1987-02-25 — Page 32

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頁四第張八第日八廿月正年卯丁曆夏

WAH KIU YAT PO

that 16b3-120+a-0.

報日僑

三期星

日五廿月二年七八九一曆公年六十七國民華中育僑華

A respectively. Find the ratio of the areo OMAT. to thut of the region UMA

6. Given that (a,b) is the

mid-point of a chord

of the circle whose equation ia

2 2 2

* Yur. Show that

the equation of the chord is.

ax+by~(a+b2)=0.

(T and AT

are tangents at

and

Body length

In the fig., UABC is tetrahedron. 0-0,

UR-b, OC=c and LAOC-A Z100 and LAOB. D is

*. (7x-1) (2x+1) 23(2x+1)2 - (7x-1) ( 2x + 1) −3 ( 2x+1),

* . (2x+1)((7x−1)~5(?)

(2x+1)(x-1) 70

the mid-point of BC,

a) If Gia the centroid

of the face ABC,

prove that 306=a+b+c:

b) If u mã, bosh and

c, find "b, b'c and ca

e) Find (306)2' in terms:

of a, b and .

d) Using the results of

(b) and (c), prove

that

90Gab +c^+2lecóBL

+2cacoß+2abcost",

MB/0B~~ON

10 -184

When x, the fraction 7x-1 2x+1

4√21

Length of AB-MB

-18.33 units (Ans.)

is undefined,

therefore x-

•`.x <- or x) *

(Ans.)

1987中學會考

附加數(廿三)

試題預習專欄

DA JA JE KARE

MILL & DALE PRESS

Revision Exercises

för

Cert. Exam. (1987)

1987中學會考試題預習專欄

生物(廿三)

明流出版社

MILL & DALE PRESS

Revision Exercises

for

Cert. Exam. (1987)

BIOLOGY (23)

Unit 10: Growth and

geneties.

(1) Immediately after

birth a set of triplets, two of whom were known to be

identical twins, were separated and reared under very different conditions. The table below shows date for the triplets at the age of 20,

ADDITIONAL MATHEMATICS (23)

Exercise 12:

Paper II, Section A

1. Find the general

solution of the equation

coaбx+cos4x+cos2x+1=0

2. Given that sin30-0

and sin-b prove

one had type ( bloud, the other. type AR.

(1)Which couple do

you think were the partents?

(2)Give reasons for

your answer to (1)

iv. Both parenta were

known to have brown eyes; Mary has blue eyes. Assuming that the brown allele(B) is dominant to the blue allele (b), what were the genotypes of

(1) the mother and (2) the father?

What is the probabi lity that June has blue eyes? Explain your answer by meanя of genetic disgram.

(2) The following graph

shows the average increase in height in boys and girls. from the ages of 9 to

Increase in height (cm per year)

18:

Key: Girls

Boys ***

3. Using the substitution

u«i+x*, evaluate

fax ( 1 + x 3 ) 2 dx

4. Express sin2usin40 às

a difference of two coaines and hence find the value of

(sia20-sin40)de

derived from these beans flowered, self-fertilized and fruited. He weighed

all the beans, which it produced, and

calculated the average bean mass for each

plant.

The result from two strains (A and B) were as follows:

300 400 500|600| Average masa per individual bean produced, in ng STRAIN A 480 450 450 460 Average mass per Judividual bean produced, in mg STRAIN B 560 540 555 535 Mass of gown beans in mg

In the figure, the equation of the curve is y=bx-x”

Offspring

Yellow

Green

.7.7.

102

надл

Parents

(A) Yellow * Yellow (B) Yellow Yellow. (C) Yellow

·66

(D) Yellow x Green

80.

(E) Green x Green

MARY} JUNE } ELIZABETH

1.78 1.78

135][140

Čཎྜཋཏཱ

Height in m

Blood group

1.Q.

Which tweʻgirls were identical twins?

Which characteristic shown in the table enabled you to answer (i)?

The three girls had

got together to try.

to find their

original parents,

and had narrowed the: search to four possible couples:: Couple 1

one had type A blood, the other type B;

Couple 2

both hail type AB blood;

Couple 3

one had type A

blood, the other

type AB;'

Couple 4

4 6 8 10 12 14 16

Age

Describe two

differences in the growth pattern of boys and girls from

the graph.

From the graph, what is the average age "of the ouget of

puberty in (1) boys and (2) girls?

iii. Describe the obser-

vable physical changes (other than increase in height) in (1) boys and.

(2) girls at puberty State two main factor which determine the height of an adult.

(3) Johannsen. studied the

inheritance of need. mass in the bean (Phaseolus vulgaris). In this species the fruits are pods, and contain the seeds "which are beans.

Johannsen weighed individual beans,

sowed them in soil and

allowed them to grow under identical

conditions. Each plant

Draw a curve to show the above results. Give an appropriate title to your graph in (1), y

iii. Explain what the

graph shows.

iv..

What may have caused

the difference in average seed mass between atrain. A and strain B2 Da

The beans produced by any individual plant differed consider ably in masë. Give

reasons for this varintion.

(4) Yellow seed colour of

peas is dominant to green colour. The following table shową: the crosses and S results between different parents with phenotypes shown. The inheritance of seed

colour in garden peas. follows Mendel's first

Law

旅遊介紹專楓康告

iii

Using the symbol (Y) for the dominant character and symbol y for the recessive character, write down the dost probable genotypes of each parent. Theoretically, in cross. (1) how many of the yellow off- spring produced i

P. would be expecte

to produce green progeny when self- pollinated.

(1)Make a genetic

diagram to show

the cross between

the yellow

offspring of

crosses (C) and (D)

(2)State the ratios

of the phenotypes. and genotypes of the offspring in the cross fb (1),

(5) The diagram below

shows the life history of mosquito and the graph shows the growth pattern of wriggler (larva of mosquito).

(a) Adult

(d) Pupa

Breathing

tube

Egg

A. Life History of the Mosquito

Time

3. The Growth Pattern of the Wriggler

(1)Name the type of

life history that mosquito

undergoes..

(2)What is the

advantage of this type, of life history?

Describe the feeding habits of the mule

and female mosquitoes. and account for their difference,

iii. (1)Where does

iva

mosquito luy its egga?

(2) Explain why a

large number of eggs are laid each time...

The graph shows the Atsowth pattern of the

larva Explain the growth (1) at X and (2) at Y.

With reference to the

diagram only, suggest

with reasona, two

efficient ways to

eliminate mosquitoes.

數學(廿三 )

明週出版社: MILL & DALE PRESS

Revision Exercises

[1+/HBC=L2+LIBC

£1-12. AB-BC

and/HAB=/KCB=90"

•ДЛАВ ДВКС (л.8.л)

BH-HK

Area of HBK=200cm2 .*(BII) (BK)=200.

(BK)=200

BK-400

BK=20cm

Area of square ABCD -256cm

BC2=256

BC=16cm

100-x+y=100-(x-y)

The minimum value of

x-y under the

constrainst is

(2)-(8)=-6

.The maximum value of

the expression is 100-(-6)

-106

Section B

(1) Let Sathe sample

n(5)=36

(Aug.)

space of throwing 2 dice

E-((1.1). (2.2).

((3.3), (4,4),) (5.51 (6.6)

.CK=/ BK"-BC,

-16

-12cm

(Anai)

(E)-6] P(E)

p=3cos6+2

3c080=p-2

q=5sin@+2

58ine-q-2

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