Author Topic: Suggested Answers to the HSC 2016 Maths Extension 2 Exam (+Discussion!) (Read 23823 times)

zdeathbringer · « **Reply #15 on:** October 21, 2016, 09:21:15 pm »

Quote from: RuiAce on October 21, 2016, 04:36:53 pm

Share your thoughts! How was it? Are you glad that the stupidly huge subject is now cleared?

I'll also add some of my thoughts on the paper below! Solutions will be done up over the afternoon.
_____________________________________________

First thought:
WOAH. I did not expect half of that at all. The paper looks polarised; on one end it's quite relaxed and then on the other end it feels brutal.

Not the hardest paper I've seen, but definitely looks harder than 2015 and possibly also harder than 2014.
_____________________________________________

Sample solutions:
Multiple Choice

Answer key:
1. C
CLICK HERE FOR AN EXPLANATION
1. Look out for where the symmetry is. The symmetry is about the y-axis. This means that the x-coordinates were what was tampered.
Indeed, you can't square root a negative number. But in here, it seems like you can. Hence it can only be C or D.

Finally, D is wrong as that would imply that you'd have something below the x-axis as well!
Answer: C
2. C
CLICK HERE FOR AN EXPLANATION
2. If x=1 is to be a multiple root, it must be a root to start with! Hence B is automatically out as plugging in x=1 gives -2.

And now, recall that if it is a multiple root, then it must be a root of the first derivative as well! Hence, differentiating each case:
$\begin{align*}A)&\qquad 5x^4-4x^3-2x\\ C) &\qquad 5x^4-4x^2-2x\\ D)&\qquad 5x^4-3x^2-1 \end{align*}$
It is clear that upon substituting in x=1, only option C) goes to 0.
Answer: C[/tex]
3. A
CLICK HERE FOR AN EXPLANATION
3. This question required you to recall what the eccentricity of each conic section actually is:
- Circle: e=0
- Ellipse: 0<e<1
- Parabola: e=1
- Hyperbola e>1

Of course, the eccentricity is never negative. Hence any option containing "parabola" amd/or "hyperbola" are all out! Because if you add on something that's 1 or more, how can you get 3/4?
Answer: A
4. D
CLICK HERE FOR AN EXPLANATION
4. There's a bit of a process of elimination here. You need to consider what doing _______ does.

A) is wrong as -w makes both the real and imaginary parts negated. This essentially flips w about the origin into the 4th quadrant.
B) is wrong as multiplying by 2 just lengthens the segment, and multiplying by i rotates it 90^o counterclockwise, putting it in the 2nd quadrant
C) is wrong as conjugation just flips the imaginary part only. This flips z about the x-axis into the 4th quadrant.
D) is the right answer by elimination. To show this, however, you may want to consider vector addition.
5. B
CLICK HERE FOR AN EXPLANATION
5. The answers all seem to have π. It seems that we're forced to evaluate that complex number and turn it into mod-arg form. (Or use your CASIO fx-100 AU PLUS)
$\begin{align*}\frac{1-i}{1+i}&=\frac{(1-i)^2}{(1+i)(1-i)}\\ &= \frac{1-2i+i^2}{2}\\ &= -i\\ &=\cos -\frac{\pi}{2}+i\sin- \frac{\pi}{2}\end{align*}$
And the argument tells us what happens to the complex number.
Answer: D
6. C
CLICK HERE FOR AN EXPLANATION
6. At first glance, subbing x+1 in for x looks like a tedious process. But there is a bit of a trick that we can employ.
$\text{Note that, for the }1+x+x^2+\dots+x^7\text{ portion}\\ \text{There won't even be an }x^8!\\ \text{Thus we just have to consider what's left}$
$\text{i.e. }x^8+x^9+\dots+x^{12}\\ \text{Which becomes }(x+1)^8+(x+1)^9+\dots+(x+1)^{12}$
This is now a 3U question. Just make sure we extract the correct binomial coefficients. (Reminder: Whilst it makes no difference, here we had (x+1)ⁿ and not (1+x)ⁿ)
$\begin{align*}(x+1)^8 &\to \binom{8}{0}x^8\\ (x+1)^9 &\to \binom{9}{1}x^8\\ &\vdots \\ (x+1)^{12}&\to \binom{12}{4}x^8\end{align*}$
So we just add all the required binomial coefficients.
$\binom80+\binom91+\binom{10}{2}+\binom{11}{3}+\binom{12}{4}=715$
Answer: C
7. D
CLICK HERE FOR AN EXPLANATION
7. This question required you to recall the 45^o rotation formula:
$x^2-y^2=k \stackrel{45^\circ\text{ anticlockwise}}{\longrightarrow}xy=\frac{1}{2}k$
So here, k/2 = 8 implying k = 16.
Answer: D
8. C
CLICK HREE FOR AN EXPLANATION
So long as you knew how to resolve your forces, this is BOSTES' way of giving you a free mechanics multiple choice. Everything can be easily inferred from a forces diagram.
<To come>
And then it's just a matter of resolving the forces
$\text{Horizontally:}\\ \begin{align*}-F\cos \theta+N\sin \theta &= -mr\omega^2\\ \implies F\cos \theta - N \sin \theta &= mr\omega^2\end{align*}$
$\text{Vertically:}\\ \begin{align*}F\sin \theta + N\cos \theta - mg &= 0\\ \implies F\sin \theta + N \cos \theta &= mg\end{align*}$
Answer: C

