of the sum
y
i 0
99
y
i0
n
99
99
i 0
i 0 n
i
.
= 2 n  1  [2*99*100/2] + [100] = 9900 + 100 = 10000.
n
n
i 0
i 0
(d) Let S n   xi and Tn   yi . Find the long term behaviour of the
sequences, ( Sn  Tn ), ( Sn  Tn ), ( Tn / S n ), and ( Sn / Tn ).
( Sn  Tn ) diverges to   , ( Sn  Tn ) and ( Tn / S n ) both diverge to  , while
( Sn / Tn ) converges to 0.
10