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#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2145 on: December 14, 2018, 06:58:05 pm »
+3
Hi
I have a problem that I can't seem to resolve in my head. Perhaps someone knows the answer. I want to compute the integral $\int \sqrt{1 - x^{2}} \space dx$Of course, to an extension 2 student this is not too difficult, you just use the substitution $x = \cos{\theta}$However, when you eventually get to the line $\int \sqrt{\cos^{2}\theta} \times \cos{\theta} \space d\theta$you encounter a problem, because $\sqrt{\cos^{2}\theta} = |\cos{\theta} | \neq \cos{\theta}$However, everywhere I look I find the solution continue as $\int \cos^{2} \theta \space d\theta$Is it correct to continue like this, ignoring the absolute values? If yes, why? If not, how can we make it rigorous?

The simple fix to make it rigorous is to impose the condition that $-\frac\pi2 \leq \theta \leq \frac\pi2$. This covers all values in the range of $y=\sin \theta$, that is $-1\leq y\leq 1$.

And then, note that for any angle $\theta$ in the first or negative first quadrants, $\cos\theta$ is positive. Therefore $|\cos\theta| = \cos\theta$ for this case.

In practice, it just tends to be assumed. It's nicer to use, because now we can assume $\theta = \sin^{-1}x$ without justification, thereby avoiding any general formula tediosity

Edit: You made a typo I think, but it was clear enough to me that you substituted in $\boxed{x=\sin\theta}$
« Last Edit: December 14, 2018, 07:08:06 pm by RuiAce »

#### Elias S

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##### Re: 4U Maths Question Thread
« Reply #2146 on: December 14, 2018, 08:00:12 pm »
0
Edit: You made a typo I think, but it was clear enough to me that you substituted in $\boxed{x=\sin\theta}$

Ah, my apologies for the typo. Thanks for the reply!

#### david.wang28

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##### Re: 4U Maths Question Thread
« Reply #2147 on: December 15, 2018, 05:08:12 pm »
0
Hello, I have a few problems that are bugging me. I have attempted some of my working out, but for a couple of the questions I don't know where to go. Can anyone please help me with Q3, Q4, Q8 and Q10? It would be greatly appreciated
HSC 2019: English Advanced(77) (gonna repeat), Chemistry, Physics, Maths Extension 1(35) (gonna repeat), Maths Extension 2, Business Studies(80) (screw this)

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#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2148 on: December 15, 2018, 05:57:15 pm »
+3
Hello, I have a few problems that are bugging me. I have attempted some of my working out, but for a couple of the questions I don't know where to go. Can anyone please help me with Q3, Q4, Q8 and Q10? It would be greatly appreciated
Starting points:

Q3 and Q4 both represent standard loci that you are expected to memorise or identify on the spot.
- $|z-1| = |z-i|$ is the perpendicular bisector of the points representing $1$ and $i$ (i.e. the perpendicular bisector of $(0,1)$ and $(1,0)$ Factoring in the inequality, we want the region cut off by this line that includes the point $(1,0)$.
- $|z-2-2i| \leq 1$ is the circle centred at $(2,2)$ with radius $1$, and the inequality suggests that we want its interior.
- $\arg(z+3) = \frac\pi3$ is the ray starting at $z=-3$ (i.e. the point $(-3,0)$, inclined at an angle of $\frac\pi3$ and excluding the point $(-3,0)$ itself. The equation of the ray itself is not of interest.

Note however that Q8 is a special locus. You should watch Eddie Woo's video to understand this locus if you were not introduced to it at school to get started on this question.

