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I think you'd need to justify your second line. Whilst it is true for all non-negative x (well, put ≥ signs), the reasoning is different based on whether x is bigger than 1 or smaller than 1.

I also thought it looked like a 98+ (for state ranking), but I guess it depends on how many silly mistakes were made etc. Probably a lower mark could also nab a state rank, given good enough internals etc.

Whilst we don't need absolute values on the second integral (as the integrand is non-negative over that interval), putting the absolute values isn't incorrect (just redundant), and so what was done is still correct (and the final answer came out correct too, i.e. didn't make a silly mistake).

I think it depends in part on what you want to do in University. If you want to do something mathematical, then it would probably be best to keep 2U Maths.

I think it depends in part on what you want to do in University. If you want to do something mathematical, then it would probably be best to keep 2U Maths.

I'm not sure but that might depend on whether you tried to claim you'd proved it for all x ≥ 0 (that's why I asked if you'd made it clear you weren't).

Basically your best hope is to hope the markers assumed you were trying to prove it only for x >= 1 and that they'll give some marks for that (because that's what you actually did, mathematically speaking.).

Well done! (And good job on testing the answer, a lot of students don't seem to think to do these kinds of checks (maybe because teachers don't appear to emphasise it).)

$\noindent Did you say clearly in your working that you were only proving it for $x \geq 1$?$

You'll probably get 1-2 marks (I think 1 because 2 seems too much, but not sure.) $\noindent You know, the result for $0 < x < 1$ actually follows immediately from what you proved for $x \geq 1$. Since we showed $\chi\sqrt{\chi}+ 1 \geq \chi +\sqrt{\chi}$ for all $\chi \geq 1$, the substitution...

$\noindent Alice starts tossing a magical coin and wins if she eventually lands a heads. The coin initially starts off with some probability $p$ ($0<p<1$) of showing heads, and changes its heads probability according to the rule: if the result of the $n$-th toss ($n=1,2,3,\ldots$) was tails, the...

Re: MATH1231/1241/1251 SOS Thread No, converge means the partial sums tend to a real number (can be any real number, yes). What I was talking about was the terms of the series going to 0. If a series is convergent, then its k-th term must tend to 0 as k -> oo (so if the k-th term of the series...

Re: MATH1231/1241/1251 SOS Thread No as in, no it doesn't converge. Remember, for a series to converge, it is necessary that the k-th term tends to 0 as k -> oo. In this case, we have a_k -> 1 (not 0) as k -> oo, so the series can't converge (i.e. it's divergent).

You could also do the Q in about one line by noting that a concave down parabola is maximised halfway between the roots.

Re: MATH1231/1241/1251 SOS Thread $\noindent I guess they're just leaving it in the form $\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}\left(x-a\right)^{k}$, because that's what a Taylor polynomial about $a$ is (no point in expanding it out).$

$\noindent We have$ $$\begin{align*}\left(2 + \frac{x}{2}\right)^{n} &= \sum_{k = 0}^{n}\binom{n}{k}\frac{x^{k}}{2^{k}}2^{n-k}\\ \Rightarrow \left(\frac{4+x}{2}\right)^{n} &= \sum_{k = 0}^{n}\binom{n}{k}x^{k}2^{n-2k}.\end{align*}$$ $\noindent Now divide both sides by $2^{n}$. The $2^{n-2k}$ in...

Re: ATAR Notes vs Bored of Studies Yeah, that does seem to be the case. Perhaps after such a long time, people just got bored of Bored of Studies.

Re: ATAR Notes vs Bored of Studies So the main reason for ATAR Notes' apparent increase in recent popularity vs. BOS would be advertising superiority?

Re: ATAR Notes vs Bored of Studies I guess BOS's popularity has declined over the years? What's the reason for this?

Here is a solution in spoilers: