Solution++MaCNVC010+K1+Algebra+&+Functions

8/3/2019 Solution++MaCNVC010+K1+Algebra+&+Functions

1/17

Suggested Solutions Test MaC1NVC10 K1 Algebra & Functions NV-College

[email protected] for sale. Free to use for educational purposes 1

Test MaC1NVCO10 on Algebra and Functions:

Quadratic Equations, Exponential Equations, Logarithms, Simultaneous Equations

Instructions

Test period Oct 6, 2011; 13:00-15:35

Part I : You are not allowed to use calculator in part I.

Part II: You may use calculator in part II. You are not allowed to share yourcalculator during test with another student. Phones are not allowed to be

used as a calculator.

Resources Formula sheet, your personalised formula booklet, ruler, protractor and a

graphic calculator.

The test For all items a single answer is not enough. It is also expected

that you write down what you do

that you explain/motivate your reasoning

that you draw any necessary illustrations.

Score and The maximum score is (30G/36VG/37VG) points depending on the

mark levels level of problems selected to solve)

The maximum number of points you can receive for each solution is indicated after each

problem. If a problem can give 2 Pass-points and 1 Pass with distinction-point this is

written [2/1]. Some problems are marked with, which means that they more than other

problems offer opportunities to show knowledge that can be related to the criteria for Pass

with Special Distinction in Assessment Criteria 2000.

Lower limit for the mark on the test

G: Pass: 20 points

VG: Pass with distinction: 30 points (minimum of 15 VG points) or 25 VG.

MVG: Pass with special distinction: 30 points (minimum of 25 VG points) and .Prob. 1a 1b 1c 1d 1e 1f 2a 2b 2c 2d 2e 2c 2d 2e

G 2 3 1 2 1 2 2

VG 2 3 1 1 3 2 2 2 2

MVG

Prob. 2f 2g 2h 3a 3b 4a 4b 5 6 7 8a 8b 9 10

G 2 2 2 4 1 2

VG 3 3 2 2 4 4 3 4

MVG

Prob. 11a 11b 11c 12 13a 13b 13c 13d 14 sum limits

G 1 1 2 30 20

VG 4 1 1 1 1 4 36 30/(V15)

MVG 37 30/(V25)

Prob. 1a 1b 1c 1d 1e 1f 2a 2b 2c 2d 2e 2c 2d 2eG 2 3 1 2 1 2 2

VG 2 3 1 1 3 2 2 2 2

MVG

Prob. 2f 2g 2h 3a 3b 4a 4b 5 6 7 8a 8b 9 10

G 2 2 2 4 1 2

VG 3 3 2 2 4 4 3 4

MVG

Prob. 11a 11b 11c 12 13a 13b 13c 13d 14 sum limits Grade

G 1 1 2

VG 4 1 1 1 1 4

MVG


2/17



Note that most of problems have alternative. You will solve only one of those problems. In

case you solve both alternatives, you will get best of two alternative grades, but not both.

Part I: No calculator may be used in this part.

The solution for the part I must be written on a separate paper. You may start working on part II without the access to any calculator. Only after submission of the solution for the part I you may use your

calculator.

1. Simplify the following expressions as far as possible:a) (G-alternative to 1b)

36

36122

2

x

xx[2/0]

or

b)

(VG-alternative to 1a)22

22

11

121

yx

yxyx

[0/2]

Solution 1a: (G alternative)

6

6

66

66

66

6

36

36122

2

2

x

x

xx

xx

xx

x

x

xxAnswer:

6

6

36

36122

2

x

x

x

xx

Solution 1b: (VG alternative) Answer:xy

xy

yx

yxyx

22

22

11

121

xy

xy

xy

xy

xy

xy

x

x

yy

y

x

x

x

yy

y

x

yx

yx

yxyx

yx

yx

yxyx

11

11

11

11

1111

11

11

1212

22

22

c) (G-Alternative to 1d)x

x

x

x 2

2

[3/0]

d) (VG-Alternative to 1c)y

x

x

y

y

x

x

y

y

x[0/3]

Solution 1c: (G-alternative)

xxxx

xx

x

xx

xxx

xx

xxx

xx

xx

x

x

x

x

x

x

x

x

x

x

x

x

2

2

22

2

42

2

42

2

42

2

2

2

22

2

2

2

22

2222

Answer:xx

x

x

x 22

2


3/17



Solution 1d: (VG-alternative)First method:

