Maths ch15 key

p.1

xa a

When a is negative,

x

a a

When a is positive,

Chp 15 Key Prepared by C.Y. So

e.g. 1 p.3

(a) 22 ax

When a is positive, axa

When a is negative, axa

(b) 14 2 x

4

12 x

2

1x Or

2

1x

(Reflection result) (Simple observation)

(c) xxx 511052

01110

0151052

2

xx

xxx

The auxiliary equation gives two roots 1 and 11.

Thus, for 011102 xx ,

111 x .

(d) xx 52 2

052 2 xx

The auxiliary equation gives two roots 0 and 2

5.

Thus, for 052 2 xx ,

0x Or 2

5x

p.2

(e) Assume 01x (i.e. 1x )

01

1582

x

xx

can be changed to )1(01

158)1(

2

xx

xxx

01582 xx

The auxiliary equation gives the roots 3 and 5.

Thus, the roots for 01582 xx

3x or 5x

However, the above solution set is made on the assumption 1x

Thus, 31 x or 5x .

Assume 01x (i.e. 1x )

01

1582

x

xx

can be changed to )1(01

158)1(

2

xx

xxx

01582 xx

The auxiliary equation gives the roots 3 and 5.

Thus, the roots for 01582 xx

53 x

However, the above solution set is made on the assumption 1x

Thus, no solution set can be made.

Finally, the overall solution for 01

1582

x

xx is

31 x or 5x .

e.g. 2 p.4

(a) (i) 3216 xx (ii) 0)3)(2( xx

14

4

4

4

44

3126

x

x

x

xx

23

0)3)(2(

)1(0)3)(2()1(

x

xx

xx

(b) Based on the range 23 x , a condition x1 makes the final solution 21 x .

p.3

e.g. 3 p.4

32 x and 0)2)(3( xx

5.12

3

32

)1(3)2()1(

x

x

x

x

23

0)2)(3(

x

xx

Based on the range 23 x , a condition x 5.1 makes the final solution

25.1 x .

e.g. 4 p.4

(i) 03 x or 05 x

3x 5x

All real numbers can be either greater than 3 or smaller than 5. Just think of some examples

yourself. For example, your age must be greater than 3. For those numbers which are not greater

than 3, they must be very small. In this case, they are smaller than 5.

(ii) 03 x and 05 x

3x 5x

Those numbers that are smaller than 3 are really quite small. How can those small numbers be

greater than 5? Therefore, no real solutions can be found.

(iii) 02 x and 0122 xx

For all x except zero, 02 x .

Thus, 02 x means 0x .

0122 xx

43

012

)1(0)12()1(2

2

x

xx

xx

Now we have 0x and 43 x .

Thus, the final solution is 03 x or 40 x .

p.4

e.g. 5 p.8

(a)

Amount of coffee Amount of coffee mate

1 kg of Brand A

5

4 kg

5

1 kg

1 kg of Brand B

5

2 kg

5

3 kg

Amount available 10 kg 9 kg

(b) 1 kg of Brand A gives a profit of $30.

x kg of Brand A gives a profit of $30x.

1 kg of Brand B gives a profit of $40.

y kg of Brand B gives a profit of $40y. Therefore, Profit yxP 4030 in dollars.

(c)

95

3

5

105

2

5

4

0

0

yx

yx

y

x

(d) Before constructing the straight lines and locating the region, we should simplify the set of

inequalities.

453

252

0

0

yx

yx

y

x

(e) We have to find a point (x, y) in the shaded region, so that yxP 4030 will be a maximum.

The point chosen is (6, 13) which gives a maximum profit of $700.

Therefore, 6 kg of Brand A and 13 kg of Brand B should be produced.

p.5

Ex15.5 1. p.11

Ex15.52. 2. p.12

(a) (b) (i) From “the total length of the four sides …

not less than 20”, we have 2022 yx

which can be simplified as 10 yx .

From “twice the length of the flower bed

should not less than three times its width”,

we have yx 32 .

From “no contractor will build the fences

if their total length is less than 12 metres”,

we have 122 yx

Thus, we have

yx

yx

yx

32

122

10

.

(ii) Let $C be the total cost of building the wall and the fence.

yxC

yxxC

600800

)2(300500

We have to minimize the cost C with (x, y) in the region defined by b(i).

It could be (6, 4), (8, 2) or (12, 0). By checking the value, the point (6, 4) gives the

minimum value $7200 of the cost.

p.6

Ex15.5 3. p.15

(a) Brand X : 1 box costs $25 with 20 pieces of chocolate.

