Analisis Bertopik Form 5 Chap1 5 r

Number

KERTAS 1

:t'r " ': :

lilr

;i Convert a number in a certainii Menukar nombor dalnm suatu asasij

i;.li-- .-,. -_.

SPM m Question 4I 1100012 - 10112 =

A 1001002B 1001102c 1110102D 1111002

SPM M Question 52 What is the value of the digit 2, ln base ten,

in the number 324Ls?Apakah nilai bagi abil Z, dalam a|as sepuluh,)^l^* -^^L^- 2-a/.1 .'dalam nombor 3241s?

850D 250

to a number in another basenombor dalam asas yang lain

in the number 1203s?Apakah nilai bagi digitdalam nombor 12035?

A10850c 200D 250

SPM m Questrbn 64 110012 - 1012 _

A 10012B 11012

c 100002D 101002

N.uffiher in futequif 'bmeNafiibat ,ilafu.aias, y Xi,l, .,ldikehendaki

al

;1:i t)!

lllrl:iij,riiiiii,'.,..... , , ' .'::

Le digit 2, rn base ten,

2, dalam asas sepuluh,

FORM 5

l. ,,Nurrtber in base two,eight or fiveNombor dalam asas dua,lapan atau lima

2. Number in ha*e fur{r, :

Numher iRr,fo,gss 16n,,

Nombor dalan asas

stpAlih'" ', '. "''':'',,'' '',,,, , +

Number ifi,base two,:Nor.nbor daW'atat dua

I>fr

Nfiurber,rin,,b'ry.eight,' .

Nom.bor dalam asas lapan

Number in base eightNornbor dalam asas lapan 0 I 2 3 4 5 6 7

Number'in bsse twoNombor dalam asas dua

000 001 010 011 100 101 110 lll

:riqr;::i::;:i:rit,.;ti.irr.tit.i;i::I.:_.+riil;.:ilijr;i1l;ii+:lai:.:ri,;tiilri;::.r::i:r.i:r,:::i,i,::i1+iii..ij:::

Numbers in Bases Two, Iand Five

SPM m Question 53 What is the value of the digi

A25c 200

139

SPM m7 Question 45 1100102 - 111012 =

A 1011012B I 10002c 1001012D 101012

SPM m7 Question 56 Given 3 x 53 + 4 x 52 + 5y = 3420s,find the

value of y.Diberi 3x53 +4x52nilai y.

+ 5y r 3420s, cari

24

SPM m Questron 57 101012 - ttz -

A 100002B 100102c 101002D l0l102

SPM m Question 68 Express 204s as a number in base five.

Ungkapkan 204s sebagai nombor dalam asaslima.

A t0t2sB 1304sc 21015D 4031s

SPM m Question 59 State the value of the digit 5 in the number

15368, in base ten.Nyatakan nilai digit 5 bagi nombor 15368, dalamasas sepuluh.

BD

AOC3

A40864c 320D 500

140

SPM mg Question 610 111002 - 10102 =

A 100002B 100102c 101002

D 101102

SPM nlO Questrbn 511 Express 5(52 + 3) as a number in base five.

Ungkapkan 5(52 + 3) sebagai satu nombor dalamasas lima.

A 30105B 3001sc 1030sD 1003s

SPM nlo Question 612 110012 - 1102 =

A 100012B 100112c 101012

D t0t1t2

SPM 2oll Questrbn 5

13 What is the value of the digit 3, in base ten,of the number 1302s?Apakah nilai digit 3, dalam asas sepuluh, baginombor 1302s?

.{25875c t25D 375

SPM 2oll Question 614 l0l102 + lll2 -

1010121100121101121trot2

ABCD

REVIEWAND

KERTAS 1

.ffi . Alternatively, use a scientific cal, answer directly.

l1o \\ 2

1 \O.8.U 12

- 10112l00tt02

Answer: B

. ldentity the place value ofnumber first. "'i

i1ffi

SWERS

or to get the

h digit in the

3 ,.,2 4 15

5: ,s2., 5r 5o

Value of the digit 2 = 2 x=50

Answer: B

-15'

1203s5: ,52 51 50

Value of the digrt 2 = 2 x 52

=50Answer: B

l:l'1 . Alternatively, use a scientific calculator to get thei answer directly. ]

I

021\0 0 12

- 1012101002

Answer: D

101012

2t0 u}2 0 2\\ee\&

111012

Answer: D

t4t

3420s = 3 x 53 + 4x 52 + 2x 5r + 0 x 50

=3x53+4x52+5y.'.5y -2x51

=10!=2

Answer: B

o210\s. 12

- llz100r02

Answer: B

ry . Convert 2O4sto the number in base ten first.' . Then, perform repeated division by 5 until thei quotient is 0., . Write the remainder in each division in sequence, from bottom to top to form a number in base

five.

204s-2x82+0x8r+4x80=128+0+4- 132rc

.'. 2048 = 132rc

= 1012s

Answer: A

1 ,,5,,,: 33r gz gr

Value of the digit 5

=5x82- 320

Answer: C

slttzsl 26

-5 I ssE0

68

3o

10 02I 1*0,02

- 10102100102

Answer: B

ffi . Expand 5(52 + 3) first.' . 53 + 3 x 5 is a number in base five written

expanded notation.i. -,',

5(52 + 3)

=53+3x5=1x53+0x52+3x51+0x50- 1030s

Answet: C

12 0t21\S812

I 102

1001 12

Answer: B

1

5:

')_L5

5n

l3 3052 , 5t

14 1t101102

+ ll12

Value of the digit 3^ -1=JX)'

=75Answer: B

111012

Answer: D

53 -'l)' 5t 50

I 0 3 0s

142

f Functions ll

KERTAS 1

fl"**+* 'Hffi .:::T iiti,J,i

,ffi ,'ffit,2.1 Graphs of Functions 1 1 1 1 1 1 1

2.2 Solution of an Equation bY theGraohical Method

2.3 Region Representing lnequalitiesin Two Variables

lathematical Ft

'#,> S

LinearLinear

j=aX*c

QuadraticKuadratik

!=ax2+c

!=ai*bxic

'Funsfron-'i ,.,fuagii'cr,,, :

d,> O fi<0

CubicKubik

!=ax3

!=ax3+c

ReciprocalSalingan

av--x

143

W Graphs of Functions

SPM ffi Question 28t Which of the following graphs represents

!=x2-4?Antara berikut, yang manakah rnewakili grafy=x2-4?AB

v\t,$',

lT*'v

t\l+J_I\

D

f1V'\/ul '.

DC

+. 1AruSPM mO Question 282 Which of the following

)= 10 -2x3?Antara yung berikut, yanggrafbagi ) = 10 -2x3?

graphs represents

manakah mewakili

BA

144

SPM 2(n7 Question 283 Which of the following graphs represents

Y=-x2+5?Antara yang berikut, yang manakah mewakili

SrafY=-*+5?AB

Yyh4lt

SPM m Question 284 Which graph represents y = -x2 - x + 2?

Graf manakah yang mewakili ! = -x2 - x + 2?

DC

BA

SPM m Question 30

B

D

1,45

SPM nlo Question 3o6 Which graph represents y = 5 - x3?

Graf manakah yang mewakili 1l = 5 - x3?

SPM frll Question 287 Diagram 10 shows a

plane.Rajah l0 menunjukkanCartesan.

graph on a Cartesian

suatu graf pada satah

"';:;:fri'Which of the following is the equation ofthe graph?Antara yang berikut, yang manakah adalahpersamaan bagi graf itu?

A !=x3-27C t=3*-27

B y=-x3-27D !=4*-27

KERTAS 2

ith:fu iI I : I il?"{rlllsErIlf f Ll:] ilIilfal:tEF:, ffi

t: t'.

!i i ffi #+2.1 Graphs of Functions

,*_**.*-**".134

34

34

34

34

34

341.1 Dotuuon oT an trquauon oy tne i

*--* -9rap[ica! \4e!hod i".j

1

41

T1

41

41

41

41

v2.3 Hegton Hepresenting lnequalities

in Two Variables 1 1 1

:i: i:i:::ll :ti

1. ^lL.

,,,,,,,,,,,,,,.,,. y = ax + c

:,!:t:,i::,

x

y>ax+c

Gffi Region Representing lnequatities in Two variablesSPM M Question 3I On the graph in the answer space, shade the region which satisfies the three inequalities

y2-2x + 1o,x< 5 andy< 10. 13 markslPada graf di ruang jawapan, lorekkan rantau yang memuaskan ketiga-tiga ketaksamaan y 2 -2x + 10,x<Sdany<lo. f3markahl

Answer/Jawapan:

:t: i;.i: i:ir::r::i:i:1:ir::ittlil!:-:ii:'ii:.;,;i:;:i:::;:iIl:!:;:;lli;ii::i:l.ii rr,-ra-,r,r,r=aa,rrji

146

SPM m7 Question 1

which satisfy the three inequalities y < -x + 6,

[3 marks]yang memuaskan ketiga-tiga ketaksamaan

[3 markah]

SPM mg Question I3 On the graph in the answer {pu.", shade the region which satisfies the three inequalities

y<-x + 10, y <x andy> 1. f3 marksTPada graf di ruang jawapan, lolek rantau yang memuaskan ketiga-tiga ketaksamaan y < -x + 10,

y<xdany>1,. L3markahl

Answer/,/awapan:

147

ffiffi

Graphs of FunctionsSolution of an Equation by the Graphical Method

SPM mS Question 12

4 (a) complete Table 1 in the answer space for the equation ! = 2* - x - 3. 12 markslL2 markahl

(b) For this part of the question, use graph paper. You may use a flexible curve rule.q, Untuk ceraian soalan ini, gunakan kertai gri|. lndo boleh *Lrggrrokan pembaris Jteksibel.

