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1770英文中學會考試題預習專欄
堅道英文書院主編
學豐
也或假包店 衛
(八)
中文中學 會考試題預習專欄 數學科 (八)
第七次預習題解答
生物科
BIOLOGY (3)
banez to the questions or Last WOOL
Distinguish between transpiration in flowering
又其根為何?
Ins.
What do you understand by transpiration?
plants and perspiration in mammals,
(b) What external factors determine the rate
transpiration?
(a) Transpiration is a process by which plants 1086 their water as water vapour into the atmosphere This process is brought about by the evaporation taking place in stams and Leaves aapeolally in the atomata, Consequently. excessiva heat in plant is released and the "transpiration pull" created facilitates the transpiration of mineral salts in the vascular bundle.
Distinction of transpiration from perspiration:
Transpiration
(i) Evaporation of water in the form of water vapour from the
atomata in the leaves of plants.
Has the function 01. forming a tranapir- ation stream which helpa the conduction' of water in xylem Vessels:
(iii) It occurs in the
parts of the body of a plant;
iv) Loss of water by
trana piration
Ang.
Para piration
(a) Release of water
and wastes of
matabolism through
the sweat pores of the sweat glands.
of maza or some
̇other mammalia
ekin.
Has funotions of:
excretion and
body temperature. regulation which keep the body. temparature
"conatant in a ho
environment.
It accurs in the
akin of a man or the skins of some other mammala.
(iv) Loss of water ans
by-products of
metabolism by
perspiration,
External factors that influence the rate or. transpiration :-
(1) Light - the influence of light on
transpiration is mainly indirect as
brings about the opening of the stomata.
which are the chief apertures of water loss in transpiration.
(1) Temperature - the higher the temperature::
is, the greater is the rate of transpiration. Usually an increase in air temperatura result in an increase of the temperature of a leaf. In turn the increase in the temperature of water in the leaf apeede the rate at which water molecules escape as water vapour from the leaf surface: through the atomata. Therefore an increas in leaf temperature favoura an increased rate of transpiration because water may evaporate more rapidly from the leaf.
Relative humidity - It is the amount of water vapour in the atmospheric air at a. stated temperature.
The higher the relative humidity the lesser
the rate of transpiration. Usually the rate
of transpiration varies inversely with the relative humidity of the air.
iv) Moving air (wind)
In still air the rate.
of transpiration is reduced while in moving air the rate of transpiration is increased.
(v) The moisture content of the soil - It affecte
the rate of transpiration indirectly by
altering the absorption rate of water by the roota of planta
As far as we know low scil temperatures
markedly reduce the absorption of water:
roots when there is an optimum amount of water in the soil.
What is meant by the term "asmosis"? With the ai of labelled diagrame, describe an experiment to show osmosis in living tissues.
Osmosis is a physical process in which the molecules of a solvent pass from either a dilute solution or a pure solvent to a more concentrated solution through a semi-permeable membrane.
Ia plants. the absorption of water by root: hairs is the result of pemonis. The plagna. membrane of each epidermal cell in root hair is a semi-permeable membrane which separates a dilute solution, the soil water, from a more concentrated solution, the cell-nap. A difference of osmotic pressure le developed and water is drawn into the root-hair cells,
(未完轉入第四張第三頁)
喬仲强,
=○之两根相等求是ż值:
(解)原方程式可化高(花+2)X2+(k~3)x+(2號-3)=0之二次 方程式的範式,因两根相等,故根之判别式為零
824AC=(R-3)−4(k+2)(2k−3)=0____
k=6k+9−4(2k+k-6)=0, Bq 7k +10k-
> (7k-11)(k+3)=0__ £= 1⁄2 * £=-3
若是=号,代入原方程式,(u)差k=-3, 代入, 2x−3x-3+(x+x+2)=0¦
25x=10x+}=C:
(5-1)
Be
3(x2+x+2)
−(x+3)=0
x=-3(18)
答:(优) €=号,則x=吉;或(C)k=-3,则x= (2)溢ax2+28x+C=0*两根為虛數,試証
ax*+2(a+b)x+a+28+C=0之两根亦潺度數 (註)因兩根為虛數,故判别式小於零。
.. (28)_4a_c<0, B4 4f-4ac 現冊求證方程式的判别式
AB=4AC=[2(a+b)]=40(a+25+c)
=(4a2+8ab+48)–(4a+8ab+4
4b-4ac<0.
