頁四第張六第二日九廿月八年巳丁夏
WAH KIU YAT PO
育教僑華
Section B
9. Tet P(u)=2• 4?
2n+1 3+1
+
The statement to be provėd is
P(n) is. divisible by
報日僑華
The cost required
v = = 3 ; ( 4 ) 3 + 2 ( 4 ) 2
48
-
#2.5
When n=1
The cost is $2.5.
(Ans.)
1978
P(1) 2.4
2+1 3+1
13.
⇒x=2x+1=9
additional linear
Y÷9 is added... The points of intersection are (-2,9) and (4,9)--
日一十月十年七七九一屉公年六十六二民華中 ★教健康
As the graph shown.
期星
y=x-2x-1 is drawn.
(1)
-2x-8=0
2
Area of 4DEF:Aree of ABC
Solution: Join AG.
中學會考試題預習專欄
128+81
4*=-2 nr x=k
209 19x11.
(ii) x-x-6=()
明德社主機
新數學 (二) 番榮家
Modern Mathematics (2)
Solutions to Test One.
Section A
“The statement is true
Assume it is true, for n=k,
P(k)=11N where N is an
31⁄2 +1 +3⋅
integer
11N
2. K÷l-11N-3 3k+1
P(k+1)=2 · 32k +3+3
2,4
2 2k+1)+27:3
(ane.)
y=-x+7 is added
the points of intersecting
are (-2) and (3,4)
x=-2 or x=3 (Ans.)
2 5
A Section B
9. Given: QR=ST, PQ=PS
ZPQR=ZPST.
To prove :
(a) PRXT is a cyclic quad. (b) PX bisects ZXT÷-
y=2x+7 is added
When nk+I
2k +3, 33k+!
(4) Th
241
+27.334 +1
and plane When is the angle VAG
angle between Line VA
2
3k+1
ACLAR+BC
#10cm.
=16-11N-16′ 3′′
3k+1
3k+1
AG=¦AC
+2743
3k+1
=5cm.
tanz
Area of // KHBC
x7 cm
36cm
Area of the shaded region PUBE
Area of APUH+Arda of long Area ofAKOU-Area of 100B
Area of <RIBC)
4x36cm
2
2
18cm2 (Ans.)
3c 2b
a+2) a+3e b÷
Ph=(a+30)x.
a=(2b+3c) x;
(1)-(2)_3c-2h=(2b=3¢)
x=-1
(Ans
(1)+(2)+(3)
a+2b+3c=2{a
The radius of the ci
=16-118+1:
#11 [168*33221]
P(k+1) is also divisible
by 11.
Since the statement is true for n=1, and if it is true for usk, then it. is also true for n=k+1} therefore it is tru for all natural numbers n. Let the circle be
22
x+y+2x+2[y+c=0 ;
the centre of the, ei rel 45 (-g-1).
Since the centre is 2x-3y+6=0
-2(-e)-3(-1)+6=0 24-31-6=0.
(1,1) is on the circle:
1 +1 +2g(1)−2f{1}+c=0) 22g+2f+c+2=0,
(1,3) is on the circle. 1^ +3 ̃ ̄ +2g(1) +~f{3}+c=0; *£g+ft+c+10=0, .(5) Solve for gu and e
· ZVAG=63°261
Let M, N be the mid-point points of AB and AD,
The angle between the planes VAR and ABCE is angle Wis.
Turf A "MG:
VG-10cm, GN=1BC 3cm
VG 10
5333
LVMG=73°19%
)The augle between the
pines: VAD and ABCD anin J.
Int4YNG
V=10cm, NG-LAB=4cm
10.
As a po
rele
(ans.)
The circle
22.
x + - +2=0 the centre is (0,2)
the radia
+1 -0
The points of intersection
(-3) and th,
(ins.)
(4) AP -- 16.
(Ans.)
BP-GT-CH
=TP+AP
× (3−t) + (C+3)
:(ars.)
1)From part
(ius.
LVNG=68"129 (Ans.) A Venn Diagram can be eanstructed as fallow
30
20
S
5
(Ans.)
40.
20
70
passed
25 - 26P 1-201
of the
er of the circle
AB Therefore, as I moves. vill move on the circumference of the circle Vith centre at B and radius equal to the length of diameter of the given ciruje
(T1)AB= √(6-2)2+(0-5)2.
The equation of the Focus of Iis:
(x-6)2+(y~0)2=5°
2
4y-12x+11=0 (äns.)
Proof:
(a) R-ST, PU=PS, (Given)
¿PQR=4PST
LAPIR APST (8.4.S.) 2Q0P=ZPTS (Corris As)
< PRX+2QRP=2xt is
(Adj on st. line) PRX+ PTS 2rts (Subat.) PRXT is a cyclic quad.
(opp. 45 supp.)
(b) Join RT.
ZRNT-ZPXT
<HXP=4RTP
(4s in the same
segment)
Since POR APST
(proved)
PHPT (corr. sidesz4)
4 FRT=/PTN(Base 4 1809 A
S41XP & CPXT (subst.)
i.e. PX 18 the bisector
of <EXT.
10. Solution:
Let x and ym be the length and the breadth
of the plot.
42(x+y)=700..
