BoS Trials Maths, Physics and Business Studies 2023 (1 Viewer)

unofficiallyred12 · Oct 6, 2023

first note 1+x+x^9+x^10 = (1+x)(1+x^9)
then we split the integrand into x^2ln[(1+x)(1+x^9)]/(1+x^3) = x^2(ln(1+x) + ln(1+x^9))/(1+x^3)
now we split the integrals up so we have integral from 1 to 0 of x^2ln(1+x)/(1+x^3) + integral from 1 to 0 of x^2ln(1+x^9)/(1+x^3)
Then we do a u =x^3 for the 2nd integral, then dummy variable it back to in terms of x.
Afterwards, if u integrate integral of x^2ln(1+x)/(1+x^3) by parts with u = ln(1+x) and v' = x^2/(1+x^3), some cancellation should occur nicely and u should get the answer.

unofficiallyred12 · Oct 6, 2023

sorry if that's hard to read, idk how to latex

member 6003 · Oct 6, 2023

unofficiallyred12 said:
sorry if that's hard to read, idk how to latex

so the answer to the indefinite integral is $\frac{\ln(x+1) \ln(x^3+1)}{3}$ ?
Can you differentiate to check?

unofficiallyred12 · Oct 6, 2023

member 6003 said:
so the answer to the indefinite integral is $\frac{\ln(x+1) \ln(x^3+1)}{3}$ ?
Can you differentiate to check?

the differentiation calculator says it's not, I think this is one of those integrals where u have to rely on the bounds being a certain way to properly solve it.

member 6003 · Oct 6, 2023

unofficiallyred12 said:
the differentiation calculator says it's not, I think this is one of those integrals where u have to rely on the bounds being a certain way to properly solve it.

Try solving it without considering the limits at all. You get the same answer, maybe its wrong since doing u=x^3 changes the limits so that you can't cancel out the integrals but I don't know how you properly account for that. Since wolfram alpha can't get it the antiderivative is probably non-elementary. Anyway that's enough time spent on this question.

unofficiallyred12 · Oct 6, 2023

member 6003 said:
Try solving it without considering the limits at all. You get the same answer, maybe its wrong since doing u=x^3 changes the limits so that you can't cancel out the integrals but I don't know how you properly account for that. Since wolfram alpha can't get it the antiderivative is probably non-elementary. Anyway that's enough time spent on this question.

Well if u do it without the limits, then after u do the sub u cannot just use a dummy variable to change the integral from in terms of u back to in terms of x. Thus when u integrate by parts the first integral u can't just cancel out the integrals cos one of them is in terms of u and the other in terms of x

Sam14113 · Oct 6, 2023

Just quickly putting in enough LaTeX so that I and others can read a bit easier. Thanks for the solution.

unofficiallyred12 said:
first note $1+x+x^9+x^{10} = (1+x)(1+x^9)$
then we split the integrand into $\frac{x^2ln[(1+x)(1+x^9)]}{1+x^3} = \frac{x^2(ln(1+x) + ln(1+x^9))}{1+x^3}$
now we split the integrals up so we have $\int_0^1\frac{x^2ln(1+x)}{(1+x^3)} + \int_0^1\frac{x^2ln(1+x^9)}{1+x^3}$
Then we do a $u =x^3$ for the 2nd integral, then dummy variable it back to in terms of x.
Afterwards, if u integrate integral of $\frac{x^2ln(1+x)}{1+x^3}$ by parts with $u = ln(1+x)$ and $v' = \frac{x^2}{1+x^3}$ , some cancellation should occur nicely and u should get the answer.

unofficiallyred12 · Oct 6, 2023

thanks

SadCeliac · Oct 6, 2023

member 6003 said:
How do you do this?
$\int_0^1 \frac{x^2 \ln(1+x+x^9+x^{10})}{1+x^3}dx$
Wolfram alpha can't get a closed form answer for this but I think it's $\frac{\ln^2{2}}{3}$

Can you send your working I have no clue where to start

Sam14113 · Oct 6, 2023

SadCeliac said:
Can you send your working I have no clue where to start

See two messages above

Thanks @unofficiallyred12

its_ace21 · Oct 6, 2023

temporarylol said:
oh my god the business paper...

reading the MC itself makes me want to give up

the first 5 mc took me like 10 minutes to do coz i literally gave up halfway thru the text and had to start over again reading

carrotsss · Oct 6, 2023

temporarylol said:
oh my god the business paper...

reading the MC itself makes me want to give up

i was honestly not sure how jimmy would manage to make a bos-level difficulty business studies paper but it seems like hes somehow done it

its_ace21 · Oct 6, 2023

we love q25's business name

synthesisFR · Oct 6, 2023

dont let nesa see these papers

011235 · Oct 12, 2023

How do you get part ii? Particularly, how does one obtain the product of tans? I can't see any straightforward way to get a product of tans from the complex roots (of unity)

member 6003 · Oct 12, 2023

011235 said:
View attachment 40438

How do you get part ii? Particularly, how does one obtain the product of tans? I can't see any straightforward way to get a product of tans from the complex roots (of unity)

