Maths Ex 123!@#abc

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Solution to Mathematics Revision Exercise

Q1

Let x= 19.5 .

x= 5.911 11(i)

10x= 59.111 11..(ii)

2.539(i)(ii) x

90

532x 1A

Let y= 82.4 .

y= 4.282 828(iii)

100y= 428.282 828(iv)

42499(iii)(iv) y

99

424y 1A

Let z= 21.2 .

z= 2.121 212. (v)

100z= 212.121 212..(vi)

21099(v)(vi) z

99

210z 1A

21.282.419.5

99210

99424

90532 1M

318

2952 1A

2.

i

ii

84

)72()53(

i

ii

84

3510216

1A

i

i

84

1141

)84)(84(

)84)(1141(

ii

ii

6416

)84)(1141(

ii 1M+1A

80

8844328164

ii

80

284252 i 1A

i80

284

80

252

i20

71

20

63 1A

3.

4(a+ bi)(2 + 5i)

= 4(2a+ 5ai+ 2bi5b)

= 8a+ 20ai+ 8bi20b

= (8a20b) + (20a+ 8b)i 1M+1A

The result of4(a+ bi)(2 + 5i)is a real number. 20a+ 8b= 0 1M+1A

5

2

b

a

a: b= 5:2 1A

4.

(a) 2 is a solution ofx2+ bx+ c= 0.

0)2()2( 2 cb 1M+1A

0)2(2

0)2()2( 2

cib

cibi

0)2()2( ibc 1M

02nda02 bc 2c 0b

1A+1A

(b) (b+ 2i)(3 + ci)

= (0 + 2i)(3 + 2i) 1M

= 2i(3 + 2i)

= 6i+ (4)i2 1M

= i64 1A

5.

(a) (1 + 2i)is a solution ofax2+ bx+4

5c= 0.

a(1 + 2i)2+ b(1 + 2i) +4

5c = 0 1M

a(1 + 4i4) + b+ 2bi+4

5c = 0

a + 4ai4a + b + 2bi+ 4

5c = 0

(3a+ b+4

5c) + (4a+ 2b)i = 0 1M

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04

53 cba and4a+ 2b = 0 1A

2a+ b = 0 1A

(b) (1 + 2i)is a solution ofbx2+ cx(a+ 3) =0.

b(1 + 2i)2+ c(1 + 2i)a3 = 0 1M

b(1 + 4i4) + c+ 2cia3 = 0

b+ 4bi4b+ c+ 2cia3 = 0

(3b+ ca3) + (4b+ 2c)i = 0 1M

3b+ ca3 = 0and4b+ 2c = 0 1A2b+ c = 0 1A

(c)

iv).........(..........02

iii).........(..........033

(ii)....................02

.(i)....................0

4

53

cb

acb

ba

cba

From (iv),2b+ c= 0b=

2

c....(v) 1M

From (ii),2a+ b= 0

a= 2

b

By (v),a=

22

1 c

a=4

c....(vi) 1M

Substitute (v) and (vi) into (iii):

033 acb

03

42

3

cc

c 1M

34

9

0342

3

c

cc

c

3

4c 1A

3

2

3

4

2

1

b 1A

3

1

3

4

4

1

a 1A

6.

37

530424

37

)(530424

136

)6()54(

)6(

)6(

)6(

54

6

54

2

ii

iii

ii

i

i

i

i

i

i

37

2629 i 1M+1A

i37

26

37

29

26

1628

125

210630

)5()5(

)5()26(

5

26

i

ii

ii

ii

i

i

13

814 i 1M+1A

i13

8

13

14

i

i

i

i

5

26

6

54

i

ii

13

8

37

26

13

14

37

29

13

8

13

14

37

26

37

29

i481

42

481

895 1M+1A

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Chapter 2

1.

(a) L1// y-axis

.2isofequationThe 1 xL 1M+1A

(b) L2//x-axis

.6isofequationThe 2 yL 1M+1A

(c)

).6,2(areandofonintersectiofpointtheofscoordinateThe 21 LL

1M+1A

(d) (a, 2a) lies on L1.

Puttingx= ainto the equation of L1, we have

a= 2 1M

4).,(2arescoordinaterequiredThe 1A

2.

(a) Mid-point of PQ=

2

04,

2

30 =

2,

2

3

1A

Slope of PQ=3

4

30

04

1A

Slope of =4

3

3

4

1

1M

The equation of is

2

3

4

32 xy 1M

i.e. 0786 yx 1A

(b) Puttingx= 0 into 6x 8y+ 7 = 0, we have

078)0(6 y 1M

8

7y

.8

7,0areofscoordinateThe

R 1A

(c) Area of PQR

=21 (PR)(OQ) 1M

=

8

74

2

1(3) 1M

=16

114 1A

3.

(a) OABCis a parallelogram.

BC// y-axis

The equation of BCisx= 2. 1M+1A

(b) The equation of the straight line passing

through the origin and C(2 , 4) is

20

)4(0

0

0

x

y

i.e. xy 2 1M+1A

(c) OABCis a parallelogram.

OA= BC= 6 1M

The coordinates ofAare (0 , 6). 1A

(d) The coordinates of Bare (2 , 4 + 6) =

(2 , 2). 1M+1A

The equation ofABis

20

26

2

2

x

y 1M

i.e. 062 yx 1A

4.

(a) Let mbe the slope of the perpendicular

bisector ofAB.

39

51ofSlope

AB

3

2 1M

1

3

2

m 1M

2

3m 1A

i.,2

15,

2

93areofpoint-midtheofscoordinateThe

AB

1M

The equation of the perpendicular

bisector is

)6(2

33 xy 1M

)6(3)3(2 xy

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18362 xy

i.e. 3x2y12 = 0 1A

(b) OC//AB

ABOC ofslopeofSlope 1M

3

2

The equation of OCis

)0(3

20 xy

i.e. xy3

2 1A

(c)

(ii)........................

