Q56 on GRE practicing to take the mathematics test

[huayualice]

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?

[DDswife]

Which number is this GRE exam?

[DDswife]

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

[thomasmgill]

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.

[huayualice]

Thanks! I was overthinking, now it seems so obvious...