Benodigdhede vir hierdie vraestel/Requirements for this paper:
Multikeusekaarte/
Nie-programmeerbare sakrekenaar/
Multi-choice cards:
Non-programmable calculator:
Grafiekpapier/
Draagbare Rekenaar/
Graph paper:
Laptop:

EKSAMEN/TOETS
EXAM
KWALIFIKASIE/
EXAMINATION/TEST:
SECOND
OPPORTUNITY
QUALIFICATION:
MODULEKODE/
WISN 112
MODULE CODE:
MODULEBESKRYWING/
MODULE DESCRIPTION:
EKSAMINATOR(E)/E
Oopboek-eksamen/
Open book examination?
B Sc IT
TYDSDUUR/
ADVANCED MATHEMATICAL
TECHNIQUES
ME E R VAN DER MERWE
MAKS/
3 hours
100
MAX:
DATUM/
03/07/2013
DATE:
TYD/TIME:
DR M PLOTZ
14:00-17:00
12-G02
Study units 1 - 7.
INSTUCTIONS / INSTRUKSIES:
Answer all questions. / Beantwoord alle vrae. / Show all calculations. / Wys alle bewerkings.
Plan your time. / Beplan u tyd. / GOOD LUCK! / STERKTE!
Formulas/ Formules
F
x[(1  i ) n  1]
i
If / As y  F (g ( x )) , then / dan is
P
x[1  (1  i ) n ]
i
dy
 F (g ( x ))g ( x )
dx
Dx (e f ( x ) )  e f ( x ) .f ( x )
i  (1 
NO
DURATION:
EXAMINER(S):
MODERATOR:
NEE/
i nm
) 1
m
1
Question 1 / Vraag 1
[22]
Choose the correct answer; write down the corresponding letter only. / Kies die
korrekte antwoord; skryf slegs die ooreenstemmende letter neer.
1.1
x 1 x  3
x 5
is / are: / Die oplossing(s) van


x  2 x  4 x 2  6x  8
x 5
The solution(s) of
x 1 x  3
is:


x  2 x  4 x 2  6x  8
2
x
A)
only / slegs x  3
5
x  3 only / slegs x  3
B)
5
x
C)
or / of x  3
2
D)
1.2
C)
D)
5 x  15
x2
is: / Die definisieversameling van f ( x ) 
(2)
5 x  15
x2
is:
x  2; x  
x  2; x  
x  2; x  
x  3; x  2; x  
(2)
If f ( x )  5 x  10 then: / As f ( x )  5 x  10 dan is:
A)
B)
C)
D)
1.5
1
1
only / slegs x 
4
4
1
x  2 or / of x 
4
5
5
only / slegs x 
x
4
4
x 
The domain of f ( x ) 
A)
B)
C)
D)
1.4
(2)
The root(s) of 1  5x  1  2x is / are: / Die wortel(s) van 1  5x  1  2x is:
A)
x  2 only / slegs x  2
B)
1.3
No solution exists. / Geen oplossing bestaan nie.
f 1( x )  10  5x
y  10
5
10 x  5
f 1 ( x ) 
5
x  10
f 1 ( x ) 
5
f 1 ( x ) 
(2)
If f ( x )  2x 2  7 and g( x )  x  1, then: / As f ( x )  2x 2  7 en g( x )  x  1, dan is:
A)
(f  g )(x )  2x 2  6
B)
(f  g )(x )  2( x  1) 2  7
C)
(f  g )(x )  (2x  2) 2  7
D)
(f  g )(x )  2x 2  x  6
(2)
2
1.6
The gradient of the tangent line to the curve f ( x )  4x 3  3x  2 where x  1 , is: /
Die helling van die raaklyn aan die kromme f ( x )  4x 3  3x  2 waar x  1 , is:
A) 11
3
B)
C) 9
D) 12
1.7
1.8
log 5 125  log 7 49  log 20 20 equals: / log 5 125  log 7 49  log 20 20 is gelyk aan:
A)
B)
log 2
C)
log
D)
51,0204
lim
x 
A)
B)
C)
D)
1.9
(2)
2
B)
(2)
8x 4  5x  2
x 3  4x 4
1