By using a scale of 2 cm to 1 unit on the x-axis and2 cm to 5 units on the y-axis, draw thegraph of y - 2x2 - x - 3 for -2< x < 5.Dengan menggunakan skala 2 cm kepada I unit pada paksi-x dan 2 cm kepada 5 unit pada paksi-y,lukiskan graf y = 2x2 - x - 3 bagi -2 < x < 5.

14 marksl14 markaltl

(c) From your graph, findGA Daripada graf anda, carikan

(i) the value of y when x = 3.6,nilaiyapabilax=3.6,

(ii) the value of x when ! = 37 .

nilaixapabila!=37. 12 marksl[2 markah)

(d) Draw a suitable straight line on your graph to find all the values of .r which satisfy the equation@ 2*2-3x = 10 for-2<x(5.

State these values of x. 14 markslLukiskan satu

^garis lurus yang sesuai pada graf anda untuk mencari semua nilai x yang memuaskan

persamaan 2x2 - 3x = l0 bagi -2 < x < 5 .

Nyatakan nilai-nilai x itu. 14 markahl

Answer/"/awapan:

(a)

Table 1

Jadual I

(b) Refer to your graph.Rujuk graf anda.

(c) (i) y -(ii) x =

x 1 -1 -0.5 1 2 3 4 4.5 5

v 7a, 1-/- 3 l2 33 42

(d)x=

148

5 (a) Complete Table I in the answer space for the equation y = + by writing down the values@l of y when x = -3 and x = ll5. x

12 marksl

Lengkapkan Jad.ual I di ruang jawapan bagi persamaan y = + dengan menulis nitai-nilai y apabila

SPM mG Question 13

x = -3 dan x = 1.5. 'L [2 markah)

(b) For this part of the question], use graph paper. You may use a flexible curve rule.q, Untuk ceraian soalan ini, gunafan kertas gri|. lndo boleh *bnggurokan pembaris fleksibel.By using a scale of 2 cm t[ I unit on the x-axis and 2 cm to 5 units on the y-axis, draw

thegraphofy- !rcr4i x<4. f|marks)Dengan menggunakan skala 2 cm kepada I unit pada paksi-x dan 2 cm kepada 5 unit pada paksi-y,

lukiskan graf y = +bagi4<x<4. 15 markahl

(c) From your graph, find


(ii) the value of x when )l = -13.nilaixapabilay=-13. [2 marks]

12 markah)

which satisfies the equation(d) Draw a suitable straight lin! on your graph to find a value of x@ 2*2 +5,r -24for4Sxa4.

State this value of x.Lukis satu garis lurus yang spsuai pada graf anda untuk mencari satu nilaipersamaanlx2 *_t = 24bagi43x{4.Nyatakan nilai x itu.

13 markslx yang memuaskan

13 markah)


(c) (i) y -(ii) x =

Answer/"/awapan:

(d)x=

149

SPM m7 Question 12

6 (a) complete Table I in the answer space for the equation ! = 6 - 13 by writing down the values@ of y when x = -L and x = 2. t2 marksf

Lengkapkan Jadual I di ruang jawapan bagi persamaan ! = 6 - x3 dengan menulis nilai-nilai y apabilax = -l dan x = 2-

12 markahl(b) For this part of the question, use graph paper. You may use a flexible curve rule.q, Untuk ceraian soalan ini, gunakan kertas graf. Anda boleh *bnggrnokan pembaris fleksibel.

By using a scale of 2 cm to I unit on the x-axis and,2 cm to 5 units on the y-axis, draw thegraphof y- 6-x3 for-3 <x<2.5.Dengan menSSunakan sknla 2 cm kepada I unit pada paksi-x dan 2 cm kepada 5 unit pada paksi-y,lukis graf ! = 6 - x3 bagi -3 S x < 2.5.

14 marksl[4 markah]

(c) From your graph, find@ Daripada graf anda, cari

(i) the value of y when x = 1.5,nilaiyapabilar= 1.5,

(ii) the value of x when y = 10.nilaixapabila)= 10.

12 marksl[2 markah)

(d) lraw a suitable straight line on your graph to find the values of r which satisfy the equation@ *3- 8x - 6 -0 for -3 < x <2.5.

State these values of x.Lukis satu garis lurus yang sesuai pada graf anda untuk mencari nilai-nilai xpersamaan x3 -8x - 6 = 0 bagi -3 < x < 2.5.Nyatakan nilai-nilai x itu.

Answer/,Iawapan:


(c) (i) y -(ii) x =

f4 markslyang memuaskan

14 markahl

(a) x -3 -2.5 -2 -1 0 1 2 2.5

v.l .lJJ 21.63 t4 6 5 -9.63

Table 1

Jadual I

(d)r=

150

SPM m Question 12

7 (a) Complete Table 12 in the a4swer space for the equation y = + by writing down the values@ofywhenx=6andx=10. - "r'

tzmarks)

Lengkapkan Jadual 12 di ruan$iawapan bagi persamaan y = + dengan menulis nilai-nilai y apabilax=6danx=lo. * l2markahl

(b) For this part of the questionl use graph paper. You may use a flexible curve rule.@ Untuk ceraian soalan ini, gunapertas graf. Ania boleh guna-pembaris fleksibel.

Using a scale of 1 cm to I u{rit on the .r-axis and l cm to I unit on the y-axis, draw the graph

of v=#".2<x<r4- l4marks)

(c)

@

Menggunakan skala I cm kepa(a I unit pada paksi-x dan I cm kepada I unit pada paksi-y, lukis grafbagiy=+untuk2<x<14.

From the graph in (b), findDari graf di (b), cari


(ii) the value of x when y = 5.nilaixapabila!=5.

(d) Draw a suitable straight linQ on the graph in (b)@

"Oration +.x-14=0fof 2<x<14.State these values of x.Lukis satu garis lurus yang sQsuai pada graf di (b)

persamaa"+ + x-14 = 0 un4t 2<x<14.Nyataknn nffoi-rlri x ini.

14 markahl

[2 marks]12 markah)

to find the values of x which satisfy the

14 marksluntuk mencari nilai-nilai x yang memuaskan

14 markahl

AnswerlJawapan:

Table 12Jadual 12


(c) (i) y -(ii) x =

(d)r=

1s1

SPM m Question 12

8 (a) complete Table 12 in the answer space for the equation I = f - 9x + 4 by writing down the@ values of y when x = -l and x = 2. [z marks]

Lengkapkan Jadual 12 di ruang jawapan bagi persamaan ! = f - 9x + 4 dengan menulis nilai-nilai yapabilax=-1 danx=2. 12 markahl

(b) For this part of the question, use graph paper. You may use a flexible curve rule.q, Untuk ceraian soalan ini, gunakan kertas graf. Anda boteh menggunakan pembaris fleksibel.By using a scale of 2 cm to I unit on the x-axis and 2 cm to 5 units on the y-axis, draw thegraphof y- x3*9x+4for-3< x<4.Dengan menggunakan skala 2 cm kepada I unit pada palcsi-x dan 2 cm kepada 5 unit pada paksi-y,lukis graf ! = x3 -9x + 4untuk-3<x<4.

[4 marksl14 markahl

(c) By using the graph drawn in (b), find3ll Menggunakan graf yang dilukis di (b), cari

(i) the value of y when r = -2.5,nilaiyapabilax=-2.5,

(ii) the value of x when ! = 20.nilaixapabila!=20. [2 marks]

[2 markahT

(d) lraw a suitable straight line on your graph to find the values of x which satisfy the equation@ *3 - l5x t 4 = 0 for -3 < x < 4.

State these values of x.Lukis satu garis lurus yang sesuai pada graf anda untukpersamaan x3 - 15x + 4 = 0 bagi -3 1 x 1 4.Nyatakan nilai-nilai x itu.

Answer/,Iawapan:

[4 marks]mencari nilai-nilai x yang memuaskan

14 markahl

(a)

Table 12Jadual 12


(c) (i) y =(ii) r =

x -3 -2 -l 0 I 2 3 3.5 4

v 4 t4 4 -4 4 15.4 32

(d)x=

152

SPM nlo Question 12

9 (a) complete Table 12 in the {nswer space for the equation } = g - 3x - z* by writing down@l tne valuesof ywhenr=-4and x=l [Zmarks1t@ me values or y wnen x = -ft &nct "f = l. [2 marksl

Lengkapkan Jadual 12 di ,ronlg io*apan bagi persamaan) = 8 - 3x - 2x2 dengan menulis nilai-nilai yapabila x=4 danx= 1. [2 markah]

(b) For this part of the questiorf, use graph paper. You may use a flexible curve rule.@ By using a scale of 2 cm tf 1 unit on the x-axis and} cm to 5 units on the y-axis, draw the

graph of y = 8 - 3x -Zx2 f{r -5 1x 13 and -27 <y 59. [4 markslUntuk ceraian soalan ini, gunfkan kertas graf. Anda boteh menggunakan pembaris fleksibel.Dengan menggunaknn skala 2 cm kepada I unit pada paksi-x dan 2 cm kepada 5 unit pada palai-y,lukis graf y - 8 - 3x-* untukls <, <3 dan-27 <y <9. 14 markahl

_e From the graph in (b), findiA Dari graf di (b), cari

[2 marks][2 markah)

the graph in (b) to find the values of x which satisfy the(x(3and-27<y39.14 markahl

pada graf di (b) untuk mencari nilai-nilai x yang memuaskan(x(3dan-27<y<9.14 markah)

Answer/"Iawapan:

(a)y-8-3x-2x2


(c) (i) y -(ii) "r =

(d)

Table 12Jadual 12

r53

SPM ?off Question 12

10 @ complete Table 12 in the answer space for the equation ) = -x3 + 3x +ill the values of y when r = *2 and x = 0.

Lengkapkan Jadual 12 di ruang jawapan bagi persamaan! = -x3 + 3x +yapabilax--2danx-0.

