1978-11-27 — Page 29

育融情帶其一第張八部日七廿月十年午戊睡

WAH, KIU: YAT PO

郭日橘

-

Section B

1979

中學會考試題預習專機

Modern Hathematics (8)

Suggested

test 3.

Section

'As shown in the figure,

0 is the centre of the

circle.

The area of the v-uliaped

polygon

={Area

of trapezium (ABFD).

*(Area of trapezium BCEF)

-3(2)(10×14)+}(2)(10614)

(Ans.)

2. Let the length of a side of

the cube be a

(1) Initial area of a face

of the cubewa

Final

of

24

06

men of the fuen

cube(142

*100 -(1.96)2 -(1.4a)

final length

of

side of the cube

2 CFD=180′′-(60°+55°)

4EOD-2BFD

=130°

But ZEOD+ZACB=180°

ZACN=180′′- LEOD

-180°-130"

(anu,)

30°

Let BD-xen, Aboliem

Inert. AABD

40°

In rt. BACD

h tan30. -(2)

Substitute (1) into (2)

10-htan40.

percentage change

1.40

1004

(a) Let T(n) and S(n) denote the nth tern and the sum

of first n terms of the

sequence repectively.

(1) _ _T(1)-8(1)

−(1)[2(1)+1)

(11) T(2)=$(2)-s(t):

-2{2(2)+1}-3

T(3)-8(3)-s(2)

-(2(3)-1)-2(2(2)+1)

-11

(414) The progression is on

arithmetic progression

becouse

T(n)=5(n)−8(n−1)

n+1)=(n-1) [2 (n-i) 2n2+n-(2n2-Jn+1)

▼{r}~T{r-1) -(hr-1)-[1(x-1)-1)

sconstant (meÐLIMON

difference)

(b) Assume a teras aust he

taken

(2n+1)> 171 +n÷17150 (2n+19)(n=9)> (

<} (Fbjected)

>9 (accepted)

6日七十月一十年八七九一公年七十六翼民豪中

-=2(15.5889)x(9.38)

Ans: The cost for 320

students is $2440,

(Ane.)

12.

14.

(a) From the given tal

huve

(a) Since the circle passes

through (1, 1).

sumber: probabilitý

1203 1000 25

100

1000

200

1000

100

1000:50

150 1000 20

250 1000 T

The required probability

-( { b ) ( t ) + ( } ) ( {} _ ) + ( } )2

The required probability

(중국) (조각)+(중국) (룩)+(주) (초급) +({}) (~ 5 ) + ( } ) ( } ) + (})(?)

+(_)(2)

2200 10000

(Aus)

The required probability

128

1~+12+2!(1)+2g(1)+3=0

21+2g+6+2=0~~~~~~(1) Also, the circle passes through (1, 3)

12+32+2? (1)+2g(3)+c=0

• 2f+6g+c+10=0---- (2)

The centre of the circle is (−f, -g) and is on 2x-3y+6=0.

2(−1)~3(−g)+6=0

21-32-6-0- (2)-(1) 4g+8=0

-(3)

g=-2 (ARS.)

substitute g=-2 into (3)

Lm0⋅ (Ans.)

substitute fTM0) and into (1),

2(0)+2(-2)+c+2=0

(Ans.)

(b) The centre of the circle

in (0, 2).

15.

radius- {{ 1-0)2+(1-2)2

-√2 (Ans.) (*1.4142)

(i) From the given table.

(ii) Final veline

cube #f1.4a)-

-2.744.3

The perCED!

change

sino.co

100%

-174.4%. An*,)

-0,7276(Aus

J

吉

m0.7276 (Ans.)

(1) Since T(n)sthe nth teru of

the A.P.

=5n-3.

~T(1)-5(1)~3

the first term 2.(ans) T(2)=5(2)~3

10tan30

ëtank0

10(0.5774)

*1+(0,8391) (0.5774)

#3.8805

AD=3.8895cm

(AMK,)

Since d and p are the roots

the equation

(1) Idiop

(ANS,)

(11)-

10 terms must be ken? (Ans.

(b) coxZIIT-

A?•¥Þ«(1)(−1)+(2)(1)

-cos EPF=0

[PT=94" (Ans.)

(a) Let C-cuat of the school

pienic

ninumber af students

3(3) 2(4)+3(+)

(Ans.)

vhe

kand

constanta.

When C-935, n=195

(0)

Since ya(ax-2)

97*»k, +105k^~~~~(1)

y=k(8x=2) where k in constant

When C-844, n#92.

-(2)

2=k(Ja=2)...

-(1)

(1)-(2).

When xeh, yuk

b=k(ba+2) Lək(2a-

(1) + (2) 1.

corimon, difference

-T(2)-T(1)

(AUB).

Substitute k„-7 into (2)

8b4=k,+7(92)

#7-2

(Ann.)

(ii)Substitute awl into (1)

2-k(3-2)

3-200

(11)5(10) the sum of first,

ten törm® --

-29(2(2)+(10−1)(5)

245 (Ans.)

(a) As shown in the figure,

is the centroid of AABC.

VO perpendicular to

plane ABC.

Since AABC is an

equilateral tria

BDLAC,

OL-D

In rt, AABD.

BDABOG 30°

Marks | Midɩvalue | Frequency

20-30

30-40

35

40-50

45

50-60

60-70-

GGGG

25.

10.

55

20-

28

70-80

00-100

mean score

~{4x25+6x33+10x45+20x55+28

(11)

x65+20x75+8x85+5x95)÷100

(Ans.)

(-5.196cm)

•*.

OD=3(3/3)

ii)From the graph,

(−1.7321cm)

the pass mark is 52.5

Inert. AADV AD=1zfcn=3c#

(Ans.

(n) (3x-4y-14:

-(1):

VD-JAVI-AD

- (2).

-91cm (*9.5394cm)

(2)x4 Rx-4y+16z=0-

-(3)

(3)-(1)

5x+30x-0

the formula is C=200+7B. (Ann.)

C=200+7(320)

#2440*

5

0.1816

ZODV=79°32' (Ans.)

(b) aren of & ABC

►ģ{AB} {a€) { gim£BAC) ➡k(6)(6)nin60o „9/Jcm3 (Ans.) (-15.5889cm VU-√VD -OD -√91–3

-9.35cm

Volume of the wolid

(1)x2

6x-8y-28z- (2)x3

6x-3y+12z=0.

-(5)

(5)-(4) 5y+40x=0

y;28:1 x:y:2=-6;−8:1 (Ans,);

(or 6:8:-1)

(b) Let x--6k, y=-8k, z-k. when kisa non-zero constant

3(~6k)3~(~8k)3+8(k)3 -16 -128k3-16 k3--

x=-6(-1)-3 y=−8(-4)-4

*--*

(Ans.)

(b) When n=320

When x3

[y=2(5=2}

(ang. )