additional mathematics form 4 and 5 notes (1)

7242019 Additional Mathematics Form 4 and 5 Notes (1)

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1Paper 11 The relation in the given graph can be representedusing the following arrow diagramBased on the above arrow diagram(a) the object of 40 is 3(b) the type of the relation is many-to-manyrelation2(a) The above relation is a many-to-one relation(b) The function which represents the aboverelation is f(x) = x23 f2(x) = ff (x)= f (px + q)= p (px + q) + q= p2x + pq + qIt is given that f

2(x) = 4x + 9By comparisonp2= 4 pq + q = 9p = plusmn 2 plusmn2q + q = 9plusmnq = 9q = plusmn91694plusmn 4

plusmn 3plusmn 2234123410203040

A B4 (a) gf x regx2+ 6x + 2gf (x) = x2+ 6x + 2g(x + 4) = x2+ 6x + 2



Let x + 4 = ux = u plusmn 4g(u) = (u plusmn 4)2+ 6(u plusmn 4) + 2= u2plusmn 8u + 16 + 6u plusmn 24 + 2= u2plusmn 2u plusmn 6g(x) = x2plusmn 2x plusmn 6(b) fg(4) = f[42plusmn 2(4) plusmn 6]= f(2)= 2 + 4= 65 Let g

plusmn1(x) = yg(y) = x3y + k = xy =y = x plusmngplusmn1(x) = x plusmnIt is given that gplusmn1(x) = mx plusmnHence by comparison

m= and plusmn = plusmn rArrk =5256k31356k3

13k313x plusmn k3SPMZOOMplusmnINForm 4 Chapter 1 Functions



The questionrequires p lt 02Paper 21 (a) f x regf (x) =Let fplusmn1(x) = yf(y) = x= xhy = x (y plusmn 3)hy = xy plusmn 3x3x = xy plusmn hy3x = y(x plusmn h)y = fplusmn1(x) =But it is given that fplusmn1(x) = x sup1 2Hence by comparison h = 2 and k = 3(b) gf

plusmn1(x) = g[fplusmn1(x)]= g ==gfplusmn1

(x) = plusmn5x= plusmn5xx plusmn 2 = plusmn15x215x2+ x plusmn 2 = 0(3x plusmn 1)(5x + 2) = 0x = or plusmn2513

x plusmn 23xx plusmn 23x13xx plusmn 2) (3xx plusmn 2



kxx plusmn 23xx plusmn h3xx plusmn hhyy plusmn 3hxx plusmn 3hxx plusmn 32 (a) Let fplusmn1(x) = yf(y) = xplusmn 2 = x= x + 2y = 2(x + 2)y = 2x + 4 fplusmn1(x) = 2x + 4 f

plusmn1(3) = 2(3) + 4 = 10(b) fplusmn1g(x) = fplusmn1[g(x)]= fplusmn1(3x + k)= 2(3x + k) + 4= 6x + 2k + 4But it is given that

fplusmn1g x reg 6x plusmn 4fplusmn1g (x) = 6x plusmn 4Hence by comparison2k + 4 = plusmn42k = plusmn8k = plusmn4(c) hf(x) x reg 9x plusmn 3h[f(x)] = 9x plusmn 3h

plusmn 2= 9x plusmn 3Let plusmn 2 = u= u + 2x = 2u + 4h(u) = 9(2u + 4) plusmn 3= 18u + 33 h x reg18x + 33



x2x2x2y2y23SPM Zoom-InForm 4 Chapter 2 Quadratic EquationsPaper 11 12x2plusmn 5x(2x plusmn 1) = 2(3x + 2)12x2plusmn 10x2+ 5x = 6x + 412x2

plusmn 10x2+ 5x plusmn 6x plusmn 4 = 02x2plusmn x plusmn 4 = 02 Sum of roots = plusmn +plusmn= plusmnProduct of roots =

plusmnplusmn=The required quadratic equation isx2+ x + = 0251915

253523191535



23plusmnb b2 plusmn 4ac2aplusmn(plusmn1) (plusmn1)2plusmn 4(2)(plusmn4)2(2)1 334x =x =x =x = 16861 or plusmn118614 x2+ 2x plusmn 1 + k(2x + k) = 0x2+ 2x plusmn 1 + 2kx + k2= 0

x2+ 2x + 2kx + k2plusmn 1= 0x2+ (2 + 2k)x + k2plusmn 1 = 0a = 1 b = 2 + 2k c = k2plusmn 1

If a quadratic equation has two real and distinctroots then b2plusmn 4ac gt 0b2plusmn 4ac gt 0(2 + 2k)2plusmn 4(1)(k2plusmn 1) gt 04 + 8k + 4k

2plusmn 4k2+ 4 gt 08k + 8 gt 08k gt plusmn8k gt plusmn15 3(x2+ 4) = 2mx



3x2+ 12 = 2mx3x2plusmn 2mx + 12 = 0a = 3 b = plusmn2m c = 12If a quadratic equation has equal roots thenb2plusmn 4ac = 0b2plusmn 4ac = 0(plusmn2m)2plusmn 4(3)(12) = 04m2plusmn 144 = 04m2= 144m

2= 36m =

615x2+ 19x + 6 = 03 3x2+ 4p + 2x = 03x2+ 2x + 4p = 0a = 3 b = 2 c = 4p

If a quadratic equation does not have real rootsthen b2plusmn 4ac lt 0b2plusmn 4ac lt 022plusmn 4(3)(4p) lt 04 plusmn 48p lt 0plusmn48p lt plusmn4p gt

p gt112plusmn4plusmn48x2plusmn (sum of roots)x + (product of roots) = 046 x



2+ 2x plusmn 8 = 0a = 1 b = 2 c = plusmn8The roots are p and qSum of roots = plusmnp + q = plusmnp + q = plusmn2Product of roots =pq = plusmnpq = plusmn8The new roots are 2p and 2qSum of new roots= 2p + 2q= 2(p + q)= 2(plusmn2)= plusmn4Product of new roots= (2p)(2q)= 4pq= 4(plusmn8)= plusmn32The quadratic equation that has the roots 2p and2q is x2

+ 4x plusmn 32 = 081ca21ba7 x2plusmn (k + 2)x + 2k = 0a = 1 b = plusmn(k + 2) c = 2k

If one of the roots is a then the other root is 2aSum of roots = plusmna + 2a = plusmn 3a = k + 2a = frac14Product of roots =2a2=a2= k frac14

Substituting into 2= k= k(k + 2)2= 9kk2



+ 4k + 4 = 9kk2plusmn 5k + 4 = 0(k plusmn 1)(k plusmn 4) = 0k = 1 or 4(k + 2)29k + 232 122k1c

1k + 23plusmn(k + 2)1b

5P

per 21 (2x plusmn 1)(x + 3) = 2x plusmn 3 plusmn k2x2+ 6x plusmn x plusmn 3 = 2x plusmn 3 plusmn k2x2+ 3x + k = 0

= 2 b = 3 c = kThe roots re plusmn2 nd pSum of roots = plusmnplusmn2 + p = plusmn

plusmnp = plusmn + 2p =Product of roots =plusmn2p =plusmn2 =k = plusmn22 2x2+ (3 plusmn k)x + 8m= 0

= 2 b = 3 plusmn k c = 8mThe roots re m nd 2m

Sum of roots = plusmnm+ 2m = plusmn6m = k plusmn 3 frac14Product of roots =m(2m) =2m2= 4mm2



= 2mm2plusmn 2m = 0m(mplusmn 2) = 0m = 0 or 2m = 0 is not cceptedm = 28m2c

13 plusmn k2b

k212k2c

123232b

From When m = 26(2) = k plusmn 3k = 12 + 3

k = 153 ( ) 2x2+ px + q = 0 = 2 b = p c = qThe roots re plusmn nd 2Sum of roots = plusmnplusmn + 2 = plusmn= plusmnp = plusmn1Product of roots =plusmn 2 =q = plusmn6

