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The 2021 HSC Mathematics Extension (3 Unit) exam paper solutions are here!
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In this post, we will work our way through the 2021 HSC Maths Extension 1 (3 Unit) paper and give you the solutions, written by our Head of Mathematics Oak Ukrit and his team. (Doing practice papers? See the solutions for the 2020 HSC Maths Ext 1 Exam here.)
Read on to see how to answer all of the 2021 questions.
| Question | Answer | Solution |
| 1 | C | \begin{align*} \overrightarrow{PQ} &= \overrightarrow{OQ} – \overrightarrow{OP} \\ &= \binom{2}{5} – \binom{-3}{1}\\ &= \binom{5}{4} \\ \end{align*} |
| 2 | B | \begin{align*} \text{cos}6x &= 1- 2\text{sin}^23x \\ \text{sin}^2 3x &= \frac{1-\text{cos}6x}{2} \\ ∴ \int \text{sin}^2 3x \ dx &= \int \frac{1 – \text{cos}6x}{2} \ dx \\ \end{align*} |
| 3 | D | \begin{align*} \text{Use the Remainder Theorem:} \\ P(-2) &= -(-2)^3 -2(-2)^2-3(-2)+8 \\ &= 14 \\ \end{align*} |
| 4 | A | \begin{align*} \int y \ dy &= \int x \ dx \\ 2 \int y \ dy &= 2 \int x \ dx \\ y^2 &= x^2 + C \\ \end{align*} |
| 5 | B | \begin{align*} \overrightarrow{OA} · \overrightarrow{OB} =|\overrightarrow{OA}||\overrightarrow{OB}| \text{cos} \theta &< 0 \\ ∴ \text{cos} \theta &< 0 \\ ∴ \theta & \text{ is obtuse}\\ \end{align*} |
| 6 | A | \begin{align*} r &= 1-P(X=0) \\ &= 1- \binom{10}{0} p^0(1-p)^{10} \\ &= 1 – 0.1^{10} \\ &= 0.9999999999 > 0.9 \\ \end{align*} |
| 7 | D | \begin{align*} \text{Let } f(x) = -a \text{sin}x + b \text{cos} x &= R \text{cos}(x+ \alpha). \text{ Then:} \\ -a \text{sin} x +b \text{cos} x &= R\text{cos}(x+ \alpha) \\ &= R \text{cos}(\alpha) \text{cos}(x) – R \text{sin}(\alpha)\text{sin}(x) \\ -a \text{sin} x + b \text{cos} x &= -R \text{sin} (\alpha) \text{sin} (x) + R \text{cos} (\alpha) \text{cos} (x) \\ \end{align*}Now, \(R \text{cos} \alpha = b >0\), \(R \text{sin} \alpha = a >0\). The signs of the trigonometric ratios indicate that our solution is in the first quadrant, i.e. \(0< \alpha < \frac{pi}{2}\).Therefore, \(f(x)\) is a \(\text{cos}(x)\) graph shifted left by \(\alpha\).Alternatively, consider the gradient and the sign of \(f(x)\) at \(x=0\).
Therefore, the graph of \(f(x)\) at \(x=0\) should be positive and decreasing; this matches (D). |
| 8 | C | In the domain \(– \frac{ \pi}{2} ≤ t ≤ \frac{ \pi}{2} \), \( -1 ≤ \text{sin}(t) ≤ 1 \) and \( 0 ≤ \text{cos}(t) ≤ 1 \). We note that \(y\) is always less than the centre 2; so we want \(y = 2 – \text{cos} (t)\), which is (C). |
| 9 | A | \begin{align*} \frac{dy}{dx} &= \frac{ \text{cos}x}{ \sqrt{1 – \text{sin}^x}} \\ &= \frac{ \text{cos}x}{|\text{cos} x|} \\ &= \begin{cases} 1 & \text{when } \text{cos}x > 0\\ -1 & \text{when } \text{cos}x < 0\\ \end{cases} \end{align*}As the domain is \(x∈ \bf{R}\), the solution must be (A).Alternatively, we note that the inner function \(\text{sin} (x) \) is periodic and has a domain of \(x∈ \bf{R}\), eliminating (C) and (D). Next, neither the range of \(\text{sin}\) nor \(\text{sin}^{-1}\) have any discontinuities, so option (A) fits best. |
| 10 | C | First note that \(\frac{3543}{15} = 236.2\). To minimise the votes of the winning candidate, we could have that each candidate receives 236 votes each, with 3 votes left to distribute. For a winner, they must have 2 of these 3 remaining votes, so \(236 + 2 = 238\). |
| \((\underset{\text{~}}i+6\underset{\text{~}}j)+(2\underset{\text{~}}i-7\underset{\text{~}}j)=3\underset{\text{~}}i-\underset{\text{~}}j\) |
