HSC 2015 MX1 Marathon (archive) (7 Viewers)

Paradoxica · Oct 3, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
It was k=1, my bad. Is it just me or does it not work for n=1?

I have proven what I believe to be the correct version of the problem.

LOL you can see the post before I made it

InteGrand · Oct 3, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
It was k=1, my bad. Is it just me or does it not work for n=1?

$It seems there is a typo in the question; the summand should have $(3k+1)$ in the denominator, not $(3k-1)$.$
Lol LaTeX not working again...
What I typed was:
It seems there is a typo in the question; the summand should have (3k+1) in the denominator, not (3k-1).

This is because this allows us to get a telescoping sum if we do a partial fraction decomposition, as (3k – 2) and (3k + 1) are consecutive terms in a sequence as k runs through the integers.

Edit. Q done above by Paradoxica.

Paradoxica · Oct 3, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$It seems there is a typo in the question; the summand should have $(3k+1)$ in the denominator, not $(3k-1)$.$
Lol LaTeX not working again...
What I typed was:
It seems there is a typo in the question; the summand should have (3k+1) in the denominator, not (3k-1).

This is because this allows us to get a telescoping sum if we do a partial fraction decomposition, as (3k – 2) and (3k + 1) are consecutive terms in a sequence as k runs through the integers.

Edit. Q done above by Paradoxica.

I was briefly considering telescopic sequences, but I'm not sure if that will ever appear in the HSC, Induction will and I need practice on that.

rand_althor · Oct 3, 2015

Re: HSC 2015 3U Marathon

Ah okay thanks for clearing that up.

InteGrand · Oct 3, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
I was briefly considering telescopic sequences, but I'm not sure if that will ever appear in the HSC, Induction will and I need practice on that.

Yeah telescoping is much rarer than induction in the HSC, I was just providing some intuition for why (3k +1) should be there.

rand_althor · Oct 3, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
I was briefly considering telescopic sequences, but I'm not sure if that will ever appear in the HSC, Induction will and I need practice on that.

Here are some more random ones:

Carrotsticks · Oct 3, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
I was briefly considering telescopic sequences, but I'm not sure if that will ever appear in the HSC, Induction will and I need practice on that.

Telescopic sequences was in the 2010 Extension 2 HSC Q8.

rand_althor · Oct 4, 2015

Re: HSC 2015 3U Marathon

Prove that the product of 4 consecutive integers is always one less than a perfect square.

kawaiipotato · Oct 4, 2015

Re: HSC 2015 3U Marathon

$Let the first integer be x$

$$ product of four consecutive integers $ = x(x+1)(x+2)(x+3) = (x^2 + x)(x^2 + 5x + 6)$

$= x^4 + 6x^3 + 11x^2 + 6x$

$= x^4 + 6x^3 + 11x^2 + 6x + 1 - 1$

$= (x^2 + 3x + 1)^2 - 1$

$= n^2 - 1 ( $where n is an integer $ )$

rand_althor · Oct 4, 2015

Re: HSC 2015 3U Marathon

$\int\frac{1-\sqrt{x}}{1-\sqrt[4]{x}} \, \mathrm{d}x$

kawaiipotato · Oct 4, 2015

Re: HSC 2015 3U Marathon

let u = x^(1/4) ==> x = u^4
dx = 4u^3 du
I = integration of (1-u^2)/(1-u) * 4u^3 du
I = integration of (1+u) * 4u^3 du
expand and integrate

Zlatman · Oct 4, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$\int\frac{1-\sqrt{x}}{1-\sqrt[4]{x}} \, \mathrm{d}x$

Alternatively, multiply the top and bottom by [1 + x^(1/4)]. This makes the bottom [1 - x^(1/2)] and the top [(1 - x^(1/2))(1 + x^(1/4)). So, you're just integrating 1 + x^(1/4).