I will only do Q10 for now as this is still a bombardment of questions, however that particular one is not something taught at school and expected to be memorised.
$\text{Note by de Moivre's theorem we have}\\ \frac{1}{t} = r^{-1} (\cos (-\theta) + i\sin (-\theta)) = r^{-1} (\cos \theta - i \sin \theta).\\ \text{Therefore }\boxed{z=r(\cos\theta + i\sin\theta) + r^{-1} (\cos\theta - i\sin \theta)}.$
\text{Let }z=x+iy.\text{ This now becomes}\\ x+iy = (r + r^{-1})\cos \theta + i (r - r^{-1})\sin \theta.\\ \text{Hence equating real and imaginary parts we have}\\ \begin{align*}x &= (r+r^{-1}) \cos \theta\\ y &= (r-r^{-1}) \sin \theta \end{align*}
As usual, keep in mind that we aim to eliminate any parameters when handling locus problems this way. Here, the parameter will be $\theta$ for part a), and $r$ for part b).
\text{For part a), subbing }r=2\text{ gives}\\ \begin{align*} x&= \frac52 \cos\theta\\ y&= \frac52\sin\theta\end{align*}
\text{It should be immediately obvious that this is a circle}\\ \text{centred at the origin with radius }\frac52.\\ \text{However, if we wish to confirm this:}\\ \begin{align*}x^2+y^2 &= \frac{25}4\cos^2\theta + \frac{25}4\sin^2\theta\\ &= \frac{25}4(\cos^2\theta+\sin^2\theta)\\ &= \frac{25}4 \end{align*}
\text{Whereas for part b), subbing in }\theta = \frac\pi4\text{ gives}\\ \begin{align*}x &= \frac{r+r^{-1}}{\sqrt2}\\ y&= \frac{r-r^{-1}}{\sqrt2} \end{align*}
$\text{It's not immediate obvious as to what goes on here.}\\ \text{The trick is to recall the identity}\\ \boxed{(a+b)^2 = (a-b)^2 + 4ab}$
\text{Keeping this trick in mind, we have}\\ \begin{align*}x^2-y^2 &= \frac12 \left[(r+r^{-1})^2 - (r - r^{-1})^2 \right]\\ &= \frac12 \left[ (r^2+2+r^{-2}) - (r^2 - 2 + r^{-2}) \right]\\ &= 2 \end{align*}\\ \text{which you will recognise is a hyperbola after the conics topic.}
« Last Edit: December 15, 2018, 06:00:39 pm by RuiAce »

#### david.wang28

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##### Re: 4U Maths Question Thread
« Reply #2149 on: December 15, 2018, 06:03:08 pm »
0
Starting points:

Q3 and Q4 both represent standard loci that you are expected to memorise or identify on the spot.
- $|z-1| = |z-i|$ is the perpendicular bisector of the points representing $1$ and $i$ (i.e. the perpendicular bisector of $(0,1)$ and $(1,0)$ Factoring in the inequality, we want the region cut off by this line that includes the point $(1,0)$.
- $|z-2-2i| \leq 1$ is the circle centred at $(2,2)$ with radius $1$, and the inequality suggests that we want its interior.
- $\arg(z+3) = \frac\pi3$ is the ray starting at $z=-3$ (i.e. the point $(-3,0)$, inclined at an angle of $\frac\pi3$ and excluding the point $(-3,0)$ itself. The equation of the ray itself is not of interest.

Note however that Q8 is a special locus. You should watch Eddie Woo's video to understand this locus if you were not introduced to it at school to get started on this question.

