SACE Stage 2 Specialist Maths 2023 Suggested Answers

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Remark 1. In SACE Stage 2 Specialist Maths, the following equivalent notation is used:

$\displaystyle \mathrm{cis}\, \theta \equiv \mathrm{cis}(\theta) \equiv \cos \theta + i \sin \theta.$

Furthermore, we use row vector notation for vectors:

$\displaystyle [x,y,z] \equiv \begin{bmatrix} x \\ y \\ z \end{bmatrix}.$

Question 1. Lines and Planes

Solution. (a) Substituting the parametric equations of $l$ into the equation of $P$ ,

$\displaystyle (1 + t) + 2(2 - 2t) - (3 - t) = 4 \quad \Rightarrow \quad 2 - 2t = 4 \quad \Rightarrow \quad t = -1.$

Setting $t = -1$ , we obtain $x = 0$ , $y = 4$ and $z = 4$ . Therefore, $P$ and $l$ intersect at $A(0, 4, 4)$ .

(b) The direction vector $[1, -2, -1]$ of $l$ is a normal vector of the required plane. Since the plane passes through $A$ ,

$\displaystyle x - 2y - z = 0 - 2 \cdot 4 - 4 = -12.$

Therefore, the required plane is $x - 2y - z = -12$ .

Question 2. Related Rates of Change

Solution. (a) $x = 50 \ln(9 + 1) = 50 \ln 10$ cm.

(b) (i) In the right-angled triangle with vertices $B$ , $C$ and the train, the side opposite $\theta$ has length $x$ and the side adjacent to $\theta$ has length $50$ . Since $0 \leq \theta < \frac{\pi}{2}$ ,

$\displaystyle \tan \theta = \frac{x}{50} \quad \Rightarrow \quad \theta = \arctan \left( \frac{x}{50} \right).$

(b) (ii) By the chain rule,

$\displaystyle \frac{\mathrm d\theta}{\mathrm dt} = \frac{\mathrm d\theta}{\mathrm dx} \cdot \frac{\mathrm dx}{\mathrm dt} = \frac{1/50}{1 + (x/50)^2} \cdot \frac{\mathrm dx}{\mathrm dt} = \left( \frac{50}{2500 + x^2} \right) \frac{\mathrm dx}{\mathrm dt}.$

$\displaystyle \frac{\mathrm dx}{\mathrm dt} = \frac{50}{t+1} = \frac{50}{9+1} = 5\ \text{cm/s} \quad \text{at } t = 9.$

By (a) and (b) (ii),

$\displaystyle \frac{\mathrm d\theta}{\mathrm dt} = \frac{50}{2500 + (50 \ln 10)^2} \cdot 5 \approx 0.0159\ \text{radians per second}.$

Question 3. Complex Numbers

Solution. (a) (i) Since $z_1$ is in the fourth quadrant,

$\displaystyle |z_1| = \sqrt{3 + 9} = 2\sqrt 3, \quad \tan^{-1} \frac{3}{\sqrt 3} = \frac{\pi}{3} \quad \Rightarrow \quad z_1 = 2\sqrt 3\, \mathrm{cis} \left( -\frac{\pi}{3} \right).$

(a) (ii) Similarly, since $z_2$ is in the fourth quadrant,

$\displaystyle |z_2| = \sqrt{27 + 9} = 6, \quad \tan^{-1} \frac{3}{3\sqrt 3} = \frac{\pi}{6} \quad \Rightarrow \quad z_2 = 6\, \mathrm{cis} \left( -\frac{\pi}{6} \right).$

(b) (i) Dividing moduli and subtracting arguments,

$\displaystyle \frac{z_2}{z_1} = \frac{6}{2\sqrt 3}\, \mathrm{cis} \left( -\frac{\pi}{6} + \frac{\pi}{3} \right) = \sqrt 3\, \mathrm{cis}\, \frac{\pi}{6}.$

(b) (ii) By (b) (i), $\theta = \frac{\pi}{6}$ .

