頁二第張八第 日五初月五年午戊歷夏
'WAH KIU YAT PO
報日僑華
六期星
1978
☺ (4n+1)
(n=1)π
sin230",
12.
日十月六年八七九一屦公年七十六國民華中 育教僑華
-cost-cost(1-cos
8
「中學會考試題預習專機
Putting n=0, - or I
OF
E
Ans.
i.e. 0.94, -0.17, -0.77.
09
Ans.
明排社主壽
岑俊彦
附加數學建議答案
Suggested answers for
H.K.C.E.EL, 78
Additional
Mathematics,
when Ost the required value is
10.
Ans.
Section B
3 2
(a)y=x2+px +qx+r
at x=-2, y=0,
̧0=-8+4p-2q+r
i.e. 4p-2q+r-8.
akcost-3cost
Βι
Апа
8. Given A(1,-
at x+4, y=0,
0=64+16p+4q+r
1.e.16p+4q+r=-64.
(2)
Paper I.
Since the curve touches
Section A
(1+k)x+(3−k)y=2(1+3k)
Putting k-1, we have
(1−1)x+(3+1)y=2(1–3)
y=-1
Again putting k-3, we have (1+3)x+(3-3)y=2(149)
i.e.
4x=20 x=5
The co-ordinates of P are (5,
2. Given y-axK
Zak(k−1)Ax*-2
Ana.
Substituting these values
The equation of the straight line with slope
passing through A is given by
y-3=m(x-1)
(a)i.e. y-nx−3+n=0 (b)The, distance of the
Ans.
straight line in (a) from the point (2,6) is 6-2m-3+m___±√2
√ 12 + ( − c ) 2
Squaring both sides,
(3-m)2=2(1+m2)
2
9-6m+m2 2+2m2
the x-axis,
dy_3x2+2px+q=0
dx
at x=-2,
3(4)+2p(-2)+q=0.
ie. -4p+q-12.
(3).
Solving (1), (2) and (3), we have p=0, q»–12, r--16.
(b)The given equation becomes
y=x2+(0)x2-1
2
-12x-16
-x-12x-16
-3x2-12-0
(a)L¬AB÷BC+CDE+EA
-2x+2y+Tx
Ans.
Let B be (X,
(b)Ares of the fig.-A where
2
Tx
A-2xy+
But y-(L-2x-Tx);}
ADS.
•A=x{L-2x-x)+2
-L-4x-27z+fx
dx Putting
L-4x-7x=0
dA
ax
L +4
Now,
---
-4-20
The area A is a
m2+6m-7=0
2
-4=0
(m+7}{m-1)=0
.m=-7 or 1 which are the
.y-32 or 2
Into
required alopes.
Ans.
.e. points' (2, -32), (-2, 0)
2
Maximum when
24 +214 25-0, we have
(c)
dx
+2x KATK
-6x0 when x-2
40: when x=-2
(c) If L=1+4
Ans..
and P be (h, k).
1+x1 .b-14*1 ⇒x -2h-1
k-^2-⇒y1-2k
Since (X, Y1) curve y 4,
(2k)2-4 (2h-1)
2
i.e. k-2h-1
on the
the locus of P is given
by y2=2x-1
Ans
x2 - k(k−1)Ax*
-2Ax
i.e.k(k−1)Az*+2k41.
k(k-1)+2k-2-0
k-k+2k-2-0
2
k ̄+k-2-0
(k−1)(k42)=0
„k=1, or −2
3. (1+21) 2 (1+21) 2
(1-1)
1-2i+i
1441-4
−3+41
3+41
--3;་
4.(x-2)(x+3)<2(x-2),
(x-2)(x+3)-2(x-2)40 (x-2){(x+3)−2]< 0 (x-2)(x+1)40
Ans.
Since AB-AC ADLBC
6-3
BC
equation of BC through
Ans.
D(2, 6) is
y_6--=(x-2)
.e.x+3y=20
(d)The equation of AB is
y-3=(-7)(x-1) |
1.e. 7x+y=10
Solving 7x+y=10 and
x+3y=20 we have
the co-ords. of B are
Ans)
(1, 12)
Ans
5. f(sing+cos) ax
-(ain+2sin cos+co2)
-[(1+28in cos)dx
-[(1+sin2)dx
-[(1+sinx)dx
-x-c08x+C.
6.y-x-sinx
Differentiating y with
respect to x.
dy
1-c08x
dx
For small changes of x, y,
we have
dxx
• 8y + (1 – coax) 8x
(1-cos)(0.001) when
#(1-3)(0.001)
#x-0.001
(0.001)
20.0005
The approximate change
of y is 0.0005.
7.00830-sing
i.e.cos30-cos(0)
30-2nf (1–0)
.*.30-2m+(9)
4 +
@_(n+1)
8
or 30-2nf-(1-0)
29_n-
6-(2n-1)T
Ans.
The general solution is
Now, BC2 BD
the maximum value of y=0:
and the minimum value of y=-32.
(c) -3x2-12
-36
when x=4
Ans.