(b) 2x2plusmn x plusmn 6 = k2x2plusmn x plusmn 6 plusmn k = 0

= 2 b = plusmn1 c = plusmn6 plusmn kIf the qu dr tic equ tion does not h ve re lroots then b2



plusmn 4 c lt 0When b2plusmn 4 c lt 0(plusmn1)2plusmn 4(2)(plusmn6 plusmn k) lt 01 + 48 + 8k lt 08k lt plusmn49k lt plusmnk lt plusmn618498q232c

p21

2p232b

3216SPMZOOMplusmnINForm 4 Ch pter 3 Qu dr tic Functions

P

per 11 f(x) = 2x2+ 8x + 6= 2(x2+ 4x + 3)= 2[x2+ 4x +

2plusmn 2+ 3]= 2(x2+ 4x + 22



plusmn 22+ 3)= 2[(x + 2)2plusmn 1]= 2(x + 2)2plusmn 2

= 2 p = 2 q = plusmn22 From f(x) = plusmn (x plusmn 4)2+ h we c n st te th t thecoordin

tes of the m

ximum point

re (4 h) But itis given th t the coordin tes of the m ximum point

re (k 9) Hence by comp

rison( ) k = 4(b) h = 9(c) The equ tion of the t ngent to the curve t itsm

ximum point is y = 93 ( ) y = (x + m)2+ nThe

xis of symmetry is x = plusmnm

But it is given th

t the

xis of symmetry isx = 1m= plusmn1When m= plusmn1 y = (x plusmn 1)2+ nSince the y-intercept is 3 the point is (0 3)3 = (0 plusmn 1)2+ nn = 2(b) When m= plusmn1

nd n = 2y = (x plusmn 1)

2+ 2Hence the minimum point is (1 2)4 (2 + p)(6 plusmn p) lt 712 + 4p plusmn p2plusmn 7 lt 0plusmnp2+ 4p + 5 lt 0p2plusmn 4p plusmn 5 gt 0

(p + 1)(p plusmn 5) gt 04242Hence the required r nge of v lues of p isp lt plusmn1 or p gt 55 3x2+ hx + 27 = 0



= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re

l rootsb2plusmn 4 c lt 0h2plusmn 4(3) (27) lt 0h2plusmn 324 lt 0(h + 18)(h plusmn 18) lt 0Hence the required r nge of v lues of h isplusmn18 lt h lt 186 g(x) = (2 plusmn 3k)x2+ (4 plusmn k)x + 2

= 2 plusmn 3k b = 4 plusmn k c = 2If qu dr tic curve intersects the x- xis t twodistinct points thenb2plusmn 4 c gt 0(4 plusmn k)

2plusmn 4(2 plusmn 3k)(2) gt 016 plusmn 8k + k2plusmn 16 + 24k gt 0k2+ 16k gt 0k(k + 16) gt 0Hence the required r

nge of v

lues of k isk lt plusmn16 or k gt 0kplusmn16 0

hplusmn18 18pplusmn1 57P per 2( ) f(x) = 2x2+ 10x + k= 2x2

+ 5x += 2x2+ 5x + plusmn += 2[



x +2plusmn +]= 2x +2plusmn + k(b) (i) Minimum v lue = 32plusmn + k = 32k =(ii) b2plusmn 4

c lt 0102plusmn 4(2)(k) lt 0100 plusmn 8k lt 0plusmn 8k lt plusmn100k gt

k gt(c) Minimum point isplusmn2 322 ( ) g(x) = plusmn2x2+ px plusmn 12 = plusmn2(x + q)2plusmn 4plusmn2x2

+ px plusmn 12 = plusmn2(x2+ 2qx + q2) plusmn 4= plusmn2x2plusmn 4qx plusmn 2q2plusmn 4By comp risonp = plusmn 4q frac14

nd plusmn12 = plusmn2q2

plusmn 4plusmn2q2= plusmn8q2= 4q =

2From When q = 2 p = plusmn4(2) = plusmn8 (Not ccepted)



When q = plusmn2 p = plusmn4(plusmn2) = 8 (Accepted)bec

use p gt 0

nd q lt 0)1112252plusmn100plusmn889225225252k225452

k2254254k2 52=

25412(b) g(x) = plusmn2x2+ 8x plusmn 12 = plusmn2(x plusmn 2)2plusmn 4The m ximum point is (2 plusmn4)When x = 0 y = plusmn12 (0 plusmn12)The gr

ph of the function g(x) is

s shownbelow

3 y = h plusmn 2xfrac14y2+ xy + 8 = 0 frac14Substituting into (h plusmn 2x)2+ x(h plusmn 2x) + 8 = 0h2



plusmn 4hx + 4x2+ hx plusmn 2x2+ 8 = 02x2plusmn 3hx + h2+ 8 = 0 = 2 b = plusmn3h c = h2+ 8If

str

ight line does not meet

curve thenb2plusmn 4 c lt 0(plusmn3h)2plusmn 4(2) (h2+ 8) lt 09h2

plusmn8h2plusmn 64 lt 0h2plusmn 64 lt 0(h + 8)(h plusmn 8) lt 0Hence the required r

nge of v

lues of h isplusmn8 lt h lt 8hplusmn8 82 12

1yOplusmn12(2 plusmn4) x8SPM ZOOMplusmnINForm 4 Ch pter 4 Simult neous Equ tionsP per 21 2x plusmn 3y = 2 frac14x2

plusmn xy + y2= 4 frac14From x = frac14Substituting into 2plusmn y



+ y2plusmn 4 = 02plusmn + y2plusmn 4 = 0(2 + 3y)2plusmn 2y(2 + 3y) + 4y2plusmn 16 = 04 + 12y + 9y2plusmn 4y plusmn 6y2+ 4y2plusmn 16 = 07y2+ 8y plusmn 12 = 0(7y plusmn 6)(y + 2) = 0y = or plusmn2

From When y = plusmn2 x = = plusmn2Hence the points of intersection re2 nd (plusmn2 plusmn2)2 4x + y = 2 frac14x2+ x plusmn y = 2 frac14From y = 2 plusmn 4xfrac14Substituting into

x2+ x plusmn (2 plusmn 4x) = 2x2+ 5x plusmn 4 = 0= 070156 or plusmn570156= x2(1)52 plusmn 4(1)(plusmn4) plusmn5

=

241 plusmn5

2 33 1216727



2 + 3(plusmn2)2) (When y = x = =67672 + 3167 2367y(2 + 3y)2(2 + 3y)42 + 3y22 + 3y22 33

2 + 3y2121From When x = 070156 y = 2 plusmn 4(070156)= plusmn080624When x = plusmn570156 y = 2 plusmn 4(plusmn570156)= 2480624Hence the solutions rex = 070156 y = plusmn080624 orx = plusmn570156 y = 2480624 (correct to five

decim

l pl

ces)3 ( ) Since (16 m) is point of intersection ofy = x plusmn 2

nd y2+ ky plusmn x plusmn 4 = 0 thenx = 16 nd y = ms tisfy both the equ tionsThereforem = (16) plusmn 2 = 2

ndm2+ kmplusmn 16 plusmn 4 = 022

+ k(2) plusmn 16 plusmn 4 = 02k = 16k = 8(b) When k = 8y = x plusmn 2 frac14y2+ 8y plusmn x plusmn 4 = 0 frac14From 4y = x plusmn 8



x = 4y + 8 frac14Substituting into y2+ 8y plusmn (4y + 8) plusmn 4 = 0y2+ 8y plusmn 4y plusmn 8 plusmn 4 = 0y2+ 4y plusmn 12 = 0(y plusmn 2)(y + 6) = 0y = 2 or plusmn6From When y = 2 x = 4(2) + 8 = 16When y = plusmn6 x = 4(plusmn6) + 8 = plusmn16Hence the other point of intersectionother th

n (16 2) is (plusmn16 plusmn6)32 33121

14141439SPMZOOMplusmnINForm 4 Ch

pter 5 Indices

nd Log

rithmsP per 11 2x +3

+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)



= 17(2x)2 3x + 3plusmn 3x + 2= 63x(33) plusmn 3x(32) = 627(3x) plusmn 9(3x) = 6(27 plusmn 9)(3x

) = 618(3x) = 63x=3x=3x= 3

plusmn1x = plusmn13 m = 3

n = 3blog3m= log3 n = blog3

= log3 m+ log3n4plusmn log327= log



3 m+ 4 log3n plusmn log333=

+ 4b plusmn 3mn427136182x24 5x= 32x plusmn1lg 5x

= lg 32x plusmn1x lg 5 = (2x plusmn 1) lg 3x lg 5 = 2x lg 3 plusmn lg 3x lg 5 plusmn 2x lg 3 = plusmn lg 3x(lg 5 plusmn 2lg 3) = plusmnlg 3x =x = 1875 log10(p + 3) = 1 + log10p

log10(p + 3) plusmn log10 p = 1log10 = 1= 101p + 3 = 10p9p = 3p =

6 log2 y plusmn log8 x = 1log2y plusmn = 1log2



y plusmn = 13 log2 y plusmn log2 x = 3log2y3plusmn log2 x = 3log2 = 3= 23y3= 8xx = y3

