1969-03-20 — Page 23

育教儒華買三第張六第・日三初月二年酉巴夏WAH KIU YAT PO

報日橋

四明星日十二月三年九六九一窖公年八十五國民華中

(接第六張第二頁)

på(a+c) = ac(p+r)

pr(a*+zac+c*)=ac(p+apr+影

pr(a+c3)=uc (1+r)

CC英文中學會考試題預習

1969

生物科 (二十)

*%$#$%$#5%£

夏僑

12ƒ2(a+c)2

去分母约简

#%#3%$#$%$#&%$#$%&$%&#$%#%3 #

一九六九丰水文中学会考試題預

ac

數學科 (二十)

喬仲强®

第十九次預習題解答

(1)(a)何謂等差級數?又話於等差級數各項中,来以

(或加以)同數,其结果仍成等差級數否?試證明之

(答) 一串數列中,若其相鄰两項之美恆相等,則此數列

·稱為等差級數即是有數列b Mar

the 13 $B$(A.P.).

在-等差級數:

一中各項桑以河

(5) 設 a,b,c,d成等比级數試

c+d* 亦成等比級數

Q.ED

(証)設此等比級數之公比為九,则岳二

與==

a (1+r)

treerd Ba+b, b+c, c+d" # G.P..

一般

數中则结果puipus.pus

一仍成等差級數

(剔証)

依和比定理

Bp pu, puz pilly piły. 如各項

仍成等差級數

故结果亦成等差級數(且公差不变,此笃等差級

數之特性,在證明題目時常利用之,

(6)在279之間,插

#

(b) sq a, b, c i A.P. SA ZI a2(f+c) b2(c+a) c2(a+b)

MAP.

a b c A AF, 37] Rux ab+be+ca

因。

alab+be+ca), b(ab+be+ca) c(ab+berce) ABA.P.

kz} m± abc. 1} a(ab+ca) b(ab+bc) c(be+ca)

仍成AF

Bp

af(b+c), B2(c+a),C(a+b) 亦成AP】

2)(a)有三數成A-P.其和為18,其平方和為126.求此三數

(解)設此級數之中項為x公差爲y, 则此=數备

£ (x-4), x, (x+y). ik U Ž

f(xy)+x+(x+y)=18

{(x-4)+X+ (x+4)2 = 126

3x=18

24 #18

(2)

126

此三数为3.6.9.

9! x-y=3, x+y=9

20-y=9,

+=

答:此三數3,

此三數為9.63.

(f)有四數或AP其和為20,其平方和蔫120.求此四致

(A4) Il sok is a, ard, arad ♬ ar3d, §}

pa+(a+d)+(a+2d)+(a+3d)=20

故好好成等比級數

數,使前三數成AP,後

試求此二數 成G.P. (解)設此二鸟x. y 則

由()式得

代入2

去分母

分解

善

-(21

2(x4)

4(x-1)=9x.

-17x+4=

(x-4)(4x-1)==0

做差 24=0,則x=4 代入)得=

則=女,代入(3)得

()

答:竹挿入之两數為4.6或為女一号

(7)有一等比級數,其無窮項之和為4,此级数各项

立方的餐羁項之和為19元,試求此级鼓,

(解)設此級數之首項為a公比爲九則

S~a+as+

(至無窮項)=

S=

-(至無窮項)=192. 因S'亦為 G.P, 其首項為2,公比為信故

10-20

(1)÷(2)

d

(3)

lz+(ardP+ (a+2d)+(a+3d=120

由(光得 4a+bd=20

4a2† 12 aa+ 14d2= 120.

*x(3)*4x 4a+120x20=20+14(10-24)=

Q2-102+16=

(a-2) (4-8)=

化簡得

分解

2,代入(3)得

t=3(10-4)=

L=4, 2+2d=

6,a+3d=8

(i) / 2-8, 42(3) 44

x=3(10−16)

d=b

at ad4

Q+3=兑

答:此四數笃

48

48 (1+2+1)=2

-4(-1)

(4)

48 (1+λ+)")—16(1-22+8) 去括号移項,化簡, 21⁄2 +54 +2

分解。

(1+2) (21+1)=0

The hormone secreted by the medulla, causes

tne neart to beat fast, blood pressure to

increase and more glycogen to convert to glucose.

(h) Islets of Langerhars =

Position:These glands appear in groupe embedded in the pancreas, Their secretions function: (1) to help the storage of carbohydrates in the

form of glycogen in liver and muscles,

(11) to make the constriction of blood vessels

and thus rise up blood pressure, Daficiency: of this hormone. causes diabetes.

