Mathematics Extension 2 - 2016 Post-HSC Exam Thoughts (1 Viewer)

123ryoma12

Member
how much do you think they'll scale this year?

porcupinetree

not actually a porcupine
Anyone know how to do q13d both
Has it got to do with discriminatnts
Just from quickly glancing at it (ie i could be wrong)

(i) differentiate and use the discriminant
(ii) 3

InteGrand

Well-Known Member
And 14aii
14 a(ii): cos(x) is odd about x = pi/2, so raising cos(x) to an odd power (or more generally, applying any odd function to it) will maintain this property.

In other words, (cos(x))^(2n-1) is also odd about pi/2 for any positive integer n, so integrates to 0 over 0 to pi (integrating a function over an interval where it's odd about the midpoint is 0, since the area above the x-axis is cancelled out symmetrically by that below).

$\bg_white \noindent By a function f being odd about a', I mean that f(a+x) = -f(a-x) for all x. So if \phi is an odd function (i.e. just a regular odd function, so odd about 0') and f is odd about a, then \phi\left(f(a+x)\right) = \phi \left(-f(a-x)\right) = -\phi \left(f(a-x)\right) (since \phi is odd), which implies that \phi \circ f is odd about a.$

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calamebe

Active Member
wtb a 49 lmao
Well I'm maybe not the best person to ask, but maybe mid-high 70s.

rulerpenkey

New Member
what raw do people think a state rank would need?

calamebe

Active Member
what raw do people think a state rank would need?
I'm guessing high 90's or something around that.

hedgehog_7

Member
Answer to q 10? 12 d ii), 13 c ii)?

wat what 8

InteGrand

Well-Known Member
$\bg_white \noindent Yeah, 10 is (B). Note that if we let x = \mathrm{cis}\left(\theta\right) (by symmetry it suffices to let 0 \leq \theta \leq \pi), we obtain 2\cos \theta = -1 \Rightarrow \cos \theta = - \frac{1}{2}. So \theta = \frac{2\pi}{3}. It follows that x^{2016} + \frac{1}{x^{2016}} = 2 \cos \left(2016\times \frac{2\pi}{3}\right) = 2 \cos \left(\text{whole number}\times 2\pi\right) = 2 \times 1 = 2 (noting that 2016 is divisible by 3 since its digit sum is a multiple of 3).$

calamebe

Active Member
Answer to q 10? 12 d ii), 13 c ii)?
For 12 d ii), I just subbed in y = c^2/x and rearranged to get a quadratic, then used the product of roots. 13 will take too long to type out, but just set T2 greater than T1 and solve for w.

petermaniah198

New Member
what mark do u think 84 raw will align to in terms of HSC mark?

KingOfActing

Well-Known Member
Did anyone do the cubic question with the inequality like this, or was that just me?

$\bg_white b^2 - 3ac < 0 \iff \left(\frac{b}{a}\right)^2 - 3 \frac{c}{a} < 0\\(\alpha+\beta+\gamma)^2 - 3(\alpha\beta + \alpha\gamma+\beta\gamma) < 0\\\frac12(\alpha-\beta)^2 + \frac12(\alpha-\gamma)^2 + \frac12(\beta-\gamma)^2 < 0$

And then argue from that about what that implies about the roots?

what mark do u think 84 raw will align to in terms of HSC mark?
~94

Glyde

Member
Did anyone do the cubic question with the inequality like this, or was that just me?

$\bg_white b^2 - 3ac < 0 \iff \left(\frac{b}{a}\right)^2 - 3 \frac{c}{a} < 0\\(\alpha+\beta+\gamma)^2 - 3(\alpha\beta + \alpha\gamma+\beta\gamma) < 0\\\frac12(\alpha-\beta)^2 + \frac12(\alpha-\gamma)^2 + \frac12(\beta-\gamma)^2 < 0$

And then argue from that about what that implies about the roots?

~94
you have a special brain

KingOfActing

Well-Known Member
you have a special brain
ahahah, I was about to use the derivative but then it said "If [inequality] then " so I just assumed the inequality and went from there

calamebe

Active Member
Yeah I just showed that the discriminant was equal to zero, and hence that was a double root of the derivative, and thus as it is also a root of the function, it must be a triple root.

Glyde

Member
i thought it said 'prove it is a double root', so i showed that the derivative could equal zero. will i get some marks?

InteGrand

Well-Known Member
$\bg_white \noindent For 13(d), p(x) = ax^{3} + bx^2 + cx + d, p'(x) = 3ax^2 + 2bx + c. Roots of p' are \frac{-2b\pm \sqrt{4b^2 - 4 \times 3ac}}{2\times 3a} = \frac{-b \pm \sqrt{b^2 - 3ac}}{3a}. So if b^{2} -3ac = 0 and p\left(x_{0}\right) = 0, where x_{0} = - \frac{b}{3a}, we have that x_{0} is a double root of p' and a root of p. Hence x_{0} is a triple root of p, following from the following general fact:$

$\bg_white \noindent \textbf{Fact.} Let P(x) be a polynomial of degree n \geq 2. Suppose x_{0} is a root of multiplicity m of the derivative P' and a root of P. Then x_{0} is a root of multiplicity m+1 of P.$

$\bg_white \noindent This result follows from if a is a root of multiplicity \mu of P, then a is a root of multiplicity \mu -1 of P''' as follows. Since x_{0} is a root of P, it has some multiplicity \mu \geq 1. Then x_{0} is a root of multiplicity \mu -1 of P'. But we are given it is a root of multiplicity m of P'. So \mu -1 = m \Rightarrow \mu = m+1, proving the claim.$

Edit: realised calamebe said this above.

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