求一道数学题解 ~~

holddreams

106. (DS)
  问|x|>|x^3|?

1)       x < -1

2)       |x^2| < |x^4|

选D

Here, |x|>|x^3|? 

if x>0, then x>x^3, then x^2<1 then 0<x<1

if x<0, then x<x^3 then x<-1 or x>1

So 1)=> |x|>|x^3|

For 2) |x^2| < |x^4|  then 1<x^2 then x<-1 or x>1 => |x|>|x^3|

Here 1) => |x|>|x^3| and also 2) =>|x|>|x^3|, so the answer is D. Am I right?

What confuses me is the following one:

131. （DS）X*(X+1)*(X+2)=0，问能否求出X？
        

          i）X*（X+1）=0

         ii）X*(X+2)=0

答案：in jijing the author choosed C

for i) we can say x = 0 or x=-1, and 问能否求出X？ the answer is yes

for ii) we can say x =0 or x=-2, and 问能否求出X？ the answer is yes

So the answer should be D.

My question is: X here is not a sole value but a x=0 or x=-1,

can we say we can 求出X if either x=0 or x=-1 or x=-2?

xp635

problems in GMAT like above need ONE definite answer, just ONE

so 131 was corrected by choice C

huskygal

QUOTE:

1.if x<0, then x<x^3 then x<-1 or x>1

这里推错了

2.GMAT里要求解是唯一的

holddreams

Thank you fighting127. The question 2 is clear now. :-)

Here below is my modified version of question 1:

106. (DS)
  问|x|>|x^3|?

1)       x < -1

2)       |x^2| < |x^4|

选D

Here, |x|>|x^3|? 

if x>0, then x>x^3, then x^2<1 then 0<x<1

if x<0, then x<x^3 then x^2<1 then -1<x<0

So 1)=> |x|>|x^3|

For 2) |x^2| < |x^4|  then 1<x^2 then x<-1 or x>1 => |x|>|x^3| (but how to explain x>1 situation?)

Here 1) can not get |x|>|x^3| and also 2) can not get |x|>|x^3|,

so the answer is D. Am I right?

My quesiton here is how to explain the part with red color? Thank you very much.

huskygal

 

QUOTE:

So 1)=> |x|>|x^3|

又反了...

答案是2个都推不出来所以选D

OceanBleu

是不是第一个X<-1-->所以对于题目可以回答X的绝对值不比那个后面的绝对值大，所以得解NO

第二个因为绝对值X的平方小于绝对值X的四次方，所以X有两个值域

一个是X>1或者X<-1,所以X的绝对值小于后面那个绝对值，所以也可以回答NO

最后是每个答案都可以回答，所以选D。

第二个，我觉得选C，因为我觉得求X的值应该是唯一的。所以必须COMBINE两个才可以。

holddreams

Thanks a million. I got it.