cxcDirect May2008

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CSEC MATHEMATICS Past Paper Solution – May 2008 cxcDirect Institute

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** Please see the original past paper for the questions. Only the answers will be provided as per copyright obligations. **********************************************************

2

3

4

2

2

2

ii)

6a b 12a b= 6a b b 2a 

iii)

2m 9m – 5

Q1a. s

2

a = 2, b = 9, c = -5

let pq = ac ;

p+q = b p+q = 9

3.90.27= 4.17

⇒

pq = -10;

 0.6724=0.82

⇒

p = 10, q = -1

⇒

2 2 2m 9m – 5 = 2m 10m – m− 5

=

2m  m 5−m 5

=

2m – 1 m 5

so:

4.17 0.82= 4.99

b) simplifying the numerator : 17 1 4 5 4 2 − − = = 10 2 5 2 5 . Divide Numerator by denominator: 17 3 ⇒ ÷ 10 4 ⇒

17 4 × = 10 3

34 = 15

2

4 15

**********************************************************

1.b

$1Can = J$72.50

⇒

$Can 250.00 = J $250 x 72.5 = J$18,125.00

Credit limit = $J 30,000 = $Can (30,000/72.5) = $Can 413.79 Credit card balance = $413.79 – $250.00 = $Can 163.79 **********************************************************

Question 2 a = 2, b = -1, c = 3 a(b+c) = 2(-1 +3) = 4 2

2

4b – 2ac ab c

=

4 −1 – 2× 2 ×3 2−1 3

=

4 – 12 = 4

− 2

b) i)

4  x5 

ii)

ab 16

c) i) 15 – 4x = 6x +2 ⇒

15− 2=10x 4x

⇒

13=10 x

⇒

x=1.3

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Math club :

2


Q4.

Q3 Career Doctor Teacher Artist Lawyer Sales Total

# 105 189 216 240 330 1080

U

Angle 35 63 72 80 110 360

M

N

17, 19, 23

16, 18, 20, 22, 24

15,21,25

't = 1080 - ( 240 + 189 + 216 + 330) = 105 Converting to a pie chart angle: 0

360 ×

⇒

Doctor =

⇒

Teacher =

0

105 1080

360 ×

=

M = prime numbers = 17, 19, 23 N = Even numbers = 16, 18, 20, 22, 24 M ∪ N  ' = 15, 21, 25

0

35

189 = 1080

0

63

*********************************************************

b.

and so on...

C

D

Pie Chart

7 cm

60

Doctor

0

A

0

( 35 )

12.1 cm

7 cm

B

Sales 0

( 110 )

Teacher 0

( 63 )

Lawyer 0

( 80 )

Artist 0

Construction details. 1. 2. 3. 4.

( 72 )

5.

Draw straight line AB = 7cm ∢ BAD using compass Construct 600 Measure AD = 7cm Set compass to a separation of 7cm and with centre D, construct an arc above B. Set compass to a separation of 7 cm and with centre B, construct a second arc to intersect the first arc. The intersection of the two arcs is the point C.

6. Measure AC = 12.1cm


Math club :

3


Q5.

The diagrams below highlight the main areas of Q5 (see past paper for original). R

S RS = 6m

10m 2m

5m

A

B

3m

R x=4m

8m

1.)

RS = 8 – 2 = 6m

2.)

x = 10 – RS = 4m

S RS = 6m

Perimeter equal distance around the two objects. Since A and B are joined at RS, then RS is not taken into account so: Perimeter A

= 2 ( 5 + 10 ) - RS = 30 – 6 = 24m

Perimeter B

= 2 ( 8 + 3 ) - RS = 22 – 6 = 16m

Perimeter (A+B) = 24 + 16 = 40m Area ( A + B ) = 10 x 5 + 8 x 3 =

74m

2

Area of floor board = 1m x 20cm = 1m x 0.2m ( 20cm = 0.2m) =

0.2m

2

# floor boards needed to cover A =

Area A area of floor board

=

10 ×5 = 250 0.2


Math club :

4


Q6

Question 6b – Transformation From the diagram given on the Past paper, we observe the following:

The diagrams below highlight the main areas of Q6

G

1.

