Matrix Blog

Mathematics Extension 1

2019 HSC Maths Ext 1 Exam Paper Solutions

In this post, we reveal the solutions to the 2019 HSC Maths Extension 1 Exam paper.

Our Mathematics team have been hard at work putting together the 2019 HSC Maths Extension 1 Exam Paper solutions for you. Here they are!

 

2019 HSC Maths Ext 1 Exam Paper Solutions

Have a look through these worked solutions and see how you did!

Section 1: Multiple Choice

QuestionAnswerSolution
1A\(\)\begin{align*}
4-x &> 0\\
4 &> x\\
x &< 4
\end{align*}
2D\begin{align*}
(2x)^2 &= x(9+x)\\
4x^2 &= x(9+x)
\end{align*}
3C\begin{align*}
\frac{d}{dx} \tan^{-1}\frac{x}{2} &= \frac{\frac{1}{2}}{1+(\frac{x}{2})^2}\\
&= \frac{\frac{1}{2}}{\frac{4+x^2}{4}}\\
&= \frac{2}{4+x^2}
\end{align*}
4B\begin{align*}
\textrm{There is a horizontal asymptote at } y=1\\
\textrm{Hence for } y = f(x) = \frac{P(x)}{Q(x)},\\
\textrm{the leading terms of } P(x) \textrm{ and } Q(x) \textrm{ must be equal}\\
\textrm{Must be B (both leading terms are }x^2 \textrm{)}
\end{align*}
5A\begin{align*}
\textrm{Period}=2=\frac{2\pi}{n}, \textrm{ so }n=\pi, \textrm{ so must be A or C}\\
\textrm{Particle starts at rest, so must be A}
\end{align*}
6B\begin{align*}
\cos^2 x &= 1 – \sin^2 x = 1 – \left(\frac{1}{4}\right)^2 = \frac{15}{16}\\
\cos x &= -\frac{\sqrt{15}}{4}, \textrm{ as } x \textrm{ is in the 2nd quadrant}\\
\sin 2x &= 2\sin x\cos x\\
&= 2\left(\frac{1}{4}\right)\left(-\frac{\sqrt{15}}{4}\right)\\
&=-\frac{\sqrt{15}}{8}
\end{align*}
7C\begin{align*}
\textrm{Let roots be } -1, \alpha, \beta\\
\textrm{Product of roots} &= -\frac{q}{q} = -1\\
∴ -\alpha\beta &= -1\\
\beta &= \frac{1}{\alpha}
\end{align*}
8A\begin{align*}
\textrm{Group LLL as one object, rearrange with 5 other objects } 6! \textrm{ ways}\\
\textrm{A’s are duplicate}\\
∴ \frac{6!}{2}
\end{align*}
9D\begin{align*}
\cos^{-1}(-\sin x) &= \cos^{-1}(\cos(x+\frac{\pi}{2}))\\
&= x+\frac{\pi}{2}\\
∴ &\textrm{ Straight line with gradient 1, hence D}
\end{align*}
10A\begin{align*}
\textrm{Domain of }f(x) \textrm{ is } x \geq -1.\\
f(-1) = -1, \textrm{ so }f(x) \textrm{ intersects the curve at }(-1,-1)\\
f(x) \textrm{ clearly decreases as }x \textrm{ increases, hence } (-1,-1) \textrm{ must be the only point of intersection.}
\end{align*}

Section 2: Long response

Question 11 a

\begin{align*}
y&=\frac{1}{2}x + \frac{1}{2},\quad y=3x-4\\
m_1&=\frac{1}{2}, m_2=3\\
\tan\theta &= |\frac{\frac{1}{2}-3}{1+\frac{1}{2}\times3}|\\
&= |\frac{1-6}{2+3}|\\
&= 1\\
∴ \theta &= \frac{\pi}{4}
\end{align*}

 

Question 11 b

\begin{align*}
\frac{x}{x+1} &< 2\\
x(x+1) &< 2(x+1)^2\\
x^2 + x &< 2x^2 + 4x + 2\\
x^2 + 3x + 2 &> 0\\
(x+2)(x+1) &> 0 \\
∴ x \ < – 2 \ \text{or} \ x>-1
\end{align*}

graph solution for question 11b blog-2019-maths-ext-1-exam-paper-solutions-question-11b-graph-1

 

