2018 HSC Maths Advanced Exam Paper Solutions

Have you seen the 2018 HSC Mathematics Advanced Paper, yet? In this post, we will work our way through the 2018 HSC Maths Advanced paper and give you the solutions, written by our leading teacher Oak Ukrit and his team.

Read on to see how to answer all of the 2018 questions.

2018 HSC Mathematics Advanced Exam Paper Solutions

1. (B) \(7^{-1.3} = 0.07968 = 0.08 \) (2 d.p.)

2. (C) Use mid-point formula, results \(Q(13, 7)\)

3. (A) Let \(y=0\) gives \(x=-6\)

4. (D) Given centre of the circle at \(Q(3,-2)\). Use perpendicular distance formula to obtain radius of the circle, which is \(4\).From these information, we can write out the equation of the circle – which is option D.

5. (D) \(\frac{d}{dx}\sin(\ln{x}) = \frac{1}{x} \cos(\ln{x})\) using chain rule.

6. (C) P(Matching Pairs) = P(Any Shoes) \(\times P(Matching Shoe) = 1\times\frac{1}{7}  =  \frac{1}{7}\)

7. (C) Given that \(\int_0^4f(x)\,dx = 10\) which takes into account negative area between \(x=3\) and \(x=4\)
Hence \(\int_0^3 f(x)\,dx = 10+3 = 13\)
Therefore \(\int_{-1}^3 f(x)\,dx = 13-2 = 11\)

8. (D) Since \(x^{2}=4ay\), at \(y=4, x=12\) (via symmetry) Substituting values of \(x\) and \(y\) gives \(a=9\)

9. (B) Point of inflexion occurs when gradient of \(f'(x)=0\) and gradient of \(f'(x)\) changes sign. The only point satisfies both conditions is point \(x=b\)

10. (D) Consider area under the curve in each option, \(f(x)=\cos{\frac{x}{2}}\) is the only option that satisfies the given condition.

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Written Response

11. (a)

\begin{align*}
\frac{3}{3 + \sqrt{2}} \times \frac{3 – \sqrt{2}}{3 – \sqrt{2}} &= \frac{9 – 3\sqrt{2}}{9 – 2} \\
&= \frac{9 – 3\sqrt{2}}{7} \\
\end{align*}

11. (b)

\begin{align*}
1 – 3x &> 10 \\
-3x &> 9 \\
x &< -3
\end{align*}

11. (c)

\begin{align*}
\frac{8x^3 – 27y^3}{2x – 3y} &= \frac{(2x – 3y)(4x^2 + 6xy + 9y^2)}{2x – 3y} \\
&= 4x^2 + 6xy + 9y^2 \\
\end{align*}

11. (d) i.

Given \(T_3 = 8\), \(T_{20} = 59\)
\begin{align*}
a + 2d &= 8 \quad (1) \\
a + 19d &= 59 \quad (2)
\end{align*}
After solving simultaneously, \(d = 3\)

11. (d) ii.

\begin{align*}
T_{50} &= (a + 2d) + 47d \\
&= 8 + 47(3) = 149 \\
\end{align*}

11. (e)

\begin{align*}
\int_0^3 e^{5x}\,dx \ = \left[\frac{1}{5}e^{5x}\right]_0^3 \\
\ = \frac{1}{5}e^{15} – \frac{1}{5}e^0 \\
\ = \frac{1}{5}(e^{15} – 1)
\end{align*}

11. (f)

\begin{align*}
\frac{d}{dx}(x^2\tan x) &= 2x\tan x + x^2\sec^2x \quad (\text{product rule})
\end{align*}

11. (g)

\begin{align*}
\frac{d}{dx}\left(\frac{e^x}{x+1}\right) &= \frac{(x+1)e^x – e^x}{(x+1)^2} \quad \text{(quotient rule)} \\
\ = \frac{xe^x}{(x+1)^2}
\end{align*}

12. (a) i.

\begin{gather*}
\angle ABN = 130° \\
\angle ABC + 130° + 120° = 360° \\
\text{therefore} \  \angle ABC = 110°
\end{gather*}

12. (a) ii.

