Solutions+Test+MaBNVC05 VG MVG Level+Algebra Geometry Functions

8/3/2019 Solutions+Test+MaBNVC05 VG MVG Level+Algebra Geometry Functions

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Solutions Test MaBNVC05-VG-MVG_Level Algebra-geometry-Functions NV-College

Instructions

Test period 115 minutes. We recommend that you reserve about 30 minutes to work with item 8.

Tools Calculator, formula sheet and ruler.

The test The test includes 10 items.

For most 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.

For some items only an answer is required. They are marked with, Only answer is

required .

After every item is given the maximum mark your solution can receive. (2/3) means that

the item can give 2 g-points (Pass level) and 3 vg-points (Pass with distinction level).

Items marked with give you a possibility to show MVG-quality (Pass with specialdistinction quality). This means e g that you use generalised methods, models and

reasoning, that you analyse your results and account for a clear line of thought in a

correct mathematical language.

Item number 8 is a larger item that demands a longer time to solve than other items. It is

important that you make a try to solve this item. Above the item is written what the

teacher has to consider at the assessment.

Mark limits The test gives totally at the most 59 points, out of which 26 vg-points. To get the test

character Pass you must have at least 18 points and to get the test character Pass with

distinction you must have at least 33 points out of which at least 12 points on Pass with

distinction level.

Write your name on all sheets of paper you hand in.

Test Ch2 MaBNVC05 1a 1b 2 2 2 3 4 4 5 5 6a 6b 6b Totalt Sum Sum Betyg

Name G G G G MVG VG G VG G VG G VG MVG VG MVG

2 2 2 3 1 2 1 3 2 3 1 2 1 25 17 2

8 G Min

17 6 VG Min

19 11 ¤ MVG Min

Not for sale. Free to use for educational purposes.© [email protected]





1. The enlargement or reduction on a copying machine is given as a percentage, which

refers to length. How many times smaller will the area of a picture be, if it copied twice

using %85 ? (1/1)

Suggested solution: Dtata: 085.0ll =

; area: ?=

A Answer: 052.0 A A ≈

First copy: ( ) ( ) 00

22

0

2

11 72.085.085.0 A A A ≈=== ll [1/0]

Second copy: ( ) ( )( ) ( ) 00

42

0

22

1

2

22 52.085.085.085.0 A A A ≈==== lll [0/1]

2. Find the angle α if °+= 72 x β and °−= 115 xγ . (1/2)

Suggested solution: Answer: °= 25 x , °== 57 β α , °=114γ

⎪⎪

⎩

⎪⎪

⎨

⎧

⎩⎨⎧

°−=

=

⎩⎨⎧

°+==

115

2

72

x

x

γ

α γ

β

β α

⎩⎨⎧

°−==

°+==

1152

72

x

x

γ α

β α

( °+=°− 722115 x x ) [0/1]

°+=°− 144115 x x

°=°+°=− 25111445 x x

°= 25 x [1/0]

( ) °=°+°=°+°=°+== 57750725272 x β α ⇔ °== 57 β α [0/1]

( ) °=°−°=°−°=°−= 1141112511255115 xγ ⇔ °=114γ

which certainly agrees with γ α =2 .

3. In a right-angled triangle the hypotenuse is cm5.8 and one of the shorter sides iscm7.1 longer than the other. Find the area of the triangle. (1/2)

Suggested solution: Dtata: cmc 5.8= ; xa = , cm xb 5.1+= ;2

5.17 cm A Area =≡

Using Pythagoras theorem, i.e. 222 cba =+ ⇔ ( ) ( )222 5.85.1 =++ x x [0/1]

( ) ( ) ( )2222 5.85.15.12 =+++ x x x ⇔ 07032 2 =−+ x x ⇔ 0355.12 =−+ x x

( ) 3575.075.02+±−= x ⇔ 96.575.0 ±−= x [1/0] cm x 21.5=

α

β

γ




( )( ) 25.17

2

5.121.521.5cm

cmcm A Area =

+=≡ [0/1]

25.17 cm A Area =≡

4. For what range (s) of values of x are the following equations valid?

a. 29351132 −≥−≥− x x x (0/2)

b. 2142 ≤+ x x (0/2)

Suggested solution: Answer: (a) 52 ≤≤ x ; (b) 37 ≤≤− x

a. 29351132 −≥−≥− x x x

x x 51132 −≥−

31152 +≥+ x x

147 ≥ x

7

14≥ x

2≥ x [0/1]

293511 −≥− x x

x x 532911 +≥+

x840 ≥

x≥8

40

x≥5

52 ≤≤ x [0/1]

b. 2142 ≤+ x x

02142 ≤−+ x x

( )( ) 037 ≤−+ x x

Plot ( )( )37 −+= x x y [0/1]

The graph cuts the x-axis at

7−= x , 3= x

02142 ≤−+ x x requires that:

37 ≤≤− x [0/1]

-30

-20

-10

0

10

20

30

-9,0 -7,0 -5,0 -3,0 -1,0 1,0 3,0 5,0

x

y = ( x + 7 ) ( x - 3 )






5. Find the range of the real constant a such that the quadratic equation

( ) ( ) a xa xa 4411 22 −=−+−

a.

has two distinct real roots. (0/3)

b. has only one real root. (0/2)

c. has no real solution.

