Suggested+Solutions+NPMaDHT2000

8/3/2019 Suggested+Solutions+NPMaDHT2000

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Suggested solutions NPMaDHT2000 NV-College, Sjdalsgymnasiet

NATIONAL TEST IN MATHEMATICS COURSE D

Mathematics D

Autumn 2000This test will be re-used and is therefore protected by chapter17 paragraph 4 of the OfficialSecrets Act. The intention for this test is to be re-used until 2010-12-31.This should be considered when determining the applicability of the Official Secret Act.

Ume University

Department of Educational

Measurement

Skolverket

Students Name

(National Agency for Education)

Skolverket [email protected] to use for educational purposes. Not for sale. 1/19


2/19


NATIONAL TEST IN MATHEMATICS COURSE D AUTUMN 2000

Directions

Test time 240 minutes without a break.

Resources Graphic calculators and Formulas for the National Test in MathematicsCourses C, D and E.

Test material The test material should be handed in together with your solutions.Write your name, the name of your education program/ adult education on allsheets of paper you hand in.

The test The test consists of a total of 16 problems.

For some problems (where it says Only answer is required) it is enough togive short answers. For the other problems short answers are not enough. Theyrequire that you write down what you do, that you explain your train ofthought, that you, when necessary, draw figures. When you solve problemsgraphically/numerically please indicate how you have used your resources.Problem 16 is a larger problem which may take up to an hour to solvecompletely. It is important that you try to solve this problem. Adescription of what your teacher will consider when evaluating your work isattached to the problem.Try all of the problems. It can be relatively easy, even towards the end of thetest, to receive some points for partial solutions. A positive evaluation can begiven even for unfinished solutions.

Score and The maximum score is 48 points.mark levels

The maximum number of points you can receive for each solution is indicatedafter each problem. If a problem can give 2 Pass-points and 1 Pass withdistinction- point this is written (2/1). Some problems are marked with,which means that they more than other problems offer opportunities to showknowledge that can be related to the criteria for Pass with SpecialDistinction in Assessment Criteria 2000.Lower limit for the mark on the test

Pass (G): 14 points Pass with distinction (VG): 26 points of which at least 7 Pass withdistinction- points.



3/19


Regarding the suggested solutions:Please note that you have to t r y to so l ve th e p rob lems you rse l f before checking your solutions against mine.

To solve a problem you may follow the following necessary steps:

Read the problem.

Think!

Read the problem again!

Have you understood what the problem is looking for?

Make a plan.

Perform the plan.

Evaluate your results.

Are your results logical and acceptable?

Obviously, if you are stock and do not know how to solve the problem youmay read my solutions. But, after reading the solutions, even if you aresure that you understood my solutions, you should try to solve theproblem by yourself without checking your steps against mine.Only, after you solve the problem yourself, you may have understoodhow to solve similar problems (but not necessarily another type.) The

ability to solve problems increases by training to think logical, solve moreproblems by yourself or in group, and discussing the problem in a grouppreferably at the presence of a pedagogue mathematician.

My solutions are just suggested ones. Usually there are more than onemethods of solving a given problem.

Warning: Just reading the solutions can never replace your own strugglein solving a given problem. By just reading the solutions you may not beable to understand the mathematics of the problem deep enough andtherefore, it will not help you to solve a new problem by yourself.

Your comments and possible corrections are deeply appreciated.

Behzad Massoumzadeh, Ph.D.

Sjdalsgymnasiet; Paradistorget 4;

141 85; Huddinge; Sweden

[email protected]

January 2011

mailto:[email protected]:[email protected]:[email protected]


4/19



P.S. I would appreciate your comments greatly. Please inform me of anypossible error as soon as possible so I correct it immediately! Behzad

1. Give the primitive functions ( )xF to ( ) 10010 2 += xxf

Only an answer is required [2/0]

Suggested solution: Answer: ( ) CxxxF ++= 1003

10 3 [1/0]

( ) ( ) ( ) CxxxFCxxxFxxf ++=+++

=+= + 1003

10100

12

1010010 3122 [1/0]

2. The figure below shows a unit circle.

a. Find vsin


b. Find ( )v

180sin Only an answer is required [2/0]

Suggested solution:

a)Answer: 8.04

sin = [1/0]5

=v

b)Answer: ( ) 8.05

4sin = [1/0]sin180 == vv


5/19



3. The graph to the linear function ( )xf isdrawn in the figure. Find the primitive

function to ( )xf [2/0]

Suggested solution:

The y-intercept of the linearfunction is( )xf 3=m , and the

slope of the function is2

3=k .

