2022 HSC Maths Advanced Exam Solutions

\(A=98+62=160 \\ B=\frac{192}{200}\times 100=96\)

\[\begin{align} \int_{0}^{2} \sqrt{1+x^2} \ dx &\approx \frac{b-a}{2\times n} [f(a)+f(b)+2(f(x_1)+\ldots+f(x_{n-1}))] \\ &= \frac{2-0}{2\times 2}[ \sqrt{1+0}+\sqrt{1+2^2} +2(\sqrt{1+1}) ] \\ &= 3.03 \ \text{(2 d.p.)} \end{align}\]

The value of \(k\) represents the amplitude of our sinusoidal function, which is evidently \(4\). The value of \(a\) is linked to the period \(T\) of the wave through \(a=\frac{2\pi}{T}\). From the graph we see that \(T=6\pi\), so \(a=\frac{2\pi}{6\pi}\). Therefore, \(a=\frac{1}{3}, \ k=4\).

\[ \begin{align} \mathbb{P}(\text{rolls a 5})&=\mathbb{P}(\text{rolls a 5 through special die})+\mathbb{P}(\text{rolls a 5 through ordinary die}) \\ &=\frac{1}{3}\times \frac{3}{6}+\frac{2}{3}\times\frac{1}{6}=\frac{5}{18}. \end{align} \]

\[ \begin{align} \mathbb{P}(\text{special die | rolls a 5})&=\frac{\mathbb{P}(\text{special die }\cap\text{ rolls a 5})}{\mathbb{P}(\text{rolls a 5})} \\ &=\frac{\frac{1}{3}\times\frac{3}{6}}{\frac{5}{18}} \\ &= \frac{3}{5}. \end{align} \]

The area enclosed is given by \( \int_{a}^{b} y_{furthest} – y_{closest} \ dx\), where \(y_{furthest}=2x+3\) is the function furthest from the \(x\)-axis in our region, \(y_{closest} = x^2\) is the closest function, and \(a=-1, \ b=3\) are the points of intersection between the two curves. So, \[ \begin{align} \text{Area} &= \int_{-1}^{3} \ 2x+3-x^2 \ dx \\ &= \left[x^2+3x-\frac{x^3}{3}\right]_{-1}^{3} \\ &= \left[3^2+3(3)-\frac{3^3}{3}\right] – \left[(-1)^2+3(-1)-\frac{(-1)^3}{3}\right] \\ &= \frac{32}{3} \ u^2. \end{align} \]

The house of cards represents an arithmetic progression with initial value \(a=3\) at the first row, and common difference \(d=3\). So, for the house of cards with \(n=12\) rows, we will have \(S_{12} = \frac{12}{2}[2\times 3+3\times(12-1)] = 234\) cards.

We wish to solve \(S_n = \frac{n}{2}[2\times 3+3\times(n-1)]=828\). That is, \(6n+3n^2-3n=1656 \Rightarrow n^2+n-552=0\). We can solve this by using the quadratic formula \(n=\frac{-1\pm \sqrt{1^2-4(1)(-552)}}{2(1)}=23\text{ or } -24\). Since we cannot have negative rows, we conclude this house of cards has 23 rows.

\(y’=4\times2x\times(x^2+1)^3=8x(x^2+1)^3\)

\(\int x(x^2+1)^3 \ dx= \frac{1}{8} \int 8x(x^2+1)^3 \ dx = \frac{1}{8} \ (x^2+1)^4 + C \)

We start by noting that since \(f\) was translated down by 5 units, we can write \(g(x)=(3x^2-12x+12) – 5\). Next, we factorise the brackets to give \(g(x)=3(x^2-4x+4)-5=3(x-2)^2-5=kf(x-m)-5\), so \(k=3\) and \(m=2\).

The initial bacteria population is \(N(0)=200e^{0.013\times 0}=200\).

The bacteria population after 24 hours is \(N(24)=200e^{0.013\times 24}\approx 273\) (nearest whole).

We wish to find \(\frac{dN}{dt} \vert_{t=24} \). Note \(\frac{dN}{dt} = 0.013\times (200e^{0.013t})=0.013N(t)\), so \[ \frac{dN}{dt} \vert_{t=24} = 0.013\times N(24)=0.013\times200e^{0.013\times 24}\approx 3.552 \text{ (3 d.p.)} \] Rate of increase in the number of bacteria after 24 hours is approximately 3.552 bacteria per hour.

