# Math Ext 1 Predictions/Thoughts (1 Viewer)

#### tito981

##### 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.
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

##### Moderator
Moderator
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

##### Active Member
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

##### Moderator
Moderator
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

$\bg_white \pi\int x^2\ dy$

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

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

#### tywebb

##### dangerman
Matrix put their solutions up now at https://www.matrix.edu.au/2021-hsc-maths-ext-1-exam-paper-solutions/

Interestingly they are saying both one eight seven's solution and andi!'s solution for 12a would be accepted.

But I think most people will agree with andi!'s (like the blue one).

They also have y on vertical axis and x on horizontal axis.

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#### yly417692

##### New Member
Your following comment "the exam was kinda simple" makes me say this ahaha, think we're on the same page here. Not only ruse, but practically 90% of all selective schools, hell, anyone good at x1 (who cares enough to read the question properly lol), is gonna get a raw 68+/70 in this.
Ruse kid here– I'll be happy if I can even get a raw mark of 60. Some of my classmates said that they only did the prove |v|^2=v•v question in Question 14. My internal rank isn't great but it's still passable (barely).

#### Siwel

##### Active Member
Ruse kid here– I'll be happy if I can even get a raw mark of 60. Some of my classmates said that they only did the prove |v|^2=v•v question in Question 14. My internal rank isn't great but it's still passable (barely).
really? they just skipped majority of q14?

#### notme123

##### Well-Known Member
Ruse kid here– I'll be happy if I can even get a raw mark of 60. Some of my classmates said that they only did the prove |v|^2=v•v question in Question 14. My internal rank isn't great but it's still passable (barely).

#### yly417692

##### New Member
really? they just skipped majority of q14?
They skipped some parts and made some unsuccessful attempts for other parts.

The ones who struggled with Q14 were around the triple digit ranks.

#### CM_Tutor

##### Moderator
Moderator
Matrix put their solutions up now at https://www.matrix.edu.au/2021-hsc-maths-ext-1-exam-paper-solutions/

Interestingly they are saying both one eight seven's solution and andi!'s solution for 12a would be accepted.

But I think most people will agree with andi!'s (like the blue one).

View attachment 34083
They also have y on vertical axis and x on horizontal axis.
Interesting, thanks for posting it.

I have doubts about the red pathway. The pattern of dashes, from -1 to -0.8 to -0.6 along y = +0.6 and y = +0.4, looks consistent to me with a curve that is decreasing but concave up.

I would, however, draw a curve similar to the red one if I started from x = +1.8 and something like y = -1.2.

And, shifting the blue curve in quadrant 1 down a little and I think we'd get a relation with a vertical tangent on / close to the vertical axis.

It certainly is an interesting slope field along y = +0.2, moving from (something like) 0- for x < -2 to -0.25 at the edge of the slope field to maybe -0.75 by x = -0.6 then increasing again but remaining negative for x = -0.4 and x = -0.2, then suddenly switching to more like 2 at the y-axis and then decreasing again, presumably towards 0+. We see a similar trend along y = 0, y = -0.2, and y = -0.4. I wonder what the underlying DE is being represented here.

##### -insert title here-
Interesting, thanks for posting it.

I have doubts about the red pathway. The pattern of dashes, from -1 to -0.8 to -0.6 along y = +0.6 and y = +0.4, looks consistent to me with a curve that is decreasing but concave up.

I would, however, draw a curve similar to the red one if I started from x = +1.8 and something like y = -1.2.

And, shifting the blue curve in quadrant 1 down a little and I think we'd get a relation with a vertical tangent on / close to the vertical axis.

It certainly is an interesting slope field along y = +0.2, moving from (something like) 0- for x < -2 to -0.25 at the edge of the slope field to maybe -0.75 by x = -0.6 then increasing again but remaining negative for x = -0.4 and x = -0.2, then suddenly switching to more like 2 at the y-axis and then decreasing again, presumably towards 0+. We see a similar trend along y = 0, y = -0.2, and y = -0.4. I wonder what the underlying DE is being represented here.
Well the vector field doesn't appear to be nice at the origin, so the uniqueness and existence theorem doesn't apply here.

#### one eight seven

##### New 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​
I wrote the set of solutions that you are referring to here.

As for Q12a), I think NESA should accept both solutions. A slope field is not sufficient to define the equation of a line passing through a point as there is not enough detail on the slope field to accurately plot it. I can see why it could be either, and I have seen several solutions of it going up and down from both teachers and students. I'm kinda on the fence about this question, I think it was a bad question from NESA and they should have just provided one with a more clear-cut answer as I don't see what is being assessed if NESA decides to lean a certain way.

For Q12b), I have mimicked a solution from the Cambridge textbook (page 582 of Extension 1 textbook) which redefines the variable A from positive only to non-zero as a result of removing the absolute value. This is why I wrote "now the domain of A is A=/=0", since previously it was A>0. The Cambridge solutions does something similar, writing that "A is positive" and upon removing the absolute value, "A is positive or negative". To be more precise, I should have redefined a new variable B and have said that $\bg_white T-25=\pm Ae^{kt}=Be^{kt}$ for B=/=0 which I have now done in an updated set of solutions.

For Q12c), there isn't really a need to write integers since n has already been defined to be an integer. This was seen in the 2020 Extension 1 solutions and was also extensively discussed on the 2005 Extension 2 solutions, which said that concluding statements of many students were too long and suggested writing "Hence the statement is true for all n>=0 by induction". I have told students whose teachers were pedantic with concluding statements to refer to official solutions as teachers cannot refute a three paragraph discussion and recommendation by an official set of solutions.