I don't know if this was how they intended but basically my thought process is if it looks like product of roots, just make an equation with roots of tan to get the product of roots. It doesn't really use part i) and is slow so it's not the best solution:

$\text{let } z=\tan{\frac{m\pi}{2n+1}}, \text{where } m=1,2,...,2n$

$\text{for clarity let } \alpha = \frac{m\pi}{2n+1}$

$\therefore z=\frac{\sin{\alpha}}{\cos{\alpha}}$

$z= \frac{\cfrac{1}{2i}(e^{i\alpha}-e^{-i\alpha})}{\cfrac{1}{2}(e^{i\alpha}+e^{-i\alpha})}$

$z= \frac{-ie^{2i\alpha}+i}{e^{2i\alpha}+1}$

$ze^{2i\alpha}+z=-ie^{2i\alpha}+i$

$e^{2i\alpha}(z+i)=i-z$

$e^{2i\alpha}=\frac{i-z}{z+i}$

$e^{i\frac{2m\pi}{2n+1}}=\frac{i-z}{z+i}$

$e^{2i\pi m}= \left( \frac{i-z}{z+i}\right)^{2n+1}$

$1=\left( \frac{i-z}{z+i}\right)^{2n+1}, \text{where } z \neq 0$

$(z+i)^{2n+1}=(i-z)^{2n+1}$

$\binom{2n+1}{2n+1} z^{2n+1}+i\binom{2n+1}{2n} z^{2n} + ... + \binom{2n+1}{1}i^{2n}z + \binom{2n+1}{0} i^{2n+1} =$

$-\left(\binom{2n+1}{2n+1} z^{2n+1}-i\binom{2n+1}{2n} z^{2n} + ... + \binom{2n+1}{1} i^{2n}z - \binom{2n+1}{0} i^{2n+1}\right)$

$2\binom{2n+1}{2n+1} z^{2n+1}+ ... + 2\binom{2n+1}{1}(-1)^{n}z = 0$

$z^{2n}+...+(2n+1)(-1)^n=0 \text{ as } z\neq0$

$\therefore \text{using product roots, }$

$\tan{\frac{\pi}{2n+1}}\tan{\frac{2\pi}{2n+1}}\tan{\frac{3\pi}{2n+1}}...\tan{\frac{2n\pi}{2n+1}}=(-1)^n(2n+1)$

Trebla · Oct 12, 2023

011235 said:
View attachment 40438

How do you get part ii? Particularly, how does one obtain the product of tans? I can't see any straightforward way to get a product of tans from the complex roots (of unity)

It can be shown that
$\tan\theta = \dfrac{1-e^{2i\theta}}{i(-1-e^{2i\theta})}$

Let

$\theta=\dfrac{k\pi}{2n+1}$

so the LHS is

$\dfrac{\left(1-e^{i\frac{2\pi}{2n+1}}\right)\left(1-e^{i\frac{4\pi}{2n+1}}\right)\left(1-e^{i\frac{6\pi}{2n+1}}\right)...\left(1-e^{i\frac{4n\pi}{2n+1}}\right)}{i^{2n}\left(-1-e^{i\frac{2\pi}{2n+1}}\right)\left(-1-e^{i\frac{4\pi}{2n+1}}\right)\left(-1-e^{i\frac{6\pi}{2n+1}}\right)\dots\left(-1-e^{i\frac{4n\pi}{2n+1}}\right)}$

From part (i), this is actually

$\dfrac{P(1)}{i^{2n}P(-1)}$

and the result follows

saltshaker · Oct 14, 2023

Will solutions be released before the 4u HSC?

synthesisFR · Oct 14, 2023

Trebla said:
It can be shown that
$\tan\theta = \dfrac{1-e^{2i\theta}}{i(-1-e^{2i\theta})}$

Let

$\theta=\dfrac{k\pi}{2n+1}$

so the LHS is

$\dfrac{\left(1-e^{i\frac{2\pi}{2n+1}}\right)\left(1-e^{i\frac{4\pi}{2n+1}}\right)\left(1-e^{i\frac{6\pi}{2n+1}}\right)...\left(1-e^{i\frac{4n\pi}{2n+1}}\right)}{i^{2n}\left(-1-e^{i\frac{2\pi}{2n+1}}\right)\left(-1-e^{i\frac{4\pi}{2n+1}}\right)\left(-1-e^{i\frac{6\pi}{2n+1}}\right)\dots\left(-1-e^{i\frac{4n\pi}{2n+1}}\right)}$

From part (i), this is actually

$\dfrac{P(1)}{i^{2n}P(-1)}$

and the result follows

I’m dropping

yanujw · Oct 14, 2023

synthesisFR said:
I’m dropping

FWIW, there were only a few attempts that moved past 1 or 2 lines, and very, very few marks were awarded across the cohort.

Make a Difference – Donate to Bored of Studies!

BoS Trials Maths, Physics and Business Studies 2023 (1 Viewer)

Active Member

Active Member

Member

Active Member

Member

Active Member

Member

Active Member

studying

Member

/æɪs/

New Member

/æɪs/

afterhscivemostlybeentrollingdonttakeitsrsly

Active Member

Member

Administrator

Member

afterhscivemostlybeentrollingdonttakeitsrsly

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)