# 2018 HSC Maths Ext 1 Exam Paper Solutions

In this post, we reveal the solutions to the 2018 HSC Maths Extension 1 Exam paper.

Have you seen the 2018 HSC Mathematics Extension 1 Exam Paper, yet? In this post, our Maths team share their completed solutions to the 2018 HSC Maths Extension 1 Exam Paper.

2018 HSC Mathematics Extension 1 Exam Paper Solutions

### Multiple Choice

1. (B)

From the polynomial:
$$a = 2 \quad b = 6 \quad c = -7 \quad d = -10$$
\begin{align*}
\alpha\beta\gamma &= -\frac{d}{a} = -\frac{-10}{2} = 5 \\
\alpha + \beta + \gamma &= -\frac{b}{a} = -\frac{6}{2} = -3 \\
\text{therefore} \ \alpha\beta\gamma(\alpha + \beta + \gamma) &= -15
\end{align*}

2. (A)

\begin{align*}
\tan\theta &= \left|\frac{m_1-m_2}{1 + m_1m_2}\right| \\
&= \left|\frac{5-3}{1+5\cdot 3}\right| = \frac{1}{8}
\end{align*}

3. (A)

\begin{align*}
\lim_{x\rightarrow 0}\frac{\sin 3x\cos 3x}{12x} &= \lim_{x\rightarrow 0}\frac{\frac{1}{2}\cdot 2\sin 3x\cos 3x}{12x} \\
&= \lim_{x\rightarrow 0}\frac{\frac{1}{2}\cdot \sin 6x}{12x} \\
&= \frac{1}{4}\lim_{x\rightarrow 0}\frac{\sin 6x}{6x} = \frac{1}{4}
\end{align*}

4. (D)

From the graph we have:
\begin{gather*}
\begin{cases}
\text{Single roots at } x = -2,-1 \\
\text{Double root at } x = 1
\end{cases} \\
\text{therefore} \ b = 2, c= 1, d = -1
\end{gather*}
As the y-intercept is at $$abcd^2$$, we have that $$abcd^2 = -6 \Rightarrow a = -3$$

5. (A)

After substituting $$t = 0$$ in all four equations, the initial value is only 3000 for (A) or (D). Out of these options, limiting $$t \rightarrow \infty$$ gives us that asymptote is only 1500 for option (A).

6. (C)

$$x_1 = a$$ will approach the wrong root. $$x_1 = b$$ or $$c$$ will not approach anywhere as they are stationary points. Thus it must be $$x_1 = c$$.

7. (C)

\begin{align*}
a &= \frac{d}{dx}\left(\frac{1}{2}v^2\right) \\
&= \frac{d}{dx}\left(\frac{1}{2}(x^2 + 2)^2\right) \\
&= (x^2+ 2)(2x)
\end{align*}
$$\text{therefore} \ \text{at } x = 1, a = \text{6}{ms^{-2}}$$

8. (B)

There are $$5!$$ ways to arrange the men. Now the women must be positioned within the gaps. There are $$6!$$ ways for the women to be arranged amongst those 6 positions. Thus, there are $$5!\times6!$$ ways in total.

9. (D)

The $$\sin$$ general solution formula gives us that
\begin{align*}
2x &= n\pi + (-1)^n\left(-\frac{\pi}{6}\right) \\
&= n\pi + (-1)^{n+1}\frac{\pi}{6} \\
x &= \frac{n\pi}{2} + (-1)^{n+1}\frac{\pi}{12}
\end{align*}

10. (B)

After letting $$v = 0$$, we have that $$x = 0$$ or $$2k$$. For SHM, this means that the particle changes direction $$x = 0$$ and $$x = 2k$$. This implies that the particle moves between $$x = 0$$ and $$x = 2k$$. Therefore the centre of motion is $$x = k$$ and the amplitude is $$k$$. This only holds for option (B).

