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Some GRE Math subject exam questions

  1. Aug 9, 2008 #1
    1) For all positive functions f and g of the real variable x, let ~ be the relation defined by:

    f ~ g iff (if and only if) lim x --> (infinity) [tex]\frac{f(x)}{g(x)}[/tex] = 1. Which of the following is NOT a consequence of f~g?

    A. f^2 ~ g^2
    B. sqrt(f) ~ sqrt(g)
    C. e^f ~ e^g (e as in the natural logarithm)
    D. f + g ~ 2g
    E. g ~f

    So I am really curious why the relationship in C DOES NOT hold. Since
    if e^f ~ e^g, then

    lim x --> infinity [tex]\frac{e^f(x)}{e^g(x)}[/tex] = lim e^(f - g) = lim e^(0) since f and g have a common limit as x approaches infinity, and hence e^0 = 1.

    2) What is the minimum value of the expression x+4z as a function defined on R^3, subject to the constraint that x^2 + y^2 + z^2 [tex]\leq[/tex] 2?

    A. 0
    B. -2
    C. -sqrt(34)
    D. -sqrt(35)
    E. -5sqrt(2)

    - For this one I tried using Lagrange multiplier method to solve it. I switched to an equality since we want the minimum value, and that will occur on the surface of the sphere of radius sqrt(2). So I get (1, 0, 4) = L(2x, 2x, 2z), but the numbers dont quite work out. I get something like -17*sqrt(2), not quite the right answer C, but close. I don't know why I'm not getting the right value.

    3) For how many positive integers k does the ordinary decimal representation of the integer k! end in exactly 99 zeroes?

    A. None
    B. One
    C. Four
    D. Five
    E. Twenty four

    - I was a little confused by this one. I didn't quite know which way I was supposed to approach it. I tried working from the fact that 100! ends in 24 zeroes, but I didn't get very far.

    Last edited: Aug 9, 2008
  2. jcsd
  3. Aug 9, 2008 #2
    Well for number one, a rational function such that f(x) and g(x) have the same leading terms satisfies the limit part. Then from deduction A,B,D, and E are consequences.

    Following your derivation then to the point where you claimed lim e^(f-g) = lim e^(0). This is not necessarily true. If the leading terms cancel, then the rest of the terms can result in a number of different limiting values.

    Anyways someone more acquainted with higher math could probably give a more formal/correct argument.

    3) That's a pretty useless fact to just know. I'm sure you know how it was determined right? If you do the problem should just come down to casework.
    Last edited: Aug 9, 2008
  4. Aug 9, 2008 #3


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    Was C given as the answer? I would be more inclined, off the top of my head, to say B because of the possiblility that f and g are negative for large x.

    How in the world would you get -17*sqrt(2)? Using the Lagrange multiplier method, we get [itex]2x= \lambda[/itex], [itex]2y= 0[/itex], [itex]2z= 4\lambda[/itex]. dividing the last equation by the first, z/x= 4 or z= 4x and y= 0. Putting that into x2+ y2+ z2= 2, x2+ 16x2= 17x2= 2 so
    [tex]x= \sqrt{\frac{2}{17}[/tex] and [tex]z= 4\sqrt{2}{17}[/tex]
    [tex]x+ 4z= \sqrt{\frac{2}{17}}+ 16\sqrt{\frac{2}{17}}= 17\sqrt{\frac{2}{17}}= \sqrt{34}[/tex]

    In order that k! end in 99 zeroes, it must have 99 factors of 10 and so 99 2's and 99 5's. Since there are going to be fewer 5's than 2's, k! must have exactly 99 5's. We first get that for k= 99*5= 495 and then get 100 5's for k= 100*5= 500. Any number from 495 to 499 will have exactly 99 5's in its factorial. There are 5 such numbers.
  5. Aug 9, 2008 #4
    But the problem with that is you're not considering the powers of 5 are you? Take 495! as your example. First consider the multiples of 5, we'll have 495/5 = 99 of them. But then we have to consider 5^2 = 25 and its multiples. For each of these, we get ANOTHER zero. We have to consider all n such that 5^n <= k!.
  6. Aug 9, 2008 #5


