## Practice Questions, really needs help. THANKS.

Forum for the GRE subject test in mathematics.
speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Practice Questions, really needs help. THANKS.

1. What is the least upper bound of the set of all numbers A such that a polygon with area A can be inscribed in a semicircular region of radius 1?
(A)4/5 (B)2/sqrt(5) (C)1 (D)pi/2 (E)2
I though the ans should be pi/4, but there is no such option, why is that?

2.If $f(x)=x^{\frac{1}_{x-1}}$ for all positive x!=1 and if f is continuous at 1, then f(1) is
(A)0 (B)1/e (C)1 (D)e (E)none of the above

3.Of the following equations, which has the greatest number of roots between 100 and 1,000?
(A)sin(x)=0 (B)sin(x^2)=0 (C)sin(|x|^(1/2))=0 (D)sin(x^3)=0 (E)sin(x^(1/3))=0

4.The order of the element $\sigma =
\left( \begin{array}{ccccc}
1 & 2 & 3 & 4 & 5 \\
\downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\
4 & 5 & 1 & 3 & 2
\end{array} \right)$
of the symmetric group S_5 is
(A) 2 (B) 3 (C) 6 (D) 8 (E) 12

P.S. I'm sorry about the latex, but I have no idea how to make it right..

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

### Re: Practice Questions, really needs help. THANKS.

First: the area of a circle of unit radius is, of course, pi*(1)^2=pi. The area of semi-circular region is half of that, that is, pi/2, and that is your answer - polygon with greater area wouldnt fit in.

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: Practice Questions, really needs help. THANKS.

EugeneKudashev wrote:First: the area of a circle of unit radius is, of course, pi*(1)^2=pi. The area of semi-circular region is half of that, that is, pi/2, and that is your answer - polygon with greater area wouldnt fit in.

Ooops...I'm taking it as a regular polygon, if it is, is the ans pi/4?

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

### Re: Practice Questions, really needs help. THANKS.

Second: for a function to be continuous at x=x_0, left-hand limit at x_0 should be equal to the right-hand limit at x_0 and both should be equal to f(x_0). Let's find the limit at x=1, then:

$x^{\frac1{x-1}}=e^{\log x^{\frac1{x-1}}}=e^{\frac1{x-1}\log{x}}=e^{\frac{\log{x}}{x-1}}$
here we'll use the L'Hopital rule, because if you take the limit at $x\rightarrow1+ or x\rightarrow 1-$ this becomes 0/0
$=e^{\frac{1/x}{1}}=e^{\frac1{x}}$
as x approaches 1 either on the left or on the right, the 1/x tends to 1, thus e^(1/x) tends to e, and that is your answer.

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

### Re: Practice Questions, really needs help. THANKS.

speedychaos4 wrote:
EugeneKudashev wrote:First: the area of a circle of unit radius is, of course, pi*(1)^2=pi. The area of semi-circular region is half of that, that is, pi/2, and that is your answer - polygon with greater area wouldnt fit in.

Ooops...I'm taking it as a regular polygon, if it is, is the ans pi/4?

I'm not following what the 'regularity' has to do with the question. You have to find the maximum area of some figure (forget about the type of it) that is to be inscribed into a semi-circle. So this upper bound is the area of your semi-circle. Does that help?

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### Re: Practice Questions, really needs help. THANKS.

EugeneKudashev wrote:First: the area of a circle of unit radius is, of course, pi*(1)^2=pi. The area of semi-circular region is half of that, that is, pi/2, and that is your answer - polygon with greater area wouldnt fit in.

Ques2)

f(1) = y = Limit (X-> 1) x^(1/(x-1))

take log both side

log y = Limit (X-> 1) [ log x / (x-1) ]

0/0 form, differentiate

log y = Limit (X-> 1) ( (1/x) /x)

log y = 1

so y = e ==> ANS

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: Practice Questions, really needs help. THANKS.

EugeneKudashev wrote:
speedychaos4 wrote:
EugeneKudashev wrote:First: the area of a circle of unit radius is, of course, pi*(1)^2=pi. The area of semi-circular region is half of that, that is, pi/2, and that is your answer - polygon with greater area wouldnt fit in.

Ooops...I'm taking it as a regular polygon, if it is, is the ans pi/4?

I'm not following what the 'regularity' has to do with the question. You have to find the maximum area of some figure (forget about the type of it) that is to be inscribed into a semi-circle. So this upper bound is the area of your semi-circle. Does that help?

I totally understand the question, I'm asking if all the edges of the polygon are equal in length, is the ans going to be pi/4? Thx.

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

### Re: Practice Questions, really needs help. THANKS.

I totally understand the question, I'm asking if all the edges of the polygon are equal in length, is the ans going to be pi/4? Thx.

I highly doubt. Check for square by yourself, for instanсe.

Concerning #4: the order of an element is the order of a cyclic subgroup generated by the element, if I recall the def correctly. Thus, you have just to apply this permutation as many times as required to get the initial 12345 element. here are the computations:

12345
45132 (a^1)
32415 (a^2)
15342
42135
35412
12345 (a^6) - voilà, the order is 6.

enork
Posts: 33
Joined: Fri Sep 18, 2009 3:16 am

### Re: Practice Questions, really needs help. THANKS.

For a regular n-gon, the area goes to pi/4 as n goes to infinity, but the area doesn't necessarily increase as n does. For example you can fit in a square that's larger than pi/4. Don't know what the max is though.

For #3, sin(f(x)) has a zero whenever f(x) crosses a multiple of 2pi, so the choice with the most zeros is the one with f that grows the fastest, which is f = x^3.

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: Practice Questions, really needs help. THANKS.

enork wrote:For a regular n-gon, the area goes to pi/4 as n goes to infinity, but the area doesn't necessarily increase as n does. For example you can fit in a square that's larger than pi/4. Don't know what the max is though.

For #3, sin(f(x)) has a zero whenever f(x) crosses a multiple of 2pi, so the choice with the most zeros is the one with f that grows the fastest, which is f = x^3.

Ah! I did not see the word NUMBER when I saw the question, I thought it was asking which equation has the biggest root!

Really a lesson to learn, be carefully while reading the question.