Math Ext 1 Predictions/Thoughts (6 Viewers)

BAYPALS · Nov 20, 2021

stupidpersonlol said:
lol what is a good raw mark for e4

safe to say 75-78 for 90. 85-90 should be 94-95

BAYPALS · Nov 20, 2021

tywebb said:
Looks like Jojofelyx was right and one eight seven wrong.

The direction field with answer is here (thanks to andy!):

View attachment 33974

is this the correct answer forsure?

Life'sHard · Nov 20, 2021

Was this year’s more difficult in comparison to last year? I’ve got a bias to say this years paper was more difficult but what do you guys think. I’d like to see what @Trebla or @CM_Tutor thinks about this years exam.

hatterene · Nov 20, 2021

imo it was much harder - but for what

BAYPALS · Nov 20, 2021

hatterene said:
imo it was much harder - but for what

i dont think it was harder - but i would say the room for errors/sillies was much greater in this paper.

This resembled a 2U paper more than a 3U paper with all the areas and volumes and shit it was weird

Trebla · Nov 20, 2021

Life'sHard said:
Was this year’s more difficult in comparison to last year? I’ve got a bias to say this years paper was more difficult but what do you guys think. I’d like to see what @Trebla or @CM_Tutor thinks about this years exam.

My personal view was it more like what you “expected” from the HSC i.e. pretty stock standard questions until you dive into Q13 onwards or last couple of multiple choice (I was very surprised they went very easy on perms/combs though). The thing about last year’s paper was that the combinatoric proof question with binomial theorem was unexpected because binomial theorem and combinatoric proofs is a very small part of the new syllabus (whereas in the old syllabus it was dedicated as a whole topic). Kinda like how this year’s Advanced and (to a lesser extent) Extension 2 turned out a little bit unexpected and caught people off.

Trebla · Nov 20, 2021

Also, slightly controversial, I quite liked the sample proportion question as it really tested whether you knew your stuff conceptually. Most sample proportion questions are pretty generic just blindly plugging numbers into formulae.

hatterene · Nov 20, 2021

um within the last 5 seconds of the test i had a lot of doubt with mc and i changed a question 4 times... looking like this. do you think they'll recognise my answer.

Trebla · Nov 20, 2021

Imagine showing that picture out of context...

hatterene · Nov 20, 2021

naurrrrr my friends all say it looks like titties

Life'sHard · Nov 20, 2021

Trebla said:
Also, slightly controversial, I quite liked the sample proportion question as it really tested whether you knew your stuff conceptually. Most sample proportion questions are pretty generic just blindly plugging numbers into formulae.

The wording of that sample proportion question was a real pain trying to dissect in the heat of the moment but looking at it now and really taking a good amount of time to comprehend it, I do see what you mean.

stupidpersonlol · Nov 20, 2021

Life'sHard said:
Was this year’s more difficult in comparison to last year? I’ve got a bias to say this years paper was more difficult but what do you guys think. I’d like to see what @Trebla or @CM_Tutor thinks about this years exam.

i feel like the time was super tight this year, but content wise this year was easier lol...i just made many mistakes anyways

notme123 · Nov 20, 2021

Trebla said:
Also, slightly controversial, I quite liked the sample proportion question as it really tested whether you knew your stuff conceptually. Most sample proportion questions are pretty generic just blindly plugging numbers into formulae.

do you think getting the answer of 6000 would be sufficient for 3 marks regardless if we had any errors with inequalities in our working??

CM_Tutor · Nov 20, 2021

Looking at the solutions provided above:

there is a small error in Q5 where $\theta \in \big[0,\ \pi\big]$ as $\cos\theta < 0\ \implies\ \theta \in \left(\cfrac{\pi}{2},\ \pi\right]$ , but the solution writer has stated that $\theta \in \left[\cfrac{\pi}{2},\ \pi\right]$ . This is because $\cos\theta = \cfrac{pi}{2}\ \implies \cos\theta = 0$ . Thus, the angle $\theta$ must be obtuse or $\theta = \pi$ - the solution writer is correct that NESA have made a mistake in including $\theta = \pi$ in the category of "obtuse" angles.
In q11(h), though it is not in the syllabus, a potentially simpler solution is:

$\begin{align*} \text{If $x = \alpha,\ \beta,\ \gamma$, and $\delta$ are the solutions of} \quad x^4 - 3x + 6 &= 0, \quad \text{then, by setting $u = \cfrac{1}{x}$,} \\ \text{the resulting equation in $u$ will have solutions $u = \cfrac{1}{\alpha},\ \cfrac{1}{\beta},\ \cfrac{1}{\gamma}$, and $\cfrac{1}{\delta}$:} \quad \left(\cfrac{1}{u}\right)^4 - 3\left(\cfrac{1}{u}\right) + 6 &= 0 \\ 1 - 3u^3 + 6u^4 &= 0 \\ \text{So, sum of roots} = \quad \cfrac{1}{\alpha} + \cfrac{1}{\beta} + \cfrac{1}{\gamma} + \cfrac{1}{\delta} &= \cfrac{-(-3)}{6} \\ \cfrac{1}{\alpha} + \cfrac{1}{\beta} + \cfrac{1}{\gamma} + \cfrac{1}{\delta} &= \cfrac{1}{2} \end{align*}$

