sol_5a01.pdf

38 Solutions

Chapter 1 More about Equations

Skills Assessment (P.4)

1. From the figure, the point of intersection is

(1 , 2.5).

` The solution of the simultaneous equations

is x = 1, y = 2.5.

2. x y

y x

2 1

3 3

=

=

+

-*

Substituting (ii) into (i), we have

2x = 3x - 3 + 1

x = 2

Substituting x = 2 into (ii), we have

y = 3(2) - 3 = 3


is x = 2, y = 3.

3. x y

x y

3

3

4 4

1

- =

+ =*

(i) + (ii) : 5x = 5

x = 1

Substituting x = 1 into (ii), we have

1 + 3y = 1

y = 0


is x = 1, y = 0.

4. x y

x y

2 14

4 5 0

- =-

+ =*

(i) # 2 : 4x - 2y = -28 .............................. (iii)

(ii) - (iii) : 7y = 28

y = 4

Substituting y = 4 into (i), we have

2x - 4 = -14

2x = -10

x = -5


is x = -5, y = 4.

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

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

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

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

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

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

5. x2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

` x = 4 or x = 1-

6. x2 - 13x + 40 = 0

(x - 5)(x - 8) = 0

` x = 5 or x = 8

7. x2 - x - 6 = 0

x = 2

( )

( ) ( ) ( )( )

2 1

1 1 4 1 6!- - - - -

= 1 25

2!

= 5

21 !

= 2- or 3

8. 2x2 - 15x - 8 = 0

x = 2

( )

( ) ( ) ( )( )

2 2

15 15 4 2 8!- - - - -

= 15 289

4!

= 17

415 !

= 21

- or 8

9. 9x2 - 12x + 4 = 0

x = 2

( )

( ) ( ) ( )( )

2

12 12 4 9 4

9

!- - - -

= 0

1812 !

= 32

(repeated)

10. From the graph, the x-intercepts are -1 and 3.

` The roots of the equation are -1 and 3.

01 NSS TM 5A (E)-2P-KJ.indd 38 2010/7/20 4:36:17 PM

Chapter 1: More about Equations 39

Exercise 1A (P.12)

1. (a) From the figure, the points of intersection

of the two graphs are (-3 , 5) and (2 , 5).

` The solutions of the simultaneous

equations are x = -3, y = 5 and

x = 2, y = 5.

(b) From the figure, the points of intersection

of the two graphs are (0 , 2) and (4 , 18).


equations are x = 0, y = 2 and x = 4,

y = 18.

2. (a) From the figure, the point of intersection

of the two graphs is (-2 , 4).

` The solution of the simultaneous

equations is x = -2, y = 4.

(b) From the figure, the point of intersection

of the two graphs is (3 , -6).


equations is x = 3, y = -6.

3. (a) From the figure, the two graphs have no

points of intersection.

` The simultaneous equations have no

real solutions.

(b) From the figure, the two graphs have no



real solutions.

4. Add the graph of y = 20 to the figure.

x

y

–1 1

5

10

15

20

25

–5

0

y = 8x2

y = 20

From the figure, the solutions of the simultaneous

equations are x = -1.6, y = 20 and x = 1.6,

y = 20.

5. Add the graph of y = 10 + x to the figure.

y = 10 + x

x -6 -4 -2 0

y 4 6 8 10

x

y

–2–4–6 –1–3–5

5

10

0

y = –x2– 6x

y = 10 + x

From the f igure , the so lu t ions o f the

simultaneous equations are x = -5, y = 5 and

x = -2, y = 8.


40 Solutions

6. (a) Draw the graph of y = x2 - 3x from

x = -1 to x = 4.

y = x2 - 3x

x -1 0 1 2 3 4

y 4 0 -2 -2 0 4

x

y

–1 1 2 3 4–1

1

2

3

4

–2

–3

–4

0

y = x2– 3x

y = x – 3

(b) Add the graph of y = x - 3 to the figure in

(a).

y = x - 3

x -1 1 3 4

y -4 -2 0 1

From the figure, the solutions of the

simultaneous equations are x = 3, y = 0

and x = 1, y = -2.

7. (a) Draw the graph of y = 1 - 2x2 from

x = -3 to x = 3.

y = 1 - 2x2

x -3 -2 -1 0 1 2 3

y -17 -7 -1 1 -1 -7 -17

x

y

–1–2–3 1 2 3–2

2

4

–4

–6

–8

–10

–12

–14

–16

–18

0

y = 2x – 2y = –1 – 4x

y = 1 – 2x2

(b) (i) Add the graph of y = 2x - 2 to the

figure in (a).

y = 2x - 2

x -3 -1 1 3

y -8 -4 0 4

From the figure, the solutions of

the simultaneous equations are

x = 0.8, y = -0.4 and x = -1.8,

y = -5.6.



(ii) Add the graph of y = -1 - 4x to the

figure in (a).

y = -1 - 4x

x -1 0 1 3

y 3 -1 -5 -13

From the figure, the solutions of

the simultaneous equations are

x = 2.4, y = -10.6 and x = -0.4,

y = 0.6.

8. (a) No. The graphs drawn by Jacky and Sally

are incomplete. They should draw the

graphs for -2 G x G 5.

(b) By combining the graphs drawn by Jacky

and Sally, the solutions of the simultaneous

equations are x = -1.2, y = 0.8 and

x = 4.2, y = 6.2.

9.

x

y

–3

–4

–1

–2

1

2

0 1 2 3 4 5

y = –x2+ 5x – 4

x + y = 1

y = (x - 1)(4 - x)

= 4x - x2 - 4 + x

= -x2 + 5x - 4

Add the graph of x + y = 1 to the figure.

x + y = 1

x 0 1 3 5

y 1 0 -2 -4

From the figure, the solutions of the simultaneous

equations are x = 1, y = 0 and x = 5, y = -4.

10. (a) From the figure, the two graphs have no



real solutions.

(b) From the figure, the solutions of the

simultaneous equations are x = -2.7,

y = -8.4 and x = 1.7, y = 0.4.

(c) From the figure, the solution of the

simultaneous equations is x = 1, y = 4.

11.

x

y

–100

–150

–50

100

50

150

0–10–15 –5 5 10 15

y = x2+ 2x – 120

y = x + 10

5x + y = –150

(a) Add the graph of y = x + 10 to the figure.

y = x + 10

x -10 0 10 20

y 0 10 20 30


simultaneous equations are x = 11, y = 20

and x = -12, y = 0.


42 Solutions

(b) Add the graph of 5x + y = -150 to the

figure.

5x + y = -150

x -20 -10 0 10

y -50 -100 -150 -200

From the figure, the two graphs have no



real solutions.

12. (a) Draw the graph of y = 4x2 + 8x from

x = -2.5 to x = 0.5.

y = 4x2 + 8x

x -2.5 -2 -1.5 -1 -0.5 0 0.5

y 5 0 -3 -4 -3 0 5

x

y

–0.5–1–1.5–2–2.5 0.5–2

2

4

6

–4

–6

–8

–10

0

y = 4x2+ 8x

4x + y + 9 = 0

(b) Add the graph of 4x + y + 9 = 0 to the

figure in (a).

4x + y + 9 = 0

x -2.5 -2 -1 0

y 1 -1 -5 -9

From the figure, the solution of the

simultaneous equations is x = -1.5,

y = -3.