32

112

1

16

52

1

2

1

8

3

2

1

2

1

2

1

4

1

2

1

2

1

2

1

2

1

2

1

y

x

x

y

y

x

y

x

x

y

y

x

x

y

y

x

x

y

y

x

y

x

x

y

y

x

x

y

y

x

y

x

x

y

y

x

x

y

y

x

Answer:32

11

y

x

y

x

x

y

y

x

x

y

y

x

Using:

4

12

1

2

12

1

2

1

x

y

x

y

y

x

x

y

832

1

432

1

41

2

1

2

1

21

y

x

y

x

x

y

y

x

y

x

x

y

y

x

16

52

1

8

52

1

8

3

2

1

2

1

2

1

2

1

x

y

x

y

y

x

x

y

y

x

x

y

y

x

x

y

32

112

1

16

112

1

16

5

21

2

1

2

1

2

1

2

1

y

x

y

x

x

y

y

x

y

x

x

y

y

x

x

y

y

x


4/17



Second method:

32

11

32

10

32

21

16

5

32

21

16

14

32

1416

16

1

4

1

32

1

8

1

2

1

32

1

16

1

8

1

4

1

2

1

y

x

x

y

y

x

x

y

y

x

x

y

y

x

x

y

y

x

y

x

x

y

y

x

x

y

y

x

y

x

x

y

y

x

x

y

y

x

Answer:32

11

y

x

y

x

x

y

y

x

x

y

y

x

e) 125log81log29log5 593 [1/1]Suggested solution 1e:

934105log39log223log25

5log9log23log5125log81log29log5

593

3

5

2

9

2

3593

Using 1log aa , and ana bn

b loglog . Answer: 9125log81log29log5 593

f)

10000log

1000log100log

4

2

2 x

xx [2/1]

Suggested solution 1f:

xxxx

xxxx

xx

log494log4log23log22

10logloglog10loglog10log10000

log1000

log100log442322

4

2

2

Using 110log , xnxn loglog , baba logloglog and bab

alogloglog

.

Answer: xxx

x log4910000

log1000

log100log4

2

2


5/17



2. Solve the following equations. Show the detail of your solutions. Give the answers inthe surd (exact) form.

(a) (G-alternative to 2b) 0652 xx [1/0](b)(VG-alternative to 2a)

3

101 x

x 0/3]

Solution 2a: (G-alternative)

303

2020320652

xx

xxxxxx Answer: 3,2 21 xx

Solution 2b: (VG-alternative)

013

10

3

101

3

101

3

101 22

xxxxxx

xxx

x

0136

100

6

10

016

10

6

10

013

10222

2

xxxx

36

64

6

100

36

36100

6

1001

36

100

6

10222

xxx

36

18

6

8

6

10

6

8

6

10

3

1

6

2

6

8

6

10

6

8

6

10

6

8

6

10

36

64

6

10

36

64

6

102

xx

xx

xxx

Answer: 3,3

121 xx

(c) (G-Alternative to 2d, 2e) 00 0150 05

1

x [2/0]

(d)(VG-Alternative to 2c, 2e) xx 7532 [0/2]

(e) (MVG-Alternative to 2c, 2d) 049log3loglog 75727 xx [0/2/]Solution 2c: (G-alternative)

04.0251

51

51

51

51

2

50 0

00 0150 0

100 01

50 0

150 0

00 01

50 0

xxxxx

Answer: 04.025

1x

Solution 2d (VG-alternative):

5

2log

3

7log

5

2

3

7

5

2

3

7

3

7

5

2

35

75

35

327532

xx

x

x

x

x

x

x

x

xxx


6/17



3log7log

5log2log

3

7log

5

2log

5

2log

3

7log

5

2log

3

7log

xxx

x

Answer:3log7log5log2log

x

Solution 2e (MVG-alternative):

049log3loglog 75

7

2

7 xx

07log3log5log049log3loglog 2772

77

5

7

2

7 xxxx

Using 67log237log349log3 72

77 , 17log7

We may make change of variables to make the problem more clear:tx 7log

03

0203206507log23log5log 2772

7t

tttttxx

34373log303

4972log202

3

7

2

7

xxtt

xxttAnswer:

343

49

2

1

x

x

(f) (G-alternative)7

5

5

7

x

x[2/0]

(g)(VG-alternative) 7log54log54log xx [0/3](h)(MVG-alternative) 5log610 xx x [0/3/]

Suggested solution 2f:7

5

5

7

x

x

757577557

5

5

7 22

xxxxx

x

125775

25775

xxx

xxx

Answer:

12

2

2

1

x

x

Suggested solution 2g: 7log54log54log xx Note the domain of x :

