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$\noindent \textbf{Q1.} $h(x) = f(x)g(x) \\ \begin{align*}f \text{ and } g &\text{ both even} \\ h(x) &= f(-x)g(-x) \\ &= h(-x) \end{align*} \\ \begin{align*}f \text{ and } g &\text{ both odd} \\ h(x) &= -f(-x) \times -g(-x) \\ &= f(-x)g(-x) \\ &= h(-x) \end{align*} \\ \begin{align*}f \text{ and...

$\noindent If $2 < x < 4$ then $4 < x^2 < 16$ and thus $7 < x^2 + 3 < 19$ $\noindent If $-1 < x \leq 3$ then $0 \leq x^2 \leq 9$ and thus $3 \leq x^2 + 3 \leq 12$

$\noindent $\begin{align*}(n+2)! + (n+1)! + n! &= (n+2)(n+1)n! + (n+1)n! + n! \\ &= n![(n+2)(n+1) + (n+1) + 1] \\ &= n!(n^2 + 4n +4) \\ &= n!(n+2)^2 \end{align*}

$\noindent \textbf{Q9.} The parabola $x^2 = -6y$ has focus $S\left(0, -\frac{3}{2}\right)$. The focal chord is simply a chord of the parabola which passes through the focal point, and so the chord passes through $(6, - 6)$ and $S\left(0, -\frac{3}{2}\right)$. From here, the equation of the focal...

Re: HSC 2017 Maths Marathon $\noindent \textbf{(a)} Now $0.52^{\circ} \equiv \frac{13\pi}{4500}$ radians. If we construct an altitude from earth viewpoint, this angle is halved, i.e. $\frac{13\pi}{9000}$ radians. \\\\ Now $\tan{\left(\frac{13\pi}{9000}\right)} = \frac{r_{moon}}{385000}...

$\noindent \textbf{Q9.} $ V = \pi \int_1^2(e^{-x}+1)^2\mathrm{d}x = \pi \int_1^2(e^{-2x} + 2e^{-x} + 1)\mathrm{d}x = \pi \left[-\frac{1}{2}e^{-2x} -2e^{-x} + x \right]^2_1 = \pi \left(-\frac{1}{2}e^{-4} - 2e^{-2} + 2 + \frac{1}{2}e^{-2} + 2e^{-1} - 1\right) = 4.8$ units$^3$ (correct to 1 dp.)...

$\noindent In that case, the area in Q15 is just the area of a quarter circle radius 2, minus the area of a right-angled triangle with base and height 2. \\\\ That is, $A = \frac{\pi r^2}{4} - \frac{1}{2}bh = \frac{\pi (2)^2}{4} - \frac{1}{2}(2)(2) = (\pi - 2)$ units$^2$

$\noindent For Q10, the area required is given by $A = \int_2^3(x-2)^2\mathrm{d}x + \int_3^4(x-4)^2\mathrm{d}x.$ However, by symmetry these two areas are equal and thus we can say $A = 2\int_2^3(x-2)^2\mathrm{d}x$. \\\\ Now, this can be evaluated as it is, but, to avoid more substitution, notice...

$\noindent \textbf{Q15.} $A = \int_{-2}^1(4-x^2-(x+2))\mathrm{d}x = \int_{-2}^1(4-x^2-x-2)\mathrm{d}x = \int_{-2}^1(2-x^2-x)\mathrm{d}x = \left[2x - \frac{1}{3}x^3 - \frac{1}{2}x^2\right]^1_{-2} = 2 - \frac{1}{3} - \frac{1}{2} - 2(-2) + \frac{1}{3}(-2)^3 + \frac{1}{2}(-2)^2 = 4\frac{1}{2}$...

$\noindent \textbf{(7a)} Now the intersection point is at $x = \frac{\pi}{4}$ and so the area $A = \int_0^{\frac{\pi}{4}}\sin{x}\mathrm{d}x + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos{x}\mathrm{d}x.$ But by symmetry, these two areas are equal and thus we can just write $A = 2 \times...

$\noindent There are multiple ways to do the second question, one being taking the difference of areas by integrating with respect to $x$. \\\\ Alternatively, we can use the fact that if $y=9-x^2$ then $x = \pm \sqrt{9-y}$. By symmetry, the required area $A = 2 \times \int_5^9...

$\noindent For the first one, the area $A$ is simply $A = \int_{-1}^1 \mathrm{d}x - \int_{-1}^1x^2 \mathrm{d}x$ since the points of intersection are $(-1, 1)$ and $(1, 1)$. This can be written as $A = \int_{-1}^{1}(1-x^2)\mathrm{d}x$ by combining the integrals. Now by symmetry, this becomes $A =...

$\noindent $y = x^4 + 1$ means $x = \pm (y-1)^{\frac{1}{4}}$. But since we are dealing with Quadrant I, then we take $x = (y-1)^{\frac{1}{4}}$. \\\\ Now $A = \int_1^3 (y-1)^{\frac{1}{4}} \mathrm{d}y = \left[\frac{4(y-1)^{\frac{4}{5}}}{5}\right]^3_1 = \left(\frac{4(3-1)^{\frac{5}{4}}}{5}\right) -...

$\noindent If $\frac{1}{x} + x = A$ then $x^2 - Ax + 1 =0$. Solve for $x$ and substitute into $x^7 + \frac{1}{x^7}$ to get $x^7 + \frac{1}{x^7} = \left(\frac{A \pm \sqrt{A^2-4}}{2}\right)^7 + \left(\frac{2}{A \pm \sqrt{A^2 - 4}}\right)^7

$\noindent Well, $\sin{A} = \sin{C}$ as $\angle A = \angle C$, being opposite angles in a parallelogram. And $\sin{A} = \sin{(180-A)}$. But $180 - A = \angle B = \angle D$, being co-interior angles; supplementary. Thus, $\sin{(180 - A)} = \sin{A} = \sin{B} = \sin{C} = \sin{D}$, as required.

You substitute x = a into f'(x). The exponents simply become a/a = 1 and -a/a = -1

$\noindent Given $f(x) = \frac{a}{2} \left(e^{x/a} + e^{-x/a}\right)$ \\ Well $\frac{\textrm{d}}{\textrm{d}x}f(x) = \frac{a}{2}\left(\frac{1}{a} \cdot e^{x/a} - \frac{1}{a} \cdot e^{-x/a}\right) = \frac{1}{2}\left(e^{x/a} - e^{-x/a}\right).$ \\ Thus, $f'(a) = \frac{1}{2}\left(e-e^{-1}\right)

When we isolate the terms we have 2xy - xy = xy.

$\noindent A nice way of simplifying through some denominator manipulation, \\ \begin{align*} \frac{x+y}{x-y} + \frac{x}{y-x} - \frac{y}{y^2-x^2} &= \frac{x+y}{x-y} - \frac{x}{x-y} + \frac{y}{(x+y)(x-y)} \\ &= \frac{y(x+y)+y}{x^2-y^2} \\ &= \frac{y(x+y+1)}{x^2 - y^2} \textrm{ or }...

$\noindent Let the roots be $\alpha , \beta , \gamma \Rightarrow \alpha = \beta + \gamma.$ Now $\alpha + \beta + \gamma = 2\alpha = -8 \Rightarrow \alpha = -4.$ Thus, $(x + 4)$ is a factor of $4x^3 + 32x^2 + 79x + 60 = 0$. The other two roots can be found through long division and solving a...