## 9768 #55 #57

Forum for the GRE subject test in mathematics.
silveer
Posts: 4
Joined: Thu Oct 08, 2009 3:29 am

### 9768 #55 #57

#55.
Suppose f is a twice-differentiable function on the set of real numbers and that f(0), f'(0), and f''(0) are all negative. Suppose f''has three of the following properties.

I. It is increasing on the interval [0, inf).
II. It has a unique zero in the interval [0, inf).
III. It Is unbounded on the interval [0, inf).

Which of the same properties does f necessarily have?

(A) I
(B) II
(C) III
(E) I, II and III

#57.
Let R be the field of real numbers and R [x] the ring of polynomials in x with coefficients in R. Which of the following subsets of R [x] is a subring of R [x]?

I. All polynomials whose coefficient of x is zero
II. All polynomials whose degree is an even integer, together with zero polynomial
III. All polynomials whose coefficients are rational numbers

(A) I
(B) II
(D) II and III
(E) I, II and III

mtey
Posts: 22
Joined: Tue Oct 27, 2009 12:30 pm

### Re: 9768 #55 #57

Today I managed to get them both wrong however after the test I solved them and they are actually not so tough.

55

Since f'' has only one root - $\alpha$ and it is increasing, f' is decreasing in $[0, \alpha )$ and increasing in $[\alpha, \infty]$. Thus from f'(0) < 0 follows that f' has only one root(it is continuous since f'' exists) and it is unbounded from the mean-value theorem. The same argument is valid for f. So f is unbounded and has a unique solution, but it is NOT always increasing (just like f' too).

57

I. You can easily verify that this is a subring by checking that 1, 0, addition and multiplication of elements are in it.

II. This is what I got wrong. Here is a simple counter example p1(x) = -x^2 + x, p2(x) = x^2; p1(x) + p2(x) = x is not in this subset of R[x]

III. Now it is a well known fact that Q[x] is ring of the polynomials with rational coefficients, but if you don't know that you can check it easily.
We cannot go out of the ring by using multiplication and addition and it contains 1 and 0 too.

feroz_apple
Posts: 30
Joined: Sat Oct 17, 2009 1:20 pm

### Re: 9768 #55 #57

mtey wrote:Today I managed to get them both wrong however after the test I solved them and they are actually not so tough.

55

Since f'' has only one root - $\alpha$ and it is increasing, f' is decreasing in $[0, \alpha )$ and increasing in $[\alpha, \infty]$. Thus from f'(0) < 0 follows that f' has only one root(it is continuous since f'' exists) and it is unbounded from the mean-value theorem. The same argument is valid for f. So f is unbounded and has a unique solution, but it is NOT always increasing (just like f' too).

57

I. You can easily verify that this is a subring by checking that 1, 0, addition and multiplication of elements are in it.

II. This is what I got wrong. Here is a simple counter example p1(x) = -x^2 + x, p2(x) = x^2; p1(x) + p2(x) = x is not in this subset of R[x]

III. Now it is a well known fact that Q[x] is ring of the polynomials with rational coefficients, but if you don't know that you can check it easily.
We cannot go out of the ring by using multiplication and addition and it contains 1 and 0 too.

I dont think you need to show the existence of 1 for it to be a subring

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

### Re: 9768 #55 #57

If 1 wasn't in the set, it wouldnt be a ring, and therefore not a subring.

And it is possible to have 0, addition, and multiplication to be defined on a subset of a ring but not include 1, such is the case for any proper ideal.

silveer
Posts: 4
Joined: Thu Oct 08, 2009 3:29 am

### Re: 9768 #55 #57

Thank you, now I see my mistakes))

mtey
Posts: 22
Joined: Tue Oct 27, 2009 12:30 pm

### Re: 9768 #55 #57

feroz_apple wrote:I dont think you need to show the existence of 1 for it to be a subring

http://en.wikipedia.org/wiki/Subring

Well, I believe that there is a certain misunderstanding It depends on whether one defines a ring to have a multiplicative identity or not. So if we have rings on the actual test it will be defined explicitly, or we should check whether 1 is in the possible subring or not.

feroz_apple
Posts: 30
Joined: Sat Oct 17, 2009 1:20 pm

### Re: 9768 #55 #57

origin415 wrote:If 1 wasn't in the set, it wouldnt be a ring, and therefore not a subring.

And it is possible to have 0, addition, and multiplication to be defined on a subset of a ring but not include 1, such is the case for any proper ideal.

A ring doesn't require the unity to exists.
Only when its either a division ring(skew field), a field or an integral domain, do we need to check the existence of unity.
eg:
the set of matrices {( 2a 0 )
( 0 2b) : where a,b are integers is a ring without unity and containing divisors of zero}

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

### Re: 9768 #55 #57

In my abstract algebra class, rings always had unity, I've never heard it any other way.

It appears the Princeton Review book doesn't require unity, so I suppose the GRE doesn't either, my mistake.

mtey
Posts: 22
Joined: Tue Oct 27, 2009 12:30 pm

### Re: 9768 #55 #57

origin415 wrote:In my abstract algebra class, rings always had unity, I've never heard it any other way.

It appears the Princeton Review book doesn't require unity, so I suppose the GRE doesn't either, my mistake.

Uhhm yes, it appears so. I asked an algebra professor from my university and he said that according to the definition a ring is not required to have a unity. Although, he said, some people define rings to have a unity.

breezeintopl
Posts: 16
Joined: Sun Jul 19, 2009 12:29 pm

### Re: 9768 #55 #57

[quote="mtey"]Today I managed to get them both wrong however after the test I solved them and they are actually not so tough.

55

Since f'' has only one root - $\alpha$ and it is increasing, f' is decreasing in $[0, \alpha )$ and increasing in $[\alpha, \infty]$. Thus from f'(0) < 0 follows that f' has only one root(it is continuous since f'' exists) and it is unbounded from the mean-value theorem. The same argument is valid for f. So f is unbounded and has a unique solution, but it is NOT always increasing (just like f' too).
quote]

Why can you assume that f'' has one root?

mtey
Posts: 22
Joined: Tue Oct 27, 2009 12:30 pm

### Re: 9768 #55 #57

breezeintopl wrote:Why can you assume that f'' has one root?

Well this is the II condition for f''

"II. It has a unique zero in the interval [0, inf)."