1970-02-26 — Page 29

第二第張七第日一廿月正年戌庚醫夏 WAH KIU YAT PO

郭日僑單

四期星1日六廿月二年〇七九一届公年九十五國民華育教

夏

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170英文中學會考試題預習專欄

(TE)

堅道英文書院主編

生物科

BIOLOGY (17)

anover to sin questions or last MOC

1. (a) The nature of an inasot's diet is often

reflected in the structure of its scuth-parts. Show how far this is true of tha. cookrosoh and the honey bio.

(b) By describing the essential parts of their

life-histories, explain what is meant when the cockroach is said to have direct development whilst the housefly is said to have indirect development,

(a) According to the diet of a cockroach it is

described as omnivorous, for it eats any type: of available food, whether animal or vegetable

The mouth-parts of a cookroach possess movable labrus behind whion ara mandibles, maxillae and labium which are suited to an omnivorous habit. Each mandible is a single segment with sharp outting and grinding testa along the midial edge. The maxilla is a jointed atructure bearing a movable palp which is sensitive to taste, and thus assists in the selection of food. The palps of the labina are also sémitive. The procesace on mezillas and labium, together with the labrum hold food for the mandibles.

In the south-parts of a honey bee the labial palp) and maxillas are fused to form a suoking tube containing a tongue formed from the middle portion of the latius. When this tube in folded back against the body the shoz! sandibles cia still be used an jaws, and the

beo in thus one of the few inssats that can both suck and chew.

(b). Unlike the majority of insects, there is no

-Laïval stag in the 1ɛfe-history of the

cockroach, for the young cockroach esosping from the egg case on hatching is in an advansed, miage of development and closely zusembles the adult. To distinguish it from

·larva it il termed a nymph; The nymph which axoepas for the lack of vings and its small vica is a ainiature adult. Eedysis occurs. seven times during its devalogment and growth which takes about a year to complete. This form of devlopment is thus called incomplete metamorpholis or direct development.

The homefly is said to have indirect development, for its egg develops into an adult, it fases through a larval stage anu papal stag, finally it attains its adult stage which is quite different from its

The firra of the housefly is legiese aad: sallad i maggot.. In less than a vook's feeding the maggot renohen fuil mizo, It pupaten iniide its larval skin, which hardens

·and feras i pupal case. Inside this onse motanerpholis takes place fairly quickly an Marn conditièns and in' less than seven days the image teaks out of the oase. This. form of development is known as completo metamorphosio or indirect development.

(a) Write a short essay on the habits of

mosquitoes,

(b) How are the mouth-parts of a female anophaline

mongniše udspted to piercing and blood sucking?

There are tiras important types of mosquitoes, namely sulsi, anopheles and audes. Anopheline mosquitoes are the carriers of malaria, yellow fever, elephantiasis and dengue fever all of which affect man in many cases can ba fatal. The organisms causing the ditensos are carried fros ons host to another by the fauale mosquitoes which are blood-suokers and disease

The danlopment of mosquitoes is called complete metamorphosis, The female mosquito kaya ngga ir vater in suaner. The agga batoh in wat, and warm weather, the young panu the larval and japal stages in water, and leave it when they develop into adult mosquitoes,

The malsamosquitoes feed by piercing plant tissus and sucking the juices or EOTe commonly on the neotar of flowers. The females feed on the blood of animals and in the process frequently spread dinoRBOS.

(1) In the mouth-parts of the female anopheline

Ronquite the labrus and the epsþžurynz combine! fors a smoking tube; the mandibles and maxillas are piercing organa; the hypopharyar carries saliva; and the labium constitutes shauth in which the other mouth parts lie when not in we. When the female bites, the forked and of the shoath is prassed on

the

animalig skin, which is then puzotured by the mandibleg and maxillas, wo that the sucking taba may be inserted into the wound, Saliva passou down the lower part of the sucking tube, and the blood issuoked up the upper part of the tube leading into the gut of th inesot.

Questions for this week

Multiple Choice Exercises-

2. Which pair of terms about innets is incorrectly

matched?

A. Air-kolesi epizaOLOS

· B. Foeleres antennas -

G. Stabilisera : halteraj

D. Breathing tubesz traoišao

E. Abdomanz neta-thoraX

Which of the following statements £s not apply. to a mesquito?

4. Its body consists of a heid, a thorax and

abdomen.

B

· It has pair of membranous vinge,

C. It has a pair of antennae and two compo

oyen.

Its life-history is called incomplete metamorphosis.

It has a tubular mouth-pars fox suoking plant juice or blood.

3. The smallest feathers on the body of a bird

the

Aquill Teathers

B. down feathers.

Cu contour feathers

D. coverts

E. filoplum Sa

Which item does not balong to the group?

A. Eosquito

B. Termite

G. Housefly

D. Grasshopper.

E. TaperezÚ

Which one of the following characteristics doon"

not apply to a larva of a housfly?

A. It has no lege on the body -

B. Its head has two hooks used for soyusont,

0% Its body has spiracles for broathing,

Do It has a pair of eyes at the narrow end of

125 body.

5. Each of the last eight abdominal seguents

a pad for orawling..

4 bony fish can detect the vibrations by the water current by means of its AB Rwin-bladder

B.oparsulua

C. lateral lines

D. dorsal fin

B. nostrile

7. The terms below are apply to the parts

bird's feathers.

