## GR9768 problems 23, 29, 32, 36, 50, 63

Forum for the GRE subject test in mathematics.
fullofquestions
Posts: 18
Joined: Sun Oct 07, 2007 1:49 pm

### GR9768 problems 23, 29, 32, 36, 50, 63

Would you guys mind giving me your thoughts regarding the following problems? If you prefer to read the actual questions, the pdf for this test is located at http://www.ets.org/Media/Tests/GRE/pdf/Math.pdf. Thanks a bunch.

23.
In the euclidean plane, point A is on a circle centered at point O, and O is on a circle centered at A. The circles intersect at points B and C. What is the measure of the angle BAC?
ans. 120 degrees

29.
Assume that p is a polynomial function on the set of real numbers. If p(0) = p(2) = 3 and p'(0) = p'(2) = -1, then integral(0 to 2) {xp''(x)dx =?
ans. -2

32.
When 20 children in a classroom line up for lunch, Pat insists on being somewhere ahead of Lynn. If Pat's demand is to be satisfied, in how many ways can the children line up?
ans. 20!/2

I know the answer is not 20!

36.
For each real number x, let mu(x) be the mean of the numbers 4, 5, 7, 9 and x; and let n(x) be the median of these five numbers. For how many values of x is mu(x) = to n(x)?
ans. 3

mu(x) = (25 + x)/5

50.
How many continuous real-valued functions f are there with domain [-1,1] such that (f(x))^2 = x^2 for each x in [-1,1]?
ans. four

63.
At how many points in the xy-plane do the graphs of y = x^12 and y = 2^x intersect?
ans. Three

I tried to solve this and get to the point, 12 log x = x log 2.

Zeek
Posts: 2
Joined: Fri Nov 02, 2007 7:20 pm
23 Is fairly apparent if you draw if carefully. Both circles have the same radius, r, which is the distance from A to O. B and C also lie along both circles, so the lines AO, AB, AC, OB, and OC all have length r. Triangles ABO and ACO are thus both equilateral triangles. Therefore angle BAO and angle CAO are both 60 degrees, and angle BAC is 120.

32. There are 20! ways for the children to line up without restriction. In half of them, Pat is ahead of Lyn and in the other half Lyn is ahead of Pat. So only half of the 20! ways are acceptable to Pat, i.e., 20!/2.

36. Here, the mean is (4+5+7+9+x)/5 = 5 + x/5. The median can only be 5 (if x < 5), 7 (if x>7) or x (if 5<x<7). Therefore, to force the median to equal the mean, you must solve on of the 3 equations
5 + x/5 = 5
5 + x/5 = 7
5 + x/5 = x
Each of these equations has only 1 solution, each different from the other, so there are exactly 3 possible values of x such that mean = median.

63. It's easier to do this graphically than to solve it. x raised to any even power graphs as a parabola-like figure, with its base centered at (0,0). The higher the power, the flatter the base of the "parabola." 2^x is a curve running from 0 to infinity, which must cross such a "parabola" at exactly two points, once on each prong of the "parabola."

I'm really confused by 50. Maybe I am missing something very obvious, but it seems to me that any constant function, f(x) = c for all x, would fulfill this requirement, so there should be infinitely many such functions.

Zeek
Posts: 2
Joined: Fri Nov 02, 2007 7:20 pm
Just realized that the answer I came up with for 63 isn't the correct one. Sorry, guess I'm confused on that one too.

29 is just integration by parts. Let f(x) = x and g(x) = p'(x), then integral(0 to 2) {xp''(x)dx} = integral(0 to 2) {f(x)g'(x)dx} = [f(x)g(x)](evaluated from 0 to 2) - integral(0 to 2){f'(x)g(x)dx} = [2(-1) - 0(-1)] integral(0 to 2){p'(x)dx} = -2 - [p(2) - p(0)] = -2 - (3 - 3) = -2

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
I'm really confused by 50.

Here I tried to sketch out how I see the solution.
Apparently, the set of points (x,y) that would satisfy such conditions could be represented by two lines:

Nevertheless, there are just four ways to establish the continuous function on the given [-1,1] interval. Here they are:

f(x) = |x|
f(x) = x
f(x) = -|x|
f(x) = -x

I am not sure whether my solution is correct. Please let me know if you have any ideas.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
Just realized that the answer I came up with for 63 isn't the correct one. Sorry, guess I'm confused on that one too.

We all (or almost all ) do that mistake considering that there are only two points of intersections here. I did either. And I think it is kind of natural and normal mistake. Trying to solve it we wonder how the graphs would like and get the picture like that:

Since the graph of power function x^12 looks so strenght near the origin it obfuscates us and makes to believe that exponential function will never catch up with it. But lets try to see what will really happen as we go to the infinity:

Exponential function literally devoured the power function. It is indeed powerful, and we made a mistake underestimating it. The real graph would be like this one:

freddie
Posts: 2
Joined: Mon Oct 27, 2008 5:24 am
Correct solution lime!

From f^2(x)=x^2, one gets |f(x)|=|x|, so, the solutions are x, -x, |x|, -|x|