Using cosine rule,
\begin{align*}
AC^2 &= AB^2 + BC^2 – 2AB\cdot BC\cdot \cos 110° \\
&= 320^2 + 190^2 – 2(320)(190)\cos 110° \\
\text{therefore} \ AC &\approx 420\text{km} \quad \text{(nearest 10km)}
\end{align*}

12. (b)

Given \(y = \cos 2x\),
\begin{gather*}
\frac{dy}{dx} = -2\sin(2x) \\
\text{at } x = \frac{\pi}{6}, \quad \frac{dy}{dx} = -\sqrt{3} \\
\text{therefore} \  y – \frac{1}{2} = -\sqrt{3}\left(x – \frac{\pi}{6}\right)
\end{gather*}
Rearrange,
$$y = -\sqrt{3}x + \frac{\pi\sqrt{3} + 3}{6}$$

12. (c) i.

\begin{gather*}
\text{In } \triangle ADF, \triangle ABE \\
AD = AB \quad \text{(sides of square are equal)} \\
\angle ADF = \angle ABE \quad \text{(angles of square is 90°)} \\
\text{note: } EC = FC, DC = BC \\
DF = DC – FC = BE \\
\text{therefore} \ \triangle ADF \equiv \triangle ABE\,\,(SAS)
\end{gather*}

12. (c) ii.

\begin{align*}
A_{ACEF} &= 14^2 – 2\left(\frac{1}{2} \times 10\times 14\right) \\
\ = 196 – 140 \\
\ = 56 \text{cm}^2
\end{align*}

12. (d) i.

\begin{gather*}
x = \frac{t^3}{3} – 2t^2 + 3t \\
v = \frac{dx}{dt} = t^2 – 4t + 3 \\
\text{at } t = 0, v = 3 \text{ms}^{-1}
\end{gather*}

12. (d) ii.

\begin{gather*}
\text{stationary when } v= 0 \\
t^2 – 4t + 3 = (t-3)(t-1) = 0 \\
\text{therefore} \ \text{at } t= 1,3\,\,s \text{ particle is stationary}
\end{gather*}

12. (d) iii. Acceleration is zero at \(v_\text{max}\)
\begin{align*}
t &= -\frac{b}{2a} \text{ from quadratic } t^2 – 4t + 3 \\
&= \frac{4}{2}\\
&=2\,\,s
\end{align*}
$$\text{therefore} \ \text{at } t = 2, x = \frac{8}{3} – 8 + 6 = \frac{2}{3}\,\,m$$

13. (a) i.

Given \(y = 6x^2 – x^3\), \(\frac{dy}{dx} = 12x – 3x^2\) Let \(\frac{dy}{dx} = 0\), and solve for \(x\). This gives \(x = 0, 4\). Find associated \(y\)-value and test using \(\frac{d^2y}{dx^2}\). Thus we have,

minimum stationary point at \((0,0)\)
maximum stationary point at \((4,32)\)

13. (a) ii. Show that \(\dfrac{d^2y}{dx^2}\) at \(x = 2\) is zero, and \(\dfrac{d^2y}{dx^2}\) changes sign at \(x= 1\) and \(x = 3\).

13. (a) iii

13. (b) i.

\begin{gather*}
\text{In } \triangle CBD, BC = CD \quad \text{(isosceles triangle)} \\
\text{Simiarly, In } \triangle ABC, AB = CD \quad \text{(isosceles triangle)} \\
\text{Since } \angle ABC \text{ is common, } \triangle CBD \, ||| \, \triangle ABC \,\,\text{(Equiangular)}
\end{gather*}

13. (b) ii.

\begin{align*}
\frac{BD}{BC} &= \frac{DC}{AB} \quad \text{(matching sides in ratios of similar triangles)} \\
\frac{BD}{2} &= \frac{2}{3} \\
BD &= \frac{4}{3} \\
\text{therefore} \ AD+ BD &= AB \\
AD &= AB – BD \\
&= 3 – \frac{4}{3} \\
&= \frac{5}{3} \text{ units}
\end{align*}

13. (c) i.

\begin{align*}
P(50) & = 184 \quad \quad \text{* units in millions} \\
92e^{50k} & = 184 \\
k &= \frac{1}{50}\ln\left(\frac{184}{92}\right) \\
&\approx 0.0319 \quad \text{(4 d.p.)}
\end{align*}

13. (c) ii.

\begin{align*}
P(110) &= 92e^{0.0139(110)} \\
&= 424 \text{ million} \quad \text{(nearest million)}
\end{align*}

14. (a) i.