Size and Orientation of Object and Image are maintained P  4,2  Q 6,−5  P  7,2  Q 9,−5 

2. 3.

32 H

0

27 J

K

Q 9 −5

⇒

GH HJ GH 0 tan 32

-

2 −7

P 7 2

   =  

If GH = 12m 0

P 4 2

   =  

where :

and : Then

Q 6 −5

0

-

2 −7

tan 32 =

HJ =

=

12 = 19.2m 0.6249

The object is transformed by a translation T where :

⇒

 

T=

similarly

0

tan 27 =

GH HK

********************************************************* 6 b contd.

⇒

so HK =

GH tan 270

=

2 −7

12 = 23.55m 0.5095

y 7

Now JK = HK – HJ = 23.55 – 19.2 = 4.35m *********************************************************

y=x

( 2 ,7 ) ( 4 ,6 )

6 5

S

4

(6,4)

( 2 ,4 )

( 4 ,2 )

3

P (7,2)

2

( 4 ,2 )

1 0


Math club :

1

2

3

4

5

6

7

x

5


Question 7

(-2, k) … w he re k = 14

A(0, 7)

mid(1, 3.5) B(2, 0)

-2

0

2 y = x -2

-2

General equation = y = mx + c c = y intercept = 7 m = slope =

y 2 – y2 x2− x1

=

0– 7 = 2– 0

so equation of line =

1.

−3.5

y= − 3.5 x 7

2. *********************************************************

For a Point ( -2, k ) on this line

⇒

k = y, and x = - 2

substituting these values in the equation : ⇒

⇒

⇒

y= − 3.5 x 7 k =−3.5 −2 7 k = 14

*********************************************************

If a second line y = x – 2, intersects the first line, the point of intersection is found by making the two equations equal: ⇒

x – 2 = −3.5 x 7

by solving this equation we get ⇒

x= 2, and y = 0

so coordinates of intersection = ( 2, 0)


Math club :

6


Question 8

Section II 2

3

x ×x ÷ x $cost -stamps

Quantity

1 1.2 2.5 4 Total

4

=

x

2 3−4 

)=x

$Total 6 6 6 6 24

6 7.2 15 24 52.2

a

3/2

b ×  ab 5/2

=

a

=

a

3/2

5/ 2

b ×a

3 / 2 1 / 2 

b

1 /2

b

3/2

5/ 23 /2

=

2

a b

4

*********************************************************

f  x= 2x – 3

if then

Total cost of stamps = $52.20

3

−1

f

x 3 2

=

x

so f(2) = 2(2) – 3 = 1 Parcel postage cost = 25.70 1.

−1

and

f

also

f

03 2

0 =

Using as many $4 stamps as possible

= 5 x $4 + 1 x $2.5 + 1 x $1.2 + 2 x $1 = $25.70 −1

f  x =

=

3 2

2x −3 3 = x 2

2. Using all the $1 stamps

so

= 6 x $1 + 4 x $4 + 1 x $2.5 + 1x $1.2 = $25.70

*********************************************************

f

−1

=2

f 2

Q 9.c 3. Largest # stamps

⇒

use all the smallest value stamps

= 6 x $1 + 6 x $1.2 + 5 x $2.5 = $25.70 largetst # = (6+6+5) = 17 stamps, with values as shown above

y 80 70 60 50 slope = 14/20 = 0.7 degree / min

40 30

14

20

20

10 0

10

15

20

30 40

1. Temperature after 15 min =

50

60

70

min

0

50 C

2. Rate of cooling at t = 30 min = gradient of the tangent a t

t = 30min =


Math club :

14 20

=

0

0.7 C / min

7


Q10.a

Q11 – Geometry & Trig

If

The figs below are used to show the theorems used for this problem. ( see past paper for original diagram)

y 4x= 27

... 1

y=27 −4x

⇒

also if

... 2

xy x=40

S 26

then Substituting 2 into 1 O 128

⇒

x 27 −4x  x= 40

⇒

4x − 28x 40=0

⇒

x −7x 10= 0

26 26

⇒

 x− 2  x−5 = 0

P

R

2

2

so

x= 2 ; or

x=5

*********************************************************

10.b From the diagram ( see Past paper)

