Author Topic: Suggested Answers to the HSC 2016 Maths Extension 2 Exam (+Discussion!) (Read 23782 times)

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.

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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

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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

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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

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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

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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

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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

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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

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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

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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.

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

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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.

$\text{By using similar triangles in both cases:}\\ \begin{align*}\frac{h}{4}&=\frac{x}{6}\\ &\implies x=\frac{3h}{2}\\ \therefore a=\frac{3h}{2}+2\end{align*}$
$\begin{align*}\frac{h}{4}&=\frac{y}{4}&\\ \implies y&=h\\ \therefore b=h+3\end{align*}$
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

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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

<An alternate answer may be found in the comments by Ali>

Question 11
(a)

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(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)

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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)

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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)

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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
- Square rooting makes it increase slower, hence the graph bloats outwards a bit.

(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)

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(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.

Component graphs[/spoiler

$\text{Note that }f(0)=0\\ \text{and }f(2)=\pi\\ \text{Hence the range is }0\le y \le \pi$

Question 12
(a)

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(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)

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(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)

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(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)

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(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)

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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)

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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.
$\begin{align*}\text{Let }\angle BPC &= \theta\\ \text{Then }\angle BCP &=\theta \text{ (base angles on isos. TriBPC})\end{align*}$
For a more clever approach, consider the quadrilateral OAQC in a different way. Firstly, note that $\angle BAC=\theta $ by the alternate segment theorem.
$\text{OC and OA are both radii, and both CQ and AQ are tangents}\\ \text{Hence }\angle OCQ = \angle OAQ = 90^\circ\text{ (tangent perp. radius at pt. of contact)}\\ \therefore \angle OAQ+\angle OCQ=180^\circ\\ \therefore OAQC\text{ is cyclic (pair of opp. angles supplementary)}$
So since we have a cyclic quadrilateral, (\angle OQC=\theta) by angles in the same segment
$\therefore \angle OPQ=\angle OQP\\ \therefore \triangle OPQ\text{ is isosceles with }OP=OQ$

(c)

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(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)

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(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)

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(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 therefore look somewhat like this:

(The graph itself isn't important, but you should understand why there's a point of inflexion at $ \frac\pi2$ regardless).

Indeed, the same idea happens; the areas cancel out. Hence, the integral is effectively equal to 0.

(iii)

$\text{The volume of a typical shell is}\\ \delta V = 2\pi r h \, \delta x = 2\pi x \sin^3x\, \delta x\\ \text{for some thickness }\delta x\text{, using the diagram.}$
$\text{Hence our desired volume is:}\\ \begin{align*}V&=2\pi \int_0^\pi x\sin^3x\, dx\\ &= \left[2\pi x\left(\frac{1}{3}\cos^3x-\cos x\right)\right]_0^\pi-2\pi \int_0^\pi \left(\frac{1}{3}\cos^3x-\cos x\right) dx\text{ (by parts, using (i))}\\ &= \frac{4\pi^2}{3}-2\pi(0-0)\text{ (by (ii)}\\ &=\frac{4\pi^2}{3}\end{align*}$

(b)

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(i) Heaps of substitutions can be used. Here, for sake of simplicity, we will use the reasonable $x=\tan \theta \implies dx=\sec^2\theta \, d\theta$
$\begin{align*}I_0&=\int_0^1\frac{1}{(x^2+1)^2}dx\\ &= \int_0^{\frac{\pi}{4}} \frac{1}{\sec^4\theta}\sec^2\theta\, d\theta\\ &= \int_0^{\frac{\pi}{4}}\cos^2\theta\, d\theta\\ &= \int_0^\frac{\pi}{4}\left(\frac12+\frac12\cos 2\theta\right)d\theta\\ &= \left[\frac{\theta}{2}+\frac{\sin 2\theta}{4} \right]_0^{\frac{\pi}{4}}\\ &= \frac{\pi}{8}+\frac{1}{4}\end{align*}$
(ii) This is only 1 mark for a reason. Simply combine the integrals to get something trivialised.
$\begin{align*}I_0+I_2&=\int_0^1\frac{1}{(x^2+1)^2}dx + \int_0^1\frac{x^2}{(x^2+1)^2}dx\\ &= \int_0^1 \frac{x^2+1}{(x^2+1)^2}dx\\ &= \int \frac{dx}{1+x^2}\\ &= [\tan^{-1}x]_0^1\\ &=\frac{\pi}{4}\end{align*}$
(iii) We will apply that same trick in part (ii) to part (iii). Note that this time, a bit more algebra is required.
$\text{First note that }I_2=\frac{\pi}{4}-I_0=\frac{\pi}{8}-\frac14$
$\begin{align*}I_2+I_4&=\int_0^1\frac{x^2+x^4}{(x^2+1)^2}dx\\ &= \int_0^1 \frac{x^2}{x^2+1}dx\\ &= \int_0^1 \left(1-\frac{1}{x^2+1}\right)dx\\ &= [x-\tan^{-1}x]_0^1\\ &= 1-\frac{\pi}{4}\\ \therefore I_4&=1-\frac{\pi}{4}-\frac{\pi}{8}+\frac{1}{4}\\ &= \frac{5}{4}-\frac{3\pi}{8}\end{align*}$

