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Mat 1032 Summer Exam 2012

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0% found this document useful (0 votes)
110 views7 pages

Mat 1032 Summer Exam 2012

Uploaded by

joshua lunan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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UNIVERSITY OF TECHNOLOGY, JAMAICA

FACULTY OF SCIENCE AND SPORT

SCHOOL OF MATHEMATICS AND STATISTICS

Final Examination, Summer

Module Name: Engineering Mathematics 1

Module Code: MAT 1032

Date: July 2012

Theory / Practical: Theory

Groups: CM1, SE1, Eng 1(Elec/Mech)

Duration: Two (2) hours

Instructions:

1. This question paper consists of five (5) printed pages, which includes a cover page,

six(5) questions and a formulae sheet.

2. You are required to ANSWER ANY FOUR (4) questions in the answer booklet provided.

3. Full marks will be awarded for full workings / explanations.

4. You are allowed to use silent electronic calculators.

5. Begin the answer to each question on a fresh page and number your solutions carefully.

DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO.

1
QUESTION# 1

a) Rationalise the denominator of

5
[3 marks]
2− 5

b) Determine the set of values of x that satisfy

x 2 + x − 12  0

Express your answer in interval notation. [4 marks]

c) A polynomial is defined by

P( x) = 2 x 3 + kx 2 − 11x + 60 .

Given that ( x − 3) is a factor of P(x):

(i) Determine the value of k

(ii) Factorize 2 x3 + kx 2 − 11x + 60 completely.

(iii) Hence solve 2 x3 + kx 2 − 11x + 60 = 0 [2+4+2 marks ]

QUESTION# 2

a) A and B are real intervals defined by (−, −2] and [−5, 0] , respectively. Using interval notation,
find:

(i) A'

(ii) ( A  B) ' [2+3 marks]

b) Find the sum to infinity of the following series:

3 + 1.2 + 0.48 + ..... [4 marks]

c) Solve the following system of linear equations:

xy = 14
[6 marks]
4 x + 3 y = 29

2
QUESTION# 3

a) The standard form of a straight line is 5 x + 3 y − 4 = 0 .

Determine

(i) The gradient of the straight line.

(ii) The equation of a line perpendicular to the line 5 x + 3 y − 4 = 0 and passing through

the point (−2,5) .

[2+3 marks]

b) The table given below shows experimental values of two quantities x and y which are known to be
1
connected by an equation of the form = a x +b.
y

x 0.50 1.00 1.50 2.00 2.50 3.00

y 1.61 0.83 0.61 0.50 0.42 0.38

1
Plot against x and use the graph to estimate the values of ‘a’ and ‘b’.
y

[10 marks]

QUESTION# 4

1
a) Compute  (−1) ( n − 2 ) .
n =−1
n +1
[3 marks]

b) Solve the following equations, for the unknown variable:

(i) 52 x−9 + 2 = 627

(ii) 7 2 x − 8(7 x ) + 7 = 0

2
(iii) log 2   = 3 + log 2 x
x

[3+4+5marks]

3
QUESTION# 5

a) Two functions, f and g, are defined as follows:

x−2
f ( x) = 3x + 4 g ( x) =
3

Find an expression for:

(i) fg (x)

(ii) g −1 ( x) [2 +3 marks]

4
5 
b) Expand and simplify  2 + 3x  using the binomial theorem. [5 marks]
x 

c) Seats in an Auditorium were arranged as follows: first row, 12 seats; second row, 15 seats;

third raw ,18 seats; etc.

(i) Find the number of seats in the auditorium if there are 30 rows?

(ii) Find the number of seats in the 20th row?

[3 +2 marks]

***END OF PAPER***

4
USEFUL FORMULAE

−b  b 2 − 4ac
Quadratic Formula , x =
2a

x m x n = x m+n

x m  x n = x m−n

(x )
m n
= x mn

log( xy ) = log x + log y

x
log   = log x − log y
 y

log( x) p = p log x

Arithmetic series

an = a + (n − 1)d

n
sn = [2a + (n − 1)d ]
2

Geometric series

an = ar n −1

 1− rn 
sn = a  
 1− r 

a
S =
1− r

5
ANSWER KEY

MAT 1032 summer 2012

Question#1

(x+4)(x+3) 1

x = -4,3

Test intervals 1

Answer (-4,3) 2

b.

k=-9 2

division 2

(x-4)((2x+5) 2

X=3,4,-5/2 2

c.

Question #2

a.

b.

Answer =5 3

c.

4x2-29x+42=0 3

x= 21/4,y=8/3

x=2,y=7 3

Question #3

a.

(i) m=-5/3 2

(ii)5y-3x-31=0 3

b.

Root(x) 0.71 1 1.22 1.41 1.58 1.73

1/y 0.62 1.2 1.64 2.00 2.38 2.63

Graph 4
6
c=-0.75 1

m= 2.03 2

Question #4

a.

2 +-1+0 2

answer =1 1

b.

x=6.5 3

(ii)

y=7,1 3

x=0,1 1

(iii)

(2/x2)=8 3

x=1/2,-1/2 2

Question #5

fg=x+2 2

g-1(x) =3x+2 3

b.

(625/x2)+(1500/x5)+(1350/x2)+540x3+81x4 5

c.

30=12+(n-1)3 1

n=7 2

a20= 30 2

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