2
equals: / lim
8x 4  5x  2
is gelyk aan:
x 3  4x 4
x 
0
f (x) 
A)
125  20
49
(2)
1
x2
Is discontinuous where x  0 because f (0) is not defined where x  0 . / Is
diskontinu waar x  0 want f (0) is nie gedefinieer waar x  0 is nie.
Is discontinuous where x  0 because lim f ( x ) does not exist. / Is
x0
diskontinu waar x  0 want lim f ( x ) bestaan nie.
x0
C)
Is discontinuous where x  0 because lim f ( x )  f (0) . / Is diskontinu waar
x 0
x  0 want lim f ( x )  f (0) .
x 0
D)
Is continuous where x  0 . / Is kontinu waar x  0 .
1.10 If f ( x ) 
A)
B)
C)
D)
ex
4
2 x
x2  x 1
f ' (x) 
f ' (x) 
f ' (x) 
f ' (x) 
ex
4
2 x
then: / As f ( x ) 
( x 2  x  1)  e x
( x  x  1)
2
e
x 42 x
4
ex
4
2 x
x2  x 1
2 x
dan is:
(2 x  1)
2
( 4 x  2)( x 2  x  1)  e x
( x 2  x  1)
3
(2)
4
2 x
(2x  1)
( x 2  x  1)e x
4
2 x
4
2 x
(2x  1)
( x 2  x  1)e x
4
2 x
4
2 x
(2x  1)
( 4 x 3  2)  e x
( x 2  x  1) 2
( 4 x 3  2)  e x
( x 2  x  1) 2
(2)
3
1.11 Determine the equation of the horizontal line that passes through (2;2) .
Bepaal die vergelyking van die horisontale lyn wat deur (2;2) gaan.
A)
x2
B)
y 2
C)
y  2x  2
D)
y  2
/
(2)
Question 2 / Vraag 2
[4]
The profit function per week of a factory that manufactures dvd’s is given by
p(q)  20q 2  8000q  2500 . / Die winsfunksie per week van ’n fabriek wat dvd’s
produseer word gegee deur p(q)  20q 2  8000q  2500 .
2.1
2.2
Find the level of production to maximize the total profit per week. / Vind die
vlak van produksie per week om die totale wins per week te maksimeer.
(2)
Find the maximum profit per week. / Vind die maksimum wins per week.
(2)
Question 3 / Vraag 3
[3]
The income for a certain company is given by R  p 2  23p , where p is the price
of the product that they sell. Determine the price value(s) for which the income
will be R50. / Die inkomste vir ’n sekere maatskappy word gegee deur R  p 2  23p ,
waar p die prys is van die produk wat hulle verkoop. Bepaal die prys waarde(s)
waarvoor die inkomste R50 sal wees.
(3)
Question 4 / Vraag 4
[7]
4.1
Suppose consumers will demand 250 units of a product when the price is R 40
per unit and 400 units when the price is R 35 per unit. / Veronderstel
verbruikers vra 250 eenhede van ’n produk aan wanneer die prys R 40
per eenheid is en 400 eenhede wanneer die prys R 35 per eenheid is.
4.1.1 Determine the demand equation, assuming that it is linear. Let q be the
number of units and p be the price. / Bepaal die vraag-vergelyking. Neem aan
dat dit lineêr is. Laat q die aantal eenhede en p die prys wees.
(3)
4.1.2 Determine the price per unit when 120 units are demanded. / Bepaal die
prys per eenheid wanneer 120 eenhede aangevra word.
(1)
4.2
The demand function for a product is given by p  21 0,7q and the supply
function by p  14  2,8q . Determine market equilibrium. /
Die vraagfunksie vir ’n produk word gegee deur p  21 0,7q en die
aanbodfunksie deur p  14  2,8q . Bepaal markewewig.
(3)
4
Question 5 / Vraag 5
5.1
Solve for c: / Los op vir c:
x
5.2
[7]
ac
ab  a
(2)
Solve for x: / Los op vir x:
log x  log( x  2)  3 log x
(5)
Question 6 / Vraag 6
[9]
The fixed cost for a brick company that manufactures special paving bricks,
is R7 920 and the variable cost is R3. / Die vaste koste van ’n