Brand Y : 1 box costs $37.5 with 40 pieces of chocolate.

Brand X : x box costs $25x with 20x pieces of chocolate.

Brand Y : y box costs $37.5y with 40y pieces of chocolate.

Now, Mrs Chiu buys x boxes of Brand X and y boxes of Brand Y.

Thus, she has yx 4020 pieces of chocolate.

She needs to pay $ )5.3725( yx

(i) From “the total number of chocolates is at least 240”,

we have 2404020 yx which can be simplified as 122 yx .

(ii) From “the total cost is no more than $300”,

we have 3005.3725 yx which can be simplified as 2432 yx .

(iii) From “the total number of boxes is not more than 10”,

we have 10 yx .

(b) The ordered pairs are (7, 3), (6, 3), (6, 4), (5, 4), (4, 4), (4, 5), (3, 5), (3, 6).

(c) After testing the above ordered pairs (4, 4) gives the minimum of the cost which is $250 .

(d) $ )5.3725( yx = $ 300

Thus, 2432 yx

(i) Possible combinations of x and y are the points (3, 6) and (6, 4).

(ii) For (3, 6), the number of pieces of chocolates = 20(3)+40(6)

= 300

For (6, 4), the number of pieces of chocolates = 20(6)+40(4)

= 280

The maximum number of pieces of chocolates = 300.

Ex15.5 4. p.18

(a) 1 tray of cookies requires 0.32 kg of flour, 0.24 kg of sugar and 2 eggs.

x trays of cookies requires 0.32x kg of flour, 0.24x kg of sugar and 2x eggs.

1 tray of cakes requires 0.28 kg of flour, 0.36 kg of sugar and 10 eggs.

y trays of cakes requires 0.28y kg of flour, 0.36y kg of sugar and 10y eggs.

p.7

As Miss Chan has 4.48 kg of flour, 48.428.032.0 yx which can be simplified as

11278 yx .

As Miss Chan has 4.32 kg of flour, 32.436.024.0 yx which can be simplified as

3632 yx .

As Miss Chan has 100 eggs, 100102 yx which can be simplified as

505 yx .

Moreover, 0x and 0y .

x and y are integers.

(b) Let $P be the total profit.

yxP 12090

The possible points in the region R are (0, 10), (4, 9), (6, 8), (8, 6), (9, 5), (10, 4), (14, 0).

You may construct a straight line cyx 43 and moves it as high as possible as long as it

touches the region R. You will find that the last integral point is (6, 8).

Thus, when 6x and 8y , the profit is the maximum and it is equal to )8(120)6(90 .

The maximum profit is found to be $1440.

p.8

Ex15.5 5. p.21

(a) 1 packet of Brand A mixed nuts contains 40 g of peanuts and 10 g of almonds.

1 packets of Brand B mixed nuts contains 30 g of peanuts and 25 g of almonds.

1000 packets of Brand A mixed nuts contains 40 kg of peanuts and 10 kg of almonds.

800 packets of Brand B mixed nuts contains 24 kg of peanuts and 20 kg of almonds.

1000x packets of Brand A mixed nuts contains 40x kg of peanuts and 10x kg of almonds.

800y packets of Brand B mixed nuts contains 24y g of peanuts and 20y g of almonds.

As the company has 2400 kg of peanuts, 24002440 yx


As the company has 1200 kg of almonds, 12002010 yx


As the company has 70 carton boxes, 70 yx

Note that x and y are both non-negative integers.

Thus, we have the shaded area for

0

0

70

1202

30035

y

x

yx

yx

yx

.

Let the total profit be $P. yxP 1000800

From the graph below, there are four possible points in the enclosed region:

(0, 60), (20, 50), (45, 25), (60, 0).

Of them, (20, 50) gives the maximum value of P. The maximum profit made is $66000

when 20 boxes of brand A and 50 boxes of brand B are produced.

p.9

(b) When the number of boxes of brand B is smaller than the number of boxes of brand A,

yx .

From the graph below, there are four possible points in the enclosed region:

(36, 34), (45, 25), (60, 0). Of them, (36, 34) gives the max of P. Thus, 36 boxes of brand A and

34 brand B have to produced in order to provide a maximum value of the profit $62800.

p.10

Ex15.5 6. p.23

(a) The quickest way of forming an equation with known x-intercept and y-intercept is

1intercept-intercept-

y

y

x

x.

