HSC Mathematics Marathon (2 Viewers)

Drongoski · Sep 27, 2011

dionb2014 said:
Yeah I have done that it's just I fail when I try and solve it as a quadratic.

You can't find the x-intercept. The parabola corresponding to x^2+ x + 1 and 5(x^2+x+1) are concave up and lie entirely above the x-axis. In 2U maths you would say the quadratics are positive definite (i.e. never 0 nor negative). Their discriminants are negative.

dionb2014 · Sep 27, 2011

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh.
So this is concave up with a positive y intercept. I am officially retarded.

Nerchio1 · Sep 27, 2011

enjoy!

dionb2014 · Sep 27, 2011

I got lost

apollo1 · Sep 27, 2011

dionb i thought u said u were going to top maths ext 2 in year 12. so r u trollin here?

dionb2014 · Sep 27, 2011

Haha. I'm in year 9 apollo. Technically I am trolling because I posted a question that is not extension 2. I'm not trolling in that I know the questions I'm asking because I genuinely don't but clearly it's a long while before I even pass ext 2.

Nerchio1 · Sep 27, 2011

since theres a new page I thought I should repost my question for easier referencing

hup · Sep 27, 2011

Nerchio1 said:
since theres a new page I thought I should repost my question for easier referencing

fuaark

dionb2014 · Sep 27, 2011

Jut for interest does anyone no what grade that question I asked would be

hscishard · Sep 27, 2011

dionb2014 said:
5x^2+5x+5=0 factorize

Solve or factorise? Factorising is a step in solving equations.
5(x^2 + x +1) is the LHS factorised

@ nerchio what paper was that from?

cutemouse · Sep 28, 2011

hscishard said:
@ nerchio what paper was that from?

Q8 HSC 1989. Stupid question IMO.

hup · Sep 28, 2011

here's a good one $\\The\ n+1^{th}\ term\ of\ a\ sequence\ is\ given\ by\\\\ u_{n+1}=u_{n}+u_{n}^2\\\\ and\ if\\\\ u_{1}=1/3\\\\ Evaluate\sum_{n=1}^{\infty }\frac{1}{1+u_{n}}$

largarithmic · Sep 28, 2011

Nerchio1 said:
since theres a new page I thought I should repost my question for easier referencing

[a] Yeah just do it on your calculator

Let 4223a = K + F(4223a) where K is an integer.
$1.989 < 10^{F(4223a)} < 1.990$
$1.989 \times 10^K < 10^{K + F(4223a)} < 1.990 \times 10^K$
$1.989 \times 10^K < 10^{4223 \times 2136 log_{10} 2} < 1.990 \times 10^K$
$1.989 \times 10^K < 2^{4223 \times 2136} < 1.990 \times 10^K$
So we can set M = 4223 \times 2136 = 9 020 328.

[ii][a] If F(mr) = 0, then mr = K for some integer K, so r = K/m, so r is rational.

If F(mr) = F(nr) = s, let mr = K + s, nr = L + s, where K, L are integers.
Then r(m-n) = K-L, so r = (K-L)/(m-n) so r is rational.

[iii][a] Suppose for some integers 0 < i < j <= N+1, F[ib] = F[jb] (i.e. not distinct).
Then from [ii], b is rational, a contradiction.

Now there are N subintervals length 1/N that partition 0 <= x < 1: the interval 0 <= x < 1/N, 1/N <= x < 2/N, 2/N <= x < 3/N, ..., (N-1)/N <= x < 1.
There are also N+1 numbers we consider: F, F[2b], F[3b], ..., F[(N+1)b]. Now each of these numbers must lie within one of the above intervals, as they all lie within 0 <= x < 1 and the subintervals partition this interval.

Now if we are putting N+1 objects into N spaces, it logically follows that there is one space with at least two distinct object in it. I.e., You have an interval
k/N <= x < (k+1)/N and F[mb], F[nb] where m=/=n (and lets say without loss of generality that F[mb] > F[nb]; note by part [a] they are distinct)
such that F[mb] and F[nb] are both in the interval k/N <= x < (k+1)/N.

Now if you think about it visually, they're both within an interval length 1/N: so the distance between them is at most 1/N.
I.e. F[mb] - F[nb] < 1/N.
Now F[(m-n)b] = (m-n)b - A for some integer A = (mb) - (nb) - A = (B+F[mb]) - (C+F[nb]) - A for some B, C integers = B-C-A + F[mb] - F[nb]
Now F[mb] > F[nb] by our assumption without loss generality above, so 0 < F[mb] - F[nb] < 1/N.
Now if B-C-A =/= 0, it is either >=1, in which case F[(m-n)b] >= 1, contradiction; or it is <= -1, in which case F[(m-n)b] < 0, contradiction.
So we require B-C-A = 0 so F[(m-n)b] = F[mb] - F[nb] < 1/N, as required.