stenemaatskappy wat spesiale plaveisel stene maak, is R7 920 en die
veranderlike koste is R3.
6.1
6.2
6.3
6.4
6.5
Write an equation for the cost function C(q). / Skryf ’n vergelyking vir die
kostefunksie C(q).
(1)
If the selling price of a paving brick is given by p  5q  1 ,where p is the price
and q the number of units, write an equation for the revenue, R(q ) . / Indien die
verkoopsprys van die plaveisel stene gegee word deur p  5q  1, waar p die
prys en q die aantal eenhede is, skryf ’n vergelyking vir inkomste, R(q ) .
(1)
How many paving bricks must be sold for the company to break even? /
Hoeveel plaveisel stene moet verkoop word sodat die maatskappy gelykbreek?
(3)
Determine the marginal revenue function. /
Bepaal die marginale inkomstefunksie.
(1)
Determine the marginal revenue if q  10 . Interpret the answer. / Bepaal die
marginale inkomste as q  10 . Interpreteer die antwoord.
(3)
Question 7 / Vraag 7
[5]
The average marginal cost for Aluminium door frames is given by
C (q )  4q 2  18q  2 . Determine the equation for the average cost of the
Aluminium door frames if the average cost for 2 Aluminium door frames is
R 2 500. / Die gemiddelde marginale koste vir Aluminium deurrame word gegee
deur C (q)  4q 2  18q  2 . Bepaal die vergelyking vir die gemiddelde koste
van die Aluminium deurrame indien die gemiddelde koste vir 2 Aluminium
deurrame R2 500 is.
(5)
5
Question 8 / Vraag 8
[17]
8.1
Find the present value of R 12 000 due 7 years from now if the interest is
compounded continuously at the annual rate of 5,5%. / Vind die huidige waarde
van R 12 000 in 7 jaar van nou af as die rente kontinu bygevoeg word teen
’n jaarlikse koers van 5,5%.
8.2
Furniture loan:
A Person purches furniture for R 110 000 and agrees to pay off this
amount by monthly payments of R 1 500. If the interest is charge at the rate
of 15% compounded monthly, how many FULL payments will there be?
What is the amount of the last payment? /
(3)
8.3
8.4
8.5
Meubel Lening:
’n Persoon koop meubels vir R 110 000 en onderneem om die bedrag
te delg in maandelikse paaiemente van R 1 500. As die rentekoers wat
vereis word, 15% maandeliks saamgestel is, hoeveel VOL paaiemente
sal daar wees? Wat is die bedrag van die laaste paaiement?
(7)
An annuity consisting of equal payments semi-annually for five years is
to be purchased for R 30 000. If the interest is 6% compounded semi-annually,
how much is each payment? / ’n Annuiteit bestaan uit gelyke paaiemente
wat elke halwe jaar vir vyf jaar betaal word vir R 30 000. As die
rentekoers 6% halfjaarliks saamgestel word, hoeveel is elke paaiement?
(3)
Convert a nominal interest rate of 12% compounded monthly to an
effective interest rate. / Herlei ’n nominale rentekoers van 12% maandeliks
saamgestel na ’n effektiewe rentekoers.
(3)
The population of a small town is 100 000. The population increases
by 3% per year. Write an equation for the population (P) against time (t). /
Die populasie van ’n klein dorpie is 100 000. Die populasie groei teen
3% per jaar. Skryf ’n vergelyking vir populasie (P) teenoor tyd (t).
(1)
Question 9 / Vraag 9
9.1 Determine AB if it exists: / Bepaal AB as dit bestaan:
0
 1

