育教僑華頁三第張五第日六初月五年巳丁屦夏
1977中學會考試題預習專欄
明德社主編
WAH KIU YAT PO
-1000[1-(0.98)}_m_
=1.000 - 1000(0.98)") m
(c) The total distance. the
train travelled until
it came to a stop
│數學 - ®
同等課程:
※和建議答案室
20 1-0.98
1000 m
女長波
Suggested Solations
"For HKCEE 1977
Mathematics 28
Section B
11. Soltuion(a) AOB 15°x2
=30°=
The length of the radius DA
inte
30
cm
b) Area of the shaded sector
AOB-
1
2
sq.cm
75
sq.cm
(c) Area of BOC=
sin
sq.cm
225
T
Sq.cm
Area of sector BOC=
m
14.Given ABCD is a cyclic
quad. AD//EC
To find (a) which two
triangle are similar to PAD (b)If PB-8cm,PC
10mm, Find BE
Solution:(a) Since EC//AD
(Given)
ECP PAD
corr. s EC//AD
ECP ADP
corr. s EC
EPC APD
Commom)
PAD PEC
are similar (Equiangu-
lar)
PBC=LPDA (Ext, ¿cyclic quad) ZBCP PAD (Ext./cyclic quad) ZBPC=/APD (proved)
PAD are similar (Equiangu-
lar)
PCB
APEC and PBC are the tri-
andles dimilar toSPADI
(b) Since PEC,PCB abd PAD
are similar to each other.
PB PC RC
=
PE (Equiangular
10.78 10
TO
PE =
7=3+k
1.e. k=-2
郭日僑華
(b) since 2x2+x-8=0
1.e. 2x-8-x=0 i.e.
i.e. x2-2-2- ~
The equation of the line added to the graph is
-y=2-
The solution of 2x+x=8=0· are x=-2.3 or 1.80
ー數學試卷建議答案全卷完
岑俊彦·
Suggested Solutions:
For HKCEE 1977:
Additional Mathematics. 9(a)
·P63,00
From the diag
w=
三期星
(a) y=x(x -7).
At A,D, y=0·
x=0,± √%
日二十月六年七七九一屦公年六十六國民中
For stationary points,
x = (x2 -7)3 +x•3(x −7)2 .(2x)
= (x2-7 )? ¦ ×2 −7+6x2 }
=7\x2 -7)2 (x2-1)=0
17 or+1
y=0 or+216
A(−17,0),B(-1,216),
C(1,-216),D(7,0)
(b)
Let I = f x(x-7) dx
If u=x-7
du-2xdx
1 (x^-7) xx
= = U+C
du
=}(x2-7} +C
Ans.
Ans
(c) Since the regions ABO and
OCD are the same, we have
Area of shaded regions
dx
=2 fyax=2 ["x(x2 -7.)3 d
(b) AX +Bxy+Cy ±4.
Since it passes through
p(1, 13) and Q(1, - £3)
A(1) +B(1)(-2)+C(-)-4
A+
√3B+20=4
and A(1)2+B(1) (~~~) +0(−−−2}
1.e. A-
and 3 give B=0
Now, AX + Cy=4
An's.
©
Differentiate with respect to x:
2Ax+20y
*
A(1)+c()=0 asp(12)
A-12 C=O and X
4A-C-0
also p(1,2) lies on
A(1) +C(3)2=4
A+ 2/2 C-4
4A+3C =16
Solving (5 and 6, we have
A-T
C=4
The curve Ax +cy.
TC-radius
=r
The equation of the circle.
-2. (x-7)" from resultof (b) 금 (-7)
i.e. x2+4y2=4
is an ellipse.
is (x−m)2+(y—n) =m2
Ans.
and t
sq.cm
Area of the segment BDC.
375
225
150
sq.cm
Solution
BE =
122
8cm
4 cm
15. Given:AS ABD, OBE are.
equilateral tri- angles
To prove:(a)ABES DBC
(b) A,B,C,D,K,
are conecyclic (c) Calculate
LAKC
Since it passes through P(2,0),
• (2-m) + (0-n)2 -m2
4-4m+m2+n2=m2
4m=n2+4
(b) m=
n=
Ans.
(a) DAF CBF 50°
Draw AG DF
DAG FAG-25°
DG=1xsin25°m=0:4226m
DP=2x014@ptim= 1
(b) BD= 2 +1. m
(c) Cin/DBG=
AN,
0.4226
2.236.
LDBG 10°54
LDBF=2x1.0° 54
21°48'
13.Solution(a) The distance that the train travelled
in the nth sec...
=20(0.98)
FIL
(b) The total distance that
the train travelled in
the first n seconds.
20(1-(0.98)")
1-0.98
m:
Proof:(a) Since AsABD, CBE
are equilaterala
(Given)
AB=BD; BE BC
LABD = LOBE 7·60°. (properties of eg.. ZABD+LABC= 2GBE+ ZABC 1.e.4ABE=4DBC
ABEDBC(S.A.s)
(b) ¿CDB=¿BAE (corr..s=15)
A,B,D,K,are concyclic (converses in the same.
segment)
:(0)| ÷ AKD=LABD (is in the same
segment)
Since ZABD=60° (proved)
AKD=60°
1.e.4AKC=180°-60° {Adj...... S
on st.line)
=120°
16.Solution:(A)Since P(3,7)
lies on the graph y=x +k
2401 sq.units.
11-4)y (2x2+3x+1)=1
Differentiate with respect
12. (A)
/ (b)
tox:
Ans
2m-2
Ans
2y¶x(2x3 +3x+1)+y^(6x+3=0
when?
x=0,y=1
t=2n
Ans.
Ans.
(c) Substituting m=
11
into 4n+4
have 4(2+2)=()+4
2(2+3)+4
16+85 t +16 t2 8s
the loucés of Q is given by
y =8x
Ans.
Differentiate again with respect to x:
20. (2x2+3x+1)+/
2yX(2x2 +3x+1)+
2y(6x+3)+2yx(6x+3)+
y2(12x)=0
Substituting dy ==, x=0,
y=1, we have
2 (−3 ) ( − } ) ( 1 ) + 2( 1 ) <¥(1)+
2(1) (~~) (3)+2 ( 1 ) (−) (3)+0=0
-9-9-0
dy
Lxx = 27
Ans.
t=coso+isino·
t =(cose+ising)
=cosno+isinnd- =cos(-n)+isin(-n
cosisin no
-n
=2cos no -2isin no
8cos 0=2(16cos 0)
=(2cosa)
= (t+t!).
= (t*+4t+6+4t+t^"")
An's
= 1⁄2 ((t"+t")+4(t+t3) +6
[2cos40+4.200320+6]
=cos40+4cos20+3 Comparing with
8cos" =cos40+AC0520+B
A=4
B3
(c) From (b),
8cos &=cos40+4cos28+3
Bcoso -2=c0s46+4cos20+1
put Bcost-2 =0, this becomes the given equation
cos40+4co520+1 =0
Now,
8cos Q-2=0
cos"Q
cose-1 or —
possible)
Q=45°,135°,225,or 315°.
Ans.
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