Parathyroid gland:

Position: There are 4 glands lying behind the thyroid gland and are partially embedded in it. Function: Their secretions serve to regulate the content of calcium and phosphorous in the blood. Deidolency of parthyroid hormone lowers the concentration of calcium in the blood, decreases, the excretion of phosphorous and thus imreassa

the amount of phosphorous in the blood. This

condition is known as teteny. Over-production or

this hormone causes as a result of soft and

easy broken bones.

Pineal body

Position: It is a stalk-like body which lies

between the cerebrum and the mid-brain, just:im the front of the cerebellum.

Function: The pinsal body seems as a reduced

organ having no known secretion. It is considered as an inhibitor for sexial maturity;

Make a comparison between an enzymes and a normone.

Enzyme

(f) an organic

catalyst

(11) La produced by

living cells, chiefly by that of the alimentary canal (tract)、

(111) Each specific enzyme acts only a specific substance or group of substancea and has no antagonistic action to each other.

Questions for this week:

(d) an organic reagent. (ire. chemical neasenger, not necessarity containing protein).

(1) Is secreted by.

endocrine organs and limited in production" by the nervous. system or the next. hormone.

·Certain hormones will stimilate or

inhibit each other,

Make a drawing to show the structure of a namaz. tooth of a named mammal and describe the functions performed by the various types of teeth in that

(註)四數成等差級數時,如其和為已知,恆設其首 項為2-3d 公差為2山,則其餘三項分别為and,

a+d>a+3d.讀者試依此設題以解之

(3)酸a,x,y8成 AP, a, U' v, 长成H.P(調和级數

Xua & * x, y £s of xv=yu-ab.

(解)因ax,长成AP誤此级數之公差為比

則

但因無窮等比级數之公比 11<1.始馬收歛級

=-2不合题意,棄之

代入(4)式

x (-2) == - 3

2=(1+支

答:此级數爲6-3孟

(8)某人欠債 1.10,000元訂明分十年攤還,年利率6%,

以每年一期的複利計算,問每年年終 逸若干元?

(解)設每年年終還 A元,則第一年末开A元之現值為

元,P=什俱

(a) What are the functions of the skeletons in a

mamma 17. Gave « labelled drawing of the skeletons of a fore-limb in a rabbit. Describe carefully now its movements are carried out.

(b) Describe briefly the following types or none

jm: hinge joint (ii) e all ano socket, sm (111) * sliding joint, giving tw examples of each. in the rabbitla skeleton.

+

7:08

1.082 ·1.08.

3600

1-(48)

ሪ

尚

3000

3000 10810

NO.

答:

·TAP

又誤此AR之公署為D. 则

将此三式相士+=30

-Camp

e+ak, zab

ab

=yu-ab

(4) a, b, c & A.P.; 4, q r s H.P. ap. bg, cr

品

(*) œ a, &, c AP

ap. be, en de G.I.

(bq) = ap cl---<3)

+

[

#1 S=10.0007 L=0.06

10,000 =A[1-(b)"]

aagx(1-0.5585)

Arx 0.4415

0.06

10,000x0.06

0.4415

NO.

LOG.

0.0000:

1:06: 0.02534

鴉好

10

A5585 77470 2.3782

0.4415 7 6449

1359

3.1333

答:每年年终应道:1359元

(9)某人購屋-間,即付20,000元,以後每年末付款

3,000元,十年付清,关照年利率 8%複利計算(每年-期) *間此屋現價為若干元?

(解)第一年末所付3000元,現付為 1.08

3000 元 第二年:

第十年”

故+需好付*現值

3000

= 3000x [1-(-/-)"]

a08x(1-0.46348.

3000x0.5366

0:08

0,730

20,000 + 20,130—40,130

答:該屋現價 40.130元

第二十次預習題

7.02.

204

0.0000.

1.08 0.03346

7.9666

0.4634 T. 6660.

3000 3.4771

0.5366 17297 &

3.20.68

0.08 €.9031 (-

20/30 4.3037.

(1) A,B 为直线XY同側两定点,求在XY上取一奌C而 使∠ACX=<BCY

(2)設ㄗ為銳角三角形ABC AB边上一定求在

BC, CALL & Q, R TE MIE 4PQB=LCQR B LARP=LCRQ..

(3)於走直线 XY同側有足园O及定奌P求於XY5 ∵取-奌C使直线CP及由C至0园之切线典XY成

等角

(4)-定奌A介於兩名知园0.02之間,求過A作 一直线典两园各安於 B, C 而使BA=AC.

(5) 0, 0酋直线 XY 同側两定园,求作一直线典 XY平行而各交两园於A, B桌并使AB等於定長Q. (6)邊街上一已知点,求作一弦使為其同心园时

三等分

(7)在乙知角XOY内有一定奌P,求遏下作=垂直线

·使典 OX.DY 各支拎A及B而使PA=PB:

(8)遇0园上两定桌.A.B.球作两年行弦使其和