University of Warwick
Coventry

# Warwick First Year Maths-Past paper questions

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exactly the same but more.

don't understand most of foundations. left it all a bit late. hopefully the whole exam will be on truth tables or sommat!
University of Warwick
Coventry
amo17
exactly the same but more.

don't understand most of foundations. left it all a bit late. hopefully the whole exam will be on truth tables or sommat!

lulz.

but they take FOREVAR.
I bring with me the hell that is the answers to some of analysis 2006. Specifically 2) b) and e), and 3) c) and d)
Checks pleaseee, 2b), the third one of 3c) and 3d) are my most unsure areas.

Thanks.
benwellsday
I bring with me the hell that is the answers to some of analysis 2006. Specifically 2) b) and e), and 3) c) and d)
Checks pleaseee, 2b), the third one of 3c) and 3d) are my most unsure areas.

Thanks.

2(b) I did pretty much same method, but chose $b_n> \frac{2C}{a}$ for any C>0 then $a_n.b_n>C$.

2(e) I did your second proof

3(c) final one: integral test

(d) same again, not sure what else we could write for 4 marks.
I get the same as you guys, using the second proof as Mr ninja said. i thought that question with the a_nb_n tending to infinity was a right bitch to put in there. I had to properly think about that. grrr.

and for 3 e did you guys just remember that inequality?
Totally Tom

and for 3 e did you guys just remember that inequality?

Was 3e) the one that was like $\sum \frac{n^n}{n!}x^n$ ?
If so, ratio test gives you the answer.

silent ninja

3(c) final one: integral test

Oh yeah that was in the workbook wasn't it? It looks like a horrible integral but it's actually not that bad with substitution.
When a question asks for an unbounded sequence. does unbounded mean not bounded ie just doesnt have upper AND lower bound, or does it mean that it cant have either an upper or a lower bound?
01perryd
When a question asks for an unbounded sequence. does unbounded mean not bounded ie just doesnt have upper AND lower bound, or does it mean that it cant have either an upper or a lower bound?

I think unbounded technically means you can use something like a_n = n, even though it's bounded below, but to be safe I usually put something like a_n = n(-1)^n. If the definition in the workbook says unbounded means either not bounded above or not bounded below then they can't drop you marks for using something that fits that.
Foundations 2007
2(c) Find the $lcm(\frac{1}{7}, \frac{2}{9})$ and $lcm(\frac{3}{4}, \frac{5}{6})$ (2+3 marks)
Find a formula for lcm(q_1, q_2) where $q_1=\frac{m_1}{n_1}, q_2=\frac{m_2}{n_2}$. Your answer can include expressions of the form lcm(m,n) or hcf(m,n) where m,n are integers (8 marks)

I get 2 and 15/2 for the first bit.

Second bit, I just wrote $lcm(\frac{m_1}{n_1}, \frac{m_2}{n_2)})= \frac{lcm(m_1n_2, m_2n_1)}{n_1n_2}$ They are all integers. Clearly not gonna get me 8 marks
i get a different answer to that.

$\displaystyle lcm(\frac{m_1}{n_1},\frac{m_2}{n_2})=\frac{m_1m_2}{hcf(m_2n_1,m_1n_2)}$...?

lulz no i don't

but i think that would get full marks.
silent ninja

Second bit, I just wrote $lcm(\frac{m_1}{n_1}, \frac{m_2}{n_2)})= \frac{lcm(m_1n_2, m_2n_1)}{n_1n_2}$ They are all integers. Clearly not gonna get me 8 marks