Question 11c

\begin{align*}
\angle ACB &= 90° \textrm{ (angle on a semicircle is 90 degrees)}\\
\angle BDC &= 90° \textrm{ (given)}\\
∴ \angle ACB &= \angle BDC\\
\angle BCD &= \angle CAB \textrm{ (alternate segment theorem)}\\
∴ \Delta ABC\,|&||\,\Delta CBD \textrm{ (equiangular)}\\
∴ \angle ABC &= \angle CBD \textrm{ (corresponding angles in similar triangles)}
\end{align*}

 

Question 11d

\begin{align*}
x^3 + 2x^2 – 3x \ – 7 &= (x-2)Q(x) + 3\\
x^3 + 2x^2 – 3x \ – 10 &= (x-2)Q(x)\\
x^3 + 2x^2 – 3x \ – 10 &= (x-2)(x^2 + 4x + 5) \textrm{ (by inspection)}\\
∴ Q(x) &= x^2 + 4x + 5
\end{align*}

 

Question 11e

\begin{align*}
\int 2\sin^2 4x\,dx &= \int (1-\cos 8x)\, dx\\
&= x – \frac{1}{8}\sin 8x + C
\end{align*}

 

Question 11f

(i)

\begin{align*}
(1-0.05)^8 = 0.95^8
\end{align*}

 

(ii)

\begin{align*}
\textrm{P(at least two winning) } &= \textrm{ P(no winning) } – \textrm{ P(one winning)}\\
&= 1- (0.95)^8 – (0.95)^7(0.05){8\choose1}
\end{align*}

 

Question 12a

\begin{align*}
\frac{dA}{dt} &= \frac{dA}{db}\times\frac{dB}{dt}\\
\frac{dB}{dt} &= 0.2\\
A &= \frac{9}{B}\\
∴ \frac{dA}{dB} &= \frac{-9}{B^2}\\
\textrm{When }A = 12, B &= \frac{9}{12} = \frac{3}{4}\\
∴ \frac{dA}{dB} &= \frac{-9}{(\frac{3}{4})^2}\\
&= -16\\
∴ \frac{dA}{dB} &= -16\times 0.2\\
&= \frac{-16}{5} \textrm{ ms}^{-1}
\end{align*}

 

Question 12b

(i)

\begin{align*}
x = \sqrt2\cos3t + \sqrt6\sin3t &\equiv R\cos(3t-\alpha)\\
&= R\cos3t\cos\alpha + R\sin3t\sin\alpha\\
R\cos\alpha &= \sqrt2\quad\quad(1)\\
R\sin\alpha &= \sqrt6\quad\quad(2)\\
(1)^2 + (2)^2: R^2\cos^2\alpha + R^2\sin^2\alpha &= 2 + 6\\
R^2 &= 8\\
R = 2\sqrt2\\
\frac{(1)}{(2)}: \tan\alpha &= \sqrt3\\
\alpha &= \frac{\pi}{3}\\
∴ x=2\sqrt2\cos(3t-\frac{\pi}{3})
\end{align*}

(ii)

\begin{align*}
\textrm{Particle is at rest at peaks of }x\\
∴ x=2\sqrt2, -2\sqrt2
\end{align*}

 

(iii)

\begin{equation*}
\frac{dx}{dt} = -6\sqrt2\sin(3t – \frac{\pi}{3}) = 3\sqrt2 \textrm{ or } -3\sqrt2 \textrm{ (as we are unsure which comes first)}
\end{equation*}
\begin{align*}
-6\sqrt2\sin(3t-\frac{\pi}{3}) &= 3 \sqrt2 & -6\sqrt2\sin(3t-\frac{\pi}{3}) &= -3\sqrt2\\
\sin(3t-\frac{\pi}{3}) &= -\frac{1}{2} & \sin(3t-\frac{\pi}{3}) &= \frac{1}{2}\\
3t-\frac{\pi}{3} &= -\frac{\pi}{6} & 3t-\frac{\pi}{3} &= \frac{\pi}{6}\\
3t &= \frac{\pi}{6} & 3t &= \frac{\pi}{2}\\
t &= \frac{\pi}{18} & t &= \frac{\pi}{6}\\
\end{align*}
\begin{equation*}
∴ t=\frac{\pi}{18}
\end{equation*}

 

