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### Question of a Sub question 2004

Posted: **Sun Sep 23, 2012 1:02 pm**

by **26186514**

if f is strictly increasing, then which is necessarily WRONG

A.xf(2x)=2f(x)

B.integral f(x)dx from 0 to 1=integral fx(dx) from 1 to 2

### Re: Question of a Sub question 2004

Posted: **Sun Sep 23, 2012 2:27 pm**

by **vonLipwig**

What are your thoughts?

### Re: Question of a Sub question 2004

Posted: **Sun Sep 23, 2012 2:37 pm**

by **26186514**

I'd say II is wrong. But I don't know if there is an counterexample.

### Re: Question of a Sub question 2004

Posted: **Sun Sep 23, 2012 5:17 pm**

by **vonLipwig**

Have you tried to construct a function satisfying the first condition? What makes you suspicious of the second?

### Re: Question of a Sub question 2004

Posted: **Mon Sep 24, 2012 4:09 pm**

by **26186514**

I think since strictly increasing functions are integratable. So their numbers should be able to be compared?

In terms of 1, I haven't really think through it. What's your conclusion?

### Re: Question of a Sub question 2004

Posted: **Mon Sep 24, 2012 8:53 pm**

by **vonLipwig**

For the second, how should you be able to compare them?

For the first, I think you should try it. It is highly unlikely that I will be present to assist you in your GRE.

### Re: Question of a Sub question 2004

Posted: **Thu Sep 27, 2012 11:38 am**

by **tarheel**

dumplinghao123 wrote:if f is strictly increasing, then which is necessarily WRONG

A.xf(2x)=2f(x)

B.integral f(x)dx from 0 to 1=integral fx(dx) from 1 to 2

The first one is necessarily wrong.

Let x=2, then A gives 2f(4)=2f(2), which is f(4)=f(2). But since f is strictly increasing, f(4)>f(2). Therefore A is wrong for all f.

### Re: Question of a Sub question 2004

Posted: **Thu Sep 27, 2012 11:58 am**

by **26186514**

Thanks!

But is there an couterexample for 2?

tarheel wrote:dumplinghao123 wrote:if f is strictly increasing, then which is necessarily WRONG

A.xf(2x)=2f(x)

B.integral f(x)dx from 0 to 1=integral fx(dx) from 1 to 2

The first one is necessarily wrong.

Let x=2, then A gives 2f(4)=2f(2), which is f(4)=f(2). But since f is strictly increasing, f(4)>f(2). Therefore A is wrong for all f.

### Re: Question of a Sub question 2004

Posted: **Thu Sep 27, 2012 3:53 pm**

by **tarheel**

dumplinghao123 wrote:Thanks!

But is there an couterexample for 2?

Just make sure B says

Right?

If my interpretation is correct, it should be false. Since

is strictly increasing,

### Re: Question of a Sub question 2004

Posted: **Thu Sep 27, 2012 4:03 pm**

by **26186514**

Yea. The question ask what is necessarily wrong. So why not choose 2?

tarheel wrote:dumplinghao123 wrote:Thanks!

But is there an couterexample for 2?

Just make sure B says

Right?

If my interpretation is correct, it should be false. Since

is strictly increasing,

### Re: Question of a Sub question 2004

Posted: **Thu Sep 27, 2012 4:08 pm**

by **tarheel**

dumplinghao123 wrote:Yea. The question ask what is necessarily wrong. So why not choose 2?

Well, I think both are necessarily wrong.

### Re: Question of a Sub question 2004

Posted: **Fri Sep 28, 2012 4:48 pm**

by **Legendre**

tarheel wrote:

The first one is necessarily wrong.

Let x=2, then A gives 2f(4)=2f(2), which is f(4)=f(2). But since f is strictly increasing, f(4)>f(2). Therefore A is wrong for all f.

Alternatively, try y = x. xf(2x) = 2x^2 = 2x => x^2 = x. Clearly wrong for large x.

What are the other options?

### Re: Question of a Sub question 2004

Posted: **Sun Aug 17, 2014 12:01 am**

by **DDswife**

Maybe when they say "necessarily wrong" they mean for all x, while not necessarily wrong would be "there is an x that"...

If this is the case, then any stricktly increaing f so that f0) = 0 would make I true