## Question of a Sub question 2004

Forum for the GRE subject test in mathematics.
dumplinghao123
Posts: 24
Joined: Sat Sep 22, 2012 4:34 pm

### Question of a Sub question 2004

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

vonLipwig
Posts: 51
Joined: Sat Mar 17, 2012 9:58 am

### Re: Question of a Sub question 2004

What are your thoughts?

dumplinghao123
Posts: 24
Joined: Sat Sep 22, 2012 4:34 pm

### Re: Question of a Sub question 2004

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

vonLipwig
Posts: 51
Joined: Sat Mar 17, 2012 9:58 am

### Re: Question of a Sub question 2004

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

dumplinghao123
Posts: 24
Joined: Sat Sep 22, 2012 4:34 pm

### Re: Question of a Sub question 2004

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?

vonLipwig
Posts: 51
Joined: Sat Mar 17, 2012 9:58 am

### Re: Question of a Sub question 2004

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.

tarheel
Posts: 10
Joined: Wed Sep 26, 2012 4:37 pm

### Re: Question of a Sub question 2004

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.

dumplinghao123
Posts: 24
Joined: Sat Sep 22, 2012 4:34 pm

### Re: Question of a Sub question 2004

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.

tarheel
Posts: 10
Joined: Wed Sep 26, 2012 4:37 pm

### Re: Question of a Sub question 2004

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

dumplinghao123
Posts: 24
Joined: Sat Sep 22, 2012 4:34 pm

### Re: Question of a Sub question 2004

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

tarheel
Posts: 10
Joined: Wed Sep 26, 2012 4:37 pm

### Re: Question of a Sub question 2004

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

Well, I think both are necessarily wrong.

Legendre
Posts: 217
Joined: Wed Jun 03, 2009 1:05 am

### Re: Question of a Sub question 2004

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?

DDswife
Posts: 153
Joined: Thu Aug 14, 2014 5:29 pm

### Re: Question of a Sub question 2004

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

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