\(FV_{\text{Opt1, 10y}} = 40000(1+\frac{0.012}{12})^{12\times 10}\approx $45097.17\) (nearest cent).

\(FV_{\text{Opt2, 10y}} = 1000 \times 45.05630 \approx $45 056.30\), using the intersection value in the future value table corresponding to interest rate \(\frac{0.024}{4}=0.006\) and number of periods \(10\times 4=40\), as our annuity is compounded quaterly. \( \\ \) So the difference in future values is \(FV_{\text{Opt1, 10y}} – FV_{\text{Opt2, 10y}} \approx $40.87\) (nearest cent), with Option 1 giving the greater future value.

First, we check for local extrema, as these are potential candidates for the global extrema. By differentiating and solving for zero, we have \(y’=3x^2-12x=0 \Rightarrow x(x-4)=0 \Rightarrow x=0,\ 4\). Use the second derivative test to determine their nature, \(y”=6x-12 \Rightarrow y”(0)=-12, \ y”(4)=12. \) Since \(y”(0)<0\), we have concave down curvature, so a local maximum, and \(y”(4)>0\), we have concave up curvature here and so a local minimum. \(\\\) We now check the function’s values at the local extrema, and also at the end points of its domain to find the global extrema: \[ \begin{align} y(-1) &= 1 \\ y(0) &= 8 \\ y(4) &= -24 \\ y(7) &= 57. \end{align} \] As our function is a polynomial, and therefore continuous on its domain without any cusps, asymptotes or discontinuities, we conclude the global minimum is \(-24\), obtained at the stationary point \(x=4\) and the global maximum is \(57\) obtained at the right end point of the domain \(x=7\). The graph of this function is given below.

Low tide will occur when \(\cos\left(\frac{4\pi}{25}t\right)=1\) and similarly high tide will occur when \(\cos\left(\frac{4\pi}{25}t\right)=-1\), so, \(d_{low}=1.3-0.6=0.7\) metres and \(d_{high} = 1.3+0.6=1.9\) metres.

The period of this sinusoidal function will give us the interval between successive low tides. This is given by \(T=\frac{2\pi}{n}\) where \(n=\frac{4\pi}{25}\) is the coefficient of \(t\) in the function. So \(T=\frac{2\pi}{\frac{4\pi}{25}}=12.5\), meaning the time interval between successive low tides will be 12.5 hours.

We wish to solve \(d(t) \geq 1\) for \(0 \leq t\leq 12\) as a low tide occurs at \(t=0\) and another at \(t=12.5\), so analysing the duration of the depth being at least 1 metre in this interval will generalise to the duration between any successive low tides. \[ \begin{align} 1.3-0.6\cos\left(\frac{4\pi}{25}t\right) &\geq 1 \\ 0.6\cos\left(\frac{4\pi}{25}t\right) &\leq 0.3 \\ \cos\left(\frac{4\pi}{25}t\right) & \leq \frac{1}{2} \\ \end{align} \] \[ \begin{align} \Rightarrow \frac{\pi}{3} \leq \frac{4\pi}{25} &t \leq 2\pi-\frac{\pi}{3} \\ \frac{25}{12} \leq &t \leq \frac{125}{12} \end{align} \] Consequently, the depth of water will be at least 1 metres between successive low tides for \(\frac{125}{12}-\frac{25}{12}=8\frac{1}{3}\) hours.

Firstly, the correlation coefficient at \(r=0.4564\) suggests a moderate positive correlation between the age of a character and the age of the actor. \(\\\) The scatterplot indicates a gradually increasing age count of actors, compared to the age of character, predicted by the positive correlation coefficient. However, as the magnitude of the coefficient is closer to zero, and given the variance in age of actor at each character age, arguing an entirely linear relationship between the two variables is not highly supported by the data. \(\\\) Graphically, we can also infer that the character age of 15 has the widest range of ages for the actor, as the range in the y-variable here is greatest. \(\\\) Note that given Jo’s research question, they did not collect data to sufficiently represent all teenage characters, as their range is between 14 and 17, whereas teenagers vary from 13-19. To this end, while interpolation of information using the line of best fit may be useful, extrapolating will not give a reliable insight into the relationship (if any) between ages of characters larger than 17 or below 14, and the age of their actors, further restricted by the fact that ages can’t be negative and that beyond teenage characters, into adulthood, we may require a different model to represent the relationship with possible actor ages.

