F.3 MATHS 15 points

1. 4a^2-8ab+4b^2
= 4(a^2-2ab+b^2)
= 4(a-b)^2

2a)9m^2-6mn+n^2
= (3m)^2-2(3m)(n)+n^2
= (3m-n)^2

b) 9m^2-6mn+n^2+6mp-2np+p^2
= (3m-n)^2+6mp-2np+p^2
= (3m-n)^2+2p(3m-n)+p^2
= [(3m-n)+p]^2
= (3m-n+p)^2

3. 8x^3-(x+y)^3
= (2x)^3-(x+y)^3
= [2x-(x+y)][(2x)^2+(2x)(x+y)+(x+y)^2]
= (x-y)[4x^2+(2x^2+2xy)+(x^2+2xy+y^2)]
= (x-y)(7x^2+4xy+y^2)

4. (x-3/5)-(2+5x/3)≦9
-2x/3-13/5<=9
-2x/3<=58/5
x>=58/5*(-3/2)
x>=-87/5

5.3/4(x+1)+2/5(4-x)≧2
7x/20+47/20>=2
7x/20>=-7/20
x>=-1


6.Let x be the largest odd integer.

x+(x-2)+(x-4)>=100
3x-6>=100
3x>=106
x>=106/3
>=35 1/3

The least possible value of the largest odd integer is 36

Proof
x=36: 36+34+32=102>=100
x=34: 34+32+30=96<100


7. Let a and b be the prices of an apple and orange respectively.

5a+9b>50 ... (1)
a<=3.5 ... (2)

From (1), 9b>50-5a
b>(50-5a)/9 ... (3)

(50-5a)/9 will be min. when a=3.5
When a=3.5, b>[50-5(3.5)]/9=3.6111

The min. price of an orange will be $3.6.


8. Let a and b be the increase of Jenny's and Raymond's pocket moneys respectively.

After increase, Jenny has 40+a pocket money and Raymond has 25+b.

40+a<=1.5(25+b) ... (1)
a+b=12 ... (2)

From (2), b=12-a ... (3)

Putting (3) in (1), we have
40+a<=1.5[25+(12-a)]
<=1.5(37-a)
<=55.5-1.5a
2.5a<=15.5
a<=6.2

The max. value of a is 6.2

Because the max. increase in Jenny's pocket money is $6.2,
her max. possible daily pocket money is 40+6.2=$46.2

Proof
When a=6.2,
b=12-6.2=5.8

After increase, Jenny has $46.2 daily pocket money and Raymond has 25+5.8=$30.8

Jenny's money is 46.2/30.8=1.5 times Raymond's money.