I will only do Q10 for now as this is still a bombardment of questions, however that particular one is not something taught at school and expected to be memorised.
$\text{Note by de Moivre's theorem we have}\\ \frac{1}{t} = r^{-1} (\cos (-\theta) + i\sin (-\theta)) = r^{-1} (\cos \theta - i \sin \theta).\\ \text{Therefore }\boxed{z=r(\cos\theta + i\sin\theta) + r^{-1} (\cos\theta - i\sin \theta)}.$
\text{Let }z=x+iy.\text{ This now becomes}\\ x+iy = (r + r^{-1})\cos \theta + i (r - r^{-1})\sin \theta.\\ \text{Hence equating real and imaginary parts we have}\\ \begin{align*}x &= (r+r^{-1}) \cos \theta\\ y &= (r-r^{-1}) \sin \theta \end{align*}
As usual, keep in mind that we aim to eliminate any parameters when handling locus problems this way. Here, the parameter will be $\theta$ for part a), and $r$ for part b).
\text{For part a), subbing }r=2\text{ gives}\\ \begin{align*} x&= \frac52 \cos\theta\\ y&= \frac52\sin\theta\end{align*}
\text{It should be immediately obvious that this is a circle}\\ \text{centred at the origin with radius }\frac52.\\ \text{However, if we wish to confirm this:}\\ \begin{align*}x^2+y^2 &= \frac{25}4\cos^2\theta + \frac{25}4\sin^2\theta\\ &= \frac{25}4(\cos^2\theta+\sin^2\theta)\\ &= \frac{25}4 \end{align*}
\text{Whereas for part b), subbing in }\theta = \frac\pi4\text{ gives}\\ \begin{align*}x &= \frac{r+r^{-1}}{\sqrt2}\\ y&= \frac{r-r^{-1}}{\sqrt2} \end{align*}
$\text{It's not immediate obvious as to what goes on here.}\\ \text{The trick is to recall the identity}\\ \boxed{(a+b)^2 = (a-b)^2 + 4ab}$
\text{Keeping this trick in mind, we have}\\ \begin{align*}x^2-y^2 &= \frac12 \left[(r+r^{-1})^2 - (r - r^{-1})^2 \right]\\ &= \frac12 \left[ (r^2+2+r^{-2}) - (r^2 - 2 + r^{-2}) \right]\\ &= 2 \end{align*}\\ \text{which you will recognise is a hyperbola after the conics topic.}
Thanks for the help kind sir!
HSC 2019: English Advanced(77) (gonna repeat), Chemistry, Physics, Maths Extension 1(35) (gonna repeat), Maths Extension 2, Business Studies(80) (screw this)

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#### 006896

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##### Re: 4U Maths Question Thread
« Reply #2150 on: December 23, 2018, 08:42:22 pm »
0
Hi,
I have a question about doing 4U Maths for my HSC.
I will graduate in 2019, so I've already started by Year 12 year. I also accelerated Maths 2U and 3U, in which I received full marks for both subjects. I also state ranked in 2U Maths (in the top 3). I love maths and am currently doing 4U maths, but I don't see myself getting better than a mid-top-band for 4U maths. So many people have been telling me that it's crazy for me to do 4U maths, because it will make my 2U marks redundant. So what should I do? Will state ranking in 2U help increase my ATAR, in comparison than, say, getting a 95 in 4U maths? What should I do to get the highest ATAR - should I drop 4U maths to keep my 2U marks, or keep 4U maths and ditch my 2U marks? To be honest, my aim is to get the highest ATAR, not to gain skills and learn more (sorry!).
I am doing 14 units right now, including my 3 units of maths, which I have already completed.
Thanks

#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2151 on: December 23, 2018, 09:04:10 pm »
+2
Hi,
I have a question about doing 4U Maths for my HSC.
I will graduate in 2019, so I've already started by Year 12 year. I also accelerated Maths 2U and 3U, in which I received full marks for both subjects. I also state ranked in 2U Maths (in the top 3). I love maths and am currently doing 4U maths, but I don't see myself getting better than a mid-top-band for 4U maths. So many people have been telling me that it's crazy for me to do 4U maths, because it will make my 2U marks redundant. So what should I do? Will state ranking in 2U help increase my ATAR, in comparison than, say, getting a 95 in 4U maths? What should I do to get the highest ATAR - should I drop 4U maths to keep my 2U marks, or keep 4U maths and ditch my 2U marks? To be honest, my aim is to get the highest ATAR, not to gain skills and learn more (sorry!).
I am doing 14 units right now, including my 3 units of maths, which I have already completed.
Thanks
If your final result for 2U and 3U were literally 100 and 50 (100 if you do end up continuing 4U) then it becomes pretty hard to say. Because of course something like a 95 in 4U would be the minimum to have a chance of topping off 100 in 2U.