[iv]Now for any such integer N,
$\frac{1}{N} < log_{10} \frac{1990}{1989} = log1990 - log1989 = log(1.990 \times 1000) - log(1.989 \times 1000) = log(1.990) - log(1.989)$
Now let $b = Klog_{10}2$ for some arbitrarily large K. In fact choose any arbitrary K > a trillion, for instance. It follows obviously that b is irrational.
Now from (iii), there are integers 1<= m,n <= N+1 such that
$F[(m-n)b] < \frac{1}{N} < log_{10}1.990 - log_{10}1.989$

Now consider F[(m-n)b], F[2(m-n)b], F[3(m-n)b], F[4(m-n)b], ...
Each new member of that sequence is F[(m-n)b] distance from the one before. If you think about that visually, that sequence must then either at some point have a member within the interval log1.989 < x < log1.990
OR it must "jump over" that interval: i.e. you have some i s.t. F(i(m-n)b) <= log1.989 and F((i+1)(m-n)b) >= 1.990.

Now it clearly can't "jump over" the interval log1.989 < x < log1.990 coz then you would have F((m-n)b) >= log1.990 - log1.989; essentially you would need the length of the "jump" to be bigger than the length of the interval you're jumping over. That's a contradiction.

So you must have some i such that log 1.989 < F(i(m-n)b) < log 1.990

Now we do the same procedure as we did in part :
Let i(m-n)b = S + F[i(m-n)b], put the inequality above to the power of 10 and subbing b = Klog2 and get
$10^S \times 1.989 < 2^{i(m-n)K} < 10^S \times 1.990$

Now sub M = i(m-n)K to get another different possible M from that in part : it's clearly different because we made K so massive (bigger than a trillion) that its much larger than 9020328.

So there do exist other such M.

cutemouse said:

Q8 HSC 1989. Stupid question IMO.

Click to expand...

And why is it a stupid question, because it's unconventional? The best questions that can be asked at HSC level are those which are unexpected. If you ask me, anything that could possibly feature in a textbook as an exercise should not be included in q7/8 of the HSC; and from what I've seen of past papers that's certainly true of q8. The last questions of 4U should not test your knowledge of drilled exercises or standard problems, but your actual mathematical ability. That question isn't actually very hard if you can convert the notation into some mental understanding of what its asking, and that's what the question is testing.

The question is a bit fiddly algebraically I guess, except it's actually a really nice question for the concepts involved. They're concepts that aren't really in the course, but are things that everyone would think intuitively anyway. Stuff like putting N+1 objects into N spaces, and stuff like the jumping over the intervals. The question's beauty comes from that anyone pretty much who read a proof closely enough of it could easily follow, and if you had a visual aid to explain it (in doing this in an exam, youd draw LOTS of diagrams) it's actually really intuitive. And the result at the end is actually really neat when you think about it. If you ask me it's an excellent question.

apollo1 · Sep 28, 2011

largarithmic said:
[a] Yeah just do it on your calculator

Let 4223a = K + F(4223a) where K is an integer.
$1.989 < 10^{F(4223a)} < 1.990$
$1.989 \times 10^K < 10^{K + F(4223a)} < 1.990 \times 10^K$
$1.989 \times 10^K < 10^{4223 \times 2136 log_{10} 2} < 1.990 \times 10^K$
$1.989 \times 10^K < 2^{4223 \times 2136} < 1.990 \times 10^K$
So we can set M = 4223 \times 2136 = 9 020 328.

[ii][a] If F(mr) = 0, then mr = K for some integer K, so r = K/m, so r is rational.

If F(mr) = F(nr) = s, let mr = K + s, nr = L + s, where K, L are integers.
Then r(m-n) = K-L, so r = (K-L)/(m-n) so r is rational.

[iii][a] Suppose for some integers 0 < i < j <= N+1, F[ib] = F[jb] (i.e. not distinct).
Then from [ii], b is rational, a contradiction.

Now there are N subintervals length 1/N that partition 0 <= x < 1: the interval 0 <= x < 1/N, 1/N <= x < 2/N, 2/N <= x < 3/N, ..., (N-1)/N <= x < 1.
There are also N+1 numbers we consider: F, F[2b], F[3b], ..., F[(N+1)b]. Now each of these numbers must lie within one of the above intervals, as they all lie within 0 <= x < 1 and the subintervals partition this interval.