3

2

(i).............01223

xy

yx

1M

Substituting (ii) into (i), we have

13

36

3613

03649

012)3

2(23

x

x

xx

xx

Substituting1336x into (ii), we have

13

24

13

36

3

2

y

.13

24,

13

36areofscoordinateThe

C

1M+1A

5.

(a) Ais the mid-point of RQ.

Coordinates ofA=

1),2(2

35,

2

04

1A

(b) (i) Slope of RQ= 24

8

04

35

1A

(ii) Slope of PA=2

1

2

1

1M+1A

(iii) The equation of PAis

)2(2

11 xy 1M

i.e. 042 yx 1A

Puttingx= 0 into the equation of PA, we

have

x2y+ 4 = 0

(0)2y+ 4 = 0y= 2

The coordinates of Pare (0 , 2). 1A

(c) (i) The equation of QSis

0

1

x

y =

04

15

1M

0

1

x

y = 1

i.e. 01 yx 1A

(ii) Bis the point of intersection of QSand

PA.

ii).........(..........042

(i)....................01

yx

yx 1M

(i) (ii):

3y5 = 0

y=3

5

Putting y=3

5 into (i), we have

3

2

013

5

x

x

.3

5,

3


B

1M+1A

6.

(a) Slope ofPQ= 320

06

1A

(b) (i) Slope ofRS=3

1

3

1

1M+1A

(ii) The equation ofRSis

)4(310 xy 1M

x+ 4 = 3y

i.e. 043 yx 1A

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(c) (i) Puttingx= 0 into the equation of RS, we

have

0430 y 1M

3

4y

.

3


T 1A

(ii) (1)

Slope ofRQ=2

3

40

06

1A

Slope ofPA=3

2

2

3

1

1M

The equation of PAis

)2(3

20 xy 1M

i.e. 0432 yx 1A

Puttingx= 0, y=3

4 into 2x+ 3y

4, we have:

L.H.S. = 2(0) +

3

43 4 = 0 =

R.H.S.

Tlies onPA. 1A (2) The three altitudes of the

triangle are PA, RSand QO.

The three altitudes of PQR

pass through the same point T. 1A

7.

(a) Slope of L= 3

AC

is perpendicular toL,

slope ofAC slope of L=1

slope ofAC 3 = 1 1M

slope ofAC= 3

1

Blies onAC,

the equation ofACis

y2 =3

1(x3) 1M

3y6 = x+ 3i.e. x+ 3y9 = 0 1A

(b)

ii)...(..............................093

)i.....(........................................3

yx

xy

Substituting (i) into (ii), we have

x+ 3(3x)9 = 0 1M

10x= 9

x= 10

9

Substitutingx=10

9 into (i), we have

y= 3

10

9 =

10

27

.10

27,

10


A 1A

Substituting y= 0 into the equation ofAC, we

have

x+ 3(0) 9 = 0 1M

x= 9

The coordinates of Care (9 , 0). 1A

(c) OC= 9

Distance fromAto OC=10

27

Distance fromBto OC= 2

Area of AOB= area of AOCarea of BOC

= 292

1

10

279

2

1 1M+1A

=20

63 1A

Alternative method:

AO=

22

10

27

10

9

= 10

9

AB=

22

10

272

10

93

=

10

7

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Area of AOB

=10

7

10

9

2

1 1M+1A

=20

63 1A

8.

(a) Slope of CD=1

1 =1

ABis perpendicular to CD,

slope ofAB slope of CD= 1

slope ofAB (1) =1 1M

slope ofAB= 1

The equation ofABis

y5 =x2 1M

i.e. xy+ 3 = 0 1A

(b)

)ii.........(..............................037

)i..(........................................03

yx

yx

(i)(ii): 6y+ 6 = 0 1M

6y=6

y=1

Substituting y=1 into (i), we have

x(1) + 3 = 0

x=4

The coordinates ofBare (4 , 1). 1A

(c) When y= 0,

x7(0) 3 = 0

x= 3

The coordinates ofEare (3 , 0). 1A

Let (c, d) be the coordinates of C. AE= EC

3 =2

2 c 1M

6 = 2 + c

c= 4

0 =2

5 d

d= 5

The coordinates ofCare (4 , 5). 1A

The equation of BCis

)4(

)1(

x

y =

)4(4

)1(5

1M

4

1

x

y=

2

1

2y+ 2 =x4

i.e. x+ 2y+ 6 = 0 1A

9.

(a) Coordinates of C= (3 , 12) 1AThe equation ofACis

12

3

x

y=

123

312

1M

12

3

x

y=

5

3

5y15 = 3x+ 36

i.e. 3x+ 5y51 = 0 1A

(b) Substituting B(2 , 9) into the equation ofAC,

we have

L.H.S. = 3(2) + 5(9) 51 = 0 1M

L.H.S. = R.H.S.

Blies onAC.

i.e. A, Band Care collinear. 1A

Alternative method:

Slope ofAB=12239

=

53 1M

Slope ofAC=5

3

Slope ofAB= slope ofAC

A, Band Care collinear. 1A

(c) ODis perpendicular toAC,

slope of OD slope ofAC=1

slope ofOD

53

=1 1M

slope ofOD=3

5

The equation of ODis y=3

5x.

)ii..(..................................................3

5

)i.........(..............................05153

xy

yx


3x+ 5

x

3

551 = 0 1M

34x= 153

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x=2

9

Substitutingx=2

9 into (ii), we have

y=

2

9

3

5 =

2

15

.2


D 1A

10.

(a) Slope of L1=7

3 , y-intercept of L1=

7

p . 1M

Slope ofL2=

q

4 =

q

4, y-intercept ofL2=

q

11 = q

11

. 1M

L1and L2have infinitely many points of

intersection,

slope of L1= slope of L2and y-intercept of

L1= y-intercept of L2.