(c)

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$\text{Since }x\ge 0\text{ we have }\sqrt{x}\ge 0$
There are many ways of doing this. An interesting way is to split up the cases here.
$\begin{align*}\text{If }x&\ge 1\\\text{Then }\sqrt{x}&\ge 1\\ \text{And }(x-1)&\ge 0\end{align*}$
$\begin{align*}\therefore \sqrt{x}(x-1)&\ge (x-1)\\ x\sqrt{x}-\sqrt{x}&\ge x-1\\ x\sqrt{x}+1&\ge x+\sqrt{x}\end{align*}$
$\text{If }0\le x \le 1\\ \text{Then }x-1\le 0\\ \text{But }\sqrt{x} \le 1$
$\text{So from }(x-1)=(x-1)\text{, noting }x-1\text{ is negative:}\\ \begin{align*}|x-1|&=|x-1|\\ |\sqrt{x}(x-1)|&\le |x-1|\\ \therefore \sqrt{x}(x-1)&\ge (x-1)\end{align*}$
The remainder is similar to the other case.

Question 15
(a)

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Fairly standard. Just replace x with sqrt(x) and be clever with algebra.
$\begin{align*}x &\mapsto \sqrt{x}:\\(\sqrt{x})^3-3\sqrt{x}+1&=0\\ \sqrt{x}(x-3)&=-1\\ x(x-3)^2&=1\\ x^3-6x^2+9x-1&=0\end{align*}$

(b)

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(i) We are given the acceleration, in terms of displacement. This immediately hints at the formula $\ddot{x}=\frac{d}{dx}\left(\frac{v^2}{2}\right) $
$\begin{align*}\frac{d}{dx}\left(\frac{v^2}{2}\right)&= -\frac{\mu^2}{x^2}\\ {v^2}&= \frac{2\mu^2}{x}+C\end{align*}$
We know that initially, x=b and v=0. Hence $C=-\frac{2\mu^2}{b} $
$\begin{align*}\therefore \left(\frac{dx}{dt}\right)^2&=\frac{2\mu^2}{x}-\frac{2\mu^2}{b}\\ &= 2\mu^2 \left(\frac{b-x}{bx}\right)\\ \therefore \frac{dx}{dt}&= \pm \mu\sqrt{2}\sqrt{\frac{b-x}{bx}}\end{align*}$
The question says that the particle then moves in a straight line, towards O. Because it is initially to the right, it must be moving to the left. Hence we take negative roots and we are done.

(ii) Because we want time to be the subject, we first reciprocate:
$\frac{dt}{dx}=-\frac{1}{\mu\sqrt{2}}\sqrt{\frac{bx}{b-x}}$
Now initially, the particle is at b. We want to bring the particle over to d.

Since the particle never changes direction (we are told it moves in a straight line), we can safely just assume that the time taken to get there is given by the integration:
$t=-\frac{1}{\mu \sqrt2}\int_b^d \sqrt{\frac{bx}{b-x}}$
$\text{Let }x=b\cos^2\theta\text{ as instructed.}\\ \text{Then, }dx=-2b\sin\theta\cos\theta\, d\theta$
$\begin{align*}\therefore t&=-\frac{1}{\mu \sqrt2}\int_0^{\cos^{-1}\sqrt{\frac{d}{b}}}\sqrt{\frac{b\cos^2\theta}{\sin^2\theta}}(-2b\sin\theta\cos\theta\, d\theta)\\ &= \frac{b\sqrt{2b}}{\mu}\int_0^{\cos^{-1}\sqrt{\frac{d}{b}}}\cos^2\theta \, d\theta\end{align*}$
(iii) Essentially, we want to let d approach 0. This is because d is the displacement we are interested in; mu is just a constant and b is just the initial position.
$\begin{align*}\text{As }d&\ to 0\\ t&\to \frac{1}{\mu}\sqrt{\frac{b}{2}} (\sqrt{0-0^2}+b\cos^{-1}0)\\ &= \frac{\pi b}{2\mu}\sqrt{\frac{b}{2}}\end{align*}$

