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Maths Extension 2 Predictions/Thoughts (3 Viewers)

tywebb

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well it only took 108 years for that to happen.

better late than never
 

tywebb

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i wanna know when nesa doing presentation for hsc marking.

usually it is last saturday in february of following year, and in recent years it is 9am, guthrie theatre at UTS.

ain't it sus that there were 3 questions saying or otherwise - all 3 of which are cross product?

they obviously knew that or otherwise meant cross product.

so why put or otherwise if they don't want cross product

3 TIMES FOR GOODNESS SAKE!

fair enough once in a blue moon like 1 question in 2023 ext 1

but 3 in same exam?

that's very contrived i think.
 
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BionicMango

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this question came up in the 2023 ext 1 exam, they had some vectors and asked for area of triangle spanned by the vectors.

not everyone in nsw have been in schools in nsw for 13 years. some come from overseas already well versed in linear algebra, cross products, scalar triple products, determinants, matrices etc.

for this question it comes naturally to them as half the magnitude of the cross product of the vectors, much neater solution

so in the nesa presentation for hsc marking they said the question said "or otherwise" and so would accept oos method, incuding cross product.
I remember that question, I was thinking that half the cross product is the most elegant and easy solution as well. Or at least using the determinant of a 2x2 matrix with the entries being the two vectors (which essentially is the cross product).
To answer that question I considered the rectangle and subtracted triangles, as that’s how I first learnt that the determinant of a 2x2 matrix represents the area of the unit square after being transformed. Thanks Khan Academy
 

tywebb

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there is yet a 4th application of cross product - and this also relates Q7 to Q16ciii in a rather subtle way.

7 alternative sol.png
so since it is not A, B nor C then answer is D
 

coolcat6778

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this question came up in the 2023 ext 1 exam, they had some vectors and asked for area of triangle spanned by the vectors.

not everyone in nsw have been in schools in nsw for 13 years. some come from overseas already well versed in linear algebra, cross products, scalar triple products, determinants, matrices etc.

for this question it comes naturally to them as half the magnitude of the cross product of the vectors, much neater solution

so in the nesa presentation for hsc marking they said the question said "or otherwise" and so would accept oos method, incuding cross product.
I don't get why they don't just add cross product in the syllabus. Every other state covers it, and brits do as well.

It's not going to dilute the stupid question 16s (which I can't do anyway) all it does is give us more options to solve a question
 

tywebb

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one possible criticism of my solution to question 7 is non-zero bilinear cross products are only possible in 3 and 7 dimensions so how can i do it like that.

firstly the way a and d are defined in Q16c it is clear they are already in ℝ3

secondly you just let the other components of a and d be 0 like this



this is in fact how calculators with vector functionality deal with cross products of vectors that seem to have 1 or 2 dimensions.
 
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coolcat6778

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bro what is the big deal of question 7 question? it's just a sketching question lol
1762102849493.png
1762103164187.png
I did all these questions without the use of calculations lol
 
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coolcat6778

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Process of elimination is a very good skill to develop for HSC mcq of any subject. You WILL thank me later.

Anecdotal evidence:
In '24 HSC Physics, my mcq mark was 12/20 before I developed "systematic guessing skills"
In '24 HSC MADV, my mcq mark was 7/10 before I developed "systematic guessing skills"

In '25 HSC Chem, my mcq mark was 16/20 and I literally guessed half of the questions lol
In '25 HSC MX2, my mcq mark was 8/10 and I literally guessed some

"Teach me your ways, how do I systematically guess?"
You just choose the most logical answer. Instead of doing A B C D and guessing B for no reason, think about why? Is it the odd one out?

Is it the odd one out?
This is the question you need ask yourself every time you're stuck on an apparently impossible question. This is your next best chance of scoring it right. Deductive reasoning doesn't even have to involve pens down, it can also just involve pattern recognition like the following two examples.

Example 1
Just look at this question (don't worry you don't require chem knowledge to answer it like I did)
Bla bla bla, get to the fucking point. Don't bother reading a single line. Go straight to the options.
1762105601815.png
Notice how straight off the bat, B is the odd one out; followed by C; finally followed by Z? This gives D as the final answer
(which was correct)


Example 2
Bla bla bla keep fucking yapping. (I didn't read a single word of the question in the exam)
1762106794094.png
A a rectangle is a parallelogram, which is like a square. Odd one out is a fucking rhombus (which was correct)
 
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Luca26

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Example 2
Bla bla bla keep fucking yapping. (I didn't read a single word of the question in the exam)
View attachment 50383
A a rectangle is a parallelogram, which is like a square. Odd one out is a fucking rhombus (which was correct)
Re example 2, a rectangle is a parallelogram with a 90 degree angle or equal diagonals

A rhombus is a parallelogram with equal sides or perpendicular diagonals

A square is a rhombus that is also a rectangle, or a rectangle that is also a rhombus

I don't see why rhombus is an "odd one out" option here
 

BionicMango

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bro what is the big deal of question 7 question? it's just a sketching question lol
View attachment 50376
View attachment 50377
I did all these questions without the use of calculations lol
How did you do q7 without sketching? I feel like I wouldn’t trust myself with sketches for questions like q7 tbh
I drew a triangle with each side 3,5,7 then cosine ruled it
 

BionicMango

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Taking advantage of multiple choice is definitely worth it, but I don’t get the rhombus one, since they are all parallelograms? I drew one, and noticed that the diagonals are perpendicular but unequal (the i and the k respectively) = rhombus

edit: q9 not q7
 

BionicMango

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Actually for question 16 what I did was prove the shortest distance was perpendicular FIRST by drawing a right triangle and saying that the legs are always shorter than the hypotenuse (you can prove it using Pythagoras’ theorem), thus perpendicular lines represent the smallest distance.
I then used that idea in part a and saved a whole lot of time.
 

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