Extension 1(1)

8
2012 Preliminary Course Half Yearly Examination MATHEMATICS EXTENSION 1 Name: Maths Teacher: Maths Line: Set by: Carmen Lloyd General Instructions Writing time — 1 hour Write using blue or black pen Calculators may be used. NOTE: QUESTIONS 1 — 8 are not Attempt questions 1 - 8 all of EQUAL VALUE START EACH QUESTION ON A NEW PAGE Total marks (46) The following outcomes are examined in this paper: PE2: uses multi-step deductive reasoning in a variety of contexts. PE3: solves problems involving permutations and combinations and inequalities. Outcome Multiple Choice Q6 Q7 Q8 Outcome Total PE2 /4 /17 /10 /31 PE3 /1 /14 /15 Total /5 /17 /14 /10 /46

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maths 3u

Transcript of Extension 1(1)

Page 1: Extension 1(1)

2012 Preliminary Course

Half Yearly Examination

MATHEMATICS EXTENSION 1

Name: Maths Teacher:

Maths Line:

Set by: Carmen Lloyd

General Instructions

• •

Writing time — 1 hour Write using blue or black pen

• Calculators may be used. NOTE: QUESTIONS 1 — 8 are not

• Attempt questions 1 - 8 all of EQUAL VALUE

• START EACH QUESTION ON A NEW PAGE Total marks (46)

The following outcomes are examined in this paper:

PE2: uses multi-step deductive reasoning in a variety of contexts. PE3: solves problems involving permutations and combinations and inequalities.

Outcome Multiple Choice

Q6 Q7 Q8 Outcome

Total

PE2 /4 /17 /10 /31

PE3 /1 /14 /15

Total /5 /17 /14 /10 /46

Page 2: Extension 1(1)

Multiple Choice Marks Outcome

1. Which expression is not equivalent to sin 150° ?

A. sin 30° B. —sin 210° C. cos 60° D. —cos 60°

2. The expression tan 0

is equivalent to:

1

1

1

1

1

PE2

PE3

sec 0

cos t 0 sin 0 A. B. C. 61 D.

sin 8 cos 2 0 cos sin°

x2 -9 3. Which expression is equivalent to ., 2x` +5x - 3

x —3 B.

x + 3 C.

x —3 D.

x +3 A.

2x —1 2x +1 2x —3 2x +3

4. Solve ,h + 2sin x = 0 for 00 __ x 360°

A. 60°, 120° B. 240°, 300° C. 30°, 150° D. 30°, 330°

5. How many arrangements of the word KITCHEN are possible if the vowels must stay together? A. 2 15!2 ! B. 2!5! C. 2!6! D. 7P2 x 5/35

Question 6 Marks Outcome

a)

b)

c)

d)

e)

f)

Solve 12x +51= 3x +9

1-7- Find b if

—5 b,fi

3

2

3

3

3

3

PE2

a and , = a+ —2 +3../7

1 1 Express fraction in form. as a single simplest 2

x-4 2 +3x+ 2

xy = Solve simultaneously

4

2x-3y-5=0

Prove 1 sin3 0 + cos38

9 61

= sin cos sin 0 + cos B

olve 2 cos30 +1 = Ofor 0 0 0 360°

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Question 7

Solve x-7 3

1

PE3 a) <3

x+4

b) David's playlist consists of 12 different songs that can be arranged in any order.

(i) How many arrangements are there for the 12 songs?

(ii) David decides that he wants to play his three favorite songs first, in any order. How many arrangements of the 12 songs are now possible?

c) How many arrangements of the word TRIGONOMETRY are possible if the letters M and E are to be together?

d) Seven people are to be seated at a round table.

(i) How many seating arrangements are possible?

1

2

1

(ii) Two people, Stephanie and Mary, refuse to sit next to each other. How many seating arrangements are now possible?

e) There are four women and five men on the local council. A committee of four councilors is to be formed to investigate the need for additional bicycle tracks.

(i) How many committees are possible if there are no restrictions?

(ii) If the committee is to include 2 women and 2 men, how many committees can be formed?

(iii) Two of the councilors are Mr Smith and Mrs Jones. If the committee is to include two women and two men, and Mrs Jones refuses to serve on the committee if Mr Smith is also on the committee, how many possible committees can be formed?

2

1

2

Exam continues on next page

Page 4: Extension 1(1)

Question 8

a) From a yacht that is sailing due north a lighthouse is seen on a bearing of PE2

030°. After 2km of sailing, the lighthouse is now on a bearing of 048° from the yacht.