Thus, the equation of 1L is and the equation of 2L is

kyx

kk

yk

k

xk

k

y

k

x

4559

1459

455

45

195

kyx

kk

yk

k

xk

k

y

k

x

60125

1605

6012

60

1512

(b) Line A : 45 man-hours to produce 1 article, releases 50 units of pollutants, profit $3000.

Line B : 25 man-hours to produce 1 article, releases 120 units of pollutants, profit $2000.

Line A : 45x man-hours to produce x article, releases 50x units of pollutants, profit $3000x.

Line B : 25y man-hours to produce y article, releases 120y units of pollutants, profit $2000y.

(i) As the factory has 225 man-hours available, 2252545 yx which can be simplified as

4559 yx .

As the total amount of pollutants discharged must not exceed 600 units, 60012050 yx


The total profit is yx 20003000 .

There are a few possible points to

consider: (0,5), (2,4), (3,3), (4,1)

and (5, 0). Of them, both (3, 3) and

(5, 0) gives the maximum profit

$15000.

(ii) As the factory has 450 man-hours available, 4502545 yx which can be simplified as

9059 yx .

As the total amount of pollutants discharged must not exceed 1200 units, 120012050 yx


p.11

Putting k = 2, the graphs are shown

in the figure.

There are a few possible points to

consider: (0, 10), (4, 8), (6, 7), (10, 0).

Of them, (6, 7) gives the maximum profit

$32000.

Ex15.5 7. p.25

54

13

x

x

3

37

7

7

21

721

34201

20413

5444

134

x

x

x

x

xx

xx

xx

012 x

5.0

12

x

x

Thus, the inequalities 54

13

x

x and 012 x

is just the same as 3x and 5.0x .

It means all the real numbers between 0.5 and 3.

Thus, the integers satisfying 54

13

x

x and 012 x means

the integers satisfying 35.0 x . Those integers are 2, 1, 0.

p.12

Ex15.5 8. p.25

(a) 6

251

xx (b) The greatest integer satisfying 8.3x is 3.

8.35

19

5

5

195

2566

25666

256166

x

x

x

xx

xx

xx

Ex15.5 9. p.25

022 xx

Thus, 21 x .

Now, for 02)100()100( 2 yy ,

Let, 100 yx . Thus, we have

10299

1002100)100(1001

21001

21

022

y

y

y

x

xx

Ex15.5 10. p.25

(a) 725

14 x

x (b) The least integer which satisfies 75.8x is 9.

75.84

35

4

4

354

351014

351014

75255

145

x

x

x

xx

xx

xx

p.13

Ex15.5 11. p.27

(a)

(b) (i) 100 cba

bac 100

(ii) 1 kg of Tablet A costs $6.

a kg of Tablet A costs $6a.

1 kg of Tablet B costs $5.

b kg of Tablet B costs $5b.

1 kg of Tablet C costs $4.

c kg of Tablet C costs $4c.

However,

ba

bac

44400

)100(44

The cost of the mixture = 6a + 5b + (400 4a 4b)

= 400 + 2a + b

p.14

(iii) 1 kg of Tablet A has 400 units of Calcium and 800 units of Potassium.

a kg of Tablet A has 400a units of Calcium and 800a units of Potassium.

1 kg of Tablet B has 600 units of Calcium and 200 units of Potassium.

b kg of Tablet B has 600b units of Calcium and 200b units of Potassium.

1 kg of Tablet C has 400 units of Calcium and 400 units of Potassium.

c kg of Tablet C has 400c units of Calcium and 400c units of Potassium.

As the mixture must contain at least 44000 units of Calcium,

20

402

4404440064

440)100(464

440464

44000400600400

b

b

baba

baba

cba

cba

As the mixture must contain at least 48000 units of Potassium,

402

240222004

240)100(24

24024

480428

48000400200800

ba

baba

baba

cba

cba

cba

As 0c .

0100 ba

100

100

ba

ba

Therefore, we have

100

402

20

ba

ba

b

.

(iv) The cost of the mixture is 400 + 2a + b.

Putting x = a, y =b and using the shaded region, a = 30, b = 20 gives the minimum of the

cost. When 30a and 20b , 50c and the minimum cost is $480.

p.15

Ex15.5 12. p.29

(a) (i) As 1L passes through (8, 16) and (12, 24),

the equation of 1L is

02

161621

2

8

164

8

8

16812

1624

8

16

yx

yxx

yx

yx

y

As the equation of 1L is 02 yx and 1L 2L ,

let the equation of 2L be kyx 2 where k is a constant.

As 2L passes through (12, 24), k )24(2)12(

60k Thus, the equation of 2L is 602 yx .

