4U Maths Question Thread

[RuiAce]

That one doesn't use by parts. (Any integral that can be converted to \(I_n = \int \tan^n u \,du\) through substitution doesn't use by parts either.)
\begin{align*} I_n&= \int \tan^n x\,dx\\ &= \int \tan^{n-2}x\tan^2x\,dx\\ &= \int \tan^{n-2}x\sec^2 x\,dx - \int \tan^{n-2}x\,dx \tag{Pythagorean identity}\\ &= \frac{\tan^{n-1}x}{n-1} - I_{n-2} \end{align*}

[Fatemah.S]

That one doesn't use by parts. (Any integral that can be converted to \(I_n = \int \tan^n u \,du\) through substitution doesn't use by parts either.)
\begin{align*} I_n&= \int \tan^n x\,dx\\ &= \int \tan^{n-2}x\tan^2x\,dx\\ &= \int \tan^{n-2}x\sec^2 x\,dx - \int \tan^{n-2}x\,dx \tag{Pythagorean identity}\\ &= \frac{\tan^{n-1}x}{n-1} - I_{n-2} \end{align*}

Oh that makes sense. Would you please be able to explain how you got from the third to the fourth row? Thank you so much!

[RuiAce]

Oh that makes sense. Would you please be able to explain how you got from the third to the fourth row? Thank you so much!

That \(I_{n-2}\) bit should be clear. As for the first integral I basically just reversed the chain rule. However if you're uncomfortable with this, observe that the substitution \(u=\tan x \implies du = \sec^2 x\,dx\) gives
\[ \int \tan^{n-2}x\sec^2 x\,dx = \int u^{n-2}\,du = \frac{u^{n-1}}{n-1}+C = \frac{\tan^{n-1}x}{n-1}+C \]

[Fatemah.S]

That \(I_{n-2}\) bit should be clear. As for the first integral I basically just reversed the chain rule. However if you're uncomfortable with this, observe that the substitution \(u=\tan x \implies du = \sec^2 x\,dx\) gives
\[ \int \tan^{n-2}x\sec^2 x\,dx = \int u^{n-2}\,du = \frac{u^{n-1}}{n-1}+C = \frac{\tan^{n-1}x}{n-1}+C \]

Got it! Thanks

[kaustubh.patel]

Hey rui please help with these two questions, I have no absolute idea on how to attempt it except making the denominator 2 for the first one

[RuiAce]

Hey rui please help with these two questions, I have no absolute idea on how to attempt it except making the denominator 2 for the first one

Where did this come from? It looks like competition level maths instead of 4U. But what vexes me more is that a quick check on Wolfram shows that \( \sqrt{\frac{2009\times2010\times2011\times2012+1}{4}} \) is actually a rational number, and the 50th digit after the decimal is (not coincidentally) zero.

One approach of starting Q2 using competition level mathematics

Edit: Later found out that I made a small mistake at the start. I think the expression was \(\sum_{k=0}^\infty \frac{F_{k+1}}{2^{k+2}} \) instead. The final answer should end up being \(1\), not \( \frac12\). Won't bother editing this though - I think all the working out should be reasonably adaptable.
\[ \text{Let }\phi = \frac{1+\sqrt5}{2}, \, \psi = \frac{1-\sqrt5}{2}\\ \text{and recall that }F_n = \frac{\phi^n - \psi^n}{\sqrt{5}} \]
\begin{align*} \sum_{k=0}^\infty \frac{F_k}{2^{k+2}} &= \frac{1}{4\sqrt5} \sum_{k=0}^\infty \frac{\phi^k - \psi^k}{2^k}\\ &= \frac1{4\sqrt5}\left(\sum_{k=0}^\infty \left( \frac\phi2\right)^k - \sum_{k=0}^\infty \left( \frac\psi2 \right)^k \right)\\ &= \frac{1}{4\sqrt5}\left(\frac{1}{1-\frac\phi2} - \frac{1}{1-\frac\psi2} \right) \end{align*}
because each of the sums are (infinite) geometric series. It can be checked through various non-calculator approaches that the common ratio is certainly less than 1.

One approach to commence proving the rationality of the expression in Q1

\[ \text{Let }x=2010+\frac12.\text{ Then,} \]
\begin{align*}\sqrt{\frac{2009\times2010\times2011\times2012+1}4}&= \sqrt{\frac{\left(x-\frac32\right)\left(x-\frac12\right)\left(x+\frac12\right)\left(x+\frac32\right)+1}4}\\ &= \sqrt{\frac{\left(x^2-\frac94\right)\left(x^2-\frac14\right)+1}{4}}\\ &= \sqrt{\frac{(4x^2-9)(4x^2-1) + 16}{64}}\\ &= \frac{\sqrt{16x^4-40x^2+25}}8\\ &= \frac{\sqrt{(4x^2-5)^2}}{8}\\ &= \frac{4x^2-5}{8} \end{align*}
noting that we expect our answer to remain positive.

[kaustubh.patel]

I actually found it on this book website, i was just scrolling and these ques were really interesting but they didnt have answers.https://www.scribd.com/document/99166274/153-TG-Math-Problems-E-book
and theres one with the answers but i need a paid subscription for it
https://www.scribd.com/doc/188084024/153-TG-Math-Problems

[louisaaa01]

Can someone please help with this question?