First note that \(f'(x)=2\cos(2x)\) and \(f”(x)=-4\sin(2x)\). So, \[ \begin{align} 2\cos(2x)=-\sqrt{3} \Rightarrow 2x &= \frac{5\pi}{6}, \ \frac{7\pi}{6} \\ x &= \frac{5\pi}{12}, \ \frac{7\pi}{12}, \\ -4\sin(2x)=2 \Rightarrow 2x &= \frac{7\pi}{6}, \ \frac{11\pi}{6} \\ x &= \frac{7\pi}{12}, \ \frac{11\pi}{12}. \end{align} \] Therefore, the common solution is \(x=\frac{7\pi}{12}\).

We’re given that \(\mu=840\) and \(\sigma = 80\). We’re also told that \(\mathbb{P}(X<860)\approx 0.6\). Note that \(860=\mu+20\), so by symmetry of the normal curve about the mean, we also know that \(\mathbb{P}(X>\mu-20)\approx 0.6 \Rightarrow \mathbb{P}(X>820)\approx 0.6\), this is illustrated below as both coloured areas are equal.

Now notice that \(920=\mu+\sigma\), so using the empirical rule for normal distributions, we have that \(\mathbb{P}(X>920) = \frac{1}{2}(1-\mathbb{P}(\mu-\sigma<X<\mu+\sigma))\approx \frac{1}{2}(1-0.68)=0.16\). \(\\\) Consequently \(\mathbb{P}(820<X<920) = \mathbb{P}(X>820)-\mathbb{P}(X>920)\approx 0.6-0.16=0.44\). Therefore, approximately 44% of batteries have a life span between 820 and 920 hours.

\(f”(x)=-2e^{-2x}-2(-2xe^{-2x}+e^{-2x})=4xe^{-2x}-4e^{-2x}=4(x-1)e^{-2x}.\)

To find stationary points, investigate \(f'(x)=0\). That is, \[e^{-2x}(1-2x)=0 \Rightarrow (1-2x)=0 \Rightarrow x=\frac{1}{2},\] as \(e^{-2x}\neq 0\). To determine the nature of this point, we use the second derivative test, \[f”\left(\frac{1}{2}\right)=4\left(\frac{1}{2}-1\right)e^{-2\times\frac{1}{2}} < 0,\] so we have concave down curvature here, therefore a local maximum stationary point at \(x=\frac{1}{2}\). To find the point itself, substitute to \(f(x)\), giving us \(\left(\frac{1}{2}, \ \frac{e^{-1}}{2}\right)\) as the point.

To find the inflection, solve \(f”(x)=0\), meaning \(4(x-1)e^{-2x}=0 \Rightarrow x=1\). Note, this does not prove that we have a point of inflection here, we must check for a change in concavity on both sides of \(x=1\). For values \(x<1\), we have that \(f”(x)<0\) since \(x-1<0\), \(e^{-2x}\) is always positive, and similarly for values \(x>1\) we have \(f”(x)>0\), therefore the concavity changes from concave down to concave up at \(x=1\), therefore we have a point of inflection here. \(\\\) The graph below sketches the curve showing stationary points, inflections, intercepts with the axes and a horizontal asymptote.

\(\text{Shaded area}=\text{Area of sector}-\text{Area of triangle} \\\) \(\tan\theta = \frac{1}{1} \Rightarrow \theta = \frac{\pi}{4} \\\) \(\text{Circle radius}=\sqrt{2} \\\) \(\text{Area of sector}=\frac{1}{2}r^2\theta = \frac{\pi}{4} \\\) \(\text{Area of triangle}=\frac{1}{2}bh = \frac{1}{2}\times 1 \times 1 = \frac{1}{2} \\\) \(\therefore \text{Shaded area}=\frac{\pi}{4} – \frac{1}{2} = \frac{\pi – 2}{4} \ u^2 \)

The points \((0, \ 0)\) and \((1, \ 1)\) lie on the hyperbola. So, we have two equations: \[ \begin{align} 1 &= \frac{a}{b-1}-1 \\ 2b-2 &= a, \ \ \ \ \ \ \ \ \ \ (1) \\ 0 &= \frac{a}{b-0}-1 \\ b &= a. \ \ \ \ \ \ \ \ \ \ (2) \end{align} \] Substituting Equation (1) into (2) we have that \(2b-2=b \Rightarrow b=2\) and by Equation (2), \(a=b=2\).