1 by writing down[2 marks]

I dengan menulis nilai-nilai12 markahl

12 marksl12 markah)

which satisfy the

14 markslyang memuaskan

[4 markah)

(b) For this part of the question, use graph paper. You may use a flexible curve rule.@l ny using a scale of 2 cm to I unit on the x-axis and} cm to 10 units on the y-axis, draw the

graphof y= -x3 +3x+ l for-3 < x<4and-51 (y< 19. 14marks)Untuk ceraian soalan ini, gunakan kertas graf. Anda boleh menggunakan pembaris fleksibel.Dengan menggunakan skala 2 cm kepadn I unit pada paki-x dnn 2 cm kepada 10 unit padn paksi-y,lukis graf y = -I3 +3x + l untuk-3<x<4dan-51 < y.l9. t4iarkahl

(c) From the graph in (b), findGEI Dari graf di (b), cari

(i) the value of y when r = -2.5,nilaiyapabilax=-2.5,

(ii) the value of x when )l = -10.nilaixapdbila)=-10.

(d) Draw a suitable straight line on the graph in (b) to find the values of x@ equation _x3 + l3x - 9 = 0 for -3 I x 14 and -51 < y < lg.

State these values of x.Lukis satu garis lurus yang sesuai pada graf di (b) untuk mencari nilai-nilai xpersamaan--x3 + I3x-9 - 0untuk-3<x34 dan-51 < y<lg.Nyatakan nilai-nilai x ini.

Answer/"/awapan:

(a)y=-x3+3x+l

Table 12Jadual 12


(c) (i) y -(ii) x =

(d)

The equation of the straight line:Persamoan garis lurus;

x a_J 1 -1 0 1 2 -tJ 3.5 4

v t9 -1 3 -l -17 -31.4 -51

154

REVIEW AND

KERTAS 1

ff 'civen y:x-4.' Y-intercePt: -4. Since the coefficient of x2

The graph in A represents y - x2 - 4.Answer: A

.E:! 'Civen Y:10-2x3.' y-intercept : l0. Since the coefficient of x3 < 0,

The graph in B represents ) = 1

Answer: B

' Civen Y: -x2 + 5.

'Y-intercept : 5. Since the coefficient of x2 < O, isA.

The graph in B represents y =Answer: B

x{! . Since the coefficient of x2 < o, its ghape is A.. Find the x-intercepts by substituting y : O into the

function.

l

!=-x2-x+2Wheny=0, 0=(-r+ lXx+2)

x=1,-2

its shape is \.L

I

" 3',

The graph in C represents ! = -x2 - x + 2.

Answer: C

.: ., ri. .:'5' 'Civen Y:2x3 - 16.. y-tntercept : -.l6. Since the coefficient of x3 > O, its shape is ,/.

The graph in A represents ! = 2x3 - 16.Answer: A

ry 'Civen Y:5-x3.'Y-intercePt : 5. Since the coefficient of x3 < O, its shape is \.

The graph in D represents ! = 5 - x3.

Answer: D

. From the graph, y-intercept : -27.

. Since its shape is \ , the graph is a cubic functionand the coefficient of x3 < O. Thus, the equationof the graph is y : *3 - 27

Answer: B

155

'ffi

KERTAS 2

ffi . The region which satisfies the inequality of the, form y 2 ox + c is the region that lies on and' above the line y : ox + c.

' . Thus, the region that satisfies the inequality, y 2 -2x + I O is the region on and above the

line Y:-?-x + ,10.

, . The region that satisfies the inequality x < 5 is on, the left of the line x:5. Since x < 5, the points

on the line x: 5 are not included. Thus, a dashed,, line needs to be drawn for x : 5.. . The region that satisfies the inequality y < 10 is, the region on and below the line y : 1O.

-x + 6 is the region on and below the line

Y:-x+6.. The region that satisfies the inequ ality y > 2x - 4

is the region on and above the line y :2x - 4.. The region that satisfies the inequality x > I is

the region on the right of the line X : 1. Sincex > 1, the points on the line x: I is not included.Thus, a dashed line needs to be drawn for.,- 1A

- t-

#::

v

bc\//,/

)'I +

o \

4x I

156

f:._: . The region that satisfies the inequality y < -x + lOis the region on and below the line y: -x + 10.

. The region that satisfies the inequ ality y < x is theregion below the line y : x. Since inequality sign"<" is used, the points on the line y: x is notincluded. Thus, a dashed line needs to be drawnfory:v.

. The region that satisfies the inequ ality y > I is theregion above the line y : 1 drawn as a dashedline.

(a) . To complete the table, substitute X : -1and x : 4 into y : 2x2 - x - 3 to get thecorresponding values of y.

. Use a scientific calculator to find the values.(b) . Use the given scales to draw the graph of

y:2x2 - x - 3 for -2 1x a 5 on graphpaper.

. Make sure that your graph is smooth andpasses through all the points plotted.

(c) . Use your graph in (b) to find the valuesrequired in (i) and (ii). No marks will begiven if the values are found by calculation.

(d) . To solve the equation 2rz - 3x : 10 usingthe graph of y : 2rz - X - 3, we need tofind the straight line that needs to be drawnon the graph.

. Draw the straight line y : 2x + 7 on thegraPhY:2x2-x-3'

. The points of the intersection of the graphy : 2r2 - x - 3 and the straight liney :2x + 7 give the values o x

(a) !=2x2-x-3When x = -1, y = 2(-l)' - (-t) - z

-0When x=4, y-2(4)2-4-3

_a<-/l

,4, ,

x -l 4

v 0 25

(c) (i) !=19(ii) r = 4.7

(d) !=2x2-x-3 .....p0=2x2-3x-10.....6

@-@: !=2x+7From the graph, x = -l .6,3.1

ffi (a) . To complete the table, substitrlrte x: -3 and

x: 1.5 into y :; to get the correspondingll

uulues of y.' (b) . U;; ,n. giu.n scales to ar.uuJ the graph of' 2!

for -+ I x 1+ on gr{ph paper.y:,tOr-4<x<4O' . Make sure that your graph i$ smooth and, passes through all the points plotted., (c) . Use your graph in (b) to firfd the valuesI required in (i) and (ii). No marks will be

given if the values are found py calculation.(d) . To solve the equation2x2 + 5x124 using the

graph of y: ?, we need to find the straight

, line that needs to be drawn Qn the graph.o Divide the equation 2x2 + 5x + 24 by x, we

get2x + 5:2!. Obviously the right hand

, side of the equation represenfis y : + .

: ' Draw the straight line y: 2x + 5 on the graph.. The point of intersection of tt"ie strarght line

y : 2x +5 and the graph of y :{ eir",

the value of x.

157

(a) _._24-xWhen x=-3, y=+

--8Whenx= 1.5, y=4

1.5

=16

(i) t = 8.25(ii) x = -7.9

2x2 + 5x = 24 ..... O

e+x: 2x+5-24x

2x+5=!The suitable straight line is ! = 2x + 5.

From the graph, x - 2.45

!=6-x3When x=-l, r:Ur-(-1)3

When x=2, ,=1;r'

(c)

(d)

6 (a)

x -3 1.5

v -8 t6

x -1 2

v 7.)

(b)

I i, :,:. ,i 'i. , i i"rl: ,I-+.ipt- l-il=. i1',,, i,';,,'-;,'l:if ',-,'i;''.+-,; :,Ii-p*j. :i., :: :

'.. .-r,- . .1.- .:,,*:l..*..:-.;.,.*. .--...J. ...*i - i.,. .[. i.-,.i-- .,. . ..

(c) (i) !=2.5 (ii)x=*1.55(d) !=-x3+6 .....@

0=x3-8x-6 .....@(D+@:tl=-8x

The suitable straight line is -I = -8x.From the graph,x = -2.35, -0.8

7 (a) 36v --

When x=6, y-

Whenx=10,y=

36--= b6

ff=ru

(b)

x 6 r0v 6 3.6

1s8

(c) (i) y-12.8(ii) x = 7.2

..... o

t4 ..... @

@-@: y=-x+14The suitable straight line is ! = -x + 14ory-14-x.From the graph,x - 3.4, 70.6

8 (a) !=x3-9x+4when x=_1, != (_1)3_g(_1)+4

=12When x=2, !=23-9(2)+4

--6

(i) )=11(ii) x = 3.7

\'=.r3 -9-r+40=x3 -l5x+4

e-@: !=6xThe suitable straight line is y

= 6x.

From the graph,x - 0.25,3.J5

(d) ,._36'x0=39 +x-

x

(c)

(d)

x -1 2

v 12 -6

of x is

on the

the positivf value

liney:]+sxl

- 3(1) - 2(4)2t2

- 3(1) - hrt9

ll

(c) (ii) . Note that onlyrequired.

(d) . Draw the straightgraph rn (b)

(a) ) = 8 -3x-2x3When x=4, r=:

Whenx=1, -y=8-3mm

(b)

r

(i) ) = 3.5(ii) Positive value of x - 2.45

(c)

(d) )'=8- 3x-2x2 .....e0=5 -8x-2x2 .....@

on(b).

e-@: y=3+5xThe equation of the straight

!=3+5xor)=5x+3From the graph,

x = -4.5, 0.55

x -4 I

v -12aJ

w thisight linegraph in

159

10 (a) !=-x3+3x+1When x = -2, y - -(-Z)3 + 3(-2) + I

-3Whenr= 0, y = -(0)3 + 3(0) + 1

. =l

(i) ) = 8.5(ii) x - 2.7

!=-x3+3x+1 .....e0=-r3 +13x-9.....@

e-@: y=*10x+ 10

The equation of the straight line:

)=-10x+ 10 Draw this straightline on the graphin (b).