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    This is the same as asking for how many k, k! has exactly 99 factors of 5 (as 2's occur more often, it will automatically have more than 99 factors of 2). If k is the first such value, then it must be a multiple of 5.
    Then, as k+5 will also be a multiple of 5, n! will end in at least 100 zeros for any n >= k+5. Therefore k,k+1,k+2,k+3,k+4 are the only ones ending in 100 0's and the answer is D:Five. *Unless* there are no such k. The number of zeros could jump from less than 99 to more than 99 if k is a multiple of 25, in which case there will be no such k and the answer is A:None. So, you need to say which of these cases happens, and it is a bit trickier than it looks at first.
  7. Aug 9, 2008 #6


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    A formula for the number of factors of p occuring in n! (where p is prime) is
    \left\lfloor \frac{n}{p}\right\rfloor
    +\left\lfloor \frac{n}{p^2}\right\rfloor
    +\left\lfloor \frac{n}{p^3}\right\rfloor
    Here [itex]\lfloor\cdot\rfloor[/itex] is the floor function.
    If n=52x16 then the number of factors of 5 in n! is 5x16 + 16 + 3 = 99. So the correct answer to (3) is D:Five.
    Note that if it had asked for 98 0's then the answer would have been A:None.
  8. Aug 9, 2008 #7
    The ETS practice book says 5 is the correct answer. Thank everyone for replying and giving me some guidance.
  9. Aug 9, 2008 #8


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    And following on to snipez90 first post, with a simple concrete example. If f=1+x and g=x then f~g, but exp(f-g)=e.
  10. Sep 13, 2008 #9
    Another GRE Math question

    Hi, new to the forums and haven't seen this question posted yet... Here's a GRE math subject test question from the sample ETS exam (Form GR0568). I'm having trouble and wondering if I'm missing something obvious.

    24. Let h be the function defined by [tex]h(x) = \int^{x^2}_{0}\\e^{x+t} dt[/tex] for all real numbers x. Then h'(1) =

    (A) [tex]e - 1[/tex]
    (B) [tex]e^2[/tex]
    (C) [tex]e^2 - e[/tex]
    (D) [tex]2e^2[/tex]
    (E) [tex]3e^2 - e[/tex]

    The answer given by the answer key is
    , but I'm not getting the right numbers. My calculations are the following:

    h(x) = \int^{x^2}_{0}\\e^{x+t} dt
    = e^{x} \int^{x^2}_{0}\\e^{t} dt
    So using the Fundamental Thm of Calculus,
    [tex] h'(x) = (e^x) (\frac{d}{dx}\int^{x^2}_{0}\\e^{t} dt) + (\int^{x^2}_{0}\\e^{t} dt) (e^x) [/tex]
    [tex] = (e^x) (2x(F'(x^2)) - F'(0)) + (e^x) (e^{t}\mid^{x^2}_{0}) [/tex]
    [tex] = (e^x) (2x e^{x^2} - e^0) + (e^x) (e^{x^2}-e^0) [/tex]
    [tex] = (e^x) (2x e^{x^2} - 1) + (e^x) (e^{x^2}-1) [/tex]
    [tex] = (e^x) ((2x+1) e^{x^2} - 2) [/tex]
    [tex] h'(1) = e (3 e - 2) = 3 e^{2} - 2 e [/tex]

    This answer does not match the correct answer (or even any of the other choices).
    Where did my calculations go wrong? Any help would be much appreciated.
  11. Sep 13, 2008 #10
    You applied the product rule correctly there. However, in subsequent calculations, you seem to have misapplied the rules to this term:

    [tex]\frac{d}{dx}\int^{x^2}_{0}\\e^{t} dt[/tex]

    which is equal to [tex]2xe^{x^2}[/tex] by the fundamental theorem. So I think you'll get

    [tex]h'(x) = (e^x)(2xe^{x^2}) + (e^{x^2} - 1)(e^x)[/tex]

    and substituting x = 1 should give you the answer you are looking for.

    But the very first step, when you pulled out the [tex]e^x[/tex], is that because we're treating it as a constant? I know that kind of function lends itself to applying the fundamental theorem to cover the original function in the integrand.
  12. Sep 13, 2008 #11
    Ah ha! Right. The derivative of F(0) (F being the antiderivative of the integrand) is 0 and not F'(0). Thanks snipez, I thought I was going crazy there.

    Yep, I was treating it as a constant (under the integral). I did at first try to use the fundamental thm with the original function, but since I was having problems getting the right answer I thought I'd try to simplify the integrand part of it as much as possible.

    Anyhow, thanks again!
  13. Oct 27, 2008 #12

    f(x)=x^2+2, g(x)=x^2+1
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