questions from GR8767

Forum for the GRE subject test in mathematics.
Post Reply
weilan8
Posts: 8
Joined: Sun Aug 16, 2009 12:19 pm

questions from GR8767

Post by weilan8 » Mon Aug 31, 2009 9:06 pm

GR8767
14 A new cast contained the statement that the total use of electricity in city A had declined in one billing period by 5%, while household se had declined by 4% and all other uses increased by 25%. Which of the following must be true about the billing period?
answer: "the statement was in error"
why the statement was in error?

47 Suppose that the space S contains exactly eight points. If g is a collection of 250 distinct subsets of S, which of the following statements must be true?
answer: "g has a member that contains exactly one element"
how to get this result?

58 If f(z) is an analytic function that maps the entire finite complex plane into the real axis, then the imaginary axis must be mapped onto
answer: "a point"
does someone know why?

64 Let S be a compact topological space, let T be a topological space, and let f be a function from S onto T. Of the following conditions on f, which is the weakest conditon sufficient to ensure the compactness of T?
(A) f is homeomorphism; (B) f is continuous and 1-1; (C) f is continuous; (D) f is 1-1; (E) f is bounded
answer: (C)
why f is 1-1 is not the weakest sufficient condition? and does f bounded imply T is compact?

mk
Posts: 13
Joined: Fri May 22, 2009 1:14 am

Re: questions from GR8767

Post by mk » Tue Sep 01, 2009 11:49 am

14: The total decreased by 5% but no individual part of the total decreased by more than 4%, so the statement must be in error.

47: If a set S contains exactly 8 points then it has 2^8 = 256 subsets, which means g omits only 6 subsets. But 8 of those subsets are 1 point subsets, so g must contain at least 2 of those (thus g has a member that contains only one element).

58: Write f(z) = u(x,y) + iv(x,y) where u,v are real valued. Then since f is real valued, v = 0 identically. Since f is analytic it satisfies the Cauchy Riemann equations, but since v = 0, this means du/dx = du/dy = 0. So u must be constant. Thus u maps EVERY complex number into a point, so it must map the imaginary axis into a point.

64: Answered in your other topic.

Hope this helps!



Post Reply