\(\text{Total shaded area} = \displaystyle \int_{0}^{1} \frac{2}{2-x} – 1 \ dx + \text{Shaded area}_{\text{ part (a)}} \\\) \(\displaystyle \int_{0}^{1} \frac{2}{2-x} – 1 \ dx = \left[-2\ln|2-x|-x\right]_{0}^1 = 2\ln (2)-1 \\\) \(\therefore \text{Total shaded area} = 2\ln(2)-1+\frac{\pi-2}{4}=\frac{8\ln(2)+\pi-6}{4} \ u^2 \\\)

Area of the first rectangle is \(a=1\), while each successive rectangle shrinks in area by a factor of \(r=\frac{1}{2}\). Since we have \(|r|<1\), the sequence of partial sums of this geometric series in fact converges when we take the limit as \(n \to \infty\). This is given by \[ \lim_{n \to \infty} S_n = \frac{a}{1-r} = \frac{1}{1-\frac{1}{2}} = 2.\]

\(\displaystyle \int_{0}^{4} 2^{-x} \ dx = -\left[ \frac{2^{-x}}{\ln2} \right]_0^4 = -\left(\frac{1}{16\ln 2}\right)+\left(\frac{1}{\ln 2}\right) = \frac{15}{16\ln 2}. \)

From part (a), we see that the rectangles are an over-estimation of the area between the curve and the \(x\)-axis. Therefore, the limiting sum of the rectangles’ areas, \(2\), is greater than the integral \(\int_{0}^{\infty} 2^{-x} \ dx\), which in turn is greater than \(\int_{0}^{4} 2^{-x} \ dx \). So, \[ \begin{align} \frac{15}{16\ln 2} &< 2 \\ 15 &< 32\ln 2 \\ e^{15} &< e^{\ln\left(2^{32}\right)} \ \text{(as the exp function is strictly increasing, the inequality is preserved)} \\ \Rightarrow e^{15} &< 2^{32}. \end{align} \]

As this is a cumulative distribution function (cdf), we expect its corresponding pdf to satisfy \(f(x)=F'(x)=\frac{1}{kx}\) and that \(\int_{1}^{e^3} f(x) \ dx = 1\). Using these conditions, we have that \[ \begin{align} \int_{1}^{e^3} \frac{1}{kx} \ dx &= 1 \\ \Rightarrow \frac{1}{k} \left[\ln|x| \right]_{1}^{e^3} &= 1 \\ \Rightarrow k &= \ln e^3 = 3. \end{align} \]

Note that \(\mathbb{P}(X<c)=2\mathbb{P}(X>c) \Rightarrow \mathbb{P}(X<c)=2(1-\mathbb{P}(X<c))\), so \(\mathbb{P}(X<c)=\frac{2}{3}\). So we are trying to solve \(\mathbb{P}(X<c)=F(c)=\frac{1}{3}\ln c = \frac{2}{3}.\) \(\\\) Therefore, \(\ln c = 2 \Rightarrow c=e^2\).

Since \(X, \ P, \ Y\) all lie on the same line with the same gradient, \(m_{YP} = m_{PX}\). That is, \[ \begin{align} \frac{y-2}{0-1} &= \frac{0-2}{x-1} \\ \Rightarrow y-2 &= \frac{2}{x-1} \\ \Rightarrow y &= \frac{2(x-1)}{x-1}+\frac{2}{x-1} \\ \Rightarrow y &= \frac{2x}{x-1}. \end{align} \]