7

3 =

q

4 1M

3q= 28

q=328 1A

and7

p =

q

11 1M

7

p =

3

28

11

p=4

33 1A

(b) L1and L2have no points of intersection,

slope of L1= slope of L2and y-intercept of

L1y-intercept ofL2.

.3

28and

4

33 qp 1A+1A

(c) L1and L2have one point of intersection,

slope of L1slope of L2.

.3

28andnumberrealanyis qp 1A+1A

11.

(a) Slope of L=1

1 =1

PTand Lhave no points of intersection,

slope of PT= slope of L

0

5

k

p =1 1M

p= 5 k 1A

(b) AP= PB

22 )2()]7([ pk =

22 )10()3( pk 1M

(k+ 7)2+ (5 k2)2= (k3)2+ (5

k10)2

(k+ 7)2

+ (k+ 3)2

= (k3)2

+ (k5)2

k2+ 14k+ 49 + k26k+ 9 = k26k+ 9 +

k2+ 10k+ 25 1M

4k= 24

k= 6 1A

p= 5 (6) = 11

The coordinates ofPare (6 , 11). 1A

-- ans end --

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Chapter 3

1.

Since andare the roots of the equationx2px

+ q= 0,

+=p 1A

= q 1A

))(( 2233 1M

]3)2)[(( 22

]3))[(( 2 1M

)3( 2 qpp 1M

pqp 33 1A

2.

Since the graph of y= ax2+ (a1)x+ (a+ 2)

touches thex-axis at only one point, the

discriminant = 0. 1A

i.e. (a1)24a(a+ 2) = 0 1M

a22a+ 14a28a = 0

3a2+ 10a1 = 0 1M

a =

(3)2

)1)(3(41010 2 1M

=

6

)12(10010

=6

11210 1M

= 3.43 or

0.097 2,cor. to 3 sig. fig.1A+1A

3.

(a) 9(x2) = 2xx2

9x18 = 2xx2

9x18 2x+x2 = 0

x2+ 7x18 = 0 1M

(x2)(x+ 9) = 0 1M

x2 = 0 or x+ 9 =

0

x = 2 or x =

9 1A+1A

(b) 07347 2 xx

0)37)(7( xx 1M+1A

07x or

037 x

x= 7

or x=7

3

1A+1A

4.(a) 9x2kx+ 1 =x

9x2kxx+ 1 = 0

9x2(k+ 1)x+ 1 = 0 1M

Since the equation has two equal real roots,

the discriminant = 0. 1A

i.e. [(k+ 1)]24(9)(1) = 0 1M

(k+ 1)236 = 0

(k+ 1)2

= 36k+ 1 = 6

k= 5 or 7

1A+1A

(b) When k=7, the equation is 9x2+ 6x+ 1 = 0.

1M

)(3

1

)9(2

6repeatedx 1A+1A

5.

Since andare the two roots of the equationx2

+ 3x4 = 0, we have

+ = 3 1A

= 4 1A

22 2222

2)( 2 1M

89

)4(2)3( 2

17 1A

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9/24

25

)4(217

2)(

2)(

22

222

)(5

))((22 1M

)5)(3( 15 1A

Sum of roots = 17 + (15)

= 2 1A

Product of roots = 17(15)

= 255 1A

The required equation isx22x255 = 0. 1A

6.

(a) Since ,are the roots of the equation,

3

1

3

)1( kk

1A

43

12

1A

922

9(b)

22

22

92)( 2 1M

9)4(23

12

k 1M

912

9)1(

989

)1(

2

2

2

kk

k

k

0822 kk 1M

0)2)(4( kk 24 kork 1A+1A

(c) Substituting k= 4 into the equation, we have

01233 2 xx

042 xx 1M

)1(2

)4)(1(411 2 x 1M

2171

2

171

2

117

xorx 1A+1A

7.

(a) Let be the smaller root, and 2be the other

root. 1A

1

)4(2

k 1M

34

43

k

k

.3

4isrootstheofOne

k 1A

9

52

3

22(b) 2 k 1M

9

52

3

22 22 k

9

52

3

2

3

42 2

2

k

k 1M

04282

263168

263)4(

9

52

3

2)4(

9

2

2

22

22

22

kk

kkk

kk

kk

02142 kk 1M0)3)(7( kk

37 kork 1A+1A

(c) (i) Substituting k= 3 into the equation, we

have

09

22 xx 1M

a+ b= 1

ab=9

2

abbaba 2)( 222

9

1

9

81

4)(42

2

22

abbaabbaba

ab=3

1 ( a b)

))((22 bababa 1M

311

3

1 1A

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(ii) From (a), we know the smaller root is

3

4 k(*)

substituting k= 3 into (*), we have

3

1

3

34

1M

3

2

,3

1

ab 1A+1A

8.

(a) Whenx= 0, y= 10,

10 = a(0)2+ b(0) + c

c= 10

Whenx= 2, y= 0,

0 = a(2)2+ b(2) 100 = 4a+ 2b10...........(i)

Whenx= 5, y= 0,

0 = a(5)2+ b(5) 10

0 = 25a+ 5b10

0 = 5a+ b2...(ii)

)25(210240(ii)2(i) baba 1M

66

660421010240

a

ababa

1a 1A

2)1(50 b

7b 1A

(b) (i) Consider ax2+ bx+ c= 2.

Substitute the values of a, band cfrom

(a) into the equation.

i.e. x2+ 7x10 = 2

x2+ 7x12 = 0

x27x+ 12 = 0

7 1A

12 1A

42

2)(onsiderC

22

222

4)( 2 1M

1

)12(4)7( 2

)(1 1A

)2)((

))(((ii)

22

2233

]))[(( 2 1M

)127)(1( 2

37 1A

0127

2107(iii)

2

2

xx

xx

01272 xx 1M

)1(2

)12)(1(4)7()7(

2 x

1M

2

17

34or 1A+1A

9.

