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

123ryoma12 · Oct 21, 2016

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

porcupinetree · Oct 21, 2016

petermaniah198 said:
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 · Oct 21, 2016

petermaniah198 said:
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).

$\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 $a$.$

Carrotsticks · Oct 21, 2016

Uploaded only Q15 for now until I return home and complete the rest.

http://community.boredofstudies.org...ticks-2016-hsc-mx2-solutions.html#post7206077

Apologies for disappointing all this year. But this is an event I cannot miss!

calamebe · Oct 21, 2016

Glyde said:
wtb a 49 lmao

Well I'm maybe not the best person to ask, but maybe mid-high 70s.

rulerpenkey · Oct 21, 2016

what raw do people think a state rank would need?

calamebe · Oct 21, 2016

rulerpenkey said:
what raw do people think a state rank would need?

I'm guessing high 90's or something around that.

hedgehog_7 · Oct 21, 2016

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

severus snake · Oct 21, 2016

hedgehog_7 said:
Answer to q 10? 12 d ii), 13 c ii)?

Question 10 was b

calamebe · Oct 21, 2016

severus snake said:
Question 10 was b

Can confirm

petermaniah198 · Oct 21, 2016

wat what 8

InteGrand · Oct 21, 2016

calamebe · Oct 21, 2016

hedgehog_7 said:
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 · Oct 21, 2016

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

KingOfActing · Oct 21, 2016

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

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

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

petermaniah198 said:
what mark do u think 84 raw will align to in terms of HSC mark?

~94

Glyde · Oct 21, 2016

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

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

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

~94

you have a special brain

KingOfActing · Oct 21, 2016

Glyde said:
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 · Oct 21, 2016

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 · Oct 21, 2016

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 · Oct 21, 2016

$\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:$

$\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 $P$.$

$\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.

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

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