頁二第張九第 日八廿月五年寅丙磨夏 WAH KIU YAT PO
報日僑華
>3.
Ans.
1986
When x 0, (x−3)(x+1)·
<0, -1<x<3,
{n}f(x)=x3+bx*skx+2
(c)
五期星
附加數學建議答案
日四月七年六八九一层公年五十七國民中育教僑華
二
中學會考試題預習專欄
Sam of rest
附加數學建議答案
Suggested Solution to
Additional Mathematics
IT 1986
Since C is the mid-
sinsdy
point of AB, vhere:
1. P.
*2-3*
When n
{n+1}
* are the x-co- ords, of A and B.
Saggen LeA
Additional
T. 1986
1. Let re
-(2)-(1):
Ay=x+3x Ax+3x(6x)
AX
+ (b =)3
Ax¬u(1x)-3x2
loga+(x+1)logb=0
(Inga) (logb)x+logb
0
For equal roots,
(logb)2-4(logn)(logb)
Togb-41oga=0
logbaloga
For turning points',
drg 2
3x^42hx+k=0
As the turning points
are distinct,
(4b)2=4(3)(k)>0
i.e. 3k..
Now, y-2;
2=x+hx
+hx+k=0 |
As the roots are not real,
(#)0C=0A+AG
>8+2b" As AC-20B
"2
=a+b
TOB+100
(a+b)
(b)Given:PR÷Rq=h: (1–h)
h00+(1-h)op
-_-3 ( 2+4b) + (1~})}
→(3a-ha+8hb)
b-a. Ans
(ii)OR-KOC
f(x)=4k+18x-kx“
df ()-18-2kx
́dx.
For stationary
values,
-2kx=0
and
11 1111
Ak <0
h4k
3k < b2 < 4k (b)For (-2, 0),
ADS.
(-2)2+h(-2)2+k(~2)
-8+4h−2k+2=0
k-25-3
From 3k <h2 <4 64-94124
81-12 *.6h-9<12 and b2¿ 8h-12
--6k+970 and1 à -8h+1240
As
Considering sinflar
PM OM
10pine
110gine
210sine
2cose
But-P-PH⋅
-10sine
20cong from (=))· kožne
11.
issuming that P
true for nek,
11.
For nek+1,
1-212-3 *k{k+1} +(k+1}(k+2)
"k+1" (k+1)(k+2)
(k+1)
k+1
dx
For stationary valu ar(x).o
(x+1)(3x+5)=0
20(cos"0-1
(sin942coal)
(sip0+2ca89)– (20ningcost)
(siu@+2cose)
(cos0–2aine):
For stationary points,
· (k+1)+1
.By the principl
mathematical induction, P is
for any positive integer n
For (a+b)", the (r+1)th term
For (42)
the 3rd
tera
ane
(istan38)
=tan 0-1
y- Stap cao-6+C
• f tan^edo-y+0+Cg
where CC
-tan30-tane 494€
Alternativ
ftan ede
Ans.
Stan (sec e-1)de Stan 0 sec ade- Stan ode Stan ed(tans)- {sec^0−1)dg
and
(h-3)2>\0 (h-6)(h−2) <0
*h is any intoger
except 3 and 24k<6;
„b=4 or 5°
Por h=b, k-5,
a@=20sin@cose
Zsing 20sinecose°
Area of aquare
But
d? de
(3k+L)x-(2k−1}y+ +(k−11)=0.........(1) alope-2k-- (887)
For Lx-2y+lim0p.
slope
For m
3k+2
tan45°-1
9k+6««£k+1
The equation of the
(~152)x-(-11-1)+
(7-11)-0 x+3y-18=1)
Ans.
For 7,
3k+2
•3 ⇒ 3k+2=6k-3
2k-1
Atia.
The equation of the second line:
Now,
d2f(x).
ax
}Given: OF AD
màxim when
4k 18()-k()-45
4k+162-81-45k
+81-0
k-z or 9
2
#Differentiating the
expression v.zat, zi
-†(8b-a) ́ ́
Ang
(=-1 or
and y=0
Since ų lies on FT
FT UPO Xb~a~ μ•z (8b~g) ~=p • (8)
The equation
tangent in
y-1-(—)(x-2)
5x+4y-14-0
the
X=3(†)(8)
3. (1+j) {(c+4)is(c-4}}}
| 2+2||(c+4)1+(x-4);}
~(√1~+1^) ( ) (c+4)2
√(24) −(√2)(√202+32)
Using cose ab Where O-angle between
and b.
20
240 +16
1-2n
neany integer,
(b)(i)z»r(conûsisine),
Zer(con9-isinė)
(Cosmo+isione)
(COAND—isimmo)
Anx
(a) Iz-21-1
Let z-xsiy
{{(x-2)+iy| −1 (x-2)2+y2-12
(b) {(x−1)+1y] = {(x-3)+iy}{
Putting 2 into
d2t(x)
x+870
when
giving a min. pt.
=6x+8 <'0' when
x=—- giving a max
NO
S-10core
S*20cos0 d5-20(-nine);
at
de
(ii)}
d£. 20sinO.
√2
20√1-c03
(cose+ising)TM
+(coso-isine) ={2, Where G-
20
(10)
√2 using (b) (1) cosae-√2
Tr=1:
5
=-2x=1 by (a)
1 or 8n1 where
is any (+)▼¢
integer
(c)(i)(1+i)P-(1-1-)P-0
(141)-(1-1)P
[+y.
1-(1)
The points of
intersection are:
241 and 2-1. Ans.
paz
vkøren in any (+)ve integer.
(11)(141
(1-1)
NR=5cose h-ON-S-NR
420co80-5cond=15cond
k=5aino
The locús.
1-1
sinecos 0+2cos
42811
̈0 when.
0-51.56
Now,
0-51.56
G
(a)Area A-(y-23)(x-16).
But xy=3600
******=(3600_25)(x−16)
Aw
-3600 16x3600
57600
4000.
-25x4400
(b) d4 −57600(-1)_25
n(n=1)2=40
-n-200 (x+4)(x-5)=0;
ди5 Ans
The coefficient of
Since the roots gre equal,
(-4cose)
4(3)(2sine)
con 8-6ginė×0
(sing+2)(2)
or-2.
(rejected)
0-150 as.0 in ahtuse
Ane.
sin2g+sin4u
20440
Jcos
2sin30cos@=cose
To prove
fof(x)dx= fof(a-x)dx
Let -AA
x--du
1..5,- ) of(x)dx
-- fof(a-u)du
fof{o-x)dx
using (a)
-25x Ans.
Q(2sin30−1)-0 cose-0 or win30-h
20+1
30+n't
(11) 。rsin2xdx
8-3+(-1)", where
Por stationary valuen,
57600-25% -0
When I«AS,
3600 48
(48-16)
1600
Now, 4-(-2)(37690)
dr
40 when x=48.
The largest value of
the area, A=1600
(c)(i)As A decreases
increases, then
(a)Slope of AB
(23+2)x−(20−1)y+
(-11)~0 32-y-4=0
(ii)Since L, is parallel
to (1), 3k42
The line Lis
(-11)-0 x-2y+7=0
- fo(7-x)xîn2(t-x)dx
using: (•)
- J (1-x)sin xda
sinh sinh
Tip
idr.
Ô sin xin gæsin_xdx
Take the point (0, 7)
cance from (0, 2)
0_2(7)
Distance from (0,7)
57600
25 <0.
57600-25x-
x-2304>0
(x−48)(x+48)>0.
x L-48 or ≥ >48
Адв:
* >48 112 ≥50, the largest value of A
57600
4000~~(50)-25(50)
decreasing,
-1598
Ans.
(d) Gives:
3600
1600 €150
5
3600:
Ans.
Aree of A APC
→(Ga−3n+30-6s-10s+9)
(39-138) -(39–138)--13
136-26
138-42
(1-1) ·(1-1)-1
AL
1+i)(1−i)
(1)(1–i from (1)
Алв.
本
(15,0);
Ris in the 1st quadrant, the locus
ashöv¡!
1600 Cz. É 2025
40 X645
Ans.
When x45, (ver
(48), A 18
increasing.
57600
-1595
The largest value of
25(45)
Wher
>0, (x-3)(x+1)
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Equation of AB:
y-2-(x-3).
*=siz=3m+2
(x-2)
報日僑
(1-cos2x)dx
-
(x-12x
6-0]
Ans)
(rejected)
10.
x-2y+10~0 ▲as,
(b) At x-axis, y=0,
+(k+11) --11-k
3k+2 3142:
•*. Aren, A-(1)(-)
For
+11)**
• 2(3k+2) (2k−1)
[2(k-11) (3k+2)
(3k+2)" (2k-1)
(2E-1)-
(k-11) (12k+1) (3k+2), (2k−1)
ationary painte,
sinxdx
(*ain(h-x)ax
(-x)+cos(x)
using (a)
sinzuz
Bînx+COST
-(k~11) [12k2+2k-lym 12k +131k+11 -(k−11)(133k+7)=0
kall or t
0
Since x and y-axes are positive,
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and
sing+cos
When
19 the
is minimum.
And
(c)The required line is
3x=2y+1=0% TÄDA.
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