### Written Response

11. (a) i.

\begin{gather*}
P(1) = 1-2-5+6=0 \\
\Rightarrow x=1 \;\; \text{is a zero}
\end{gather*}

11. (a) ii.

\begin{gather*}
P(x) = (x-1)(x^2+bx+c) \\
\end{gather*}
Equating constants: $$P(x) = (x-1)(x^2+bx-6)$$
Equating coefficients of $$x^2$$: $$-2 = b / – 1 \Rightarrow b=-1$$
\begin{align*}
P(x) &= (x-1)(x^2-x-6) \\
&=(x-1)(x-3)(x+2)
\end{align*}
So zeroes are 1, 3, and -2

11. (b)

\begin{gather*}
\log_2{(5x-10)} = 3 \\
5x-10 = 2^3 = 8 \\
x = \frac{18}{5}
\end{gather*}

11. (c)

\begin{gather*}
\sqrt{3}\sin x + \cos x \equiv R\cos\alpha\sin x + R\sin\alpha\cos x \\
\begin{cases}
R\cos\alpha = \sqrt{3} \\
R\sin\alpha = 1
\end{cases} \\
R^2 = 4 \Rightarrow R = 2 \\
\tan\alpha = \frac{1}{\sqrt{3}} \Rightarrow \alpha = \frac{\pi}{6} \\
\Rightarrow \sqrt{3}\sin x + \cos x = 2\sin\left(x + \frac{\pi}{6}\right)
\end{gather*}

11. (d) i.

$$x \neq \frac{1}{4} \Rightarrow D:x\in \mathbb{R};x\neq \frac{1}{4}$$

11. (d) ii.

\begin{gather*}
\frac{1}{4x-1} < 1 \\
\Rightarrow (4x-1) < (4x-1)^2 \\
\end{gather*}
Moving terms to the right:
\begin{gather*}
(4x-1)[4x-1-1] > 0 \\
(4x-1)(4x-2) > 0 \\
(4x-1)(2x-1)>0 \\
x > \frac{1}{2} \;\; \text{or}\;\; x < \frac{1}{4}
\end{gather*}

11. (e)

\begin{gather*}
u=1-x \Rightarrow du = -dx \\
x=-3 \Rightarrow u=4 \\
x=0 \Rightarrow u=1 \\
\end{gather*}
\begin{align*}
\int_{-3}^{0}{\frac{x}{\sqrt{1-x}}dx} &=-\int_4^1{\frac{1-u}{\sqrt{u}}du} \\
&= \int_4^1{u^{\frac{1}{2}}-u^{-\frac{1}{2}}du} \\
&= \left[\frac{2}{3}u^{\frac{3}{2}}-2u^{\frac{1}{2}}\right]_4^1 \\
&=\left(\frac{2}{3}\times 1\sqrt{1} – 2\sqrt{1}\right) – \left(\frac{2}{3}\times4\sqrt{4} – 2\sqrt{4}\right) \\
&= \frac{2}{3} – 2 – \frac{16}{3} + 4 \\
&= -\frac{8}{3}
\end{align*}

12. (a)

\begin{align*}
\int \cos ^2(3x)\,dx &= \int \frac{(1+\cos 6x)}{2}\,dx \\
&= \frac{1}{2}\left(x + \frac{1}{6}\sin 6x\right) + C \\
&= \frac{x}{2} + \frac{1}{12}\sin 6x + C
\end{align*}

12. (b) i.

\begin{gather*}
\sin\theta = \frac{h}{20} \Rightarrow h = 20\sin\theta \\
\text{therefore} \ \frac{dh}{d\theta} = 20\cos\theta
\end{gather*}

12. (b) ii.

\begin{align*}
\text{Rising speed } = \frac{dh}{dt} &= \frac{dh}{d\theta}\times \frac{d\theta}{dt} \\
&= 20\cos\theta \times 1.5 = 30\cos\theta
\end{align*}
When $$h = 15$$, the diagram gives us that
\begin{align*}
\cos\theta &= \frac{\sqrt{20^2 – 15^2}}{20} = \frac{\sqrt{7}}{4} \\
\text{therefore} \ \frac{dh}{dt} &= 30\times \frac{\sqrt{7}}{4} \\
&= \frac{15\sqrt{7}}{2} = {19.8} \text{m/min} \quad \text{(1 d.p.)}\\
\end{align*}

12. (c) i.

$$f'(x) = \frac{1}{\sqrt{1-x^2}} + \left(-\frac{1}{\sqrt{1-x^2}}\right)$$

12. (c) ii.

\begin{align*}
f(x) &= \int f'(x)\,dx \\
&= \int 0\,dx = 0+ C = C \\
f(0) &= \sin^{-1}(0) + \cos^{-1}(0) \\
&= \frac{\pi}{2} = C
\end{align*}
$$\text{therefore} \ \sin^{-1}x + \cos^{-1}x = f(x) = C = \frac{\pi}{2}$$

12. (c) iii.

