Math Ext 1 Predictions/Thoughts (1 Viewer)

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

icycledough · Nov 20, 2021

notme123 said:
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 · Nov 20, 2021

icycledough said:
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 · Nov 20, 2021

icycledough said:
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 · Nov 20, 2021

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 · Nov 20, 2021

tywebb said:
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 · Nov 20, 2021

Trebla said:
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.

tywebb · Nov 21, 2021

deanjacobson_ said:
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

I doubt it. Here are another set of solutions, by Nash: https://cdn.discordapp.com/attachme...021_HSC_Mathematics_Extension_1_Solutions.pdf

And you can see they used the same diagram in 12a.

notme123 · Nov 21, 2021

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.

tito981 · Nov 21, 2021

notme123 said:
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.

it is important that its less than 2.5, because the question states that 'the sample size is chosen so that the chance of shutting down the machine unnecessarily is less than 2.5%'. So this means since you use z=2, the upper 2.5% will never cause the machine to shut down because it is at the upper end where the machine will never shut down. This is because when you think about p, you want the machine to be in the upper 2.5% because that means it is working to near perfect standard.

CM_Tutor · Nov 21, 2021

tywebb said:
I doubt it. Here are another set of solutions, by Nash. And you can see they used the same diagram in 12a.

Nash's solution is missing the sketch in q12(d)(iii) and has the sane problem as the other solution in handling the absolute value in q12(b). It does use the vector approach that I believe was sought in q14(a), though.

Siwel · Nov 21, 2021

CM_Tutor said:
Nash's solution is missing the sketch in q12(d)(iii) and has the sane problem as the other solution in handling the absolute value in q12(b). It does use the vector approach that I believe was sought in q14(a), though.

what about q13a) why is it square root y? thought it should just be y

CM_Tutor · Nov 21, 2021

Siwel said:
what about q13a) why is it square root y? thought it should just be y

You are correct, and well caught! Nash's solution to q13(a) is indeed incorrect. The solutions posted earlier answered this question correctly. Volumes rotated about the y-axis are indeed of the form

$\pi\int x^2\ dy$

which, in this case of one volume subtracted from another, gives

$\pi\int_0^2 y\ dy - \pi\int_1^2 y - 1\ dy$

Math Ext 1 Predictions/Thoughts (1 Viewer)

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