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