F.4 M2 Pre-Revision

To solve this kind of question, you must consider different cases.

a < bThere are three cases：
i) 1 < 2 (two positive)
ii) -1 < 1 (one positive; one negative)
iii) -2 < -1 (two negative)
Moreover, you must also consider that k can be even number and odd number .

First, let's try A:

A. -a>-b
-1>-2 (correct)
1>-1 (correct)
2>1 (correct)

Luckily, you don't need to try B, C and D, because A can satisfy all the cases.
Therefore, the answer is A.

2013-11-30 19:57:21 補充：
B. 1/b<1/a
1/2<1 (correct)
1<-1 ((incorrect))
-1<-1/2 (correct)


C. ka










2013-11-30 19:57:50 補充：
D. a^k>b^k
when k is 1
1>2 ((incorrect))
-1>1 ((incorrect))
-2>-1 ((incorrect))

when k is 2
1>4 ((incorrect))
1>1 ((incorrect))
4>1 (correct)

∴only A and C are the answers.

2013-11-30 20:02:36 補充：
C. ka










2013-11-30 20:02:50 補充：
C. ka < kb
when k is 1
1<2 (correct)
-1<1 (correct)
-2<-1 (correct)

when k is 2
2<4 (correct)
-2<2 (correct)
-4<-2 (correct)

2013-11-30 20:07:43 補充：
∴only A and C are the answers.

a<b

When both sides are multiplied by (-1), the inequality sign should be changed.
That is -a>-b, so (A) is true.

When both sides are divided by (ab), as we don't know where (ab) is positive or not, so the inequality sign may be wrong.

When both sides are multiplied by k, as k is positive, so no need to change the inequality sign.
That is ka<kb, so (C) is also true.

As k can be odd or even, so we can't determine whether a^k is greater or smaller than b^k.

So, only (A) and (C) are always true.

A,B,C all are correct.
Algebra manipulation shows

A) a < b
multiply both sides by -1 (note that signs will change whenever you multiply by negative number)
so:
-a > -b ( true )

B) a < b
divide both sides by a you will get
1< b/a
divide both sides by b you will get
1/b < 1/a ( true)

C) a < b
multiply both sides by k you will get
ka < kb (true)

D) if a=1 b=3 k=2
1 < 3
1^2 > 3^2 (not true)

Jei 元朗 Math & English