"Find the Cartesian equation of the locus of z such that arg(z-2) = arg(z²). Describe the locus geometrically, noting any restrictions"

I was okay with finding the locus, but I'm stuck on how to determine the restrictions.

Thanks!

[goodluck]

Hey, I'm not sure if this is part of the syllabus but my teacher gave us some extension questions to do (apparently our school doesn't teach harder 3u saying we ought to have already have learned it...) but could someone help me with this:

By considering the cases x ≥ 0, −1 < x < 0 and x ≤ −1 (or otherwise) prove that 1+x+x^2+x^3+x^4 is always positive.

[AlphaZero]

Hey, I'm not sure if this is part of the syllabus but my teacher gave us some extension questions to do (apparently our school doesn't teach harder 3u saying we ought to have already have learned it...) but could someone help me with this:

By considering the cases x ≥ 0, −1 < x < 0 and x ≤ −1 (or otherwise) prove that 1+x+x^2+x^3+x^4 is always positive.

I'm not too familiar with the HSC courses yet (I'll be reading through the study designs so I can help in the HSC boards in the future). Nonetheless, here's a start to your question.

Case 1: \(x\geq 0\)
This is the simplest of the three cases. For all \(x\geq0\), we have \(x^n\geq 0\) for all natural numbers \(n\).
\[\text{So, }\ 1+x+x^2+x^3+x^4\geq 1+0+0+0+0>0.\]

Case 2: \(-1<x<0\)
This case is a little trickier, but the result becomes obvious to prove if we write \begin{align*}1+x+x^2+x^3+x^4&=(1+x)+x^2(1+x)+x^4\\
&=(1+x)(1+x^2)+x^4.\end{align*}For \(-1<x<0\), we have \(0<x+1<1\) with \(x^2+1>1\) and \(x^4>0\). \[\text{So, }\ 1+x+x^2+x^3+x^4>0\times 1+0=0.\]

Case 3: \(x\leq -1\).
There are a few ways to solve this case. I'll let you try this one on your own.

If you still need help, don't hesitate to ask

[RuiAce]

Can someone please help with this question?

"Find the Cartesian equation of the locus of z such that arg(z-2) = arg(z²). Describe the locus geometrically, noting any restrictions"

I was okay with finding the locus, but I'm stuck on how to determine the restrictions.

Thanks!

If you've managed to find both the circle and the ray, the restrictions are that the circle only excludes \( (0,0) \), but the ray is essentially \( \arg (z-2) = 0 \)

I'll let someone else go into more depth though - don't quite have the time right now to figure this locus out again yet sadly

Note

The relevant circle is centred at \( (2,0) \) with radius \(r=2\).

[AlphaZero]

Hey, I'm not sure if this is part of the syllabus but my teacher gave us some extension questions to do (apparently our school doesn't teach harder 3u saying we ought to have already have learned it...) but could someone help me with this:

By considering the cases x ≥ 0, −1 < x < 0 and x ≤ −1 (or otherwise) prove that 1+x+x^2+x^3+x^4 is always positive.

Alrighty... I'll give the explanation a go, but it's been a very long time since I've any complex analysis this difficult. (VCE complex numbers is so much easier lol).

I'm assuming that you've already described the locus. Mainly,
> circle of radius 2 centered at \(2+0i\)
> line \(\text{Im}(z)=0\) (essentially the real axis)

All the points on the circle are fine except one, and we'll see why this is in a second.

Let's consider individually different sections of the line \(\text{Im}(z)=0\). For our convenience, let's write \(z=a+0i\).

If  \(a<2\)  (\(a-2<0\)),  then we have  \(\arg(z-2)=\arg(a-2)=\pi\)  and  \(\arg(z^2)=\arg(a^2)=0\)  (since \(a^2>0\)).

Note that obviously we cannot have \(z=2+0i\) since \(\arg(0)\) is undefined.

If \(a>2\)  (\(a-2>0\)),  then we have  \(\arg(z-2)=\arg(a-2)=0\)  and  \(\arg(z^2)=\arg(a^2)=0\)  (since \(a^2>0\)).

Therefore, what is essentially left of the line is the ray given by  \(\arg(z-2)=0\).


Altogether, the locus of \(z\) is given by \[\left\{z\mid \arg(z-2)=0\right\}\cup\left\{z\mid\ |z-2|=2\right\}\setminus\{0\},\] which can be read as "the set of values \(z\) that satisfy  \(\arg(z-2)=0\)  or  \(|z-2|=2\),  and  \(z\neq0\)"

[goodluck]

Hey, could someone explain to me how to do this:
Prove using vector methods that the midpoints of the sides of a convex quadrilateral form a parallelogram.

I'm not getting the same results for both the sides to prove it's parallel and I don't really get the internet explanations, I think I might need it to simplified a lot!

[RuiAce]

Hey, could someone explain to me how to do this:
Prove using vector methods that the midpoints of the sides of a convex quadrilateral form a parallelogram.

I'm not getting the same results for both the sides to prove it's parallel and I don't really get the internet explanations, I think I might need it to simplified a lot!

Is this supposed to be under complex numbers? No other use of vectors appears in the current MX2 course.

[goodluck]

Is this supposed to be under complex numbers? No other use of vectors appears in the current MX2 course.

I'm not quite sure, my substitute teacher gave us a bunch of maths questions to do, saying her son said was appropriate for us (I think some of it is uni maths since there were some she marked wasn't in our syllabus). I thought it was just vector addition like in physics, but I'm not sure how to go about it