5

44554054 xxxx

5

445054 xxx

Answer: Domain of x is:5

4

5

4 x

7log5454log7log54log54log xxxx


7/17



725167log2516log 22 xx

5

3

25

9

25

99252571672516

2222 xxxxx

Answer:

5

3

5

3

2

1

x

x

Note that both answers

5

4

5

3

5

4

5

3

2

1

x

x

are in the domain ofx :

5

4

5

4 x

Suggested Solution 2h: 5log610 xx x

06506log5log6log5log

10loglogloglog10

loglog10

10

222

65

6

5log

6

5log5log6

ttxxxx

xxxx

xx

xxx xxx

1000103log303

100102log202032065

3

2

2

xxtt

xxtttttt

Using:

xx log3log

3

, 2100log , xnxn

loglog which implies xxx x logloglog log .Lets change the variable and rename: tx log

Answer: 1001 x and 10002 x

Check:

100

1010100

101010100101001010

100

10 105255

10462610 0log6log6

5log6

x

x

x

x

xx

x

x

Checks.

1000

10101000

10101010001010001010

1000

10 155355

1596361000log6log6

5log6

x

x

x

x

xx

x

x

OK!


8/17



Solve only one of problems#3 or #4. In case you solve both, you will receive grade only

for one of them:

3. (G-Alternative to problem #4) The graph of a quadratic function xf is plotted inthe figure. As illustrated in the figure The function crosses y-axis at 4,0 , and x-axis at 0,2 and 0,3 .a) Find analytically the equation of the quadratic function xf . [2/0]b) Find algebraically the coordinates of the minimum point of the function. Reading

from the graph is not accepted. [2/0]

-6

-4

-2

0

2

4

6

-4 -3 -2 -1 0 1 2 3 4 5

Suggested solution #3:Lets 21 xxxxAxf Functions roots are 0,2 and 0,3 . Therefore, 21 x and 32 x :

3232 xxAxfxxAxf The function passes 4,0

3

2

6

44630204 AAA Answer: 32

3

2 xxxf

Minimum point of the function lies on the symmetry line of thefunction which is:

2

1

2

32

2

21 xx

xsym

6

25

2

61

2

41

3

23

2

12

2

1

3

2

2

132

3

2

fxxxf

Answer: The coordinates of the minimum point of the function is

6

25,

2

1.


9/17



4. (VG-Alternative to problem #3) The graph of a quadratic function xf is plotted inthe figure. As illustrated in the figure The function crosses y-axis at 3,0 , and istangent to the x-axis at 0,2 .a) Find analytically the equation of the quadratic function

xf . [0/2]

b) If cbxaxxf 2 find conditions for the constants cba &,, such that thefunction has: [0/2]

i. two different solutions (roots.)ii. one double solution (root.)

iii. no real solution.Note that function cbxaxxf 2 has

a

acbbx

2

42 as its roots.

-2

0

2

4

6

-1 0 1 2 3 4 5

Suggested solution #4:Lets 2111 xxAxxxxAxf Functions double root is 0,2 .Therefore, 221 xx :

221 2 xAxfxxAxf The function passes 3,0 :

4

32032

22 aAxAxf Answer: 22

4

3 xxf

Second method: cbxaxxf 2

The function passes 3,0 : 330002

ccbaf

The function passes 0,2 : 032403222 2 babaf Due to the fact that the function has only one double root, 042 acb

12344404

22222 b

ab

ac

babacacb

030960323

03212

4032422

22

bbbbb

bb

ba

2222

24

333

4

3

4

3

12

3

123 xxfxxxf

bab

The function cbxaxxf 2 hasa

acbbx2

42 as its roots.


10/17



Depending on the value of acb 42 the function has two, one or notany real roots:

i. If 042 acb , the function has two different real roots.ii. If 042 acb , the function has one real double root.iii.

If 04

2 acb , the function does not have any real root.

Solve only one of problems#5, #6 or #7. In case you solve more than one problem, you

will receive grade only for one of them:

5. (G-alternative to problems 6 and 7) Solve the following simultaneous equations:

12loglog2

1log2log

yx

yx[4/0]

6. (VG-Alternative to problems 5 and 7) Solve the following simultaneous equations:

14log

27log

7

3

y

x

yx

[0/4]

7. (MVG-Alternative to problems #5 & 6 V1) The equation 032 3log2log xxx

has a rational root. Find it in the form q

p

where p and q are integer numbers.