2
5
A
 5 3

 7 1
9.2
3  2

6  1
5 3 

0 0 
 1

1
B
0

2
[12]
2

2
3

0
(4)
Solve for x1 , x 2 and x 3 by using Gauss Jordan elimination. /
Los op vir x1 , x 2 en x 3 met behulp van Gauss Jordan eliminasie.
x1  x 2  2x 3  2
2x1  x 2
 4
(8)
4 x1  x 2  5 x 3  2
6
Question 10 / Vraag 10
[14]
A soap factory manufactures two types of soap, type P and type Q. The factory can
only produce a maximum of 12 cases of P and 8 cases of Q per day. It takes 30
minutes for the plant to produce a case of P and 1 hour to produce a case of Q. The
plant operates for a maximum of 9 hours per day and can produce a maximum of 14
cases in total during this time. SOLVE GRAPHICALLY!
’n Seepfabriek produseer twee tipes seep, tipe P en tipe Q. Die fabriek kan alleenlik ’n
maksimum van 12 kaste van P en 8 kaste van Q produseer per dag. Dit neem 30
minute vir die aanleg om ’n kas van P te produseer en ’n uur om ’n kas van B te
produseer. Die aanleg funksioneer vir ’n maksimum van 9 ure per dag en kan in totaal
’n maksimum van 14 kaste produseer gedurende daardie tyd. LOS GRAFIES OP!
10.1 Write down the inequalities in terms of x and y to show the constraints on the
production. / Skryf die ongelykhede neer in terme van x en y om die beperkings
van die produksie te wys.
(5)
10.2 Sketch the system of inequalities on a graph. Show the feasible region clearly. /
Skets die stelsel van ongelykhede op ’n grafiek. Dui die gangbare gebied
duidelik aan.
(5)
10.3 If the profit on P is R 60 per case and on Q is R 80 per case, write down an
equation which expresses the profit. / As die wins op P R 60 per kas is en op Q
R 80 per kas is, skryf ’n vergelyking vir die wins.
(1)
10.4 Determine the number of cases of each to give a maximum profit subject to the
constraints. / Bepaal die aantal kaste van elk wat ’n maksimum wins sal gee
onderhewig aan die beperkings.
(1)
10.5 If the profit on both types is R 80 per case, would the best option be different? /
As die wins op albei tipes R80 per kas is, sal die beste opsie verskil?
TOTAL: / TOTAAL: 101
(2)
Maximum: Maksimum [100]
7
WISN123
Question 1 / Vraag 1
1.1
C 
1.2
A 
1.3
C 
1.4
D 
1.5
B 
1.6
C 
1.7
B 
1.8
C 
1.9
A 
1.10 D 
1.11 D 
MEMO SECOND OPPOTUNITY
[22 ]
Question 4 / Vraag 4 [7]
4.1.1 (250;40) (400;35)
p  40 
40  35
(q  250 ) 
250  400

1
(q  250 )
30
1
145

p
q
30
3
p  40  
(3)
Or
(22)
(250;40)
(400;35 )
p  35 
Question 2 / Vraag 2
[4]
2.1 p(q)  20q 2  8000q  2500 .
35  40
(q  400 ) 
400  250

1
p  35  
(q  400 )
30
1
145

p
q
30
3
b
2a
 8000 

x
2 20 
x 
x  200
 200
(2)
2.2
p(200 )  20(200 ) 2  8000(200 )  2500 
p(200)  802500
(2)
 R802 500 
Question 3 / Vraag 3
29/11/2012
[3]
R  p  23p
4.1.2
p
(3)
1
120   145
30
3
 R43.33
4.2 p  21  0,7q

(1)
p  14  2,8q
21 0,7q  14  2,8q 
7  3,5q
q  2
2
p  21  0,7q
50  p 2  23p 
p  21  0,72
0  p 2  23p  50
0  ( p  25 )( p  2) 
p  25 or p  2
 R25
p  19,6
or
p  14  2,8q
p  14  2,82
p  19,6 
(3)
 R19,60
(3)
8
Question 5 / Vraag 5
[7]
Question 7 / Vraag 7
C (q )  4q 2  18q  2
ac
ab  a
abx  ax  ac 
x
5.1
2
 (4q  18q  2)dq 
a(bx  x )  ac
bx  x  c 

