2002 paper (1 Viewer)

[paper cup]

Um, I don't have solutions to this paper and have basically worked from the markers notes.
A couple of problems....if anyone could help it would be appreciated, in form of rep points or whatever else (shut up Rorix)
Why is it that for Q3c) x = rt two cos (4t – pi/4), not +? And how do you determine whether to subtract or add the angle?
Q6 ii) How do you twiddle with the result to get the required equation?
Q7b) I have substituted x = 1 into the expansion of (1 + x)n like the markers said to but it still doesn't work...
cb thanks you in advance.

 

[ToO LaZy ^*]

q3c) if you're dealing with cos, it will always be the opposite sign used in the question
eg cos@ + rt3sin@ = 2cos(@ - x)
if sin, the signs are the same

 

[Rorix]

i was going to answer your questions but instead ill shut up as instructed

 

[einsteinthe2nd]

http://members.swiftdsl.com.au/~einsteinthe2nd/2002_3unitsol_page1.JPG
http://members.swiftdsl.com.au/~einsteinthe2nd/2002_3unitsol_page2.JPG
http://members.swiftdsl.com.au/~einsteinthe2nd/2002_3unitsol_page3.JPG
http://members.swiftdsl.com.au/~einsteinthe2nd/2002_3unitsol_page4.JPG
http://members.swiftdsl.com.au/~einsteinthe2nd/2002_3unitsol_page5.bmp
http://members.swiftdsl.com.au/~einsteinthe2nd/2002_3unitsol_page6.bmp
http://members.swiftdsl.com.au/~einsteinthe2nd/2002_3unitsol_page7.bmp

Enjoy!

 

[mojako]

Q4(c)(iii):
x=acos(nt+@)
v=-ansin(nt+@)
put "t=0 and x=1" and "t=0 and v=4" into the two expressions and you know the quadrant of @

Q6(iii)
1/(n+1) < [lnx]_nⁿ⁺¹ < 1/n
1/(n+1) < ln([n+1]/n) < 1/n
Noting that [n+1]/n = 1 + 1/n, and considering the two inequalities separately,
Inequality#1:
1/(n+1) < ln(1 + 1/n)
e^1/n+1 < 1 + 1/n
e < (1 + 1/n)ⁿ⁺¹

Inequality#2:
1/n > ln(1 + 1/n)
e^1/n > (1 + 1/n)
e > (1 + 1/n)ⁿ

Then you can combine them.

Q7(ii)
Differentiate both sides then put x=1.
Then you can see you need extra
"c0 + c1 + c2 + ... + cn"
which are the coefficients of (1+x)ⁿ
and adds to up 2ⁿ

So you add
n(2)^n-1 + 2ⁿ
and it gives you the result.

For part (ii), integrate two times (don't forget the constant)
then put x=-1 into the second integral expression

 

[Rorix]

Why is it that for Q3c) x = rt two cos (4t – pi/4), not +? And how do you determine whether to subtract or add the angle?
Q6 ii) How do you twiddle with the result to get the required equation?
Q7b) I have substituted x = 1 into the expansion of (1 + x)n like the markers said to but it still doesn't work...
cb thanks you in advance.

3c is growth/decay?

6ii), integrate, 1/n+1 < ln (1 + 1/n) < 1/n
recriprocal, n+1> 1/ln(..) > n
but 1/ln(..) = log (base 1 + 1/n) e (change of base)
then,
(1+ 1/n) ^ n+1 > e > (1+ 1/n) ^n

7) you need to integrate twice and evaluate constants first

 

[Rorix]

goddamn speed typers

 

[mojako]

goddamn speed typers

well u used a different approach for the inequality


uh that BIG picture is really annoying!!
and it doesnt answer any question.
EDIT: the picture is not replaced by links

 

[Rorix]

oh, it's probably because i was doing this bitchy 2unit question yesterday involving a lot of base changing.


changing laaaaanesssss

 

[mojako]

oh, I meant its good to see a different approach
not to criticise or to say that mine is more normal

 

[Rorix]

the last guy that said "its good to see a different approach" to me...


well...........we had a bit of a misunderstanding

 

[paper cup]

i was going to answer your questions but instead ill shut up as instructed

oh you're so mean I don't want you any more.
but for answering my questions I will refrain from making sarcastic comments in news and current affairs
Thank you mojako..
and einstein I love you thank you so much, I will make it up to you...somehow...
EDIT: oh my god I love you I love you I love you I love you thank you so much.
I'm not going to get over this for a while now...