(MVG-Alternative to problems #5 & 6 V2) The equation

0117 11log7log xxx

has a rational root. Find it in the formq

pwhere p and q are integer numbers.

[0/4/]

Suggested Solution #5:

12loglog2

1log2log

yx

yx

We may multiply both sides of the second equation by 2 to eliminatey :

5105log25log524log2log4

1log2log

12loglog2

1log2log

2

xxx

yx

yx

yx

yxadd

Substitute 510x in the second equation to calculate y .

12log1012log10log5212log10log212loglog2

10 55

yyy

yx

x

2102loglog121012log10 yyyy Answer: 510x , 210y


11/17



Suggested solutions # 6 Answers: 211

10x , 27

10y ;& 211

10x , 27

10y

First Method:

2

9

7

3

10

10

27

14log

93

27log

14log7

27log3

14log

27log

y

xyx

y

x

yx

y

x

yx

y

x

yx

2

112

2

7

2

7

2

2

2

7

2

7

72

92

2

10101010

10

1010101010

10

x

yx

yyyyy

yx

Second Method Prob. #6:

2loglog

9loglog

27

14log

93

27log

14log7

27log3

14log

27log

7

3

yx

yx

y

x

yx

yx

yx

y

x

yx

2

11

102

11log1129log2

2loglog

9loglog

xxx

yx

yxadd

Substitute 4log x in 5loglog yx :

2

7

2

4112

2

11log2log

2

112log10log

2loglog

10 211

2

11

yyyyx

x

Answers: 211

10x , 27

10y

Note: We may note that 211

10x , 27

10y are also simultaneous

solutions of the simultaneous equations. This set of answers may beunnoticed and therefore missed in this method.


12/17



Suggested Solution Prob. #7V1: 032 3log2log xxx Take logarithm of both sides:

3log2log 3log2log xx

xx 3log3log2log2log

xx log3log3loglog2log2log

xx log3log3loglog2log2log 22

xx log2loglog3log3log2log 22 2log3loglog3log2log3log2log x 3log2loglog3log2log3log2log x

3log2log3log2loglog

3log2log

3log2log3log2log

x

6

1loglog

6

1log6log6log32log3log2loglog 1 xx

Answer:6

1x

Suggested Solution Prob. #7V2: 0117 11log7log xxx Take logarithm of both sides:

011log7log 11log7log xxx xx 11log11log7log7log

xx log11log11loglog7log7log

xx log11log11loglog7log7log 22

xx log7loglog11log11log7log 22 7log11loglog11log7log11log7log x 11log7loglog11log7log11log7log x

11log7log11log7loglog

11log7log

11log7log11log7log

x

77

1log77log77log117log11log7loglog

1x

Answer:

77

1x


13/17



Part II:

You should submit your solutions to part I problem before having

access to your calculator!

Note: Most problems come in a group of two or three alternative

problems. Choose only one of problems in the group to solve.



8. (G-alternative to problems 9 and 10) xxxf x log5232 3 i. Find 1f [1/0]

ii. Find the domain and range of the function. Why? Explain. [2/0]9. (VG-Alternative to problems 8 and 10) 532 xxxf . Find and simplify as far as

possible: hfhf 22

. [0/3]

10.(MVG-Alternative to problems #8 & 9) caxxf 2 is given. Determine theconstants a and c , if 15248 24 xxxff [0/4/]

Suggested Solution prob. # 8: xxxf x log5232 3

80621log523121 13 f The domain of the function is all positive real numbers, i.e. 0x . This isdue to the fact that logarithm is defined only or positive real numbers.

The range of the function is all real numbers larger than -492.Even though xxxf x log5232 3 is an increasing function ofx , butfor small 10 x , the logarithm term increases negatively.

Suggested Solution prob. # 9: 532 xxxf

35364452322

356452322

222

2

hhhhhhhhf

f

1

13322 2

h

h

hh

h

hh

h

fhf

Suggested Solution prob. #10: caxxf 2 , 15248 24 xxxff

ccaxaxffcxafxffcaxxf 2222 caccxaxaccacxxaaccaxaxff 22243224222 22

1533215215

38

242442242

28

15248

2

222

2

3

24

22243

cccac

ccca

aa

xxxff

caccxaxaxff

Answer: 3,2 ca


14/17





11. (G-alternative to problems 9 and 10) The formula krP t82.0000375 describesthe value of a car which decreases as a function of time (year).a) What is the price of the car as new? [1/0]b) What does 82.0 as the base represent? [1/0]Calculate after how many years the value of the car is reduced to kr000150 . [2/0]

Suggested Solution #11:a) The car costs kr000375 as

new.b) 82.0 as the base of the

exponential means that thecars value is decreased at

the rate of %18 per year.c) krP t82.0000375

krt82.0000375000150

t82.0375150

375

15082.0

t

yearsyearsttt

56.482.0log

375

150log

375

150

log82.0log375

150

log82.0log

Answer: The value of the car is dropped to 150 000 kr five years afterits purchase at 375 000 kr.


