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2018 HSC Maths MX1 Discussion (1 Viewer)

Ace0804

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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
 

phunkyy

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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
 

toxicdonut

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

1729

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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)
 

supR

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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

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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
 

InteGrand

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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

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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

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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

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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

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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

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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

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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
 

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