1969-01-30 — Page 14

CER FORCE ORD

WAP KI YAT PO

郭日橋警

四期星E卅月一年九六九一公年八十五國民華中有教備载

罗僑

3#2%$#$%$#TES ## $%3 #5°45 #5%$ #$%$#$%

《一九六九丰中文中学會考試題預習!

1969

題預習

· Prove 1':{A-

Proof :

TTT

TFF

現代數學科

(13)

FTT

F TF

Solutions.

数昂科 (13)

33.

第十二次預習題解答

(1)証明下列恒等式:

(a) (1+ sino+co18) = 2(1+sing) (1+C010).

(o) £3 = (1+sin 0)2 + 2 ( 1 + sinf) cos ( + cos30

== |+ 2 sing + sin § + 2 cos 0 (1 + sint

1+2 sind+|+ 200s A (1+ sin() =2(1+sino) + 2e010 ( 1+ sin O). 2(1+ sing) (1+ coro) [s]

(b) (It seca tam 8) (it tamp sin^"^()

= (1+ sec's tam 8) (14 ton'd sin2 () (CE) Ebs at tand, sec's 24 x to 123 1+tand tamp hit 3 secs -]. H^

*3= [1 + (1+ ṭan_x) tan0] { |+ (sac2p-1) sin^8]

= (1 + LanD + tanx Lan 0) (1 + sec'p sin ()-sin §)

= ( sec2 0 + tana tan2 0 ) ( cos10) + sec3ß sin^8)

CHA

+ tarx cin @ ) ( cas ^8 + secß aini^6)

It tand sin 0

• (cor20 + sec2ß sin3D)

=(1+ tand sin 8), cos20+ Lee's,

-(I+ tano sin^0) (1+ sec2p tan'0) [ts] D.E.D

()+cotx) + (1-cotx)2.

(*x)/3

(I+ 2 cot x+ cot'x) +

tan x)

2(1+

X+siny

(d) CASX - Cody

正左方

tanx [tos]

(sinx+sing) (cxx+co+y)

[B cox in

2 co

cory + sin y == }}}

(e) cos×+ cos A+ cosy + cos (A+B+7)

= 2+ 2 001 (4+B) cos (B+7) cos (Y+4)

ED

=2(1+ CO124), cors = 2 (1+ cos ap) cas2 (A+B+Y) -=[1+008 (23+28+20)]

* j = {( 1 + css.za) + ¦ (1+c=2p)+ 2 (1+ cos 2Y)

+£ [I+cos(24+28+21}]

2 [ COLAR +eN2ß+ 001 (24+28+29)+6w$27]

•2+ £ [2008(4+p) cat (amp) +2 cos (2+p+2+) car (+ß)] 2+cos(x+p) { am (q+p+xx) + cos (a-p)]

cos(8+4)]

=2+cm(d+p) [zan (d+7) cos

b=2+2008 (d+ß) cos (8+9) cos(7+4) [* ́s] Q.ED.

(2)一索餐繞二輪其半抂各第13呎両輪中に

相距5呎,求此震之長

(解)韻如右图:0.0'為两輪

中心,則此索之長為

AB+BD}} 36+¢D+AL

LE 04FOPHIAB, #I

MODERN MATHEMATICS (13)

It is not true that if prime numbers are odd," ther groupe have identities. If groups do not have identities, then there are non commutative rings." Also if there are non factorable expression, there must be non commutative rings, therefore, there are no nonfactorable expressions.

Solution:

Let p be prime numbers that are odd, g be groups have identities, r be rings that are commeat and a be nonfactorable expressions. Symbolic form of the argument.

The left hand member IAŞ

KPA Alq var)] A (AVN) -

[(p ^~q ng) y (paw game)}^{~sver);

→

[F V (DANGANP)] ^{~S W~T)

(panan MT)^(~SINI)

(PAN QA IANS) V(PANQANI ANT); {{PAN QA~r) A^~^S] V (PANGANI) PANGAN P

There

the left hand member implies PANQA I, but not us. Hence the argument in valid.

2. If rich people are unhappy, then money is evil. Ir medical care is expensive, then money is not evil. As a matter of fact: medical care is expensive and rich people are unhappy. Therefore medical care is both axpensive and inexpensive, Solution:

Let p be rich people that are happy, q be money that is evil, and r be medical care that is expensive,

Then the sentences are represented by

The left hand proposition:

p^p) y_{r^N DANI) EVARANTAN D)..

V

The right hand proposition=F Hence the argument is valid. That is

Frice rises, only when wages rism. If employes fringe benefits are not increased, wages rise, conversely, Since prices have risen, it follows that employee fringe benefits have not been increased.

Solution:

Let p be price rise, o be wages rise, and z be employes fringe benefits are increased, Than va have

Left side

·QA

་་་་

(و)

9. Butor and q are equivalent. There-fors the argument is valid."

11.1. Prova : MU (A BAY B BAVNE FTTTT "FF F ལག་པ་བ:T FATT

I FET TETTE TTT

Since the biconditional is a tautology, the relation is proved.