Therefore ( ) 32

3= xxf .

The primitive function of is:( )xf

( ) Cxx

xF += 322

3 2 [1/0]

According to the figure

( ) 32

322

0

=/

/= dxxf

( ) ( ) ( )022

0

FFdxxf =

( ) ( ) 3632324

302 2 === FF

Answer: ( ) CxxxF += 34

3 2 [1/0]

4. Calculate the integral by using a primitive function. [2/0]( ) 4

1

34 dxx

Suggested solution: Answer: [1/0]( ) 75.5144

1

3 = dxx

( )

=

=

+=

+

41

144

444

44

1344

444

1

44

1

134

1

3 xx

xxdxx [1/0]

( ) ( ) 75.5175.3484

1464164

4

1

3 ==

= dxx


6/19



5. In the triangleABCis cm , cmAB 4.36= AC 2.25= and the angle = 0.120C .

What is the length of side BC? [3/0]

Suggested solution:Data: , ,cmcAB 4.36= cmbAC 2.25= = 0.120C ; ?= aBC

We may use the sine rule to solve the problem:SinC

c

SinB

b

SinA

a==

We may first calculate the angle B using the sine ruleSinC

c

SinB

b= , and

therefore the A . Then may either use the sine rule, or cosine rule tocalculate the length of the side aBC .

( )

( 600.0sin600.04.36120sin2.25

sin

sin

sin

1

=

)

=

== BBc

Cb

BSinC

c

SinB

b

= 16.230.1208.361808.36 AB [1/0]

First method:

Now, we may use the cosine rule Acbcba cos2222 += to calculate theside :aBC

( ) ( ) ( ) ( ) ( +=+= 16.23cos4.362.2524.362.25cos2 222222 aAcbcba )

( ) ( ) ( ) ( ) ( ) 3.27316.23cos4.362.2524.362.25 222 =+=a [1/0]

Answer: cmaBC 5.163.273 = [1/0]

Second method:

We may instead use the sine ruleSinC

c

SinA

a= to calculate the side :aBC

( ) ( ) cmaaC

Aca

SinC

c

SinA

a5.16

0.120sin

16.23sin4.36

0.120sin

16.23sin4.36

sin

sin

=

===

Answer: cmaBC 5.163.273 = [1/0]

6. The function ( )xf is defined by ( ) xexxf =

a) Find ( )xf [1/0]

b)Solve the equation ( ) 0= xf . [1/0]

Suggested solution: Answer: ( ) ( ) xexxf += 1 ; 1=x

a) ( ) ( ) ( ) ( ) xxxx exxfexexfexxf +=+== 11 [1/0]

b) ( ) ( ) ( ) 101010 ==+=+== xxexxfxf x [1/0]


7/19


7. The graph to the function ( )xfy = isreproduced in the figure below.

a. Which of the graphs in figures A-D

is the best representation of the

derivative of the function ?


b. Motivate your answer. [0/2]

Suggested solution:

Answer: Graph C. [1/0]

The function has a local

maximum at , therefore

, and

( )xf0=x

( ) 00 =f ( ) 00


8/19



8. The graph to the three functions ( )xf , ( )xg ,and ( )xh are plotted in the accompanying figure.

a. Calculate the value of the integral

( ) ( )( )dxxfxg 4

0

only an answer is required. [1/0]

b. By using integrals write down an expressionfor the area of the shaded region in thefigure.

only an answer is required. [0/1]

a. )dxxf is the area of the shaded

region in the figure. The value of theintegral may be calculated using a simplegeometry:

( ) ( )( xg 4

0

( )auArea 12

2

439=

=

Answer: ( ) ( )( ) audxxfxg 124

0

= [1/0]

Answer: ( ) ( )( ) ( ) ( )( ) audxxhxgdxxhxfArea +=

7

4

4

0 [0/1]


9/19



9. Show that xx ee is a solution to the differential equation xe . [1/1]y += 33 yy = 43

Suggested solutions:

xx eey += 33

( ) xx eey += 133 3 xx

eey= 39 [1/0]

We may substitute by

and

yxx

eey+= 33

yby xx eey = 39 in the equation

below( )xxxx eeeeyy += 33 3393

xxxxeeeeyy

= 3993 33

xeyy

= 43 QED

[0/1]

10. In the equation =a

dxx

1

1

3

2where 1>a .