8y3xy3xlog2x3log2

xlog2813p + 3pp + 3pplusmnlg 3lg 5 plusmn 2 lg 3log

28 = log223= 310P per 11 Let point A be (0 k)AB = 10



= 1064 + k2plusmn 14k + 49 = 102k2plusmn 14k + 13 = 0(k plusmn 1)(k plusmn 13) = 0k = 1 or 13B sed on the di gr m k lt 7k = 1A(0 1)2 (

) x + 2y + 6 = 0x + 2y = plusmn6+ =+ = 1(b) mMN= plusmn = plusmnTherefore the gr dient of the perpendicul rline is 2

Hence the equ

tion of the str

ight line whichp

sses through the point N

nd isperpendicul r to the str ight line MNisy = 2x plusmn 33 plusmn = 1At point P (on the x-

xis) y = 0plusmn = 1 rArrx = 4P is point (4 0)At point Q(on the y-axis) x = 0plusmn = 1 rArry = plusmn3Qis point (0 plusmn3)y3

0403x4y3x412plusmn3

plusmn6y(plusmn3)x(plusmn6)plusmn6plusmn62y(plusmn6)x



(plusmn6)(0 plusmn 8)2 + (k plusmn 7)2 Hence the area of ∆PQR== |plusmn12 plusmn (plusmn6 + 20)|= |plusmn26|= 26= 13 units24 (a) 2y = 3x plusmn 12At point L (on the x-axis) y = 02(0) = 3x plusmn 12x = 4L (4 0)At point N(on the y-axis) x = 02y = 3(0) plusmn 12y = plusmn6N(0 plusmn6)M=

= (2 plusmn3)(b) mLN= =Gradient of perpendicular line = plusmnHence the equation of the perpendicularline isy plusmn y1= m(x plusmn x1

)y plusmn (plusmn3) = plusmn (x plusmn 2)3(y + 3) = plusmn2(x plusmn 2)3y + 9 = plusmn2x + 43y = plusmn2x plusmn 55 PA = PB=(x plusmn 1)2+ (y plusmn 2)2= (x plusmn 0)2

+ (y plusmn 3)2x2plusmn 2x + 1 + y2plusmn 4y + 4 = x2+ y2



plusmn 6y + 9plusmn2x plusmn 4y + 5 = plusmn6y + 9plusmn2x + 2y plusmn 4 = 0plusmnx + y plusmn 2 = 0y = x + 2(x plusmn 0)2 + (y plusmn 3)2 (x plusmn 1)2 + (y plusmn 2)2 232332plusmn6 plusmn 00 plusmn 40 + (plusmn6)2

4 + 0212121212400

plusmn32540SPMZOOMplusmnINForm 4 Chapter 6 Coordinate GeometryIntercept form+ = 1ybxa

m= plusmn y-interceptx-intercept11Paper 21 (a) y plusmn 3x + 6 = 0At point B (x-axis) y = 00 plusmn 3x + 6 = 0 rArrx = 2B is point (2 0)y plusmn 3x + 6 = 0



At point C(y-axis) x = 0y plusmn 3(0) + 6 = 0 rArry = plusmn6C is point (0 plusmn6)y = 3x plusmn 6mBC= 3mAC= plusmnLet A(k 0) mAC= plusmn= plusmnplusmnk = 18k = plusmn18A is point (plusmn18 0)(b) Let D(p q)Midpoint of BD= Midpoint of AC=

= (plusmn9 plusmn3)Equating the x-coordinates= plusmn9p = plusmn20Equating the y-coordinates= plusmn3q = plusmn6

Dis point (plusmn20 plusmn6)q22 + p2q22 + p20 + (plusmn6)2plusmn18 + 02

0 + q22 + p2130 plusmn (plusmn6)k plusmn 013



13(c) A(plusmn18 0) B(2 0) C(0 plusmn6) D(plusmn20 plusmn6)Area of ABCD== | plusmn12 plusmn (120 + 108)|= 240= 120 units22 (a) (i) y plusmn 3x + 6 = 0At point P (on the y-axis) x = 0y plusmn 3(0) + 6 = 0 rArry = plusmn6P is point (0 plusmn6)(ii) The coordinates of point S are= (3 3)(b) Area of ∆QRS = 48 units2= 4815k + 21 plusmn (45 + 3k) = 9612k plusmn 24 = 9612k = 120

k = 10(c) S(3 3) Q(10 0) T(x y)TS TQ= 2 3=3TS = 2TQ9(TS)2= 4(TQ)29[(x plusmn 3)2+ (y plusmn 3)2

] = 4[(x plusmn 10)2+ (y plusmn 0)2]9(x2plusmn 6x + 9 + y2plusmn 6y + 9) =4(x2plusmn 20x + 100 + y

2)9x2plusmn 54x + 81 + 9y2plusmn 54y + 81 =4x2plusmn 80x + 400 + 4y



25x2+ 26x + 5y2plusmn 54y plusmn 238 = 023TSTQ12k071533k04(plusmn6) + 3(15)3 + 44(0) + 3(7)3 + 4

121212plusmn180200plusmn6plusmn20

plusmn6plusmn18012SPMZOOMplusmnINForm 4 Chapter 7 StatisticsPaper 11 After the given score are arranged inascending order we have6 6 6 k k 9Since the mode is 6 then k sup1 9For 7 to be the median k = 8 as shown below6 6 6 8 8 9

After two new scores 7 and 10 are added to theoriginal scores the mean of the eight scores== 752 (a) xplusmn=27 =n =n = 7



(b)3aringx2

_ = ETHETHplusmn ( x )2n5278==ETHETHETHplusmn 272725= 518927189naringxn6 + 6 + 6 + 8 + 8 + 9 + 7 + 108

Median = 7x f fx fx230 3 90 270032 5 160 512034 2 68 2312Sum 10 318 10 1324aringfxaringf22 =

10 132= plusmnplusmn10=19631810aringfxaringf22

Number 1 k 6 k + 3 11 13Frequency 2 2 1 3 1 1(a) 1 lt k lt 6 6 lt k + 3 lt 11k = 2 3 4 5 3 lt k lt 8k = 4 5 6 7Taking into consideration both casesk = 4 or 5(b) 1 1 4 4 6 7 7 7 11 13M Q



3 Q3= 7Paper 21Mass (kg) FrequencyCumulativefrequency11 plusmn 20 5 521 plusmn 30 9 1431 plusmn 40 12 2641 plusmn 50 8 3451 plusmn 60 6 40Frequency246810120105 205 305 405 505 605Mass (kg) (Mode) 35

(a)Mode = 35 kg13(b) The Q1class is given byThe Q3class is given byHence the interquartile range= Q3plusmn Q

1= 455 plusmn 261 = 194 kg(c) New interquartile range= Original interquartile range= 194 kg23(40)4T 30 T = 41 plusmn 50 =3Q405 +

348(1) = 455 kg(40) plusmn 26=΄ ΅404T 10 T = 21 plusmn 30 =



1Q205 +4049(1) = 261 kgplusmn 5=΄ ΅(a) Median = 465(b)L +plusmn Ffmc = 465n2΄ ΅395 +plusmn plusmn11k(10) = 465

26 + k2΄ ΅plusmn plusmn11k(10) = 726 + k2΄ ΅plusmn 11 = 07k26 + k226 + k plusmn 22 = 14k

04k = 4k = 10Marks fCumulativefrequency20 plusmn 29 4 430 plusmn 39 7 1140 plusmn 49 k 11 + k50 plusmn 59 8 19 + k60 plusmn 69 5 24 + k70 plusmn 79 2 26 + kMarks fMid-

fx fx2point (x)20 plusmn 29 4 245 980 24010030 plusmn 39 7 345 2415 83317540 plusmn 49 10 445 4450 198025050 plusmn 59 8 545 4360 237620060 plusmn 69 5 645 3225 208012570 plusmn 79 2 745 1490 111005036 1692 86199



