Math Ext 1 Predictions/Thoughts (2 Viewers)

hatterene

Active Member
naurrrrr my friends all say it looks like titties

Life'sHard

Well-Known Member
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

Well-Known Member
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

Well-Known Member
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

Moderator
Moderator
Looking at the solutions provided above:
• there is a small error in Q5 where $\bg_white \theta \in \big[0,\ \pi\big]$ as $\bg_white \cos\theta < 0\ \implies\ \theta \in \left(\cfrac{\pi}{2},\ \pi\right]$, but the solution writer has stated that $\bg_white \theta \in \left[\cfrac{\pi}{2},\ \pi\right]$. This is because $\bg_white \cos\theta = \cfrac{pi}{2}\ \implies \cos\theta = 0$. Thus, the angle $\bg_white \theta$ must be obtuse or $\bg_white \theta = \pi$ - the solution writer is correct that NESA have made a mistake in including $\bg_white \theta = \pi$ in the category of "obtuse" angles.
• In q11(h), though it is not in the syllabus, a potentially simpler solution is:
\bg_white \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 $\bg_white T$ is increasing towards 25, hence $\bg_white |T - 25| = 25 - T$ and the equation is $\bg_white T = 25 - Ae^{kt}$ where $\bg_white A > 0$ and $\bg_white k < 0$. The solutions state that $\bg_white A = e^C > 0$ and then find $\bg_white A = -20$, which is a contradiction. They should have $\bg_white A = 20$ and $\bg_white k = \cfrac{1}{8}\ln\cfrac{3}{4} = -\cfrac{1}{8}\ln\cfrac{4}{3} < 0$ to give $\bg_white t = \cfrac{16\ln{2}}{2\ln{2} - \ln{3}} \approx 38\ \text{min}\ 33\ \text{s}$.​
• The resulting equation is better written as $\bg_white T = 25 - 20e^{\frac{-t}{8}\ln{\frac{4}{3}$ so that the $\bg_white t$ term does not appear to be part of the log. Alternatively, it can be written as $\bg_white T = 25 - 20\left(\cfrac{4}{3}\right)^{-\frac{t}{8}}$ or as $\bg_white T = 25 - 20\left(\cfrac{3}{3}\right)^{\frac{t}{8}}$.​
• For 12(c), the conclusion should be for integers $\bg_white n \geqslant 1$ or for $\bg_white n \in \mathbb{Z}^+$.​
• With q13(b), I wonder if the marking will require a justification that $\bg_white h > 0$ 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

Well-Known Member
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

Well-Known Member
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

Member
Looking at the solutions provided above:
• there is a small error in Q5 where $\bg_white \theta \in \big[0,\ \pi\big]$ as $\bg_white \cos\theta < 0\ \implies\ \theta \in \left(\cfrac{\pi}{2},\ \pi\right]$, but the solution writer has stated that $\bg_white \theta \in \left[\cfrac{\pi}{2},\ \pi\right]$. This is because $\bg_white \cos\theta = \cfrac{pi}{2}\ \implies \cos\theta = 0$. Thus, the angle $\bg_white \theta$ must be obtuse or $\bg_white \theta = \pi$ - the solution writer is correct that NESA have made a mistake in including $\bg_white \theta = \pi$ in the category of "obtuse" angles.
• In q11(h), though it is not in the syllabus, a potentially simpler solution is:
\bg_white \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 $\bg_white T$ is increasing towards 25, hence $\bg_white |T - 25| = 25 - T$ and the equation is $\bg_white T = 25 - Ae^{kt}$ where $\bg_white A > 0$ and $\bg_white k < 0$. The solutions state that $\bg_white A = e^C > 0$ and then find $\bg_white A = -20$, which is a contradiction. They should have $\bg_white A = 20$ and $\bg_white k = \cfrac{1}{8}\ln\cfrac{3}{4} = -\cfrac{1}{8}\ln\cfrac{4}{3} < 0$ to give $\bg_white t = \cfrac{16\ln{2}}{2\ln{2} - \ln{3}} \approx 38\ \text{min}\ 33\ \text{s}$.​
• The resulting equation is better written as $\bg_white T = 25 - 20e^{\frac{-t}{8}\ln{\frac{4}{3}$ so that the $\bg_white t$ term does not appear to be part of the log. Alternatively, it can be written as $\bg_white T = 25 - 20\left(\cfrac{4}{3}\right)^{-\frac{t}{8}}$ or as $\bg_white T = 25 - 20\left(\cfrac{3}{3}\right)^{\frac{t}{8}}$.​
• For 12(c), the conclusion should be for integers $\bg_white n \geqslant 1$ or for $\bg_white n \in \mathbb{Z}^+$.​
• With q13(b), I wonder if the marking will require a justification that $\bg_white h > 0$ 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

Moderator
Moderator
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 $\bg_white x$-axis

notme123

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

tito981

Well-Known Member
yes but unlikely for maths

icycledough

Well-Known Member
can you still get a scaled mark of 100 if you didnt get 100%? e.g. 69.5 or 69/70
Just for clarification, I believe there is no half marks awarded in the HSC, but someone can let me know if they do in fact provide half marks.

Trebla

Just for clarification, I believe there is no half marks awarded in the HSC, but someone can let me know if they do in fact provide half marks.
Converting a mark out of 70 into a mark out of 100 will lead to decimal places.

notme123

Well-Known Member
Just for clarification, I believe there is no half marks awarded in the HSC, but someone can let me know if they do in fact provide half marks.
its true, my english tutor got half marks for mod c and unseen but idk for maths. probs not.

so you basically get no returns from scaling if you get 69/70. well a 69 is like a 98.6% or something which rounds to 99 ig

icycledough

Well-Known Member
Like for example, if there was a 1 mark question in a maths exam (not MC), then you would either get 0 or 1 for it; half marks aren't actually awarded in the marking of the exam. Obviously, the percentage conversion will be decimals (like if you get 67,68,69), but the raw mark will be a whole number. I was saying this as for internals for a language I did last year (which involved translation into English), our teacher would deduct a quarter mark for every mistake.

Trebla

A bit out of order.

I've reordered it.

Also part of 12a is missing (direction field on writing paper)

EDIT: 12a picture is here (thanks to pokipoki beanz):

View attachment 33972
Just realised the x and y axes are differently orientated. Is that a typo? Or is that deliberate?

tywebb

dangerman
Just realised the x and y axes are differently orientated. Is that a typo? Or is that deliberate?
It is not specified in the question that y is a function of x or vice versa. I agree that it's a bit strange. But I don't think it makes any difference.

deanjacobson_

Member
Just realised the x and y axes are differently orientated. Is that a typo? Or is that deliberate?
I could be wrong but I don't think that diagram was the exact one on the writing booklet. We'll have to wait and see when the actual paper is posted tomorrow

notme123

Well-Known Member
for that sample proportion q was it even important that it was less than 2.5?? because every solution ive seen so far doesnt use that fact. i think the z score has to be greater than 2.