Using sine rule,
\begin{align*}
A_{\triangle KLM} &= \frac{1}{2}(3)(6)\sin60° \\
&= \frac{9\sqrt{3}}{2} \text{ unit}^2
\end{align*}

14. (a) ii.

\begin{align*}
A_{\triangle KLN} + A_{\triangle NLM} &= \frac{9\sqrt{3}}{2} \\
\frac{1}{2}(3x)\sin30° + \frac{1}{2}(6x)\sin30° &= \frac{9\sqrt{3}}{2} \\
\frac{3x}{4} + \frac{6x}{4} &= \frac{9\sqrt{3}}{2} \\
9x &= 18\sqrt{3} \\
x &= 2\sqrt{3} \text{ units}
\end{align*}

14. (b)

Rearrange the equation to make \(x\) the subject.
\begin{align*}
x &= (y – 1)^{\frac{1}{4}} \\
V &= \pi\int_1^{10}(y – 1)^{\frac{1}{2}}\,dy \\
&= 18\pi \text{ units}^3
\end{align*}

14. (c)

If \(f(x)\) has no stationary points, \(f'(x)\) has no roots.

No roots if \(\Delta < 0, B^2 – 4AC < 0\). Therefore
$$-3 < k < 3$$

14. (d) i.

DAY 1: \(n = 1 \Rightarrow 2^{1} + 1 = 3\) downloads
DAY 2: \(n = 2 \Rightarrow 2^{2}+ 2 = 6\) downloads
DAY 3: \(n = 3 \Rightarrow 2^{3} + 3 = 11\) downloads

14. (d) ii.

Consider AP & GP separately.
GP for \(2^n\):
\begin{gather*}
a = 2, r = 2, n = 20 \\
S_{20} = \frac{a(r^n-1)}{r-1} = 2097150 \\
\end{gather*}
AP for \(n\):
$$ a = 1, d = 1, n = 20 $$
\begin{align*}
S_{20} &= \frac{n}{2}(a+l) \\
&= \frac{20}{2}(1 + 20) \\
&= 210
\end{align*}
Therefore, the Total number of downloads is \(2097150 + 210 = 2097360\).

14. (e) i.

\begin{align*}
\text{P(at least one fault)} &= 1 – \text{P(no fault)} \\
&= 1 – (0.9 \times 0.95) \\
&= 0.145
\end{align*}

14. (e) ii.

\begin{align*}
\text{P(A, NF, NF)} + \text{P(B, NF, NF)} &= \frac{1}{2}\left(\frac{9}{10}\right)\left(\frac{9}{10}\right) + \frac{1}{2}\left(\frac{19}{10}\right)\left(\frac{19}{10}\right) \\
&= \frac{137}{160}
\end{align*}

15. (a) i.

let \(t = 0\), \(L(0) = 12 + 2\cos(0) = 14 \text{hrs}\)

15. (a) ii.

As \(-1 \leq \cos\left(\frac{2\pi t}{366}\right) \leq 1\), it is least when \(\cos\left(\frac{2\pi t}{366}\right) = -1\). Thus \(\min [L(t)] = 12 – 2 = 10 \text{hrs}\).

15. (a) iii.

Let \(L(t) = 11\)

\begin{align*}
11 &= 12 + 2\cos\left(\frac{2\pi t}{366}\right) \\
-\frac{1}{2} &= \cos\left(\frac{2\pi t}{2}\right) \\
\frac{2\pi t}{366} &= \frac{2\pi}{3},\frac{4\pi}{3} \\
t &= 122, 244 \\
\end{align*}

15. (b)

\begin{align*}
\int_{0}^k\frac{1}{x+3}\,dx &= \int_{k}^{45} \frac{1}{x+3}\,dx \\
[\ln(x+3)]_0^k &= [\ln(x+3)]_k^{45}\\
\ln(k+3) – \ln 3 &= \ln 48 – \ln(k+3) \\
2\ln(k+3) &= \ln 144 \\
k+3 &= 12 \quad (\text{because} \ k > 0) \\
k &= 9
\end{align*}

15. (c) i.

\begin{align*}
\text{Area} &= -\int_0^3(x^3 – 7x)\,dx + \int_0^32x\,dx \\
&= \int_0^3(2x + 7x – x^3)\,dx = \int_0^3(9x- x^3)\,dx \\
&= \left[\frac{9}{2}x^2 – \frac{1}{4}x^4\right]^3_0 \\
&= \frac{81}{2} – \frac{81}{4} = \frac{81}{4} \text{units}^2
\end{align*}

15. (c) ii.

\begin{gather*}
\text{let } f(x) = 2x – (x^3 – 7x) = 9x – x^3 \\
f(0) = 0, f(1.5) = 10.125 = \frac{81}{8}, \ f(3) = 0 \\
h = \frac{b-a}{2} = 1.5
\end{gather*}
\begin{align*}
\text{Area} &= \frac{h}{3}\left(1\times 0 + \frac{81}{8}\times 4 + 1 \times 0\right) \\
&= \frac{1}{2} \times 4 \times \frac{81}{4} = \frac{81}{4} \text{units}^2
\end{align*}

15. (c) iii.