(c) Since $\frac{3\theta}{2} = \frac{\pi}{4}$ , scaling by $\frac 12$ and rotating anticlockwise through $\frac{\pi}{4}$ amounts to multiplying by $\frac 12\, \mathrm{cis}\, \frac{\pi}{4}$ . Hence,

$\displaystyle z_3 = \frac 12\, \mathrm{cis}\, \frac{\pi}{4} \cdot 6\, \mathrm{cis} \left( -\frac{\pi}{6} \right) = 3\, \mathrm{cis} \left( \frac{\pi}{4} - \frac{\pi}{6} \right) = 3\, \mathrm{cis}\, \frac{\pi}{12}.$

Question 4. Mathematical Induction

Solution. (a) For the base case $n = 1$ , $4^1 + 15 - 1 = 18 = 9 \cdot 2$ , which is divisible by $9$ . Fix $k \in \mathbb N^+$ such that $4^k + 15k - 1 = 9m$ for some integer $m$ , i.e. $4^k = 9m - 15k + 1$ . Then

$\displaystyle \begin{aligned} 4^{k+1} + 15(k+1) - 1 &= 4 \cdot 4^k + 15k + 14 \\ &= 4(9m - 15k + 1) + 15k + 14 \\ &= 36m - 45k + 18 \\ &= 9(4m - 5k + 2), \end{aligned}$

which is divisible by $9$ since $4m - 5k + 2$ is an integer. Since the statement holds for $n = 1$ and it holds for $n = k+1$ whenever it holds for $n = k$ , it holds for all $n \in \mathbb N^+$ by mathematical induction.

(b) By (a), $4^n + 15n - 1 = 9m$ for some integer $m$ . Therefore,

$\displaystyle 4^{n+1} + 60n - 1 = 4(4^n + 15n - 1) + 3 = 36m + 3 = 3(12m + 1),$

which is divisible by $3$ since $12m + 1$ is an integer.

Question 5. Systems of Linear Equations

Solution. (a) (i) The augmented matrix is

$\displaystyle \left[ \begin{array}{ccc|c} 1 & 1 & -3 & 4 \\ 1 & -1 & -a & 2 \\ 3 & a & -1 & 8 \end{array} \right].$

(a) (ii) Applying $R_2 \to R_1 - R_2$ and $R_3 \to 3R_1 - R_3$ ,

$\displaystyle \left[ \begin{array}{ccc|c} 1 & 1 & -3 & 4 \\ 0 & 2 & a - 3 & 2 \\ 0 & 3 - a & -8 & 4 \end{array} \right].$

Applying $R_3 \to 2R_3 - (3-a)R_2$ , the third row becomes

$\displaystyle \begin{aligned} &[\, 0,\ 2(3-a) - 2(3-a),\ -16 - (3-a)(a-3) \ :\ 8 - 2(3-a) \,] \\ = \ &[\, 0,\ 0,\ a^2 - 6a - 7 \ :\ 2a + 2 \,] \\ = \ &[\, 0,\ 0,\ (a-7)(a+1) \ :\ 2(a+1) \,], \end{aligned}$

as required.

(b) Figure 4 shows three planes, no two of which are parallel, with no common point of intersection. This corresponds to an inconsistent system, which occurs when $(a-7)(a+1) = 0$ but $2(a+1) \neq 0$ , i.e. $a = 7$ . (Indeed, for $a = 7$ the normal vectors $[1,1,-3]$ , $[1,-1,-7]$ and $[3,7,-1]$ are pairwise non-parallel.)

$\displaystyle y = 1, \quad z = -\frac 12, \quad x = 4 - 1 - \frac 32 = \frac 32.$

Therefore, the three planes intersect at the single point $\left( \frac 32, 1, -\frac 12 \right)$ .

(d) (i) $a = -1$ , since then the third row is entirely zero.

(d) (ii) For $a = -1$ , the second row reads $2y - 4z = 2$ . Setting $z = t$ ,

$\displaystyle y = 1 + 2t, \quad x = 4 - (1 + 2t) + 3t = 3 + t.$

Therefore, the line is

$\displaystyle x = 3 + t, \quad y = 1 + 2t, \quad z = t, \quad t \in \mathbb R,$

or in vector form, $\mathbf r = [3, 1, 0] + t[1, 2, 1]$ .