(c) (i) New median = Original median + 10= 465 + 10= 565(ii) New variance = Original variance= 18542aringfxaringf2=86 199= plusmnplusmn36Variance=18542169236aringfxaringf22

14SPMZOOMplusmnINForm 4 Chapter 8 Circular MeasurePaper 11BOC = p plusmn AOB plusmn COD= 3142 plusmn 09 plusmn 09= 1342 radArea of sector BOC= 821342= 4294

12OC09 rad8 cm 8 cmp plusmn 1809 radBDA r =sq

2 BOC = 20ordm =20 rad3 Area of the shaded region= Area of sector OAB plusmn Area of sector OXY= 384 plusmn 10= 284 cm2



= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa

er 21(a) MO= r plusmn 6In ∆OMB using Pythagoras theoremMO2+ MB2= OB2(r plusmn 6)2

+ 82= r2r2plusmn 12r + 36 + 64 plusmn r2= 0plusmn12r + 100 = 0r = 8(b) In ∆BOMsin BOM =

BOM = 1287 rad AOB = 2 1287 = 2574 rad(c) Area of the shaded region= 822574 plusmn sin 2574



r= 7071 cm22 (a) BOA = = 1160 rad BOQ= p plusmn 1160 = 1982 rad(b) Area of the shaded region= r2(q plusmn sin q)= (10)2(1982 plusmn sin 1982r)= 5327 cm21212p plusmn 082221

3122425sin BOM =881313O

ACBr cm(r plusmn 6) cm8 cm 8 cm6 cmM(c) Perimeter of the shaded region= 2r sin + rq= 2(10) sin r

+ 10(1982)= 1673 + 1982= 3655 cm3(a) Since ∆ADB is inscribed in a semicircleit is a right-angled triangleIn ∆ADBcos ABD =ABD = 06435 rad(b) AOD = 2 ABD



= 2 06435= 12870 rad Length of the arc AD= 5 12870= 6435 cm(c) Area of ∆ODB= 8 3= 12 cm2Area of sector BOC= 5206435= 804375 cm2Hence the area of the shaded region= Area of ∆ODB plusmn Area of sector BOC= 12 plusmn 804375= 3956 cm21212

810O ADBC5 cm3 cm4 cm4 c

m5 cm19822q2MB = AB = 16 = 8 cm1212The angle at the centreis twice the angle at

circumference16Pa

er 11 f (x) = = (5x plusmn k)plusmn2f cent (x) = plusmn2(5x plusmn k)plusmn3(5)=f cent(1) = 10



= 10(5 plusmn k)3= plusmn15 plusmn k = plusmn 1k = 62 y = (x + 1) (2x plusmn 1)2= (x + 1)[2 (2x plusmn 1)1(2)]+ (2x plusmn 1)2(1)= (2x plusmn 1)[4(x + 1) + (2x plusmn 1)]= (2x plusmn 1)(6x + 3)3 y = 2x3plusmn 4x + 5= 6x2

plusmn 4Gradient at the

oint (plusmn1 7)= 6 (plusmn1)2plusmn 4= 2E

uation of the tangent isy plusmn 7 = 2[x plusmn (plusmn1)]y plusmn 7 = 2(x + 1)y plusmn 7 = 2x + 2y = 2x + 9dydx

dydxplusmn10[5(1) plusmn k]3plusmn10(5x plusmn k)31(5x plusmn k)24 z = xyz = x(30 plusmn x)

z = 30x plusmn x2= 30 plusmn 2xWhen z has a stationary value= 030 plusmn 2x = 0x = 15= plusmn2 (negative)Hence the maximum value of z= 30(15) plusmn 15



2= 2255 y = = (2x plusmn 5)plusmn3= plusmn3 (2x plusmn 5)plusmn4(2) =raquody raquo dy= (301 plusmn 3)= 001= plusmn 0066 A = 2pr2+ 2prh= 2pr2+ 2pr(3r)= 8pr2= = 16pr 01= 16p (5) 01= 8p cm

2splusmn1 r t

A r A tplusmn 6[2(3) plusmn 5]4plusmn 6

(2x plusmn 5)4

y x

y xdydxplusmn 6(2x plusmn 5)4 y

x

1(2x plusmn 5)3

2z x2 z

x



z

xSPMZOOMplusmnINForm 4 Cha ter 9 Differentiation17Pa er 21 y = plusmn = xplusmn2plusmn xplusmn3= plusmn2xplusmn 3+ 3xplusmn 4= plusmn += 6xplusmn 4plusmn 12xplusmn 5= plusmnx4+

+ x2y + 5 = 0x4plusmn + + plusmn+x2

plusmn+ 5 = 0plusmn2x + 3 + 6 plusmn + 1 plusmn + 5 = 0plusmn2x + 15 plusmn = 0plusmn2x2+ 15x plusmn 13 = 02x2plusmn 15x + 13 = 0(2x plusmn 13)(x plusmn 1) = 0x = or 1

2 (a) y =

x3+ kx= 3 x2+ kAt (1 1) x = 1 an m= = plusmn5 3 x2+ k = plusmn5



3 (1)2+ k = plusmn53 + k = plusmn5 frac14The curve asses through oint (1 1) 1 = (1)3+ k(1)

+ k = 1 frac14plusmn 2 = plusmn6 rArrp = plusmn3From plusmn3 + k = 1 rArrk = 4 22 121dydxdydx13213x1x

12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y



dx23x42x3dydx1x31x2(b) When p = plusmn3 and k = 4y = plusmn3x3+ 4x= plusmn9x2+ 4= plusmn18x

At turning points= 0plusmn9x2+ 4 = 0x2=x =

When x = y = plusmn3 3

+ 4 = 1= plusmn 18 = plusmn12 (lt 0) 1is a turning point which isa maximumWhen x = plusmn

y = plusmn3plusmn3+ 4plusmn= plusmn1



= plusmn18plusmn= 12 (gt 0)plusmn plusmn1is a turning point which isa minimum792323d2ydx279

232323792323d

2ydx2792323232

349dydxd2ydx2



dydx183(a) Volume of the cuboid = 5832 cm3(6x)(6x)(y) = 583236x2y = 5832x2y = 162y =L =Area of ABCD+ 4 (Area of GBCH)+ 4 (Area of VGH)L =(6x)2+ 4(6xy) + 4 (6x)(5x)L = 36x2+ 24xy + 60x2L = 96x

2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED

y m4x m5x m3x m6x m6x m(b) L = 96x2+ = 96x2



+ 3888xplusmn1= 192x plusmn 3888xplusmn2= 192x plusmnAt stationary point= 0192x plusmn = 0192x =x3=x3= 2025x = 273= 192 + 7776xplusmn3= 192 + (gt 0) L is a minimum4 y = = h(1 + 2x)plusmn2= plusmn2h(1 + 2x)plusmn3

(2) = plusmndy = dxplusmn = plusmn cplusmn = plusmn cplusmn = plusmnh = h = 18274834hc27

8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

19Pa er 21 (a) UST = 180ordm plusmn 65ordm = 115ordmSUT = 180ordm plusmn 43ordm plusmn 115ordm = 22ordmIn ∆UST using the sine rule=US = sin 43ordm= 16385 cm(b) In ∆USR using the cosine ruleUR2= 72

+ 163852plusmn 2(7)(16385)cos 65ordmUR2= 2205238UR = 1485 cm(c) Area of ∆RSV = 4136 cm27 12 sin RSV = 4136sin RSV = 098476Basic = 7998ordmRSV = 180ordmplusmn 7998ordm

= 10002ordmIn ∆RSV using the cosine ruleRV2= 72+ 122plusmn 2(7)(12)cos 10002ordmRV



2= 22223064RV = 1491 cm12 cm7 cmS RV10002deg12UR S T22deg65deg115deg16385 cm

43deg7 cm 9 cm9sin 22ordm9sin 22ordmUSsin 43ordm2 (a) In ∆PQS using the sine rule=sin QSP = 8sin QSP = 065552QSP = 4096ordm

PQS = 180ordm plusmn 35ordm plusmn 4096ordm= 10404ordmHence the area of ∆PQS= 8 7 sin 10404ordm= 2716 cm2(b) This problem involves the ambiguous case ofsine rule The sketch of ∆QRS1is as shownbelowIn ∆QRS using the sine rule=

sin QSR = 10sin QSR = 0974283Basic = 7698ordm QSR = 7698ordm or QS1R = 10302ordmIn ∆QS1RRQS



