1970-06-15 — Page 22

頁二梦張六第二日二十月五年戌庚座复 WAH KIU YAT PO

郭日橋

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中中會考高級數學(一)答案

$#$%5#5#$#%5%S #

僑

(續完)

喬仲强

英中會考現代數學(二)答案(讀完)髙鐶葵

(8,50,6.75) (500, 5oo) Hence or otherwise draw

the graph of the function F. laking Linth as

for

both

x and y axes.

(c) By drawing suitable tangents, find the coord

I correct to 1 place of decimals) of points at which the gradients of the graph are 0, +11e grectively

(d) by scinting squares, find the area (correct to / fo

of dicimals) founded by the curve and the line {(x, y) y^s; X, YER}.

By drausing a suitable horizontal straight line,

find the Values if x such that (2,6:00) CF

Solution

8+7 = 6 x [2.45

「星日五十月六年〇七九一屣公年九十五國民華中育教僑華

( 8 ) — [ 3 2 + < AŽ ABCD, EFGH IZAE, BF CG, HD各長7cm,其他各边均長4cm. 今使長方体之对 角线 AG與地面垂直而A在地上,求

(4) E, F, G, HOZ.

(L) 平面 ABFE與地面

成之角,

(AG) ABCD, EF G H I J π

A

AC=√ 42+42 = 4√2 (cm): 側面ABFE, ADHE-等長方形

<HA AF-√7+42 - 165 (cm)

AG=AC2+AG2

4 +4 +72

-√81 ~9:(cm.)

*AAEG + ELAEG=90"

ANLAGE= AE-1=07778

AG

**24 LAGE=51°4′

7cm

EJ + ££ £ £ EP=h1, B| CEPA = 90°

LGAP=90° (B GA, EP 1 K+eã).

故AE與地面之傾角

¿EAP= 51°4′ (AA4GAEA),

•h,= AE sin 51°4′

-7x2-54 (cm).

7A AFG 4,

AG=9, GF=4

AF÷JZ5

AF+GF

=65+

H

LFIEFRLAK, 剣

FR=FK sin & FKR = = X sin 60°15

— 65x0.8682–8.062 (cm)

[ sin 60°15' I. FR=45x7-125-8.062] </

me FRQ=

165

018958

**/<FRQ=63:37

:(1)平面(ABFE)典地面成63°37′之角

(9) of tan(# +x) − tan

Rta tan(I+x) – tan(π-x)= 間之值。

(EE) 14 tan(A±B)

tan (I+x)-tan (I-x).

tan # + tan x

1- tan I tanx

It tanx

Itanx

*x 4012211

tan A±tan B

IF tan Atan B.

Tan 4-tan x It tan tanx

It tanx

(tanx) ( tanx)

(1 tenx) (1+ tanx)

依倍角公式,

2 tanzx=2)

左方一右方

[|| tan ====]]

,900)

y gradient so

on makes angle with

•AG2 the LAFG=90°

sin LAGF-

THE

0.8958.

* 2 <AGF=63°37′

F +7 AF be đu Z1th A <FAQ=63°37′ FJR W @ < ZL #FQ=h2 = AF on 63°37′

73 (cm)

#AAGH

COLLAGH===

"AG+GHAH2

2XAGXGH

81+16~ 65. 2x9x4

sin LAGH=

BAH 165-

AQ

故AH與地面之傾角亦為6337.

is one

fan and the point

f contact. =(3,4).

draw a line xy.

# LANG=90°

day Using a

set aquares, draw

Inn and find out

the point of contact

-(2.5,8.8)

(d) the area bounded

by the curve and the

line {(x,y) Y=s, XiYER}

10.7 square comits

Draws y=6.

而从典地面之距離亦馬.7cm.

(1) / E Be teño 54 cm, Gele te to 9 cm.

F, H BE Wein by # 73 cm

設手面(ABFE)與

地面之支线為AK

則因 AEII BF,故BF 典地面之傾角,亦為

0=51°4′

The points of

【解)由上証得代入求解之方程式

別以

tan 22=

tan 2x==

∙14°2

4 tasex

tan X

QED

nx 90°+7°}′ (111)

代入,即得

-71 971, 4871, 277° 1

x=7°1′ 97°1′, 187°1' a 2771.

(10)試绘出下列两方程式之图线:

y=costx

y=2rin (X+30°) [0°<~=180]

INAAB*# **£* ∞2x= [3 sioux-1-cos Xin 【解)先製成下列两表:

Cotx

() y cos? 0° 369 6.0° 10.866 0.500

0.25.

90% 720° 156° $180°%

Ö -0.500 -0.866 -1

O

2 sin (X+30°)

0.25 0.751

180°

60° 90° 120° 150° 50°

0.75

(u) y=

X

300

X+30° 30°

60°

sin(x+30) 0.5 0.866

7,000 0.864 0.500

0 -0.5.00

100 473 2.00 173

100

-800

1200 750° 1805 ··2/00:

分别描出各桌,柃同一图中,给出两相图像如 下图示

KF

由图线(C)y=cosx,以

1x to st y m sta

Aino

求解之方程式,得

65-92 (cm.) 因AEFK笃直角梯形 故在AABK中

Fa BK=97—7—27(cm)

•2 (sin x co≤ 30°+ cos x sin. 30° =2sin(x+30)

比马图线(i)之方程式,故两图线),(W)之百奌

ten & AKB-

AB

OK =1.75

B = **k.

由图中看得

x=135

X=135

x=47 or 2

9 (9) ABC i QD M is a point on BC such that M(BY),

piz Denote AB by x and Ac by ÿ Express

BC, BM and hence AM in terms of

X and

** LAKB=60°15′

b) In the geven gigare, M(BM) ?? (HC) -p1q, MNHIBH, and

(未完轉入第六强第三頁】

APMN and ANMC between

Height in

(mm)

same parable!

Frequency

DPMN ANMC = p z

PNUBE Find the following

Ratios.

area & AMEN area of AADC

11) area of 4 APN area of A ABC

W)) area of APMN area of 4 ABC

Solution

(a) 02-7-Z87 - pg (7-2)

AM - 17+87

A MCN similar to A BCA

AMCN LABC

A APN similar to LABE

A APN: BABC = p2 (pty)

from (1) DPMN DABC-PZ PIEL

The following table shows the heights

60 children

ه کانیان

43.8 22.7

42.8 437 43.6 435 04

03

43.5 44.2

چو ہو ہے میری مجھے ہوس کو ہے مدد کے کمرے کیا کہ جو ہونے کو ہونے

41.2 4.4 43,0 23.0 430 439 434 43.8 aig as 03.5 43.4 43.2 45.5 43,4 427 436 43.6 441 444 432 437 43.2 466 415 419 414 420 347

complete the following prequency distribution loble

[ 6) Construct a cumulative grequency table using the

intervals as in (a).

(6) Display

the data of (6) in a polygon and estimate from at the median Solution

(a) (6)

424-42.6 427-42.9

43.0 42.3

43.6

442-444 44.5-447 To Lal

43.2

42.0