15/17



12. (VG-Alternative to problems 11 and 12) An archeologist believes that he has foundthe remnant of Noahs Ark. According to bile Noahs Ark was build 4000 BC. How

many percent of the normal amount of C-14 should be expected in the ark? Half-life

of C-14 decay is years5730 . [0/4]

Suggested Solution prob. # 12

73 05

0

73 05

1

173 0573 05

00

0 2222

1

2

t

tNNaaaN

NaNN

73 05

0 2

t

NN

4000 BC is about 6000 years ago,i.e. yearst 0006

073 050 48.02 NNN

t

Answer: Only 48% of the normalvalue of C-14 must be left in theark!


16/17



13. (MVG-Alternative to problems #11 and 12) [0/4/]In the begging of year 1900 a laboratory got mg0.65 Radium-226 which is

radioactive with half-life of 1600 years.

a) Set up a mathematical which demonstrates how Radium decays as a functionof time. [0/1/]

b) Show that the amount of Radium mgy left over after yearsx may beexpressed as T

x

yy

20 [0/1/]

where0

y is initial amount of the Radium , and T is the half-life of Radium.

c) Calculate the amount of the Radium year 2000. [0/1]d) Which year the amount of Radium in the laboratory is decreased by 25%?

[0/1/]

Suggested Solution prob. .# 13

a) xaCy where mgC 0.65 is the initialamount of the Uranium (It is thevalue of y at 0x : i.e.

mgCaC 0.650.65 0 . The half-

life Uranium 226 is 1600 years.This means that after

yearsx 6001 , mgC

y 5.322

0.65

2

6001

0.655.32 a

1600

1

60 01

2

1

2

1

aa

Answer: mgy

x

1600

2

10.65

T

xxx

yymgymgy

220.65

2

10.65 0

16001600

QED

b) Year 2000, yearsx 10019002000 mgymgy 2.62

2

10.65

1600

10 0

Answer: mgy 2.62

c) 75.0log2

1log

2

175.0

2

10.65

160016001600

xxx

CCmgy

2

1log

75.0log160075.0log1600

2

1log75.0log

2

1log

1600xx

x

yearsx 664

2

1log

75.0log1600

Answer: yearsx 664 , Year 2675.


17/17


behzad massoumzadeh@huddinge se Not for sale Free to use for educational purposes 17

14.Newton's Law of Cooling states that the rate of change of the temperature of an objectis proportional to the difference between its own temperature and the ambient

temperature (i.e. the temperature of its surroundings).

Let us assume that T is the temperature of the body and0

T is the temperature of the

surrounding such that TT 0 . If the surrounding temperature 0T is constant and theventilation is good, the cooling takes place such that the temperature difference

0TTD is a decreasing exponential function of time.

A high ranked secret police clerk is found death in her well air-conditioned office. The

body temperature of the death secret police was C5.31 at the time of discover of thebody at 00:14 . Two hours and fifty minutes later, at 50:16 the temperature of the

body was dropped to C5.27 . The temperature of the office was C0.21 . Estimate thetime of the murder. Normal body-temperature is C0.37 . [0/4/]

Suggested Solution prob. #16.

0TTD taCTD 0.21

17 00.215.27

0.215.31

aC

aCt

Where t is the time in minutes

counted from 00:14 .i.e.: we may

set 0t at 14:00. Therefore:

5.100.215.31 CaC t 1705.100.215.27 a

5.65.10 170a

21

13

105

65

5.10

5.6

5.10

5.10 17 0

a

17 0

1

17 0

21

13

21

13

aa

17 0

1

21

13

a

Newtons cooling temperature law applied to estimate the time of death

of the secret police:17 0

21

135.100.21

t

T

2132

105160

5.100.16

21130.16

21135.10

21135.100.210.37

17 017 017 0

ttt

min149min

21

13log

21

32log

17021

32log

21

13log

17021

32log

21

13log

17 0

t

tt

Answer: The secret police was murdered 149 minutes before 14:00. Shewas murdered at 11:31 AM.

Date post:	07-Apr-2018
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Solution++MaCNVC010+K1+Algebra+&+Functions

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