1= 180ordm plusmn 43ordm plusmn 10302ordm = 3398ordmIn ∆QS1R using the sine rule==RS1= sin 3398ordm= 5737 cm10sin 10302ordm10sin 10302ordmRS1sin 3398ordm10sin RS1QRS1

sin RQS1sin 43ordm7sin 43ordm7sin QSR1043degQS1R S

10 cm7 cm 7 cm12sin 35ordm7sin 35ordm7sin QSP8SPMZOOMplusmnINForm 4 Chapter 10 Solution of Triangles20

Paper 21 (a) 100 = 120x = 480y = 100 = 105100 = 110z = 600(b)plusmnI = 115= 115



= 11515 200 + 130m = 16 100 + 115m15m = 900m = 60(c)plusmnI2006(based on 2002)= plusmnI2004= 115= 14375(d) Total yearly cost in 2006= 5 500 000= RM7 906 250214375100125100100 + 25

10015 200 + 130m140 + m(120 20) + 130m+ (105 80) + (110 40)20 + m+ 80 + 40660z525500x400SPMZOOMplusmnINForm 4 Chapter 11 Index Numbers

I = 100P2004P2002Health I2004(based I2006(basedWeightagesupplement on 2002) on 2004)A 115 150 3

B 120 130 2C 105 120 x(a) Supplement AI2004(based on 2002) = 115100 = 115 100 = 115P2002



=P2002= RM6000(b) Supplement BI2006(based on 2002)= 100= 100= 100= 156(c)plusmnI2004(based on 2002) = 111= 111= 111585 + 105x = 555 + 111x30 = 6xx = 5(d)plusmn

I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P



2004P2002P2006P2004P2006P200269 10011569P2002P2004P200221Paper 11 (a) T6

= 38a + 5d = 38a + 5(7) = 38a = 3(b) S9plusmn S3= [2(3) + 8(7)] plusmn [2(3) + 2(7)]= 279 plusmn 30= 2492 (a) T2

plusmn T1= T3plusmn T22h plusmn 1 plusmn (h plusmn 2) = 4h plusmn 7 plusmn (2h plusmn 1)h + 1 = 2h plusmn 6h = 7(b) When h = 7 the arithmetic progression is 513 21 frac14 with a = 5 and d = 8S8

plusmn S3= [2(5) + 7(8)] plusmn [2(5) + 2(8)]= 264 plusmn 39= 2253 ==(x= (x plusmn 4)(9x + 4)x



2+ 4x + 4 = 9x2plusmn 32x plusmn 168x2plusmn 36x plusmn 20 = 02x2plusmn 9x plusmn 5 = 0(2x + 1)(x plusmn 5) = 0x = plusmn or 512x plusmn 4x + 2x + 29x + 4T3T2T2

T1328232924 T3plusmn T

2= 3ar2plusmn ar = 34r2plusmn 4r = 34r2plusmn 4r plusmn 3 = 0(2r + 1)(2r plusmn 3) = 0r = plusmn or

5 0242424 frac14= 024 + 00024 + 0000024 + frac14====6 The numbers of bacteria form a geometricprogression 3 6 12 frac14The number of bacteria after 50 seconds= T



11= ar10= 3(210) = 3072Paper 21 (a) The volumes of cylinders arepr2h pr2(h + 1) pr2(h + 2) frac14T2plusmn T1= pr2(h + 1) plusmn pr2h

= pr2h + pr2plusmn pr2h= pr2T3plusmn T2

= pr2(h + 2) plusmn pr2(h + 1)= pr2h + 2pr2plusmn pr2h plusmn pr2

= pr2Since T2plusmn T1= T3plusmn T2



= pr2 thevolumes of cylin ers form an arithmetic

rogression with a common

ifference of pr283324990240990241 plusmn 0013212SPMZOOMplusmnINForm 5 Cha ter 1 ProgressionsSyen=

a1 plusmn r22(b) a = pr2h

= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2



h + 3r2= 52r2(2h + 3) = 52 frac14 ==16h + 24 = 13h + 393h = 15h = 5From r2(5 + 3) = 32r2= 4r = 211382h + 3h + 3

5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3

plusmn T2= 45ar2plusmn ar = 45ar (r plusmn 1) = 45 frac14 ==3 + 3r = 10r



2plusmn 10r10r2plusmn 13r plusmn 3 = 0(2r plusmn 3)(5r + 1) = 0r = or plusmn(b) For the sum to infinity to exist plusmn 1 lt r lt 1Thus r = is not acce te

Therefore r = plusmnFrom a1 plusmn= 150a = 187 a1 plusmn r= Syen 156 1

4=211871 plusmn plusmn=511215

1153215321031 + rr (r plusmn 1)

15045a(1 + r)ar (r plusmn 1)122123SPMZOOMplusmnIN



Form 5 Cha ter 2 Linear LawPa er 11 y = + qx= + q= 2 ( ) + q23 plusmn ax =xy2plusmn ax3= bxy2= ax3+ b(plusmn1 10) 10 = a(plusmn1) + b frac14(5 plusmn2) plusmn2 = a(5) + b frac14plusmn 12 = plusmn6aa = plusmn2From plusmn2 = plusmn2(5) + bb = 82

2 121bx2y2x=

yxk

lg = lg

yxklg y plusmn lg k = lg

xlg y plusmn x lg k = lg

lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2



(3 )yx1x2yx2x2yx2xPa er 21 (a)(b)y = hxklog10y = log10

(hxk)log10y = log10h + log10xk12

log10y = log10h + klog10xlog10y = 2log10h + 2klog10

xlog10y = 2k log10x + 2 log10h12



x3xy2x 15 20 25 30 35y 142 338 660 1348 1995log10x 018 030 040 048 054log10y 215 253 282 313 330log10 xOlog10 y0105101520

25303515502 03335 plusmn 175= 16055 plusmn 006 = 049Gra h of log10 y against log10 x

04 05 06The gra h of log10y against log10x is asshown below24(c) 2k = Gra

ient2k =2k = 32653k = 1632 log

10h = Yplusmninterce t2 log10h = 155log10h = 0775h = 5962 (a)



335 plusmn 175055 plusmn 006x 01 03 04 05 07 08y 078 060 054 050 044 042164 278 343 400 517 5671y2xO0105101520253035404550551146

02 0346 plusmn 2 = 2606 plusmn 016 = 04404 05 06 07 08Gra h of against xy21y21(b) (i)=

= x +Gra ient == 591q = 017Y-interce t = 11= 11= 11

= 019(ii) When x = 06 from the gra h= 46y2= 02174

y = 0471y2

017

q1q



46 plusmn 206 plusmn 016

q1q1y2x +

q1y2=1yx +

qSquaring both si es25SPMZOOMplusmnINForm 5 Cha ter 3 IntegrationPa er 1

1 k5(y plusmn 5) y = 8[plusmn5y]k5= 8plusmn 5k plusmnplusmn 5(5)

= 8plusmn 5k + = 8k2plusmn 10k + 25= 16k2plusmn 10k + 9 = 0(k plusmn 1)(k plusmn 9) = 0k = 1 or 92

2plusmn13g (x) x +423g (x)

x= 3[2plusmn1



g(x)

x +42g(x) x]= 3[4plusmn1g(x)

x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2

xy = plusmn 8 + 6 + cy = plusmn 2x4

+ 2x3+ cSince the curve asses through the oint1 plusmn 1plusmn = plusmn2 + 2 + cc = plusmn2Hence the equation of the curve isy = plusmn 2x4

+ 2x3plusmn 2x551595



45x55x33x44x55

y x252k22522

k22y224 Area of the sha e region=2plusmn1y x=2

plusmn1(x2plusmn 2x + 1) x=[plusmn x2+ x]2plusmn1= plusmn 4 + 2 plusmn

plusmn plusmn1 plusmn 1= 3 units2Pa er 21Area P0plusmn1



y

x=0plusmn1(plusmnx3plusmnx)

x=[plusmn plusmn]0plusmn1= 0 plusmn[plusmn plusmn]= +=Area Q20y

x=

20(plusmnx3plusmn x) x=[plusmn plusmn]20= plusmn plusmn plusmn 0= plusmn4 plusmn 2

