SOLUTIONS - Terry Lee (1 Viewer)

seanieg89

Well-Known Member
Well done on typing solutions Carrot . I have just a couple of technical issues with the last two qn's that should not matter to most people:

15biii) How do we know that the real root must have multiplicity two?

iv) Should probably consider the case where alpha is real, (although as we have seen, this means that alpha either has a double root at 1 or -1, so does not change the outcome of the calculation).

16) iii) I dont think your argument is valid...perhaps I am misinterpreting what you are saying.

And finally I am unsure if more justification is required for the last one, why does being larger than your two immediate neighbours imply that you are larger than all other P(k)?

Just nitpicking, I don't know how anal BoS are actually going to be.

seanieg89

Well-Known Member
The reason I stress the cases in iv) is that it is very wrong to think that if alpha and beta are two distinct complex roots of a real quartic, then the roots are alpha, beta, conjugate of alpha, conjugate of beta. This is only necessarily the case if alpha and beta are NOT real. The reason why this question still works out okay is that the symmetry of coefficients forces a pretty rigid structure on any real roots that P can have.

traiwit

Member
Yes

I would think to get into a band 6 is around a raw mark of 70ish
Do u mean e4 or really band6 ?

Carrotsticks

Retired
Thanks Sean, going to start addressing them now. A cup of coffee does wonders.

Hop on FB perhaps, much easier that way.

Life4Never

New Member
Do u mean e4 or really band6 ?
Opps The extension system slipped in my mind for that moment. E4 is what I meant which is 90% anyways. Same as a band 6

Zeroes

Member
Damn only 62%
Me too haha, but that was actually more than I was expecting

OH WELL come at me 3u

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traiwit

Member
Me too haha, but that was actually more than I was expecting
Hahaha yeah, btw will we get E4 ?

Zeroes

Member
I'm hoping my assessment mark will pull me up to an E4 but HSC mark may be just below. I think cutoff is around 65% going from raw marks database.

Nws m8

Banned
59% .... is that band e4?

Zeroes

Member
^doubt it, but fairly close I'd think

brianphamm

Member
Hey Carrot, are you sure the height of the rectangle is bounded to x^2 + y^2 = r^2? Since it's on an angle, would it be different? (question is 14c)

seanieg89

Well-Known Member
Hey Carrot, are you sure the height of the rectangle is bounded to x^2 + y^2 = r^2? Since it's on an angle, would it be different? (question is 14c)
Carrot is correct.

Life4Never

New Member
Hey Carrot, are you sure the height of the rectangle is bounded to x^2 + y^2 = r^2? Since it's on an angle, would it be different? (question is 14c)
Thats correct as the height of the rectangle is the y value of the circle at each value of x.

dx is so small there is no difference if its "sloped" as exaggerated in the diagram if thats what you are trying to say.

lolcakes52

Member
looking at a 70-80, maybe a bit over 80. With 14a, I split it into 3x^2+4/x(x^2+4) + 4/x(x^2+4). Im not sure if that was valid but I thought I got the answer in the exam.

seanieg89

Well-Known Member
Last two lines:

P(k) is greatest when k is the integer closest to $\bg_white \sqrt{n}+1/2$ so P(k) is greatest when k is the integer closest to $\bg_white \sqrt{n}$

Nice try terry.

Nws m8

Banned
Last two lines:

P(k) is greatest when k is the integer closest to $\bg_white \sqrt{n}+1/2$ so P(k) is greatest when k is the integer closest to $\bg_white \sqrt{n}$

Nice try terry.
is he wrong?

tomp1612

Member
would 53% get an e3??