Now: Lines OS and PR are parallel equal: ∢ RPS =∢ OSP= 26

⇒

⇒

alt. Angle as are

0

girls

2. Triangle OPS is isosceles

B(5,8) - profit = $55 (6, 6)

6 5

⇒

Base angles are equal

0

⇒

∢ OSP =∢ OPS =26

⇒

∢ POS =180 – 2 26 =128

0

(10 ,5) S

T

64 O 128

A(1 ,2) - $13

C(12, 2)- $46

1.

2.

boys

10

6

10 boys and 5 girls falls outside the feasible region and therefore cannot be members of the club at the same time. 6 boys and 6 girls falls inside the feasible region and therefore can be members at the same time.

P

Angle at centre = twice angle at circumference

Vertex A

⇒

3 x 1 + 5 x 2 = $13

Vertex B

⇒

3 x 5 + 5 x 8 = $55

Vertex C

⇒

3 x 12 + 5 x 2 = $46


Math club :

128 0 =64 2

S

O

The three equations that define the region are: 1. y≥2 4 y≤− x 12 2. 5 3. y≤2x

If Profit equation = 3x + 5y, then max profit occurs at one of the three vertices.

∢ PTS =

⇒

R

26 26

90

28 Q

P

Angle between Radius (OP) and Tangent (PQ) = ( max Profit)

⇒

0

90

0

∢ RPQ =90 –  2626 = 28

Q11 – B 8


N N

75 cos 10 = 73.9 km

X

T 10

O 58.5 117

A

7.25

47

2600

33

75 km

r = 8.5

0

112

x

R

B

7.25

N

68

33

56 km 68

S

The diagram above is drawn to illustrate the principles used to solve this problem, ( see past paper for original drawing) Define x as the perpendicular bisector of AB.

1. The completed bearing diagram is shown above.

∢ XOB=½ ∢ AOB

⇒

where:

sin ∢ XOB=

=

∢ RST = 6833=101

0

7.25 = 0.85294 8.5

−1 sin 0.85294 =

⇒

∢ XOB

⇒

∢ AOB=2 ×58.5=117

0

58.5

2. To find

0

************************************************************

Area of triangle AOB =

2

0

½ r sin 117

= ½ 8.5 2× 0.8906

=

32.17 cm

we use the sine rule:

∢ RTS

⇒

56km sin ∢ RTS

=

75km sin 1010

⇒

sin ∢ RTS

=

sin 101 ×

=

sin 0.7329

2

∢ RTS

⇒

0

56 = 0.7329 75

−1

=

47

0

************************************************************

Area of shaded section = Area of Sector AOB – Area of triangle AOB

3. Bearing of R From T =

18033 47= 260

0

0

area of sector AOB 117 = area of circle 3600

now : so :

2

× 8.5 ×

= ⇒

2

r ×

Area of sector =

Shaded area =

4. To find the distance TX, we note that Triangle XTR is a right angled triangle, and:

117 360

117 360

=

73.73 cm

73.73 – 32.17=41.56 cm

2

2

************************************************************

Length of minor arc = length of major =

2  r×

2 r×

∢ XTR =90 –  4733 =

117 360

Hence: ⇒

0

cos 10 =

TX =

0

10

TX 75 0

75Cos10

= 73.9km

360− 117 = 36cm 360

Q12.a © cxcDirect Institute cxcDirect.org;

Math club :

9


Vector & matrices

Given

O ⃗A = a

and

O⃗ B =b

a)

now if:

O ⃗A = a

then

⃗ =− a AO

B

(reverse the sign)

M A

b

⃗ AB =

so

−a + b

=

P

b − a

a

now: OP : PA = 2 : 1 O

so ;