Yet at the same time it's still possible to get a 100 in 4U regardless if you've already managed a perfect score twice. Won't say that's easy obviously but it's still theoretically possible.

I'm still gonna stick to my same old advice of making the decision to drop after the half yearly. If after the half yearly you feel that there might not be a chance of gunning for your 95+, then you can go for it. But otherwise the gamble may pay off for all we know.

The only thing that takes me aback a bit is that last bit with the gain skills and learn more. If your maths potential is already that great and you have a passion for it, there's no reason for you not to. Whilst it's not as great as rank 3, I know someone who came rank 15 in 2U maths and still made the decision to carry through 4U anyway. But because I can understand the desire to get the maximum ATAR, you really should consider still learning the 4U content but then dropping it before, say, the trials. You don't need to necessarily put effort into 4U, so that your ATAR doesn't get hit by anything, but at least you come out with the skills set everyone else did. (Trying to put myself in your shoes, if I were having the debate over keeping my 2U mark and somewhat favouring that option, that's what I would've done.)

Don't blame you at all for your predicament, especially since you have 14 units in total, but again that's probably my suggestion. Take as you will, or ask more if you wish to

#### terassy

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##### Re: 4U Maths Question Thread
« Reply #2152 on: January 07, 2019, 08:49:31 pm »
0
Question from Terry Lee's textbook:

The normal at an end P(x1,y1) of the latus rectum of the ellipse x^2/a^2+y^2/b^2=1 meets the y-axis in M, and PN is the abscissa of P (i.e. PN is perpendicular to the y-axis). Prove that MN=a.

I know how to do this but I'm getting confused whether to use the ± sign. For example when finding the x coordinate of P it would be ±ae right? But the answers only use ae. Same when finding the y coordinate of P, which I substitute ae into x^2/a^2+y^2/b^2=1 to find the y value which would be ±b2/a. But again, the answers only use b2/a. Please help sorry to disturb your holidays

#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2153 on: January 07, 2019, 09:05:25 pm »
+2
Question from Terry Lee's textbook:

The normal at an end P(x1,y1) of the latus rectum of the ellipse x^2/a^2+y^2/b^2=1 meets the y-axis in M, and PN is the abscissa of P (i.e. PN is perpendicular to the y-axis). Prove that MN=a.

I know how to do this but I'm getting confused whether to use the ± sign. For example when finding the x coordinate of P it would be ±ae right? But the answers only use ae. Same when finding the y coordinate of P, which I substitute ae into x^2/a^2+y^2/b^2=1 to find the y value which would be ±b2/a. But again, the answers only use b2/a. Please help sorry to disturb your holidays
Well, most likely in the exam they'd give you a diagram so that you only have to consider one of the endpoints of one particular latus rectum.

But for the textbook question, you're right that in theory you should consider the four cases of $P\left(\pm ae, \pm a(1-e^2) \right)$ and $P\left( \pm ae, \mp a(1-e^2) \right)$. And you could do this and it should work in every case (provided you can juggle the plus/minus and the minus/plus, else you'd have to do all 4 cases by hand). But at the same time I could argue via the symmetry of the ellipse that this will work for all four cases?

(Remark: you could rewrite the two subcases as $\left(ae, \pm a(1-e^2)\right)$ and $\left(-ae, \pm a(1-e^2) \right)$, but you may still bump into some minus/plus symbols.)

So I could say something like, "suppose without loss of generality that $P$ is the endpoint $(ae, a(1-e^2) )$ of the right-hand latus rectum", then do yadayadaya, and justify by symmetry why it must also hold for the other three endpoints as well.