(c)

(d)

From the graph,x - 0.1, 3.2

x a,_L 0

v 3 I

KERTAS 2

PA'. f. Scale factor. k = "-r

" PA'Faktorskala,k- _PA

2. Area of image - k2 x area of objectLuas imej = l? x luas objek

Behagian ffi

Mathematical Formulae and Facts ' j ', ' ::

3. If f una g ,r. *" oansformations,Jika A, dan B ialah dua penjelmaan,

" AB = transformation B followed by transformation A,

penjelmaan B diikuti dengan penjelmaan A,

BA = transformation A followed by transformation B,penjelmaan A diikuti dengan penjelmaan B,

A2 = AA = transformation A carried out twice.penjelmaan A dilakukan dua kali.

SPM M Questron 13

1 (a) Diagram 6 shows two points, M and N, on a Cartesian plane.Rajah 6 menunjukkan dua titik, M dan N, pada suatu satah Cartesan.

Transformation T is a translation

Transformation R is an anticlockwcentre (0,2).

ffi Gombination of Two Transformations

)_4

Diagram 6Rajah 6

(j,ise

I

rotation of 90" about the

PenjelmaanT ialah satu transla / 3 \

"(-,/x PenjelmaanR ialah satu putaran 9O" lawan arahjam pada pusat (0, 2).

(i) State the coordinates of the image of point M undertransformation R.Nyatakan koordinat imej bagi titik M di bawah penjelmaanR.

(ii) State the coordinates of the image of point N under thefollowing transformations :

Nyatakan koordinat imej bagi titik N di bawah penjelmaan berikut:

(a) T2,

15 marksl15 markahl

v

A

N")

4 -2 o 2

,) M

160

(b) TR.

Answer/"Iawapan:(a) (i)

(ii) (a)

(b) Diagram 7 shows three qua{rilaterals, ABCD, EFGH and PQRS, on a Cartesian plane.Raiah'7 menunjukkan tiga sisi tmpat, ABCD, EFGH dan PQRS, pada suatu satah Cartesan.

v

Dn1ram 7

EFGH is the image of ABC| under transformation V.PQRS is the image of EFGI{ under transformation W.EFGH ialah imej bagi ABCD d) bawah penjelmaan Y.PQRS ialah imej bagi EFGH di bawah penjelmaanW.

(i) Describe in full the tralrsformation:Huraikan s e len gkapny a p f nj e lmaan:

(a) V, (b) W.(ii) Given that quadrilateral PQRS represents a region of area 45.6 cm2, calculate the atea,

in cm2, of the region rqpresented by the shaded region.Diberi bahawa sisi emppt PQRS mewakili suatu kawasan yang mempunyai luas 45.6 cm2,hitungkan luas, dala* cmP, kawasan yang diwakili oleh rantau berlorek.

l7 marks)17 markahl

Answet/Jawapan:(b) (i) (a)

(b)

(b)

(ii)

161

SPM mO Question 12

2 (a) Transformation T is a translatio " f:) urO transformation P is an anticlockwise rotarion of90" about the centre (1, 0). \ - /

PenielmaanT ialahffanslasi(;') danpenjelma.anP ialahputarangYo lawanarah jampad.apusat(I,0).

State the coordinates of the image of point (5, 1) under each of the following transformations:Nyatakan koordinat imej bagi titik (5,1) di bawah setiap penjelmaan berikut:(i) Rotation P, (ii) Translarion T,

Putaran P, Translasi T,(iii) Combined transformation T2.

Gabungan penj elmaan T2.

14 marksl14 markah)

(b) Diagram 6 shows three quadrilaterals, ABCD, EFGH and JKLM, drawn on a Cartesian plane.Rajah 6 menuniukkan tiga sisi empat, ABCD, EFGH dan IKLM, dilukis pada suatu satah Cartesan.

Diagram 6Rajah 6

(i) JKLM is the image of ABCD under the combined transformation VU.Describe in full the transformation:JKLM ialah imej bagi ABCD di bawah gabungan penjelmaanYlJ.H uraikan s e lengkapny a p e nj e lmaan:

(a) fJ, (b) V.(ii) It is given that quadrilateral ABCD represents a region of area 18 m2.

Calculate the area, in m2, of the region represented by the shaded region.Diberi bahawa sisi empat ABCD mewakili suatu kawasan yang mempunyai luas 18 m2.Hitungkan luas, dalam m2, kawasan yang diwakili oleh rantau berl.orek.

l8 marksl[8 markahT

Answer/,/awapan:(a) (i) (ii) (iii)

(b) (i) (a) U:

(b) v:

(ii)

D

-4 1*L o 4 8-

L62

SPM 2@7 Question 15

3 Diagram 8 shows quadrilaterals ABCD, EFGH and, JKLM drawn on a Cartesian plane.Rajah 8 menuniukkan sisi empat ABCD, EFGH dan JKLM yang ditukis pada satah Cartesan.

Diagram 8Rajah 8

(a) Transformation R is a rotatipn of 90' anticlockwise about the centre (0, 2).Transformation P is a reflecfion in the straight line x - 2.Penjelmaan R ialah putaran 90" lawan arah jam pada pusat (0,2).PenjelmaanP ialah pantulan pyda garis lurus x = 2.

State the coordinates of the lmage of point A under each of the following transformations:Nyatakan koordinat imej titik A di bawah setiap penjelmaan berikut:(i) R,(ii) RP.

13 marks)[3 markah]

(b) EFGII is the image of ABCD under the combined transformation MN.EFGH ittlah imej bagi ABCD dl bawah gabungan penjelmaan MN.Describe in fullHuraikan selengkapnya

(i) the transformation M,penjelmaan M,

(ii) the transformation N.penjelmaan N.

16 marks)[6 markah]

(c) JKLM is the image of E(Gllunder an enlargemenr ar centre (3, 0).JKLM ialah imej bagi EFGH di bawah satu pembesaran pada pusat (3, 0).

(i) State the scale factor of the enlargement.Nyatakan .faktor skala penpb e s aran itu.

(ii) Given that EFGH reprdsents a region of area ll2 m2,, calculate theregion represented by JtXfU.Diberi bahawa EFGH mewakili suatu kawasan yang mempunyai luasdalam m2, kawasan yang diwakili oleh JKLM.

area, in m2, of the

ll2 m2, hitung luas,

13 marks)13 markah)

L63

Answer/,Iawapan:(a) (i)

(ii)

(b) (i)

(ii)

(c) (i)

(ii)

SPM m Question 13

4 (a) Transformation P is a reflection in the line x - k.Transformation R is a clockwise rotation of 90' about the centre (0, 4).

Transformation T is a translatio, (: )PenjelmaanP ialah pantulan pada garis lurus x = k.PenjelmaanR ialah putaran 90o_ikut arah jam pada pusat (0, 4).

PenjelmaanT ialah translc i -) \

'ri (_: i.(i) The point (8, 6) is the image of the point (-4, 6) under the transformation P.

State the value of k.Titik (8,6) adalah imej bagi titik (-4,6) di bawah penjelmaanp.Nyatakan nilai bagi k.

(ii) Find the coordinates of the image of point (2, 8) under the following combinedtransformations:Cari koordinat imej bagi titik (2,8) di bawah gabungan penjelmaan berikut:(a) T2,(b) rR.

15 marksl15 markahl

164

(b) Diagram 13 shory9 three pentagons, ABCDE, FGHIJ and KLMNJ,drawn on a Cartesian plane.Raiah 13 menunjukkan tiga pentagon, ABCDE, FGHIJ dan KLMNJ, dilukis pada suatu satah Cartesan.

-10 -8 -6 4 ) lO 2 4 6 8 10 12I ll"-T-t

Diagram 13Rajah 13

(i) KLMNJ is the image of ABCDE under a combined transformation VU.KLMNJ ialah imej bagi ABCDE di bawah gabungan penjelmaanylJ.

Describe in full, the ffansformation:H uraikan s e I e n gkap ny a pp nj e lmaan:

(a) u,(b) v.

(ii) It is given that pentagon KLMNJ represents a region of area 42.5 m2.Calculate the area, i, +', of the region represented by the coloured region.Diberi bahawa pentagon KLMNJ mewakili luas 42.5 m2.Hitung luas, dalam m2, kawasan yang diwakili oleh kawasan yang berwarna.

Answer/Jawapan;(a) (i)

17 marksll7 markahl

(ii) (a)

(b) (i) (a)

(b)

(b)

(ii)

16s

SPM m9 euestion t35 Diagram 13.1 shows rhe point

Rajah 73.1 menunjukkan tii (_2, Iine r = I drawn on a Cartesian plane.dilukis pada suatu satah Cartesatn.

"';:;:il:'r'(a) Transformation T is a transl / -5 \

rransrormarion p is a ."o";:; ,j;'Jr,raight rine x = l.state the coordinates or the image of point 7-2, quro", ih" rottoring transformations:PenjelmaanT ialah satu translc . / 5 \penjetmaan p iatah satu pant,;::t;J raris turus x = t.Nyatakan koordinat i*"ioogiiiit'r-2, i aiur*rn penjermaan berikut:(i) r,(ii) TP.

(-2, 4) and the srraighr4) dan garis lurus r = I

Answer/Jawapqn:(a) (i)

[4 marks][4 markah]

(ii)

166

vl6

E D

C

4/

A B

2

Jo

U0

.)_L

R s N

li :.1:.i

/L M

,:{

-6/K

(b) Diagram 13.2 shows pentagpns ABCDE, JQRSU and JKLMN, drawn on a Cartesian plane.Rajah 13.2 menunjukkan penta'gon ABCDE, JQRSU dan JKLMN, dilukis pada suatu satah Cartesan.

Diagram 13.2Rajah 13.2

(i) JKLMN is the image of ABCDE under the combined transformation HG.Describe in full, the transformation:JKLMN ialah imei bagi AtsCDE di bawah gabungan penjelmaarz HG.H uraikan s e I e n g kap ny a p eni e lmaan:

(a) G,(b) H.

(ii) It is given that pentagon ABCDE represents a region of area 80 m2.

Calculate the area, in n[2, of the region represented by the coloured region.Diberi bahawa pentagon ABCDE *rrikili tuitu kawasan yang mempunyai luas 80 m2.