Question 11b
| \begin{align*} (2a-b)^4 &= (2a)^4-4(2a)^3b+6(2a)^2b^2-4(2a)b^3+b^4 \\ &= 16a^4-32a^3b+24a^2b^2-8ab^3+b^4 \\ \end{align*} |
Question 11c
| Let \(u=x+1\). \(\frac{du}{dx}=1\), \(du=dx\).\begin{align*} \int x \sqrt{x+1} \ dx &= \int (u-1) \sqrt{u} \ du \\ &= \int u^{\frac{3}{2}} – u^ \frac{1}{2} \ du \\ &= \frac{2}{5}u^{\frac{5}{2}} – \frac{2}{3} u^{\frac{3}{2}} + C \\ &= \frac{2}{5} (x+1)^2 \sqrt{x+1} – \frac{2}{3}(x+1) \sqrt{x+1} + C \\ \end{align*} |
| We choose the 5 men from the 10 first, then choose 3 women from the 8 to obtain \(\binom{10}{5} \times \binom{8}{3} = 14112\) |
| We have that \( \frac{dr}{dt}= 0.2 \ \text{mm/s}\), so we hope to obtain the change of the volume through the chain rule:\(\frac{dV}{dt} = \frac{dr}{dt} \times \frac{dV}{dr} = 0.2 \times \frac{dV}{dr}\) It remains to find \(\frac{dV}{dr}\) by noting that \(V= \frac{4}{3} \pi r^3\) and so \( \frac{dV}{dr} = 4 \pi r^2 = \frac{36}{25} \pi\) at \(r=0.6 \ \text{mm}\). Substituting into the change rule above, we arrive at: \( \frac{dV}{dt} ≈ 0.9 \text{mm}^3 \text{/s}\) |
| This is a standard integral: \begin{align*} |
| We factorise first:\begin{align*} 2\sin^3x + 2\sin^2x-\sin x-1 &=0 \\ 2\sin^2x(\sin x+1) – (\sin x + 1) &= 0 \\ (\sin x+1)(2 \sin^2 x-1) &= 0 \end{align*}to obtain that\begin{equation*} \sin x = -1 \qquad \mathrm{or} \qquad \sin x = \pm \frac{1}{\sqrt{2}}. \end{equation*}The first solution gives \(x=\frac{3\pi}{2}\) only, whereas the second solution consists of the four quadrant solutions corresponding to the related angle of \(\frac{\pi}{4}\): \(x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4}\) and \(\frac{7\pi}{4}\). In total, we have five solutions for \(x\):\begin{equation*} x = \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{3\pi}{2} \ \mathrm{and} \ \frac{7\pi}{4}. \end{equation*} |
| We express the quantity \(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\delta}+\frac{1}{\gamma}\) in terms of the various sums and products of roots:\begin{align*} \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\delta}+\frac{1}{\gamma} &= \frac{\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta}{\alpha \beta \gamma \delta} \\ &= \frac{-d/a}{e/a} \\ &= \frac{-(-3/1)}{6/1} \\ &= \frac{1}{2}. \end{align*}where the original quartic is of the form \(ax^4+bx^3+cx^2+dx+e\) with \(a=1,b=c=0,d=-3,e=6\). |
Various solutions will likely be permitted – as long as the curve generally conforms to the direction field, and goes through the point P. Two alternatives are presented below.
| We solve the differential equation by separating variables and integrating both sides. \begin{align*} Initially (\(t=0\)), the water is at \(5 ° \textrm{C}\), so \(T-25 < 0\) and we flip the terms in the absolute value. Substituting this point yields: \begin{align*} Now we can find \(k\). \begin{align*} At \(t=8\), we are given that \(T=10\). Hence \begin{equation*} We keep writing \(k\) until the last step of the calculation in order to avoid propagating round-off errors. Now we determine the solution to the differential equation. \begin{align*} Substituting \(k\) in and solving for \(T=20\), we have: \begin{align*} A more efficient method: To find the time at which \(T=20\), we may return to the earlier step \((*)\): \begin{equation*} Therefore to the nearest minute. |
We plot the function \(T(t)\) below, showing the horizontal asymptote.