(ii) The shaded region is formed by

02

602

10

8

yx

yx

y

x

.

(b) Let x = number of square tables and y = number of round tables

As the manager wants at least 8 square tables and 10 round tables, we have

8x and 10y .

As the number of round table is not more than 2 times that of the square time,

we have xy 2 which is just the same as 02 yx .

As the floor area occupied is at most 240,

24084 yx which can be simplified as 602 yx

Let $P = total profit on a day. yxP 60004000 where (x, y) is in the shaded region.

The point (40, 10) gives the maximum value of P.

The maximum value of P is 4000(40) + 6000(10) = 220000. Thus, the total profit cannot exceed

$230000 that day.

p.16

Ex15.5 1. p.31

The region formed by 40 x is

The region formed by

40

40

y

x is


6

40

40

yx

y

x

is


62

40

40

yx

y

x

is

(ANS : E)

p.17

Ex15.5 2. p.32

When we are asked about the maximum

value of yx 3 , there are a few points to consider:

(1, 2), (2, 2), (3,1).

However, (2, 2) must outdo (1, 2).

Thus, we need to consider (2, 2) and (3, 1) only.

When 2x , 2y ,

8

)2()2(3

3

yx

When 3x , 1y

10

)1()3(3

3

yx

Therefore, the maximum value of yx 3 is 10.

(ANS : D)

Ex15.5 3. p.32

The shaded region is on the left of

the straight line 6x .

Thus, we have 6x .

The shaded region is above

the straight 4 yx .

Thus, we have 4 yx .

The shaded region is below

the straight yx .

Thus, we have yx .

I. yx is false.

II. 4 yx is false.

III. 6x is true.

(ANS : C)

p.18

Ex15.5 4. p.33

The shaded region is a triangle which is formed by three sides.

They are 6x , yx and 6 yx .

The set of inequalities is

6

6

yx

yx

x

.

(ANS : A)

Ex15.5 5. p.34

Consider a straight line with equation 0 cbyax

When both a and b are positive, the straight line is one going down. One of the examples

is 06 yx .

When a is positive and b is negative, the straight line is one going up. One of the examples is

0 yx

In Q5, it is given that b is negative and the straight line is 0 cbyx .

Thus, the straight line is going up. This means the answer is A, B or C.

Let’s consider the x-intercept and y-intercept.

For x-intercept, put y = 0. Thus, we have 0)0( cbx

cx

As c is negative, c is positive. Thus, the straight line has a positive x-intercept.

This means the answer is either B or C.

As the inequality is 0 cbyx and b is negative, it refers to the area below the straight line.

Thus, the answer is C.

(ANS : C)

Ex15.5 Q6. p.35

Note that for any two real numbers a and b.

ababba 22

In order to get the greatest value of 3 aybx , we may substitute numbers into the brackets in the

following expression. 3) () ( ab

p.19

It would be good if b is put into the first pair of brackets and a is put into the second.

In this case the result will be 322 ab which is the greatest value.

Luckily, there is such a point ) ,( ab which is the answer.

(ANS : D)

Ex15.5 Q7. p.35

The shaded region is formed by

023

10

0

yx

yx

y

(ANS : D)

Ex15.5 Q8. p.36

First consider the straight line 2 yx .

When 2y , 0x . This means that the line cuts the positive y-axis.

Thus, the answer is either C or D.

Consider the inequality 2 yx .

We choose the region above the straight line. Thus, the answer is D.

(ANS : D)

Ex15.5 Q9. p.36

Actually, all the four regions satisfy the inequality 2x .

However, only II and III satisfy 2 yx .

For the inequality 0 yx , it refers to the region below it. Thus, only III satisfy it.

(ANS : C)

Ex15.5 Q10. p.37

The inequalities are

9

2

0

yx

yx

y

.

(ANS : B)

p.20

Ex15.5 Q11. p.37 (Cancelled)

Ex15.5 Q12. p.38

012

3122 xx

x

means 312 x and 0122 xx

312 x

1

22

132

x

x

x

1

0)1(

0122

2

x

x

xx

Thus, 312 x and 0122 xx

can be rewritten as 1x and 1x .

Therefore, the solution is 11 x or x1 .

(ANS : C)

Ex15.5 Q13. p.38

1523

251

x

x

means 3

251

x and 152

3

25

x

x.

4

352

2533

251

x

x

x

x

5

26455

45625

1523

25

x

xx

xx

xx

Thus, 1523

251

x

x

means 4x and 5x

which can be rewritten as 45 x .