Hitung luas, dalam m2, kawasan yang diwakili oleh kawasan yang berwarna.

[8 marks][8 markah)

Answer/-Iawapan:(b) (i) (a)

(b)

(ii)

167

SPM 2010 Question /36 Dragram 13.1 shows points .I and K drawn on a Cartesian plane.

Raiah 13.7 menunjukkan titik J dan titik K ditukis pada suatu satah Cartesan.


(a) Transformarion T is a translatio, (T)Transformation R is an anticlockwise rotation of 90" about the centre K.State the coordinates of the image of point -/ under the following transformations:

PenjelntaanT ialah satu trans, ' l -Z\,rr\ t /.Penjelmaan R ialah satu putaran 90" lawan arah jam pada pusat K.

Nyatakan koordinat imej bagi titik J di bawah penjelmaan berikut:

(i) T2,,

(ii) TR.

[4 marks)[4 markah]

Answer/"Iawapan:(a) (i)

(ii)

J

K

o

168

(b) Diagram 13.2 shows two he4agons, ABCDEF,and peRAl{- drawn on square grids.Raiah 73.2 menunjukkan dr;i;b"";;;;;;;; dan pe7iii)i,trr,,

pada g,ri ,egi empcn sama.i

"';:,':f,:tr,(i)

if*A#il iff1ffi1,"f,fl'f?tr.under the combined ransrormation vui#xtr"':';:;#:i,:;:'#f,i#::;;wahgabunsqnpenietmaanytr(a) [J,(b) v.

(ii) It is given that AB,DEF represents a region of arca 45 mz.s?;::,:y,;:Izffi?;;;;;:xF jir;;r::r*,en,edb{r"shadedregionHituns tuas, data*

^1 ko*osan yqns o,*I{,f,'li"l##:#;:i,:::{o' +s fr)2.

Answer/Jawapan: t8 marksl

(b) (i) (a) " ' [8 markah]

(b)

(ii)

.sT',i j":'!4:a

i,i-! jrii:.tr'..ii'!:,ii

:l"l'+,:lili,.'.:,l;,1 .-i.rtl

.5r, ffi ll::i;ii}ri.,,..:i

;iiii:'1+!i:i!141 P

,.,:

:]:::I ffi ;ri':i-lJ i;ii#

irr',il1i,: 'rl v;rr

tit!:j;ii

ffi WP ru ,/A,affi il

\ i'f.,'!tia

iirff, ii

ltjr,rilii,rg 'h"\ ffi jlii::jt!:118

rr1Lli;*lill':$

r#:lrl:fi:li.iilil

|ffi'.\o R

169

v

Z.

\ B(2, s)

.|

o

SPM frll Question 13

7 (a) Diagram 13.1 showspointBandstraightliney +x=5 drawnonaCartesianplane.Rajah l3.I menuniukkan titik B dan garis lurus y + x = 5 dilukis pada suatu satah Cartesan.

Diagram 13.1Rajah 13.l

Transformation T is a translatio. f 1).\-JlTransformation R is a reflection at the line y * x = 5.

PenjelmaanT ialah satu trans, ' l Z \or,

\ _l /.PenjelmaanR ialah satu pantulan pada garis ! * x = 5.

State the coordinates of the image of point B under each of the following transformations:Nyatakan koordinat imej bagi titik B di bawah setiap penjelmaan berikut:(i) r,(ii) TR.

13 marksl[3 markahl

Answer/,/awapan:(a) (i)

(ii)

170

E

/,/%

z%tru,Fru

A wa

YC D

o

(b) Diagtam l3-2 shows trapepium ABCD and trapezium FCDE drawn on a Cartesian plane.Raiah 13.2 menuniukkan traPlzium ABCD dan trapezium FCDE dilukis pada suatu satah Caitesan.

v


(i) FCDE is the image of ABCD under the combined transformation VU.Describe, in full, the transformation:FCDE ialah imej bagi ABCD di bawah gabungan penjermaanylJ.Huraikan se len gkapnya p enj e lmaan:(a) IJ,(b) v.

(ii) It is given that ABCD represents a region of area 60 m2.Calculate the area, in m2, of the region represented by the shaded region.Diberi bahawa ABCD mircwakili suatu io*otoi yong mempunyai luas 60 m2."Hitung luas, dalam m2, llawasan yang diwakili otei kawaion yong berlorek.

Answer/,Ia wapan:(b) (i) (a)

19 marks)[9 markah]

(b)

(ii)

t7t

KERTAS 2tffi (a) (i)

(ii)

(b) (i)

(i)

(ii)

(b) (i)

(ii)

. The image of point M under a

transformation R can be determined bylooking at the rotation of the line joiningpoint M and the centre (O,2) under thetransformation R.

(a) . T' is a transformation T followed bya transformation T again.

. TR is a transformation R followed bya transformation T.

. The description of the transformations Vand W must be complete in order notto lose marks.

. To describe V, focus on the reflection ofABCD to EFGH. To describe W, focus onthe enlargement of EFCH to PQRS.

. Area of image : k2 x area of object

. Area of the shaded region: area of PQRS - area of EFGH

M(2, -t) -3-, Q, 4)

(a) N(-4,3) -I-, er,2) -L (2, t)

(b) N(-4,3) 5 er, -2)l- (2,-3)

(a) V : A reflection in the line x = 3

(b) Scale factor = g\ = ? = |FGI^W : An enlargement of scale factor

2 with centre (1,2)

Area of PORS a.,

Area of EFGH = :45.6 _ A

Area of EFGH - -

Area of EFGH = 45:'6

4

= 11.4 cmz

(a) v

A ,€ T

E

4 ", ;( t) 4

T

(ii)

172

:'15il$all'*::':i EFGH

ry (a) (i) . To find the image of point (5, I ), rotate, the line joining point (5, 1) and the

6entre of rotation (O,2) under rotation p.

' (iD . Since the transtation T ,, (;), add -jto the x-coordinate and add 2 to the:. y-coordinate of point (5, I )., (iii) . With.the result from (ii), repeat the steps

: In (ii) to get the image under T2.(b) (i) - To describ-e U, focus on the factthatEFCH

is a reflectio n of ABCD.To describe V, focus. on the fact that JKLM is an enlargement' of EFCH.

(ii) . Area of EFCH: dreo of ABCD. Since JKLM is the image of of EFCH, Under an enlargement of scale factor 3,, the area of JKLM is 32 x area of EFGH.. Area of the shaded region

: area of JKLM - area of EFCH

(s, 1) -I, (0, 4)

(5, l) -I. (2,3)

(s, l) -I, (2,3) l, (-1,5)

(a)

(i)

(ii)

(iii)

(b) (i) (a) U: A reflection in the line x = -l(b) V: An enlargement of scale factor

3 with centre (2,4)

(ii) Area of EFGH = area of ABCD = 18 m2

Area of JKLM = 32 x area of EFGH=9x18= 162 m2

)-6

(- ,5(0, )

I ,1)

')II

I (s, )r

1 a 2 6

='P:'J:1':IEFGHffi (a) (i) . To find the image, rotate the line joining

, point A and the centre df rotation (O, 2j, gnder a transformation R.

(ii) . RP is a transformation P followed byr a transformation R.

(b) . lt is important to describe in full the, transformations involved in order not to lose: markS.

, . lt can be seen that ABCD is rotated I8O", about the centre (2, 4) to JKLM, then

enlarged to EFGH at the centre (3, 0) witha scale fador of 2.

(c) (i) . Since JKLM is the image of EFCH under, an enlargement at cerltre (3, 0), the

scale factor is *.2

, (ii) . Area of JKLM: f+)' x area of EFCH\21

A(3, 5) -L (-3, 5)

A(3,5) -f, (1, s) -\ (-3, 3)

(a) M : An enlargement of scale factor2 with centre (3, 0)

(b) N : A rotation of 180' about thecentre (2,4)

Scale factor = #aJ

61

2

Area of JKLM = f1)' X area of EFGH\2t

(b)

(i)

(ii)

(i)

(c) (i)

=f,xnZ=28m2

(ii)

173

(a) (i) . From the coordinate axes, it can be seenthat point (-4, 6) is reflected to point(8, 6) in the line x: 2.

(ii) (a) . T2 is a transformation T followed bya transformation T again.

(b) . TR is a transformation R followed bya transformation T.

(b) (i) . lt can be seen thatABCDE is reflected inthe line x: 1 to FGHll, then enlargedlo KLMNJ at the centre J(8, I ) with

a scale factor of a.2

(ii) . Area of KLMNJ: /+)'x area of FGHTJ\2). Area of the coloured region

: area of FGHIJ - area of KLMNJ

(a) (i) (1,6) -L (8, 6)

. t._ _4+8 _,"A-

2 -L

(a) (2, 8) g (7, s) *I- 1rr2, 2)

(b) (2, 8) -3, (4,2) g (e, -1)(b) (i) (a) U: A reflection in the line r = 1

(b) Scale factor = ry- = ? = !GH42V: An enlargement of scale factor

7 with centre "r(8, 1)

(ii) Area of KLMNJ = (+)'x area of FGHIJ

1

42.5=TxarcaofFGHIJ

Area of FGHIJ = 42.5 x 4= l7O m2

.'. Area of the coloured region

= area of FGHIJ - area of KLMNJ= l7O - 42.5

= 127.5 m2

s (a)

6 (a)

(b)

(i) (-2, 4) -L (3,2)

(ii) (-2, 4) g (4, 4) l- e,2)(i) (a) G :Aclockwiserotationof90oabout

the centre (0,2)

(b) Scalefactor =#=*=+H : An enlargement of scale factor

{ witt centre J(2, O)2

(ii) Area of JQRSU = area of ABCD=10m2

Area of JKLMN

= (+)' x area of JQRSU\2t

=2x804

= 180 m2


= area of IKLMN - area of JQRSU

=180-80= 100 m2

v

(i) /(s, 3) -L (2,4)1, (-1, s)

(ii) J(s, 3) R , (1, 3) -L e2, 4)

I s)

( 4 T

T J

K

o

174

(b) (i) (a) U: A reflection in the line EC

(b) Scalefactor =#=l=lV: An enlargement of scale factor

3 with centre P

(ii) Area of PQRATS

= 32 x area of ABCDEF=9x45= 405 m2


= area of PQRATS - ilrea of ABCDEF

-405-45= 360 m2

7 (a)

(i)

(ii)

(b) (i)

(ii)

B(2,5) g @,2)

B(2, s) g (0, 3) -I- (2,0)

(a) U: An anticlockwise rotation of 90o

about the centre B(3,3)

(b) Scalefactor =FC -!-2-ABV: An enlargement of scale factor

2 with centre F(3, 5)

Area or FCDE

=i'j;r*a

or ABCD

.'. Area of the shaded region

= area of FCDE - area of ABCD

=240-60= 180 m2

v

\ B( s)

\ T

@.