= plusmn6222244x22x44

341214(plusmn1)22(plusmn1)



44x22x44O 1 2yP Qxy = plusmnx3 plusmn x1383x3326Hence the total area of the sha e region

= Area P + |Area Q|= + |plusmn6|= 6 units22 (a) y = hx2+ k= 2hxAt the oint (plusmn2 8) the gra ient of thecurve is plusmn 4 = plusmn 42hx = plusmn 42h (plusmn2) = plusmn 4

plusmn 4h = plusmn 4h = 1The curve y = hx2+ k asses through the

oint (plusmn2 8)8 = h(plusmn2)2+ k8 = 4h + k8 = 4(1) + kk = 4

y

x

y x3434(b) When h = 1 an

k = 4 y = x2+ 4



Volume generate

Vx= Volume generate

by the curve plusmnVolume generate by the straight line PQ(from x = 0 to x = 2)= p30y2

x plusmn pr2h= p30(x2+ 4)2

x plusmn p(4)2(2)= p

30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30

plusmn p= p[+ (3)3+ 16(3) plusmn 0] plusmn p= 157 p units3 141532383

3553238x33x5



53231313yOxQPy = x2 + 4x = 32 3427Pa er 11 (a) EAreg= DCreg

= (12

_ ) = 9

_ (b) EQreg= EDreg=EAreg+ AB

reg+ BCreg+ CDreg=9

_ plusmn 6r _ plusmn 9q _ plusmn 12

_ =plusmn 3

_ plusmn 6r _ plusmn 9q _ 2 XYreg= XD



reg+ DYreg= BDreg+ DCreg=BAreg+ ADreg+ ABreg=plusmn6b _ + 2a _ + (6b _ )= plusmn3b _ + a _ + 4b _

= a _ + b _ 3 (a) a _ + b _ + 2c _ = 7j _ + (10i _ plusmn 5j _ ) + 2(plusmn4i _ + j _ )= 7j _

+ 2i _ plusmn j _ plusmn 8i _ +2j _ =plusmn 6i _ + 8j _ (b) |a _ + b _ + 2c _| =Hence the unit vector in the irection ofa _ + b _ + 2c _ =plusmn6i _ +8j _

= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434

4 (a) If the vectors a _ an

b _ are arallel thena_ = hb _ (h is a constant)2i _ plusmn 5j _= h(ki _ plusmn 3j _ )2i _ plusmn 5j _

= hki _ plusmn 3hj _ Equating the coefficients of j _ plusmn3h = plusmn5h =Equating the coefficients of i _ hk = 2k = 2k =



(b) a _ = hb _ a _ = b_5 (a) ACreg= ABreg+ BCreg= 9i _ plusmn 4j _ + (plusmn6i _ + mj _ )= 3i _ + (m plusmn 4) j _ (b) If ACregis arallel to the x-axis thecoefficient of j _ equals zeromplusmn 4 = 0m = 4|a|

|a| |b| = 5 3=|b|5353655353

SPMZOOMplusmnINForm 5 Cha ter 4 Vectors28Pa er 21 (a) OTreg= OAreg+ ATreg= 4x _ + AQreg= 4x _ + (AO

reg+ OQreg)= 4x _ + (plusmn4x _ + 6y _ )= x _ + 2y _ (b) OS



reg= OQreg+ QSreg= 6y _ + hQPreg= 6y _ + h(QOreg+ OPreg)= 6y _ + h(QOreg+ 4OAreg)= 6y

_ + h[plusmn6y _ + 4(4x _)]= (6 plusmn 6h) y _ + 16hx _ 831313

13(c) Since the oints O T an

S are collinearthen OTreg= kOSreg where k is a constantOTreg= kOSregx _ + 2y

_= k [(6 plusmn 6h)y _ + 16hx _]x _ + 2y _= k (6 plusmn 6h)y _ + 16hkx _ x _ + 2y



_= (6k plusmn 6hk)y _ + 16hkx _ Equating the coefficients of x _= 16hk1 = 6hkhk = frac14Equating the coefficients of y _ 6k plusmn 6hk = 2 frac14Substituting into 6k plusmn 6 = 26k = 3k =From hk =h =h =1

3161216212162 1

21168383838329

2 (a) (i) OMreg= OBreg= (14y _ ) = 10y _ (ii) AKreg





232721272121414145757(c) AKreg= AL

reg+ LKregplusmn x _ + y _ = plusmn2 x _ + 10 y _ +qx _ + qy _plusmn x _ + y _ =

plusmn2

+ qx _ +10 + qy _ plusmnx _ + 7y _ = (plusmn4 + 3q)x _ + (20 + 7q)y _ Equating the coefficients of x _ plusmn 4 + 3q = plusmn1 frac14

Equating the coefficients of y _ 20 + 7q = 7 frac14plusmn20 + 15q = plusmn5 frac14 520 + 7q= 7 frac1422q = 2q =From plusmn 4 + 3 = plusmn1



plusmn 4 = plusmn =722141111111112 +121723272127

2327212KAreg= plusmnAKreg= x _ plusmn y _

721230Pa

er 11sin (90ordm plusmn q)= cos q2 plusmn 10 tan x = 03 sec2

x plusmn 10 tan x = 03(tan2x + 1) plusmn 10 tan x = 03 tan2x + 3 plusmn 10 tan x = 03 tan2x plusmn 10 tan x + 3 = 0



(3 tan x plusmn 1)(tan x plusmn 3) = 0tan x = or tan x = 3When tan x = x = 1843ordm 19843ordmWhen tan x = 3x = 7157ordm 25157ordmx = 1843ordm 7157ordm 19843ordm 25157ordm13133cos2x 1 +

2

= plusmnOq 11 +

2plusmn

3 3 tan q = 2 tan (45ordm plusmn q)3 tan q = 2 3 tan q = 2 3 tan q + 3 tan2q = 2 plusmn 2 tan q3 tan2q + 5 tan q plusmn 2 = 0(3 tan q plusmn 1)(tan q + 2) = 0tan q = or tan q = plusmn2

When tan q = Basic = 1843ordmq = 1843ordm 19843ordmWhen tan q = plusmn2Basic = 6343ordmq = 11657ordm 29657ordm q = 1843ordm 11657ordm 19843ordm 29657ordm13131 plusmn tan q1 + tan q

tan 45ordm plusmn tan q1 + tan 45ordm tan qSPMZOOMplusmnINForm 5 Cha

ter 5 Trigonometric FunctionsWebsiteZI F505_4th 101508 940 AM Page 3031Pa

er 21 (a) LHS ===



== tan x= RHS(b) (i) (ii)The gra

h of y = |tan x| is as shown belowplusmn = 0=|tan x| =Number of solutions= Number of

oints of intersection= 4x2px2p1 plusmn cos 2xsin 2xx2p1 plusmn cos 2xsin 2xxyO

32y =(2p 1)pppp psinxcos x2 sin2x

2 sin x cos x1 plusmn (1 plusmn 2 sin2x)2 sin x cos x1 plusmn cos 2xsin 2x2 (a) (b)3 sin + = 23 sin = 2 plusmnSketch the straight line y = 2 plusmnNumber of solution= Number of intersection

oint

= 12xp2xpx22xpx



2yxO 2y = 3 sin x2 32plusmn2y = 2 plusmn2x2Sketch thestraight liney =x2px 0 2py 2 plusmn2WebsiteZI F505_4th

101508 940 AM Page 3132Pa

er 1

1Hence the number of 4-digit odd numbers greaterthan 2000 but less than 3000 that can be formed= 3 3P2= 182 Each grou

of boys and girls is counted as oneitemOuml Ouml This gives 2Ouml Ouml Ouml

At the same time B1 B2 and B3can be arrangedamong themselves in their grou

This gives 3Ouml Ouml Ouml In the same way G1 G2

and G3can also bearranged among themselves in their grou

Thisgives another 3Using the multi

lication

rinci

le the totalnumber of arrangements= 2 3 3 = 72Number of arrangements2 1



3P22 33P22 53P23 Number of different committees that can beformed=4C17C28C

3= 4704SPMZOOMplusmnINForm 5 Cha

ter 6 Permutations and CombinationsB1 B2and B3 G1 G

2and G3Choosing a femalesecretary and a femaletreasurer from 7 femalesChoosing 3subcommitteemembers from 8(males or females)Choosing a male

resident from 4 malesWebsiteZI F506_4th

101508 940 AM Page 32

33Pa

er 11 P(Not a green ball) ==5h + 25 = 3h + 3k + 152h = 3k plusmn 10h =2 P (not getting any

ost)= =



3 There are 3 `E and 4 `Eplusmn in the bag(a) P(EE) = =(b) P(EEplusmn) = =274637172637835473

5233k plusmn 10235h + 5h + k + 535SPMZOOMplusmnINForm 5 Cha

ter 7 Probability

34Pa

er 11 X plusmn Number of

enalty goals scoredX sim BnP(X = 0) =nCo 0

n=(1)(1) n= n=



4n = 42 X sim N(55 122)Area of the shaded regionP (X lt 37)= PZ lt= P (Z lt plusmn15)= 0066837 plusmn 55122525166252516