(c)

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If we are to use partial fractions, we must let
$\begin{align*}\frac{3!}{x(x+1)(x+2)(x+3)}&=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+\frac{D}{x+3}\\ \therefore 3! &=A(x+1)(x+2)(x+3)+Bx(x+2)(x+3)+Cx(x+1)(x+3)+Dx(x+1)(x+2)\end{align*}$
By successively doing let x = ... we have our required values.
$\begin{align*}x=0 &\implies A=1\\ x=-1&\implies B=-3\\ x=-2&\implies C=3\\ x=-3&\implies D=-1\end{align*}$
$\therefore \frac{1}{x}-\frac{3}{x+1}+\frac{3}{x+2}-\frac{1}{x+3}$
(ii) This is a very famous ongoing partial fractions decomposition. We will put x=-k. The point is that by doing so, every other term vanishes.

After that, we will just need to do some factorial tricks and algebraic manipulation.
$\begin{align*}n!&=a_0[...(x+k)...]+a_1[...(x+k)...]+\dots+a_k[x(x+1)\dots(x+k-1)(x+k+1)(x+k+2)\dots(x-n)]+\dots+a_n[...(x+k)...]\\ \therefore n!&=a_k[[-k(-k+1)\dots(-1)][(1)(2)\dots(-k+n)]]\\ &= (-1)^k[[k(k-1)\dots (1)][(1)(2)\dots(n-k)]]\\ &= (-1)^k k! (n-k)!\\ \therefore a_k&=(-1)^{-k}\frac{n!}{k!(n-k)!}\\ &=(-1)^k\binom{n}{k}\end{align*}$
(iii) Now, however obvious it was that we had to let x=1 is will depend on you, because the above computation was demanding. However, the fact that not only are the signs alternating, but the fact the denominators all go up by 1 (starting from 1) just implies we need to sub this and the above result in.
$\text{Put }x=1\\ \begin{align*}\frac{n!}{(1)(2)\dots (1+n)}&= a_0+\frac{1}{2}a_1+\frac{1}{3}a_2+\frac{1}{4}a_3+\dots+\frac{1}{n+1}a_{n}\\ &\frac{1}{n+1}= 1-\frac{1}{2}\binom{n}{1}+\frac13\binom{n}{2}-\frac14\binom{n}{3}+\dots+\frac{(-1)^n}{n+1}\end{align*}$
The limiting sum is essentially what happens as $ n\to \infty$. It is clear that the sum tends to 0.
$\lim_{n\to \infty}\frac{1}{n+1}= 1-\frac{1}{2}\binom{n}{1}+\frac13\binom{n}{2}-\frac14\binom{n}{3}+\dots+\frac{(-1)^n}{n+1}=\lim_{n\to \infty}\frac{1}{n+1}=0$

Question 16
(a)

CLICK HERE FOR AN EXPLANATION

(b)

CLICK HERE FOR AN EXPLANATION

(c)

CLICK HERE FOR AN EXPLANATION

`

_____________________________________________

Final thoughts
Alright, breaking it down:

- Multiple choice was fairly relaxed, however full of curveballs. Even Q3, as easy as it seems, really toys around with stuff that people can forget.
- Q11 was full of standard questions. e) was a new case but still nothing too big of a deal
- Q12 was easy for the most part. However d) would've thrown people off. Quadratics are some of the biggest ninjas in the HSC.
- They added some flavour into Q13. Whilst conceptually all of them can be completed by any average student, the way the questions are worded was a bit interesting. Basically, it wouldn't have been too painful if you could work your way around the wording; the maths remained the same difficulty

- Probably one of the more 'fun' Q14s I've seen. Time to split the breakdown up:

The first mark is probably a freebie, but then your knowledge of the cosine curve and graphs is tested immediately. A 4u student knows the definition of integrals as areas to a deeper level than other students so the mark wasn't too hard to obtain, but you need to know how to sketch it. Of course, if you knew how to apply parts (i) and (ii), then (ii) was just your standard volumes methods and using what you know.