13. (a) Draw the graph of y = -x2 + x - 2 from

x = -2 to x = 3.

y = -x2 + x - 2

x -2 -1 0 1 2 3

y -8 -4 -2 -2 -4 -8

x

y

–1–2 1 2 3–2

2

4

–4

–6

–8

0

y = –x2+ x – 2

y = 4 – 3x

(b) Add the graph of y = 4 - 3x to the figure in

(a).

y = 4 - 3x

x 0 1 2 3

y 4 1 -2 -5




real solutions.



14. (a) Draw the graph of y = (20 - x)(10 + x)

from x = -20 to x = 30.

y = (20 - x)(10 + x)

x -20 -10 0 10 20 30

y -400 0 200 200 0 -400

x

y

–10–20 10 20 30–100

100

200

300

–200

–300

–400

0

y = (20 – x)(10 + x)

y = 300 – 10x

(b) Add the graph of y = 300 - 10x to the

figure in (a).

y = 300 - 10x

x 0 10 20 30

y 300 200 100 0


simultaneous equations is x = 10, y = 200.

Exercise 1B (P.20)

1. y x

y x

2

2=

= +*

Substitute (i) into (ii):

x2 = x + 2

x2 - x - 2 = 0

(x - 2)(x + 1) = 0

` x = 2 or x = -1

Substitute these values of x into (ii):

When x = 2, y = 2 + 2 = 4.

When x = -1, y = -1 + 2 = 1.

` The solutions of the simultaneous equations

are x = 2, y = 4 and x = -1, y = 1.

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

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

2. y x x

y x

2= +

=-*


x2 + x = -x

x2 + 2x = 0

x(x + 2) = 0

` x = 0 or x = -2


When x = 0, y = 0.

When x = -2, y = 2.


are x = 0, y = 0 and x = -2, y = 2.

3. y x x

y x

2 3

1

2= - -

= +*


x2 - 2x - 3 = x + 1

x2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

` x = 4 or x = -1


When x = 4, y = 4 + 1 = 5.

When x = -1, y = -1 + 1 = 0.


are x = 4, y = 5 and x = -1, y = 0.

4. x x

y

y

x

4 0

6 3

2- - =

= -*

Substitute (ii) into (i):

x2 - 4x - (6 - 3x) = 0

x2 - x - 6 = 0

(x - 3)(x + 2) = 0

` x = 3 or x = -2


When x = 3, y = 6 - 3(3) = -3.

When x = -2, y = 6 - 3(-2) = 12.

` The so lu t ions o f the s imul taneous

equations are x = 3, y = -3 and x = -2,

y = 12.

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

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

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

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

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

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


44 Solutions

5. x y

y x

0

2 8

2- =

- =*

(i) + (ii):

x2 - 2x = 8

x2 - 2x - 8 = 0

(x - 4)(x + 2) = 0

` x = 4 or x = -2

From (i): y = x2 ............................................. (iii)

Substitute these values of x into (iii):

When x = 4, y = 42 = 16.

When x = -2, y = (-2)2 = 4.


equations are x = 4, y = 16 and x = -2,

y = 4.

6. 2 0y x

x y3 2

2- =

=+*

(ii) - (i):

3x + 2x2 = 2

2x2 + 3x - 2 = 0

(2x - 1)(x + 2) = 0

` x = 21

or x = -2

From (ii): y = 2 - 3x .................................... (iii)


When x = 21

, y = 2 - 321a k =

21

.

When x = -2, y = 2 - 3(-2) = 8.


equations are x = 21

, y = 21

and x = -2,

y = 8.

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

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

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

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

7. x x

x y

y

6

1

5 0

2- =

- - =

+*

From (ii): y = 5x - 6 .................................... (iii)

Substitute (iii) into (i):

x2 - x = 5x - 6 + 1

x2 - 6x + 5 = 0

(x - 5)(x - 1) = 0

` x = 5 or x = 1


When x = 5, y = 5(5) - 6 = 19.

When x = 1, y = 5(1) - 6 = -1.


equations are x = 5, y = 19 and x = 1,

y = -1.

8. x y x

x y

4

020

22

+ - =

+ - =*

From (ii): y = 20 - x .................................... (iii)


4x + 20 - x - 2 = x2

x2 - 3x - 18 = 0

(x - 6)(x + 3) = 0

` x = 6 or x = -3


When x = 6, y = 20 - 6 = 14.

When x = -3, y = 20 - (-3) = 23.


are x = 6, y = 14 and x = -3, y = 23.

9. Rewrite the given equation as:

x y

x x y

1

22

=

=

+

- -*


x2 - x - 2 = x + 1

x2 - 2x - 3 = 0

(x - 3)(x + 1) = 0

` x = 3 or x = -1

Substitute these values of x into (i):

When x = 3, y = 3 + 1 = 4.

When x = -1, y = -1 + 1 = 0.


are x = 3, y = 4 and x = -1, y = 0.

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

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

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

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

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

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




2 x

y

y

x

15

2 32

=

- = -

- -*


3 - x = 15 - x2

x2 - x - 12 = 0

(x - 4)(x + 3) = 0

` x = 4 or x = -3

From (i): y = 5 - x ....................................... (iii)


When x = 4, y = 5 - 4 = 1.

When x = -3, y = 5 - (-3) = 8.


are x = 4, y = 1 and x = -3, y = 8.


x x

x y x

x y 7 30

12 202

=

= + -

+ +

+*

(ii) - (i):

x2 + 7x - 30 - 12x - 20 = 0

x2 - 5x - 50 = 0

(x + 5)(x - 10) = 0

` x = -5 or x = 10

From (i): y = 11x + 20 ................................. (iii)


When x = -5, y = 11(-5) + 20 = -35.

When x = 10, y = 11(10) + 20 = 130.


are x = -5, y = -35 and x = 10, y = 130.

12. q p

qp

4

5 02 11

2= -

=+ +*


5p + 2(p2 - 4) + 11 = 0

2p2 + 5p + 3 = 0

(2p + 3)(p + 1) = 0

` p = -23

or p = -1

Substitute these values of p into (i):

When p = -23

, q = 2

23

-a k - 4 = -47

.

When p = -1, q = (-1)2 - 4 = -3.

` p = -23

, q = -47

or p = -1, q = -3.

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

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

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

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

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

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

13. ( )( )a b

a b

0

5

1 3

2 =

=+ -

+*

From (i), b = 5 - 2a .................................... (iii)

Substitute (iii) into (ii):

(a + 1)(5 - 2a - 3) = 0

(a + 1)(2 - 2a) = 0

` a = -1 or a = 1

Substitute these values of a into (iii):

When a = -1, b = 5 - 2(-1) = 7.

When a = 1, b = 5 - 2(1) = 3.


equations are a = -1, b = 7 and a = 1,

b = 3.

14. y x

y x

1

2

2= +

=-*


x2 + 1 = -2x

x2 + 2x + 1 = 0

The discriminant of the equation is:

D = (2)2 - 4(1)(1)

= 0

` There is one solution for x.

i.e. The simultaneous equations have one set of

solution.

15. y x

y x

x3

2 1

2= +

= -*


3x2 + x = 2x - 1

3x2 - x + 1 = 0


D = (-1)2 - 4(3)(1)

= -11

1 0

` There are no real solutions for x.

i.e. The simultaneous equations have no real

solutions.