(i) Draw a diagram showing the above information. 1

(ii) Calculate the distance the yacht is currently from the lighthouse, correct to 2 1 decimal place.

b)

F

Alli.

i North

7North

A

Points A, B and C lie on horizontal ground. A and B are 300 metres apart and C is due North of A. The bearing of C from B is 240°. Point F is at the top of a flagpole h metres high with its base at C. The angles of elevation of F from A and B are 16° and 14° respectively.

(i) Copy the diagram into your writing booklet and mark the given information on it.

1

(ii) Show that ZACB =120° . 1

(iii) Find expressions for AC and BC in terms of h. 2

(iv) Hence, find the height of the flagpole, h, correct to the nearest metre.

3

End of Examination

Page 5: Extension 1(1)

-fG n e- & (Ds. Sec. e-

- cos e- cos &-

S~n

li - r3 Pi-5c

I I — 131)7 q= i i - -=

59 59 59

cix -/- 2x 2

3xz - 3) 3. -

(a. 2.)(- 4-

MI — — 2)

(x1-2) -2y-x.

LI 4 3 sir)

pk. - avute_ = db CQ3 `1

= (08, 3(cci' -

3o6'

5 vows ns 1 9,,-,DL 6 gv-of_s

cv) 2.).—

Tess = RNs = 3 (-++)-r

-3 .s.tp-tcs, 42.1-ts i s LHS rnss 6c ("OS .-111.-e.

- 4 is riot 0, solvf-loy■

21+ 2,17—i5-1=7----10

2Z+ S-Dc -3 -I) Jc 2 —

+5 3Dt. + 9

2.)L-1-- 5 =

— I-4

- (3Dt_-r ot)

- 3x -9 _-i ii - 5

3(-z+),

5 / 1-1- is -fig/. 0,0 t.f

so/.-r, 0 ,

! x a

awes+IO✓l 4z,

= 4-4

Li j = )

d") - - 5 = c

(-±)

Mt-H !pie CVio1ce=

I. 3 I 15-0° = sin 3C5'

--5-

CDS 60 - COS 436- = = -

T.

a 1=-7 -

LI-IS = FT- 3 2

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< 3

/ volt;

arisvuer

=

-1

C

CDkpes---1101/-)

<3 -17-es ;) )-= 0 < 3

(

, l(-z_ < — a S oval z >— (-1-

ic_s4 =-10 -5 -S- -2

-r2 c3

2_ < 3

c 3

si n COS

LHS = r —

Sir) 63' C cDS

(Sly-)6- 4 c_(:),s.,e s _sole cos- & zer

(sonzo –son cos

-1 1{-1 p7,7). r

5 .1 ,n e--

1)--)0---i- COS O-

2 Licas 3e- I = o

Cos & a

CoO ° CD. -2 3

O°‹ e < c, o

o° c- toe.o

Ee- 120 , 2 L O zr 8-0, GO Q° 1 e c Co 8° 2_0e 3:46'

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C ) ' ̀-F7 (X>? (00C)

ey-op 3 -Pckyou6 -1-e SO,IT 10 r X 3 2i 7 2P.,;oo

GT) 9Y-040 (fl ofrxi . 2'

9 9-7G, D.c,o

2f 2-! / R

cf)

(10 -rs.eph # n'brti S'11 --icte-Attc2-r- = 5 x 2

f-r St -f ctprl ----- 720

e) (') 12(o

(n) 4 c 2 y 5c2 = (,0

(1 II) If bC51 h Crn r-} d- 1/4_3(7.1ruLc 0Y1 COVVVVI rfripe

3 c- x 4c-- 1 — 12 7

I ca-I connyv, Ilie e -4-93c -t h _c_Ar D — 12

(-V? /

Ou-esii cvl

0) (0

2

son3-ot'

Sun IS'

= 3oc'

3- 2 k_ ry--) / 71".a._ ctC_ I/1 t 5 -11 3. 2_ ty-v-1 -ry-pr}-1 -11etz_ ( 9 1.r -t kl 0%.)k

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F F

I 1 — 2 c_cS i 20°

+(Jai 4°);'z 4 4q 1,, )(; vl g0 000

H (D Is

Q )

ci+if■.&1-

L,ec4 v 0'11

► t-nla 1 rtotcAr o-( rc vol o rs 1%)

= 3 : 20-

/LAC(?) = i 2d) (0 1-f t7 4:5

re1.1.: 0 CrSicS on pow-cill1 It vvS)

rroty, 0 -ion 10° _, in Friory, 0 aci-)

Ac =.1-Cr (°

Ec -flay,

Pylorv, 3002, (

)z-4- 1—\ ) I b --61-11(0 -}qv, ty-

in a • 9" 0 c00

1 -4 -

•-fror? -by?. 1 1+ -iofroto

ry-N

)--c S I rule-1

cos 20