1977-04-20 — Page 26

頁二第張七第日三初月三年巳丁瑟夏

WAH KIU YAT PO

郭日偏華

三期星

OA = O

tang +

tang

*1977中學會考試題預習專擇

明德社主編!

i.e. D

=

(Subst.):

POA = 4BOP (Common)

POA

· (canữ) ( time)

From (5),

tane

·(4)·

.().

(Ans.)

育教师: 日十二月四年七七九一瑟公年六十六國民都中

(iii) The increased percentage Z

E

n+1

(b) Let the number of students

in the college be N

* 100%

As

數品(廿四)·交長波

BOP are similar (ratio of 2 sides. inc. 4)

Mathematics 24.

Solution to exercise 10

Section B.

1.1.

Solution: Let x,y and z

be the rate of working of man, woman and boy respectively.

ZOPA =LOBP (equianular As) OP touches the circle APB

at B. (Converse, £ in alt segment).

(b) 4CFB = 4OBE- LOBE

(Ext. 4 of 4)

4 Arc = Lorc- 4OFA.

Since OPO- LOCK

(Base Zs isos 4

GARC= ZOB (Subst)

i.e. cr bisects 4AMB.

FROM

k(1.15)n+1

sing

si nücost

sinëcose

From (3)

sinße ose

Im the set of students in the

college

li - the set of studenta

ing brown hair

- the set of students hav-

blue eyes

1)

x100%

k(1.15)

15 x 100%

15%

(Ans.)

14

where the estimated

enrolment a years after 1970) - k(1,15)"100%

k(1.15)*

k(1.15)*(1,15

x+3y+42-6

(1)

2x+8y=30

(2)

(c) rAirB = AC÷CB (4

2x+3y+2= 720

... (3)

bisector properties)

Since A,c and B are the

(1)x22x+6y+8x=1

• (4)

fixed points

(4)-(2)

6y= 120

y= 720

Subst. y into (3)

2x+2=240 (2)-(5) 7z=

subst.

=X

.(5)

120

2= 840

840 into(5)

872

The work can do by 5 men and 12 boy in

hours or

828-

46 hours.

12. Solution: Let xm be the distance that the stone reached. then

u2singe - (40)2since

29

=80sin20%

2x10

0 15 30 45 60 75 90

* 040 69.80 69 40 0

The ratio AC:CB is a constant i.e. The ratio PA:PB is a

constant.

15. Solution: Since (x+1)(2x

-3) and (x-1)(3x-5) are either consecvtive odd or even numbers (x+1)(2x-3)-(x−1 ) (3x-5)=2

(1)

or (x-1) (3x-5 )−(x+1 ) ( 2x−3)

2(2).

from(1)

2

2x2-x-3-3x2+8x

... 2

x=7x+10=0 (x-2)(x-5)=Q

x=2 or 5.

When x 2, the numbers are 1 and 3

When x=5, the numbers are40 and 42.

from (2) 3x2-6x+5−2x2+x

x2-7x+6=0

(x-1)(x-6)=0

x=-1 or 6

n(1)

N

therefore, the required

[n(i);}

equation is:

n(E)

18

n(H) B)

18x + 7

(Anm.)

n(HVE) »

• n(E) + n{H) -N(E

11.

2N + N - 2ON

IN

(1) The required probability

n(H) (3/20)N

(2/5)N

} (Anm.)

+3=2

area” of meni-circle AGB

(11) The required probability

When x=1, the numbers are -2 and 0.

When x=6, the numbers are 63 and 65.

Ans. The pair of numbers are 1 and 3, 40 and 42, -2 and0 or 63 and 65.

16. Solution: Let a be the

common difference of the A.P. then

9+39- 9+7d

454a+9d2

d=0 or

9430

The 2nd term of the A.P. =9+1=10

The sum of the first 2n

terms of the A.P.

-22 [2x9+ (2n−1)x1]

In (48) 2

(A)

area of semi-circle AEG

= area of semi-circle BDC

TBC

zrea of rt. ABC

• † (AC)(BC)

Area of the shaded region

+

(Ag). (^

n(2)

38 -rôས

(Ann.)