As the domain of $$\sin^{-1}x$$ and $$\cos^{-1}x$$ are both $$-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$$, $$f(x)$$ must have the same domain. Thus from part (ii), we have the following

12. (d)

\begin{align*}
P(\text{10 finish}) &= \binom{12}{10}(0.75)^{10}(1-0.75)^2 \\
P(\text{11 finish}) &= \binom{12}{11}(0.75)^{11}(1-0.75) \\
P(\text{12 finish}) &= \binom{12}{12}(0.75)^{12} \\
\text{therefore} P(\text{at least 10 finish}) &= P(\text{10 finish}) + P(\text{11 finish}) + P(\text{12 finish}) \\
&= \binom{12}{10}(0.75)^{10}(1-0.75)^2 + \binom{12}{11}(0.75)^{11}(1-0.75) + \binom{12}{12}(0.75)^{12} \\
&= \frac{3^{10}}{4^{12}}\left(\binom{12}{10} + 3\binom{12}{11} + 9\binom{12}{12}\right) \quad \text{(simplification not necessary)}
\end{align*}

12. (e) i.

If we let $$q = 0$$, point $$Q$$ will coincide with point $$O$$ and thus point $$T$$ will coincide with point $$A$$. Thus using the provided formula for $$T$$ we have that
\begin{gather*}
A = T = (a(p+0),ap\cdot 0) = (ap,0) \\
\text{therefore}
\begin{cases}
m_{SA} = \dfrac{a – 0}{0 – ap} = -\dfrac{1}{p} \\
m_{AP} = \dfrac{0 – ap^2}{ap – 2ap} = p
\end{cases} \\
m_{SA} \times m_{AP} = -1
\end{gather*}
Therefore $$SA \perp AP$$ and thus $$\angle PAS = \text{90°}$$

12. (e) ii.

Similarly to part (i), we have that $$\angle QBS = \text{90°}$$.
$$\text{therefore} \begin{cases} \angle PAS = \angle TAS = {90°}\\ \angle QBS = \angle TBS = {90°} \end{cases}$$
Therefore $$ST$$ subtends equal angles at $$A$$ and $$B$$. Hence $$S,B,A,T$$ are concyclic.

12. (e) iii.

As $$\angle PAS$$ is $$\text{90°}$$, $$ST$$ must be the diameter of circle $$SBAT$$. By the distance formula we have
\begin{align*}
ST^2 &= \left(0 – a(p+q)\right)^2 + (a – apq)^2 \\
&= a^2(p^2 + 2pq + q^2 + p^2q^2 -2pq + 1) \\
&= a^2(p^2q^2 + p^2 + q^2 + 1) \\
&= a^2(p^2 + 1)(q^2 + 1) \\
ST &= a\sqrt{(p^2 + 1)(q^2 + 1)} \quad (\text{because} \ ST \geq 0)
\end{align*}
Thus, the diameter is $$a\sqrt{(p^2 + 1)(q^2 + 1)}$$

13. (a)

\begin{align*}
\text{Base case: $n=1$} \\
\text{LHS} &= 2(-3)^{1-1} \\
&= 2 \\
\text{RHS} &= \frac{1-(-3)^1}{2} \\
&= 2 \\
&= \text{LHS}
\end{align*}
\begin{gather*}
\text{Now assume that the statement is true for $n=k$} \\
2-6+ \dots + 2(-3)^{k-1} = \frac{1-(-3)^k}{2} \\
\text{So as to prove the statement for $n=k+1$} \\
2-6+ \dots + 2(-3)^{k-1} + 2(-3)^k = \frac{1-(-3)^{k+1}}{2} \\
\text{Using the assumption: } \\
\end{gather*}
\begin{align*}
\text{LHS} &= \frac{1-(-3)^k}{2} + 2(-3)^k \\
&= \frac{1-(-3)^k + 4(-3)^k}{2} \\
&= \frac{1 + 3(-3)^k}{2} \\
&= \frac{1 – (-3)(-3)^k}{2} \\
&= \frac{1 – (-3)^{k+1}}{2} \\
&= \text{RHS}
\end{align*}
Hence proven by the principle of Mathematical Induction

13. (b) i.

