Practice Questions, really needs help. THANKS.

 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 $$f(x)=x^{\frac{1}_{x1}}$$ 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..
(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 $$f(x)=x^{\frac{1}_{x1}}$$ 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..

 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 semicircular region is half of that, that is, pi/2, and that is your answer  polygon with greater area wouldnt fit in.

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: Practice Questions, really needs help. THANKS.
Ooops...I'm taking it as a regular polygon, if it is, is the ans pi/4?EugeneKudashev wrote:First: the area of a circle of unit radius is, of course, pi*(1)^2=pi. The area of semicircular region is half of that, that is, pi/2, and that is your answer  polygon with greater area wouldnt fit in.

 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, lefthand limit at x_0 should be equal to the righthand limit at x_0 and both should be equal to f(x_0). Let's find the limit at x=1, then:
$$x^{\frac1{x1}}=e^{\log x^{\frac1{x1}}}=e^{\frac1{x1}\log{x}}=e^{\frac{\log{x}}{x1}}$$
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.
$$x^{\frac1{x1}}=e^{\log x^{\frac1{x1}}}=e^{\frac1{x1}\log{x}}=e^{\frac{\log{x}}{x1}}$$
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.

 Posts: 27
 Joined: Tue Apr 06, 2010 8:22 am
Re: Practice Questions, really needs help. THANKS.
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 semicircle. So this upper bound is the area of your semicircle. Does that help?speedychaos4 wrote:Ooops...I'm taking it as a regular polygon, if it is, is the ans pi/4?EugeneKudashev wrote:First: the area of a circle of unit radius is, of course, pi*(1)^2=pi. The area of semicircular region is half of that, that is, pi/2, and that is your answer  polygon with greater area wouldnt fit in.
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 semicircular 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/(x1))
take log both side
log y = Limit (X> 1) [ log x / (x1) ]
0/0 form, differentiate
log y = Limit (X> 1) ( (1/x) /x)
log y = 1
so y = e ==> ANS

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 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.EugeneKudashev wrote: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 semicircle. So this upper bound is the area of your semicircle. Does that help?speedychaos4 wrote:Ooops...I'm taking it as a regular polygon, if it is, is the ans pi/4?EugeneKudashev wrote:First: the area of a circle of unit radius is, of course, pi*(1)^2=pi. The area of semicircular region is half of that, that is, pi/2, and that is your answer  polygon with greater area wouldnt fit in.

 Posts: 27
 Joined: Tue Apr 06, 2010 8:22 am
Re: Practice Questions, really needs help. THANKS.
I highly doubt. Check for square by yourself, for instanсe.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.
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.
Re: Practice Questions, really needs help. THANKS.
For a regular ngon, 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.
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.

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: Practice Questions, really needs help. THANKS.
Ah! I did not see the word NUMBER when I saw the question, I thought it was asking which equation has the biggest root!enork wrote:For a regular ngon, 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.
Really a lesson to learn, be carefully while reading the question.