The area of the triangle is given by \(A=\frac{1}{2}xy = \frac{x^2}{x-1} \) using part (a). So, we wish to find the global minimum on \(A(x)\) with the restriction that \(x>1\). We begin by setting \(A'(x)=0\) to find the stationary points, then determine their nature using a table of values: \[ \begin{align} A'(x) &= \frac{2x\times (x-1)-1\times x^2}{(x-1)^2} \\ &= \frac{x^2-2x}{(x-1)^2} \\ &= \frac{x(x-2)}{(x-1)^2} \\ &= 0, \end{align} \] implying that \(x(x-2)=0 \Rightarrow x=0, \ 2\), as the denominator cannot be zero. Furthermore, we’re interested in \(x>1\), so our only stationary point candidate is \(x=2\). \[ \begin{array}{|c|c|c|c|} \hline x & 1.9 & 2 & 2.1 \\ \hline A'(x) & -0.23 & 0 & 0.17 \\ \hline \text{Gradient} & \searrow & \rightarrow & \nearrow \\ \hline \end{array} \] The table of values confirms that we in fact have a local minimum at \(x=2\). Since the function restricted to \(x>1\) only has a single turning point, which is a local minimum, it in fact must be the global minimum. So the minimum area of the triangle \(XOY\) is \(A(2)=\frac{2^2}{2-1} = 4 \ u^2\).

We know that \(n=180\) and that \(P=2\times 10^5\), and that \(A_{180}\) must be zero, that is we no longer owe the lender any money after 180 months. So by rearranging the given equation, we have \[ \begin{align} 0&=2\times 10^5(1.0025)^{180} – M(1+(1.0025)^1 + (1.0025)^2+\ldots + (1.0025)^{179})\\ \Rightarrow M &= \frac{2\times 10^5(1.0025)^{180}}{1+(1.0025)^1 + (1.0025)^2+\ldots + (1.0025)^{179}} \\ &= \frac{2\times 10^5(1.0025)^{180}}{\frac{1.0025^{180}-1}{1.0025-1}} \ \text{(using the partial sum of a GP with }a=1\text{, }r=1.0025\text{)} \\ &\approx $1381.16 \ \text{(nearest cent)} \end{align} \]

After 100 repayments, we currently have an outstanding amount of \($100 032\). We can take this value as our new principal \(P=100032\), taking \(M=1381.16\), then after some unknown \(k\) number of months, we will have paid this principal off. We also have an interest rate of 0.35% per month, so we will replace the \(1.0025\) terms with \(1.0035\) in the equation given by the question. That is, \[ \begin{align} A_k &= 100032(1.0035)^{k} – 1381.16(1+(1.0035)^1 +\ldots + (1.0035)^{k-1}) \\ \Rightarrow 0&=100032(1.0035)^{k} – 1381.16\frac{1.0035^{k}-1}{1.0035-1} \\ & \text{(using the partial sum of a GP with }a=1\text{, }r=1.0035\text{)} \\ \Rightarrow -\frac{1381.16}{0.0035} &= (1.0035)^{k}(100032 – \frac{1381.16}{0.0035}) \\ \Rightarrow (1.0035)^k &= \frac{-\frac{1381.16}{0.0035}}{(100032 – \frac{1381.16}{0.0035})} \\ \Rightarrow k &= \frac{\ln\left( \frac{-\frac{1381.16}{0.0035}}{(100032 – \frac{1381.16}{0.0035})} \right)}{\ln 1.0035} \\ &\approx 83.674 \ \text{(3 d.p.)} \end{align} \] So Jane will need to make 83 more full monthly payments of $1381.16.

We found in part (b) that \(k\) is slightly larger than 83, but less than 84, so we deduce that even after the 183rd monthly payment of $1381.16, there will be some amount owing left over. To find this amount, we need to calculate \(A_{83} = 100032(1.0035)^{83} – 1381.16(1+(1.0035)^1 +\ldots + (1.0035)^{82})\) using the model from part (b). That is, \(A_{83} = 100032(1.0035)^{83} – 1381.16 \frac{1.0035^{83}-1}{1.0035-1} \). So Jane will need to pay \(A_{84} = A_{83} \times 1.0035 = \left(100032(1.0035)^{83} – 1381.16 \frac{1.0035^{83}-1}{1.0035-1}\right)\times 1.0035 \approx $931.54 \) (nearest cent) in the final payment, on the 184th month to pay off the loan entirely.