| If \(n=1\), then the right hand side is\begin{equation*} \frac{1}{4} – \frac{1}{2\times 2\times 3} = \frac{2}{3\times 4} = \frac{1}{6} \end{equation*}while the left hand side is simply \(\frac{1}{1\times 2\times 3} = \frac{1}{6}\). Hence the base case is true.Now assume the statement holds for some integer \(n\ge 1\). We need to prove that\begin{equation*} \frac{1}{1\times 2\times 3} + \frac{1}{2\times 3\times 4} + \ldots + \frac{1}{n(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} = \frac{1}{4} – \frac{1}{2(n+2)(n+3)} \quad (\star) \end{equation*}Starting with the left hand side, we obtain\begin{multline*} \frac{1}{1\times 2\times 3} + \frac{1}{2\times 3\times 4} + \ldots + \frac{1}{n(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} \\ = \frac{1}{4} – \frac{1}{2(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} \end{multline*}by the induction assumption. Now we calculate \begin{align*} Therefore equation (\(\star\)) holds, which completes the induction step. |
We may first consider the basic parabola defined by \(y = (1-\frac{x}{2})^2\) for \(x\) in the domain \((-\infty, 2]\), which looks like this:
Then \(y = 4-(1-\frac{x}{2})^2\) is obtained by reflecting the above graph about the \(x\)-axis, and then shifting the vertex of the parabola up by 4 units:
| We consider the function \(f(x) = 4-(1-\frac{x}{2})^2\) with domain \((-\infty,2]\) and range \((-\infty, 4]\). Thus the domain of the inverse function \(f^{-1}(x)\) is \((-\infty, 4]\). Now we can determine \(f^{-1}\) by first writing \(x = 4-(1-\frac{y}{2})^2\), and then solve for \(y\): \begin{align*} Finally we need to determine the correct choice of sign. Since \(f\) passes through the point \((x,y) = (-2,0)\), it follows that \(f^{-1}\) passes through \((0,-2)\). Plugging in \(x=0\) into \(2 \mp 2\sqrt{4-x}\), we find that \(–\) is correct. Hence the inverse function is \begin{equation*} with domain \((-\infty, 4]\). |
We sketch \(f^{-1}\) (red) along with \(f\).
| The volume of the sculpture is calculated as the difference in volume between two paraboloids. The larger paraboloid, formed by the curve \(y=x^2\) for \(0 \leq y \leq 2\), has volume\begin{equation*} V_1 = \pi \int_0^2 x^2 \,dy = \pi \int_0^2 y \,dy = \pi \left[ \frac{y^2}{2} \right]^2_0 = 2\pi. \end{equation*}On the other hand, the smaller paraboloid is formed by the curve \(y=x^2+1\) for \(1 \leq y \leq 2\). We notice that \(x^2 = y-1\) in this case, hence the volume is given by\begin{equation*} V_2 = \pi \int_1^2 x^2 \,dy = \pi \int_1^2 y-1 \,dy = \pi \left[ \frac{y^2}{2}-y \right]^2_1 = \frac{\pi}{2}. \end{equation*}Therefore \begin{equation*} V_{\text{sculpture}} = V_1 – V_2 = 2\pi – \frac{\pi}{2} = \frac{3\pi}{2} \text{ units}^3. \end{equation*} |
We are given the following initial data:
Hence we find that \(Vt \cos\theta = 6\sqrt{3}\, t\) and \(Vt\sin\theta = 6t\). The trajectory of the ball is therefore parametrised by Now if we set \(x(t) = 10\), then \(6\sqrt{3} t = 10\), which gives |
| We apply the area bounded between two curves formula, noting that we have an even function and so we can simplify things by only finding the area from \(x=0\) to \(x=2\) and doubling our answer. \begin{align*} A &= \int_{-2}^{2} 2 – |x| – \left(1-\frac{8}{4+x^2}\right) \ \mathrm{d} x\\ &= 2 \int_0^2 1-x+\frac{8}{4+x^2} \; \mathrm{d}x\\ &= 2 \left[x-\frac{1}{2} x^2 + 4\tan^{-1} \frac{x}{2}\right]_0^2 \\ &= 2\pi \ \mathrm{units}^2. \end{align*} |
| If \(A=B-d\) and \(C=B+d\) then \begin{align*} \frac{\sin A+\sin C}{\cos A + \cos C} &= \frac{\sin(B-d)+\sin(B+d)}{\cos(B-d)+\cos(B+d)} \\ &= \frac{2\sin B \cos d}{2\cos B \cos d} \\ &= \tan B \end{align*} where we make the use of the sum to product formulae in the first step. |
| If we let \(B=\frac{11\theta}{14}\) and \(d=\frac{\theta}{14}\) in (i) then we get:\begin{equation*} \frac{\sin \frac{5\theta}{7} + \sin \frac{6\theta}{7}}{\cos \frac{5\theta}{7} + \cos \frac{6\theta}{7}} = \tan \frac{11\theta}{14} \end{equation*}and so we simplify our problem to solving the equation\begin{equation*} \tan \frac{11\theta}{14} = \sqrt{3} \qquad \mathrm{where} \qquad 0\leq \theta \leq 2\pi ⇒0\leq \frac{11\theta}{14}\leq \frac{11\pi}{7}. \end{equation*}Looking at first and third quadrant solutions, we obtain\begin{align*} \frac{11\theta}{14} &= \frac{\pi}{3}, \frac{4\pi}{3} \\ ∴\theta &= \frac{14\pi}{33}, \frac{56\pi}{33} \end{align*} |
| Consider the following diagram, where the 175km/h vector is placed tip-to-tail with the 42 km/h wind vector. By alternate angles, the angle between the wind vector and the destination vector is \(63^\circ\).
