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