Alternatively, this could've been done with a bit of intuition. Note that the friction and normal reaction forces are both working against gravity. Hence, the signs should both be positive for that one.
Whereas they work against each other horizontally, so their signs must be the opposite there.
9. A
CLICK HERE FOR AN EXPLANATION
9. This is the very classic volume of a trough question. The method of considering trapeziums and similar triangles is presented below. You could also have done it by other methods such as fitting a line through relevant points.

Keep in mind that the boundaries of 0 and 4 are taken care of already for us. We note that the cross section is of a rectangle. Let the side lengths of the rectangle be a and b, and hence consider the following analysis with trapeziums.
<IMAGE TO COME>
$\text{By using similar triangles in both cases:}\\ \begin{align*}\frac{h}{4}&=\frac{x}{6}\\ &\implies x=\frac{3h}{2}\\ \therefore a=\frac{3h}{2}+2\end{align*}$
$\begin{align*}\frac{h}{4}&=\frac{y}{4}&\\ \implies y&=h\\ \therefore b=h+3\end{align*}$
So by using our very classic volumes methods, we arrive at
$V=\int_0^4 (h+3)\left(\frac{3h}{2}+2\right)dh$
Answer: A
10. B
CLICK HERE FOR AN EXPLANATION
10. You may have unintentionally spent too much time trying to figure out how to manipulate algebra. But in reality, it's best to go back to basics and solve that quadratic.
$\begin{align*}x+\frac1x&=-1\\ x^2+x+1&=0\\ x&=\frac{-1\pm \sqrt{1-4}}{2}\\ &= -\frac{1}{2}\pm \frac{\sqrt3}{2}i\\ &= \cos \left(\pm \frac{2\pi}{3}\right)+i\sin \left(\pm \frac{2\pi}{3}\right)\end{align*}$
And this is where the fun is. We can just take the positive value because it won't matter once we use De Moivre's theorem.
$\begin{align*}x^{2016}+\frac{1}{x^{2016}}&= \left(\text{cis }\frac{\pi}{6}\right)^{2016}+ \left(\text{cis }\frac{\pi}{6}\right)^{-2016}\\ &= \text{cis}1344\pi + \text{cis}(-1344\pi)\\ &= \cos 1344\pi + i \sin 1344\pi + \cos (-1344\pi) + i\sin (-1344\pi)\\ &=2\end{align*}$
Answer: B
Question 11
(a)
CLICK HERE FOR AN EXPLANATION
(i) This is just your regular conversion. 1 mark for the modulus, and 1 mark for the argument. (Some calculators may have done it for you.)
$z=\sqrt{3}-i=2\left(\cos \left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)$
(ii) The modulus-argument form allows us to use De Moivre's theorem to evaluate z⁶
$\begin{align*}z^6&=\left[2\left(\cos \left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)\right]^6\\ &= 64(\cos \pi + i\sin \pi)\\ &= -64\end{align*}$ Which is clearly real.