(ANS : A)

p.21

Ex15.5 Q14. p.38

Consider Option B, which says abba

2

.

Suppose, abba

2

.

Then, we have abba 2 Then, we have abbaba 4))((

Then, we have abbaba 42 22

Then, we have 042 22 abbaba

Then, we have 02 22 baba Then, we have 0))(( baba

which is usually wrong (it holds only when a equals b).

Thus, Option B is incorrect.

Now, consider Option D, which says 2a

b

b

a.

Suppose, 2a

b

b

a

Then, we have 2ba

bb

ba

aa

Then, we have 222

ab

ba

Then, we have abba 222

Then, we have 02 22 baba Then, we have 0))(( baba

which is always true.

In fact, we can start from 0))(( baba to work out 2a

b

b

a.

(ANS : D)

Ex15.5 Q15. p.39

It is given that ba and dc , can we make them to form I, II and III?

For I, we have ba

and dc ,

Thus, dbca

cdba Therefore, I is true.

p.22

12 2

For II, we have ba

and dc ,

ac may or may not be greater than bd . Think about 52 and 81 and …

For III, It may hold or may not. It depends on the values of a, b, c and d.

(ANS : A)

Ex15.5 Q16. p.39

03)12(2)12( 2 xx

Let 12 xy . We have

0322 yy

Thus, 3y or 1y .

312 x or 112 x .

1

22

x

x

1

22

x

x

Thus, finally, 1x or 1x .

(ANS : A)

Ex15.5 Q17. p.39

024102 xx

12x or 2x

(ANS : A)

Ex15.5 Q18. p.40

4 x can be simplified as 4x .

23

162

x can be simplified as )2(3

3

1623

x

6162 x

1662 x

102 x

5x

The result of 4x and 5x

is 5x .

(ANS : D)

p.23

Ex15.5 Q19. p.40

When ba ,

ba . Thus, option A is wrong.

When 8a and 2b ,

14 b

a Thus, option B is wrong.

When 8a and 2b ,

642 a and 42 b . Thus, option C is wrong.

Note that x10 is an increasing function. It means that when ba ,

ba 1010 .

Thus, option D is true.

(ANS : D)

Ex15.5 Q20. p.40

0log a and 0log b

For I,

b

alog

ba loglog

= (a positive number) (a negative number)

= The sum of two positive numbers

> 0 Thus, I is true.

For II, 2logb

blog2

= 2 (a negative number)

< 0 Thus, II is false

For III,

a

1log

a

a

a

log

log0

log1log

Thus, III is false

(ANS : A)

p.24

Ex15.5 Q21. p.41

Given that BC is 01232 yx (slope is 0.66)

and AB is 0122 yx (slope is 2)

Let 43 yxP 。 (slope is 0.33)

Construct a “profit line” which has x-intercept

and y-intercept in the ratio 3 : 1 。

Obviously, when the line passes through point C,

it gives the maximum value of P.

The coordinates of C can be obtained by putting 0x into 01232 yx 。

C is (0, 4).

Thus, the maximum value of P is 4)4(3)0(

= 16

(ANS : C)

Ex15.5 Q22. p.41

Of regions I, II, III and IV, only regions I and II satisfy the condition

4y .

Both regions I and II satisfy the condition 8 yx .

Only region II satisfies the condition 82 yx .

(ANS : B)

Ex15.5 Q23. p.41

Those points on the x-axis with y-values < 3 are clearly

in the range between a and d.

(ANS : A)

p.25

Ex15.5 Q24. p.42

Given that RQ is 20 yx (slope is 1)

and QP is 363 yx (slope is 3)


In order to minimize P, put the line as high as possible.

Obviously, when the line passes through point R,

it gives the least value of P.

P is (0, 20).

The least value of P = 180)20(3)0(2

= 120

(ANS : B)

Ex15.5 Q25. p.42

xx 15

3

62

51

x

x

xx

As x is a positive integer, x may be 3, 2, or 1.

The least value of x is 1.

(ANS : B)

Ex15.5 Q26. p.42

Given that AB is 183 yx (slope is 0.33)

and BC is 162 yx (slope is 2)


In order to maximize P, put the line as low as possible.

Obviously, when the line passes through point C,

it gives the greatest value of P.

C is (8, 0).

The greatest value of P = 3(8) (0) + 16

= 40

(ANS : C)

p.26

Ex15.5 Q27. p.43

30

60

y

x represents the region on the right.

yx

y

x

2

30

60

represents the region on the right.

(ANS : D)

Maths ch15 key

Education

Transcript of Maths ch15 key