1968-12-26 — Page 14

decaying organic matters, Generally; the min constituents of soil are air, water, Hums

買二第張四第

日七初月一十年申戊屡夏

WAH KIU YAT PO 報日橋

四期星日六十月二十年八六九一般公年七十五國民華中

橋

1969

ST

一九六九率

考試題預

數學科 (凡)

齹仲強

Ans.

第七次預習題解答

(1)(a)試述餘數定理,並證明之

答:馀数定理茑”若有x之多项式 f(x),除以x一长,則其 餘數為f($)

()設其商代為g(x)則g(x)之次數毂 f(x)之次數低一 次又因除式為x之一次式,故餘數(較除式次數為低必 為常數項R(附R不為加之函數式)依:

被除数≡除數*商數 + 餘數:

泛間侄得、f(x)=(x-b) g(x)+R

因上式為「恒等式」,故 f(b)=(1-8).g(B)+R

R= f(b)

(**)

(6)利用餘數定理求2x2+4X2+5x+6除以x-3∗餘數

(解)由餘數定理得知餘數

·R = {(3) = 2x 3 + 4× 3-5×

=54+36-15+6.

答:餘數萬8!

(註)本題不能用综合除法以求餘數 R. 因為未曾利

用‘餘数定理,與題意不符

(2)設 f(x)=x+ax

尾,以x+1除之餘1

(解)由餘數定理得

f(2)=2*+

x2 + bx+b. 2.除之餘 求af之值

- 8x2+6=16

·4a+

e f(1) ===(-1)+ax(-1)+5x(-1)+ bx (1) + £ 化簡之

(1)-(-)

答:

#

=2

(3) 設 ax+bx2+1触被(*小旺整除,求ab之值 (解)由餘数定理得 f(1)=〇| 即a+8+1=0 祇得-個二元方程式,不能求得。B值,故有重複的 子時,即不能利用餘數定理,故本题祇有用「特定係 敷法;或「直接相除法,解之:

(待定係數法)因被除式為四次式除式 (X-2x+1)為二 次式,故商式為二次式,且首項為ax末項為」故可 設商式為:ax2+ px+1 <形式(其中中為特定之係數畝

ax+fx2+] = (X 2x+1)(ax+ fx+1)

Ins

C文 學會考試題預習

生物科

BIOLOGY (8)

to the questions of last week:

Why is it necessary that planta should scatter their seeds? Describe the chief methods of seeds or fruits dispersal, and give one example of wach:

The Important reason for the dispersal of seeds is to give the seeds a better chance of successful life. Hy. being dispersed. they grow away from the competition which would face them if they fell to the ground immediately around the parent. The process also serves to spread the species over a wide area. The chief methods of seeds or fruits dispersal are outlined as follows::

(a) By animals – Animals assist in the distribution

of many fruits and seeds, usually by the

development on the seeds or fruits of processes (e.g. hooks or stiff haira) which cause them to adhere to the bodies as the coats of the animals Example is provided by goose-grass. The collection of nuts by squirrels results in dispersal of seeds, Mary fruits are fleshy ae that animals, would like to eat them. In this case, the seeds are either rejected by the animals, or they may pass out undigested with the animals faecen. 8.g. orange,

(b) By wind: - In this case, the pericarp shows.

modifications which will result in the fall of the Pruit to the ground being delayed in order to give the fruit time to blow away from the parent plant. Wings or plumas formed from extensions of the pericarp are the acat favoured. In cases where the seeds are dispersed after leaving the fruit, the testa may be extended into wings or plumes. $.g. Beeds of cotton, fruits of dandelion,

(c) By water-The fruit or seed is provided tough" and fibrous pericarp or testa which protects it from damage in water and enables it to float on water for a long distance;;s.g. coooout.

{d} By self-dispersal - The seeds of any dry fruita

may be distributed by other means... în măny cases, the seeds are shot out of the fruit with some violence as a result of the strain set up t the wall of the fruit, which eventually bursts, The pea is a good example of this type of dispersal..

What is soil composed of? List the physical, chemical and biological factors that determine the character of a soil. Deacribe an experiment which will help you to analyse mechanically the size of the soil particles.

Soil is a mixture of mineral parbicles of different sizes together with a certain amount

有教信慢

cro-organisms, minerals, sands, clays and silts,

The physical, chemical and, biological factors that

etermine the quality of a soil are:-

a) The amount of humus which supplies the soil with

decaying organic matter,

(b) The mature of the rock from which the solu La

formed.

The size or the component particles, Jană diw clay differ only in the size of their particles, Clay being smaller in size, is important for the retention of water, whithe sand being porous, is fmportant to give the, soil good aeration.

