Mathematics Extension 1 HSC thoughts (1 Viewer)

[tgone]

About q9, I felt (A) was the right answer but I heard my friends arguing it was (D), could you consider why?

consider the example y=-x^3

inverse function is y=(-x)^1/3

They intersect at three places, but only one of them is along y=x.

 

[ExtremelyBoredUser]

consider the example y=-x^3

inverse function is y=(-x)^1/3

They intersect at three places, but only one of them is along y=x.

Ah shit yea that makes sense

 

[tgone]

oh you know what. D is correct

consider f(x) = 1/x
f-1(x) is 1/x
they intersect everywhere, not only on y=x

answer is D

thanks for that ill change it and upload new ones later

this example is not correct, see the one i posted before:

the question specified both the functions were differentiable and real across all x and y, which hyperbolas are not.

 

[tgone]

Ah shit yea that makes sense

although, i think if f(x) is strictly increasing then (A) is correct.

 

[Trebla]

I think UAC uses only the HSC aligned mark and then scales that based on their own thing, which is skewed non-linearly from the very high marks (which is what adjusts it so that a student who gets 98-->100 is rewarded over the 67-->90), which is what counts to your aggregrate:

I'm not sure nesa even gives UAC the raw marks? Not sure of this though, don't quote me on that

View attachment 36772

Scaling is applied on raw marks. See below for a high level summary of the process:

Flowchart of how HSC marks and ATAR are determined

Hi all, We have received a lot of questions on the aspects of the HSC and ATAR calculation process ranging from scaling, moderating and aligning. It has become apparent that some confusion exists on how the process works overall. This is important to understand before diving into the finer...

boredofstudies.org

 

[ExtremelyBoredUser]

although, i think if f(x) is strictly increasing then (A) is correct.

Nah nah pre sure D is right because it doesnt specify it

 

[tgone]

Nah nah pre sure D is right because it doesnt specify it

yeah D is right, but if it did it would be A haha

 

[Run hard@thehsc]

Wait so scaling obs applies on the corresponding raw of your average HSC mark right?

 

[tgone]

Scaling is applied on raw marks. See below for a high level summary of the process:

Flowchart of how HSC marks and ATAR are determined

Hi all, We have received a lot of questions on the aspects of the HSC and ATAR calculation process ranging from scaling, moderating and aligning. It has become apparent that some confusion exists on how the process works overall. This is important to understand before diving into the finer...

boredofstudies.org

thanks, that's very helpful!

 

[yanujw]

oh you know what. D is correct

consider f(x) = 1/x
f-1(x) is 1/x
they intersect everywhere, not only on y=x

answer is D

thanks for that ill change it and upload new ones later

Just another change to bring to your attention is that 6 is (C) because the projection onto that vector must have equal i and j components.

Also, 8 is (A) by considering an isosceles triangle with unit length and requiring the joining side to have length to be less than 1, which can occur up to the point that the triangle is equilateral, therefore less than pi/3 angle within the vectors. Nvm you were correct, welp just discovered another question I got wrong lol.

 

[011235]

Two thoughts on notme's answers;

• I'm pretty sure that 13c is false. It says is g the inverse of f, NOT the inverse function of f. Isn't the inverse simply the reflection in y=x? This would mean that arcsin(x) is not the inverse as it doesn't cover the full range of the inverse.
• For 14d, isn't the number of no shows dependent on the number of tickets that are sold? n isn't 350 (the number of seats), it's the number of tickets sold, as each ticketholder can be considered a bernoulli trial of whether they will show up or not. (Is how I read the question)

I could be completely wrong though

(btw, thanks notme for the very fast answers )

 

[Daedalus13]

Nah nah pre sure D is right because it doesnt specify it

If you consider f(x)= x^3 and its inverse, the intersection at the origin has the tangents to the graph be perpendicular. So D is also incorrect?

Edit: I realised that a vertical tangent means it f'(x) is not defined for the inverse for all real x.

 

[notme123]

If you consider f(x)= x^3 and its inverse, the intersection at the origin has the tangents to the graph be perpendicular. So D is also incorrect?

inverse of x^3 does not have a defined derivative at x = 0

 

[100orbust]

Just another change to bring to your attention is that 6 is (C) because the projection onto that vector must have equal i and j components.

Also, 8 is (A) by considering an isosceles triangle with unit length and requiring the joining side to have length to be less than 1, which can occur up to the point that the triangle is equilateral, therefore less than pi/3 angle within the vectors. Nvm you were correct, welp just discovered another question I got wrong lol.

how would you reason it isn't B by that judgement? mid-exam i intuitively ruled off C because the vector u was in the opposite direction, thus there could be no component acting in the same way (which is what projection vectors are)

 

[Daedalus13]

how would you reason it isn't B by that judgement? mid-exam i intuitively ruled off C because the vector u was in the opposite direction, thus there could be no component acting in the same way (which is what projection vectors are)

Also the diagram was 'DRAWN TO SCALE', so 3.2 is much too large.

 

[mmmmmmmmaaaaaaa]

HOw did people do 14b? the vector proof

 

[tgone]

Two thoughts on notme's answers;

• I'm pretty sure that 13c is false. It says is g the inverse of f, NOT the inverse function of f. Isn't the inverse simply the reflection in y=x? This would mean that arcsin(x) is not the inverse as it doesn't cover the full range of the inverse.
• For 14d, isn't the number of no shows dependent on the number of tickets that are sold? n isn't 350 (the number of seats), it's the number of tickets sold, as each ticketholder can be considered a bernoulli trial of whether they will show up or not. (Is how I read the question)

I could be completely wrong though

Your comment on 13c, this was what I wrote as well. Something like g(x)=arcsinx is defined only for x [-1,1], but f(x)=sinx is defined for x across the reals. They even specified x across all real for f(x)=sinx, which to me made it seem clear they wanted you to conclude false due to domain/range non-equivalence.

 

[notme123]

Just another change to bring to your attention is that 6 is (C) because the projection onto that vector must have equal i and j components.

you're right lol that completely escaped me.

I'm doing more correct ones now sorry for the inconvenience guys

 

[tgone]

HOw did people do 14b? the vector proof

do the regular projectile motion things to find time of flight (let y=0 etc), solve simultaneously for x and rearrange, you're left with:



Then, optimise the function on the LHS and you get the maximum value of is equal to 0.369..., so then all other d must be below

 

[Daedalus13]

you're right lol that completely escaped me.

I'm doing more correct ones now sorry for the inconvenience guys

I believe 6 C is incorrect, it should be B (see above).

Also the diagram was 'DRAWN TO SCALE', so 3.2 is much too large.