625253516625353 (a) X plusmn Mass of a crab in gX sim N(175 15)Z ==

= 1(b) P(175 lt X lt 190)= Plt Z lt= P (0 lt Z lt 1)= 05 plusmn 01587= 034134P(Z gt plusmn 09) = 08159k = plusmn09190 plusmn 175

15175 plusmn 17515190 plusmn 17515X plusmn microsSPMZOOMplusmnINForm 5 Chapter 8 Probability Di

tribution

plusmn15



0066801587O 108159plusmn090184135Paper 21 (a) X plusmn Number of blue bead

drawnX sim B10X sim B10(i) P(X sup3 3)= 1 plusmn P(X = 0) plusmn P(X = 1) plusmn P(X = 2)= 1 plusmn10C0 0

10plusmn10C1 1 9plusmn10C2

2 8= 07009(ii) Mean = n

= 10 = 3(b) X plusmn Lifes

an of a s

ecies of dogX sim N(12 s2)(i) P(X gt 8) = 90PZ gt

= 09PZ gt= 09plusmn = plusmn1282s = 31201 year

4



splusmn12820109plusmn4s8 plusmn 12s= npq Standard deviation= 149= 10 313213132313

2313231313618(ii) P(10 lt X lt 13)

= Plt Z lt= P(plusmn0641 lt Z lt 0321)= 1 plusmn 02608 plusmn 03741= 036512 (a) X plusmn Number of cu

tomer

requiring a

upplementary cardX sim B7

X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2

5= 01402(b) X plusmn Time taken to settle invoicesX sim N(30 52)(i) P(28 pound X pound 36)= P pound Z pound= P(plusmn04 pound Z pound 12)

= 1plusmn 03446 plusmn 01151= 05403plusmn04 1203446 0115136 plusmn 30528 plusmn 3051125142511

25142511251425112514



25142528050013 plusmn 123120110 plusmn 1231201plusmn0641 032102608 0374136(ii) P(X lt 22)= PZ lt= P(Z lt plusmn16)= 00548Hence the ex

ected number of invoices which aregiven discounts= 00548 220= 1222 plusmn 30

537Pa

er 21 (a) For

article A at maximum velocity= 01 plusmn 2t = 0t == plusmn2 (negative)Hence vmax= 12 + plusmn 2

= 12 m splusmn1(b) sB= 2t3plusmn 7t2plusmn 15tWhen article B returns to OsB= 0

2t3plusmn 7t2plusmn 15t = 0t(2t2plusmn 7t plusmn 15) = 0t(2t + 3)(t plusmn 5) = 0t = 0 plusmn or 5





2vAdt212dvAdt(c) When

article A reverses its directionvA= 012 + t plusmn t2= 0t2plusmn t plusmn 12 = 0(t + 3)(t plusmn 4) = 0t = plusmn3 or 4t = plusmn3 is not acce

tedt = 4vB

=vB= 6t2plusmn 14t plusmn 15aB=aB=12t plusmn 14When t = 4

aB=12(4) plusmn 14 = 34 m splusmn22 (a) a = 12 plusmn 6tv =

a dtv =

(12 plusmn 6t) dtv = 12t plusmn 3t2

+ cWhen t = 0 v = 15 Thus c = 15 v = 12t plusmn 3t2+ 15At maximum velocity= 012 plusmn 6t = 0t = 2When t = 2



v = 12(2) plusmn 3(2)2+ 15 = 27 m splusmn1= plusmn6 (lt 0)Therefore v is a maximumd2vdt2dvdtdvBdtdsBdtSPMZOOMplusmnINForm 5 Cha

ter 9 Motion Along a Straight Line38(b) s =

v dts =

(12t plusmn 3t2+ 15) dts = 6t2plusmn t3+ 15t + cWhen t = 0 s = 0 Thus c = 0 s = 6t2plusmn t

3+ 15tAt maximum dis

lacement= 012t plusmn 3t2+ 15 = 03t2plusmn 12t plusmn 15 = 0t2plusmn 4t plusmn 5 = 0

(t plusmn 5)(t + 1) = 0t = 5 or plusmn1t = plusmn1 is not acce

tedt = 5When t = 5s = 6(5)2plusmn 53+ 15(5) = 100 m



= 12 plusmn 6tWhen t = 5 = 12 plusmn 6(5) = plusmn18Therefore s is a maximumd2sdt2d2sdt2dsdt(c) When the

article travels to the rightv gt 012t plusmn 3t2+ 15 gt 03t2plusmn 12t plusmn 15 lt 0t

2plusmn 4t plusmn 5 lt 0(t + 1)(t plusmn 5) lt 0plusmn1 lt t lt 5Since the values of t cannot be negativetherefore 0 pound t lt 5plusmn1 5 t39Pa

er 21 (a) I 180x + 90y pound 54002x + y pound 60II 3x + 4y pound 120III y pound 2x

(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest

oint on the straight line y = xinside the feasible region R is (2015)xmax= 20 ymax= 15(ii) Profits = 200x + 150yDraw the straight line200x + 150y = 3000The o

timal

oint is (24 12)Hence the maximum

rofit= 200(24) + 150(12)= RM66002 (a) Mixing30x + 10y pound 15 603x + y pound 90Baking40x + 40y pound 26 60x + y pound 40Decorating10x + 30y pound 15 60x + 3y pound 90

23343443xySPMZOOMplusmnINForm 5 Cha

ter 10 Motion Along a Straight Linex 0 30

y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040

(b)(c) (i) When y = 17 xmax= 23(ii) Profits = 5x + 10yDraw the straight line 5x + 10y = 50From the gra

h the o

timal

oint is(15 25)Hence the maximum rofit= 5(15) + 10(25)



= RM325x O201030405030 40607080903x + y = 90x + 3y = 90yx + y = 40Max (15 25)5x + 10y = 50R102317520 50 60 70 80 90



Let x + 4 = ux = u plusmn 4g(u) = (u plusmn 4)2+ 6(u plusmn 4) + 2= u2plusmn 8u + 16 + 6u plusmn 24 + 2= u2plusmn 2u plusmn 6g(x) = x2plusmn 2x plusmn 6(b) fg(4) = f[42plusmn 2(4) plusmn 6]= f(2)= 2 + 4= 65 Let g

plusmn1(x) = yg(y) = x3y + k = xy =y = x plusmngplusmn1(x) = x plusmnIt is given that gplusmn1(x) = mx plusmnHence by comparison