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

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

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

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

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

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


46 Solutions

16. x y x

x y 2

12+ = +

+ =*

(i) - (ii):

x2 - x = x + 1 - 2

x2 - 2x + 1 = 0

(x - 1)2 = 0

` x = 1 (repeated)

Substitute x = 1 into (ii):

1 + y = 2

y = 1


is x = 1, y = 1.

17. x y x

y x

3

2 10

2+ =

= -*

(i) # 2 - (ii):

6x = 2x2 - x + 10

2x2 - 7x + 10 = 0


D = (-7)2 - 4(2)(10)

= -31

1 0



solutions.


( )

y

y x x x

x2

2 4

4 8

8

- =

- - + =*

From (i): y = 4 + 2x ..................................... (iii)


4 + 2x - 2x - x(x + 4) = 8

x2 + 4x + 4 = 0

(x + 2)2 = 0

` x = -2 (repeated)

Substitute x = -2 into (iii):

y = 4 + 2(-2)

= 0


is x = -2, y = 0.

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

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

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

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

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

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


( )x x x

x

y

y

3 2 5 12

6 4 2 12

2- =

- =

-

+*

From (ii), 3x - 2y = 5 .................................. (iii)


5x - 5x2 = 12

5x2 - 5x + 12 = 0


D = (-5)2 - 4(5)(12)

= -215

1 0



solutions.


x

x y

y

2

4

4

2 2+ =

- + =*

From (ii), x = y + 2 ...................................... (iii)


(y + 2)2 + y

2 = 4

2y2 + 4y = 0

2y(y + 2) = 0

` y = -2 or y = 0

Substitute these values of y into (iii):

When y = -2, x = -2 + 2 = 0.

When y = 0, x = 0 + 2 = 2.


equations are x = 0, y = -2 and x = 2,

y = 0.


x y

x y3

7

7

2 2=

+ =

-*

From (ii), y = 7 - 3x .................................... (iii)


x2 - (7 - 3x)

2 = 7

4x2 - 21x + 28 = 0


D = (-21)2 - 4(4)(28)

= -7

1 0



solutions.

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

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

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

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

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

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



22. x y

x y

2 1

3 4 1

2 2=

=

-

+*

From (ii), y = 41

- 43

x ................................. (iii)


x2 - 2 x

2

41

43

-a k = 1

x2 - 6x + 9 = 0

(x - 3)2 = 0

` x = 3 (repeated)

Substitute x = 3 into (iii):

y = 41

- 43

(3) = -2


is x = 3, y = -2.

23. x xy y

x y

4 4

2 0

2 2+ + =

=- -*

From (ii), x = y + 2 ...................................... (iii)


(y + 2)2 + (y + 2)y + 4y

2 = 4

6y2 + 6y = 0

6y(y + 1) = 0

` y = 0 or y = -1

Substitute these values of y into (iii):

When y = 0, x = 0 + 2 = 2.

When y = -1, x = -1 + 2 = 1.


are x = 2, y = 0 and x = 1, y = -1.

24. x y

x y c

c2 2 2+ =

+ =*

From (ii), y = c - x ...................................... (iii)


x2 + (c - x)

2 = c

2

2x2 - 2cx = 0

2x(x - c) = 0

` x = c or x = 0


When x = c, y = c - c = 0.

When x = 0, y = c - 0 = c.


are x = c, y = 0 and x = 0, y = c.

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

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

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

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

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

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

Exercise 1C (P.34)

1. (a) 2x - 3 = x2

2x2 - 3x = 2

2x2 - 3x - 2 = 0

(2x + 1)(x - 2) = 0

` x = 21

- or x = 2

(b) x2

+ x6

= 4

x2 + 12 = 8x

x2 - 8x + 12 = 0

(x - 6)(x - 2) = 0

` x = 6 or x = 2

2. (a) x - x 3

8-

= 1

( )

x

x x

3

3 8

-

- - = 1

x(x - 3) - 8 = x - 3

x2 - 3x - 8 - x + 3 = 0

x2 - 4x - 5 = 0

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

` x = 5 or x = 1-

(b) x 1

6+

= 5 - 2x

6 = (5 - 2x)(x + 1)

6 = 5x - 2x2 + 5 - 2x

2x2 - 3x + 1 = 0

(2x - 1)(x - 1) = 0

` x = 21

or x = 1

3. (a) x1

+ x 2

3+

= 2

( )x x

x x2

2 3+

+ + = 2

x + 2 + 3x = 2x(x + 2)

4x + 2 = 2x2 + 4x

x2 - 1 = 0

(x - 1)(x + 1) = 0

` x = 1 or x = 1-


48 Solutions

(b) x1

4-

+ x9

= 1

( )

( )

x x

x x

1

4 9 1

-

+ - = 1

4x + 9(1 - x) = x(1 - x)

4x + 9 - 9x = x - x2

x2 - 6x + 9 = 0

(x - 3)2 = 0

` x = 3 (repeated)

4. (a) Let u = 2x,

then the original equation becomes

u2 - 5u + 4 = 0

(u - 4)(u - 1) = 0

` u = 4 or u = 1

When u = 4, 2x = 4

` x = 2

When u = 1, 2x = 1

` x = 0

(b) Let u = 7x, then u

2 = (7

x)2 = 7

2x.

The original equation becomes

u2 - 56u + 343 = 0

(u - 49)(u - 7) = 0

` u = 49 or u = 7

When u = 49, 7x = 49

` x = 2

When u = 7, 7x = 7

` x = 1

5. (a) Let u = 3x, then u

2 = (3

x)2 = 3

2x.


32:u

2 - 10u + 1 = 0

9u2 - 10u + 1 = 0

(9u - 1)(u - 1) = 0

` u = 91

or u = 1

When u = 91

, 3x =

91

` x = 2-

When u = 1, 3x = 1

` x = 0

(b) Let u = 4x, then u

2 = (4

x)2 = (4

2)

x = 16

x.


u2 - 20u + 64 = 0

(u - 16)(u - 4) = 0

` u = 16 or u = 4

When u = 16, 4x = 16

` x = 2

When u = 4, 4x = 4

` x = 1

6. (a) log x + log (x - 3) = log 4

log x(x - 3) = log 4

` x(x - 3) = 4

x2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

` x = 4 or x = -1 (rejected)

(b) log (x - 4) + log (x + 4) = log 9

log (x - 4)(x + 4) = log 9

` (x - 4)(x + 4) = 9

x2 - 16 - 9 = 0

x2 - 25 = 0

(x - 5)(x + 5) = 0


7. (a) log (x + 3) + log (x + 12) = 1

log (x + 3)(x + 12) = log 10

` (x + 3)(x + 12) = 10

x2 + 15x + 36 = 10

x2 + 15x + 26 = 0

(x + 2)(x + 13) = 0

` x = 2- or x = -13 (rejected)

(b) log2 (x + 1) + log2 (x - 1) = 3

log2 (x + 1)(x - 1) = log28

` (x + 1)(x - 1) = 8

x2 - 1 - 8 = 0

x2 - 9 = 0

(x - 3)(x + 3) = 0




8. (a) Let u = sin i, then u2 = sin

2i.