"Reduction" was different though. This year it wasn't focused on establishing an actual formula, but rather knowing how to play with the algebra.

There's probably many ways of doing the inequalities, including working backwards. I just thought of it via the somewhat obvious rearrangement-and-factorisation.

And whilst that was setting up for some interesting stuff, Q15 was where it really spiked. a) was just a classic, however you had to be careful with your algebra in b). You needed to know when and how to use what.

c) was a huge surprise and tested your algebra even further. The trick was to not work on all terms, but just the one they gave you. Then you had to realise how (-1)^k could actually appear, and not forget that there was an n! on the LHS!
Part (iii) of c) reminds me of binomial theorem though; it's just knowing when to sub what.

This makes two years in a row where complex numbers have had a MAJOR appearance in Q16. Part a) baffled me the most, because I was hoping that there were ways around without explicitly finding theta (became salty for a night but then I just admitted to it). b) was more chill if you knew what you were doing; equilateral triangles and Argand diagrams have appeared before but whether or not you've encountered it is your story.

c) was just weird. Recurrence is the stuff I learn about in first year uni, not high school. But you could do with combinatorics anyway - you just needed to figure out the outcomes. The way they worded it may have been confusing, but you have to think about how many hats (and hence people) are actually left.

Sandwiched in c), though, (iii) was a free mark. Always worth looking out for. And then (iv) was the somewhat classic repeated usage.

Whoever thought induction was to be 2 marks though... I have no idea. Unless they didn't care about you testing n=1.

If you want to know how i felt about the difficulty, try plotting something like y=5tan^-1(x-4)+4 for x > 0. There were still a lot of easily obtainable marks, but then the other half was questionable.

Whilst there was a small shocker in Q12, the paper's difficulty was still ascending. No unrealistic trends were broken here.

Whether or not I'd call this paper fair is a bit questionable. I reckon yes in that it was a good paper and it really tests many of the things overlooked in 4U, but that being said some questions were still really hard to work around. Apparently Q16b) was also in a first year uni maths paper...
_____________________________________________

Mapping
I reckon the following questions targeted E4:
Q10, Q14 c), Q15 b) (ii), Q15 c) (ii), Q16 a), Q16 c) (i), (ii), (v)
This adds up to 21 marks

Some candidates for E4 that I don't really think made it there however definitely targeted upper E3:
Q9, Q12 d) (ii), Q13 c) (i), Q13 d) (i), Q13 d) (i), Q14 b), Q15 b) other bits, Q15 c) (iii), Q16 b), Q16 c) (iv)

Which sounds fair. Every year I can roughly single out 20 marks that specifically hit band E4. This implies that the alignment algorithm will be, for the most part, similar to previous years.

However, it might be a bit more generous than 2015. With that Q16 there, I feel as though the 2015 paper was definitely easier. Note that in 2015, a 73 became an 89.

So probably E4 cutoff would be around 71. E3 is a bit more blurred out to me because there was a huge overlap between what would be E3 and what would be E2, as well as what might've been E4.

birdwing341 · « **Reply #1 on:** October 21, 2016, 06:15:08 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 First Half
Answer key:
1.C 2.C 3.A 4.D 5.D

Explanations
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. 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. 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. 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. 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)
$\begin{align*}\frac{1-i}{1+i}&=\frac{(1-i)^2}{(1+i)(1-i)}\\ &= \frac{1-2i+i^2}{2}\\ &= -i\\ &=\cos \frac{-\frac{\pi}{2}}+i\sin \frac{-\pi}{2}\end{align*} And the argument tells us what happens to the complex number. Answer: [b]D[/b]$

Twas a bit of a shock for me, but it was a test I know that i did worse the first sections than in the last ones. Unfortunately I rely on finishing the test early to check for silly mistakes and I know I've made at least three. That said, I'm pretty happy with the result, although like Rui said, the test was very polarised and had the potential to be very difficult.

Hope everyone was happy with how they went

ladyofathena · « **Reply #2 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

RuiAce · « **Reply #3 on:** October 21, 2016, 06:37:31 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

Ouch! Well when I mean polarised I really am referring to the second half of the paper. That was insane.

Definitely harder than 2015. But yeah, really sorry to hear that!