(a) Since andare the roots of the equation,

42

8

1A

23 1A

22(i))(b 2222

2)( 2 1M

2

32)4( 2 1M

316

= 19 1A

42

2)(onsiderC(ii)

22

222

4)( 2 1M

2

34)4( 2 1M

616

= 22

)(22 1A

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11/24

))(((iii) 22 1M

)22)(4( 1M

224 1A

(c) From (b), we know that 1922 ,

22422 .

Sum of roots = )22419( 1M

Product of roots = 2276 1M

The quadratic equation with roots22

and 22 is

02276)22419(2 xx 1A

10.

(a) Consider the discriminant = 524(1)(2) 1M

= 25 + 8

= 33 0 1A

Since 0,

the equation has two distinct real roots. 1A

Therefore andare distinct real

numbers. 1A

(b) (i) 5 1A2 1A

1339)13)(13(

1)(39 1M

1)5(3)2(9

1M

11518 32 1A

(ii) ))(( 2233

)32)(( 22

]3))[(( 2 1M

)]2(3)5[(5 2 1M

)625(5

=155 1A

025(c) 2 xx

)1(2

)2)(1(455 2 x 1M

2

533

2

533

2

335

xorx

2

533

2

533

1A+1A

Special Topic : Applications of Quadratic Equation

1.

Let the length of BDbexcm.

BC= BD+ 4 =x+ 4

62

)4( xx 1M+1A

x2+ 4x= 12

x2+ 4x12 = 0 1M

(x+ 6)(x2) = 0 1M

x+ 6 =0 or x2 = 0 1M

x=6 (rejected) or x= 2

Area of squareABDE= 22cm2= 4 cm2

Area of trapeziumACDE= (4 + 6) cm

2

2

cm10 1M+1A

2.

(a) By using Pythagoras' theorem,

height of the top end above the ground

m610 22 1M

= 8 m 1A

(b) Since the length of the ladder is unchanged,

by using Pythagoras' theorem,

(8x)2+ (6 +x)2= 102 1M+1A

x216x+ 64 +x2+ 12x+ 36 = 100

2x24x= 0 1A

x(x2) = 0 1M

x=0 or x2 = 0 1M

x=0 (rejected) or x= 2 1A

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3.

(a) (i) Sides of hexagon = cm)232( x 1A

(ii) By using Pythagoras' theorem,

x2+x2= (32 2x)2 1M

2x2= 4x2128x+ 1 024

2x2128x+ 1 024 = 0

)2(2

)0241)(2(4128128 2 x 1M

= 37.9 , cor. to 3 sig.

fig. or 54.6 (rejected) 9.372 6 1A

(b) (i) Width of rectangle = 2xcm

= 2 9.372 6 cm

= 18.745 2 cm 1A

Area of rectangle = 32 18.745 2

cm2 1M2cm600 , cor. to 3

sig. fig. 599.85 1A

(ii) Area of right-angled triangle =2

2x

cm2

22

cm2

6372.9

sig. fig.cor. to 5,cm923.43 2 1A

Area of hexagon = (599.85 4

43.923) 2cm 1M

sig. fig.cor. to 3,cm424 2 1A

4.

(a) BM= (x+ 2) cm

BC= 2(x+ 2) cm =AB 1M

cm)2(3

cm)2()]2(2[

2

22

22

x

xx

BMABAM

cm)2(3 x 1A

Area of ABC= 2cm2

)2(32)2( xx

1M22 cm)2(3 x 1A

22(b) BDBGGD 1M

cm3

cm)2(

2

22

x

xx

cm3x 1A

(c) Area of DEFG= x322 cm22cm34 x 1A

xx 342)2(3 2 1M

xxx

xx

844

8)2(

2

2

0442 xx 1M

0)2(

2

x x= 2 (repeated) 1A

5.

(a) InABCandADE,

CBA= EDA= 90

BAC= DAE

ABC~ADE

given

vert. opp.s

AAA

(b) LetAD=xcm.

AC= (13 x) cm 1A

AB=AE= )12(2

1cm = 6 cm 1A

From (a),ABC~ADE.

AD

AB =

AE

AC 1M

x

6 =

6

13 x

36 = 13xx2

x213x+ 36 = 0

(x4)(x9) = 0 1M

x = 4 or x= 9

AD

7/25/2019 Maths Ex 123!@#abc

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6.

(a) Volume = 3 900 cm3

2

)12)(28( x(3x2) = 3 900 1M

2

)23)(4(2 xx= 325

(x+ 4)(3x2) = 3253x2+ 10x8 = 325

3x2+ 10x333 = 0 1M

(x9)(3x+ 37) = 0 1M

x= 9 orx=3

37

(rejected) 1A

(b) Whenx= 9,

DC= 2(9) cm = 18 cmCH= [3(9) 2] cm = 25 cm

DX=2

818 cm = 5 cm 1A

InADX,AD = 22 DXAX

= 22 512 cm 1M

= 13 cm

The required total surface area

=

)2(2

)12)(188()25)(1318138(

cm2 1M

= 1 612 cm2 1A

7.

(a) 4x2+ 404x1 035 = 0

(2x5)(2x+ 207) = 0 1M

x= 2

5 orx= 2

207

1A+1A

Alternative method:

4x2+ 404x1 035 = 0

x=

)4(2

)0351)(4(4404404 2 1M

=8

424404

=8

424404 or

8

424404

=2

5 or

2

207 1A+1A

(b) Letx% be the annual interest rate of the first

year.40 000(1 +x%)[1 + (x+ 1)%] 40 000(1 +x%)

= 1 435 1M+1M+1A

40 000(1 +x%)[1 + (x+ 1)% 1] = 1 435

40 000

100

1

1001

xx = 1 435 1M

4(100 +x)(1 +x) = 1 435

4(100 + 101x+x2) = 1 435

400 + 404x+ 4x2 = 1 435

4x2+ 404x1 035 = 0

From (a),

x=2

5 orx=

2

207 (rejected)

The annual interest rate of the first year

is 2.5%. 1A

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Chapter 4

1.