(iii) There are heaps of valid answers to this question. n=3 will do, as
$\begin{align*}z^3&=\left[2\left(\cos \left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)\right]^3\\ &= 64\left(\cos \frac{\pi}{2} + i\sin \frac{\pi}{2}\right)\\ &= 8i\end{align*}$
(b)
CLICK HERE FOR AN EXPLANATION
This is a simple integration by parts. By the rule of LIATE we choose to differentiate the x.
$\begin{align*}\int xe^{-2x}dx&= -\frac{1}{2}xe^{-2x}+\int \frac{1}{2}e^{-2x}dx\\ &= -\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C\end{align*}$
(c)
CLICK HERE FOR AN EXPLANATION
Simply differentiate, however being careful with the product rule.
$\begin{align*}x^3+y^3&=2xy\\ 3x^2+3y^2\frac{dy}{dx}&=2x\frac{dy}{dx}+2y\\ \frac{dy}{dx}(3y^2-2x)&=2y-3x^2\\ \frac{dy}{dx}&=\frac{2y-3x^2}{3y^2-2x}\end{align*}$
(d)
CLICK HERE FOR AN EXPLANATION
Graphs to come.
(i) Key points:
- You can't square root anything negative, so whatever is below the x-axis there is gone.
- Your y-intercept also gets square rooted

(ii) Key points:
- Your y-intercept and horizontal asymptote always gets reciprocated
- The nature of stationary points reverses
- Vertical asymptotes become discontinuities along the x-axis (no x-intercept as we don't have a rule for f(x))
(e)
CLICK HERE FOR AN EXPLANATION
(i) Note that the x has no effect whatsoever on the domain, as the domain of g(x)=x is all real x. It is the sine inverse that imposes the restriction.
$-1 \le \frac{x}{2} \le 1\\ -2 \le x \le 2$
Now, we need to be careful here. f(x) = x.sin^-1(x/2) is actually an even function because it's two odd functions multiplied together.
This means that it is symmetric about the y-axis.

The minimum is now at x=0. This can be seen by separately sketching the two curves and multiplying ordinates. <TO COME>
$\text{Note that }f(0)=0\\ \text{and }f(2)=\pi\\ \text{Hence the range is }0\le y \le \pi$

Question 12
(a)
CLICK HERE FOR AN EXPLANATION
(i) This is quite simple. $\frac{x^2}{9}+\frac{y^2}{4}=1$
(ii) Apply a formula that you know.
$\begin{align*}4&=9(1-e^2)\\ e&=\frac{\sqrt{5}}{3}\end{align*}$
(iii) Further formula application
$S\text{ is the point }(ae,0) = (\sqrt{5},0)\\ S^\prime\text{ is the point }(-ae,0)=(-\sqrt5,0)$
(iv) $x=\pm \frac{9}{\sqrt5}$
(b)
CLICK HERE FOR AN EXPLANATION
(i) Remember that when differentiating something integrated, it cancels back out to the original function. So we just use this and the product rule.
$\begin{align*}\frac{d}{dx}\left(xf(x)-\int xf^\prime(x)\,dx\right)&= f(x)+xf^\prime(x) - xf^\prime(x)\\ &= f(x)\end{align*}$
(ii) The otherwise method is a classic integration by parts. It turns out to yield the same result using the hence method.
$f(x)=\tan x \implies f^\prime(x)=\frac{1}{1+x^2}$
$\begin{align*}\therefore \int \tan^{-1}x\, dx&=x\tan^{-1}x - \int \frac{x}{1+x^2}dx\\ &= x\tan^{-1}x - \frac12\ln (1+x^2)+C \end{align*}$
(c)
CLICK HERE FOR AN EXPLANATION
(i) Applying De Moivre's theorem with the binomial theorem
$\begin{align*}z^4&=(\cos 4\theta + i\sin 4\theta)\\ &= (\cos\theta+i\sin \theta)^4\\ &= \cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta \sin^2\theta-4i\cos\theta\sin^3\theta+\sin^4\theta\end{align*}$
Hence, considering the real parts as told, upon equating:
$\cos 4\theta = \cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta$
(ii) This is now just brute force applying a Pythagorean identity and expanding.
$\begin{align*}\cos 4\theta &= \cos^4\theta - 6\cos^2\theta (1-\cos^2\theta) + (1-\cos^2\theta)^2\\ &= \cos^4\theta - 6\cos^2\theta + 6\cos^4\theta + 1 - 2\cos^2\theta + \cos^4\theta\\ &=8\cos^4\theta - 8\cos^2\theta + 1\end{align*}$
(d)
CLICK HERE FOR AN EXPLANATION
(i) A classic procedure.
$xy=c^2\implies y=\frac{c^2}{x}\implies \frac{dy}{dx}=-\frac{c^2}{x^2}$ $\text{When }x=cp, \frac{dy}{dx}=-\frac{1}{p^2}\text{ giving us }m=p^2$
$\text{Using the point-gradient formula}\\ \begin{align*}y-\frac{c}{p}&=p^2(x-cp)\\ p^2x-y&=c\left(p^3-\frac{1}{p}\right)\\ px-\frac{y}{p}&=c\left(p^2-\frac{1}{p^2}\right)\end{align*}$
(ii) Since Q is where they re-intersect, we are tempted to just substitute the coordinates of Q in and do battle with algebra. However there is a neater trick.

