1977-05-04 — Page 19

日六月六年七七九一服公年六十六國民體中

Profit 20x + 12y

育教僑華 頁三第張五第 日七十月三年巳丁歷夏

WAH KIU YAT PO

報日僑

AB

1977中學會考試題預習專欄

明德社主編

PR PA

PB

新數學(十六),菊榮家。

Altan45° ADtan30°

ADtan45° |

tan45° tan30"

tan450

11

MODERN MATHEMATICS (26)

Answers to Revision Exercise

Section

(-5)

8.

since

5eon

sin20

+

Jain cose

-

(= 0.4227) (Ans.)

eos 9 -3sinëcost. —

sine)

+ sine - 0

tinh

M

10-11

210 (Ane)

Звілосове

- 135° or 315" (Ans).

cose-4sine

三期星

the tank

721875 22-70

18.75 cm

The wetted surface area

area of the wetted side

wall base area

2(4)(7)(18,75) +(2)(20)

7975 cm (Ans.)

11. (a)

tang

E,F are mid-points of AC and

2(FA + FC)

B is the mid-point of AC

- 2FB

|(1) From the graph above

75 catties of mixture X.

and 195 catties of mixture

should be made to give maximun profit.

(ii) Maximum profit

20(75) + 12(195)

$.3840. (Ans.) 14.(a) f(x)= x2-5x

always positive implies

does not touch the axis

and has no intersection with

the x-axis,

Hence,

52 - 4(1)k < 0

25

(Ans.)

As figure shown; ABC is an

equilateral triangle inscrib-

ed in the circle with centre

or 194

HCX - DCY - ÷ XAY. -

(ext. Z of cyclic quad.)

50 + x

50

(Ans.)

Section B

BD

.(1)

Area of the circle

π (OA)2 = 48 πT

d 50+

(2)

ext.

▲ of triangle)

DA = 48 units

(1)

(2)

9.(a) Let Q(a,b) be an arbit-

rary point on the parabola

(1).

Since P is the mid-point

AB +

CB + CD

2CF

AB

+ AD +

"In rt. ADAD

AD - OAcos LOAD.

but

b + d = 100 + (x + y)

b • d 180.

of QA, therefore p(x,y) is

2AF

207

.(2)

48 coa30° (~ 48x})

opp. 4s of cyclic quad.)

6)

6 units

From (2) and (3)

AB

2AD

12 units

100

peri

Meter of ▲ ABC aides;

Square both

5x

4x

80

12 = 36 units (Ans.)

Y 80

(Ans.)

Let the equation of the

circle be

Substitute into (1)

- 6(2x

CB CD

-2(2FE)

ABP

(Ans.)

(b) HD - DE - EF ■ FG - GC,

(b) f(x) 2c

always negative mean there

is no intersection between

f(x) and the x-axis

Therefore,

2.2/

+ ( x − 1)

( x − h)2 + (y - k)2 = p2

Since (0,0) is on the

the equation of

Iocus of Pis

BE

EC 2:3

4AC < 0

AB

BAR + 2AC

32 - 4(−1)(20) <0

12(x-1)2 (Ans.)

circle

-

.. (1)

Square both sides

(r-

5)

(x − 1) {(x − 1)

1)2 - (x − 1)

1]-

the centre of the circle is

(h,k) and is on 3r

(b) (i) y =

pendicular to z

alope of r

is per

2;

2

遗

JAB+

3DA

12.(1) AG -

+

= 1 (Ans.)

(x − 1)(x − 2) = 0

3h 4k -

·(2)

or x = 2 (Ans,)

From (2)

(ii) hence, y = x + k is

the tangent of

+

*

2

be rejected

.(3)

AC

(x + k)2 - 2

35

are the roots of

4x+3=0

n

- 2

.(1)

Substitute into (1) (*)2 + 12

k = ± 3

Substitute into (3)

. (1)

- 25

has repeated roota.

From (1)

AH -

2x2+ 2kx +

2AC

SAE

2CA - 5EA (Ans.

BC CG

√162. 122 + 102

J500 en

22.36 cm (Ans.)

+BC

-√162

+ 122 20 cm

JAD2

+ DH2 -√122 + 102

(Ane.)

(c) If f(x)

(Ans.)

11-10x

is real valued, then

the expression inside the

square root must be positive

11 10元

(11 + x)(1

x) 0

-11

or x > 1 (Ans,)

-4AC-

15.62cm

(Ans.)

(2k)2

4(2)(k2-2) -

+ n(n+ 2

(2+1) (m+2)

2.

*+2(n+n)

mn + 2(m+n) +

(m + n)2 2mn + 2(m + 1)

-

mn +2 {m + n) 4 k

+

-22-2(3/2) + 2(2)

(3/2)

(Ans.)

In rt.4PAD;

PA *

Antan30°

In rt.APBC ;

PB = BCtank 5°

CD РВ

therefore, the circle in

2

( x −4) ► (y - 3)2 = 25.

(x + 4)2 + (y + 3)2

- 25

(Ans.)

7. Since

I.

10

+

D(n+1)

10x11

11 x12

:

12X13 12

+

19x20

+) 20121

=

+

19

+

13

10311-11-12 12:13

k = 2 or k = -2

(ii) The angle between planes

HAB and ABCD 18: Z. HAD

(Ans.

In rt. A HAD

tan¿HAD

AD

端

-0.8333

10. (i) The capacity of the

cylindrical water tank

(22)(2)2(300) en3

1155000 cm

3

The rate of the pipe

- 30 x 10 x 22 x(1·75)

cm3/min.

- 7218.75 cm

Time required

-

1155000 min.

7218.75

160 minutes

2

2 hours and 40 minutes

2. - 2 hr.s)

(ii) If the pipe is turned

on lo minutes the volume

of water in the tank is

7218.75 em

HAD = 39°481 (Ana.)

(iii) The angle between CE and

(i) Yes, the rate of flow is

constant for the graph is

straight.

(ii). The constant

-

plane BCGF is ECF

In rt.A EGF

tan 4ECF -

EF

CF

16

From

the graph

1

√122+ 102

* 1.0243

45°41 ZBCF - 45 41* (Ans.).

13. Let x catties of mixture X

and y catties of mixture Y

should be made,

(2)

1

+

"20 x 21

Q

www

ADtan450

+

12x

X 4

< 105

+

y

100

(2)

+

· -·} --

wwww

-

72187.5cm3

I

190

+(-27) + (28-21

The height of water inside

x 2 0

y > 0

m3/min. x 10 min

rate

the slope of the graph

10 gallons/min,

t-σ ▼ 10

there is 10 gallone

of liquid fuel already

in the tank.

(iii) Let the required time

be t minutes from the start.

750

-

10- 10t

t = 74 min.

1 hour and 14 mins.

(Ans.)