Math GRE subject test practice questions

I would appreciate it if anyone could shed some light as to the answer/
procedure to reach an answer for the following problems. Anything to
get me started will do.


I
If c > 0 and f(x) = e^(x) - cx for all real number x, then the minimum
value of f is

a) f(c)
b) f(e^c)
c) f(1/c)
ans d) f(log c)
e) nonexistent

y' = e^x - c, when y' = 0 c = e^x
y'' = e^x
Don't see a way to discern the min value...


II
Suppose that f(1 + x) = f(x) for all real x. If f is a polynomial and
f(5) = 11 then f(15/2) is

a) -11
b) 0
ans c) 11
d) 33/2
e) not uniquely determined

Don't know how to get this one started


III Other than trying out the answers, is there a quicker way?
Let x and y be positive integers such that 3x + 7y is divisible by 11.
Which of the following must also be divisible by 11?

a) 4x + 6y
b) x + y + 5
c) 9x + 4y
ans d) 4x - 9y
e) x + y - 1


IV This one seems common...
If a polynomial f(x) over the real numbers has the complex numbers 2 +
i and 1 - i as roots, then f(x) could be

a) x^4 + 6X^3 + 10
b) x^4 + 7x^2 +10
c) x^3 - x^2 + 4x +1
d) x^3 + 5x^2 + 4x +1
ans e) x^4 - 6x^3 + 15x^2 - 18x +10

I realize that these roots occur in complex conjugate pairs...
Checking the answers seems like too much work...


V
In a game two players take turns tossing a fair coin; the winner is
the first one toss a head. The probability that the player who makes
the first toss wins the game is

a) 1/4
b) 1/3
c) 1/2
ans d) 2/3
e) 3/4

This one might be about the wording. I don't see why the answer is 2/3


VI
Let x(sub 1) = 1 and x(sub n + 1) = sqrt(3 + 2*n(sub n)) for all
positive integers n. If it is assumed that {x(sub n)} converges, then
lim x -> infiniti x(sub n) =

a) -1
b) 0
c) sqrt(5)
d) e
ans e) 3

I get this somewhat nasty nested function of square roots and
basically I do not see how it may simplify...

Thanks again

.

Relevant Pages

  • Re: Symbolic Mathematics quesetion
    ... what i get is 'Unable to find closed form solution'. ... is that show i can get the ans? ... roots of the resulting polynomial expression. ...
    (comp.soft-sys.matlab)
  • Re: fsolve
    ... Shaun wrote: ... rmsearch will find them. ... The solutions that remain are the roots. ...
    (comp.soft-sys.matlab)
  • Re: Polynomial roots
    ... to get the roots of this function use the roots command ... The next MATLAB function, which I remember is the fzero command. ...
    (comp.soft-sys.matlab)
  • Re: cubed root
    ... supposed to take uneven roots of negative numbers and get the ... Because there are three cube roots ...
    (comp.soft-sys.matlab)
Quantcast