Supposed we wanted to find the solutions. Then we would solve simultaneous equations.
$xy=c^2\text{ and }px-\frac{y}{p}=c\left(p^2-\frac{1}{p^2}\right)\\ \text{Hence solving and rearranging:}$ $\begin{align*}px-\frac{c^2}{px}&=c\left(p^2-\frac{1}{p^2}\right)\\ p^2x^2-cpx\left(p^2-\frac{1}{p^2}\right)-c^2&=0\end{align*}$
With plenty of studying, you will realise that the sum/product of roots loves to creep their way in when you least expect it. Note that x=cp and x=cq are the roots of this quadratic equation. Hence, by the product of roots
$(cp)(cq)=-\frac{c^2}{p^2}\implies q=-\frac{1}{p^3}$
Question 13
(a)
CLICK HERE FOR AN EXPLANATION
Taking the hint and differentiating ln(f(x)), clearly manipulating a log law (and using implicit differentiation to tidy things up)
$\begin{align*}\ln f(x) &= \ln x^x\\ &= x\ln x\\ \frac{1}{f(x)}f^\prime(x)&= 1+\ln x\end{align*}$
Because we want to maximise, we will set f'(x) = 0. Note that x^x is never equal to 0.
$0=1+\ln x \implies x = \frac{1}{e}\approx 0.3678$
Test a bit to both sides to verify that it is a min:
$\begin{align*}f^{\prime}(x)&=x^x(1+\ln x)\\f^\prime(0.367)&-1.65...\times 10^{-3} < 0\\ f^\prime(0.368)&=2.26...\times 10^{-3}> 0 \end{align*}$
So f(x) decreases and then increases, hence we do have a minimum at x=1/e.
(b)
CLICK HERE FOR AN EXPLANATION
Upon joining OQ (which you have to), it becomes clear that an isosceles triangle may be floating around. Hence, attempt to use base angles.