Question 12c

\begin{align*}
x^2 &= 4ay\\
y &= \frac{x^2}{4a}\\
\frac{dy}{dx} &= \frac{x}{2a}\\
\textrm{at } x&=2ap\\
\frac{dy}{dx} &= p\\
∴ y-ap^2 &= p(x-2ap)\\
\textrm{at } x=0:\\
y – ap^2 &= -2ap^2\\
y &= -ap^2\\
∴ R(0,-ap^2)\\
SP &= PM \textrm{ (by the locus definition of the parabola)}\\
&= a + ap^2\\
&=a(p^2+1)\\
SR &= a + ap^2\\
&= a(p^2+1) \\
&= SP\\
∴ \Delta SPR &\textrm{ is isoceles}\\
∴ \angle SPR &= \angle SRP \textrm{ (base angles of isoceles triangle)}
\end{align*}

 

Question 12d

(i)

\begin{align*}
T &= 3 + Ae^{kt}\\
\frac{dT}{dt} &= kAe^{kt}\\
&= k(T-3)
\end{align*}

(ii)

\begin{align*}
T &= 3+Ae^{kt}, \textrm{ at }t=0, T=30\\
∴ 30 &= 3 + A\\
A &= 27\\
∴ T &= 3 + 27e^{kt}\\
\textrm{at } t&= 15, T= 28\\
∴ 28 &= 3 + 27e^{15k}\\
25 &= 27e^{15k}\\
e^{15k} &= \frac{25}{27}\\
15k &= \ln(\frac{25}{27})\\
k &= \frac{1}{15}\ln(\frac{25}{27})\\
\frac{dT}{dt} &= kAe^{kt}\\
&= \frac{1}{15}\ln(\frac{25}{27})(27)e^{\frac{1}{15}\ln(\frac{25}{27})t}\\
\textrm{at } t &= 60:\\
\frac{dT}{dt} &= \frac{27}{15}\ln(\frac{25}{27})e^{4\ln(\frac{25}{27})}\\
&= \frac{27}{15}\ln(\frac{25}{27})(\frac{25}{27})^4\\
&= (\frac{25}{27})^3\ln(\frac{25}{27})\,\,° \textrm{C}/\textrm{minute}
\end{align*}

 

Question 13a

\begin{align*}
\textrm{let } u &= \cos^2x\\
\frac{du}{dx} &= -2\cos x\sin x\\
du &= -\sin2x\,dx\\
x &= 0\rightarrow\frac{\pi}{4}\\
u &= 1 \rightarrow \frac{1}{2}\\
∴ \int_0^{\frac{\pi}{4}} \frac{\sin2x}{4+\cos^2x}\,dx &= \int_1^{\frac{1}{2}} \frac{-du}{4+u}\\
&= \int_{\frac{1}{2}}^1 \frac{du}{u+4}\\
&= \left[\ln(u+4)\right]_{\frac{1}{2}}^1\\
&= \ln(5) – \ln(\frac{9}{2})\\
&= \ln(\frac{10}{9})
\end{align*}

 

Question 13b

\begin{align*}
T_{r+1} &= {20\choose r}(5x)^r(2)^{20-r}\\
&= {20\choose r}5^r2^{20-r}x^r\\
\textrm{when } r&=k \textrm{ and } r=k+1, \textrm{ coefficients are equal}\\
∴ {20\choose k}5^k2^{20-k} &= {20\choose k+1}5^{k+1}2^{20-(k+1)}\\
\frac{20!}{k!(20-k)!}5^k2^{20-k} &= \frac{20!}{(k+1)!(20-(k+1))!}5^{k+1}2^{19-k}\\
\frac{\frac{20!}{k!(20-k)!}5^k2^{20-k}}{\frac{20!}{(k+1)!(20-(k+1))!}5^{k+1}2^{19-k}} &= 1\\
\frac{\frac{2}{20-k}}{\frac{5}{k+1}} &= 1\\
\frac{2k+2}{100-5k} &= 1\\
2k+2 &= 100-5k\\
7k &= 98\\
k &= 14
\end{align*}

 

Question 13c

(i)

\begin{align*}
a &= -2e^{-x}\\
\frac{vdv}{dx} &= -2e^{-x}\\
\int vdv &= \int -2e^{-x}dx\\
\frac{1}{2}v^2 &= 2e^{-x} + C\\
\textrm{when }x&=0, v=2\\
2 &= 2 +C\\
C &= 0\\
∴ \frac{1}{2}v^2 &= 2e^{-x}\\
v^2 &= 4e^{-x}\\
v &= \pm 2e^{-\frac{x}{2}}\\
\textrm{Initially, } v &= 2\\
\textrm{Thus } v &\textrm{ is positive}\\
∴ v &= 2e^{-\frac{x}{2}}
\end{align*}

 