(2.

o

KERTAS 1

ll I Ll I'I+l t'ItI I I : I :lt {rlt}5ErIrr I al:] fi I rrt:E : tr:+rji

il|:

4.1 Matrices

4.2 Equal Matrices

4.3 Addition and Subtraction ofMatrices

4.4 Multiplication of a Matrix bya Number

1 1 1 1 1 2 1

4.5 Multiplication of Two Matrices 1 1 1 1 1 1

4.6 ldentity Matrix

4.7 lnverse Matrix

4.8 Solve Simultaneous LinearEquations Using Matrices

W Mu[iplication of a Matrix bya Number

SPM ffi Questrbn 39

1'('^:) (1 l)=/4 2\A l,,0 4)/4 3\B\z 4)

c(;3)/s 2\D lq s)

SPM m Question 3E)

2 (; ?).'(1 _")-(_l 1)=/14 l \A (,ir -n)/13 4\B [,ro 41/8 4 \c l" -B)/7 3\D \,0 o)

17s

SPM m7 Question 403 Find the value of m in the following matrix

equation:Cari nilai m dalam persamaan matriks berikut:

=(3 ?)

SPMmQuestion& ,

4 (6 -s 2)-(r 4 -7)+3(-l 3 2)=A(48ls)B (4 0 1)c (2 8 r)D(20ls)

G ?)-,(3 ))A1 83C4 D6

SPM mg Question 3!)

s (i 2)-,(?;)=I -2 ll\A[-,0 6) B

c (:,1, :) D

SPM nlo Question 39

6,(1 i i) -(1A (j, ; ls)

B (i, -J s)

c (-i, 'i r,,,)

D (-i, '-!, -f')SPM 2010 Question 407 Given:

Diberi'.

( ; ) - ,(-;) = ( i)Find the values of r and y.Cari nilai x dan y.

A x=-1,)=-10B x=-1,!=-ZC x=3,!=6D x=5,!=L4

/-2\-r o

(-3

42

T)

:)

2\3 )=

L76

SPM 2oll Question 3!l

8 (?)-(i) .+(-3) =

A (j.)

c (;)B (i)D (-;)

B2D+

A1c+

A (r,) B (1;)

/rs 20\ lrs -8\c (,z -q) Dl-s -q)

s\

;l2 g\12 ol4 rl

A (#)

c (i'i)

@ Muftiplication of Two Matrices

SPM M Question &s Given (fr ,,(:- ?) = ,ro 5), calculare

the value of k.

Diberi(k t)(j* ?) = ,ro 5), hitungtcnnnitaik.

A3 B8c-l D-r?

SPM n6 Question 4

10 rr(3 ,r(o;) =,r,,rhenp=/ 4o\

Jika (3 ,)\; ) = (r), maka p -

SPM 2(n7 Question 39

,, (:, ;)(:,) =

SPM W Question 39

,z(i i)l:l =

SPM ru Question tlo

13 Given (2w -,(jr) - (16), find the value

of w.

Diberi (2w ,(1) = 1rol, cari nitai w.

A1B2C3D4

SPM frll Question 40

I s l\t4 (1 -l oll -z ol=

\ I 4lA(3 1)B(7 1)

c(?)D (1)

KERTAS 2

i,,, rr a = (i uo),rn",A-r = #6e,-:),where ad - bc * 0.

','1 trko n = (X uo), *,*A-'| = ,a+(:, -:), o"nro, ad - bc * 0.

ii

li S. A-r does not exist when ad - bc = 0.ji *' tidak wuiud apabila ad - bc = o.

i,',4' *+bv=t1 + (:'r)(;) =(?) * (;)=#r(!,-:X?)'r',, '* + dY = fta:

lJtrll' .-I f I r I i I :I{rlrl5Er,Irr f LJ :l rtr I gl,l :J 5 :k

m-: "s00,' 20fO'" .stt&. Fon ,A

lrrin lB. ,l[ .8, A. 8.,' A,t- .B ii.4,, B A B

4.1 Matrices

4.2 Equal Matrices

4.3 Addition and Subtraction ofMatrices

4.4 Multiplication of a Matrix bya Number

4.5 Multiplication of Two Matrices

4.6 ldentity Matrix

4.7 lnverse Matrix 1

2

'l

2

1

T1

21

T1

T1

z4.8 Solve Simultaneous Linear

Equations Using Matrices1

z'l

71

21

T1

21

21

T

177

Bahagia*ffiW lnverse Matrix

f;Er solve simultaneous Linear Equations using Matrices

SPM m Question 11

I It is given rhat matrix , = (? ?) *o matrix O = k(:r l) *.n that pe= (; ?)

Diberi bahaw / 2 -5 \,a matriks, = l; i 1 a* ma*iks O = k(:t !) orrrrn keadaan pO = (10 ?)

(a) Find the value of t and of h.@ Carikan nilai k dan nilai h.

(b) Using matrices, find the value of x and of y that satisfy the following simultaneous linear@ equations:

Dengan menggunakan kaedah matrilcs, hitungkan nilai x dan nilaiy yang memuaskan persamaan linearserentak berikut:

2x-5Y=-17x+3Y=$

l7 markslU markahl

Answer/,Iawapan:(a) O)

SPM m Question 1l

2 (a)It is give, *" (i ;),, the inverse matrix "(j, ;)@ \^'-t

Find the value of n.

Diberi urrr*, ( | '\ *^, nratriks sonssans urr, (1 ;I\, "tCarikan nilai n.

(b) Write the following simultaneous linear equations as matrix equation:@ Tulis persamaan linear serentak berikut dalam bentuk persamaan matriks:

3u-4v=-5-u+2v=2

Hence, using matrices, calculate the value of u and of y.

Seterusnya, dengan menggunakan kaedah matriks, hitungkan nilai u dan nilai v.

[6 marks]16 markahl

Answer/,Iawapan;(a) (b)

178

SPM m7 Questrbn 9

, BGiven

*(7 iy(: i) = (; ?), t*o the varue or m and or n

Diberi *(7 1)(\ I ) = (l ?), ,,,,nitai m dan nitai n

(b) Using matrices, calculate the value of x and of y that satisfy the following matrix equation:lftilll Menggunakan kaedah matriks, hitung nilai x dan nilai y yang memuaskan persamaan matriks berikut:

E 3X;)=(l)U marksl[7 markah]

Answer/,Iawapan:(a) (b)

SPM m Question 8

4 rhe inverse matrix *(1 ?)r' +(l ;)Matriks sonssans b"st (24

t) ,^"^ +(l ;)

(a) Find the value of m and of k.l@ Cari nilai m dan nilai k.

(b) Write the following simultaneous linear equations as matrix equation:Gl Tulis persamaan linear serentak berikut dalam bentuk persamaan matriks:

2x+3Y=-l4x+7Y=5

Hence, using matrix method, calculate the value of x and of y.Seterusnya, menggunakan kaedah matriks, hitung nilai x dan nilai y.

[6 marks][6 markah]

Answer/Jawapan:(a) G)

L79

SPM mg Question 7

5 It is given that matrix A = (? -: )'t'^ - \4 -5 )'

Diberi bahawa matriks A - (? -1)\4 -5/

(a) Find the inverse matrix of A.@ Cari matriks songsang bagi A.

(b) Write the following simultaneous linear equations as matrix equation:@ Tulis persamaan linear serentak berikut dalambentuk persamaan matriks:

2x-3Y=74x-5y=13

Hence, using matrix method, calculate the value of x and of y.Seterusnya, dengan menggunakan kaedah matriks, hitung nilai x dan nilai y.

16 marksl16 markahl

Answer/,/awapan:(a) O)

SPM nlo Question 11

6 rhe inverse matrix " (; ? ) r. r(:, l)

Matriks sonssans b"si (: -]) nun r(i ,_)

(a) Find the value of m and of k.@ Cari nilai m dan nilai k.

(b) Write the following simultaneous linear equations as matrix equation.@ Tulis persamaan linear serentak berikut dalam bentuk persamaan matriks.

4x-!=72x+5y=-2

Hence, using matrix method, calculate the value of x and of y.Seterusnya, dengan menggunakan kaedah matriks, hitung nilai x dan nilai y.

[7 marks][7 markah)

Answer/,Iawapan:(a) (b)

180

SPM z'/-ll Question 8

7 (a) Itisgiventhat -(Z 3)=(l f),*nr.eMrsa2x2matrix.@ pira u.Diberi bahawa -(', 3) = (l ?),orrron keadaan M iatah matriks 2x2.Cari M.

(b) Write the following simultaneous linear equations as matrix equation:@ Tulis persamaan linear serentak berikut dalam persamaan matriks:

';:'lr=',Hence, by using matrix method, calculate the value of r and of 1l.Seterusnya, menggunakan kaedah matriks, hitung nilai x dan nilai y.