It is given that AB is a diameter, hence join AC to use the angle in a semicircle.
$\begin{align*}\text{Let }\angle BPC &= \theta\\ \text{Then }\angle BCP &=\theta \text{ (base angles on isos. TriBPC})\\ #therefore \angle ACQ&=90^\circ - \theta \text{ (ACB angle in semicircle, angles on straight angle PCQ)}\end{align*}$
To be finished
(c)
CLICK HERE FOR AN EXPLANATION
(i) Somehow we need to relate T₁ and T₂.
$\text{Note that by right angled triangle analysis, the hypotenuse is }0.5\\ \text{Hence }\sin \theta = \frac{4}{5}, \cos \theta = \frac{3}{5}$
The vertical resolution is straightforward.
$T_1\sin \theta - Mg = 0 \implies T_1 = 10M \left(\frac{5}{4}\right)=12.5M$
For the horizontal resolution, note that Mg obviously plays no role in it. Hence, it must be that if we sub in the above result into the horizontal resolution, we get what we require.
$\begin{align*}T_2+ T_1\cos \theta &= Mr\omega^2\\ &=0.3M\omega^2\\ \therefore T_2 + 12.5M\left(\frac{3}{5}\right)&=0.3M\omega^2\\ T_2&=0.3M\omega^2-7.5M\\ &=0.3M(\omega^2-25)\end{align*}$
(ii) We can sub T₁ back into the above equation and hence reach an answer.
$\begin{align*}M&=0.08T_1\\ \therefore T_2&=0.024T_1(\omega^2-25)6\end{align*}$
$\text{So if }T_2>T_1\text{ then }\\ 1> 0.024(\omega^2-25)\\ \text{Doing some rearranging and noting }\omega > 0\\ \text{This solves to give }\omega > 10{\sqrt\frac{2}{3}}$
(d)
CLICK HERE FOR AN EXPLANATION
(i) Since this is just a cubic polynomial, there is a classic trick.

Note that they don't want you to prove that p(x) cuts the x-axis only once IF b²-3ac < 0.
They want you to prove the converse, which is an other way around analysis.
$\text{Since }p(x)\text{ is a cubic}\\ \text{if }p(x)\text{ is strictly increasing it DEFINITELY cuts the }x\text{-axis only once.}\\ p^\prime(x)=3ax^2+2bx+c$
$p(x)\text{ is strictly increasing }\iff p^\prime(x) > 0\\ \text{Since }p^\prime(x)\text{ is quadratic, we want it to be positive or negative definite}\\ \therefore \Delta < 0$
$\Delta < 0 \implies (2b)^2-4(3a)(c) < 0 \implies b^2-3ac < 0$
(ii) We just keep testing derivatives. First note that we rearrange the given equation to $c=\frac{b^2}{3a} $
$\begin{align*}p^\prime(x)&=3ax^2+2bx+\frac{b^2}{3a}\\ p^\prime\left(-\frac{b}{3a}\right)&=0\end{align*}\\ \text{So we keep going.}$
$\begin{align*}p^{\prime\prime}(x)&=6ax+2b\\ p^{\prime\prime}\left(-\frac{b}{3a}\right)&=0\end{align*}\\ \text{So we keep going, but }p^{\prime\prime\prime}(x)=6a\neq 0\text{ as }a\neq 0$
Hence it is a root of multiplicity 3.

(Or we can use the fact that a cubic cannot have a root of more than multiplicity 3, because it can only have 3 roots altogether!)

Question 14
(a)
CLICK HERE FOR AN EXPLANATION
(i) It can probably be done by differentiation, however this isn't too hard to integrate. Use a Pythagorean identity, and if required use a u-substitution.
$\begin{align*}\int \sin^3x\,dx&=\int (1-\cos^2 x)\sin x\, dx\\ &= \int (\cos^2x-1)(-\sin x)dx\\ &= \frac{1}{3}\cos^3x-\cos x + C\end{align*}$
(ii) First, sketch y=cos(x). From the graph, it is clear that the areas cancel out, verifying the result $ \int_0^\pi \cos x\, dx = 0$
Now, 2n-1 is an odd number. Observe that the graph of y=cos^2n-1(x) must look somewhat like this:
<IMAGE TO COME>
Indeed, the same idea happens; the areas cancel out. Hence, the integral is effectively equal to 0.

(iii) $\text{The volume of a typical shell is}\\ \delta V = 2\pi r h \, \delta x = 2\pi x \sin^3x\, \delta x\\ \text{for some thickness }\delta x\text{, using the diagram.}$
$\text{Hence our desired volume is:}\\ \begin{align*}V&=2\pi \int_0^\pi x\sin^3x\, dx\\ &= \left[2\pi x\left(\frac{1}{3}\cos^3x-\cos x\right)\right]_0^\pi-2\pi \int_0^\pi \left(\frac{1}{3}\cos^3x-\cos x\right) dx\text{ (by parts, using (i))}\\ &= \frac{2pi^2}{3}-2\pi(0-0)\text{ (by (ii)}\\ &=\frac{2\pi^2}{3}\end{align*}$

For question 14 a iii- should the volume be 4pi^2/3, not 2pi^2/3

RuiAce · « **Reply #16 on:** October 21, 2016, 10:27:28 pm »

Quote from: huwsername on October 21, 2016, 09:00:44 pm

IDK if you wanted anyone else's opinion, but I agree with the general impression that it was more 'polarised', as in q10-14 were probably easier than 2010, but q15-16 were a bit harder (around the early 00's kind of level).