a. Calculate a . [2/0]

b. The integral in the equation may be interpreted as an area. Draw a figure thatrepresents this area. [0/2]


a. 13

1

323

21

3

2 22

1

2

1

==/

/

=ax

dxx

aa

3

4

3

13

3

1

1313

1

3

22

=

+

=+==

aa

243

4

3

22

=== aaa

[1/0]

Answer: a [1/0]2=

b. 2

13

2dx

xmay be interpreted as the

area bounded by the curve of

the function 3

2x

y=

, x-axis, andthe lines 1=x , and 2=x . Thecolored area in the figure above

is an interpretation of 2

13

2dx

xas

an area. [0/1]

Since 13

22

1

= dxx

, the area of the

region is necessarily 1. This may beindependently verified. Using a

simple geometry, the area of thetrapezoid is:

auArea 12

2

23

6

23

4

3

2

===+

= QED

[0/1]


10/19



11. The angle is indicated in the figure.Calculate an exact value of vcos .

[0/2]


We may first calculate thehypotenuse of the triangle, usingPythagoras theorem:

139432 22 =+=+=c

( )13

132

13

2180cos == v [0/1]

( )v= 180coscos

Answer:13

132cos = [0/1]

12. Linus is moving along a 60 m long line. To be able to describe where Linus is on the line,the line has been graded from -30 to 30 as can be seen in the figure below:

Linus starts at time . His position

on the line is determined by the

time according to the equation

.

0=t( ) mtx

st

( ) ( ) ( )tttx = 62 2

a. Where on the line is Linus at the time

0=t ? only an answer is required.[1/0]

b. Find an expression for Linus velocityat the time t. [0/2]

c. When Linus turns the velocity is zero.At what times does this happen? [1/0]


a. ( ) ( ) ( )tttx = 62 2

( ) ( ) ( ) 246406200 2 ===x

Answer: [1/0]( ) mx 240 =

( ) ( ) [ ] 031420 == tttv

b. ( ) ( ) ( )tttx = 62 2

( ) ( ) ( ) ( ) ( )12622 2 += ttttx

( ) ( ) ( ) ( )[ ]2622 = ttttx [0/1]

( ) ( ) ( ) [ ]22122 +== ttttxtv

Answer: ( ) ( ) [ ] smtttv /3142 = [0/1]c. ( ) ( ) [ ] 031420 == tttv

Answer: st 2= and s7.43

14=t

[1/0]

30 20 10 0 10 20 30

2

3


11/19



13. It seems as if the x-axis is a tangent line to

the curve )xsin at the origin(see the accompanying figure.)

(xy sin=

Find an expression for the derivative and

use that to investigate if the x-axis reallyis a tangent line to the curve at the origin.

[0/2]

Improved version [0/2/M2, M3, M5]

-1,5

-1,0

-0,5

0,0

0,5

1,0

1,5

-3,14 -1,57 0,00 1,57 3,14

x rad

y(x)

( )xxy sinsin =


( )xxy sinsin =

Lets make a change of variable:

=

==

xdx

dzxxz

zdz

dyzy

cos1sin

cossin

( )xzdx

dz

dz

dy

dx

dyy cos1cos ===

( ) ( xxxy sincoscos1 = )

)

[0/1]

If the x-axis is a tangent to the curve

of the function at theorigin, then both the function andthe tangent must have an identicalgradient at the point, i.e. they musthave identical . Due to the fact that

the slope of the x-axis is zero, wemust show that

( xxy sinsin =

y

( ) 00 =y share thepoint a

( ) ( )( ) ( ) ( )( )0sin0cos0cos10 =y

( ) ( ) ( )00cos110 =y

( ) 00 =y [0/1]

We must also check if the functionpasses through the origin, i.e. ( ) 00 =y

( ) ( ) ( ) 00sin00sin0sin0sin0 ====y

Therefore, x-axis is a tangent to thecurve of the function

at the origin.

( )xxy sinsin =

MVG-Quality: The student demonstrate the highest quality

(MVG) in solving problem 13 throughM2: Analyses and interprets the results,

concludes and evaluates if they are

reasonable.

Finds the derivative of the functionand shows that

( )( )

=

=

00

00

y

y

M3: Carries out mathematical proof, or

analyses mathematical reasoning.

Finds correctly the derivative of the

function ( ) ( xxxy sincoscos1 )=

M5: The presentation is structured, and

mathematical language is correct.

