Page 1 of 1

### Question of a Sub question 2004

Posted: Sun Sep 23, 2012 1:02 pm
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

### Re: Question of a Sub question 2004

Posted: Sun Sep 23, 2012 2:37 pm
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
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
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
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
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
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
dumplinghao123 wrote:Thanks!

But is there an couterexample for 2?

Just make sure B says

$\int_0^1 f(x) \, dx = \int_1^2 f(x) \, dx$

Right?

If my interpretation is correct, it should be false. Since $f(x)$ is strictly increasing, $\int_0^1 f(x) \, dx < \int_1^2 f(x) \, dx$

### Re: Question of a Sub question 2004

Posted: Thu Sep 27, 2012 4:03 pm
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

$\int_0^1 f(x) \, dx = \int_1^2 f(x) \, dx$

Right?

If my interpretation is correct, it should be false. Since $f(x)$ is strictly increasing, $\int_0^1 f(x) \, dx < \int_1^2 f(x) \, dx$

### Re: Question of a Sub question 2004

Posted: Thu Sep 27, 2012 4:08 pm
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
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
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