Using the sine law, \begin{align*} Hence, the bearing is \(063^\circ T + \theta = 075^\circ T\) to the nearest degree. |
| This is a classic example of a logistic equation. We first separate variables\begin{equation*} \frac{C}{P(C-P)} \,dP = 0.1\,dt \end{equation*}and then integrate both sides, using the given fact. Note that we use \(K\) for our integration constant, since \(C\) is already taken!\begin{align*} \int \frac{C}{P(C-P)} \,dP &= \int 0.1 \,dt = 0.1t + K \\ ⇒ 0.1t + K &= \int \frac{1}{P} + \frac{1}{C-P} \,dP \\ &= \ln P – \ln(C-P) = \ln\left(\frac{P}{C-P}\right). \end{align*}(We do not need absolute values here: we must have \(C \ge P(t)\) for all \(t\ge 0\), since \(C\) is the carrying capacity, i.e. the largest population the habitat can sustain).At \(t=0\) (the year 1980), we are given that the deer population was \(150\, 000\). Setting \(t=0\) and \(P=150\, 000\) in the above result, we find\begin{equation*} K = \ln\left( \frac{150\, 000}{C – 150\, 000} \right). \end{equation*}In the year 2000, which corresponds to \(t=20\), we are given that \(P = 600\, 000\). Therefore \begin{equation*} We can now eliminate \(K\) and start solving for \(C\): \begin{align*} Taking the exponential of both sides, we continue to solve: \begin{align*} The carrying capacity is therefore \(C \approx 1\, 130\, 000\), to 3 significant figures. |
| The easiest way to do this is algebraically. Let \(v= x \underset{\text{~}}{i} + y\underset{\text{~}}{j}\) then \(\underset{\text{~}}{v}\cdot\underset{\text{~}}{v} = x^2+y^2.\) We now check the right hand side, \(|\underset{\text{~}}{v}|^2 = \left(\sqrt{x^2+y^2}\right)^2=x^2+y^2\) and so we conclude that \(\underset{\text{~}}{v}\cdot\underset{~}{v} = |\underset{\text{~}}{v}|^2\). |
| Note that\begin{equation*} \overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC} = \underset{\text{~}}{a}+\underset{\text{~}}{b} \qquad \mathrm{and} \qquad \overrightarrow{BD} = \overrightarrow{AD}-\overrightarrow{AB} = k \underset{\text{~}}{b}-\underset{\text{~}}{a} \end{equation*}Using the fact that \(|\overrightarrow{AC}| = |\overrightarrow{BD}|\), or rather (in order to use (i)), \(|\overrightarrow{AC}|^2 = |\overrightarrow{BD}|^2\) to obtain\begin{equation*} \overrightarrow{AC}\cdot\overrightarrow{AC} = \overrightarrow{BD}\cdot\overrightarrow{BD} \end{equation*}Substituting for our expressions for \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\) and rearranging, we get\begin{equation*} (\underset{\text{~}}{a}+\underset{\text{~}}{b})\cdot(\underset{\text{~}}{a}+\underset{\text{~}}{b}) = (k\underset{\text{~}}{b}-\underset{\text{~}}{a})\cdot(k\underset{\text{~}}{b}-\underset{\text{~}}{a}) \end{equation*}Now we expand both sides and make use of (i) to obtain the required result:\begin{align*} \underset{\text{~}}{a}\cdot \underset{\text{~}}{a}+2\underset{\text{~}}{a}\cdot \underset{\text{~}}{b}+\underset{\text{~}}{b}\cdot\underset{\text{~}}{b}&=k^2\underset{\text{~}}{b}\cdot\underset{\text{~}}{b}-2k\underset{\text{~}}{a}\cdot\underset{\text{~}}{b}+\underset{\text{~}}{a}\cdot\underset{\text{~}}{a} \\ 2\underset{\text{~}}{a}\cdot\underset{\text{~}}{b}+|\underset{\text{~}}{b}|^2&=k^2|\underset{\text{~}}{b}|^2-2k\underset{\text{~}}{a}\cdot \underset{\text{~}}{b}\\ 2(1+k)\underset{\text{~}}{a}\cdot\underset{\text{~}}{b} &= (k^2-1)|\underset{\text{~}}{b}|^2 \\ 2(1+k)\underset{\text{~}}{a}\cdot\underset{\text{~}}{b} &= -(1+k)(1-k)|\underset{\text{~}}{b}|^2 \\ 2\underset{\text{~}}{a}\cdot\underset{\text{~}}{b} &= -(1-k)|\underset{\text{~}}{b}|^2 \quad \text{since } 1+k \neq 0\\ 2\underset{\text{~}}{a}\cdot\underset{\text{~}}{b}+(1-k)|\underset{\text{~}}{b}|^2 &= 0. \end{align*} |
| With \(p = \frac{3}{500}\), we can calculate the following quantities for \(\hat p\),\begin{align*} \mu &= p = \frac{3}{500} \\ \sigma^2 &= \frac{p(1-p)}{n} = \frac{1491}{500^2 n} \end{align*}The aim is to find the probability \(P(\hat p \geq \frac{4}{500})\), which we will change into z-scores,\begin{align*} P(\hat p \geq \frac{4}{500}) &= P \left(Z \geq \frac{\frac{4}{500}-\mu}{\sigma}\right)\\ &= P \left(Z \geq \frac{\frac{4}{500} – \frac{3}{500}}{\frac{1}{500} \sqrt{\frac{1491}{n}}} \right) \\ &= P \left(Z \geq \frac{\sqrt{n}}{\sqrt{1491}} \right) \end{align*}From the empirical rule, 95% of the data is between z-scores -2 and 2.So, 2.5% is to the right of 2.Hence, we need\begin{align*} \frac{\sqrt{n}}{\sqrt{1491}} &\geq 2 \\ n &\geq 4 \times 1491 = 5964. \end{align*} Hence, the sample size required is 6000 to the nearest thousand. |
| We have that \(g(x)=x^3+4x-2\) and \(g'(x)=3x^2+4\). In order to find the gradient of the tangent to \(f(x)=x g^{-1}(x)\) at \(x=3\) we need to find \(f'(3)\). Differentiating \(f(x)\) using the product rule, we get\begin{align*} f'(x) = g^{-1}(x) + x \frac{d}{dx}\left(g^{-1}(x)\right) \end{align*}It becomes necessary to differentiate \(g^{-1}(x)\). To find what this is, consider the function \(y=g^{-1}(x)\): we have \(x=g(y)\), and we want to find \(\frac{dy}{dx}\).Then\begin{align*} \frac{dx}{dy} &= g'(y) \\ \frac{dy}{dx} &= \frac{1}{g'(y)} \end{align*}However, note here that \(y\) in this context is not the same as \(g(x)\), because in this context, \(y = g^{-1}(x)\). We should instead try to express \(\frac{dy}{dx} = \frac{d}{dx}(g^{-1}(x))\) in terms of \(x\) only, by substituting \(y = g^{-1}(x)\):\begin{align*} \frac{dy}{dx} = \frac{d}{dx} (g^{-1}(x))&= \frac{1}{g'(y)} \\ &= \frac{1}{g’\left(g^{-1}(x)\right)} \end{align*}So the derivative of \(f(x)\) is given by \begin{equation*} Since \(g(x)\) passes through the point \((1,3)\), we have \(g(1)=3\) so \(g^{-1}(3)=1\). Now substituting \(x=3\) we have \begin{align*} We conclude that the gradient of the tangent to \(f(x)\) at \(x=3\) is \(\frac{10}{7}\). |
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