(a) 2f(1) = 6

2[a(1)2+ c] = 6 1M

a+ c= 3.(i)

f(2) = 6

a(2)2+ c= 6 1M

4a+ c= 6.(ii)

(ii)(i): 3a= 3

a= 1 1A

Substituting a= 1 into (i),

1 + c = 3

c= 2 1A

(b) f(a+ c) =f(3) = (3)2+ 2= 11

f(a) =f(1) = (1)2+ 2

= 3

f(c) =f(2) = (2)2+ 2

= 6

)()(

)(

cfaf

caf

=

63

11

1M

=911 1A

2.

(a) If f(x) =g(x),

then x2x= 4x6 1M

x25x+ 6 = 0 1M

(x2)(x3) = 0

x= 2 or x=

3 1A+1A

(b) If H(x) = 12,

then f(x)xg(x) = 12 1M

(x2x) x(4x6) =12

x2x4x2+ 6x = 12

3x2+ 5x+ 12 = 0

3x25x12 = 0

(x3)(3x+ 4) = 0 1M

x = 3 or x=4

3

1A+1A

3.

(a) x 1 (given)

The domain of E(x) is all real numbers

greater than or equal to zero and smaller

than 1. 1A

1 > E(x) 0

Jenny may not be able to escape

successfully. 1A

20

19

20

11)20((b) E 1A

(c) (i) The difference:

30

2

2

11)2()2(

SE 1M

15

1

2

1

30

13 1A

(ii) The difference:

30

10

10

11)10()10(

SE 1M

3

1

10

9

30

17 1A

(d) S(x) = E(x)

30

x =

x

11 1M

x2 = 30x30

x

2

30x+ 30 = 0x = 28.964 or x= 1.036, cor.

to 3 d.p.

When there are 28.964 or 1.036 minutes

left, the two functions will give the same

chance that Jenny can escape

successfully. 1A+1A

7/25/2019 Maths Ex 123!@#abc

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4.

(a) C(x) = $(450 000 + 2 000x) 1A

(b) (i)

10

2

%)81(

10)]9003(0002000450[509003)9003(5003

$)9003(

P

1M

10%)81(

10)8000955($

77.41360323$ ,cor. to 2

d.p. 1A

(ii)

10

2

%)81(

10)]0006(0002000450[50

0006)0006(5003

$)0006(

P

1M

10%)81(

10)0003087($

12.50068236$ ,cor. to 2

d.p. 1A

(c) (i) )9003(P

10

102

%)121(

10)04.01)](9003(0002000450[50

9003)0093(5003

$

1M

10%)121(

10)65.7841331($

13.4836503$ , cor. to 2 d.p. 1A

(ii) )0006(P

10

102

%)121(

10)04.01)](0006(0002000450[50

0006)0006(5003

$

1M

10%)121(

53.58650918$

48.5919595$ ,cor. to 2 d.p. 1A

(d) (i) )9003(P

10

102

%)121(

3(0002000450[)04.01(50

9003)9003(5003

$

1M

10%)121(

27.28843075$

05.53428624$ ,cor. to 2 d.p. 1A

(ii) )0006(P

10

102

%)121(

(0002000450[)04.01(50

0006)0006(5003

$

1M

10%)121(

5.127903115$

09.70531737$ , cor. to 2 d.p. 1A

5.(a) )25000050$()( xxC 1A

(b) Profit function:

)25000050(100

0001$

)]()($[)(

2

xx

x

xCxIxP

00050

100750$

2x

x 1A

Domains:

Sincexis the number of cattle,x 0 .

The domain of the cost function C(x) is

all real numbers greater than or equal

1A

The domain of the income function I(x)

all real numbers greater than or equal to

1A The domain of the profit function P(x)

all real numbers greater than or equal

7/25/2019 Maths Ex 123!@#abc

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1A

(c) (i) C(500) = $[50 000 + 250(500)]

= $175 000

500

000175$costaverage 1M

350$ 1A

(ii)

100

)500()500(0001$)500(

2

I

500497$ 1A

(iii) average profit 500

)500(P

500

00050100

)500()500(750

$

2

1M

500

500322$

645$ 1A

(d) C(x) = $[50 000(1 0.5) + 250(1 0.4)x] 1M

= )15000025$( x 1A

(e) The profit in future ten years

1012)15000025(100

0001$2

x

xx

120)550(15000025100

)550()550(0001$

2

1A

)00000010$(00073752$

120)500107975546$(

The farm can break even. 1A

(f) C(x) =

1012

0000001015000025$ x 1M

=

x150

3

000325$ 1A

Chapter 5

1.

(a) Putting (2, 0) into the equationy= ax2+ 2x+

c, we get

0 = a(2)2+ 2(2) + c

0 = 4a

4 + c (i) 1A

Putting (4, 0) into the equationy= ax2+ 2x+

c, we get

0 = a(4)2+ 2(4) + c

0 = 16a+ 8 + c ..(ii) 1A

(ii) (i), we get

0 = 12a+ 12

a =1 1A

Putting a =

1 into (i), we get0 = 4(1)4 + c

0 =8 + c

c = 8 1A

(b) y =x2+ 2x+ 8

=(x22x) + 8

=(x22x+ 11) + 8

=

(x2

2x+ 1) + 1 + 8=(x1)2+ 9 1M+1A

The maximum value ofyis 9. 1M+1A

2.

(a) f(x) = 2x2+ 16x28

= 2(x2+ 8x) 28

= 2(x2+ 8x+ 4242) 28

= 2(x+ 4)232 28

= 2(x+ 4)260 1M+1A

The minimum value of the function is

60. 1A

The value of the function is negative for

somex. 1A

(b) f(x) = 3x26x+ 10

= 3(x22x) + 10

= 3(x2 2x+ 1212) + 10

= 3(x1)23 + 10

= 3(x1)2+ 7 1M+1A

The minimum value of the function is

7/25/2019 Maths Ex 123!@#abc

17/24

7. 1A

The value of the function is positive for

all real values ofx. 1A

3.