2u2 - 1 = 0

u2 =

12

` u = !22

When u = 22

, sin i = 22

` i = 45c, 135c

When u = -22

, sin i = -22

` i = 225c, 315c

(b) Let u = tan i, then u2 = tan

2i.


3u2 - 1 = 0

u2 =

31

` u = !33

When u = 33

, tan i = 33

` i = 30c, 210c

When u = -33

, tan i = -33

` i = 150c, 330c

9. (a) Let u = cos i, then u2 = cos

2i.


2u2 - 3 u = 0

u(2u - 3 ) = 0

` u = 0 or u = 23

When u = 0, cos i = 0

` i = 90c, 270c

When u = 23

, cos i = 23

` i = 30c, 330c

(b) Let u = sin i, then u2 = sin

2i.


u + u2 = 0

u(u + 1) = 0

` u = 0 or u = -1

When u = 0, sin i = 0

` i = 0c, 180c

When u = -1, sin i = -1

` i = 270c

10. (a) Let u = tan i, then u2 = tan

2i.


u2 + 2u + 1 = 0

(u + 1)2 = 0

` u = -1 (repeated)

When u = -1, tan i = -1

` i = 135c, 315c

(b) Let u = cos i, then u2 = cos

2i.


2u2 - u - 1 = 0

(u - 1)(2u + 1) = 0

` u = 1 or u = -21

When u = 1, cos i = 1

` i = 0c

When u = -21

, cos i = -21

` i = 120c, 240c

11. No. The value of u obtained is the solution for

the quadratic equation in one unknown instead

of the solution for the original exponential

equation.

When u = 51

, 5x =

51

` x = 1-

When u = 1, 5x = 1

` x = 0


50 Solutions

12. When solving the equation, we cannot simplify

the equation by dividing both sides by the

common factor which involves the variable. This

will lose part of the solution.

sin i = 2sin2i

sin i(1 - 2sin i) = 0

sin i = 0 or sin i = 21

` i = 0c, 30c, 150c, 180c

13. Let the two integers be x and x + 1, then

x1

+ x 1

1+

= 127

( )x x

x x1

1+

+ + =

127

12(x + 1 + x) = 7x(x + 1)

24x + 12 = 7x2 + 7x

7x2 - 17x - 12 = 0

(x - 3)(7x + 4) = 0

` x = 3 or x = -74

(rejected)

` The two integers are 3 and 4.

14. Let the two numbers be x and x + 2, then

log x + log (x + 2) = log 35

log x(x + 2) = log 35

` x(x + 2) = 35

x2 + 2x - 35 = 0

(x - 5)(x + 7) = 0


` The two numbers are 5 and 7.

15. (a) 1x2

+` j(x + 6) = 1

( ) ( )

xx x2 6+ +

= 1

(2 + x)(x + 6) = x

2x + x2 + 12 + 6x = x

x2 + 7x + 12 = 0

(x + 4)(x + 3) = 0

` x = 4- or x = 3-

(b) x

x3+

+ 2x 9

6

- =

x 31-

- 1

2

( )

x

x x

9

3 6

-

- + = 2

2)(

x

x x

9

93

-

-+ -

x(x - 3) + 6 = x + 3 - x2 + 9

x2 - 2x - 3 = 0

(x + 1)(x - 3) = 0

` x = 1- or x = 3 (rejected)

(c) xx

3 53

-+

+ 1 = ( )

x

x

1

3 2

-

-

x

x x3 53 3 5

-+ + -

= ( )

x

x

1

3 2

-

-

xx

3 524

--

= ( )

x

x

1

3 2

-

-

(4x - 2)(x - 1) = 3(3x - 5)(x - 2)

4x2 - 6x + 2 = 9x

2 - 33x + 30

5x2 - 27x + 28 = 0

(5x - 7)(x - 4) = 0

` x = 57

or x = 4

16. (a) Let u = 4x, then the original equation

becomes

u - 3 = u10

u2 - 3u = 10

u2 - 3u - 10 = 0

(u + 2)(u - 5) = 0

` u = -2 or u = 5

When u = -2, 4x = -2 (rejected)

When u = 5, 4x = 5

log 4x = log 5

x log 4 = log 5

x = log

log

4

5

= .1 16, cor. to 2 d.p.

(b) Let u = 5x, then the original equation

becomes

2u + u1

= 3

2u2 + 1 = 3u

2u2 - 3u + 1 = 0

(u - 1)(2u - 1) = 0

` u = 1 or u = 21

When u = 1, 5x = 1

` x = 0



When u = 21

, 5x =

21

log 5x = log

21

x log 5 = log 21

x = log

log

521

= .0 43- , cor. to 2 d.p.

(c) Let u = 2x, then the original equation

becomes

82

u1` j + u

1 = 7

2u

8 + u

1 = 7

8 + u = 7u2

7u2 - u - 8 = 0

(u + 1)(7u - 8) = 0

` u = -1 or u = 78


When u = 78

, 2x =

78

log 2x = log

78

x log 2 = log 78

x = log

log

278

= .0 19, cor. to 2 d.p.

17. (a) log (x - 4) - log (3x - 10) = log x1` j

log xx

3 104

--

= log x1` j

` xx

3 104

--

= x1

x(x - 4) = 3x - 10

x2 - 7x + 10 = 0

(x - 2)(x - 5) = 0

` x = 2 (rejected) or x = 5

(b) log (x2 - 9) - log (x - 5) = log (2x - 1)

log 2

xx

59

--

= log (2x - 1)

` 2

xx

59

--

= 2x - 1

x2 - 9 = (2x - 1)(x - 5)

x2 - 11x + 14 = 0

x = 2

( )

( ) ( )( )

2 1

11 11 4 1 14! - -

= 11 65

2+

or 6511

2-

(rejected)

= .9 53, cor. to 2 d.p.

(c) log3 (4 + x2) - log3 (7 - x) - 2 = 0

log3 2

xx

74

-+

- log3 9 = 0

log3 2

xx

74

-+

= log3 9

` 2

xx

74

-+

= 9

4 + x2 = 9(7 - x)

x2 + 9x - 59 = 0

x = 2

( )

( )( )

2 1

9 9 4 1 59!- - -

= 9

2317- +

or 2

9 317- -

= .4 40 or .13 40- , cor. to 2 d.p.

18. (a) sin2i =

23

cos i

1 - cos2i =

23

cos i

2 - 2 cos2i = 3 cos i

2 cos2i + 3 cos i - 2 = 0

(cos i + 2)(2 cos i - 1) = 0

cos i + 2 = 0 or 2 cos i - 1 = 0

cos i = -2 (rejected) or cos i = 21

` i = 60c, 300c


52 Solutions

(b) 2 cos2i + 5 sin i = 5

2(1 - sin2i) + 5 sin i = 5

2 - 2 sin2i + 5 sin i - 5 = 0

2 sin2i - 5 sin i + 3 = 0

(sin i - 1)(2 sin i - 3) = 0

sin i - 1 = 0 or 2 sin i - 3 = 0

sin i = 1 or sin i = 23

(rejected)

` i = 90c

(c) 3 sin2i + 8 cos i - 7 = 0

3(1 - cos2i) + 8 cos i - 7 = 0

3 - 3 cos2i + 8 cos i - 7 = 0

3 cos2i - 8 cos i + 4 = 0

(cos i - 2)(3 cos i - 2) = 0

cos i - 2 = 0 or 3 cos i - 2 = 0

cos i = 2 (rejected) or cos i = 32

` i = 48.19c, 311.81c, cor. to 2 d.p.

