2018 HSC Maths Ext 1 Exam Paper Solutions

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.

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*}

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

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