(0/2)

Suggested solution: Answer:

a. If 3>a or 5−<a the equation

has two distinct real solutions

(roots.)

b. If 3=a or 5−=a the equation

has only one real double solutions

(root.)

c. If 35 <<− a the equation has no

real root (solution.)

( ) ( ) a xa xa 4411 22 −=−+−

( ) ( ) ( ) 01411 22 =−+−+− a xa xa

Divide both sides of the equation by

( )1−a :

( )

( )

( )

( )

0

1

14

1

122 =

−

−+

−

−+

a

a x

a

a x

( ) 0412 =+++ xa x [0/1] Eq. (2.1)

Compare the equation with:

02 =++ q px x

Withe solutions:

q p p

x −⎟ ⎠

⎞⎜⎝

⎛ ±⎟

⎠

⎞⎜⎝

⎛ −=

2

22

( 1+= a p ) , and 4=q ,

The equation (2.1) has two real roots

only, and only if:

042

1

2

22

>−⎟ ⎠

⎞⎜⎝

⎛ +=−⎟

⎠

⎞⎜⎝

⎛ aq

p

( )04

4

12

>−+a

[0/1]

( ) 01612

>−+a

( )[ ] ( )[ ] 04141 >−+++ aa

[ ][ ] 035 >−+ aa

which requires: 3>a or 5−<a [0/1]

as illustrated below:

-20

-10

0

10

20

30

40

50

-9,0 -7,0 -5,0 -3,0 -1,0 1,0 3,0 5,0 7,0

a

( a + 5 ) ( a - 3 )

For the case [ ][ ] 035 =−+ aa , i.e. for:

3=a or 5−=a the equation has

only one (double) real solution.

[0/2]

On the other hand if:

For the range: 35 <<− a ,

042

1

2

22

<−⎟ ⎠

⎞⎜⎝

⎛ +=−⎟

⎠

⎞⎜⎝

⎛ aq

p.





Therefore, if 35 <<− a the equation

has no real root (solution.) [0/2]

6. Which of the following lines are perpendicular to each other? Which are parallel? Solve

the problem analytically. Plot the lines and illustrate your findings above graphically.

(4/4)

a. ( )28

524

4

1 x y x y −+=

−−

b. ( ) ( ) y x x y 4111813 −+=+−

c. 5

4

2

7 y x x y −=

++

d. ( ) ( ) 01322 =+++ x y

Suggested solution:

Answer: Due to the fact that7

332 == k k ( ) ( ) y x x y 4111813 −+=+− is parallel to

54

27 y x x y −=++ .

Answer: is normal (perpendicular) to ( ) ( ) 01322 =+++ x y . x

0 5,00 2,00 -5,00 -3,5

1 5,67 2,43 -4,57 -5

53

2+= x y 2

7

3+= x y 5

7

3−= x y 5.3

2

3−−= x y x

-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6 8 10 12 14

x

y

a. ( )

74

524

4

1

×/

−+=

/

−− x y x y

( ) ( ) x y x y 52417 −+=−−

x y x y 584777 −+=−−





775847 ++−+=− x x y y

1523 += x y

5

3

2+= x y :

3

21 =k , 51 =m [1/0.5]

b. ( ) ( ) y x x y 4111813 −+=+−

y x x y 41111833 −+=+−

31181143 ++−=+ x x y y

1437 += x y

2

7

3+= x y :

7

32 =k , 22 =m [1/0.5]

c. 5

4

2

7 y x x y −=

++

( ) ( ) y x x y −=++ 4275

y x x y 283555 −=++

355825 −−=+ x x y y

3537 −= x y

57

3−= x y :

7

33 =k , 53 −=m [1/0.5]

Answer: Due to the fact that

7

332 == k k

( ) ( ) y x x y 4111813 −+=+− is

parallel to 5

4

2

7 y x x y −=

++. [0/1]

d. ( ) ( ) 01322 =+++ x y

03342 =+++ x y

732 −−= x y

5.32

3−−= x y :

2

34 −=k , 5.34 −=m [1/0.5]