But again, in the exam most likely they wouldn't leave it so ambiguous.
« Last Edit: January 07, 2019, 09:07:35 pm by RuiAce »

#### DrDusk

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##### Re: 4U Maths Question Thread
« Reply #2154 on: January 09, 2019, 05:27:27 am »
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I was so happy to be done with 4u maths, now starting at UNSW I found out just some time ago I need to do math1131 or math1141 which covers 4u in apparently like a month. Ah RIP to me
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#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2155 on: January 09, 2019, 07:07:59 am »
+1
I was so happy to be done with 4u maths, now starting at UNSW I found out just some time ago I need to do math1131 or math1141 which covers 4u in apparently like a month. Ah RIP to me
Maybe post this under the UNSW section next time.

Anyway, whilst I'd say MATH1141 is mildly harder than 4u, I personally reckon MATH1131 is equal in difficulty overall. There are some learning curves (e.g. definition of a limit) that you do need to wrap your head around and it takes a fair while, just as harder 3u inequalities does. But other than that a lot of the stuff is just coverage of the basics in 4u. (Note that proofs in conics and volumes are not examinable in MATH1131 anymore - only the fact that the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and similarly for the hyperbola is necessary.)

The biggest overlap in MATH1131 and 4u is really just complex numbers and polynomials.

I also feel like MATH1231 (the follow up course in term 2) is easier

#### louisaaa01

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##### Re: 4U Maths Question Thread
« Reply #2156 on: January 11, 2019, 02:48:47 pm »
0
Just in relation to the ATAR Notes MX2 Topic Tests, I came across this question in Complex Numbers:

"The complex number z satisfies |z-i| = 1. What is the greatest distance that z can be from the point (1,0) in the Argand diagram?"

Intuitively, I thought that the line from (1,0) to the circle |z-i| = 1 with greatest length must pass through the centre of the circle, since the diameter is the longest 'chord' in a circle.

I got the right answer, but I was just wondering whether this was a valid conclusion or not? Would this kind of logic occur in every case or was I just lucky in getting the answer right? The worked solutions have two alternative methods, none of which use this reasoning.

#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2157 on: January 11, 2019, 02:57:49 pm »
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Just in relation to the ATAR Notes MX2 Topic Tests, I came across this question in Complex Numbers:

"The complex number z satisfies |z-i| = 1. What is the greatest distance that z can be from the point (1,0) in the Argand diagram?"

Intuitively, I thought that the line from (1,0) to the circle |z-i| = 1 with greatest length must pass through the centre of the circle, since the diameter is the longest 'chord' in a circle.

I got the right answer, but I was just wondering whether this was a valid conclusion or not? Would this kind of logic occur in every case or was I just lucky in getting the answer right? The worked solutions have two alternative methods, none of which use this reasoning.
That's interesting; I was surprised to see that that method is not there. Your method is certainly also valid.

#### _Himani_

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##### Re: 4U Maths Question Thread
« Reply #2158 on: January 15, 2019, 04:20:56 pm »
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Hi
I'm fairly new to integration and got a bit stuck with this question type: ∫(x-1)/(x^2+1) dx. Should I be integrating these question with partial fractions or can I do this with inverse trig?
PS: This is my first post on the forum!

#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2159 on: January 15, 2019, 04:42:14 pm »
+1
Hi
I'm fairly new to integration and got a bit stuck with this question type: ∫(x-1)/(x^2+1) dx. Should I be integrating these question with partial fractions or can I do this with inverse trig?
PS: This is my first post on the forum!
$\text{Seeing as though the quadratic in the denominator is irreducible}\\ \text{i.e. it cannot be factored any further without complex numbers}\\ \text{some kind of splitting may be necessary.}$
$\text{In your case, you may wish to write the integral as}\\ \int \frac{x}{x^2+1} - \frac{1}{x^2+1}\,dx.\\ \text{You should be able to identify the inverse trig in the second integral.}$
Partial fractions would usually be the go-to if you found that the fraction on the denominator could be factorised, for example $\frac{x-1}{x^2-4}$ instead.

Note: If you look at it closely, the "split" is always done to force a log to appear, whenever there's $x$ in the top and $x^2$ in the bottom.