## Question of a Sub question 2004

### 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

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

What are your thoughts?

### Re: Question of a Sub question 2004

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

### 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?

### 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?

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

### 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.

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

The first one is necessarily wrong.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

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

Thanks!

But is there an couterexample for 2?

But is there an couterexample for 2?

tarheel wrote:The first one is necessarily wrong.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

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

Just make sure B saysdumplinghao123 wrote:Thanks!

But is there an couterexample for 2?

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

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

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

tarheel wrote:Just make sure B saysdumplinghao123 wrote:Thanks!

But is there an couterexample for 2?

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

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

Well, I think both are necessarily wrong.dumplinghao123 wrote:Yea. The question ask what is necessarily wrong. So why not choose 2?

### Re: Question of a Sub question 2004

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

What are the other options?

### 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

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