I. Baahis

11. After shaft

III. QUIL

IV. vans

V. Inferior umbilicum

Which of the above does a quill feather

bira cənşase of?

4. I, II and III

B. I, II, III and IV

C. II, III, IV and V

B. II, III, V and I

B. all of these.

· caused.

Which method is not affective for controlline mosquitosa?-?

4. Mil up unused wells and pozda

B. Well over the food in the house.

C. Mako mure that water in the drains 10.

running.

D. Kill the adults with insecticides,

5. Kosguito nets may be used to cover beds,

Which method will not be effective for controlling houss-flies?

A. Humax exoreta is not lert exposes.

B. Kasping all rübbish well covered.

C. Mll up pools of stagnant water..

D. Kill the adults with inssṣtibides or fly-

traps.

E. Human food should be kept away from flie.

The life-oyole of a Cabbage White Butterfly

in called complete metamorphosân - which involves the following stagess

11. Imago

III. Chrysalis

IV. Caté:pillar

Which of the following in the right saquenost

A. I, II, III, IV

B. I, III, II, IV. 0. II, I, III, IV II, I, III, I .I, IV, III, II

II. (a) What do you understand by the "social life

in the animal kingdom? Illustrate your andwer by referring to the hive-boo.. (b). Tiith the aid of fully laballod diagrand

describe the life-cycle of a hiva-boo.

Kake a list of Bight named multicellulax orgoniams that you have found in any one, named habit.

(b) Give a brief account of the life history

and habits of one plant aid one animal from your list.

文中學會考試題預習專欄|

數學科

(十七)

143

数学科

番仲強

第十六次預習題解答

(1)用配足方法或用其他方法,分解下列為因式:

(a) x2+40x+39%.

(A) I =(2+4077400) 9

=(x+20)2 32

=(x+20+3)(x+20-

(x+23) (x+19).

(b) +2+104-551.

(16) √ * = (42+106+25)−576

(1+5)*~24*

=(1+5+24) (4+5-24)

(4+29)(4-19). [2]

因政題之線(391,551)不易分解,故用配方法。

(2)(a) A, BIH.C.FIX = C(MA+78)✯

(3) A BH·CF % H, 9] A−Ha, B=Hb, #

中口,不為豆質式(即其公园子之两式)

#L_C=MA+n8=mHQ+nHb=H(ma+nb), 故 H亦需C式之因子。

(D) 18 A=5-13x2+23x-2).

=6x+x2-44x+2),

A&A, B÷H·C F

9X2=12X=2/X=3x(3x-7)(x+1).

SAX € 7 0 3 (3 *x) T£ A,B2LEZ.

AB483, 4# (32-7) (x+1)20.

但因A式ż符号,正

*L × 32-7 **.

15 32-7 1⁄2 A Z Z83

故亦弟B之公因子,

* X H & B 3.

3-13+23-2) 14+21

(~~) C—2A−B=-27x+90x-63=−9 (3x-10x+7)

(3)(1) 赤可得次數載城式 註二) 30-7之因子,不需再在8式檢核之

3)有等差級數差其第八項為第十三項之两倍試講 【第二項萬第十項之两倍。

(証設此等差級數之首項蒚a,公差高d,

$^=q=a+7d, }+===a+ıæd, 11

a+7d=2(9+18)d=20+24d.

Ajaxed, ard=2a+18d = (a+9dy).

但因第二項二 a+d, +2=arod, 故第二項第十項之两倍。

(2)

(4)有成等差级数÷五数,其和為10,粉和鸡60. 求此五數

(ME) B£ ££; ≥ k ≥ +>Å £a, až § d. u‡✯❤k 63a-zd, and, and a+zd.

(a-2d)+(a-d)+a+(a+d)+(a+zd]=10----(1)

(a-ad)2+(a-d}2+a+(a+d)+(a+24)=60==(2)

油(1)式

由(z)式

50-10

5a+10d=60,

d=2,此五數客莺-

d=2,此五數各為

答:此五數萬

(5)求下列各式之平方根

(a) |+√21+135.

(14) #2 £21412√3 =12+JZ

白菜21+12/3=x+2/酒+售

2)+2/10}=(x+4)+2\{%},

x+4=21 78416 2-)

Xy=108

JE+9=2/5+3 2/4/2/3 =1+2/5+8=3+1+1/3

=(3)+(17) + 2/3×1

答:方根萬+(+1)

(b) 3x+1+2][2x=x−6.

(14) πx=3x+1+2](2x+3)(x-2)

(2x+3)+(x-2)+2√(2x+3)*

-((2x+3)+(√x-2)+

答:其平方根為土(J2x+3+ [x-z).

(6)設2&c滿不盡相等之寔數,证

(b−c)(c-a)+(1-a)(a−f)+(a~f) (b−c) **¶ £¬

ax) FA=(bc-ab-c2+ac)+(ca-bc-ararı

+(ab-ac-b+be,

ab+bc+ac-a fc

=-4(2ar 26+20-2ab-abc-2ac)

=-x [(a2-2af+f3)+(b=abc+c2)+(c2=20q+

- /[(a~b)2+(b−c)+(c_a}]

At ab cha+*+ £, (a−b) 20,

(b−c)=0, (c-a)2 = 0. min 4 e=b-cs=0¥*+ 能適合,旅拍号内恒笃正數,即原式恆為頁數(在罩) (7) = -

(未完轉入第七强第三寶)