19. (a) 2 cos i = tan i

2 cos i = cossin

i

i

2 cos2i = sin i

2(1 - sin2i) = sin i

2 sin2i + sini - 2 = 0

` sin i = 2

( )

( )( )

2 2

1 1 4 2 2!- - -

= 4

1 17- + or

41 17- -

(rejected)

` i = 51.33c, 128.67c, cor. to 2 d.p.

(b) tan

3i

+ sin i = 0

sincos3

i

i + sin i = 0

3 cos i + sin2i = 0

3 cos i + (1 - cos2i) = 0

cos2i - 3 cos i - 1 = 0

` cos i = 2

( )( )( )

( )

4 1 13 3

2 1

! - --

= 3 13

2+

(rejected) or 2

3 13-

` i = 107.62c, 252.38c , cor. to 2 d.p.

(c) sin i tan i + cos i = tan

4i

2

cossin

i

i + cos i =

sincos4

i

i

2 2

coscos

sini

i i+ =

sincos4

i

i

cos

1i

= sincos4

i

i

4 cos2i = sin i

4(1 - sin2i) = sin i

4 sin2i + sin i - 4 = 0

` sin i = 2

( )( )

( )

1 1 4 4 4

2 4

!- - -

= 8

1 56- + or

1 658

- - (rejected)

` i = 61.98c, 118.02c, cor. to 2 d.p.

20. Let Jack’s original cycling speed be x km/h.

From the question,

x30

- x 330+

= 0.5

( )

( )

x x

x x

3

30 3 30

+

+ - = 0.5

( )x x 3

90+

= 0.5

0.5x(x + 3) = 90

0.5x2 + 1.5x - 90 = 0

x2 + 3x - 180 = 0

(x + 15)(x - 12) = 0

x = -15 (rejected) or x = 12

` Jack’s original cycling speed is 12 km/h.

21. (a) Let the original number of glassware bought

by the students be x.

From the question,

(x - 2) 3x48

+` j - 48 = 22

( ) ( )

xx x2 48 3- +

= 70

(x - 2)(48 + 3x) = 70x

48x - 96 + 3x2 - 6x - 70x = 0

3x2 - 28x - 96 = 0

(x - 12)(3x + 8) = 0

x = 12 or x = -38

(rejected)

` The original number of glassware bought

by the students was 12.



(b) The selling price of each piece of

glassware

= $ 31248

+a k

= $7

22. From the question,

20 000(1 - 36%)t - 11 000(1 - 20%)

t = 3 000

20 # 0.64t - 11 # 0.8

t - 3 = 0

Let u = 0.8t, then u

2 = (0.8

t)2 = (0.8

2)

t = 0.64

t.


20u2 - 11u - 3 = 0

(4u - 3)(5u + 1) = 0

` u = 43

or u = -51

When u = 43

, 0.8t =

43

log 0.8t = log

43

t log 0.8 = log 43

t = .log

log

0 843

= 1.29, cor. to 2 d.p.

When u = -51

, 0.8t = -

51

(rejected)

` The value of t is 1.29.

23. From the question, R = 25.

` 25 = 2tan

tan

1

80

i

i

+

25(1 + tan2i) = 80 tan i

25 + 25 tan2i - 80 tan i = 0

5 tan2i - 16 tan i + 5 = 0

tan i = 2

) ( )( )(

( )

16 4 5 516

2 5

! --

= 16 156

10+

or 10

16 156-

` i = 70.66c or 19.34c, cor. to 2 d.p.

` The minimum value of i is 19.34c.

Supp. Exercise 1 (P.39)

1. (a) From the figure, the points of intersection

of the two graphs are (-1.8 , -0.6) and

(0.3 , 3.6).


equations are x = -1.8, y = -0.6 and

x = 0.3, y = 3.6.

(b) From the figure, the point of intersection

of the two graphs is (-1.5 , 6).


equations is x = -1.5, y = 6.

2.

x

y

2

–2

–4

4

6

8

10

0 21 3–2 –1–4 –3

y = x2+ 2x – 3

y = –xy = 2x + 1

(a) Add the graph of y = 2x + 1 to the figure.

y = 2x + 1

x -3 -1 0 3

y -5 -1 1 7


simultaneous equations are x = -2,

y = -3 and x = 2, y = 5.


54 Solutions

(b) Add the graph of y = -x to the figure.

y = -x

x -4 -2 0 2

y 4 2 0 -2


simultaneous equations are x = -3.8,

y = 3.8 and x = 0.8, y = -0.8.

3. y x

y x

1

3

2= +

= -*


x2 + 1 = 3 - x

x2 + x + 1 - 3 = 0

x2 + x - 2 = 0

(x - 1)(x + 2) = 0

` x = 1 or x = -2


When x = 1, y = 3 - 1 = 2.

When x = -2, y = 3 - (-2) = 5.


are x = 1, y = 2 and x = -2, y = 5.

4. y x

x y

x

3

2 3

1

2=

- =

- +*

(i) + (ii):

3x = x2 - 2x + 3 + 1

x2 - 5x + 4 = 0

(x - 4)(x - 1) = 0

` x = 4 or x = 1

From (ii), y = 3x - 1 .................................... (iii)


When x = 4, y = 3(4) - 1 = 11.

When x = 1, y = 3(1) - 1 = 2.


are x = 4, y = 11 and x = 1, y = 2.

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

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

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

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

5. y

xy

x

12

4

=

= +*


x(x + 4) = 12

x2 + 4x - 12 = 0

(x - 2)(x + 6) = 0

` x = 2 or x = -6


When x = 2, y = 2 + 4 = 6.

When x = -6, y = -6 + 4 = -2.


are x = 2, y = 6 and x = -6, y = -2.

6. ( )y

x y

x x

3 2

2

0

=

- + =

+*

(i) + (ii):

3x + 2 = x(x + 2)

3x + 2 = x2 + 2x

x2 - x - 2 = 0

(x + 1)(x - 2) = 0

` x = -1 or x = 2

Substitute these values of x into (i):

When x = -1, y = (-1)(-1 + 2) = -1.

When x = 2, y = 2(2 + 2) = 8.


are x = -1, y = -1 and x = 2, y = 8.

7. x2

= 4 - x8

x2

= xx4 8-

x2 = 2(4x - 8)

x2 - 8x + 16 = 0

(x - 4)2 = 0

` x = 4 (repeated)

8. x 1

1+

+ x2

= 3

( )

( )

x x

x x

1

2 1

+

+ + = 3

x + 2x + 2 = 3x2 + 3x

3x2 - 2 = 0

x2 =

32

` x = .0 82 or x = .0 82- , cor. to 2 d.p.

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

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

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

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



9. 2x2 - 4

= 8x

2x2 - 4

= 23x

` x2 - 4 = 3x

x2 - 3x - 4 = 0

(x + 1)(x - 4) = 0

` x = 1- or x = 4

10. Let u = 5x, then the original equation becomes

u2 - 7u + 6 = 0

(u - 1)(u - 6) = 0

` u = 1 or u = 6

When u = 1, 5x = 1

` x = 0

When u = 6, 5x = 6

log 5x = log 6

x = log

log

5

6

= .1 11, cor. to 2 d.p.