Proof:{& VB)

FP

Prove: (A

Hance it is proved.

IV. Relations and Functions

Some kinds of relations

The notion of set is the most fundamental in mathematics, When we are considering the properties of objects, sets are useful. For example, to say that an integer, say 3, has no multiplicative inverse ža to say it belongs to the set 2 (set of integers). When we are comparing objects, the notion or relation becomes important. For example, to say -2 1a less than 1, is to speak in terms of a relation "less than". As a second example, to say that 2y in equal to 4x is to speak a reletion of y and x. The concepts of set and relation can serve to unify and carify all notions of mathematics. Hence it is not without reason that why the topic of relation must be presented in the study of modern mathematics. Almost all ideas in mathematics can be expressed or explained in terms of sets. No doubt, the idea of relation will be developed on the basis of set motion also.

Relations are of different orders. Relations arising from comparisons of two objects are known as binary relation. For example, the congruence relation of two polygons is a binary relation. Relations arising from the comparisons of three objects are called ternary relations, such as point bis between points A and B.In secondary school mathematics, we consider only the binary relations, If two objects x and y are ii relation with each other, then we shall designate their relation by xhy. In what follows we shall consider those relations which occur frequently in mathematics, assigning to aach of them certain nama3.

(1) Reflexive relations

There may be member of a given set for which there is no y such that either xRy or yo. In such case, x 19 said to be involved in the relation. R. The relation is denoted by xx which is reflexive.

examples:

(a) In the set of numbers, each one is equal to

itself

(b) in the set of triangles, each is similar

itself.

Symmetric relations

If a certain relation is of this type "such that if Ry, then ylix. Such relation 16 said be symmetric.

Examples: is

{•) IT AABC~A DEF, then Al

(b) If AB // CD), than 1) // li {

symbol for lines).

Antisymmetric relation

in the relation xity if x am y cân BOI DS interchanged, then the relation is said to be antisymmetric. That is if xity is true, cinza yx is not true.

Examples: the

(a) Ifab, then a 1 zut brue.

(b) If A is a proper subset of by then B-1 proper subset of A is impossible.

Transitive relation

gà relation R is called transitive if the following property holds: For any elements

x, y, and z in the given set, if xky and yiz, then xi.

Examples:

(a) For three numbers a, v; and e if a> bam

be, then ac.

(b) For three coplanar lines y

and

l1 // l 2 and 12 //ly, then (1//L G

{(5) Equivalence relations

Sometimes we may have relations which pertain all the three properties, namely the reflexive property, symmetric property and transitive property. Such relations are called equivalence relations. Relations as equality, congruence and similarity are examples of this type

Example

In a set of congruent triangles, we have

ABC ABC

If

ABC ABIC then

LA ABC = A A'B'C' and A

MABCD m2 OP 5 sin. 15× 0.9165458.

LAC

27Z=૨૩)

141′′BOD= 360–132°50′=227°10′

優雅 BD 227010

273 360

136275

340

217.99

3&=2×4:58+2.32+11.89=23.37

答:索長23.37呎.

NOT LOG 793: 219015.

0.4971.41

1080 RG334. 2.32 0.365

1363:3 1345

7 0.49710

3.6316

360 2.55636-

1.89 1.9753.

(3)一园半径6吋,由国外一桌至此园作两切线,若两切

线成55之角,求两切线典巧吏之苏弧吁圍城之 面積

(解)談如右图:因LAPB-55°

··LAPO=27.5°.

=/25

AP=OA cot 27.5

6x19210=11.53 (4)

& OA P== x 6 x 1153=

34-59 (20₤)

FATT

TFF

Since the biconditional is always

relation is proved,

Proof: (A

BANB) TFTFF T

TFFF

F

FTE F

Since the conditional is a tautolog

proved,

relation

MĦ{ A0B = 12 × π1⁄2^— 32-*π × 36 = 39. 28 (35•†),

PASMIADI)=2×34-59–39, 28=29.90

答:面積為29.90方时

(4)-倒立錐形之量杯,高8厘米,口径(直径3厘米,求

jet 里2cc間在斜高間に距離(錐体積一字どん)

(AE) Ik $5 h, cm, ja ¥1I &, chat,

#1=141* V = /c.c.

--- = h = Zh

=1, H.

$11 (/[h1) xh, == 1-

256x3

h - 3576=3.01

за

(用對數計算)

(未完轉入第四張第三頁)

1JT£ £ his 20.0.1$$¶®± 5

NO

LOG.

(The) xha=2,

0.49.7.1

उभ.

18+1.5 =√22.25 = 8.14

25622

उग

3.0J 0.4790

512 27093. 37

0.97426

512

-35123.79

s 2 42 443 21.

故

3.01

-8, 14 x 0.78 — 0.79

379 0.5784

2:183207917

(5)一塊三角形之板,各地為5吋,6吋,7吋,在水平拾上) 沿最長边韩 360°角,求其对真美(2)枱面之距離

(未完铸入第三張第三言)