1970-06-14 — Page 14

頁二筅張四第日一十月五年庚家 WAH KIU YAT PO

報日僑華

日期星日四十月六年〇七九一八年九十五國民華中育教僑華

6$#$%$#$%$#$*5#%3%3#3#£#%A $%5#$%$##TERNSHISES%$#%3 #$%G PRESI 中中會考高級數學(一)答案(續)

•$#5%$#5#$*F%3ES

百樂園

喬仲强。

(4)(i) ‡*143*<außr,#fn»J &¥*AQ

Ch. Be of logr

|英中會考現代數學(二)答案(二)高錦葵

T2-2 7-3 1-2 - 31 + 72 * 1 + 3 -

Let xy

we have

be the

Z+y=8

*1.75

(y 8 + x ) = ( 2 - 8 + y) =

55 | x=3, y=5}

The number is 358

by Mathematical induction show that mentis

durable by 2 for all R.EN.

students took part in a certain snaminate

• passed in Chinese, 49 passed in Mathematics,

47. passed in English, 23 passed in Chinese and in Maths, 22 passed in Maths, and in zng, 221

passed in Eng and in Chinese, no one facts in

all the three subject How many passed in all

the three subjects!!]

lution

(a) Lat pengo ninte

Aasume P(F) = K(K+1) in

C

(正)設首項為a,末項為见,依公式 lar

两边各取对數, log C = (n=1) logr

log C

(証) log r 说)某城市人口每年依幾何級數增加,該城人在 1965年時為3,000,000,在1970年時增至3,600,co. 求該幾何級數之公比,並估計1975年時該城之人口。 解)設每年增加率為2, N=1970-1965=5 164* A=P{}+2)"

3,600,000=3,000,000× (1+r)*

化筒 (1+r)2=

各取对数

5 log (1+2)== log).2

0.0792 0.0158.

log (1+r) = log1.2 5

查表得 1+2=1.037 (幾何級數之公比)

由1965年至1975年,共絰10年,設當時人口為N, 則

N=3,000,000 × (1+2)10.

=3,000,000 × [(1+r)*]*

-3,000,000 X 1.22.

- 4 320,000.

答:該幾何級數之公比為1.03元,

1975 23 A 073 4,320,000.

x+o, a x x x x

(5) z&• f(x)= x+1

儿)证明f(X)恒小於9(i)求x值之界限便f(x)<2 (E)

22 y 移項,整理,

yx2+y= =x+8

x≥#l3⁄4I✯_^=(-1)2-4x4x(y-8)

65—4(y—4)” 65-(24-8)*

log2 | (3x-1)(x-1)=2,

依對數定義

移項,併項

log2 (3x=4x+1)=4.

3x-4x+1=16 3x=4x-15=0

分解。 (3x+5)(x-3)=0

x=-(不合题意)或x==

x=3

(陽)利用數學歸納法証明富儿马正整數時,

恒為:13ż倍數

(SE) () 1 J(n)=33”—1.

* ƒ(1)=33-|

26=13×2,

tx=zof, †(2)=34-1=728=13×56. 故 1,20€. f(n) * 13 × 11⁄21⁄2.

(4) ke & n=k of ƒ (~) 2 132 13

f(t)=33號-}=13M(M為整數)

ƒ(k+1)−ƒ(k)=33(k+1) — 1 —(331⁄2_1)

故

• 33k+3_33k

=33ky (33-1)

— 26× 33k —13X(2x334)=13N

f(x+1)=13N+ƒ(R)=13N+13M== 18(NTM)

=13元倍數

時f(n)亦為:13ż倍數

综合以上两点,由化)証得 n=1,2時 33°C/高13 之倍数;再由(五)可証得 = (141), 3(2+1) 0$, 33x−1亦為13=倍數

如此類推下去,n=3.4 時 直至九為任何正整數 時3−1亦為13

7)(a)証明C

Q.E.D.

(証法一)因在几何不同物中,取无组之组合方法

Ch. A mit nie tot €-18 AYO, RITMB ≥ 18 合時,可有下列两種互斥情形:

(i)每次圴不包含A物即在九個不同物中取九個

之组合,英法C..

(花)通次均色金A物,即杜2-1順不同物中、萌え狙え

組合英法 CH

yx−x+(y-8)===

— 1-4(y2-84+16)+64

因為賓數,故其根之判别式 △≥0.

