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Call the males A, B, C, D. Note in the table's arrangement, it'll be: --M F---F --M, where M refers to a pair of males and F refers to a pair of females. We can wlog (by rotational symmetry) place A at the top M above. We can choose this position in 2 ways (either A will have a male on his...

Re: HSC 2016 4U Marathon $\noindent I showed above why it is $z_1$. It can't be $z_2$ because $z_2= \overline{z_1}$ (you can show this as an exercise; use $\overline{w^k} = w^{7-k}$ and the fact that the conjugate of a sum is the sum of conjugates to do this), so they have opposite imaginary...

Re: HSC 2016 4U Marathon $\noindent If we want to do it algebraically, note since $w = \mathrm{cis}\left( \frac{2\pi}{7}\right)$, we have $w^2 = \mathrm{cis}\left(\frac{4\pi}{7}\right)\Rightarrow \mathrm{Im}\left(w^2\right) = \sin \frac{4\pi}{7} = \sin \frac{3\pi}{7}$ (supplementary angles...

$\noindent Yes (and vice versa), so finding the $t$ that optimises $f$ becomes equivalent to finding the $t$ that optimises $\left(f(t)\right)^2$. Sometimes optimising the square is easier (usually the case for distance problems since it means we don't need to worry about a square root, which is...

$\noindent We can generalise this as follows. Let $f$ be a (real-valued) function defined on a subset $X$ of $\mathbb{R}$. Let $\phi$ be a (real-valued) function defined on the range of $f$. If $\phi$ is an increasing function (on range of $f$), then $t^*$ maximises $f(t)$ (respectively...

$\noindent If $f(t)$ is something that it always non-negative (e.g. a distance), then $t^*$ maximises (respectively minimises) $f$ if and only if $t^*$ maximises (respectively minimises) $f^2$, because the square function is increasing on $\left[0, \infty \right)$.$

HSC would probably first ask you to show the sum to product identity.

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent a) Consider a `concentric' $40$ m by $10$ m rectangle inside the rectangular field (so it has the same centre as the field, its edges are parallel to the field, and it is contained inside the field, so like the...

Why did they write it as 3x + 2(x-3)? Why not just immediately give it as 5x – 6? Was it a typo (should have been a quadratic term)?

$\noindent We have $\sqrt{3x+2\left(x-3\right)} = \sqrt{5x -6}$. So in addition to solving the inequation, we need to also make sure $5x-6\geq 0$, so a condition on $x$ is that $x \geq \frac{6}{5}$.$

$\noindent We can also do the Q. like this. Write $u\equiv x-1$, then $x\equiv u+1$, so $x(x-1)\equiv p+q(x-1)+r(x-1)^2 \Longleftrightarrow (u+1)u \equiv p + qu + ru^2 \Longleftrightarrow u^2 + u \equiv p+qu+ru^2$. Therefore, $r=1,q=1$ and $p=0$.$

Re: HSC 2016 2U Marathon Lol, I remember this was on Terry Lee's website. Funny Q.

Re: MX2 2016 Integration Marathon 'Tis an advanced form of 'inspection'. :p

$\noindent Note $x = Vt\cos \theta \Rightarrow t = \frac{x}{V \cos \theta}$. Since $y = Vt\sin \theta -\frac{1}{2}gt^2$, we have$ $$\begin{align*}y &= V\times \frac{x}{V \cos \theta} \times \sin \theta -\frac{1}{2}g\left( \frac{x}{V \cos \theta}\right)^2 \\ &= x\tan \theta -\frac{gx^2}{2...

Re: MX2 2016 Integration Marathon Maybe he was thinking of difference of two squares?

$\noindent First show that $I,B,O$ are collinear. To do this, show that $\angle IBD = 90^\circ$, then since $\angle OBD = 90^\circ$ by virtue of radius being perpendicular to tangent, the adjacent angles $IBD$ and $OBD$ are supplementary. To show that $\angle IBD = 90^\circ$, show that $B$...

Derivative of sec(x) is sec(x)tan(x), derivative cot(x) is -csc^2 (x) and derivative of csc(x) is -csc(x)cot(x). Reversing these relations gives us some antiderivatives. These derivatives are listed in the Year 12 3U Pender (Cambridge) textbook I believe (because they're essentially...

Yeah

For the bounds you essentially replace them with the value of the new variable when the old variable equals those bounds. Like if the bounds were 0 to 16 and we did a substitution of u = sqrt(x) (where the original variable was x), then the bounds for u would be from 0 to 4, since when x = 0, u...

$\noindent You should write `*' for the multiplication sign, otherwise it looks like an $x$. Anyway, for that integral, let $u = \cos x$, so $\mathrm{d}u =-\sin x \text{ d}x$. When $x = 0, u = \cos 0 = 1$, and when $x = \pi$, $u = \cos \pi = -1$. So $\int _0 ^{\pi} \cos^6 x \sin x \text{ d}x =...