# 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

##### lukewarm mess
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

##### lukewarm mess
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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