The domain of $$f^{-1}(x)$$ is the range of $$f(x)$$ and the range of $$f^{-1}(x)$$ is the domain of $$f(x)$$
\begin{gather*}
D: 0 < x \leq \frac{1}{2} \\
R: y \geq 1
\end{gather*}

13. (b) ii.

Reflect the graph of $$f(x)$$ for $$x \geq 1$$ about the line $$y = x$$

13. (b) iii.

\begin{gather*}
\text{ Swapping $x$ and $y$:} \\
x = \frac{y}{y^2+1} \\
xy^2 – y + x = 0 \\
y = \frac{1 \pm\sqrt{1-4x^2}}{2x} \\
\end{gather*}

But since $$y \geq 1$$:

\begin{gather*}
y = \frac{1 +\sqrt{1-4x^2}}{2x} \\
\end{gather*}
We took the positive root because taking the negative root and substituting $$x=\frac{1}{4}$$ yields $$y < 1$$

13. (c) i.

\begin{gather*}
y = 0 \\
\Rightarrow t\left(V\sin\theta – \frac{gt}{2}\right) = 0 \\
t \neq 0 \Rightarrow t = \frac{2V\sin\theta}{g} \\
\end{gather*}

Now substitute into $$x(t)$$:

\begin{align*}
x &= V\cos\theta\left(\frac{2V\sin\theta}{g}\right) \\
&= \frac{V^2}{g}(2\sin\theta\cos\theta) \\
&= \frac{V^2\sin2\theta}{g}
\end{align*}

13. (c) ii.

\begin{align*}
\frac{V^2\sin(2(\frac{\pi}{2}-\theta))}{g} &= \frac{V^2\sin(\pi-2\theta}{g} \\
&= \frac{V^2\sin2\theta}{g}
\end{align*}

13. (c) iii.

\begin{gather*}
\text{By symmetry, $h_{\alpha}, h_{\beta}$ is attained when} \\
x = \frac{d}{2} = \frac{V^2}{2g}\sin 2\alpha \\
\text{Let the time at max height be $t_m$} \\
x(t_m) = \frac{V^2}{2g}\sin 2\alpha = Vt_m \cos\alpha \\
\frac{V^2}{2g}2\sin\alpha\cos\alpha = Vt_m \cos\alpha\\
t_m = \frac{V}{g}\sin\alpha \\
\end{gather*}
So $$h_\alpha$$ is
\begin{align*}
h_\alpha = y(t_m) &= V\left(\frac{V}{g}\sin\alpha\right) – \frac{g}{2}\left(\frac{V}{g}\sin\alpha\right)^2 \\
&= \frac{V^2\sin^2\alpha}{2g} \\
\end{align*}
For $$h_\beta$$ we have,
\begin{align*}
h_\beta &= \frac{V^2\sin^2\beta}{2g}\\
&= \frac{V^2\sin^2(\frac{\pi}{2}-\alpha)}{2g} \\
&= \frac{V^2\sin^2\alpha}{2g} \\
\end{align*}
\begin{equation*}
\Rightarrow \frac{h_\alpha + h_\beta}{2} = \frac{1}{2}\frac{V^2}{2g}(\sin^2\alpha + \cos^2\alpha) = \frac{V^2}{4g}
\end{equation*}
which only depends on $$V$$ and $$g$$.

14. (a)

\begin{gather*}
\text{From $\Delta AQD$:} \angle PQR = 180 – (\alpha + \delta) \\
\text{From $\Delta BSC$:} \angle PSR = 180 – (\beta + \gamma) \\
\end{gather*}
Now,
\begin{align*}
\angle PQR + \angle PSR &= 360 – (\alpha + \beta + \gamma + \delta) \\
&= 360 – \frac{2\alpha + 2\beta + 2\gamma + 2\delta}{2} \\
&= 360 – \frac{360}{2} \\
&= 180
\end{align*}
Hence, $$PQRS$$ is cyclic.

14. (b) i.

Equating coefficients of $$x^r$$ on both sides:
\begin{align*}
\text{LHS} &= \sum_{k=0}^n \binom{n}{k}(1+x)^k (1)^{n-k} \\
&= \sum_{k=0}^n \left(\binom{n}{k}\sum_{j=0}^{k}x^j(1)^{k-j}\right) \\
\end{align*}
$$x^r$$ occurs where $$j=r$$
\begin{align*}
\binom{n}{r}\binom{r}{r} + \binom{n}{r+1}\binom{r+1}{r} + \dots + \binom{n}{n}\binom{n}{r} \\
\end{align*}
Which is the coefficient of $$x^r$$
Clearly the coefficient of $$x^r$$ on the RHS is $$\binom{n}{r}2^{n-r}$$. Hence the expression is proven

14. (b) ii.