m= and plusmn = plusmn rArrk =5256k31356k3

13k313x plusmn k3SPMZOOMplusmnINForm 4 Chapter 1 Functions


















253523191535










2= 36m =

615x2+ 19x + 6 = 03 3x2+ 4p + 2x = 03x2+ 2x + 4p = 0a = 3 b = 2 c = 4p













5P









13 plusmn k2b

k212k2c

123232b








p21

2p232b


P

per 11 f(x) = 2x2+ 8x + 6= 2(x2+ 4x + 3)= 2[x2+ 4x +

2plusmn 2+ 3]= 2(x2+ 4x + 22





tes of the m

ximum point






But it is given th

t the







= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re




nge of v



+ 5x += 2x2+ 5x + plusmn += 2[













use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal






















253523191535










2= 36m =

615x2+ 19x + 6 = 03 3x2+ 4p + 2x = 03x2+ 2x + 4p = 0a = 3 b = 2 c = 4p













5P









13 plusmn k2b

k212k2c

123232b








p21

2p232b


P

per 11 f(x) = 2x2+ 8x + 6= 2(x2+ 4x + 3)= 2[x2+ 4x +

2plusmn 2+ 3]= 2(x2+ 4x + 22





tes of the m

ximum point






But it is given th

t the







= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re




nge of v



+ 5x += 2x2+ 5x + plusmn += 2[













use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn




ter 5 Trigonometric FunctionsWebsiteZI F505_4th 101508 940 AM 031Pa

er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal
















253523191535










2= 36m =

615x2+ 19x + 6 = 03 3x2+ 4p + 2x = 03x2+ 2x + 4p = 0a = 3 b = 2 c = 4p













5P









13 plusmn k2b

k212k2c

123232b








p21

2p232b


P

per 11 f(x) = 2x2+ 8x + 6= 2(x2+ 4x + 3)= 2[x2+ 4x +

2plusmn 2+ 3]= 2(x2+ 4x + 22





tes of the m

ximum point






But it is given th

t the







= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re




nge of v



+ 5x += 2x2+ 5x + plusmn += 2[













use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal














2= 36m =

615x2+ 19x + 6 = 03 3x2+ 4p + 2x = 03x2+ 2x + 4p = 0a = 3 b = 2 c = 4p













5P









13 plusmn k2b

k212k2c

123232b








p21

2p232b


P

per 11 f(x) = 2x2+ 8x + 6= 2(x2+ 4x + 3)= 2[x2+ 4x +

2plusmn 2+ 3]= 2(x2+ 4x + 22





tes of the m

ximum point






But it is given th

t the







= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re




nge of v



+ 5x += 2x2+ 5x + plusmn += 2[













use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal















5P









13 plusmn k2b

k212k2c

123232b








p21

2p232b


P

per 11 f(x) = 2x2+ 8x + 6= 2(x2+ 4x + 3)= 2[x2+ 4x +

2plusmn 2+ 3]= 2(x2+ 4x + 22





tes of the m

ximum point






But it is given th

t the







= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re




nge of v



+ 5x += 2x2+ 5x + plusmn += 2[













use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








13 plusmn k2b

k212k2c

123232b








p21

2p232b


P

per 11 f(x) = 2x2+ 8x + 6= 2(x2+ 4x + 3)= 2[x2+ 4x +

2plusmn 2+ 3]= 2(x2+ 4x + 22





tes of the m

ximum point






But it is given th

t the







= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re




nge of v



+ 5x += 2x2+ 5x + plusmn += 2[













use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








p21

2p232b


P

per 11 f(x) = 2x2+ 8x + 6= 2(x2+ 4x + 3)= 2[x2+ 4x +

2plusmn 2+ 3]= 2(x2+ 4x + 22





tes of the m

ximum point






But it is given th

t the







= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re




nge of v



+ 5x += 2x2+ 5x + plusmn += 2[













use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal









tes of the m

ximum point






But it is given th

t the







= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re




nge of v



+ 5x += 2x2+ 5x + plusmn += 2[













use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







= 3 b = h c = 27If

qu

dr

tic equ

tion does not h

ve re




nge of v



+ 5x += 2x2+ 5x + plusmn += 2[













use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal

















use p gt 0


k2254254k2 52=



s shownbelow





str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








str




nge of v









=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal










=

241 plusmn5

2 33 1216727





decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal









decim

l pl










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal










pter 5 Indices

nd Log


+ 2x+ 16 (2x plusmn 1)= 2x23+ 2x+ 16

= 8(2x) + 2x+ 8(2x)= (8 + 1 + 8)( 2x)




) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








) = 618(3x) = 63x=3x=3x= 3














8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal















8y3xy3xlog2x3log2







Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal









Hence the equ

tion of the str

ight line whichp




0403x4y3x412plusmn3











4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal














4 + 0212121212400























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal

























(b)3aringx2

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal

























= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn




ter 5 Trigonometric FunctionsWebsiteZI F505_4th 101508 940 AM 31Pa

er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal



















= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM 32Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal













= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







= 8212 plusmn 5 41212r2q12rs12OB = = 44 cm20 15363142180) (3142180

15Pa







3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








3122425sin BOM =881313O








circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal










circumference16Pa













2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal
















2splusmn1 r t


(2x plusmn 5)4

y x


x

1(2x plusmn 5)3

2z x2 z

x



z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







z




2 (a) y =






12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal









12x1x31x212x56x

43x42x3d2ydx2dy

dx12x56x4d2y











232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal















232323792323d

2ydx2792323232

349dydxd2ydx2




2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








2+ 24xyL = 96x2+ 24x L = 96x2+ (shown)3888x162x

212162x2A BCHVG FED






8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal









8c34h[1 + 2(1)]38c34h(1 + 2x)38c3

y

x4h(1 + 2x)3 y

xh(1 + 2x)27776



x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







x3

2L

x238881923888x23888x2 L

x3888x2

L

x3888x

























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal


























I2006(based on 2004)=== 131Thus 100 = 131100 = 131P2006=P2006

= RM393131 300100P2006300P2006P2004131010(150 3) + (130 2) + (120 5)

3 + 2 + 5585 + 105x5 + x(115 3) + (120 2) + 105x3 + 2 + x120100130100P

























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal





























= pr2T4= 32pa + 3 = 32ppr2

h + 3pr2= 32pr2h + 3r2= 32r2(h + 3) = 32 frac14S4

= 104p(2a + 3

) = 104p4a + 6 = 104p4pr2h + 6pr2= 104p2r2




5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








5232r2(2h + 3)r2(h + 3)21242

12 (a) S2= 150T1+ T2= 150a + ar = 150a (1 + r) = 150 frac14T3







1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal










1153215321031 + rr (r plusmn 1)





2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








2 121bx2y2x=

yxk

lg = lg



lg y = x lg k + lg

Y = lg y X = x m = lg k c = lg 1x2(1 5)

5 = 2(1) + qq = 3q = 3

= 2(3) + 3 = 9yx1x2












ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal
















ient2k =2k = 32653k = 1632 log








y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal











y = 0471y2

017

q1q




q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








q1q1y2x +

q1y2=1yx +





x= 3[2plusmn1



g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







g(x)


x]= 3(20)= 603 = x4plusmn 8x3+ 6x2y =

x4

plusmn 8x3+ 6x2






45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







45x55x33x44x55

y x252k22522






y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







y



x=


= plusmn6222244x22x44








y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal











y

x


k = 4 y = x2+ 4



Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







Volume generate

Vx= Volume generate




30(x4+ 8x2+ 16) x plusmn p= p[+ + 16x]30


3553238x33x5




= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








= (12

_ ) = 9


reg+ BCreg+ CDreg=9


_ =plusmn 3







= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal










= plusmn i _ + j _ 4535110



15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







15 (plusmn6)2 + 82 = 1015151523122312231

2121212123434







|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








|a| |b| = 5 3=|b|5353655353













3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal















3161216212162 1

21168383838329






232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal









232721272121414145757(c) AKreg= AL


plusmn2





plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







plusmn 4 = plusmn =722141111111112 +121723272127


721230Pa






2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








2

= plusmnOq 11 +

2plusmn





er 21 (a) LHS ===








oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal












oint

= 12xp2xpx22xpx




101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








101508 940 AM Page 3132Pa

er 1







lication

rinci









101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal












101508 940 AM Page 32

33Pa


ost)= =





ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal









ter 7 Probability

34Pa



n=(1)(1) n= n=







tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal











tribution

plusmn15






2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal










2 8= 07009(ii) Mean = n


an of a s



4




2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal








2313231313618(ii) P(10 lt X lt 13)


tomer

requiring a


X sim B7(i) P(X = 3)=7C3 3



4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







4= 02304(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)=7C0 0 7+7C1 1 6+7C2 2



25142511251425112514





537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal









537Pa

er 21 (a) For










tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal











tedt = 4vB



a dtv =







v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal









v dts =












(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal











(b)x O201030405030 40(20 15)60y = 2x2x + y = 60y = x

3x + 4y = 120y34Max (24 12)200x + 150y = 3000R10 15 20(c) (i) =3x = 4y



y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal







y = xThe furthest


timal





y 60 0x 0 40y 30 0x 0 30y 0 60200 150 01 = 3000x 0 30y 90 0x 0 90y 30 0x 0 40y 40 040


h the o

timal





additional mathematics form 4 and 5 notes (1)

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