Answer/Jawapan:(a)

REVIEW AND A

[6 marks][6 markah)

(b)

KERTAS 1

'ril' lii. r r:*+:: t:'lr.r*pandz(l

l3tn matnx \4

] ) t", by multiplying each element

l) o,,

ry . r*pand s(] -o=)t*.

, . Calculate from left to right.|,.

(; ',).,(', :',) (-i=(; ?).(:

")-(_iIrt z\ /-3 t\

= \rr -s/ - \-z s)

-(tt -(-3) 2- 1\-\r: -?z) -8-s/It+ r \=\rs 4)

Answer; A

,('^ :) - ( I l) = (: ,1,) - (r, l)1 6-t 2-0\

=l I

\8 - (-1) t2 - 4)

ls z\=l I\e 8l

t)

1)

Answer: D

181

8 (?)-(j-).+(j)

=(J--,1,) .(1::])

0\r )= {z+

5(1)) = (24

-5k-24-2k = 24

k=-12

=(-J).(-? )

=(7)Answer: D

i{'-sryffi . Compare the corresponding elements to find thevaueofk

(k

(3k - sk

Answer: D

Answer: C

. Order of the product of the two matrices:(Ex2)(2 xE):(lrx tl)

s)

s)"(:*

ft(0) +:. 3k

10 (3 o(^;) =,r,

Q@il+2p)=(7):.l2P+2p=J

l4p -7p=L

2

(:, ;)(:,)

_ / tlltsl + (+)(-2) \\(-r)(s) + (2)(-2) )

/ls-8\=l I

\-s - 4/

_(7\- \-sl

Ilm'-tri,.,_ff

:i

Answer: A

182

Answer: B

1.l'.'T . Expand 3(-1 3 2) first.

,, . po the calculation from left to right.

I

(6 -s 2) - (1 4 -7) + 3(-1=(5 -9 9)+(-3 9 6)

=(2 0 15)

Answer: D

Ii'ffi . e,pand z(! j) ti*,

(:2)-\?=(i 3) (!_l-z Il\-\-ro 6)Answer: A

,(30 ?

le-t- \-rz_( s

\-1sAnswer: C

G;) -,(3 ;)=G?) (s :-)=

(:, , -',*) =

:.7 -2m=-2m =

2 -s\-2 r)2 -s\-2 t)2 -5\-2 t)

(

(

(

1

3-6

32)

-;)

;)

i) (l-:)13) -(j -; )l0\_n)

7 (;) '(?)=(;)(;: i,) = (',)

x+3=2 2-x=-l

9

J

13

I

12=l)=-10

.'. x=-l,y =-10Answer: A

re . Order of the product of the two matrices:(Ex2)(2xE):(jxIl)

Answer: A

13 (zw ,(1)= (16)

(2w(3) + 4(-2)) = (16)(6w - 8) = (16)

:.6w =24w=4

Answer: D

G r)o=(.-i!iiii!d,_(,-rz\= (,?-.ool

_ /-10\= t,;J

,

i.:Lli'J.'ll1iffi,L +i /-\

r (d/'l

a

t,

.::.:'i:

):,

i (b)..:

iiiio

since ,O : (J ?), O is the inverse of p.

Find the inverse of P to deterrnine the valuesof k and h.

rnverse matrix or (! 2): #oe, t)

Write the simultaneous linear equations inmatrix form first.

Use the information in (a) to ma (l ?) '

To solve the equation, pre-multiply bothsides of the matrix equation by p,.

'W . Order of the product of the two matrices:(Ex3)(3xE):1[x2l)

ls 1\(r -r o)(-2 o I\3 +l= (l(s) + (-1X-2) + 0(3) t(1) + (-l)(0) + 0(a))-(5+2+0 1+0+0)=(7 1)

Answer: B

KERTAS 2

183

(a) o(1, :) =_(? -:,), ( , s \

2(3) - (-sxl) \-l 2 )_ | (3 s\- tt\-t z)

Compare both sides:I:. k-ft,h=5

(b) (? ,'

...r=_1,!=3

(a) . Find the inverse , (_=.' -ro) .rO comparelt 2\ -t

it with | 1 | to find the value of n.

\z nJ

(b) . Write the simultaneous linear equations inthe matrix form first.

. Then, pre-multiply both sides of the matrix

equation by the inverse , (_', ,a) to rina

the values of u and v.

)(; ) = (-t')

(;)=(? ?) '(-;')

=+(j, ;)(-i')I I 3(-r7) + 5(8) \= l1 \r-tx-rl+2@)t l-51 + 40\

tt- l1 \ t7 + L6 )t /-tt\11\ 33 //-r \=\: /

,$i*ffi&.,:,tilr,.,ta1i

'i.i

(a, (; i)=l:, 4\-r2)

I Gi)3(2) - (-4X_1)

-L(2 +\2\1 t)

Compare both sides:

;. n =Ar:l2'',=1

2

,r),

= | (7 -3\2(7) - 3(4) \-4 2 )t_( 7 -3\-z\-+ z)

Compare both sides:.'.k=2,m=-4

+G ;)=G4 (a)

(b) (l

5 (a) a-r -

(b) (1

.'.r=-ll,y-7

')(;) =(;')

(;) = +a;)(-J)_r((t)(-l)+(-3Xs)\- z\f-+l(-t)+ <zt<s>l

t l-t - rs\=t\ a+rc)

=+(1)/-11\=l 7 )

)(--;2(-s) - (-3X4)

+(_:^ ))

li i)_3 )(;)= (1,)

(;)= +(_:- )Q,)t I ?s)(7) + 3(13) \

=_l !

z $4)(7) + 2(t3) )t (-35 + 39\

- _t I

Z \-28 + 26)

= r /4\- T\_2)_ ( 2\- \-rl

...r=2,!=-l

184

:)(i)=(r)(;)=(:, -:)'f;)

=+(? l)(;)_ t (2(-s) +a(2)\- 7\ t(-s) + 3e) )

I l-2\=;l_i )

= 1-,'\\;/

_(32x-s) \ s

:)

i)(;)) + (-2X2)$ + ?a)Q))

(b) (j,

.'. Lr = -t,, = t(a) . From the given identity matrix equation, it

can be said that *(; _1) ,r the inverse

" (:i i). Compare the inverse of (:i 3) **1 ln -2\, (.; _;) ,o find the values of m and n.

(b) . Pre-multiply both sides of the matrix

equation by the inverse * (-! 3) ,o t,nothe values of x and y.

tr..:]l

_?-)

*(; _i)=(_: :)'_l

(4)(3) _ (2

t (3 _2\= J\s -41

.'. m=-2rn=3

(_1 3)6) =(j)

(;) =+(;

=+(3[i]

=+(ri )

(i)13

"4- 2'J- 2

(a)

(b)

3)',7 (a)

(b) (;

3(s) - 2(6)

/a

M =l )\6'-) =(; l) 'r(16 (a)

(:, :)i)(_"

_ | ( s -z\:\-o z I

6i)

4(s) - (-1X2)

r ls t\= n\_z q)

I:. ls=fi,m=4

lX;)= (,)(;)= *(:,')g)

t ( 5(7) + t(-2) \- n\eDo + aGD)t I ts - z \__t

I

zz\-t+ - s I

- I /:rt= n\_zz)/3\

-l 2l-\-rl

(b) (;

3)(;)=(;)

(;)=(j l)r;l=(+(3) + (-i),r,)\ r-zlol + tg I

= (-t;tr)_ /-r\-\:/

...x=-I,!=3;. x=|,r=*,

l8s

KERTAS 1

1. Direct variationsUbahan langsung

(i) y = kx, where k = constant! = ktc, dengan k - pemalar)'r lt(11) --tv x2

Mlthematical Formulae and Facts ::

2t. I;";;r- variationsUbahan songsang

(i) y -L,whereft= constant,xky=- _. denganft=pemalar

(ii) x1y1 = xzlz

ffi Direct Variations

SPM mS Question 37I It is given that y varies directly as the square

rootof xand),= 15when x-9.Calculate the value of x when y = 30.Diberi bahawa y berubah secara langsungdengan punca kuasa dua x dan y = 15 apabila*-OA

- t.

Hitungkan nilai x apabila) = 30.

818D36

SPM 20AO Question 362 lt is given that uu varies directly with the

cube of p and w = 8 when p = 4.Calculate the value of p when w - 27 .

Diberi bahawa w berubah secara langsungdengan kuasa tiga p dan w - 8 apabila p - 4.Hitungkan nilai p apabila w = 27.

A+C6

SPM m7 Question 363 Table 1 shows some values of the variables

"r and y.Jadual I ruenunjukkan beberapa nilai bagipembolehubah x dan y.

?::::,\

It is given that y varies directly as the cubeof x.Calculate the value of n.Diberi bahawa y berubah secara langsungdengan kuasa tiga x.Hitung nilai n.

A5c2s

B+D24

A4B8ct6D30

x 2 n

v 4 32

186

SPM m Questrbn 364 Table 2 shows some values of the variables

RandLJadual 2 menunjukkan beberapa nilai bagipembolehubah R dan T.

R 54 l2T 36 v

Table 2Jadual 2

It is given that R varies directly asFind the value of -y.Diberi bahawa R berubah secaradengan T.

Cari nilai y.

T.

langsung

827D64

A24c48

SPM mg Question 375 It is given that y varies directly as the square

of x.Find the relation between y and x.Diberi bahawa y berubah secera langsungdengan kuasa dua x.Cari hubungan antara y dan x.

A y*{i B yq

C !qx2 D \x.

SPM 2o10 Question 376 Table 1 shows some values of the variables

P and Q, such that Q varies directly as thesquare root of P.Jadual I menunjukkan beberapa nilai bagipemboleh ubah P dan Q, dengan keadaan Oberubah secara langsung dengan punca kuasadua P.

Table 1

Jadual I

Calculate the value of n.Hitung nilai n.