Per the discussion above, q12(a) appears wrong
the handling of the absolute value in q12(b) is wrong as the temperature $T$ is increasing towards 25, hence $|T - 25| = 25 - T$ and the equation is $T = 25 - Ae^{kt}$ where $A > 0$ and $k < 0$ . The solutions state that $A = e^C > 0$ and then find $A = -20$ , which is a contradiction. They should have $A = 20$ and $k = \cfrac{1}{8}\ln\cfrac{3}{4} = -\cfrac{1}{8}\ln\cfrac{4}{3} < 0$ to give $t = \cfrac{16\ln{2}}{2\ln{2} - \ln{3}} \approx 38\ \text{min}\ 33\ \text{s}$ .
The resulting equation is better written as $T = 25 - 20e^{\frac{-t}{8}\ln{\frac{4}{3}$ so that the $t$ term does not appear to be part of the log. Alternatively, it can be written as $T = 25 - 20\left(\cfrac{4}{3}\right)^{-\frac{t}{8}}$ or as $T = 25 - 20\left(\cfrac{3}{3}\right)^{\frac{t}{8}}$ .
For 12(c), the conclusion should be for integers $n \geqslant 1$ or for $n \in \mathbb{Z}^+$ .
With q13(b), I wonder if the marking will require a justification that $h > 0$ at impact with the wall proves that it has yet to hit the floor...
I am confident that q14(a) is meant to be solved by vectors

stupidpersonlol · Nov 20, 2021

notme123 said:
do you think getting the answer of 6000 would be sufficient for 3 marks regardless if we had any errors with inequalities in our working??

probs not lol. maybe 2 marks

tito981 · Nov 20, 2021

the wording of the sample size question was so long that i didnt see 'round to the nearest thousand', gonna lose a mark to not rounding imagine that.

BAYPALS · Nov 20, 2021

CM_Tutor said:
Looking at the solutions provided above:

there is a small error in Q5 where $\theta \in \big[0,\ \pi\big]$ as $\cos\theta < 0\ \implies\ \theta \in \left(\cfrac{\pi}{2},\ \pi\right]$ , but the solution writer has stated that $\theta \in \left[\cfrac{\pi}{2},\ \pi\right]$ . This is because $\cos\theta = \cfrac{pi}{2}\ \implies \cos\theta = 0$ . Thus, the angle $\theta$ must be obtuse or $\theta = \pi$ - the solution writer is correct that NESA have made a mistake in including $\theta = \pi$ in the category of "obtuse" angles.

In q11(h), though it is not in the syllabus, a potentially simpler solution is:

$\begin{align*} \text{If $x = \alpha,\ \beta,\ \gamma$, and $\delta$ are the solutions of} \quad x^4 - 3x + 6 &= 0, \quad \text{then, by setting $u = \cfrac{1}{x}$,} \\ \text{the resulting equation in $u$ will have solutions $u = \cfrac{1}{\alpha},\ \cfrac{1}{\beta},\ \cfrac{1}{\gamma}$, and $\cfrac{1}{\delta}$:} \quad \left(\cfrac{1}{u}\right)^4 - 3\left(\cfrac{1}{u}\right) + 6 &= 0 \\ 1 - 3u^3 + 6u^4 &= 0 \\ \text{So, sum of roots} = \quad \cfrac{1}{\alpha} + \cfrac{1}{\beta} + \cfrac{1}{\gamma} + \cfrac{1}{\delta} &= \cfrac{-(-3)}{6} \\ \cfrac{1}{\alpha} + \cfrac{1}{\beta} + \cfrac{1}{\gamma} + \cfrac{1}{\delta} &= \cfrac{1}{2} \end{align*}$

Per the discussion above, q12(a) appears wrong

the handling of the absolute value in q12(b) is wrong as the temperature $T$ is increasing towards 25, hence $|T - 25| = 25 - T$ and the equation is $T = 25 - Ae^{kt}$ where $A > 0$ and $k < 0$ . The solutions state that $A = e^C > 0$ and then find $A = -20$ , which is a contradiction. They should have $A = 20$ and $k = \cfrac{1}{8}\ln\cfrac{3}{4} = -\cfrac{1}{8}\ln\cfrac{4}{3} < 0$ to give $t = \cfrac{16\ln{2}}{2\ln{2} - \ln{3}} \approx 38\ \text{min}\ 33\ \text{s}$ .

The resulting equation is better written as $T = 25 - 20e^{\frac{-t}{8}\ln{\frac{4}{3}$ so that the $t$ term does not appear to be part of the log. Alternatively, it can be written as $T = 25 - 20\left(\cfrac{4}{3}\right)^{-\frac{t}{8}}$ or as $T = 25 - 20\left(\cfrac{3}{3}\right)^{\frac{t}{8}}$ .

For 12(c), the conclusion should be for integers $n \geqslant 1$ or for $n \in \mathbb{Z}^+$ .

With q13(b), I wonder if the marking will require a justification that $h > 0$ at impact with the wall proves that it has yet to hit the floor...

I am confident that q14(a) is meant to be solved by vectors

Im confused. Is Q12A going up the correct solutions (always above x axis) or going down (below x axis)? which one is the correct answer

CM_Tutor · Nov 20, 2021

BAYPALS said:
Im confused. Is Q12A going up the correct solutions (always above x axis) or going down (below x axis)? which one is the correct answer

Assuming that the slope field posted elsewhere in this thread is accurate, the path will have a minimum stationary point and always be above the $x$ -axis

notme123 · Nov 20, 2021

can you still get a scaled mark of 100 if you didnt get 100%? e.g. 69.5 or 69/70

tito981 · Nov 20, 2021

yes but unlikely for maths

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