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oldmathguy

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  1. Think roughly about the shape of the graphs. There is one intersection with x higher. But we know that for very large x ("in the limit") x^12 ends up lower. So (looking at the general shape of the graphs) there must be two intersections with x > 0. So there are three intersections altogether.
  2. Here's a practical way to view a counterexample. It's a problem about slopes of pieces of the graph since (f(x)-f(y) average rate of change (f(x)-f(y))/(x-y) must be Draw a jagged continuous graph f(x)which is piecewise-linear ( made up of line segments) with infinitely many larger and larger slopes. If you can imagine this, mark option III false and go on to the next problem. If you want a rigorous definition of a function try this: f(0)=f(1) = 0, between 0 and 1/2 it has slope 1 going up to (1/4,1/4) and slope -1 down to (1/2, 0), from x=1/2 to x=3/4, it goes up to height 2/8 with slope 2 to the half-way point at x=5/8 and down with slope -2 to (3/4,0) Continue in this way, with slopes n and -n between points (1- 1/2^{n-1}, 0) and (1 - 1/2^n, 0), achieving maximal height n/2^{n+1} halfway between these two points. Notice that the limit of n/2^{n+1} as n--> infinity is 0, so the function is continuous at x=1. Then given any E there is an integer n>E and a piece of the graph with slope n. So it won't be true that the slope( f(x)-f(y))/(x-y) x and y in [0, 1].
  3. The question must have been about "homeomorphic" not "homomorphic": It's a topology question. C is homeomorphic to N, S and U since all four of these letters can't be straightened out to (are homeomorphic to) a line segment. C is not homeomorphic to J because if we take away the top middle point of the J we are left with three connected components, while this cannot happen with any point of C. (Using: If X is homeomorphic to Y and x is element of X and y is element of Y then X-{x} is homeomorphic to Y-{y}.) C is not homeomorphic O because the latter has a loop in it that can't be shrunk to a point in it. (Technically, they have different fundamental groups -- not a concept that would be on the GRE.)
  4. The group is given as a multiplicative group of order 15 (with 15 elements). So let's keep it multiplicative. Every element x has order (least power = to identity e) which divides 15. So order of x is 1, 3, 5, 15. If order is 1, 5 or 15 then {x^3,x^5,x^{9}} = {e}, {x^3, e, x^4} or {x^3, x^5, x^9} all distinct in last two cases. So, since the set has two distinct elements, the order of x is 3. Then x^3=x^9=e and x^13=(x^12)x =ex=x. So x^{13n} = x^n. Since x has order 3, x^{13n} achieves 3 values as n goes from 1 to infinity. So {x^{13n} | n is in N} has 3 elements.
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