(iii) The required probability

n(r)-a{Eų8) - 8 - 48

(Anm.)

13(a) Let the numbers be

y and s

(i) The maximum of y occurs

when

(ii)

2

2x

-2x2

+ 3(4) - 2(4)

2.125 (Anz.) -3x 30.

+ 3x + 3 = 0

+ 3x + 1 - -2

y = -2 in plotted

om the graph

(iii) 3

(Ana,)

or x m 2.2 is the roots of

=3 2

-6

2+3x+1

3r y - 31 – 5 is plotted) The intersections shows

values of (ADE.)

(−1.7% or 1.7

1 + 4 - 2 -

15. (i) Since

1 5: 11/

Let x k

5k and z -llk

* :

AC2 2 - ( — AB2 -AC-BC)

2

AC.

(AC2+ BC2- AB2)+ Acc

aneti tuke into (1)

=n(2n+17)

新數(廿四)

長榮家

MODERN MATHEMATICS. (24),

Answers to Exercise (cont.)

In rt, à ABC

AC2

BC

AB

(a) The greatest distance

that the stone can be

reached is 80m.

(b) The values of ® for which

the distance is 60m are

23 and 63.

13. Given: AP, AQ are the

tangents to the circles To prove: (a) AsPBA, ABQ are equiangular. (b) If PBQ is a st. line, then 4 PAQ=90°

10. (a) Since sine and cose

que moote

the

equation

Area of the shaded region

-

AC BC

1 + 4 + 1 - 2

the general rule is

2

*20-1

(for all nat ral numbers

2 When n = 1.

5k+3

11k+

k+ 3

2512

+ 30k + 9 =

5k+3 11k2+ 36k+9

(ii)

1442

R.S.

6k - 0

2k( 7k -3)

(rejected) or k =

L.S. = 1

2

1 L.S.

the statement is true

$1

Assuming that the statement

is true for n k

Proof: (a) Since AP,AQ are the tangents to the circles.

¿PAB=LAQB; LAPB=¿BAQ.

(2 in alt. segment) LABP-LABQ. (3rd ofA)

PBA

S

ABQ

are equiangular.

(b) ZBPA+¿PAQ +ZBQA

180° (subst.)

·ZPAC-90°

14. Given: 0 is the centre.

of the circle OA.OB=OC

To prove: (a) OP touches the circle APB at P (b) PC bisects LAPB.

(c) PA:PB is constant

for all positions of P.

Proof:(a) OP=OC (radii)

OA.OB=OC' (Given)

therefore

sinë + cos®

minßcomě -

aquare both sides of (1)

- arem of the rt. A ANG

the area ratio

4(2)

ares of shaded region

urea of rt, & ABC

Hence

$

(sine + cose)2 - 16

2

1. (Ama,)

(b) (1)

12

-23

(Ans.)

E- k(1.15)"...(1)

since +

19

+ 2sin@cos☺

12(a) Let the total number of

Since in 1970, 900 students

When n k+1

2k-1 * 20

2s in@con8

16

studenta in the college be N

60

enrolled

Number of boys ́ w

TOUN - ZN

therefore; when n

60

900

Number of girls

100

- N

Substitute into (1)

900 - k(1.15)o

k- 900 (Ans.)

SON

R-

900(1.15)TM ....(2) (ii) When n = 1980.

2

-

1970

2

The statement is true for

- 10

('." sin ́é + com

2«i nücomů –

sinëcomë m 1 ... (3)

Compare (2) and (3)

k.

3-3

Z (anm.) ·

(b) Let the required equation

2

be :

✦ px + q = 0

Hence ;

Number of boys studying

10 history - 100

I

Number of girls studying

history. N I 123-10-02

The probability required

+

N

SON

K-

900(1.15)10

900 x 4,046

-

3641 (Ans.)

the estimated enrolment

-3641

'n = 1; and if it is true for. then it is also true -k

#

for a

- k+1,

therefore, the

statement is true for all

کو

natural numbers n.

Sinhji lid