(2)
4
(2) 3  9(2) 2  2(2)  c 
3
2470,67  c 
log x( x  2)  log x 
3
4
3
 C (q )   q 3  9q 2  2q  2470 ,67  (5)
x  2x  x
x  x 2  2x  0
x( x 2  x  2)  0
x x  2x  1  0 
2
4 3
q  9q 2  2q  c 
3
2500  
log x  log( x  2)  3 log x
log x( x  2)  3 log x 
5.2
[5]
3
3
Question 8: / Vraag 8:
x  2
12 000  Pe 0,055(7) 
12 000
(5)
Question 6/ Vraag 6
C(q )  7920  3q 
(1)
6.2
R(q)  5q 2  q 
(1)
6.3
7920  3q  5q2  q 
5q  2q  7920  0 
(q  40)(5q  198)  0
198
5
 40 spesial paving bricks
(3)
R (q )  10q  1 
(1)
6.4
8.2 pord 
110000 
2
or q  
P
e 0,055 (7 )
 P  8165,4076
 Present value is R 8165,41
[9]
6.1
q  40
A  Pe rt
8.1
x  0 or x  2 or x  1
n/a
n/a
[17]
6.5 R (10 )  10(10 )  1
R (10 )  101 
If the company manufacture the 11th
Special paving brick, the revenue will be
R101 more. / As die maatskappy die 11de
Spesiale plaveisel steen vervaardig sal
die inkomste R101 meer wees.
(3)
(3)
x [1  (1  i )  n ]
i
1500[1  (1 
15 n
) ]
1200

15
1200
1500[1  (1,0125 )  n ]
110000 
0,0125
1
110000  0,0125 
 1  (1,0125 ) n
1500
1375

 1  (1,0125 ) n
1500
1375
1
 (1,0125 ) n
1500
 1375 
ln1 
  n ln(1,0125 ) 
 1500 
 1375 
ln1 

 1500   n

 ln(1,0125 )
200,0324129  n 
 200 full payments
Last payment R1500  0,0324129
=R48,62
(7)
9
x [(1  i ) n  1]
i
6
x[(1 
)10  1]
100

2
30000 
6
200


10
x[(1,03 )  1]
30000 
0,03
30000  0,03
x
[1,03]10  1
8.3
x  2616,9158
 R 2616,92
8.4
8.5
9.2
Ford 
 1

 2
 4
(3)
P  ba t
Question 9 / Vraag 9
9.1
0
 1

2
5
A
 5 3

 7 1
 5

1
AB  
 14

 8
3  2

6  1
5 3 

0 0 
11

32 
11

12 
1
5
(1)
[10]
 1

1
B
0

2
2

2
3

0

 1

R2  3
 0


 0

2 

 4
2 
1 2
2

3  4  8
3  3  6
 1

 4R1  R 3  0

 0
i  (1 
3 t
P  100000 (1 
)
100
P  100000(1,03)t 
2
0
2R1  R 2
(3)
i nm
) 1
m
0,12 12
i  (1 
)  1
12
i  0,126825 
12,68% 
1
1
1
2
1
4
3
3
3

 1

3R 2  R 3
 0


 0

1
2
1
4
3
0
1
 1 1

0
1
4R 3  R 2
 0
0
0
 1

1
R 2  R1   0
 0
0
x1  2 , x 2  0 and x 3
2R 3  R1

2 
 8 

3 

 6

0
0
1
0
0
1

2 
 8 

3 

2 

 2

0  
2 
 2

0 
2 
 2
(8)
(2)
10
Question 10/ Vraag 10
[14]
10.1 Soap P: x
Soap: Q y
x  12 
y  8
1
x  y  9
2
x  y  14

x, y  0

x, y  Z
(5)
10.2
Y
x  12
14


x  y  14
9

( 2,8 )
y  8
Soap Q
8
(10,4 )

1
(12,2 )
0
12
14
2
xy 9
X
18

Soap P
10.3 P  60 x  80 y 
(1)
10.4 10 cases of P and 4 cases of Q. 
(1)
10.5 P  80 x  80 y
12 cases of P and 2 cases of Q  yes
(2)
(14)
11