Help with the following problems greatly appreciated

 Posts: 18
 Joined: Sun Oct 07, 2007 1:49 pm
Help with the following problems greatly appreciated
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...
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...

 Posts: 2
 Joined: Tue Oct 30, 2007 1:30 am
I.
log c = x is a critical point
second derivative is always positive, therefore critical pt is a minimum
II.
This function is periodic. therefore there are infinitely many pts where f' = 0. since this its derivatives can only have a finite number of roots, as it is also a polynomial or identically 0, then f must be a constant function. i.e. f(x) = 11 for all x
log c = x is a critical point
second derivative is always positive, therefore critical pt is a minimum
II.
This function is periodic. therefore there are infinitely many pts where f' = 0. since this its derivatives can only have a finite number of roots, as it is also a polynomial or identically 0, then f must be a constant function. i.e. f(x) = 11 for all x
III)
Ok. if 3x+7y is divisible by 11 so is A=5*(3x+7y)=15x+35y. also B=11x44y is also divisble by 11. So AB=4x9y should also be divisble by 11.
IV)
since the answers should be complex conjugates, the sum of all roots will be equal to S=2+1+2+1=6. and sicne in any polynomial of order n, the sum of roots are given by S= a_{n1}/a_n, where a_n and a_{n1} are the coefficients of x^2 and x^{n1}, respectively, the correct answer should in the form of x^4+ax^3+..., where a=6. This is only the case for the last choice
V)
Ok. Let's see what the problem is about. it states that the first one receiving the head will win the game. Suppose that players are named as A and B. player A, who starts ther game, can win in the following cases,
I) A=head.
II) A=tail, B=tail, A=head
III) A=tail, B=tail, A=tail, B=tail,A=head
IV) ...
so the probability would be P=1/2+1/2*(1/4)+1/2*(1/4)^2+...=2/3
VI)
the last one! we have the following relation x_{n+1}=SQRT(3+2*x_n), for the sequence to converge as n>inf, we must have x_{n+1}=x_n, let us name limit n>inf x_n=p, then we have the following equation after raising the both sides to power of 2
p^22P3=0 ==> p=3
Ok. if 3x+7y is divisible by 11 so is A=5*(3x+7y)=15x+35y. also B=11x44y is also divisble by 11. So AB=4x9y should also be divisble by 11.
IV)
since the answers should be complex conjugates, the sum of all roots will be equal to S=2+1+2+1=6. and sicne in any polynomial of order n, the sum of roots are given by S= a_{n1}/a_n, where a_n and a_{n1} are the coefficients of x^2 and x^{n1}, respectively, the correct answer should in the form of x^4+ax^3+..., where a=6. This is only the case for the last choice
V)
Ok. Let's see what the problem is about. it states that the first one receiving the head will win the game. Suppose that players are named as A and B. player A, who starts ther game, can win in the following cases,
I) A=head.
II) A=tail, B=tail, A=head
III) A=tail, B=tail, A=tail, B=tail,A=head
IV) ...
so the probability would be P=1/2+1/2*(1/4)+1/2*(1/4)^2+...=2/3
VI)
the last one! we have the following relation x_{n+1}=SQRT(3+2*x_n), for the sequence to converge as n>inf, we must have x_{n+1}=x_n, let us name limit n>inf x_n=p, then we have the following equation after raising the both sides to power of 2
p^22P3=0 ==> p=3

 Posts: 18
 Joined: Sun Oct 07, 2007 1:49 pm
Thank you for your help
Thank you for the help mamal...
Could you please explain how you got the value
B=11x44y is also divisble by 11
in your answer to number III?
Could you please explain how you got the value
B=11x44y is also divisble by 11
in your answer to number III?

 Posts: 18
 Joined: Sun Oct 07, 2007 1:49 pm
Regarding question II
ihaveanswers,
For question II, how did you discern that f(x) is periodic. Once that is understood your answer follows completely. Thanks!
For question II, how did you discern that f(x) is periodic. Once that is understood your answer follows completely. Thanks!
Yep. you see regardless of x, 11x can be devided by 11 (since it contains 11) and also 44y can be devided by 11 (it may be written as 4*11*y and is so devisible by 11) thus the sum of 11x44y are also devisible by 11.
PS: I am assuming that you know that x and y are integers. You know that, don't you ?
PS: I am assuming that you know that x and y are integers. You know that, don't you ?

 Posts: 18
 Joined: Sun Oct 07, 2007 1:49 pm
Regarding question III
mamal,
You are totally right, we have x & y as positive integers otherwise we could have infinite solutions.
The choice for B is totally clear since AB gives 4x  9y is divisible by 11. Other than through a bit of trial and error, was there another method you used for picking B?
You are totally right, we have x & y as positive integers otherwise we could have infinite solutions.
The choice for B is totally clear since AB gives 4x  9y is divisible by 11. Other than through a bit of trial and error, was there another method you used for picking B?

