56. If k is a real number and f(x) = sin(1/x) for x not equal to 0, and f(x)=k for x=0
and if the graph of f is not a connected subset of the plane, then the value of k
(A) could be 1
(B) must be 0
(C) must be 1
(D) could be less than 1 and greater than 1
(E) must be less than 1 or greater than 1
Answer is (E)
Can anyone help on this? I can make sense of (E), but how do you prove it?
Q56 on GRE practicing to take the mathematics test

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 Joined: Thu Sep 11, 2014 7:19 am
Re: Q56 on GRE practicing to take the mathematics test
Which number is this GRE exam?
Re: Q56 on GRE practicing to take the mathematics test
You don't want a connected subset of the plane. Hence, k cannot be in [1,1]. As you can see, the function and you will see that it gets 'crazy' about 0
http://www.webalice.it/extrabyte/sin1_x.jpg
http://www.webalice.it/extrabyte/sin1_x.jpg
Last edited by DDswife on Fri Sep 12, 2014 3:41 pm, edited 2 times in total.

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 Joined: Thu Sep 11, 2014 9:36 pm
Re: Q56 on GRE practicing to take the mathematics test
Since the range of f is a subset of R, to determine if it is a connected subspace, we must find if the range of f is an interval.
Since 1<=sin(1/x)<=1 for all x not zero, and it is continuous for x not zero, then the range of f is {[1,1]Uf(0)}.
Clearly this is an interval if f(0) is in [1,1], and not an interval is f(0) is not in [1,1]. Thus the answer is E, that f(0) must be less than 1 or greater than 1.
Since 1<=sin(1/x)<=1 for all x not zero, and it is continuous for x not zero, then the range of f is {[1,1]Uf(0)}.
Clearly this is an interval if f(0) is in [1,1], and not an interval is f(0) is not in [1,1]. Thus the answer is E, that f(0) must be less than 1 or greater than 1.

 Posts: 15
 Joined: Thu Sep 11, 2014 7:19 am
Re: Q56 on GRE practicing to take the mathematics test
Thanks! I was overthinking, now it seems so obvious...