The presentation is structured, and



12/19



14. Given that )2cosx , show that 12(sin21 xy = sin2 = xy [0/2]


( )2cossin21 xxy =

( )xxxy 22 coscossin2sin21 += [0/1]( )xy 2sin121 =

xy 2sin221 +=

12sin2 = xy [0/1] QED

In the solutions above, we used thefollowing well acceptedtrigonometric relationships:

Trigonometric Unity: 1cossin 22 =+ xx

Double angle relationship:xx cossin22sin =

15. Some twenty kilometers east of Ystad, on the 42 m high ridge of Kseberga is a place calledthe Stones of Ale. The stone circle is 70 m long, 18 m wide and consists of 59 stones. Theshape of the stone circle has for a long time made people believe that it was a stone shipfrom the Viking Age. More recent research indicates that it might be a cult centre from theBronze Age.

The placing of the stones (see the figurebelow) can be assumed to follow twoopposite parabolas (= graphs toquadratic functions). Your task is to

a. find a suitable function to one of theparabolas. [0/3]

Find the function of the other parabola,

and plot your functions in a propercoordinate system. Analyze important

aspects of the function. [0/3/]

b. calculate the area enclosed by the stones.

[0/2]


13/19




a. The equation of the upperparabola may be given as:

( )( 99 += xxAy )

where A may be determined using

the information regarding the y-intercept of the parabola,i.e.: ( )m35,0 : [0/1]

( )( ) 8135

81909035 ==+= AAA

Answer: ( )( ) mxxy 9981

35+= [0/1]

Similarly, the equation of the lowerparabola may be written as

Answer: ( )( ) mxxy 9981

35+=

These functions may be plotted in aproper Cartesian-coordinates. [0/1]

Please note the scale; otherwise, itcould be misleading.

-42

-35-28

-21

-14

-7

0

7

14

21

28

35

42

-1

0

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

( )( ) mxxy 9981

35+=

( )( ) mxxy 9981

35+=

b. The area enclosed by the stones may be calculated using integration:

( )( )

+=

9

9

9981

352 dxxxArea ( )

=9

9

2 8181

70dxxArea [0/1]

9

9

12

811281

70

+

+= x

xArea

9

9

3

81381

70

= x

xArea

( )( )

( )( )

= 981

3

9981

3

9

81

7033

Area

( ) 284048648681

70mArea == Answer: [0/1]2840 mArea =

MVG-Quality: The student demonstrate the highest quality(MVG) in solving problem 14 through

M2: Analyses and interprets the results,

concludes and evaluates if they arereasonable.

Finds the equation of parabola based

on information regarding the y-intercept and x-intercept of thefunction.

M3: Carries out mathematical proof, or

analyses mathematical reasoning.

Finds correctly the equation of the

parabola, ( )( ) mxxy 9981

35+= and

plots it in a proper coordinate system

M5: The presentation is structured, and


The presentation is structured, and



14/19



When assessing your work your teacher will consider the following:

how close to a general solution you are

how well you present your work

how well you justify your conclusions

16. In this problem, your task is to investigate how large the area of the triangle ABCcan be.The first two points in this problem can be used as support for the investigation. You maychoose whether you want to do the general investigation (the third point) at once or if youwant to solve the problem step by step using all three points in sequence. [4/5/]

In the triangleABCsideAB is 3.00 cmlong and sideACis twice as long as side

BC.

Choose a value of the length of side

BCand calculate the area of the

triangleABCby first calculating the

angle C.

Find a value for the length of sideBCthat gives an area of the triangle larger than the one

you calculated in the previous point.

Investigate how large the area of the triangleABCcan be.


Data: xa , ,xb 2 cmc 00.3

First method (Generalization)

We may rename the sides as above.

We may combine the area rule,2

sinBcaArea

= , and the cosine rule,

Bacacb cos2222 += , to express the area of the triangle as a function ofx ,

where xa , and xb 2 . The basic requirement for the formation of any triangle

in this problem is , otherwise, no triangle may be formed

having sides

cmxcm 00.300.1


15/19



Bacacb cos2222 +=

( ) ( ) ( ) Bxxx cos00.3200.32 222 +=

Bxxx cos694 22 +=

222 3949cos6 xxxBx =+=

( )x

x

x

xB

/

/=

=

32

33

6

39cos

22

cmxx

xB 00.1

2

3cos

2

>

=

Use Trigonometric Identity:

