買四第張五第二日九十月五年申庚曆夏
1980
WAH KIU YAT PO
郭日僑華
二期星
dy:
(a) Let P(x,y) be the poin
中學會考試題預習專欄
,2), dy = (1)_1
equation of the
7+3
附加數學卷二(夜完)
tangent at (1,2) is
Mpp-
y-2-1-(x-1)
1.e. y-x-1-0.
Ans
c−1)2+(y−2) +k(y-
建議參考資料
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ADDITIONAL MATHS
Section A
(I):
(1+2x)3 (1+3x)“
=(1+3(1) (2x)+3(1)(2x)2 ........]
(1+4(1) (3x)+6(1)2 (5x)2 -
-[1+6x+12× 6...]{ 1+12x+54x-
-1+12x+54x+x+72x
+1+18x+138x2
Since u-x-14°
du-dx and X+2=
1-}{u+3)√d du
Diff. w.r.
4 x 3 + 2 x
1-(2xy+
x2x+1+y
-kx−k=0}
~+y ́~(2+k}x_{4-k) y
which is equat
circle because
(1).
日一月七年〇八九一层公年九十六國民南中育教僑叢
put n=1,
1-272(α-p)
U2-22(-6)
−√2))-1
Ans
-312 {(3+2√2)-(3‐2/2)m2
-kx425k-9-0
which is the required.
locusi
Ane
2 (0)×0.
of a
(ii) Since integer integer is
also an integer.
(b) The locus is
Is an
kx −25k+9=0
{n+1
(1) When k=-1,
(1)
nteger.
+2
+7-25-
coefficients of x and
are equal
coefficients of xy is
zéro..
adius=(2+k
4+4kak +16-8k+k -2044k
Substitute (1,2) into (1)
− ( 2+k) ( 1 ) − ( 4−k)(2)
C represents a circle passing through (1,2):
(11)Centre of C is (2rk, 4-k)
the required area
3-5
qunity:
Ans.
(11) When k40, and k1
is an ellipse.
(iii) when k>0, (1)
hyperbola.
If k=0, (1) is y
)-amp(Z ̧-Z
rele
-LCAB
Апе
(rif) Yes, U is an integer.
for all positive integer
Reasons:
y3 which ia a pair of parallel lines.
If
(1) is
25:
2_25(-3)+9-0
1.e. 9x-25y. =0
(3x+5y) (3x-5y)-0
i.e. 3x+5y=0 or 3x-5y
which is a pair of straight lines.
Ana.
(c) No
U are in
and if
k+1 are
then U
integer
and so
inte
is also an +2
by mathematical induction, U is an
integer for all positive n.
V is not an integer. Reasons; From (2)
▼1′′2√2(x+p)
72(1+/2+1-12]
not an integer
<amp(v)<
(a) Given: f(x)×f(a−1)
Proof: When x=0, u=8;
When x-a, umo;
du-dx
(1)
√ xz(x)dx= f (a-u)?(a-u) (-du)
√, (a−6)f(a+u) du
(a-u)f(8-u)du
* af(0-u)du-}, uf(a-u)du
u)du- ̈ ̈uf(a-u)du
7(u)du= f^uf(u)dù
Using (1)
Ans.
The above can be written
-] xx(x)dx=u], t(x)dx-] 21(2)dx
+5=0
042)
2a+1
at (2,0),
-26-2
slope of tangen (2,0) is
But
on
11. Given
When k-1, the locus is
An:
y cos(ainx)
· -· -16.
am-sin(sinx)ers
(2a+1)
4(2-2)-5
2(2a+1)
The required volume
[ensx cos(sinx) cosx +sin(sinx)(sinx)
-COS X COS(sinx) +sinx-sin(sinx) Ans
106,X-log 16×1
Using the fact. 10&h=log.
the expression become s
log, 16
-8648-5=0.
16b+6=0.
equired locus of
log
(-1,2)
M(a
(iii)
(2+k, 4-k)
10g
log, x-Ingh =log, (196,x) −105x-2=0
Let t-log
(t+1)(t−2)=0
t=-1
10 x=-1
log,
“Ans.
(1,2)
S1ape of the tangent at (1,2) is
#1
16) dy
0<amp{v}<t
Cos120'
120
(6)
16
~T(9+48~(~9–48)]
-114 cubic units.
-1-0
2=√2^-4(1)(-1)
If = -3 and amp(z ̧)=9
-1±√2
−(cisz_)(Jeisė)
Jein(e)
-(ci)(Jein(+0)
(*) Un,2-2√2(x®
(1)
(2)
from(1)
(a) The Argand diagram is
***(1+√2)2
(5+2√2)
*p*(1−√2)2
=p°(J=2√2)
V.2-272 (3627)
B(%2)
sin x cOA
sin (+)
f(x)dx. Ans
)+008 (u+1⁄2)
(3)
Bin xicOS
win K CÓ B
sin
cos xdx from(3)
sin x cos x
above
Ans
+1.272X
+2/2("+p")..(3) n+1_¡B+1) from(1)
sin
(1)dx
Ana
(c) Since LAOB-LBGC-
LAOC
21
20.1 272 x-)+
*+/2(< " + p")]
4A0B-4BOC=AAOC (SAS)
AB-BC-AC
(c)
ng (a), equation (2)
and let f(x)·
AABC is an equilateral4.
Area of ICD
1 (2+2)x6sy.units
15
299. units
Area under the curve
the required tangent at (1,2):
y-2-(1)(x-1)
y-x-1-0
Centre (2tk, kk)
(c) Since centre
lies on x-y=3,
2+k−4+k=6
Substitute kwh into (1) in
(b)
22
+3 = −(2+4)x+(4-4)+(5—4)×0
2
which is the required
rele
Section B
(72)
=3xq.units
B(-5,-3)
A(5,3)
−2}>{3(¿" -p")+2k («®+p")
Similarly,
(b).
(i) From (1) and (2),
from(3)
An's.
sin
sin
sin
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