As far a raw marks/scaling would go, I'd put this as a bit harder than 2014, and the 2014 cutoff was ~70. So you should be alright.

(But I'm a random pundit on the internet so I wouldn't quote me on it, these are touchy guesses)

Input's always welcome

Quote from: MarkThor on October 21, 2016, 08:33:54 pm

How would you compare this exam to 2010?
Because if it's the about the same difficulty or slightly harder there's still hope

About the same. 2010 was a bit harder though because it jumped into the deep end more quickly.

This one just spiked from Q14 to Q15. Q15 was draining though. Still yet to tackle that Q16...

All corrections listed above are implemented - tyvm

zoe_rammie · « **Reply #17 on:** October 21, 2016, 10:44:08 pm »

Quote from: ladyofathena on October 21, 2016, 06:33:20 pm

I found it very difficult. In my opinion it was much harder than 2015 and 2014 and I left about 40 marks
So disappointed, particularly since I got around 80% when I was doing the 2015 paper

ITS OKAY SAME SAME SAAAAAMEEEE PHEWWWWW I'M RELIEVED I'M NOT THE ONLY ONE WHO FOUND IT SUPER HARD

RuiAce · « **Reply #18 on:** October 21, 2016, 10:45:18 pm »

Quote from: zoe_rammie on October 21, 2016, 10:44:08 pm

ITS OKAY SAME SAME SAAAAAMEEEE PHEWWWWW I'M RELIEVED I'M NOT THE ONLY ONE WHO FOUND IT SUPER HARD

Nah. No way were you going to be the only one to find it hard. Although bits of it were chill, the other bulk of it was completely insane

edmododragon · « **Reply #19 on:** October 21, 2016, 11:59:05 pm »

The mechanics question was a godsend, and am relieved there was no extremely difficult Conics question near the end. Careless mistake in question 2, remind me never to sub in 1 before differentiating

RuiAce · « **Reply #20 on:** October 22, 2016, 12:00:24 am »

Quote from: edmododragon on October 21, 2016, 11:59:05 pm

The mechanics question was a godsend, and am relieved there was no extremely difficult Conics question near the end. Careless mistake in question 2, remind me never to sub in 1 before differentiating

Don't sub in 1 before differentiating for 3u (y)

edmododragon · « **Reply #21 on:** October 22, 2016, 12:01:20 am »

Hahaha I just subbed in my head 5 - 4 - 2 + 1 = 0 and missed that the constant 1 disappears rip me

RuiAce · « **Reply #22 on:** October 22, 2016, 12:15:31 am »

Gotta be careful of little things. Don't rush

MarkThor · « **Reply #23 on:** October 22, 2016, 08:44:29 am »

Quote from: huwsername on October 21, 2016, 09:00:44 pm

IDK if you wanted anyone else's opinion, but I agree with the general impression that it was more 'polarised', as in q10-14 were probably easier than 2010, but q15-16 were a bit harder (around the early 00's kind of level).

As far a raw marks/scaling would go, I'd put this as a bit harder than 2014, and the 2014 cutoff was ~70. So you should be alright.

(But I'm a random pundit on the internet so I wouldn't quote me on it, these are touchy guesses)

Thanks that is a bit relieving hahah

birdwing341 · « **Reply #24 on:** October 22, 2016, 09:26:25 am »

Quote from: edmododragon on October 22, 2016, 12:01:20 am

Hahaha I just subbed in my head 5 - 4 - 2 + 1 = 0 and missed that the constant 1 disappears rip me

Not gonna lie this was also me hahaha.....

Ty510 · « **Reply #25 on:** October 22, 2016, 09:27:21 am »

So I feel like I absolutely screwed up this exam completely.

