2 unit paper (2 Viewers)

[pikachu975]

Honestly, with this year's exam there wasn't any unexpected or 'I have never seen this before' questions if you did a lot of past papers. However, in comparison with other years, there weren't a lot of 'free mark' questions as a result in my opinion. MC, Q11, Q12 and Q13 were really straight forward and Q14, Q15 and Q16 were rather disappointing although there is always those questions that may 'overconfident' you. Based off each past HSC exam, it seems BOSTES has dimmed down the difficulty for 2U when comparing with the older papers particularly without the MC. I believe if you did do the drill, understood the content and did all past trial and hsc papers, you would've been rewarded. I am predicting a similar Band 6 cutoff roughly around 80 due to this. Otherwise, what were the questions you found the hardest?

Hardest was probability and the yabby one.

 

[photastic]

If around 80 is B6, what would you expect an 88 to be?

Roughly between 90 and 93 depending how the state went as well.

 

[photastic]

Be sure to have a look at rawmarks.info predicting a similar alignment to 2014's exam where raw marks beyond 95 did not have any additional marks added. Ofc the lower the raw mark the more alignment.

 

[mrstripedshades]

lol... I forgot to flip the inequality for the probability N>3/4 question.

 

[PC]

... and here are some worked solutions. The usual disclaimer applies. It would great if you could point out any errors.

 

[pikachu975]

... and here are some worked solutions. The usual disclaimer applies. It would great if you could point out any errors.

Looks all correct. Do you reckon we'll lose a mark in last part of yabbies for not showing it's a maximum considering it's only 2 marks? Maybe 1 for differentiating and 1 for solving?

 

[photastic]

... and here are some worked solutions. The usual disclaimer applies. It would great if you could point out any errors.

For Q 14 (d) since this is a 2U paper, although it's not incorrect, factorising with a power of 5 i've never seen for 2U. What should've been done was using the sum of the geometric series for 1 + x + x^2 + x^3 + x^4 which is equal to (x^5 - 1) / (x - 1). So in this case replace (x^5 - 1) / (x - 1) with 1 + x + x^2 + x^3 + x^4 for the limit.

 

[pikachu975]

For Q 14 (d) since this is a 2U paper, although it's not incorrect, factorising with a power of 5 i've never seen for 2U. What should've been done was using the sum of the geometric series for 1 + x + x^2 + x^3 + x^4 which is equal to (x^5 - 1) / (x - 1). So in this case replace (x^5 - 1) / (x - 1) with 1 + x + x^2 + x^3 + x^4 for the limit.

Oh wow was it that easy. What I did was changed the sum to 1 + x (x^4 - 1 / x - 1) and then subbed in 1 at the end after simplifying to get 5.

 

[candyman]

for 16 b ii if i put 10 < y < 200 would i be able to get one mark?

 

[henryg124]

quick question regarding 14 e .. if you put say 9log32 instead of 45log2, would you still get full marks for the question? it specifies a and b simply as positive integers

 

[henryg124]

... and here are some worked solutions. The usual disclaimer applies. It would great if you could point out any errors.

your working for 16 b iv) R'(y) i found hilarious, using quotient rule when you could have got in 3 less lines by just expanding the bracket . i did same method though

 

[InteGrand]

your working for 16 b iv) R'(y) i found hilarious, using quotient rule when you could have got in 3 less lines by just expanding the bracket . i did same method though

You could also do the Q in about one line by noting that a concave down parabola is maximised halfway between the roots.

 

[photastic]

Oh wow was it that easy. What I did was changed the sum to 1 + x (x^4 - 1 / x - 1) and then subbed in 1 at the end after simplifying to get 5.

Well it did say to sum the geometric series so pretty much a giveaway with what to do.

 

[PC]

For Q 14 (d) since this is a 2U paper, although it's not incorrect, factorising with a power of 5 i've never seen for 2U. What should've been done was using the sum of the geometric series for 1 + x + x^2 + x^3 + x^4 which is equal to (x^5 - 1) / (x - 1). So in this case replace (x^5 - 1) / (x - 1) with 1 + x + x^2 + x^3 + x^4 for the limit.

Yeah, I did it that way too, but it wasn't as much fun.

By the way, the method of expanding xⁿ - cⁿ IS part of the 2 unit course. It's part of the introduction to calculus and is usually done with limits and differentiating from first principles. Questions on first principles are rare, and so is this expansion I guess. It would have been interesting if the GP hint wasn't there.

 

[adibimam]

Fairly sure its n=11 because if you sub in n=10 it was like 74.7%

Regarding 16. a) iv), I tried to take the absolute value of two different integrals. is this right? https://postimg.org/image/dcm8be3ad/

 

[henryg124]

Regarding 16. a) iv), I tried to take the absolute value of two different integrals. is this right? https://postimg.org/image/dcm8be3ad/

Because its dealing with distance, you only need to take the absolute value of any integral with a negative value. this is becuse if you evaluate the enitre integral it will automatically add the negative value of the integral from 0-1. so abs from 0-1 and normal from 1-7 should give you the right answer of 10 -4ln2

 

[InteGrand]

Because its dealing with distance, you only need to take the absolute value of any integral with a negative value. this is becuse if you evaluate the enitre integral it will automatically add the negative value of the integral from 0-1. so abs from 0-1 and normal from 1-7 should give you the right answer of 10 -4ln2

Whilst we don't need absolute values on the second integral (as the integrand is non-negative over that interval), putting the absolute values isn't incorrect (just redundant), and so what was done is still correct (and the final answer came out correct too, i.e. didn't make a silly mistake).

 

[henryg124]

^yes, redundant is what i meant

 

[kashkow]

How do you do q9?

you draw the graph then calculate the areas of both the trianlges?

Oh LOL. I split them up and tried integrating both triangles. Haha.

am i the only one who didn't finish the test....

Well I thought I finished the test but then I looked at 16. a) iv) and realised I might've skipped that question entirely :O. Can't remember doing it

 

[photastic]

Alternative to Q16 (b) (iii) which is the intended way I believe.

TYPO: Should read part (i)