Selector A chooses a group of at least 4. (i.e. could choose $$4,5,6,\dots,23$$) \\
Then Selector B chooses 4 from the group. This can occur in
\begin{equation*}
\binom{23}{4}\binom{4}{4} + \binom{23}{5}\binom{5}{4} + \dots + \binom{23}{23}\binom{23}{4} \\
\end{equation*}
ways. Letting $$n=23$$ and $$r=4$$ this expression is:
\begin{align*}
\binom{n}{r}\binom{r}{r} + \binom{n}{r+1}\binom{r+1}{r} + \dots + \binom{n}{n}\binom{n}{r} &= \binom{n}{r}2^{n-r} \;\;\; \text{(from (i))}\\
&= \binom{23}{4}2^{23-4} \\
&= \binom{23}{4}2^{19} \\
&= 4642570240
\end{align*}

14. (c) i.

\begin{equation*}
BC \perp AC \Rightarrow \angle BCA = 90°
\end{equation*}
\begin{equation*}
\text{So in } \Delta ABC \ \text{and}\ \Delta ACD: \\
\end{equation*}
\begin{align*}
\angle BCA &= \angle ADC \ (\text{right-angle}) \\
\angle BAC &= \angle DAC \ (\text{common}) \\
\text{therefore} \ \Delta ABC &||| \Delta ACD \ (\text{equiangular})
\end{align*}

14. (c) ii.

\begin{align*}
\frac{AB}{AC} &= \frac{BC}{CD} \ (\text{corresponding ratios of similar triangles are equal}) \\
\frac{c}{b} &= \frac{a}{x} \\
\Rightarrow x &= \frac{ab}{c} \ \text{as required.} \\
\end{align*}

14. (c) iii.

Consider the general case:

\begin{align*}
\text{Now, } x_{n+1} &= \frac{a_{n+1}\times b_{n+1}}{c_{n+1}} \ \text{(from (ii))} \\
&= \frac{a_{n+1}}{c_{n+1}}(b_n-x_n) \\
&= \frac{a_n b_n}{c_n} – \frac{a_n}{c_n}x_n \\
\end{align*}
Noting that $$\frac{a_{n+1}}{c_{n+1}}=\frac{a_n}{c_n}$$ by similar triangles:
\begin{align*}
x_{n+1} &= x_n \left(1-\frac{a_n}{c_n} \right) \\
&= x_n \left(1-\frac{a}{c} \right) \\
&= x_n\left(\frac{c-a}{c} \right)
\end{align*}
So we have established that:
\begin{gather*}
x_1 = x\left(\frac{c-a}{c} \right) \\
x_1 = x\left(\frac{c-a}{c} \right)^2 \\
\vdots \\
x_n = x\left(\frac{c-a}{c} \right)^n \\
\vdots \\
\end{gather*}
and so on. Hence the limiting sum is
\begin{align*}
\frac{\pi}{4}(x^2+x_1^2 + x_2^2 + \dots) &=\frac{\pi}{4}x^2\left(1+\left(\frac{c-a}{c} \right)^2 + \left(\frac{c-a}{c} \right)^4 + \dots\right) \\
&= \frac{\pi a^2 b^2}{4c^2}\frac{1}{1-\frac{c^2-2ac+a^2}{c^2}} \\
&= \frac{\pi a b^2}{4}\frac{a}{c^2-c^2+2ac-a^2} \\
&= \frac{\pi a b^2}{4}\frac{a}{a(2c-a)} \\
&= \frac{\pi ab^2}{4(2c-a)} \ \text{as required.} \\
\end{align*}

14. (c) iv.

We observe that the sum of the areas of the quadrants must be smaller than the area of the triangle itself (as there are gaps). Thus:
\begin{align*}
\frac{\pi ab^2 }{4(2c-a)} &< \frac{1}{2} ab \\
&= \frac{2c-a}{b} \\
\end{align*}

Hence
$$\frac{\pi}{2}<\frac{2c-a}{b} \text{as required.}$$