Ty510 · « **Reply #4 on:** October 21, 2016, 07:09:59 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

I feel the exact same, I think I lost 40+ marks and started crying at the end of the exam.
Heavy days, those last questions were impossible.

matthewcrowther24 · « **Reply #5 on:** October 21, 2016, 07:13:34 pm »

I don't know if I'm remembering this wrong, but in question 5 isn't multiplying by -i a rotation clockwise by 90°?

rafaishere1997 · « **Reply #6 on:** October 21, 2016, 07:25:14 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. D
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)
$\begin{align*}\frac{1-i}{1+i}&=\frac{(1-i)^2}{(1+i)(1-i)}\\ &= \frac{1-2i+i^2}{2}\\ &= -i\\ &=\cos \frac{-\frac{\pi}{2}}+i\sin \frac{-\pi}{2}\end{align*}$
And the argument tells us what happens to the complex number.
Answer: D[/tex]
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. A
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.
$\begin{align*}x^{2016}+\frac{1}{x^{2016}}&= \left(\text{cis }\frac{\pi}{6}\right)^{2016}+ \left(\text{cis }\frac{\pi}{6}\right)^{-2016}\\ &= \text{cis}1344\pi + \text{cis}(-1344\pi)\\ &= \cos 1344\pi + i \sin 1344\pi + \cos (-1344\pi) + i\sin (-1344\pi)\\ &=1\end{align*}$
Answer: A
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.
$\begin{align*}\int xe^{-2x}dx&= -\frac{1}{2}xe^{-2x}dx + \int \frac{1}{2}e^{-2x}dx\\ &= -\frac{1}{2}xe^{-2x}dx -\frac{1}{4}e^{-2x}+C\end{align*}$

Rui Tong, Vansh approves

huwsername · « **Reply #7 on:** October 21, 2016, 08:11:13 pm »

Reaffirming goandgoo, this seems to suggest Question 10 = 2 (B)

http://www.wolframalpha.com/input/?i=(-(1%2F2)+%2B+(sqrt(3)%2F2)i)%5E(2016)+%2B+(-(1%2F2)+%2B+(sqrt(3)%2F2)i)%5E-(2016)

goandgoo · « **Reply #8 on:** October 21, 2016, 08:11:58 pm »

Quote from: huwsername on October 21, 2016, 08:11:13 pm

Reaffirming goandgoo, this seems to suggest it = 2 (B)

http://www.wolframalpha.com/input/?i=(-(1%2F2)+%2B+(sqrt(3)%2F2)i)%5E(2016)+%2B+(-(1%2F2)+%2B+(sqrt(3)%2F2)i)%5E-(2016)

Looks like it's fixed

RuiAce · « **Reply #9 on:** October 21, 2016, 08:24:05 pm »

Quote from: goandgoo on October 21, 2016, 08:11:58 pm

Looks like it's fixed

Yep, thanks for stating this for me (y)

Ali_Abbas · « **Reply #10 on:** October 21, 2016, 08:25:28 pm »

For anyone interested, a slightly faster way to do Question 10 of the multiple choice is as follows.

$x+\frac{1}{x} = -1 \\ \Longrightarrow x^{2}+x+1 = 0 \\ \Longrightarrow \frac{x^{3}-1}{x-1} = 0 \\ \Longrightarrow x^{3} - 1 = 0 \\ \Longrightarrow x^{3} = 1 \\ \text{Observe that 2016 is a multiple of 3 and so it follows that } x^{2016} = 1 \\ \Longrightarrow \frac{1}{x^{2016}} = 1 \\ \text{Hence, they sum to 2.}$

MarkThor · « **Reply #11 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

RuiAce · « **Reply #12 on:** October 21, 2016, 08:37:01 pm »

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

I'll get back to you on this

MarkThor · « **Reply #13 on:** October 21, 2016, 08:46:28 pm »

Quote from: RuiAce on October 21, 2016, 08:37:01 pm

I'll get back to you on this

Thanks Rui

huwsername · « **Reply #14 on:** October 21, 2016, 09:00:44 pm »

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

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)

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Suggested Answers to the HSC 2016 Maths Extension 2 Exam (+Discussion!)

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Ty510

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Re: Suggested Answers to the HSC 2016 Maths Extension 2 Exam (+Discussion!)

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huwsername

Re: Suggested Answers to the HSC 2016 Maths Extension 2 Exam (+Discussion!)

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Ali_Abbas

Re: Suggested Answers to the HSC 2016 Maths Extension 2 Exam (+Discussion!)

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