2 ⃗ OP= a 3

1 ⃗ PA = a 3

and

⃗ PM

Finding b)i

B

M A

N

b

P

1 a 3

2 a 3

B

M A

b

⃗ OB

a

⃗ OB

O

P

⃗ OA

To go from P to M :

O

b ii)

so :

 PM

= ⃗ PA + ⃗ AM

but

BM

= MA

therefore Finding

b

⃗ AB

A

then:

a

⃗ OB

and since: ⃗ PA =  PM

1⃗ AB 2

=

1 ( b−a ) 2

⃗ ( shown dotted ) is the vector going from point A Vector A B to point B. Imagine that you are at point A and you wish to get to Point B

so:

1 a 3

=

1 1 a + ( b−a ) 3 2

=

1 a 3

=

2a +3b−3a 6

=

3b−a 6

⃗ OA

O

⇒

⃗ AM =

⃗ AB B

⇒

P to A then A to M

⇒

 = PM

+

1 1 b − a 2 2

1 (3b−a ) 6

First go from A to O and then from O to B ⃗ AB

=

⃗ AO

+ O ⃗B


Math club :

10


13. c)

now note that since:

⃗ PM =

1 (3b−a ) 6

and

⃗ MN =

1 ( 3b−a ) 2

then :

⃗ MN =

3×⃗ PM

N b ⃗ MN

B

M

⃗ MB

⃗ AM

A ⃗ PM

b

P

1 a 3

2 a 3

This completes the proof that : P , M and N are on a stright line ( collinear)

O

Collinear means: On a straight line. If the three points P, M and N are collinear then, the two ⃗ and M N ⃗ that connect these three points vectors P M will also be collinear. The proof that two vectors are collinear is that one vector must be a constant ( scalar) multiple of the other vector : ( where k is a constant)

⃗ = k PM ⃗ M N

⇒

we have already found ⃗ PM so now we need to find Finding

Steps: • Find the two vectors connecting the three points • Show that one vector is a constant multiple of the other vector i.e V⃗1= k× V⃗2

1 (3b−a ) 6

=

⃗ MN

⃗ MN

⃗ MB +

=

M to B then B to N

⇒

⃗ BN

where

⃗ BN =

O⃗ B

and

⃗ MB =

1⃗ AB = 2

so

1 ⃗ ( b−a ) MN = b + 2

⇒

The proof that three point are collinear (on a straight line), is the same as the proof that the two vectors connecting the points are collinear.

⃗ MN

To go from M to N : So:

nb.

= b

1 1 b − a 2 2

=

b

=

2b+b− a 2

⃗ MN =

1 ( 3b−a ) 2

+

1 ( b−a ) 2


Math club :

11


⃗ AN :

To find the length of

Q 14: Matrices N

b

⃗ AN

⃗ MN

B

M

⃗ MB

=

⃗ AM

A ⃗ PM

b

1 a 3

P

=

2 a 3

=

O

Finding Now:

2



 

−2 0  4 −1 5 1 3 7

2 X Y =



        

−2 0 −2 0  4 −1 5 1 5 1 3 7



4 0  4 −1 −5 1 3 7

8 −1 −2 8

⃗ AN ⃗ AN

⃗ AO

=

+

⃗ ON

***********************************************************

14 c i (a)

where

⃗ ON

= 2b

and

⃗ OA

=a

so

⃗ AN

=

⃗ AN

6 2

==

Length =

2

−a

−a + 2b



now given : a =

⃗ = AO

⇒

H=

 1 2

and b =

−6 −2

 −4  2

2

1 2

+2

=

2 4

+

( ) k 0

0 k

If the point D(5, 12)

  ( ) ()   6 2

If H is a 2 x 2 transformation matrix , which represents an enlargement with scale factor k , then this may be written as:

=

4.47

→ (7.5, 18)

This means that under the transformation (H ) , the point D(2,5) maps onto its image point at D' (7.5, 18). This is written in the column matrix format as : H

D 5 12

D' 7.5 18

( )( )   k 0

−4 2

0 k

=

5 k = 7.5 12 k = 18

⇒ and

From the first equation :

k =

7.5 5

= 1.5

Note also that we will get the same result for k if we use the second equation; Hence: k = 1.5 14 c i (b)