11. Let u = 2x, then u

2 = (2

x)2 = (2

2)

x = 4

x.


u2 - 5u + 4 = 0

(u - 4)(u - 1) = 0

` u = 4 or u = 1

When u = 4, 2x = 4

` x = 2

When u = 1, 2x = 1

` x = 0

12. log (x + 2) + log x = 1

log (x + 2)(x) = log 10

` x2 + 2x = 10

x2 + 2x - 10 = 0

x = 2

( )

( )( )

2 1

2 2 4 1 10!- - -

= 2 44

2!-

= -1 + 11 or -1 - 11 (rejected)

= .2 32, cor. to 2 d.p.

13. (log x)2 - log x

2 = 3

(log x)2 - 2 log x = 3

(log x)2 - 2 log x - 3 = 0

(log x + 1)(log x - 3) = 0

` log x = -1 or log x = 3

` x = 10-1

or x = 103

= .0 1 = 1000

14. 4 cos2i = 3cos i

cos i(4 cos i - 3) = 0

cos i = 0 or cos i = 43

` i = 90c, 270c or

i = 41.41c, 318.59c, cor. to 2 d.p.

15. 6 sin2i - sin i - 1 = 0

(2 sin i - 1)(3 sin i + 1) = 0

2 sin i - 1 = 0 or 3 sin i + 1 = 0

sin i = 21

or sin i = -31

` i = 30c, 150c or

i = 199.47c, 340.53c, cor. to 2 d.p.

16. (a) 1x

x y

y

2 5

02

=

+ =

+*

(b) x y

x y

2 5

102

+ =

+ =*

From (ii), x = 10 - y2 .......................... (iii)


2(10 - y2) + y = 5

20 - 2y2 + y - 5 = 0

2y2 - y - 15 = 0

(y - 3)(2y + 5) = 0

` y = 3 or y = -25

(rejected)

Substitute y = 3 into (iii):

x = 10 - 32

= 1

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

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


56 Solutions

17. Let the denominator of the original fraction be

x, then

x 27 2

++

- x7

= 201

( )

( )

x x

x x

2

9 7 2

+

- + =

201

2x x

x

2

2 14

+

- =

201

x2 + 2x = 20(2x - 14)

x2 - 38x + 280 = 0

(x - 10)(x - 28) = 0

` x = 10 or x = 28

When x = 10,

the original fraction is 107

.

When x = 28,

the original fraction is 728

(rejected).

` The original fraction is 107

.

18. (a) The original width

= cmx224

The new width

= cm cmor1x x224

1225

+-

` aj k

(b) From the question,

(x - 1) 1x224

+` j = 224 + 1

( ) ( )

xx x1 224- +

= 225

224x - 224 + x2 - x = 225x

x2 - 2x - 224 = 0

(x - 16)(x + 14) = 0


19. (a) Draw the graph of y = 3x - x2 + 1 from

x = -1 to x = 4.

y = 3x - x2 + 1

x -1 0 1 2 3 4

y -3 1 3 3 1 -3

x

y

–1 1 2 3 4–1

1

2

3

4

5

6

–2

–3

0

y = 3x – x2+ 1

y = 2x + 4

(b) Add the graph of y = 2x + 4 to the figure

in (a).

y = 2x + 4

x -1 -0.5 0 1

y 2 3 4 6




real solutions.



20. (a) Draw the graph of y = x2 + 5x + 10 from

x = -10 to x = 5.

x -10 -8 -5 0 3 5

y 60 34 10 10 34 60

x

y

–2–4–6–8–10 2 4–10

10

20

30

40

50

60

–20

–30

–40

0

y = x2+ 5x + 10

y = –15 – 5x

(b) Add the graph of y = -15 - 5x to the

figure in (a).

y = -15 - 5x

x -10 -5 0 5

y 35 10 -15 -40


simultaneous equations is x = -5, y = 10.

21. x y

yx5 3

2 2

25

+ =

=-*


x2 + (5 - 3x)

2 =

25

x2 + 25 - 30x + 9x

2 -

25

= 0

4x2 - 12x + 9 = 0

(2x - 3)2 = 0

` x = 23

(repeated)

Substitute x = 23

into (ii):

y = 5 - 323a k

= 21


is x = 23

, y = 21

.

22. ( )( )x y

x y

8 4

2

5

6

- - =

- =

-*

From (ii): x = 2y + 6 .................................... (iii)


(2y + 6 - 8)(y - 4) = -5

2y2 - 2y - 8y + 8 + 5 = 0

2y2 - 10y + 13 = 0


D = (-10)2 - 4(2)(13)

= -4

1 0

` There are no real solutions for y.


solutions.

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

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

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

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


58 Solutions

23. x y2 11

2 2

y y x

3

3

2

+ =

=+ -*

From (ii): 2(y + 3) = 3(y - x)

y = 3x + 6 ....................... (iii)


2x2 + (3x + 6)

2 = 11

2x2 + 9x

2 + 36x + 36 = 11

11x2 + 36x + 25 = 0

(11x + 25)(x + 1) = 0

` x = -1125

or x = -1


When x = -1125

, y = 31125

-a k + 6 = -119

.

When x = -1, y = 3(-1) + 6 = 3.


are x = -1125

, y = -119

and x = -1, y = 3.


x y

y x

2 3

4 3

2 2- =

=-*

From (ii): y = 4x + 3 .................................... (iii)


2x2 - (4x + 3)

2 = 3

2x2 - 16x

2 - 24x - 9 = 3

7x2 + 12x + 6 = 0


D = (12)2 - 4(7)(6)

= -24

1 0



solutions.

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

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

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

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


9 6 9xy x

x y

y

3 2 2

222

=

=

+

+ -*

From (i): y = 1 - x23 ................................... (iii)


9x x123

-a k + 6 x12

23

-a k - 9x2 = 2

9x - 227

x2 + 6 - 18x +

227

x2 - 9x

2 - 2 = 0

9x2 + 9x - 4 = 0

(3x + 4)(3x - 1) = 0

` x = -34

or x = 31


When x = -34

, y = 1 - 34

23

-a k = 3.

When x = 31

, y = 1 - 23

31a k =

21

.


are x = -34

, y = 3 and x = 31

, y = 21

.