(a) Whenx= 0,y= 4(0) a(0)2= 0. 1A

Yes, the graph passes through the

origin. 1A

(b) (i) [y= 4xax2is a quadratic function.

Since the coefficient ofx2< 0, the

parabola opens downward. Its axis of

symmetry isx= 2, and the graph passes

through the origin.]

The required graph is:

1A+1A+1A

(ii) Let (h, k) be the coordinates of the

vertex.

The axis of symmetry isx= 2.

h= 2

If the maximum value ofyis 2, then k=

2.

The coordinates of the vertex are (2 ,

2). 1A

(c) Substituting (2 , 2) intoy= 4xax2,

2 = 4(2) a(2)2 1M

= 8 4a

4a = 6

a =2

3 1A

Special Topic 2 : Applications of Graphs of

Quadratic Functions

1.

(a) f(x) = 2x2+ 16x+ 140

= 2(x28x) + 140

= 2(x28x+ 4242) + 140

= 2(x40)2+ 32 + 140

= 2(x4)2+ 172 1A

The maximum height that the athlete can

achieve is 172 cm. 1M+1A

(b) Refer to the figure below. If the athlete plans

to achieve the maximum height, the jumping

point should be a point on thex-axis, and thebar should be placed at the axis of symmetry

of the function.

Whenf(x) = 0, 172)4(20 2 x

1M

0)864)(864(

86)4(

172)4(2

2

2

xx

x

x

864486 xorx 1A

ts t a r thewhereplacetheandbarebetween thDistance

cm)]864(864[2

1 1M

cm86 1A

7/25/2019 Maths Ex 123!@#abc

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2.

(a) Substituting t= 0 into h(t),

h(0) = 20(0) 10(0)2+ 30 1M

= 30

The height of the building is 30 m. 1A

(b) h(t) = 20t10t2+ 30

= 10(t22t) + 30

= 10(t22t+ 1212) + 30

= 10(t1)2+ 10 + 30

= 10(t1)2+ 40 1A

The maximum height of the water bomb

above the ground is 40 m. 1M+1A

(c) When h(t) = 0,

20t10t2+ 30 = 0 1M

t22t3 = 0

(t3)(t+ 1) = 0

t= 3 or t= 1 (rejected)

The time for the water bomb to reach the

ground is 3 seconds. 1M+1A

3.

(a) Substitutingx= 6 into h(x),

h(6) = 29

2(6)2+ 16(6) + k = 29 1M

k = 29 96 + 72

= 5 1A

(b) From (a), h(x) = 2x2+ 16x+ 5

= 2(x28x) + 5

= 2(x28x+ 4242) + 5

= 2(x4)2+ 32 + 5

= 2(x4)2+ 37 1M+1A

The distance between the fire engine and

the building is 4 m

so that the height of water column is

maximum. 1M+1A

(c) From (b), the maximum height of the water

column is 37 m.

The fireman cannot fight the fire. 1M+1A

4.

(a) Substitutingx= 12 intoy=x2+ 40x,

y= (12)2+ 40(12) 1M

= 336

The distance of Babove the ground is 336

cm. 1A

(b) y = x2+ 40x

= (x240x+ 202202)

= (x20)2+ 400 1A

The coordinates of the vertex of y=x2+

40xare (20 , 400). 1A

The height that Bincreases is (400 336)

cm = 64 cm. 1M+1A

(c) Acannot throw the water bomb to B. 1A

IfAchanges his/her throwing position, as the

throwing locus is unchanged,

the water bomb cannot pass through the

rocky column Q. If the throwing position

is unchanged, the water bomb will hit the

ground atx = 40,but not at the position of B. 1A+1A

5.

(a) Substituting y= 0 into y= 2x2+ 6x , 1M

i.e. 2x2+ 6x8 = 0

x2+ 3x4 = 0

(x1)(x+ 4) = 0

x = 1 or x = 4

The coordinates of pointsAand Bare

(4 , 0) and (1 , 0) respectively. 1A+1A

82

3

2

332

8)3(2

862

22

2

2

2

xx

xx

xxy

2

25

2

3

2

2

x 1A

.2

25,

2

3reapointofscoordinateThe

C

7/25/2019 Maths Ex 123!@#abc

19/24

1A

(b) Substitutingx= 2a, y= 12ainto y= 2x2+ 6x

8,

12a= 2(2a)2+ 6(2a) 8 1M

12a= 24a2+ 12a8

8a2= 8

a2= 1

a= 1 or a= 1 (rejected) 1A

The coordinates of Dare (2 , 12). 1A

(c) Length ofAB= 1 (4) = 5

ConsiderADB.

Area of ADB= 30)512(2

1 1A

Consider

ACB.

Area of ACB=4

1255

2

25

2

1

1A

4

245

4

12530alquarilatertheofArea ABCD 1A

6.

(a) (i) By Pythagorastheorem,

PQ2 =AQ2+AP2 1M

= 222 cm])2()2([ xx

= 4x2cm2

PQ= cm2x 1A

(ii) By Pythagorastheorem,

QC2= BC2+ BQ2 1M

= 222 cm])210(10[ x

= 22 cm)2220200( xx

Let the mid-point of Pand Qbe M, i.e.

PM= MQ=xcm. 1A

By Pythagorastheorem,

QC2= MC2+ MQ2

22 MQQCMC 1M

cm)210(

cm220200

cm2220200

2

22

x

xx

xxx

cm.)210(isheightrequiredThe x

1A

(b) Area of QCP = MCPQ 2

1 1M

= 2cm)210(22

1xx

= 2cm)210( xx 1A

(c) Area of quadrilateral PQCD

= area of trapeziumADCQarea of APQ

= 22 cm)2(2

1)102(

2

10

xx 1M

= 22 cm)5025( xx

= 22

cm2

2550

2

25

x

= 22

cm2

125

2

25

x 1A

.cm2

125isralquadrilateofareamaximumThe 2PQCD

1M+1A

7.