1

r-VxI

x2

B+J

D36A+ct2

PI1 n

O-J 2

187

SPM 2O11 Question 36

7 Which table represents the relation of y x a2tJadual manakah yang mewakili hubungan y o x2?

A

ffi tnverse VariationsSPM m5 Question 368 Table 3 shows some values of the variables

x and y such that y varies inversely as thesquare root of x.Jadual 3 menunjukkan sebahagian daripatlanilai-nilai bagi pentbolehubah x dan y, dengankeadaan y berubah secara songsang denganpunca kuasa dua x.

Table 3Jadual 3

Find the relation between y and x.Carikan hubungan antara y dan x.

A y=3'tT B

SPM 2NG Question 37

g It is given that), * +andy = 3 when x -36.- r/x

Calculate the value of y whefi x - 4.

Diberi bahaway * + dan y = 3 apabila x = 36." i'rHitungkan nilai y apabila x = 4.

B

C

D

c y_+*,

L2)'- ix96Y- )x'

A2B9c18D48

x 1 2 aJ 4

v I 8 27 64

x I 2.,J 4

v I L6 81 256

x I 2 --) 4

v 2 4 6 8

x 1 2 -J 4

v aL 8 18 32

x 4 t6

v 6.lJ

SPM m7 Question 3710 P varies inversely as the square root of M.

Given that the constant is fr, find the relationbetween P and M.P berubah secara songsang dengan punca kuasadua M. Diberi k ialah pemalar, cari hubunganantara P dan M.

1

A P_KMT

C P=kMz

SPM nB Question 3711 Given y varies inversely as x3, and that

!=4when *=L2'Calculate the value of x

Diberi y berubah secara

dany -4apabito*=t.Hitung nilai x apabila y -

B P=+rt

D P=4Mt

when y = *.songsang dengan x3,

I16'

A+B+C2D8

SPM nlo Question S ,

12 It is given that PQz = k, where k is a constant.Which statement is true?

I

Diberi bahawa PQz = k, dengan keadaan kadalah pemalar.Pernyataan manakah yttng benar?

A P varies directly as the square of QP berubah secara langsung dengan kuasadua Q

P varies directly as the square root of QP berubah secara langsung dengan puncakuasa dua Q

P varies inversely as the square of QP berubah secara songsang dengan kuasadua Q

P varies inversely as the square root of QP berubah secara songsang dengan puncakuasa dua Q

188

SPM 2011 Question 3713 It is given that m varres inversely with n and

m=20 when n=2.Calculate the value of n when m = 5.Diberi bahayva m berubah secara songsangdengan n dan m = 20 ctpabila n = 2.

Hitung nilai n apabila m -- 5.

1rA8 B,C2 D8

ffiffi Joint VariationsSPM NOi Question I14 Table 4 shows some values of the variables

w, x and y such that w varies directly as thesquare of x and inversely as y.Jadual 4 menunjukkan sebahagian daripadanilai-nilai bagi pembolehubah w, x dan y dengankeadaan w berttbah secara langsung dengankuasa dua x dan secara songsang dengan y.

Table 4Jadual 4

Calculate the value of m.Hitungkan nilai m.

SPM n6 Quesfion 3815 Table 3 shows two sets of values of p, q

and r which satisfy r G pQ.Jadual3 menunjukkan dua set bagi nilai p, q dan

r yang memuaskan r x pe.

Table 3Jadual 3

Calculate the value of w.Hitungkan nilai w.

B T.2

D 7.2

A90 845c30 D15

AC

0.22.4

w' x v

40 4 2

m 6 4

p 8 w

q -J 10

r 4 t2

SPM m7 Question 3816 The relation between the variables x, y and z

isxoc I rrisgiven thatx=f *n.ny=2

whenr=furoande-8.

Calculate the value of z

!=6.Hubungan antara p embolehubah-pembole hubah

x, y dan z ialah * * I. Diberi bahawa * = ]aPabila!=2danz-8.

sHitung nilai z apabila * - I dan y * 6.

A2818c32D72

SPM n@ Question 3817 It is given that P varies directly as the square

root of Q and inversely as the square of R.Find the relation between P, Q and R.Diberi bahawa P berubah secara langsungdengan punca kuasa dua Q dan secara songsangdengan kuasa dua R.

Cari hubungan antara P, Q dan R.

A P*4VR

B P*qRz

c p*L,lO

D P*qQ'

189

SPM mg Question 38

18 It is given that p * # and, pvaries directly

as the square root of q and inversely as thecube of r.State the value of m and of n.

Diberi bahawa p * # dan p berubah secara

langsung dengan punca kuasa dua q dan secarasongsang dengan kuasa tiga r.Nyatakan nilai m dan nilai n.

A m=2,n=3B m=2,n--3 1c nr=;.n_3

ID m=*,n=-3z

SPM 2011 Questron 3€l

19 R varies directly as the square root of ,S andinversely as 7.Given that the constant is k, find the relationbetween R, S and Z.R berubah secara langsung dengan punca kuasadua S dan secara songsang dengan T.

Diberi k ialah pemalar, cari hubungan antara R,S dan T.

A R_+

B R = kvs

T

R_KT^sz

R=F

REVIEW AND ANSWEH$

KERTAS 1

'r''ri ri . y varies directly as

voiv.'SubstituteY:15

find k.

the square root of x means

andx:gintoy:kfi- to

' SubstitutQ Y:30 into y:5t[x to 1nd the valueof x.

) x f/7=+ ,y = /.^ix, where k is a constant.

15 = kr/9

k=*J

-5.'. -v = 5.F

When.y = 30, 30 = 5lF

^F=+-6=62-36

Answer: D

. uz varies directly with the cube of p means

wop3.. Substitute w : B and p : 4 into ruv : kps to

find k.. Substitut e w:27 into w: Lp3 to find the value

of P 8'

w,x p3 + w = kp3, where ft is a constant.

8 = k(4)3

31r-:-64

I=-8

1,:. ty -- Tp'

When w=2J,Zl =f,n3

P3 =21 x8= 216

p -xi6_4.

Answer: C

190

'rrri;a 1r I L:

. y varies directly as the cube of x means y q x3.

' Substitute y : 4 and x : 2 into y : k3 tofind k.

. Substitute x: n and y:32 into y: *x, to findthe value of n. 2

! q x3 =+ y = kr3, where ft is a constant.4 - k(2)3

,48

_t.2

.'. r- I *'''2When x=n andy =32.

32 - Ln32

n3 =641r-

n - "'164_A-+

Answer: A

RxT iR= kT,where ftisaconstant.54 = k(36)

t._ 54It--

36aJ=-2

at

.'. R =:72

WhenR=J2and7=y,3L-_al v

2'

,-=72*1=48

Answer: C

. y varies directly as the square of x means y q x2.

The relation between y and x is y x x2.

Answer: C

Q * ^'[T + Q - k^[P, where k is a constant.

3--ftx E\4.k

2

k_6:. Q- 6{PWhen P=nandQ-2.

2 - 6,,[nl-

;=lrz,1

Answer: on =

'^v! x x' + -= = ft, where fr is a constant.x'

.'. Table D represents the relation of y o x2.

Answer: D

. y varies inversely as the square root of x meansI

V x --:-.' ^l

x

r * -L =.r,- !, where k is a constant.{x {x

L- k

k-6x2_12

12, ^/*

Answer: B

y * + ==r y - 4, *h"re k is a constant.

^/* fr'-kt

--{36

k=3x6= l8

18.. ) - t_

trxWhen x = 4,

18

' t[T18

2

-9

x I 2 J 4

x2 I 4 9 l6v 2 8 18 32

v)x"

2 2 2 2

Answer: B

19t

I '+ . P varies inversely as the square root of M means

p*!orpx-+.1M Mz

"*aM2

D_ kI

M2

Answer: B

11 v

13 n'L x

40...m__-/t

When m = 5,

Answer: D

wherekisaconstant.

y = +. where k is a constant.xt

^_ k

/ 1\3\r)l.

A-n-f - 1

8

k=4xL8

=12

oA=+x-'

12

1v-' 2xr

whenv- I I =-l .' 16' 16 2x3

2x3 - 16

x3= 81.-x=V8

-)Answer: C

I

pe1 =k+p=*=poL

.'. P varies inversely as the square root of Q.

Answer: D

L=*-L20=L

2k=40

-40n

n=8

I*1+ . rz varies directly as-the square of x and inversely

asymeanr**4..Substitute vv:

Yqo,

r: q and y:2 intokx2w:Ltofindk.Y 5x2. Substitute w : tn, X :6 and y : 4 into uz - ::-

to find the value of m. Y

i

)t)

w,x x' = w=E,wherekisaconstant.yv

40 - k(4)2

2

k=40* 2

t6

- r = j5x'' ar,

-

-

. . fv - v

Whenw=m,x=6andy={,5,6\2m=&

45x36

4_45

Answer: B

15 r x pq ) y = kpq, where ft is a constant.4 = t(8X3)

k=L6

I.-. r= -pQf)

When P=w,Q=l0andr-12,n = !@Xro)6'w -- 7.2

Answer: D

16 xo. a = x -- 9, *h.re k is a constant.ZZs _@_48

5o:: "o

5r,

Z

192

Whenr=4andy=(,-15 = 5(6)32.z=30x 3

=18 5

Answer: B

:ffitli , P varies directly as the square root of Q and

:' inversely as the square of R means p * €

The relation betwee n P, Qand R i, P *8.Answer: B

T'Stg . p varies directly as the

inversely as the cube ofI

on Q't-

rJ

'ComPare P o Q^ withrn

values of m and n.

t

0* qapxJ-= Oq-:-'r''rJ

1.'.m= r-.n=3Answer: C

square root of q and

^16rmeansDCr+ort<

t-_2

p o 9= to find therJ

.14i+ . R varies directly as the square root of S and

;fmeansR*€inversely as T

rflr k{sftz +R=TTAnswer: B

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Analisis Bertopik Form 5 Chap1 5 r