When x=4, y=(4)3-12(4)-16-0
the equation of the
tangent at x=4 is
given by
y-0×36(x-4)
1.e. y-36x+144=0
(d)The graph is shown below:
11
2(3-2)+(-6) (a), z=h+31, P2-behi
√10
{AD[=√(2−1)2+(6-3)2
10
area of LABC-|BC|· |AD||
(a) In ADBN,
DN-BDcos2
-1xco82
*.DP=2DN
-2cos20
AP-AD+DP
»1+2cos20
-1+2(1-2sin28)
-3-48in 0
(b) In AAPQ,
PQ-APsino
2
5 sq. units.
P=644i
Distance between
P1 P2 P2#P1|
10/10
Ans.
(b)z=x+iy
Ans.
Ans.
-(3-48ine)sing from (a)
-38in0-48in
In A PBQ, PQ-PBsin3
-1xsin3
(*.sin30-39in-sin3 Ans.
(c)Putting xsing into the
equation
8x3-6x-1-0, we have
Bsine-6sine-1-0
i.e.4sin3-
0–3sin6») :
3sing-sin-
Comparing this with the result of (b), sin30-
*.38-210, 330°, 570o,
690°
*. 9×70°, 110°, 190°, 230°
.the 3 roots are sin70" (or sini10), sin190°,
2+1
-32)
12~21) = |(x−4) + (y-3)il
=(x-4)2+
2+(y-3) 2
[2¬22| = |(x~6)+(y-4) il
~(x-6)2+(y-4)2 Since 2-2 | = |Z-22|
.Area of figure A
-x(L-2x-x)+
(1)((14)-2(1)-(1)) +7(1) 2
~ ( 2 + 1 ) m2
Suggested answers for H‚K.C‚É‚È.. 78 -
Ans.
Section A 3
Ans
(x-4)2+(y-3)-(x-6)2+(y-4)2
x2-8x+16+y2-6x+9=x2-12x+36
2 +y2 −8y+16
4x+2y-27=0 which is the required relation between x and y.
(c)zz=12+71
Ans.
©p(4+3i)+q(6+4i)+(12+7i)=0
(4p+6q+12)+(3p+4q+7)i=0
Equating the real and imaginary parts
4p+6q+12=0
3p+4q+7=0 Solving: p=3-
(d)7-4-31
7-6-41
7-12-71
9=-4
Ans.
0
¿(4−3 i ) + § ( 6-4 i)+(12–71)=0 (4x +66 +12) − ( 35+4)+7) 1=0 ·
.4k+65 +12m0
·3 +43 +7=0
Ans.
Ans.
Additional Mathematics, Paper 11.
2 2 3 1, xy+2x y −xy”-14
Differentiate with respect
to x:
(x2+3x2y)+2(2x2 y2+2xy2).
dx
-(3x+3)=0
at (1,-2),
(1)+3(1)(-2)+4(1)(-2) dz
+4(1)(4)-3(1)(4)+8-0
dy-18 18.
dx -19.19
which is the slope of the tangent at (1, −2) Ans. 2. y-r sin
1.
dy-sin(x)+x
+
(sin1)
-2xsin +x2 (cos) (→)
=2xsin__cos
··.d2-2uin •2x (cos) (→+p)
() ()
-2sin cos pein
3. (x32) 10
The (r+1)th: term
100x30-31
-10x30-5r
r
* 30-5r-0⇒r-b
-6x-4y+9={
centre (3, 2), radius= 32+229
Let the equation of the tangent be y=mx
i.e. mx-y=0)
The radius-22
(3)–(2) a2+(-1)2
.4(m2 +1)=(3m-2
2
5m2-12-0
(5-12)-0
m=0 or 12
the required. tangents are 5y-12x and y=0
17. Let 3x-sin
ट
..√1-9x-cos dx-co
8.
when x-0, 6-0
9xdx √1-9x 3sin cos
3cos
-I sinfide
- (cost] =
--[cos -1] -1-
(a)(1+ax+bx2)8
-(1+(8x+bx2)8]
Aus
Ans
-1+ C (ax+bx)+ C(ax+bx2)2
3
+ Cz (ax+bx2) 3+.......
=1+8(ax+bx2)+28(a2x2+2abx
+........)+56(a2x2+..
=1+8ax+(8b+28a2)x2+56(ab
3
Ans.
the term independent
of x-1006
10.9.8.7
10 9 8 7
4.3 2.1 -210
ADA.
4. Using De Moivre's Theorem,
cos30+isin30 -(cos +isint)3
-cos3+3 cos2(isinf)
+Jcos (ising)2+(ising)3 (cos-Jeossin3)
+(3 cos2sine-sin3)i Equating the real part: cosmcos30-3cos&sin2&
3
(b)(1+ax+bx2)8-1+8x+
+56(abra3)x2+.
result of (a),
Comparing this with the
Ans.
8a-8
8b+28-52
.*.b =3
a=1
The coefficient of 13
·≈56(3+1)
224
Ana
Ans.
Putting x-0.01, we have
(1.0103)8
=1+0.08+0.0052+0.000224+...
-1.085424
1.0854 correct to 4 places of decimals.
(c)(1+pr+qx2)8
=1+8px+(8q+28p2)x2
+56(pq+p3)x3
8p-1
8q+28 (1)2=
P
Ans
9--28-=-=-78
Ans.
Page 30Page 31
附加數學建議答案
寄俊彦
• ( 1 + x = 125
x2)8=1+xterms
involving x?and higher
.'. (1.016)
. ́. (1.016) -140.002–0.
1) 2 -4(1)(1).
-cosisin (polar form) Ans.
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