I got this, for the marks I would think you need some sort of justification for it. Mine was that $A\frac{m_1}{n_1} = B\frac{m_2}{n_2}$ and you want to find the least possible value for this. Hence you want the least value $Am_1n_2 = Bm_2n_1$ which kind of justifies using $\text{lcm} \{ m_1n_2, m_2n_1 \}$ as you let $A = \frac{\text{lcm} \{ m_1n_2, m_2n_1 \}}{m_1n_2}$ then $A\frac{m_1}{n_1} = \frac{\text{lcm} \{ m_1n_2, m_2n_1 \}}{n_1n_2}$
That doesn't seem to make much sense to me now but if you try to adapt the process you use to do the numerical ones then it might be easier. You could also justify it by showing that the analogy with the lcm for integers holds (let n_i = 1) and show it holds for the cases you did earlier in the question.
The way I thought of it was make the denominators common. Easiest way is to multiply by n1/n1 and n2/n2 respectively. Then the lcm is simply the lcm of the numerators divided by the new deonominator. In other words, i've kinda ignored the definition in the question so that's why I was wondering whether that would be sufficient for 8 marks.
Oh yeah that's a lot easier. It's 8 marks because it's unseen work, they have sections bookwork (stuff in the lecture notes directly), seen work (assignment question or very similar to one), and unseen work, and as you go up from each of them the marking gets nicer.
That makes sense. I suppose that question should be worth a lot because it takes a few minutes to get over, "what the hell, fractions?!?!"... and then it's okay after that lol
Has anyone done Foundation 2008, 3) iv) yet?
In $\mathbb{C}[X]$ define a relation $P_1 \sim P_2$ If P1 - P2 is divisble by X-1.
Show that ~ is an equivalence relation (I can do this). Let Q denote the set of equivalence classes of ~ in C[X]. Find an explicit bijection Q -> C.
Hint : What if the relation were P1 ~ P2 is P1-P2 is divisble by X.

I have no idea on that second part. Since it wants a bijection to C I assume it's got something to do with either the coeffecients of the polynomials or the roots.
Each equivalence class would be something like $\{ P : P \sim P_1 \}$ hence $P - P_1 = R(X-1)$ where R is just some polynomial. So what can we say about the set of these P's that'd be useful for mapping them in a one-to-one way...I have a feeling it's something to do with the coeffecients of P_1 but don't know how you'd map them...
BEN WHY AREN'T YOU ON MSN?

warggs.
what is the outline of the proof for foundations 06 4c? where you have to prove if gof is injective, f is injective. Do i start gof injective means a1 not equal to a2 implies gf(a1) does not = gf(a2)?
01perryd
what is the outline of the proof for foundations 06 4c? where you have to prove if gof is injective, f is injective. Do i start gof injective means a1 not equal to a2 implies gf(a1) does not = gf(a2)?

You could prove the equivalent f not injective => gof not injective. Pretty easy, really. Assume a_1 =/= a_2 but that f(a_1) = f(a_2). Trivially g(f(a_1)) = g(f(a_2)), which makes gof not injective because we assumed a_1 =/= a_2.

Also, I'm interesed in Ben's question. Tom will hand out Potter points on my behalf for solutions. They want a bijection of the form f:Q->$\mathbb{C}$, f([P]) = z? The hint looks like it would help significantly.
Is [P] the equivalence class of P?
It'd of been much easier if they asked for f:Q -> C[X], unless theres a bijection g:C[X] -> C, but that'd avoid the hint completely so it's probably not right.
I'm not actually sure how the hint helps, P1 - P = RX for some R, so how would you map [P] from there...I'm lost...and I'm not on campus yet, still at home, so I better start packing etc. Hopefully there's a solution by the time I'm there
benwellsday
Is [P] the equivalence class of P?
It'd of been much easier if they asked for f:Q -> C[X], unless theres a bijection g:C[X] -> C, but that'd avoid the hint completely so it's probably not right.
I'm not actually sure how the hint helps, P1 - P = RX for some R, so how would you map [P] from there...I'm lost...and I'm not on campus yet, still at home, so I better start packing etc. Hopefully there's a solution by the time I'm there

Your bijection wants Q as its domain. Each element of Q is of the form [P]. [P] is the equivalence class related to P. This is all well-defined because you know that the equivalence classes will partition Q, i.e. a ~ b => [a] = .

The point is that there's something unique about each equivalence class that you need to exploit. What that uniqueness is I don't know.

I think the hint helps because it's much easier to visualise what's going on in specific examples. If P_1 - P = RX then the LHS must look something like x^4 + x^2 + x, i.e. there can't be a constant term (so the constant terms in P and P_1 must be equal). Don't get me wrong though, I've not reallly tried the question so I could be verrrry wrong here.