(a) (i) y= h2 x2+ 16x

2

1

2

1

2

2

2

16

2

16

2

162

2

162

hhh

h

h

h

xx

xx

2

22

1 2

16

2

162

hh

hx 1A

2

16cm

2

162travelsballthat thedistanceHorizontal

1 hh

1M+1A

(ii) By (a)(i),

cm2

256cm

2

16balltheofheightmaximum

22

2

hh

1M+1A

(b) (i) Substituting h= 4 intoh

2

16,

by the result of (a),

horizontal distance that the ball travels inthe 4th bounce

cm1cm2

164

1M+1A

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(ii) Substituting h= 4 into22

256h

,

cm4cm2

256bounce4thin theballtheofheightmaximum

24

1M+1A

(c) Total horizontal distance that the bouncing

ball travels = 21 2

16

2

16 1A

Substituting a= 16, r=2

1 into

r

ar

1,

2

11

2

16

distancetotal

1M

= 16

The total distance that the bouncing ball

travels is 16 cm. 1A

Chapter 6

1.

Whenf(x) is divided byx2, the remainder is 8.

i.e. 8)2( f . 1A

Whenf(x) is divided byx+ 1, the remainder is 10.

i.e. 10)1( f . 1A

Whenf(x) is divided byx25x+ 4, the remainder is

0 and the quotient is bax .

))(45()( 2 baxxxxf 1A

Puttingx= 2, we get

)2(28

)2(]4)2(5)2[()2( 2

ba

baf

(i).......................42 ba 1APuttingx= 1, we get

)(1010

))(451(10

)1(]4)1(5)1[()1( 2

ba

ba

baf

(ii).......................1 ba 1A

Solving (i) and (ii), we get

a= 1, b= 2. 1A+1A

2.

(a) Let kxkxkxxf 3)3()12(2)( 23 .

f(1) = 2(1)3+ (2k1)(1)2(3 + k)(1)

3k 1M

=2 + 2k1 + 3 + k3k

= 0 1A

For any integer k,x+ 1 is a factor of the

polynomial. 1A

(b)

kkx

kkxxkxk

xkxk

xx

kxkxkxx

kxkx

33

33)32()32(

)3()32(

22

3)3()12(21

3)32(2

2

2

23

23

2

1

M

1

M

kxkxkx 3)3()12(2 23

]3)32(2)[1( 2 kxkxx 1M

)32)()(1( xkxx 1M+1A

3.

(a) f(2) = 0 1A

a(2)3+ 4(2)27(2) + b = 0

8a+ 16 14 + b = 0

8a+ b = 2.(i)

f(1) = 12 1A

a(1)3+ 4(1)27(1) + b = 12

a+ 4 7 + b = 12a+ b = 9(ii)

(i) (ii):

7a= 7

a= 1 1A

Substituting a= 1 into (ii),

1 + b = 9

b = 10 1A

(b) f(x) =x3+ 4x27x10

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21/24

56

105

105

126

76

2

10742

2

2

2

23

23

xx

x

x

xx

xx

xx

xxxx

f(x) = (x2)(x2+ 6x+ 5) 1M+1A

= )1)(5)(2( xxx 1A

(c) f(x) = 0

(x2)(x+ 5)(x+ 1) = 0

x2 = 0 or x+ 5 = 0 or x+ 1 = 0

x=

2 or x=

5

or x=

1 1A+1A+1A

4.

(a) Dividend = quotient divisor + remainder

5)3()()( xxgxf 1M

5)3)(27( 2 xxx

5621327 223 xxxxx 1M

11910 23 xxx

11910)( 23 xxxxf 1A

(b)

14

124

24

3

273

4

2

2

x

x

xx

xxx

x

1M

Quotient = 4x , remainder = 14 1A+1A

(c) By (b), g(x) = (x+ 3)(x+ 4)14 1A

By (a),

5)()3()( xgxxf 1M

5)3(141)3()3)(3(

5)3(14)4)(3)(3(

514)4)(3()3(

xxxx

xxxx

xxx

5)3(14)3()3()( 23 xxxxf

1M+1A

5.

(a) Remainder =

kg

1 1A

11

611

2

kkk

k

261

2 kk

1A

(b) From (a), we have:

Remainder = 261

2 kk

=2

9

k

1 6k+ 2k2 = 9

2k26k+ 1 9 = 0

2k26k8 = 0

2(k4)(k+ 1) = 0 1Mk4 = 0 or k+ 1 = 0

k = 4 or k=

1 1A+1A

6.

3

1510

393

123

3

120013

2

2

23

232

x

x

xx

xx

xxx

xxxxx

1M

Quotient =x3, remainder = 10x15 1A+1A

7.

433

72

844

164

633

03

633

107032

2

2

2

23

23

234

2342

xx

x

xx

xx

xxx

xxx

xxx

xxxxxx

1M

1M

Quotient = 3x2+ 3x+ 4, remainder =2x

7 1A+1A

7/25/2019 Maths Ex 123!@#abc

22/24

8.

(a) The remainder is 6whenf(x) is divided

byx1,

f(1) = 6

i.e. (11)(13)(1+ a) + b(1) + c= 6 1M

b+ c=

6..................................(i)

The remainder is 4whenf(x) is divided

byx3,

f(3) = 4

i.e. (31)(33)(3+ a) + b(3) + c= 4 1M

3b+ c=

4.................................(ii)

(ii) (i): 2b= 2 1M

b=

1 1ASubstituting b=1 into (i), we have

1 + c = 6

c = 7 1A

(b) f(x) = (x1)(x3)(x + a)x+ 7

= (x24x+ 3)(x + a)x+ 7 1M

Whenf(x) is divided byx24x+ 3, the

remainder is

x+ 7. f(x) is not divisible byx24x+ 3. 1A

9.