Drsoccerball · Oct 4, 2015

Re: HSC 2015 3U Marathon

Can someon do the induction question from 2015 bos trials i cant get it D: ?

$Prove using mathematical induction$

$n^{n+1} > (n+1)^{n}$

$$ For all integers $ n \geq 3$

rand_althor · Oct 4, 2015

Re: HSC 2015 3U Marathon

$$1. $\int \frac{1+\cos{x}}{1-\cos{x}} \, \mathrm{d}x$
$$2. $\int \frac{\cos{\frac{1}{x}}}{x^2} \, \mathrm{d}x$
$$3. Evaluate $ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{3^{m+n}}$
$$4. Find $\int \frac{1+2\sqrt{x}}{2\sqrt{x}(x+\sqrt{x})} \, \mathrm{d}x$ using the substitution $u=x+\sqrt{x}$
$$5. $\int \frac{1+\cot{x}}{1-\cot{x}} \, \mathrm{d}x$
$$6. $\int^\frac{\pi}{2}_0 \sqrt{1-\sin{x}} \, \mathrm{d}x$

Drsoccerball · Oct 4, 2015

Re: HSC 2015 3U Marathon

I dont think some of these are 3U difficulty... For example partial fractions in ex 2.

rand_althor · Oct 4, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
I dont think some of these are 3U difficulty... For example partial fractions in ex 2.

Don't know what partial fractions are. For question 2, might be easier if you use the substitution u=1/x.

leehuan · Oct 4, 2015

Re: HSC 2015 3U Marathon

1. Multiply top and bottom by 1+cos(x) then use Pythagorean identities everywhere.
2. Easy u=1/x
5. Similar to 1. Not exactly

rand_althor · Oct 4, 2015

Re: HSC 2015 3U Marathon

leehuan said:
1. Multiply top and bottom by 1+cos(x) then use Pythagorean identities everywhere.
2. Easy u=1/x
5. Similar to 1. Not exactly

Nice. The solutions that I saw were (in white):
1. Use cosine double angle formula to change it to cot^2(x/2), then pythagorean identities. Not sure whether your method is easier or not though.
5. Change cot to cos/sin, simplify and then realise it is f'(x)/f(x).

InteGrand · Oct 4, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$1. $\int \frac{1+\cos{x}}{1-\cos{x}} \, \mathrm{d}x$
$$2. $\int \frac{\cos{\frac{1}{x}}}{x^2} \, \mathrm{d}x$
$$3. Evaluate $ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{3^{m+n}}$
$$4. Find $\int \frac{1+2\sqrt{x}}{2\sqrt{x}(x+\sqrt{x})} \, \mathrm{d}x$ using the substitution $u=x+\sqrt{x}$
$$5. $\int \frac{1+\cot{x}}{1-\cot{x}} \, \mathrm{d}x$
$$6. $\int^\frac{\pi}{2}_0 \sqrt{1-\sin{x}} \, \mathrm{d}x$

$6. The integrand is simplified as follows. Note $\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2}$ and $1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}\Rightarrow 1 - \sin x = \sin^2 \frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}+ \cos^2 \frac{x}{2} = \left( \sin \frac{x}{2} - \cos \frac{x}{2} \right)^2 \Rightarrow \sqrt{1-\sin x} = \left |\sin \frac{x}{2} - \cos \frac{x}{2} \right | $. The rest is straightforward, noting that the expression in the absolute values is negative in the domain of integration, so flip it to integrate and remove the absolute values.$

rand_althor · Oct 4, 2015

Re: HSC 2015 3U Marathon

Some more:
$$1. $\sum_{k=2}^n \frac{1}{\log_k u}$
$$2. Evaluate $\int_{\frac{1}{2}}^{2} \frac{\ln x}{1+x^2} \, \mathrm{d}x $ using the substitution $u=\frac{1}{x}$
$$3. $\lim_{c\to\infty}\int^c_{-c}\frac{\mathrm{d}x}{e^x+e^{-x}}$

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