1980-07-01 — Page 20

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

明德出版社提供

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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