(a) Letf(x) =x3+ (1 a)x2+ 5ax6a2.

f(a) = a3+ (1 a)a2+ 5a(a)6a2 1M

= a3+ a2a3+ 5a26a2

= 0

For any constant a,xais a factor ofx3+

(1a)x2+ 5ax6a2. 1A

(b) Let a= 100.

From (a), we know that

x100 is a factor of the polynomialx3+ (1

100)x2+ 5(100)x6(100)2, 1A

i.e. a factor ofx399x2+ 500x60 000.

Using long division, we have

x3

99x2+ 500x

60 000 = (x

100)(x2+x+

600) 1M+1A

10.

(a) x+ 2 is a factor off(x),

f(2) = 0

i.e. (2)3+ 3(2)2(p2+ 2)(2) + 8p= 0 1M

8 + 12 + 2p2+ 4 + 8p= 0

2p2+ 8p+ 8 = 0

p2+ 4p+ 4 = 0

(p+ 2)2= 0 1M

p =2

(repeated) 1A

(b) From (a),

f(x) =x3+ 3x2[(2)2+ 2]x+ 8(2)

=x3+ 3x26x16

The required remainder=f(4)

=(4)3+ 3(4)26(4) 16 1M

= 64 + 48 24 16

= 72 1A

(c) Using long division, we have

f(x) = (x+ 2)(x2+x8) 1M+1A

f(x) = 0(x+ 2)(x2+x8) = 0

x= 2 1A

or x= )1(2

)8)(1(4)1(1 2 1M

x= 2

331

x

=2

331 orx =

2

331 1A+1A

11.

(a) The remainder is awhenf(x) is divided

by 2x+ 3,

2

3f =

a

i.e. 4

3

23

+ 4b

2

23

23 + 2a

10 =

a 1M

7/25/2019 Maths Ex 123!@#abc

23/24

2

27 + 9b+

2

3 + 2a10 =

a

a+ 9b22 =

0............................(i)

f(x) is divisible byx+ b,

f(b) = 0

i.e. 4(b)3+ 4b(b)2(b)+ 2a10 = 0 1M

4b3+ 4b3+ b+ 2a10 = 0

2a+b10 =

0....................................(ii)

From (i), a= 22

9b......................................................................(iii)

Substituting (iii) into (ii), we have

2(229b) +b10 = 0 1M

4418b+b10 = 0

17b=34

b= 2 1A

Substituting b=2 into (iii), we have

a= 229(2)

= 4 1A

(b) From (a),

f(x) = 4x3+ 4(2)x2x + 2(4) 10

= 4x3+ 8x2x2

f(x) is divisible byx+ 2.


f(x) = (x+ 2)(4x21) 1M+1A

= (x+ 2)(2x+ 1)(2x1) 1A

12.

(a) 2x2

3x5 = (2x5)(x+ 1) f(x) is divisible by 2x23x5,

f(x) is divisible by 2x5 andx+ 1. 1M

2

5f = 0

102

5

2

5

2

523

nm = 0 1M

2

25

4

25

8

125 nm = 0

5m2n+ 4 =

0..............................(i)

f(1) = 0

m(1)3n(1)2+ (1)+ 10 = 0 1M

mn1 + 10 = 0

n= 9

m.................(ii)


5m2(9 m) + 4 = 0 1M

5m18 + 2m+ 4 = 0

7m= 14

m= 2 1A

Substituting m= 2 into (ii), we have

n= 9 2

= 7 1A

(b) From (a),f(x) = 2x37x2+x + 10.

The required remainder=f(5)

= 2(5)37(5)2+ 5+ 10 1M

= 250 175 + 5 + 10

= 90 1A

13.

(a) h(x) = 2f(x) g(x)

= 2(x+ 2)P(x)

(x+ 2)Q(x), whereP(x)and Q(x) are polynomials. 1M

= (x+ 2)[2P(x)Q(x)]

h(x) is divisible byx+ 2. 1A

(b) f(x) is divisible byx+ 2,

f(2) = 0

i.e. (2)3+ b(2)2(2)a= 0 1M

8 + 4b+ 2a= 0

a

4b+ 6 =

0........................(i)

g(x) is divisible byx+ 2,

g(2) = 0

i.e. a(2)3+ 9(2)2+ 7(2)3b= 0 1M

8a+ 36 143b= 0

8a+ 3b22 =

0.................(ii)

(i) 8: 8a32b+ 48 =0...................................(iii)

(ii)(iii): 35b70 = 0 1M

35b= 70

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b = 2 1A

Substituting b=2 into (i), we have

a4(2) + 6 = 0

a= 2 1A

h(x) = 2(x3+ 2x2x2)(2x3+ 9x2+ 7x6)

= 5x29x+ 2 1A

(c) h(x) = 0

5x29x+ 2 = 0

5x2+ 9x2 = 0

(5x1)(x+ 2) = 0 1M

x=5

1 orx=2 1A+1A

14.(a) 4x3 is a factor off(x),

4

3f = 0

i.e. 4

3

4

3

+ m

2

4

3

n

4

3 + 6 = 0 1M

16

27 +

16

9m

4

3n + 6 = 0

9m12n+ 123 = 0

3m

4n+ 41 = 0

4

413

mn 1A

(b) n< 14

4

413 m < 14

3m+ 41 < 56 1M

3m< 15

m< 5 1A

(c) (i) From (b), mis a positive integer less than

5.

The possible values of mare 1, 2, 3

and 4.

When m= 1, n=4

41)1(3

= 11.

Whenm= 2, n=

4

41)2(3 = 11.75.

Whenm= 3, n=

4

41)3(3 = 12.5.

Whenm= 4, n=

4

41)4(3 = 13.25.

1M

m= 1, n= 11 1A+1A

(ii) From (c)(i),

f(x) = 4x3+x211x + 6


f(x) = (4x3)(x2+x2) 1M+1A

= (4x3)(x+ 2)(x1) 1A

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Maths Ex 123!@#abc