WRP Solutions Mathematics 2012

8/13/2019 WRP Solutions Mathematics 2012

1/11

WESTERN REGION

2012Preliminary Final

EXAMINATION

Mathematics

SOLUTIONS


2/11

Preliminary ExaminationMathematics 2012

Multiple Choice Answer Sheet

Name ______________________________

Completely fill the response oval representing the most correct answer.

1. A B C D

2. A B C D

3. A B C D

4. A B C D

5. A B C D

6. A B C D

7. A B C D

8. A B C D

9. A B C D

10. A B C D


3/11

Question 11 Preliminary FinalExamination- Mathematics 2012

Part Solution Marks Comment

a) 2 3 64 Letx= 03646464........

10x= 3646464....

100x= 36464646....1000x= 364646464....

990x= 361

x=361

990 2 3 64 = 2 361

990

2

1working

1 - answer

b)i.

5

+3 =3

5 = 3 + 35 = 3 + 95 3 = 9

2

= 9

= 92 = 4 12

2

Lose one mark

per mistake.

ii. 2 + 1 = 3 + 22 + 1 = 3 + 22 + 1 = 3 + 2

2 + 1 = 3 2 2 + 1 = 3 + 22 + 3 = 2 1 2 3 = 2 1

5 = 3 = 1 = 3

5 = 1

Check by substitution.

= 1is not a solution = 35is the only solution.

2

1for obtaining

solutions

1 for testing

solutions andcorrect result.

c) = 2 + 4 (1) = 2 + 1(2)Sub (2) in (1) 2 + 1 = 2 + 42 2 3 = 0 3 + 1 = 0 = 3 = 1Therefore (3, 10) and (1, 2)

2 1 for

substitution

1 for solution

d) Letxbe one side, other =x+ 6 = 12

= 12 + 6 = 216

2 + 6 = 4322 + 6 432 = 0 18 + 24 = 0 = 18 = 24Therefore = 18Three sides are 18, 24 and 30 by Pythagoras.

2

1 for value ofx

1 for the 3

sidelengths

/10


4/11

Question 12 Preliminary Final Examination- Mathematics 2012


a) i. A(3, 5) , B(2, 3)

Gradient AB = 2

1

21 = 3

5

2 3 = 8

5 1

ii. Distance AB = 2 12 + 2 12= 2 32 + 3 52= 52 + 82= 89units

1

iii. AB, m= 8

5B(2, 3)

Equation is 1 = 1

3 = 8

5 25 + 1 5 = 8 + 168 + 5 1 = 0

1

iv. Perpendicular to AB, m=5

8 C(5, 8)

Equation is 1 = 1 1 = 1 8 = 5

8 5

8 64 = 5 255 8 + 39 = 0

2

1new

gradient

1Equation

v. AB: 8 + 5 1 = 0 C(5, 8)Distance C to AB

= + + 2 + 2 =85 + 58 1

82 + 52 =7989

8989 =

798989

1

vi. Area =1

2

=1

2 89 7989 = 792 = 39 12 2 1


5/11



vii.

1

b) D = 2 12 + 2 12 52 + 32 = 12 + 2

52 + 32 = 12 10 + 2 5 + 2 6 + 9 = 2 2 + 12 6 = 8 33 2

1 for equating

distances

1 for equation

/10


6/11



a) i. 32 2 5 = 6 2 1

ii.d

dx

4 x

3

= d

dx

4x

3

2

=3

2. 4.x

1

2

= 6x

1

2

= 6 x

ii. 52 44

Let u = 5xv= 2 44u= 5v= 42 43. 2

= 82 43

= +

= 52 44 + 582 43= 52 4392 4

2

1Working

1 - Answer

iii. 2+349

Let u = 2x+ 3 v= 4x9u= 2 v= 4

=

2 =492 2+34

492

=8

18

8

12

492 =

30492

2 1Working

1 - Answer

b) = 22 4 + 5 atx= 3 = 4 4Whenx= 3, = = 43 4 = 232 43+ 5 = 8 = 18 1 2 + 5

y= 11

Normal, = 18

1 = 1 11 = 18 3

8 88 = + 3 + 8 91 = 0

2 1Gradient and

point

1 - Equation

c)lim 4

2 12 4

= lim 44+3

4 = lim 4 + 3 = 4 + 3

= 7

2 1Factorising

1 - Answer

/10


7/11



a) 1

2

+

1

2

+

1

2

= 2 + 2 + 2 = 1 + 2

= 2

2 1 off for eachincorrect step.

b)

By Pythagoras

24 7 242 = 2 + 72576 = 2+ 49

A 576 49 = 2

=

527

= =24

527In 2ndquadsec = 24527

2

1Use of

Pythagoras to

findx.

1Correct ratio

including sign

(exact form).

c) i)

1`

1diagram with

58 and 27 anglesshown


8/11



ii) Using the sin rule:

=

=

27 = 15058

= 150 2758 = 80 ()

2

1Sine Rule

1Correct

answer

d) i

1

ii. 2 = 2 + 2 22 = 362 + 242 2362478

2

= 1512 728598d = 3889381182d= 389 nautical miles (3sf)

2

1use of cosine

rule1answer

/10


9/11



a) i. In Triangle COD,

COD = 90(AD BC, given)

ODC = 36(Angle sum of COD)

ABC = BCD = 54(Alternate angles ABCD)BCA = 54(Base angles of isosceles ABC)

CAD = 180 36 118 = 36(Angle sum ofACD )

2

Other proofs

possible

Deduct 1 for

incomplete

solution, or if a

minor error

ii. In Triangle ACD,

CAD = ODC = 36 (Proven Above)

ACD is isosceles (Two equal Angles)

1

b) i. In ABC andDECC in commonABC = DEC (corresponding angles ABDE)BAC = EDC (corresponding angles ABDE)

ABC EDC (Equiangular)

2

1steps

1reasons

shown

ii. Since ABC EDCsides are in proportion.

+ 2 = 71111 = 7 + 211 = 7 + 14

4 = 14 = 14

4= 3

1

2

2

1proportion

1value ofy

c) i.1+5 = 2 + 2 5= 2 + 3

= 5

1

ii.2 = 2 5 1d) = 32 2 + 6 = 32 2+ 6 = 32 + 2 + 6

Neither odd of Even.

1

Working must

be shown for

full marks.

/10


10/11



(a)x

25x 24 0

First solve x25x 24 = 0

(x8)(x + 3) = 0x = 8 or x = 3

Test x = 0

025(0) 24 0

24 0 True Solution is 3 x 8

2 1 for values of

boundaries

1 for correct

inequality

b) 32 + 5 4 = 0,i. = = 43 1ii. + = = 53 1iii. 22 + 22 = 22 + 2

= 2 + 2 2= 2 5

32 2 4

3

= 108

9

1

iv. 3

+3

=

3+

=35

3

4

3

= 33

4

1

(c) + 12 + + 2 + = 0Equal Roots if = 0i.e. 2 4 = 0 + 22 4 + 1 = 02 + 4 + 4 42 4 = 032 + 4 = 0 32 = 4

2 = 43

= 23 = 233

2 1Working

1Answer

d) i. 2 6 4 15 = 02 6 = 4 + 152 6 + 9 = 4 + 1 5 + 9 32 = 4 + 24 32 = 4 + 6Vertex = (-6, 3)

1

ii. 4a= 4

a= 1Focus = (-6 + 1, 3)

(-5, 3)

1


11/11



/10

WRP Solutions Mathematics 2012

Documents

Transcript of WRP Solutions Mathematics 2012