Now if we are putting N+1 objects into N spaces, it logically follows that there is one space with at least two distinct object in it. I.e., You have an interval
k/N <= x < (k+1)/N and F[mb], F[nb] where m=/=n (and lets say without loss of generality that F[mb] > F[nb]; note by part [a] they are distinct)
such that F[mb] and F[nb] are both in the interval k/N <= x < (k+1)/N.

Now if you think about it visually, they're both within an interval length 1/N: so the distance between them is at most 1/N.
I.e. F[mb] - F[nb] < 1/N.
Now F[(m-n)b] = (m-n)b - A for some integer A = (mb) - (nb) - A = (B+F[mb]) - (C+F[nb]) - A for some B, C integers = B-C-A + F[mb] - F[nb]
Now F[mb] > F[nb] by our assumption without loss generality above, so 0 < F[mb] - F[nb] < 1/N.
Now if B-C-A =/= 0, it is either >=1, in which case F[(m-n)b] >= 1, contradiction; or it is <= -1, in which case F[(m-n)b] < 0, contradiction.
So we require B-C-A = 0 so F[(m-n)b] = F[mb] - F[nb] < 1/N, as required.

[iv]Now for any such integer N,
$\frac{1}{N} < log_{10} \frac{1990}{1989} = log1990 - log1989 = log(1.990 \times 1000) - log(1.989 \times 1000) = log(1.990) - log(1.989)$
Now let $b = Klog_{10}2$ for some arbitrarily large K. In fact choose any arbitrary K > a trillion, for instance. It follows obviously that b is irrational.
Now from (iii), there are integers 1<= m,n <= N+1 such that
$F[(m-n)b] < \frac{1}{N} < log_{10}1.990 - log_{10}1.989$

Now consider F[(m-n)b], F[2(m-n)b], F[3(m-n)b], F[4(m-n)b], ...
Each new member of that sequence is F[(m-n)b] distance from the one before. If you think about that visually, that sequence must then either at some point have a member within the interval log1.989 < x < log1.990
OR it must "jump over" that interval: i.e. you have some i s.t. F(i(m-n)b) <= log1.989 and F((i+1)(m-n)b) >= 1.990.

Now it clearly can't "jump over" the interval log1.989 < x < log1.990 coz then you would have F((m-n)b) >= log1.990 - log1.989; essentially you would need the length of the "jump" to be bigger than the length of the interval you're jumping over. That's a contradiction.

So you must have some i such that log 1.989 < F(i(m-n)b) < log 1.990

Now we do the same procedure as we did in part :
Let i(m-n)b = S + F[i(m-n)b], put the inequality above to the power of 10 and subbing b = Klog2 and get
$10^S \times 1.989 < 2^{i(m-n)K} < 10^S \times 1.990$

Now sub M = i(m-n)K to get another different possible M from that in part : it's clearly different because we made K so massive (bigger than a trillion) that its much larger than 9020328.

So there do exist other such M.

plz try telling me now that u arent going to top the state in 4U maffs this year. jesus christ if u can do that question u can do anything.

seanieg89 · Sep 28, 2011

apollo1 · Sep 28, 2011

seanieg89 said:
$$hups question:\\ From the recurrence relation given: $\frac{1}{1+u_n}=\frac{u_n}{u_{n+1}}=\frac{1}{u_n}-\frac{1}{u_{n+1}}.\\$Hence by telescoping: $\sum_{k=1}^n \frac{1}{1+u_k}=\frac{1}{u_1}-\frac{1}{u_{n+1}}.\\$It is easy to see that $1/u_n\rightarrow 0$ so we can conclude that: $\sum_{n=1}^\infty \frac{1}{1+u_n}=\frac{1}{u_1}=3.$

wtf. wat did u do in the first line?

hup · Sep 28, 2011

seanieg89 said:
$=\frac{1}{u_n}-\frac{1}{u_{n+1}}$

how did you see this part so quickly? It took me a while to figure out

apollo1 · Sep 28, 2011

ah i see wat u did thats smart.

seanieg89 · Sep 28, 2011

Intuition i guess, the best hope of evaluating series like these is either by relating them to a series you already can evaluate or using a telescoping sum. Recurrence relations are usually a pretty good indicator that the latter will be applicable.

apollo1 · Sep 28, 2011

wat does telescoping mean?

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