2018 HSC Maths MX1 Discussion (1 Viewer)

HoldingOn

Active Member
What would 57 align to? Would it be enough for E4?

sourish

Member
What would 57 align to? Would it be enough for E4?
Easily a E4, probabbly to 94 or 95

jjuunnee

Active Member
What would 57 align to? Would it be enough for E4?
idk what it aligns to but it would definitely be e4

Ace0804

Member
you guys think 62 will scale to 94-95

Ace0804

Member
tbh i don't know why everyone was saying question 14 was hard, the only hard part was the limiting sum question, everything else was pretty standard

Lolihentai

New Member
suck my dick

• jjuunnee

phunkyy

New Member
ughhh I forgot to change my calculator to RAD mode for the ferris wheel question, and also didn't see that the domain of the function was x>1!! Sighhh silly mistakes

• pikachu975 and supR

toxicdonut

New Member
does anyone have the paper? I want to see the total marks with solutions

1729

Active Member
looking at 68 or 69 out of 70. is it possible for a 69/70 to align to a 100?

for 14c)iii) did we have to prove that |r| < 1 or are we allowed to assume this since we are given in the wording of the question that a limiting sum exists? all i did was find x and x1 (using similar triangles), said r = x1/x and then substitued into a/(1-r)

and for 14b)ii) did we have to write the actual number or could we just leave it as 23C4 * 2^(19)

• boredofstudiesuser1

supR

Active Member
looking at 68 or 69 out of 70. is it possible for a 69/70 to align to a 100?

for 14c)iii) did we have to prove that |r| < 1 or are we allowed to assume this since we are given in the wording of the question that a limiting sum exists? all i did was find x and x1 (using similar triangles), said r = x1/x and then substitued into a/(1-r)

and for 14b)ii) did we have to write the actual number or could we just leave it as 23C4 * 2^(19)
For c iii), I reckon you had to prove x, x1 and x2 before finding the ratio, like how do you know it's a gp otherwise ~ Not sure tho

for b ii), maayyybe (I did) but surely the marks is for understanding selector A and B's role, and then using part i, not calculator shit

1729

Active Member
For c iii), I reckon you had to prove x, x1 and x2 before finding the ratio, like how do you know it's a gp otherwise ~ Not sure tho

for b ii), maayyybe (I did) but surely the marks is for understanding selector A and B's role, and then using part i, not calculator shit
you know its a GP because the question said that a limiting sum exists, and limiting sums only exists for GPs with |r|<1

so it can be assumed that x, x1, x2, ... form a GP with |r|<1 otherwise a limiting sum wouldnt exist

i think if they wanted x2 they wouldve shown x2 in the diagram, but they didnt. also this wouldve taken ages since you wouldve had to prove similar triangles again

• supR

InteGrand

Well-Known Member
looking at 68 or 69 out of 70. is it possible for a 69/70 to align to a 100?

for 14c)iii) did we have to prove that |r| < 1 or are we allowed to assume this since we are given in the wording of the question that a limiting sum exists? all i did was find x and x1 (using similar triangles), said r = x1/x and then substitued into a/(1-r)

and for 14b)ii) did we have to write the actual number or could we just leave it as 23C4 * 2^(19)
I would be quite surprised if you had to write out the actual number for 14 b) ii).

• 1729

1729

Active Member
I would be quite surprised if you had to write out the actual number for 14 b) ii).
yeah I think it was like 4 billion something idk, i just saw 2^(19) and automatically thought it would be too big so i left the answer as 29C4 * 2^(19)

jjuunnee

Active Member
ughhh I forgot to change my calculator to RAD mode for the ferris wheel question, and also didn't see that the domain of the function was x>1!! Sighhh silly mistakes
oh shit i think i forgot to change it to radian mode as well, but i don't recall having to put down anything for theta i just replaced the whole trig function with something. oh well whatever i probably screwed that up anyway

InteGrand

Well-Known Member
you know its a GP because the question said that a limiting sum exists, and limiting sums only exists for GPs with |r|<1

so it can be assumed that x, x1, x2, ... form a GP with |r|<1 otherwise a limiting sum wouldnt exist
I haven't seen the question wording, but did it say it's a GP? Non-GP sums can also have limiting sums.

Without having seen the question, I think that if you know (either via proof or by being given it) that it's a GP, and the question tells you the sum converges, then you shouldn't need to waste time proving |r| < 1.

• 1729

1729

Active Member
I haven't seen the question wording, but did it say it's a GP? Non-GP sums can also have limiting sums.

Without having seen the question, I think that if you know (either via proof or by being given it) that it's a GP, and the question tells you the sum converges, then you shouldn't need to waste time proving |r| < 1.
the question said something like 'show that the limiting sum of the areas of the quadrants is X' so you already know that the sum converges to X and as far as 3u maths goes only geometric sums converge

Ace0804

Member
oh shit i think i forgot to change it to radian mode as well, but i don't recall having to put down anything for theta i just replaced the whole trig function with something. oh well whatever i probably screwed that up anyway
na u got that right, cause thats the exact same thing i did, but the answer should be 19. something

jjuunnee

Active Member
na u got that right, cause thats the exact same thing i did, but the answer should be 19. something
oo yay i got 19. something as well! hopefully the only marks i lose in this exam are from the questions I couldnt do