(d) Soil water is an important solvent for the

soluble minerals.

The atmospheric air present in the air space. (f) Anime Is and micro-organisms, eig. earthworm,

fungi, algae and bacteria,

(8) Acidity and alkalinity.

(h) Mineral salts and other soluble materia

Humus

Suspension of fine clay

and minerali

Clay Silf

•Fame sands coarse sands

Experiment to show the various soil particles

Place a handful of garden soil in a long glass cyIdep and together..with an excess amount of water; the soil is stirred up thoroughly :by means of af glass rod, When all the components are settled and note the separation of the various. soil particles. Host of the organic matter will float on the surface of the waterr, Pine clay. particles. remain dn auspension in water while line sands and coarse sands together with sons small stones settle at the bottom.

Questions for this weeks.

(a) what do you understand by vegetative:

propagation?

(b) Give a brief account or advantages and

disadvantages of vegetative propagation.

State the type, food-storage, external feature (or

Cinternal structure), and method of reproduction :og:

each of. the following wgetative reproductive organa:

(a) Potato (b) Tulip (c) Crocus (d) Ginger (e) Str

Dahlia

比較两边係數

由(3)式得

代入 (2)

代入(0

(3)

(直接相除法)除式被除式用昇冪次序排列算式如下;

1+2x+3x

因链数為零,故

(註) ax

答!!

+fx2+ax2

+1之第一次導函數 f'(6)=4ax + 3bx+

*} {(1)=4a+3b=0 lk a+b+1=0 alight a 居之值,但此性質,神在高等代数中有汗 鉿及,並錄 之以供参考

(4)利用餘數定理,分解下列為因式:一

(Q) X*+ 2x2-11x

(解)因 4= 2

故用以試除*值為土,出土

$ f(x) = x2+ 2x2-

而f)=1+2-11

用综合除法求得商式

g(x) = x + x2 + x2=9x+b 周q(1)=1+1+1+9+6=6 故g(x)又有x-1 之因子. 再得商(x)=x

故九(x)亦有x

1+0+0−10+15-6\1

+1+1+1-9+6.

1+1+1 −9+6|L

+1+2+3 - L

|+2+3-6 ||

+1+3+6

1+3+1

10 h(1)=1+2+3−6=0

因子

答:原式=(x-1)*(x+3x+6)

(5)解下列高次方程式:—

(a) (X-1) (x»3)(x-4)(x-6)+8=6

(AE) [(X-1)(x-6)] [(x−3) (x−4)]

(X=7x+6)(x=7x+12)+8=

(x27x) + 18(x−7x) + 72+8

(x2-7x) +18 (X=7x)+80=0

此為二次三项式,因可视x-7x=y 分解

(x27x+10) (X-7x+8)=C

(6)解联立方程式:

++x=

【解)因(1)式可分解為

•g)(x+y)=

(5)

6

* 3x-4=0 (3) ☆ x+y=

故本題之解答為(二)(3)之联立方程式 及 (2) (4)之

联主方程式由 (3) 得

x2+(3x)+2x=12

10X4ZX-12=0

(x-1) (5x+b)=0

-1=0,則x=! 代入(個得g=3

則x=

北九分山川

(x-2)(x-5)=0 代入根之公式

(女)

=(7±√17)

X=2152÷(7±√77).

(8)x+3x3+2X+3X+

(解)為例數方程式,故全式除以下

13x-2+麦+ =b 集項(8++)+3(x+女)

則

(x-2

x+30,則x=

4 3-4-23.

x2xy-zy2x+4y--

-16)

3. A^(6) 19

2x-xy — y2 - x-54-6-0

(解)解此類方程式 時光觀察可否用「加減法」以踱 吉某一些项 使蒔餘之方程式能分解或能直接求失 光或者,但本題卻不是這類題,故再察看原方程式 之一能否分解為两個一次因式,現(2)式中

-xy-y―(2x+y) (x_y) X-L=43× (~~)

+3

用综合除法本出商式

q(x)=

•g(2) = 243x2-8x2-4

2+12=16-

故gr)有x-2之因子

答:原式= (x-1)(X-2)(C+5x+2)

(f) x2-10x+15x=6

(解)本題時用之誠除數為土),±2,±3±6

設

x-10x*+15x-b C=

因(1)=1-10+ 15-6=0祗役者其倍數云代數和] 敌

f(x)有X-1之因子

式可分解為

x+y+3)

(y2 = 2) + 3y

。

(4)

分解

(y+4)(y-1)

由(3)式得

3)

(5)

¡(u)% y=! & x+1=1

(2x+3】

+3)+z-4(歲x+3)-2

5x+28x

分解

2X7X/

代5

(未完轉入第四第三武