45-(24-8) ≥0.

(2)

新诚=)由组合公式 Ch

(n=1)!"

(n=!)!

(n-1)!

(n-1)!

+

(−1)!×(64) xx

(n-1)}

PIK+1) = (k+1}{K+6) = A* {K+1} + 2 (K+1)=[[K] + 2 (K+1}.

and 2(X+1) as dur

P(ATI) is diraible.

by 2

Let M = the set of students who passed on Maths

E deset

acciderela who passed in Eng

the set of students whis prosed on

NIMUEUC)=N(M)+N(E) + 1}{(C) −R (MOE)-TI (MAC)

−N(ENC) +N(Mnene)

N (MAENC) = 100-49-47 −60 + 28 1 2 3 + 2 1 =

I plicdents passed in all three

5 f (x,y) = (x,y3) and

(xy) 5 (X, Y) are transform-

ations of the plane defined by (5)=(3-0)[J

and (3) (34) (5) (-6) and (s) are then

said to be the notices of the cran

9 respectively

a) A 10,0) B(0, 1) c(1,0) are

f and

ga triang

If I transforms.

A ABC into AABC" find

the coordinates

за

and

*X(-1),

(24-8) ~65 <0...

分解, (2.9−8+165) (24-8-165)≤0.

西因子之穫為負數之必要條件,是较天之因子為正數 而较小之因子為負數,故

由(2)式

24-8+563 ≥0 1 29-8-√75 <0.

bk y<9.

(*) (*) **

-(1)_ @_4<8+,25

$+8.062 By=8.031

14 f(x) 4E1*9.

x+1

X+8<2x+2

2x

(2)

(誕畢)

@ x * £ $ # x2+1 $$, œšk

各減去x+8 分解 (2x+3)(x−2)>0, &p(x+\)(x−2) >ċ 两因子倶為正數之條件,是較小之子為正數、

x=2>0_____ #kx>z

1903 12 $1 $k<14+4, £#£<03 4 1 %

•. x+2<0;

数軸表出其结果如下:

答:

1.5

(6) (a) 8£ x>1, #4*43✯ loge [3x−1 +loge! x~} = å (*)**** loge [(3x-1)(x-1)=2.

[n=2)+=2] (N=^)}

A! (n=t)!

QED.

(七)同學9人旅行,分乘大小汽車一輛大車最多 可载7人,小車最多可第4人,求分组之辩法有我? (解)八車載7人,則小车载2人,其分组法有

C2 — 1. § — 36

六率截6人,则小草载3人,其分组法有

98827—84 7×2×3

汽車載5人,則小車载 4人,共分组法有

9 x 8×7×6-126

1 x 2 x 3 x4

此三情形為不共立事件故

C} +O!+C}=36+84+126=246.

答:只有分组法好種

米完

its If we apply f and then f

Gansformate

is denot

had is the matrix of g f

4)+(2,4).

what should be the matrix of h (x,y') + (2,5) such that the matrix of hof two (oo)"

Solution:

^ - ( ) - ) (·)-(-) --(--)(0)-19 <-(3-2) (0) (7)

A' (0,0), B (2, +4), C' (1,3)

@) j· ƒ− ( 47 )( ; + ) = (?)

(e) Let h=(ad)

we have (ab)(2)(68)

-

-(2)

7. (2) A card is drawn at random from a pack of 52 playing cards What is the probability of drawing an Ace or a Heart?

(b) Bag A contains 5 white balls and one black

Bag B contains 5 black balls and one ball Bag B red ball One Call us drawn at random from

Bag A and 4 balls from Bag B, and the balls this drawn are all put into Bage. I now one ball is drawn at random from Bag I, who bability that this ball is white

Solution

one space

ons die mond ace, one

clist

an act of a than

The probability of drawing

・

from Bag fo

the chance that drawn

white ball is t

If Bag C contains a white ball, then the chance that drawn a white ball from Bag C The probability that drawn from Bag a is white 1 x - x

•{ cx, y) y + x

YER} is a quenction

(0) Find the values of 2, and to such that 12.8.00).

(22, 2,00) are elements of F

16) Jome more elements of F are awen belon

(2.50,875), (3,20,875)