1cossin 22 =+

BB 22 cos1sin = BB 2cos1sin =

22

2

31sin

=

x

xB

2

42

4

691sin

x

xxB

+=

2

422

4

694sin x

xxxB

+=

2

42

4

910sin

x

xxB

=

2

sinBcaArea

=

2

sin3 BxyArea

=

x

xxxyArea

/

/=

2

910

2

3 42

where cmxcm 00.300.1


16/19



Second method: step-by-step method

Step 1

Lets ,cma 00.2= cmb 00.4= , cmc 00.3=

Bacacb cos2222 +=

Bcos232234 222 +=

BB cos1213cos124916 =+=

4

1

12

3

12

1613cos

=

=

=B

Trigonometric Identity: 1cossin 22 =+


4

15

16

116

16

11

4

11sin

2

=

==

=B

2

sinBcaArea

= 2

4

15

2

23cmArea

/

/=

22 90.24

153cmcmArea

Step 2Lets ,cma 50.2= cmb 00.5= , cmc 00.3=

Bacacb cos2222 +=

( ) ( ) Bcos5.2325.235 222 +=

BB cos1525.15cos1525.6925 =+=

15

75.9

15

2525.15cos

=

=B


( )2

222

15

75.915

15

75.91sin

=

=B

2

sinBcaArea

=

( ) 215

130

2

5.23cmArea

=

22 2.85

4

129.9375cmcmArea =

Similarly if we try with

cma 25.2= , cmb 00.5= ,

the area of the triangle is

cmc 00.3=

22 3.009996.2 cmcmArea

We may try few more numbers,

tabulate and plot their results:

a(cm)

area(cm2)

1,00 0,00

1,25 1,53

1,50 2,18

1,75 2,622,00 2,90

2,24 3,00

2,25 3,00

2,50 2,85

2,75 2,30

3,00 0,00

0,0

0,5

1,0

1,5

2,0

2,5

3,0

3,5

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50

x cm

Areacm2

The results are similar to the

generalized case, i.e.:

Answer: The triangle has a local

maximum of 2max 00.3 cmy = at

mx 5= .


17/19



Assessment aims at Quality Level

HigherLower

Total

score

Choice of the method and itsimplementation

In which level the student caninterpret a problem situation

and can solve different types of

problem.

How complete and how well

the student uses different

methods and procedures that

are suitable to solve the

problem.

The student

calculates thearea of a

triangle

for a chosen

lengths of sides

and .

ABC

AB BC

1-2G

The student

calculates thearea of two

triangle

for two different

selected lengths

of sides

and .

ABC

AB

BC

3G

The student finds ageneral function for

the area of atriangle ,

where

ABC

cmc 00.3

xa , ,and finds themaximum value ofthe area of the

triangle .

xb 2

ABC

3G and 1-2VG 3/2

Mathematical reasoning

The degree of the quality ofevaluation, analysis, reflection,

proof and other forms of

mathematical reasoning.

The student using at the

results of at least three

different studies, i.e.

different sides, x , draws a

conclusion regarding the

maximum area of the

triangle.

1 VG

The student introduces

proper variables, and

graphs. begins a general

strategy for solving the

problem generally.

2G2/0

Mathematical language and

the quality and clarity of thepresentation of the solution:How clear, distinct and

complete the students solution

and presentation is, and how

well the student uses

mathematical terms, symbols,

and conventions.

It is possible to follow the

solution is easy to

understand and follow. The

mathematical language is

acceptable.

1 G

The solution is well

structured, complete, clear,

easy to follow and

understand. The

mathematical language

must be correct and proper.

1G and 1VG1/1

Total sum 3/3


18/19


MVG-Quality: The student demonstrate the highest quality(MVG) in solving problem 14 through

M1:Formulates and develops the problem,uses general methods with problem solving.

Formulates and develops theproblem, uses general methods withproblem solving.

M2: Analyses and interprets the results,

concludes and evaluates if they are

reasonable.

Using generalised method, finds thean equation for the area of a generaltriangle of sides cmc 00.3

xa , xb 2 , solves , and finds

using

0=y

yyx table, the student

correctly finds the local maximumpoint of area function.

M3: Carries out mathematical proof, oranalyses mathematical reasoning.

Finds analytical a correct function for

the area of the triangle:

( ) cmxxxy 319104

32

142


19/19


M3 Carries out mathematical proof, or analyses

mathematical reasoning.

M4 Evaluates and compares different methods

and mathematical models.

M5 The presentation is structured, and



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