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In this post, we reveal the solutions to the 2019 HSC Maths Extension 1 Exam paper.
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Our Mathematics team have been hard at work putting together the 2019 HSC Maths Extension 1 Exam Paper solutions for you. Here they are!
Have a look through these worked solutions and see how you did!
| Question | Answer | Solution |
| 1 | A\(\) | \begin{align*} 4-x &> 0\\ 4 &> x\\ x &< 4 \end{align*} |
| 2 | D | \begin{align*} (2x)^2 &= x(9+x)\\ 4x^2 &= x(9+x) \end{align*} |
| 3 | C | \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*} |
| 4 | B | \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*} |
| 5 | A | \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*} |
| 6 | B | \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*} |
| 7 | C | \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*} |
| 8 | A | \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*} |
| 9 | D | \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*} |
| 10 | A | \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*} |
\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*}
\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*}
\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*}
\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*}
\begin{align*}
\int 2\sin^2 4x\,dx &= \int (1-\cos 8x)\, dx\\
&= x – \frac{1}{8}\sin 8x + C
\end{align*}
(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*}
\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*}
(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*}
\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*}
(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*}
\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*}
\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*}
(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*}
(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*}
\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*}
(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*}
\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*}
(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*}
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