故其根為虛數
(3)飯x*+(3k+2)x+龙--2克-5=0方程式中,若一根為
他根之三倍,試求龙之值,並核驗之.
(解)設一根為一,則他根為300,依根與係數之間係得
·B 42=-(3k+2)
•3x=(k*2k-5) by 34=k-2k-5
(3)
3(-35+3)在龙-5
3(9k2+12k+4)=16(K¬ak
11k+68k+92
(k+2) (11k+46)—0
核验 (i) =-之,代入符 x+(-6+2)x+4+4-
x2-4x+3=0 (x-1)(x-3)=0
分解
或x=3(滿足)
4) ₺ k=-# ^^13 x2+(-4*+z)x+2016+17
1212 1276x+2523÷0
(11%−29) (11x-87)—c
或特
--(2)
*(}}} # &=~4(3k+2)=- £[ ??? +2]=-4(-")=?
由此结果得知 2523=29x87
(4)=ax+bx+C=0px*+gx+2=0 两方程式中
★-**@m,**** (cp-ar)=(br-cg)(aq-bp).
(証)設為該两方程式相同之根,則
jax+bx+c=0
1x+gu+2=
依十字案比法得
cf-ar "cp-ar
«2(cp-ar)=(br-cg)(ag-bp) α
因20(因代入原方程式不能滿足)
·· (cp-ar)'=(br−cz)(aq-bp).
[3]
(5)溢xj 為 ax2+bx+cm)之两根,求下式之值:
(a) (ax+b) (ap+b), _____ (f) (bu+c) (bp+c} {c} (ax+b)^+ (ap+b)
(解)由根典係數閩係得<p=p=暟 (a) (ax+b)(aß+f}) ==a*«ß+a&(a+p)+b2 = ax&+ab (£) + b
ac[]
“别解)因爲ax+bx+c=0之根,故能滿足之, 代入 ax+bd+ c=0,
4元ś解 <(ax+b):
何理可能
as+b=-=-
(au+b) (a8+b) =~~(&)—_—ac. (b) (bu+c)(bp+c)=bàp+bc(«+ß)+c2
= b2 (a) + b c ( &)+c
=c[] [答]
(别解)团ax2+bd+C=0
同理
fa+c
bal + C =
(bd+c)(&s+c)=a2 (ap2) at(aps
a(&)c
(0) E a+b2+C=O
ad+f=-=
數(a+ 5 = x+ (
(ax+b)>*+(ap+b)*=+=
=[((b)=()=>
[答]
- /[(-2)2 2(&)) = ↓ (+-*c)=122ac [6] 6)設戶為ax+28x+C=0之两根,求作以及 B* 為根之方程式.
(解注一)由根典係數關係 +8
岈求方程式之两根和
又两极预
故旴求方程式為
•(a+p)=2dB
a*y*-(4f*zac)4+c*=
此方程式可化為
亦即
ay+2acy+C2
(解)設為巧求方程式之根,则
土网,以代入原方程式,
移項
a(±ly)'+2b(±/7)+c=0
ay+c=-28(417)
(ay+c) 4fy. [*]
(註)此題之第二解法,稱為方程出之变换
(Transformation of Egnation) *a****X 這一類題目有時用之,較為容易深盼费者细 意揣摩。
(7)38 a+b+c=0, * #£
+++(+)=0
(JE) & a+b+c2-3abc
=(a+b+c)(a+b+cad
02+b+c?=3alc
+(+++
be+actab
adc
fre2+2(ab+ac+bc)
30fc
[因分子 a+b+c=
第八次預習題
a+B+C2
(I) ABCD為不平行四边形,E,F分别為 AB,CD的中 EF *** £(BC+AD)._____
4,
2)設△ABC中之 AB> AC. 今由 B及C至∠A之平分线 作出直线BE及CF,其委趾各尊 E, F,又D為BC之中奌 求證 DE=DF支(AB-AC)
(3) △ABC大高為 AD, 外接园心島O試证 ∠OAB=∠PAC
(4) 已典 △ABC, 求以 A. B. C各羔园心作三因使两两 对外切
(5)已典 △ABC, 求以AB, C备高园心作三园,使BC两
因外切,且均典A目内切。
(6)内模於园而外切於园之四边形,其对边切莫之騏、 线互相垂直
Page 15Page 16
27 DEC 1969 REFERENT LIBRARY