頁二第張八第日二十月三年午戊磨惠 WAH KIU YAT PO
報日僑華
二期星
日八十月四年八七九一曆公年七十六國民華中 育教儒备
k
(1)
育教僑華
12x+20
- (2).
10.(a) Let the first three
terms of the progression
be
4. Solution:
y-5-
-(1)
(x-1)(x-3) 3-x-x-1
y
30000 30000 506----(2)
from (1) y=x+5=-
has only one solution. Substitute (1) into (2)
-(3),
(x-1)(x-3)
-4x+k=2x-12x+20
1978
- 8x + (20-k):
()
【中學會考試題預習專欄
This quadratic equation
9 ar
+a+ar = 39%
(1)
2
729
4x-x(x-1)=2(x
subst. (3) into (2)
30000 30000
HVOR MAI AM
140) % Dale, Presi
must have only one root
4(2) (20
From (2)
:3
3 729
2
4-x+x-2x+6
x+x-2-0
(x+2)(x-1)-8
‚'‚x=-2 or 1 (rejected)
Ans, x--2.
30000(x+5)-30000x-500x(x+5) 60(x+5)-60x=x(x+5)
2
300 : 52
(x+20)(x−15)-0
‚!.x=15 or -20 (rejected) y-15+5-20
.The scholarship-holders
in school A and B are 15 and 20 respectively.
(b)Each scholarship-holder
in school A
30000
15 -$2000
Each scholarship- school B
-500.
12 (ARS)
新數學 廿九
魯榮家
Since
a = mb
Modern Mathematics (29)
Answers to Revision exercise
Paper I
Section A
1. (a)
Length of are AB
60
360
x (2π-r)
x 2 x 5.14 x 8
8.373 cm
Leugth of AB.
xBD
➡ 2(8sin30°)
8cm
perimeter
8.373 +8
16.373. cm (Ans.)
A(x - 1) + B(x 1) FC
12
Ar2+(B-2A)x+(A+B+C)
+B+C 2
A 5, B 7
Substitute into (1)
9r. 39
9r + 9r* 39r
30r + 9 = 0)
2.)
3(3r
1){r
-3)- 0
or r3
(2)
When r
the first
three terms are
• m(í - 23) • n(I +
-(□ + n)ī +(12n
2.
1223
2x(1) + (2)
1en - 7
Substitute into (1)
Since
3k
6K
275 9 and 3 respect-
first
ively
When 3 (the)
hree terms are.
1. (Abs.)
where kia a constant:
greater than zero.
2
376
22) ► 2 (64) 376-
94k
(rejected)
- 3k 6 (Ans.)
k-2
(Ana.)
3,9 and 27 respectively
herefore, the first term
is 27 and the common ratio
19 1/3, or,
the first term is 3 and the
common ratio is 3;
2
(b) If the sum is greater. than 1100, then the progres
sion must be the one with
first term 3 and common ratio - 3. (ice. the progr ession is increasing)
Assume 21 terms are taken
233",
2200
1100
Solution:
Let rom be the radius of the required circle
OD=0Ccoa36°52'
~(3+r). cm DB-5-(3+r)
CD=0C-sin36°52.
(3+r) cm (°
LCDB 90°
•*. x2 - (1 } } r)2 + (}{ (3+r)]2 25r2-(13-4r)2+(9+3p) 25r2=169-104r+1Gx2+81+5%r
60r=250
The radius of the
required circle is
4cm.
30000
20
-$1500
10.Solution:
The equation
2
X-3kx+3k+8=0 has
equal roots
(−3k)2-4(3k+8)=0
2
9kTM-12k-32-0
(3k-8) (3k+4)=0
or
The equation
x=3kx+3k+8=0 has no real roots.:
(-3k)2-4(Jk+8)<0
2
9k"-12k-3240.
(3k-8) (3k+4)<0
<<
+ B(x =
6k - 12 (448)
Section B,
9. (a)
and C - 10 (Aus.)"
3. Total amount returned
from bank A
--810000 (1+ (6)* (1+200)
=$11932,56
Total amount returned
from bank" B
10000+ 10000x
-812250
Therefore, Bank B will pay
more.
$12250 - 81.1932, 56
$317.44
ize. Bank B will pay
$317,44 more than Bank A.
51)+ D{x 451)
41)(x -51) *5(1—A)1:
+25
B.S. - (141)2 - 1
the statement is
'true when n^- 1
Assume the statement is
true when we have
When uk + 1
13. 23+...+(k+1)3
2203
nlog3
log2203log3 log2203. Tog3.
log
➤ 6.0069
Ans: At least 7 terms
must be taken.
1978
「中學會考試題預習專欄
明德社主構
1)2+ {k+!)3
■(k+1)2()2 + (k + 1)]
數學
~(k+1)2{k2 • ** • 1)
(k+1)2 (k + 2)2
(k+1){(k+1) + 1]2
Hence, it is also true
for n kl
Since the statement is true
when oel,' and if it is true
a
for nk, then it is also true for n-kel; therefore, the statement is true for all
patural numbers n,
(+1)
2025
2025
[2 (n+1)]}'> 2025
(廿九)女長波
Mathmatic's 29
Solution to exercise
Section A.
1. Solution:
Since p and q are the roots
of x2+2x+6=0.
.", p+q»-2 und pqw6,
2 2
qppq
==34°
Multiply both sides by
which must
he a positive number
3)2 <h(y-3) 2(x-3)2 - 4(x-3) 40 2(x-3)((x-3) - 9}<0·
5)<0
3 y 5
6. Since the straight line.
Louches the curve, there- fore, the system
Since both sides are positive quantities, therefore it
Is reasonable to take square
root at both sides
(n+1)>45
+ - 90 >0
or 0 <-9 (rejected
__(p+q)=-2pq
pq
(-2)2 -2(6) 3
2. Solution:
Cost price of shoes =$270÷(1–10%)
-$300
Selling price of shoes if a gain of 10% is intended-$300x(1+10%) =$330
Solution:
2(y
-
3)(F
√1+siní 1+sio+ √1-sinx
1+sin30
9) > 0
because a is a natural
BA -71
number)
n = 9
B(13-1)
31.
-(3-3)
Solution:
Amount for $1000 at 3%.
pla, for 2years compounded yearly
-$10000x (1+3%)
810609
Amount for 810000 at 3%p.a.
for 2years simple interest
-£10000x(1+2x3%)
-$10600
The difference between compound and simple
interest A
-8(10609~10600)
Solution:
Since H is the orthocentre
ABFC-LEEC-90°
B, F, B, C are concyclic BM-MC and LBFC-90"
Mia the centre of circle BFEC.. ZACF-180°-90°.
-50°-40°
¿FME=24ACF-80°
Solution:
BC//DE.
LABC-LADE;
LACB-LAED.
BAC-LDAE
ABC
As
ADEare similar.
AB BC AD DE
2DD BC
3BD 6
DC=4
1.e. The length of BC in
4cm.
Section B.
9. Solution:
(a)Let x and y be the number
of scholarship-holders in
school A and B respectively.
學能推理棟習專
文字推理練習(十六)
文字推理辣┛
文肉的理解
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