For Q13b), I would doubt they'd need that much justification since the displacement equation assumes there are no walls or floor, and so a simple line stating that if it lands partway up the wall, then it couldn't have touched the floor would have been sufficient. This is kind of obvious since if it touches the floor then the height is less than or equal to zero.

For Q14a), I think this was just a terrible question. They should have explained to use a vector method, or they should have had defined it using vectors. The vector method for this question is inefficient (as numerous students who used this method have pointed out) and the sine rule is the only good way to do it. Regardless, NESA cannot refuse marks for alternative methods as I have emailed them about this issue and they have responded with

"The syllabus does not specify any specific method that students must use to solve problems. HSC markers are aware of alternative methods. Students will not be assessed on the method used but on the correct results of the process."

This opens the door for unorthodox methods and out of syllabus methods. However, I usually recommend against out of syllabus methods since nobody knows how you will be marked and whether you need to prove out of syllabus results before using.

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#### CM_Tutor

##### Moderator
Moderator
I wrote the set of solutions that you are referring to here.

As for Q12a), I think NESA should accept both solutions. A slope field is not sufficient to define the equation of a line passing through a point as there is not enough detail on the slope field to accurately plot it. I can see why it could be either, and I have seen several solutions of it going up and down from both teachers and students. I'm kinda on the fence about this question, I think it was a bad question from NESA and they should have just provided one with a more clear-cut answer as I don't see what is being assessed if NESA decides to lean a certain way.

For Q12b), I have mimicked a solution from the Cambridge textbook (page 582 of Extension 1 textbook) which redefines the variable A from positive only to non-zero as a result of removing the absolute value. This is why I wrote "now the domain of A is A=/=0", since previously it was A>0. The Cambridge solutions does something similar, writing that "A is positive" and upon removing the absolute value, "A is positive or negative". To be more precise, I should have redefined a new variable B and have said that $\bg_white T-25=\pm Ae^{kt}=Be^{kt}$ for B=/=0 which I have now done in an updated set of solutions.

For Q12c), there isn't really a need to write integers since n has already been defined to be an integer. This was seen in the 2020 Extension 1 solutions and was also extensively discussed on the 2005 Extension 2 solutions, which said that concluding statements of many students were too long and suggested writing "Hence the statement is true for all n>=0 by induction". I have told students whose teachers were pedantic with concluding statements to refer to official solutions as teachers cannot refute a three paragraph discussion and recommendation by an official set of solutions.

For Q13b), I would doubt they'd need that much justification since the displacement equation assumes there are no walls or floor, and so a simple line stating that if it lands partway up the wall, then it couldn't have touched the floor would have been sufficient. This is kind of obvious since if it touches the floor then the height is less than or equal to zero.

For Q14a), I think this was just a terrible question. They should have explained to use a vector method, or they should have had defined it using vectors. The vector method for this question is inefficient (as numerous students who used this method have pointed out) and the sine rule is the only good way to do it. Regardless, NESA cannot refuse marks for alternative methods as I have emailed them about this issue and they have responded with

"The syllabus does not specify any specific method that students must use to solve problems. HSC markers are aware of alternative methods. Students will not be assessed on the method used but on the correct results of the process."

This opens the door for unorthodox methods and out of syllabus methods. However, I usually recommend against out of syllabus methods since nobody knows how you will be marked and whether you need to prove out of syllabus results before using.

On q12(a), if I was marking the paper, I may well conclude that accepting both is the fairest outcome given the ambiguity that should have been addressed before the paper was finalised. This does not alter my view that the curve with a stationary point fits the given slope field more naturally, however.

On 12(b), I see what you mean and recognise that it is one way to handle the removal of the absolute values, though it was not clear when I first saw the solution.

On 12(c), while a definition of n as an integer earlier in the answer is sufficient, and I am aware of discussions around induction and setting out, the principle behind mathematical induction as a property of the set of positive integers is (IMO) less well understood. The omission of the word "integers" would not render the solution worthy of less than full credit, but I think the inclusion of the word is desirable.

On 13(b), I suspect marking will not require any additional statement, and agree that it could be as simple as "since the path is parabolic and is above the floor at the end of the room, it must not have reached the floor." In general, however, asserting that a curve is above the x-axis at x = a and at x = b is insufficient to prove that it is above the x-axis for all x in [a, b].

On 14(a), I am sure that the intention of the question-writers was to have students solve the problem by vector methods. You are correct that the question specified no method and so a non-vector sine rule solution is an option, and indeed simpler. I concur that the question should have been edited to direct students to a vector solution if such was the intent.

#### notme123

##### Well-Known Member
For Q14a), I think this was just a terrible question. They should have explained to use a vector method, or they should have had defined it using vectors. The vector method for this question is inefficient (as numerous students who used this method have pointed out) and the sine rule is the only good way to do it. Regardless, NESA cannot refuse marks for alternative methods as I have emailed them about this issue and they have responded with
ye i personally used vectors and not sine rule. since the bearing of 63 didn't have a length, i left it as length r in cartsian form vector. i then cancelled out the r and had to do auxiliary angle method to get a value for theta.

#### 5uckerberg

##### Active Member
Okay this is ridiculous if the students had the old version then something seriously wrong is going on.

#### notme123

##### Well-Known Member
do we lose marks for not labelling axes? like for example the inverse fuction sketch if we dont label x and y do we lose the full mark

#### 5uckerberg

##### Active Member
do we lose marks for not labelling axes? like for example the inverse fuction sketch if we dont label x and y do we lose the full mark
Transparency NESA does not like it when you are ambiguous about which is which axis they want to give marks only when you provide evidence of the correct working and answer.

#### notme123

##### Well-Known Member
Transparency NESA does not like it when you are ambiguous about which is which axis they want to give marks only when you provide evidence of the correct working and answer.
so yes or no?