HSC Extension 1 Mathematics Predictions / Thoughts (1 Viewer)

Trebla

For the last part of 14a I wrote the answer at the very start then spent probably 15 minutes trying to simplify it and equate it with part iii cause I thought thats what it wanted, but would I still get the marks for that.
The question said "By considering the reversed process and using part (ii)" so I'm not sure you will get full marks without doing that tbh

david.h9

New Member
looks like i only got the f(x)=g(x) question wrong, didn't test in separate regions

Kangaaroo

Member
is that question wrong? I swear it was P(55<x<65) not P(55<x<60) so the z score should be -1<z<1 which is 68%
I got this also

coop_163

New Member
What would a raw 47/70 align to this year do we think?

4u Boy

Member
What would a raw 47/70 align to this year do we think?
probably mid 80

Althacommie

New Member
Riuace seems to be the first to do complete solutions. He beat Terry Lee.

The format is a bit strange. Big long pictures. You'll have to zoom in.

Maybe Terry Lee still doesn't have Q6 and 7.
For q6. Is it A? I thought it was D???

shashysha

Well-Known Member
For q6. Is it A? I thought it was D???
yeah if its the a-b vector one then its D

Fabrizio

Active Member
For Q13 c). Is just proving the equations are equal with pythag enough to get full marks? I dont see why you would have to prove any points if you show they are equal

Xanthi

Active Member
For Q13 c). Is just proving the equations are equal with pythag enough to get full marks? I dont see why you would have to prove any points if you show they are equal
If you used proved across the entire domain, yes. Otherwise you risk losing substantial number of points.

New Member
One wonders why they even changed the syllabus?

They moved a few things from 4U into 3U, and moved a few things from first year university into 3U. But they pick the most trivial things to move, and they give away free marks in the exam.

Trebla

For Q13 c). Is just proving the equations are equal with pythag enough to get full marks? I dont see why you would have to prove any points if you show they are equal
It depends on whether you address the entire range of cos-1x or only the acute angle part.

tokyocalling

New Member
did anyone actually get more than 50% of Q14?
yea, i cross checked my ans with the ones posted by ngo and sons and im pretty sure i got all 15 marks

tokyocalling

New Member
smells like a state rank to me
ahaha nah, 13c ahaha im currently 15 rank at school, my schools top 5 i just hope thats enough to get an overall mark (after scaling and all that) of 98-99

tokyocalling

New Member
Hopefully just the 1 lol... anymore than that and I'll be dead xD
it'll probs be 1 and as long as the rest is correct it'd be carry on!

tokyocalling

New Member
yes! i'd be stoked ahaha

thatmailman

Member
I didnt see 14(b) until 4 minutes left and only got out part (i) and (ii), so lost 3 marks straight away. Can't even get into my course cause of that BRUH

Xanthi

Active Member
ahaha nah, 13c ahaha im currently 15 rank at school, my schools top 5 i just hope thats enough to get an overall mark (after scaling and all that) of 98-99
Should be sufficient, a 70/70 score + approx 64-67 score for internals will probs align to 98-99. So well done!

s97127

Member
its such bullshit. only reason i got the binomial question was because i did pretty much the same question in the day before otherwise i wouldn't have gotten it. They could of included the hardest questions of content from year 12 ex1 and i would of found it way easier then that question
may i ask where you got that question from? Thanks

tokyocalling

New Member
its such bullshit. only reason i got the binomial question was because i did pretty much the same question in the day before otherwise i wouldn't have gotten it. They could of included the hardest questions of content from year 12 ex1 and i would of found it way easier then that question
right! i was never expecting an 8 mark binomial question, came as a true shock during reading time. I'm not sure about yall but if there was one thing my school teachers and tutor emphasised it was to revise binomial theorem and perms and combs as its probably the hardest part of the stage 6 syllabus so that definitely helped.

Trebla

A remark on Q13c)ii).

It had a nice trap where you had to recognise that f'(x) and g'(x) are undefined at the points (1,0) and (-1,0) but these are points where the f(x) and g(x) are defined.

So technically speaking you can only say that since their derivatives are

$\bg_white -\frac{1}{x^2\sqrt{1-x^2}}$

then

$\bg_white f'(x)=g'(x)\quad\text{ for }0

Therefore you can only really claim that

$\bg_white f(x)=g(x)+c\quad\text{ for }0

This would suggest that simply substituting in say x = 1 to find c has some rigour issues as it doesn't quite fit that domain (even though it works out).

I think the intended solution was for students to recognise that since

$\bg_white f(-1) = g(-1) = 0$
$\bg_white f(1) = g(1) = 0$

then f(x) and g(x) share common points at x = -1 and x = 1.

From part (i), they have the same gradient behaviour in the domain -1 < x < 0 but have a common point at x = -1 on the "edge" of that domain. This means that

$\bg_white f(x) = g(x)\quad\text{ for }-1 \leq x < 0$

Similarly, they have the same gradient behaviour in the domain 0 < x < 1 but have a common point at x = 1 on the "edge" of that domain. This means that

$\bg_white f(x) = g(x)\quad\text{ for }0 < x \leq 1$