(220)(y-20) xy, from (2)
xy-20x-20y+400=2xy
xy-80x-H01 100 0 from (1) x+y=350
x=350-y..
.(1).
(2)
Put x=350-y into (3).
· (350-y)y-80(350-y)-80y
2
+1600=0
y -350y+26400-0
(y-110) (y-240))=0
y=110 or 240
When y=110, x=240.
y=240, x=110
The dirmension of the"
plot is 110m"
by
240m%
or BC,
數學(二)
交長波
Since
BE-TEC
Mathematic.
ABE
Soluti
ABC
Le. & ABE¦¿ ABC.
Section AB
Solution:
ABDE-4ABE
ABC.
\(x® +x+2)+(Bx+C) (x−1)
− (A+B) x2 + (A-B+C ) x «42A=C)
(A+B)x +{A=H+C}x+(2,1−C)
A−B+C=0.
2A-C=-1..
(1)+(2) 2A+C=5
(Ang.)
Substitute into (3) and.
(3)+(4) 4A=4
A=1
(1), we have
B=4 and C=3
5. 4(-2)87-18
>7-1x
8.
Multiply both sides
(x-7)(x+6)70:
3x+5
(ans.)
*.x-6 orx27
(ins.)
t-1 (~5)=2(-5)="
3 sine +4cond
5séng-beos!
co 88 (35
sin! CONS
sing
cose
3 tane + 5tene-6
6)
{m+1
(ans.)
(Ans.).
mx (x−1 ) -m(m + 1 ) = x(x−19 (m−1).
-x-m(m+1}={}
Since the equation has
equal roots, the discriminant
-- D=144m (m+1)=0
• (2m+1)=0
the Perimeter of the
triangle is, then the Feriveter of the 2nd
triangle is 18, and that
for the 3rd is 18. The sum of perimeters
༤༠༤༣༥༧-s+ཙsཝཾ
#25
Therefore, the sun of: perimeter of triangles starting with a triangle of perimeter 30cm is
2x30cm=60em. (Ang.) (b) If the area of the
triangle is then the areas of the 2nd, 3rd,
are ^, (4) 1,
respectively.
4th.
(40)
The sum of areas
= A + ¦ à + (?) ?
The area of the equilateral triangle of perimeter 30cm. =!(10)(10)sin60°
-50siu60o
2
-43.3cm2
The sum of areas
43.3cm -57.73cm
(Aus)
12. Let the radius of the beli
be rcm, the cost is Sy.
3
th
2
where ki and kg are
zero constants.
when r-12, -37.5
37.
1101-
.5=k ̧ ( 12 ) + k ̧(12) "...
. . . . - (1)
when r-8, y=13
2. 133 - k1 (8) 3 +k,, (8) 2
(2)
Solve k, and ky from (1)
and (2), we have
**
m=-}
(ans.)
when r
Where E. U and M are the Sels of denn
în English, Chinese, and Mathematyes.. respectively
(a) (4) Only 5 students passed
all 3 subjects,
the probability-200
-=(1+41og2-10g5) logx=(log10+log2"-log5)
og
(ans,}
(11)70 students failed
all subjects, "the probabili 200
70.
10gx-10g10x2
(Aus.
(ii)The number of students
passed in one subject only
=20+30+40=90
·90
«tle probabili
200
(Ans.)
(b) Number of students passed in English÷30+20+5+5.
60
Number of students passed in both Mathematics and. English
10RX-10g25
10gx-log2 x=2
Solution Let p and
xbe the roots of
?;p+q=1_and_pq»-6
2.2
p2+q ={p+q) -2pq
2
= 1 * −2 (−6 ) -
=13
*(pq)?
-(--6)2
=36
The required equation is
x~13x+3=0,
3."Solution;
2 sing tank.
2+5
=10
the probabili
2 sing
(Ans.)
sing Cose
15.
$(2cos8-1)=0
sin}=0
or
180°,360°, 60°or 300°.
4. Sofation!
5sina+6cosA
7c09a-3síða
BD-3AD or 4AB.
ABDE
AABE
Simarily 4ECFAABC;
ADF
- ABC ADEF=1-3(-Z) of 4ABC
1.P. DEF 16
F-1 of DABC
Aile Aren of DEF: Area of
4 ABC
-7:16,
6. Given the circle XYZ
touches the triangle ABC
· AB=c, BC=a, &C=b
To find: BX in terms of band
B
7. Solution:
BG
5. Given: 1-3
To find:
EC
Solution: BX=a-CX
Sa-CY
~8-(b-AY) -a-b+AZ =a-h+c-BZ
2BX-a-b+c
• BX = ( a − b + c )
'A' share :B'share
=(4000x12):(2400x4+1800x8)
=2:1
The profit of B=815090x=
=8500
8. Solution:
Let xgm be the weight of zinc that added to the alloy.
then weight of tin in the mixture is 400(65%)gm and. weight of the mixture is (400+x)gm.
2
400(65%).1-66%
400+x
260
400+x3
780=400+x
x=380
Weight of zinc added to
the alloy is 380gm.
5sinA+6:oSA
COSA
7cosa-3sinA
COSA.
Stand+6
7-3tanA
5(-4)+6
7-3(-)
2
--33333