26. y mx

xy

1

1

=

=

-*


x(mx - 1) = 1

mx2 - x - 1 = 0


D = (-1)2 - 4(m)(-1)

= 1 + 4m

Since the simultaneous equations have no real

solutions, there are no real solutions for x.

i.e. D = 1 + 4m 1 0

` m 1 -41

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

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

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

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



27. 2

2( )

x

x1 + - 4 = 0

2

2( )

x

x1 + = 4

(1 + x)2 = 4x

2

1 + 2x + x2 = 4x

2

3x2 - 2x - 1 = 0

(x - 1)(3x + 1) = 0

` x = 1 or x = 31

-

28. 2( )x 1

3

- -

x 11-

= 41

2( )

( )

x

x

1

3 1

-

- - =

41

2( )x

x

1

4

-

- =

41

(x - 1)2 = 4(4 - x)

x2 - 2x + 1 = 16 - 4x

x2 + 2x - 15 = 0

(x - 3)(x + 5) = 0

` x = 3 or x = 5-

29. xx

23

-+

- xx1 -

= 417

( )

( ) ( ) ( )

x x

x x x x

2

3 1 2

-

+ - - - =

417

2

2 2

x x

x x x x

2

3 3 2

-

+ - + + =

417

2

2

x x

x

2

2 2

-

+ =

417

17(x2 - 2x) = 4(2x

2 + 2)

9x2 - 34x - 8 = 0

(9x + 2)(x - 4) = 0

` x = 92

- or x = 4

30. 3x x2 11

2 11

++ +

a k = 10

x2 11+

:( )

x

x

2 1

1 3 2 1

+

+ + = 10

2( )x

x

2 1

6 4

+

+ = 10

10(2x + 1)2 = 6x + 4

40x2 + 40x + 10 = 6x + 4

20x2 + 17x + 3 = 0

(5x + 3)(4x + 1) = 0

` x = 53

- or x = 41

-

31. Let u = 3x, then the original equation becomes

u243

= 36 - u

243 = 36u - u2

u2 - 36u + 243 = 0

(u - 27)(u - 9) = 0

` u = 27 or u = 9

When u = 27, 3x = 27

` x = 3

When u = 9, 3x = 9

` x = 2

32. Let u = 4x, then u:4

2 = 4

x:4

2 = 4

x + 2.


u:42 + u

1 = 10

16u2 - 10u + 1 = 0

(8u - 1)(2u - 1) = 0

` u = 81

or u = 21

When u = 81

, 4x =

81

22x

= 2-3

2x = -3

` x = 23

-

When u = 21

, 4x =

21

22x

= 2-1

2x = -1

` x = 21

-

33. log (x + 1) - log (3x2 - 5) = -1

log 2x

x

3 5

1

-

+ = log

101

` 2x

x

3 5

1

-

+ =

101

3x2 - 5 = 10(x + 1)

3x2 - 10x - 15 = 0


60 Solutions

x = 2

( ) ( )( )

( )

10 10 4 3 15

2 3

! - - -

= 6

10 280!

= 5 70

3+

or 5 70

3-

(rejected)

= .4 46, cor. to 2 d.p.

34. log (102x

- 56) - x = 0

log (102x

- 56) = x

` 102x

- 56 = 10x

(10x)2 - 10

x - 56 = 0

(10x - 8)(10

x + 7) = 0

` 10x = 8 or 10

x = -7 (rejected)

x = log 8

= .0 90, cor. to 2 d.p.

35. tan i + tan

1i

= -2

tan2i + 1 = -2 tan i

tan2i + 2tan i + 1 = 0

(tan i + 1)2 = 0

tan i = -1 (repeated)

` i = 135c, 315c

36. 5 cos i + 2 sin2i + 1 = 0

5 cos i + 2(1 - cos2i) + 1 = 0

5 cos i + 2 - 2 cos2i + 1 = 0

2 cos2i - 5 cos i - 3 = 0

(cos i - 3)(2 cos i + 1) = 0

` cos i - 3 = 0 or 2 cos i + 1 = 0

cos i = 3 (rejected) or cos i = -21

` i = 120c, 240c

37. tan i = sincos

32

i

i+

cossin

i

i =

sincos

32

i

i+

3 sin2i = cos i(2 + cos i)

3(1 - cos2i) = 2 cos i + cos

2i

4 cos2i + 2 cos i - 3 = 0

cos i = 2

( )( )

( )

2 2 4 4 3

2 4

!- - -

= 2 52

8!-

= 1 13

4- +

or 1 13

4- -

(rejected)

` i = 49.35c, 310.65c, cor. to 2 d.p.

38. (a) 3 sin2x + 4 sin x cos x - 4 cos

2x = 0

2

2

cos

sin

x

x3 + 2

cos

sin cos

x

x x4 - 2

2

cos

cos

x

x4 = 0

3 tan2x + 4 tan x - 4 = 0

(b) 3 sin2x + 4 sin x cos x - 4 cos

2x = 0

3 tan2x + 4 tan x - 4 = 0　 (From (a))

(tan x + 2)(3 tan x - 2) = 0

` tan x + 2 = 0　or 3 tan x - 2 = 0

tan x = -2 or tan x = 32

` x = 116.57c, 296.57c or

x = 33.69c, 213.69c, cor. to 2 d.p.

39. (a) The original number is 10x + y and the

new number is 10y + x.

From the question,

( )x y

xy

y x

35

10 10 18

=

+ - =+*

i.e. xy

x y

35

2

=

- =*

(b) 35xy

x y 2

=

- =*

From (ii): y = x - 2 .............................. (iii)


x(x - 2) = 35

x2 - 2x - 35 = 0

(x - 7)(x + 5) = 0



y = 7 - 2 = 5

` The original two-digit number is 75.

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

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



40. (a) From the question,

( )

x y

x yx y 3 1

500

3 00

2 2+ =

+ + =-*

i.e. x y

x y

500

02 5

2 2+ =

+ =*

(b) x y

x y

500

2 50

2 2+ =

+ =*

From (ii): y = 50 - 2x ......................... (iii)


x2 + (50 - 2x)

2 = 500

x2 + 2 500 - 200x + 4x

2 = 500

x2 - 40x + 400 = 0

(x - 20)2 = 0

` x = 20 (repeated)


y = 50 - 2(20) = 10

` The length of the longest side of the

foundation

= (x + y) m

= (20 + 10) m

= 30 m

41. Let the number of members of the club be x.

From the question,

x 42 000

- - x

2 000 = 25

( )

( )

x x

x x

000 2 000 4

4

2 - -

- = 25

2x x4

8 000

- = 25

25(x2 - 4x) = 8 000

25x2 - 100x - 8 000 = 0

x2 - 4x - 320 = 0

(x - 20)(x + 16) = 0


` The number of members of the club is 20.

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

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

42. Let the time taken for the cyclist to finish the

journey by the longer route be x hours.

From the question,

x24

- .x 0 5

20+

= 4

( . )

( . )

x x

x x

0 5

24 0 5 20

+

+ - = 4

2.x x

x

0 5

4 12

+

+ = 4

4x + 12 = 4x2 + 2x

4x2 + 2x - 4x - 12 = 0

2x2 - x - 6 = 0

(x - 2)(2x + 3) = 0

` x = 2 or x = -23

(rejected)

` It takes 2 hours for the cyclist to finish the

journey by the longer route.

43. According to the question, we need to find the

value of x when P = 0.

` 11 000(2-0.02x

) - 8 000(2-0.04x

) - 3 000 = 0

11(2-0.02x

) - 8(2-0.04x

) - 3 = 0

Let u = 2-0.02x

, then u2 = (2

-0.02x)2 = 2

-0.04x.


11u - 8u2 - 3 = 0

8u2 - 11u + 3 = 0

(u - 1)(8u - 3) = 0

` u = 1 or u = 83

When u = 1, 2-0.02x

= 1

x = 0 (rejected)

When u = 83

, 2-0.02x

= 83

-0.02x = log

log

283

x = . log

log

0 02 283

-

= 70.75, cor. to 2 d.p.

` The machine can produce paint at most for

70.75 months.