Answer: Due to the fact that

12

3

3

241 −=⎟

⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ =⋅ k k

( )28

524

4

1 x y x y −+=

−−is normal

(perpendicular) to

( ) ( ) 01322 =+++ x y . [0/1]

7. The functions ( ) x f and ( ) xg are defined by:

( ) IR x x x f ∈−= ,35 ( domain of x is real numbers)

( ) IR x x xg ∈+= ,72

a. Find the range of ( ) xg . (1/0)

b. Determine ( )( ) xg f . (0/2)

c. Determine ( )( ) x f g . (0/2)

d. Determine the value of x for which ( )( ) ( )( ) x f g xg f = (0/2)





Suggested solution: Answers: (a) ( ) 7≥ xg ; (b) ( )( ) 325 2 += x xg f ;

(c) ( )( ) 163025 2 +−= x x x f g ; (d) 42.0−= x , 92.1= x

a. ( ) IR x x xg ∈+= ,72 .

The range of ( ) 72 += x xg is all

positive real numbers larger than 7.

( ) 7≥ xg [1/0]

b. ( )

( )IR x

x xg

x x f ∈

⎩⎨⎧

+=

−=,

7

35

2

( )( ) ( ) 35 −= xg xg f .

( )( ) ( ) 375 2 −+= x xg f

( )( ) 3355 2 −+= x xg f

( )( ) 325 2 += x xg f [0/2]

c. ( )

( )IR x

x xg

x x f ∈

⎩⎨⎧

+=

−=,

7

35

2

( )( ) ( )( ) 72+= x f x f g

( )( ) ( ) 735

2+−= x x f g

( )( ) 793025 2 ++−= x x x f g

( )( ) 163025 2 +−= x x x f g [0/2]

d. ( )( ) ( )( ) x f g xg f =

163025325 22 +−=+ x x x

0321630525 22 =−+−− x x x

0163020 2 =−− x x

08.05.12 =−− x x [0/1]

Compare the equation with:

02 =++ q p x x

with solutions:

q p p

x −⎟ ⎠

⎞⎜⎝

⎛ ±⎟

⎠

⎞⎜⎝

⎛ −=

2

22

5.1−= p , and 8.0−=q ,

( ) ( ) ( )80.075.075.02

−−±−−= x

17.175.0 ±= x

42.0−= x , 92.1= x [0/1]

-0,417 32,87 32,861,917 50,37 50,36

( )( ) 325 2 += x xg f ( )( ) 1630252 +−= x x x f g x

0

10

20

30

40

50

60

70

-1 -0,5 0 0,5 1 1,5 2 2,

x

f ( x ) & g ( x )

5





8. Write a formula to calculate y if the x in the rectangle below is known. Note that AE

is not necessarily perpendicular to BD . (1/3/¤)

A D

B CE

y

x

68

51

Suggested solution: Answer:( ) x

x y

−=

85

68

Lets name the intersection of AE and BD point O . Due to the fact that ABCD is a

rectangle, AD is parallel to BC . Therefore the angles DBC and ADB are alternative

angels and therefore are equal. Similarly angles AEB and EAD are equal. On the other hand

the angles AOD and BOE are vertical opposite angles and therefore are equal. Therefore,

the triangles AOD and BOE are similar triangles and therefore the fractional relationship

between the similar sides are equal: [0/1]

DO

BO

AD

BE = ⇔

( ) x

x y

−=

8568⇔

( ) x

x y

−=

85

68 [0/2]

cm DB 85685122 =+= , and cm x BO DB DO −=−= 85 [1/0]

9. A weather balloon is sent up in the sky but, due to a leak, doesn’t clime very high. Its

height above the ground (in m) can be calculated from the formula

2025.025.31 t t h −+=

where t is the time from the start (in minutes.) After how long is the balloon at a height

of m76 ? Solve the problem analytically. Plot the height of the balloon as function of

time and clearly demonstrate your solution on the graph. (1/3)

Suggested solution: Answer: min38min88.37 ≈=t , min87min12.87 ≈=t

2025.025.31 t t h −+=

2025.025.3176 t t −+=

07525.3025.0

2

=+− t t

0025.0

75

025.0

25.32 =+− t t

030001302 =+− t t

First method:

O




( )( ) 010030 =−− t t

min30=t , min100=t

Second method:

( ) 300065652−±=t

min30=t , min100=t

Third Method: graphical solution:

• Plot2025.025.31 t t h −+=

• Plot 76=h

• The solution to the problem is the

intersection of the two graphs.

Weather Baloon's height as a function of time

0

1020

30

40

50

60

70

80

90

100

110

0 20 40 60 80 100 120 140 160

t/min

h / m

2025.025.31 t t h −+=


Solutions+Test+MaBNVC05 VG MVG Level+Algebra Geometry Functions

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