Finding co-ordinate of \(P\)
\begin{align*}
y&=x^{3}-7x\\
y’&=3x^{2}-7\\
\text{Let}\,\,y’&=2\,\Rightarrow x=\sqrt{3}\,\,,y=-4\sqrt{3}\\
\text{therefore} \ P(\sqrt{3},-4\sqrt{3})\\
\end{align*}

15. (c) iv.

\begin{align*}
d&=\frac{|2\sqrt{3}-(-4\sqrt{3})|}{\sqrt{2^{2}+1^{2}}}\\
&=\frac{6\sqrt{3}}{\sqrt{5}}\\
\text{Hence OA}\,&=\sqrt{3^{2}+6^{2}}\\
&=\sqrt{45}\\
&=3\sqrt{5}\,\,\text{unit}\\
\text{therefore} \ \,\text{Area}\Delta\,OAP\,&=\frac{1}{2}\times\frac{6\sqrt{3}}{\sqrt{5}}\times3\sqrt{5}\\
&=9\sqrt{3}\,\,u^{2}\\
\end{align*}

16. (a) i.

\begin{align*}
V&= \frac{1}{3}\pi^{2}h,\\
but\,\, h^{2}+x^{2} &= 100\\
\text{therefore} \ h&=\sqrt{100-x^{2}}\\
V&=\frac{1}{3}\pi x^{2}\sqrt{100-x^{2}}
\end{align*}

16. (a) ii.

\begin{align*}
\frac{dV}{dx}&=\frac{2}{3}\pi x\sqrt{100-x^{2}} – \frac{2}{3}\pi x^{3} \frac{1}{\sqrt{100-x^{2}}}\times\frac{1}{2}\\
&=\frac{2\pi x(100-x^{2})-\pi x^{3}}{3\sqrt{100-x^{2}}}\\
&=\frac{\pi x(200-2x^{2}-x^{2})}{3\sqrt{100-x^{2}}}\\
&=\frac{\pi x(200-3x^{2})}{3\sqrt{100-x^{2}}}
\end{align*}

16. (a) iii.

\begin{gather*}
Let \frac{dV}{dx}=0\\
x=0\,(Omit),\,or\,x=\sqrt{\frac{200}{3}}\\
\text{Test the nature of the stationary point using either table method or Second derivative} \\
\Rightarrow \text{Max point at}\, x=\sqrt{\frac{200}{3}}\\
\text{Now}\, 2\pi x=10\,\theta \,\,\, \text{By equating circumference}\\
\text{therefore} \ \theta=\frac{2\sqrt{2}\pi}{\sqrt{3}}
\end{gather*}

16. (b) i.

16. (b) ii.

$$Pr(Win) =\frac{1}{36}\times(\frac{2\times20}{6})=\frac{5}{27}$$

16 (c) i.

\begin{align*}
A_0&=300000 \\
A_1&=300000\times1.04-P \\
A_2&=A_1(1.04)-P(1.05)\\
&=300000(1.04)^{2}-P(1.04+1.05)\\
&=300000(1.04)^{2}-P(1.04+1.05)\,\,\,\text{As Required}\\
\end{align*}

16. (c) ii.

\begin{align*}
A_3&=A_2(1.04)-P(1.05)^{2}\\
&=300000(1.04)^{3}-P((1.04)^{2}+1.05(1.04)+(1.05)^{2})\,\,\,\text{As Required}\\
\end{align*}

16. (c) iii.

\begin{align*}
A_4&=300000(1.04)^{4}-P(1.04^{3}+1.04^{2}(1.05)+1.04(1.05)^{2}+(1.05)^{3})\\
\vdots \\
A_n&=300000(1.04)^{n}-P((1.04)^{n-1}+(1.04)^{n-2}(1.05)+…+(1.05)^{n-1})\\
\text{Now,}\,\, 1.05^{n}-1.04^{n}&=(1.05-1.04)(1.04^{n-1}+…+1.05^{n-1})\\
(1.04^{n-1}+…+1.05^{n-1})&=(1.05^{n}-1.04^{n})(100)\\
\text{Since} A_n>0,\\
300000(1.04)^{n}-P(1.05^{n}-1.04^{n})&>0\\
\frac{300000(1.04)^{n-1}}{P}&>(1.05^{n}-1.04^{n})(100)\\
\frac{3000}{P}&>(\frac{105}{104})^{n}-1\\
\text{therefore} \ (\frac{105}{104})^{n}&<1+\frac{3000}{P}\,\,\,\text{As Required}
\end{align*}