 Posts: 8
 Joined: Tue Mar 04, 2008 1:02 am
After spending a few minutes on the ones that look easy, here are some hints I can think of:
I. Remember to solve for x. I hate it when log is used rather than ln when dealing with e.
III. In problems like this, if you cannot solve it via theorems, try plugging in some numbers. For example, within a minute you can find that if x=3, y=7,
3x + 7y is divisible by 11. Substituting these values of x,y into the answer choices shows us d is the only choice divisible by 11 when x=3, y=7.
V. Even if you have no clue about probability, you should have a 1/2 chance of answering this question correctly. First off, suppose player A and B are playing the game. Player A tosses first, and has a .5 chance of getting heads. Thus, the probability that player A wins on the first toss is .5. Choices a) and b) should be immediately discarded. Choice c) may seem plausible, but remember that player A is tossing first, so he will have an advantage over player B, so choice c) can be tossed out. Thus, two choices are left (d and e). I believe mamal explained the solution pretty well.
VI. Choice a) can be eliminated immediately. Choice c) can be eliminated since x_(n+1) will always be at least as large as sqrt(3). Thus, you are already at an advantage when it comes to blind guessing. mamals answer is pretty solid.
I. Remember to solve for x. I hate it when log is used rather than ln when dealing with e.
III. In problems like this, if you cannot solve it via theorems, try plugging in some numbers. For example, within a minute you can find that if x=3, y=7,
3x + 7y is divisible by 11. Substituting these values of x,y into the answer choices shows us d is the only choice divisible by 11 when x=3, y=7.
V. Even if you have no clue about probability, you should have a 1/2 chance of answering this question correctly. First off, suppose player A and B are playing the game. Player A tosses first, and has a .5 chance of getting heads. Thus, the probability that player A wins on the first toss is .5. Choices a) and b) should be immediately discarded. Choice c) may seem plausible, but remember that player A is tossing first, so he will have an advantage over player B, so choice c) can be tossed out. Thus, two choices are left (d and e). I believe mamal explained the solution pretty well.
VI. Choice a) can be eliminated immediately. Choice c) can be eliminated since x_(n+1) will always be at least as large as sqrt(3). Thus, you are already at an advantage when it comes to blind guessing. mamals answer is pretty solid.
hello people. let me joining your comunity be the first thing i do on the way to the GRE.
1): i think you got it in second reading.
2): i didnt get it at first, but can point that it is obvious that it must be periodical
F(x)=F(x+1) for every x (including x+1)>>>
>>>F(x+1)=F(x+1+1)>>>=F(x+k*1)
3): if (3x+7y)11 then so is c*(3x+7y). for every a E Z
by multipling this for every c=0:10 we will find all the pairs (a,b) so that a*x+b*y is also divisible by 11.
we can write a mod 11 & b mod 11 instead of a & b because from: a*x+b*y11 >>
>>>>>(ak*11)*x+(bs*11)*y11
a) 4x + 6y
c) 9x + 4y
d) 4x  9y
this are the choises for (a,b) and we check for what c we have a=4,9
fof a to be 4, c*3 mod 11 = 4, this is true for c=5>> (a,b)=(4,2) but (when working with moduls) 2=9 >>>
the true answer is C.
4): obvious by using the Wiet formulas
5): Tha famous Pascal problem that founded the study of modern probability
6): nice aproach mamal. have never done such kind of problems
1): i think you got it in second reading.
2): i didnt get it at first, but can point that it is obvious that it must be periodical
F(x)=F(x+1) for every x (including x+1)>>>
>>>F(x+1)=F(x+1+1)>>>=F(x+k*1)
3): if (3x+7y)11 then so is c*(3x+7y). for every a E Z
by multipling this for every c=0:10 we will find all the pairs (a,b) so that a*x+b*y is also divisible by 11.
we can write a mod 11 & b mod 11 instead of a & b because from: a*x+b*y11 >>
>>>>>(ak*11)*x+(bs*11)*y11
a) 4x + 6y
c) 9x + 4y
d) 4x  9y
this are the choises for (a,b) and we check for what c we have a=4,9
fof a to be 4, c*3 mod 11 = 4, this is true for c=5>> (a,b)=(4,2) but (when working with moduls) 2=9 >>>
the true answer is C.
4): obvious by using the Wiet formulas
5): Tha famous Pascal problem that founded the study of modern probability
6): nice aproach mamal. have never done such kind of problems