(ii)

\begin{align*}
\frac{dx}{dt} &= 2e^{-\frac{x}{2}}\\
\int e^{\frac{x}{2}} dx &= \int 2dt\\
2e^{\frac{x}{2}} &= 2t + C\\
\textrm{when }t&=0,x=0\\
∴ C &= 2\\
∴ 2e^{\frac{x}{2}} &= 2t + 2\\
e^{\frac{x}{2}} &= t + 1\\
\frac{x}{2} &= \ln(t+1)\\
x &= 2\ln(t+1)
\end{align*}

 

Question 13d

(i)

\begin{align*}
\textrm{Trajectory lies on } y&=-x\\
\frac{18\sqrt3t-5t^2}{18t} &= -1\\
18\sqrt3t – 5t^2 &= -18t\\
-5t^2 + (18\sqrt3 + 18t)t &= 0\\
t(18\sqrt3 + 18 – 5t) &= 0\\
18\sqrt3 + 18 – 5t &= 0\\
5t &= 18\sqrt3 + 18\\
t &= \frac{18\sqrt3+ 18}{5}\\
∴ x &= 18(\frac{18\sqrt3+18}{5})\\
\cos45° &= \frac{18\frac{18\sqrt3+18}{5}}{OA}\\
OA &= \frac{18\frac{18\sqrt3+18}{5}}{\frac{1}{\sqrt2}}\\
OA &= \frac{18\sqrt2(18\sqrt3+18)}{5}\\
&= \frac{324(\sqrt6 + \sqrt2)}{5} \text{units}
\end{align*}

 

(ii)

\begin{align*}
\frac{dx}{dt} &= 18\\
\frac{dy}{dt} &= 18\sqrt3 -10t\\
\textrm{at } t&=\frac{18\sqrt3 + 18}{5}:\\
\frac{dy}{dt} &= 18\sqrt3 – 2(18\sqrt3 + 18)\\
&= -18\sqrt3 – 36
\end{align*}
\begin{gather*}
m_1 = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-18\sqrt{3}-36}{18} = -\sqrt{3}-2 \\
m_2 = \tan (-45^\circ) = -1 \\
∴ \tan\theta = \left|\frac{m_1 – m_2}{1+m_1m_2}\right| = \left|\frac{-\sqrt{3}-1}{\sqrt{3}+3}\right| = \frac{1}{\sqrt{3}} \\
∴ \theta = 30^\circ
\end{gather*}

 

Question 14a

\begin{align*}
\textrm{(1) Prove true for }n&=1:\\
LHS = 1(1!) = 1\\
RHS = 2! – 1 = 1 &= LHS\\
∴ \textrm{ true for }n &= 1\\
\textrm{(2) Assume true for }n &= k:\\
∴ 1(1!) + 2(2!) + … + k(k!) &= (k+1)! – 1\\
\textrm{(3) Prove true for } n&= k+1:\\
RTP: 1(1!) + 2(2!) + … + k(k!) + (k+1)(k+1)! &= (k+2)!-1\\
LHS&: (k+1)! -1 + (k+1)(k+1)! \textrm{ (from assumption)}\\
&= (k+1)!(1+k+1) – 1\\
&= (k+1)!(k+2) – 1\\
&= (k+2)! – 1 = RHS\\
∴ \textrm{ true for } n &= k+1\\
∴ \textrm{True by the principle} &\textrm{ of Mathematical Induction for} \ n \geq1
\end{align*}

 

Question 14b

(i)

\begin{align*}
\textrm{Consider the points of intersection of }y&=x^2 \textrm{ and } y = \frac{1}{x-k}:\\
x^2 &= \frac{1}{x-k}\\
x^2(x-k) &+ 1\\
x^3 – xk – 1 &= 0
\end{align*}

graph solution of function y= 1/x-k for question 14b part 1 blog-2019-maths-ext-1-exam-paper-solutions-question-14bi-graph-2

\begin{align*}
\textrm{Graphically, } y=x^2 \textrm{ and } y &= \frac{1}{x-k} \textrm{ only intersect once}\\
\textrm{Hence } x^3 – xk – 1 &= 0 \textrm{ only has one real zero}
\end{align*}

 

(ii)

\begin{align*}
f(x) &= x^3 – kx^2 – 1\\
f'(x) &= 3x^2 – 2kx \\
x_2 &= x_1 – \frac{f(x_1)}{f'(x_1)}\\
&= k \ – \frac{k^3-k^3-1}{3k^2-2k^2}\\
&= k \ – \frac{-1}{k^2}\\
&= k + \frac{1}{k^2} \textrm{ as required}
\end{align*}