62 Solutions

44. From the question, H = 5.

` 5 = 12 tan i - 59

(1 + tan2i)

25 = 60 tan i - 9 - 9 tan2i

9 tan2i - 60 tan i + 34 = 0

tan i = 2

( ) ( )( )

( )

60 60 4 9 34

2 9

! - -

= 18

60 2 376!

= 10 66

3+

or 10 66

3-

` i = 32.02c, 80.60c, cor. to 2 d.p.

` The maximum value of i is 80.60c.

45. Let the speed of the cruise ship be x km/h.

From the question,

x 3840+

+ x 3840-

= 37.5

( ) ( )

( ) ( )

x x

x x

3 3

840 3 840 3

+ -

- + + = 37.5

2x

x

9

1680

- = 37.5

37.5(x2 - 9) = 1 680x

37.5x2 - 337.5 - 1 680x = 0

5x2 - 224x - 45 = 0

(x - 45)(5x + 1) = 0

` x = 45 or x = -51

(rejected)

` The time taken for the cruise ship to arrive

at town B

= 45 3

840+

hours

= 17.5 hours

` The cruise ship arrived at town B at

11:30 p.m.

46. (a) Let the time taken for tap A alone to fill up

the tank be x minutes.

From the question,

x30

+ x 25

30-

= 1

( )

( )

x x

x x

25

30 25 30

-

- + = 1

2x x

x

25

60 750

-

- = 1

60x - 750 = x2 - 25x

x2 - 85x + 750 = 0

(x - 75)(x - 10) = 0

x = 75 or x = 10 (rejected)

` The time taken for tap A alone to fill

up the tank is 75 minutes.

(b) The time taken for tap B alone to fill up

the tank

= (75 - 25) minutes

= 50 minutes

Integrated Questions (cross-chapters)

47. Substituting x = -3 into the equation

2x2 + ax - 12b = 0, we have

2(-3)2 + a(-3) - 12b = 0

i.e. 6 - a - 4b = 0 .................. (i)

Since the equation x2 - ax + b = 0 has two

equal real roots,

the discriminant D = 0

i.e. (-a)2 - 4(1)(b) = 0

a2 - 4b = 0 .............................. (ii)

(ii) - (i):

a2 + a - 6 = 0

(a - 2)(a + 3) = 0

` a = 2 or a = -3

From (i): b = 41

(6 - a) ................................ (iii)

Substitute these values of a into (iii):

When a = 2, b = 41

(6 - 2) = 1.

When a = -3, b = 41

[6 - (-3)] = 49

.

` a = 2, b = 1 or a = -3, b = 49

.



48. (a) From the figure, the y-intercept is 18.

` c = 18

(b) y = ax2 + bx + c

= a x x2

ab

+` j + c

= a x x2

2 2

ab

ab

ab

2 2++ -a ak k; E + c

= a x2

ab

2+a k + c -

2

ab4

a The vertex of the parabola is (3 , 0).

` c - 2

ba4

= 0

i.e. 18 - 2

ab4

= 0

2

ba4

= 18

b2 = 72a

(c) a The vertex of the parabola is (3 , 0).

` -a

b2

= 3

` b = -6a .................................... (i)

Substitute (i) into the equation of (b):

(-6a)2 = 72a

36a2 = 72a

a2 - 2a = 0

a(a - 2) = 0

` a = 0 (rejected) or a = 2

Substitute a = 2 into (i):

b = -6(2) = 12-

49. From the question, 21

ab = 1241 .................... (i)

The equation of the straight line can be written

as y = - ab

x + b

Given that the line passes through (2 , 3), we

have

3 = - ab

(2) + b

i.e. 3a = -2b + ab ................................... (ii)

From (i): a = b2

49 ........................................... (iii)


3b2

49a k = -2b + b2

49a kb

b2

147 = -2b +

249

4b2 - 49b + 147 = 0

(b - 7)(4b - 21) = 0

` b = 7 or b = 421

Substitute these values of b into (iii):

When b = 7, a = ( )2 749

= 27

.

When b = 421

, a = 2

49

421a k

= 3

14.

When a = 27

and b = 7,

the equation of the line AB is

y = -7

27

x + 7

i.e. 2x + y - 7 = 0

When a = 3

14 and b =

421

,

the equation of the line AB is

y = -

314421

x + 421

i.e. 9x + 8y - 42 = 0


64 Solutions

Lively Maths Problem

50. Let the number of chickens he bought be x.

From the question,

18x1 000

+` j(x - 5) +

10 5x1 000

#-` j = 1 000 + 220

(1 000 + 18x)(x - 5) +

5(1 000 - 10x) = 1 220x

1 000x + 18x2 - 5 000 -

90x + 5 000 - 50x = 1 220x

18x2 - 360x = 0

x2 - 20x = 0

x(x - 20) = 0

` x = 0 (rejected) or x = 20

` He bought 20 chickens.

Revision Test (P.47)

1. D From the figure, the points of intersection of

the graphs of y = x2 - 4 and y = -x + 2

are (-3 , 5) and (2 , 0).


equations are (-3 , 5) and (2 , 0).

2. A x x

y

y

x

1

5 3

2= + +

=

-

-*


-x2 + x + 1 = 5 - 3x

-x2 + x + 3x + 1 - 5 = 0

x2 - 4x + 4 = 0

(x - 2)2 = 0

` x = 2 (repeated)

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

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

3. D Rewrite the given equation as:

y x

x y

x3

4

1

3 2 1

2- + =

- =-*

(i) # 2 + (ii):

-2x2 + 9x - 4 = 3

2x2 - 9x + 7 = 0

(x - 1)(2x - 7) = 0

` x = 1 or x = 27

From (ii): y = 12

(3x - 5) ..................... (iii)


When x = 1, y = 12

[3(1) - 5] = -1.

When x = 27

, y = 3 521

27

-a k: D = 411

.

` y = 1- or 114

4. B x41-

+ x2

1+

= 43

( ) ( )x x

x x4 22 4

- ++ + -

= 43

( ) ( )x x4 2

6- +

= 43

2x x8 2

2

+ - =

41

8 + 2x - x2 = 8

x2 - 2x = 0

x(x - 2) = 0

` x = 0 or x = 2

5. A Let u = 2x, then u

2 = (2

x)2 = 2

2x.


u2 + 2u - 8 = 0

(u + 4)(u - 2) = 0

` u = -4 or u = 2


When u = 2, 2x = 2

` x = 1

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

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



6. C log3 (2x + 1) + log3 (x - 2) = 1

log3 (2x + 1)(x - 2) = log3 3

` (2x + 1)(x - 2) = 3

2x2 - 3x - 5 = 0

(x + 1)(2x - 5) = 0

` x = -1 (rejected) or x = 25

7. A 4 cos2i - 9 cos (180c + i) + 2 = 0

4 cos2i + 9 cos i + 2 = 0

(cos i + 2)(4 cos i + 1) = 0

` cos i + 2 = 0 or 4 cos i + 1 = 0

cos i = -2 (rejected) or cos i = -41

a 0c G i 1 180c

` i = 104.48c, cor. to 2 d.p.

i.e. The equation has 1 root.

8. B The speed of the car is (x - 5) km/h after

decreasing and the time taken will be

x 5900-

hours.

From the question, we have

x 5900-

- x900

= 2.

HKCEE Questions

9. D 10. B

11. E 12. D

13. D 14. B

15. D 16. C


sol_5a01.pdf

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