Does this mean my overall mark is gonna be awful? My marks across the year in extension 2 were alright but I'm the only one in the class. I'm worried this mess up is gonna ruin my marks overall.

RuiAce · « **Reply #26 on:** October 22, 2016, 10:08:36 am »

Quote from: Ty510 on October 22, 2016, 09:27:21 am

So I feel like I absolutely screwed up this exam completely.

Does this mean my overall mark is gonna be awful? My marks across the year in extension 2 were alright but I'm the only one in the class. I'm worried this mess up is gonna ruin my marks overall.

The annoying thing is that if you were in a cohort of 1 then your final mark probably means everything.

The paper was hard, no doubt about that. What in particular makes you think you screwed up though?

Ali_Abbas · « **Reply #27 on:** October 22, 2016, 10:52:20 am »

Here is an alternative method for Question 16 a) i). I do not claim it is better,
but am simply offering it for anyone wondering what the "otherwise" method might have been.

$\text{We have that } 1+w+z = 0 \hspace{0.6cm} (1) \\ \Longrightarrow \overline{1+w+z} = \overline{0} \text{ (taking the conjugate of both sides)} \\ \Longrightarrow 1+\overline{w} + \overline{z} = 0 \hspace{0.6cm} (2) \\ (1)\times \overline{z}: \\ \overline{z} + w\overline{z} + 1 = 0 \hspace{0.6cm} (3) \text{ (noting that } z\overline{z} = |z|^{2} = 1) \\ (2) - (3): \\ \overline{w} - w\overline{z} = 0 \\ \Longrightarrow \overline{w} = w\overline{z} \\ \Longrightarrow w = \overline{w}z \text{ (taking the conjugate of both sides)} \hspace{0.4cm} (4) \\ \text{By a similar approach, it can be shown that } z = \overline{z}w \hspace{0.4cm} (5) \\ \text{Sub (5) into (4):} \\ w = \overline{w}\overline{z}w = \overline{z} \space \space \space (w\overline{w} = |w|^{2} = 1) \\ \text{Thus, we have the following:} \\ 1+w+z = 0 \text{ (given)} \\ (1)w+1(z)+wz = w+z+1 = 0, \text{ and} \\ (1)(w)(z) = z\overline{z} = |z|^{2} = 1 \\ \text{Hence, the complex numbers } 1, w, \text{ and } z \text{ form the roots of the equation } u^{3} - 1 = 0 \text{ and are thus} \\ \text{the three roots of unity. As such, they are uniformly distributed along the circumference} \\ \text{of the unit circle and form a regular 3-gone, i.e. an equilateral triangle.} \\ \text{QED}$

Ty510 · « **Reply #28 on:** October 22, 2016, 11:19:19 am »

Quote from: RuiAce on October 22, 2016, 10:08:36 am

The annoying thing is that if you were in a cohort of 1 then your final mark probably means everything.

The paper was hard, no doubt about that. What in particular makes you think you screwed up though?

Unfortunately left a lot of it blank. Probably like 40+ marks worth

Wasn't able to answer the difficult questions.

RuiAce · « **Reply #29 on:** October 22, 2016, 11:25:56 am »

Quote from: Ty510 on October 22, 2016, 11:19:19 am

Unfortunately left a lot of it blank. Probably like 40+ marks worth
Wasn't able to answer the difficult questions.

I've heard that from several people. I could spot about 40 that were tough.

Keep your head up firstly; a raw mark of 70 aligns to about 90 and a raw mark of 60 to about 84-85. Which is still fantastic no matter what. (NO subject aligns as brilliantly as 4U)

It's entirely up to you if you want to spend time looking at the solutions, but take it just as a learning curve. By this time next year the HSC won't even matter to you. So just look for areas of improvement.

If the only area of improvement you could've had was knowing what to do then it just means to practice more. Forget about 4U for now; time to move on to other stuff. There can never be a limit to how much exposure you have to content before an exam.

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Author Topic: Suggested Answers to the HSC 2016 Maths Extension 2 Exam (+Discussion!) (Read 23823 times)

zdeathbringer

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zoe_rammie

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RuiAce

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edmododragon

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RuiAce

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