 

(iii)

\begin{align*}
f(x) &= x^3 – kx^2 – 1\\
f(x_1) = f(k) &= k^3-k^3-1 = -1\\
f(x_2) = f(k+\frac{1}{k^2}) &= (k+\frac{1}{k^2})^3 – k(k+\frac{1}{k^2})^2 – 1\\
&= k^3 + 3 + \frac{3}{k^3} + \frac{1}{k^6} – k(k^2 + \frac{2}{k} + \frac{1}{k^4}) – 1\\
&= k^3 + 3 + \frac{3}{k^3} + \frac{1}{k^6} – k^3 – 2 – \frac{1}{k^3} – 1\\
&= \frac{2}{k^3} + \frac{1}{k^6} > 0 \textrm{ (as k is positive)}\\
∴ f(x_1) < 0 &\textrm{ and } f(x_2) > 0\\
\textrm{By the Intermediate Value Theorem, }&\textrm{there exists a root between } x_1 \textrm{ and } x_2.\\
\textrm{But there is only one solution, }&\textrm{which must be } \alpha\\
\ ∴ x_1 < \alpha < x_2
\end{align*}

 

Question 14c

(i)

\begin{align*}
\textrm{Gradient of tangents at } x = x_0 \textrm{ of}:\\
y=\sin x \textrm{ is } \cos(x_0)\\
y=\sin(x-\alpha)+k \textrm{ is } \cos(x_0 – \alpha)\\
\textrm{The tangent is common, so:}\\
\cos(x_0) = \cos(x_0-\alpha) \textrm{ as required}
\end{align*}

 

(ii)

\begin{align*}
\textrm{From part (i), we have } \cos(x_0) = \cos(x_0 – \alpha)\\
x_0 \textrm{ is in the 1st quadrant, thus } \cos(x_0) > 0 \textrm{ and hence } \cos(x_0 \ -\ \alpha) > 0\\
\textrm{Hence } x_0 \ – \ \alpha \textrm{ must be in the 1st or 4th quadrant}
\end{align*}
\begin{align*}
&\textrm{If 1st quadrant:} & &\textrm{If 4th quadrant:}\\
x_0 &= x_0 – \alpha & x_0 &= 2\pi-(x_0-\alpha)\\
\alpha &= 0 & \sin(x_0) &= \sin(2\pi – (x_0 \ – \ \alpha))\\
&\textrm{Clearly false} & \sin(x_0) &= -\sin(x_0-\alpha) \textrm{ as required}\\
\end{align*}

 

(iii)

\begin{align*}
\sin(x_0-\alpha) + k &= \sin(x_0)\\
k &= \sin(x_0) – \sin(x_0-\alpha)\\
&= 2\sin(x_0) \textrm{ (using the result from part (ii))}\\
\textrm{We can divide }&\textrm{the results in parts (i) and (ii):}\\
\frac{\sin(x_0)}{\cos(x_0)} &= \frac{-\sin(x_0-\alpha)}{\cos(x_0-\alpha)}\\
\tan(x_0) &= -\tan(x_0-\alpha)\\
\tan(x_0) &= \tan(\alpha-x_0)\\
x_0 \textrm{ is in }&\textrm{the 1st quadrant (given)}\\
x_0 – \alpha \textrm{ is }&\textrm{in the 4th quadrant (from part (ii))}\\
∴ -(x_0-\alpha) &= \alpha-x_0 \textrm{ is in the 1st quadrant}\\
∴ \textrm{ both } x_0 &\textrm{ and } \alpha-x_0 \textrm{ are in the 1st quadrant}\\
\textrm{ and } \tan(x_0) &= \tan(\alpha-x_0)\\
∴ x_0 &= \alpha-x_0\\
x_0 &= \frac{\alpha}{2}\\
∴ k &= 2\sin(\frac{\alpha}{2})
\end{align*}

 

Want to know your ATAR?

Use the calculator to explore the HSC Marks you need to achieve your ATAR Goal.

 

Written by Matrix Maths Team

The Matrix Maths Team are tutors and teachers with a passion for Mathematics and a dedication to seeing Matrix Students achieving their academic goals.

 

© Matrix Education and www.matrix.edu.au, 2018. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.

Get free study tips and resources delivered to your inbox.

Join 75,893 students who already have a head start.

Our website uses cookies to provide you with a better browsing experience. If you continue to use this site, you consent to our use of cookies. Read our cookies statement.

OK, I understand