Similarly, under the transformation (H) , the image points E' and F' , are determined as shown below: ⇒

H 1.5 0 0 1.5

(

E F 2 8 7 4

)( )

=

E ' F' 3 12 10.5 6

(

)

coordinates of E' = (3, 10.5) coordinates of F' = (12, 6) © cxcDirect Institute cxcDirect.org;

Math club :

12


14 c ii (a)

  a c

So : let R =

Finding a 2 x 2 transformation matrix

R

Q( 0, 1 )

P(1, 0) 0

1

a c

b d

P' Q' 0 1 −1 0

=



x

We can now solve for the the unknowns a, b, c, and d

To Find a 2 x 2 transformation matrix that represents a clockwise rotation of 900 about the origin: Consider the Object OPQ above with coordinates P(1, 0) and Q (0, 1)

⇒

a ×1+b×0

= 0

⇒

a×0+b×1

= 1

.. so b = 1

⇒

c×1+d×0

= -1

.. so c = -1

⇒

c×0+d ×1

= 0

.. so b = 0

Giving: R = Now if rotate this triangle through 900 in a clockwise direction about the origin. We note from the graph that

(

0 1 −1 0

.. so a = 0

)

This result can be obtained faster if we observe that

y

0

1

P Q 1 0 0 1

since

( )

then:

( )( )

Q' ( 1, 0) x

R P' ( 0 , -1)

Hence: Q 0, 1  Q ' 1, 0 

The means that the object point P( 1, 0) maps unto its image P'(0, -1) , and the object point Q( 0, 1) maps unto its corresponding image point Q'( 1, 0)

Our task is now to find a 2 x 2 matrix , that will produce the exact result that we obtained graphically above. If we define this transformation matrix as R, then under this transformation, object points P( 1, 0) and Q( 0, 1) must map unto their corresponding image points at P'(0, -1) , and Q'( 1, 0)


a c

b d

is the identitiy Matrix ( I )

P Q 1 0 0 1

Math club :

Giving: R =

( )= a c

b d

(

0 1 −1 0

R

( )

=

a c

b d

P' Q' 0 1 −1 0

R

P 1,0   P '  0,−1

and

P Q 1 0 0 1

( )( ) 

then:

-1

( transformation matrix)

where a, b, c and d are unknowns

y

1

b d



)

( see matrix transformation workbook at www.cxcDirect.org for tutorial)

13


14 c ii (b)

nb: Notes on combined transformations:

Under the transformation R: The image of

D', E' F' is

D", E", F" :

If H and R represent two separate transformations

That is:

Then the combined transformation that represents H first, followed by R is writtens as : [R][H]

R 0 1 −1 0

(

)(

D' E' F' 7.5 3 12 18 10.5 6

)

And the combined transformation that represents R first, followed by H is writtens as : [H][R] Note carefully that: [H][R]

=

=

D' ' E'' F'' (0x7.5+1x18 ) (0x3+1x10.5) (0x12 +1x6) (−1x7.5+0x18) (−1x3+0x10.5) (−1x12+ 0x6)

(

)

D' ' E' ' F' ' 18 10.5 6 −7.5 −3 −12

(

)

≠

[R][H]

One way to remember the correct order, is to view the two fg ( x) means matrices as functions, and recall that function g (first) followed by function f .

so

RH ( )

⇒

H first then R

so: coordinates of D'' = ( 18, -7.5 ) coordinates of E'' = ( 10.5, -3 ) coordinates of F'' = ( 6, -12 )

14 c ii (c) If the symbol H is used to represent the Enlargement, and R used to represent the rotation, then two successive transformations, H first, followed R will map the object points DEF onto its image at D" E" F" . The 2 x 2 matrix that represent the combined transformation, H first then R is: [R][H]

=

=

( (

0 1 −1 0

)(

1.5 0 0 1.5

0 1.5 −1.5 0

)

)

END


Math club :

14

cxcDirect May2008

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