Question 6. Logistic Differential Equations

Solution. (a) (i) Combining the fractions on the right-hand side,

$\displaystyle \frac 1B + \frac{1}{K - B} = \frac{(K - B) + B}{B(K - B)} = \frac{K}{B(K-B)}.$

(a) (ii) Rewriting the differential equation as

$\displaystyle \frac{\mathrm dB}{\mathrm dt} = \frac{0.1 B(K - B)}{K}$

and separating variables, by (a) (i),

$\displaystyle \begin{aligned} \int \left( \frac 1B + \frac{1}{K - B} \right) \mathrm dB &= \int 0.1\, \mathrm dt \\ \ln |B| - \ln |K - B| &= 0.1t + C \\ \ln \left| \frac{K - B}{B} \right| &= -0.1t - C \\ \frac{K - B}{B} &= \pm e^{-C} e^{-0.1t} =: Ae^{-0.1t}. \end{aligned}$

Therefore,

$\displaystyle \frac KB - 1 = Ae^{-0.1t} \quad \Rightarrow \quad B = \frac{K}{1 + Ae^{-0.1t}}.$

(b) The slope field shows horizontal segments at $B = 600$ , so $K = 600$ for this butterfly house. The solution curve rises from $(0, 120)$ in an S-shape (steepest at $B = 300$ ) and approaches $B = 600$ from below. We sketch as follows.

$\displaystyle 120 = \frac{960}{1 + A} \quad \Rightarrow \quad 1 + A = 8 \quad \Rightarrow \quad A = 7.$

$\displaystyle \frac{960}{1 + 7e^{-0.1t}} = 600 \quad \Rightarrow \quad 7e^{-0.1t} = 0.6 \quad \Rightarrow \quad t = 10 \ln \frac{35}{3} \approx 24.6.$

Therefore, the population exceeds $600$ butterflies after approximately $24.6$ months.

Question 7. Integration and Volumes of Revolution

Solution. (a) Integrating by parts with $u = x^2$ and $\mathrm dv = e^x\, \mathrm dx$ , and using the given integral,

$\displaystyle \begin{aligned} \int x^2 e^x\, \mathrm dx &= x^2 e^x - \int 2x e^x\, \mathrm dx \\ &= x^2 e^x - 2(xe^x - e^x) + k \\ &= e^x (x^2 - 2x + 2) + k. \end{aligned}$

(b) (i) For $-2 \leq x \leq 2$ , $g(x) \geq 0$ , so that

$\displaystyle f(g(x)) = \sqrt{\frac{e^x}{4}(4 - x^2)} = \frac{\sqrt{e^x(4 - x^2)}}{\sqrt 4} = \frac 12 \sqrt{e^x(4 - x^2)}.$

(b) (ii) The curve meets the $x$ -axis at $x = \pm 2$ and passes through $(0, 1)$ . Its maximum occurs where

$\displaystyle \frac{\mathrm d}{\mathrm dx} \left( e^x(4 - x^2) \right) = e^x(4 - 2x - x^2) = 0,$

i.e. at $x = -1 + \sqrt 5 \approx 1.24$ , where $y \approx 1.46$ . We sketch as follows.

$\displaystyle V = \pi \int_{-2}^2 (f(g(x)))^2\, \mathrm dx = \pi \int_{-2}^2 \frac 14 e^x(4 - x^2)\, \mathrm dx = \frac{\pi}{4} \int_{-2}^2 e^x(4 - x^2)\, \mathrm dx.$

$\displaystyle \int e^x(4 - x^2)\, \mathrm dx = 4e^x - e^x(x^2 - 2x + 2) + k = e^x(2 + 2x - x^2) + k.$

Therefore, by (c) (i),

$\displaystyle \begin{aligned} V &= \frac{\pi}{4} \left[ e^x(2 + 2x - x^2) \right]_{-2}^2 \\ &= \frac{\pi}{4} \left( e^2 (2 + 4 - 4) - e^{-2}(2 - 4 - 4) \right) \\ &= \frac{\pi}{4} (2e^2 + 6e^{-2}) \\ &= \frac{\pi}{2} (e^2 + 3e^{-2}). \end{aligned}$

Question 8. Vector Geometry

Solution. (a) In the right-angled triangle $ABT$ , the side $AT$ is opposite $\theta$ and $BA$ is the hypotenuse, so that $|\overrightarrow{AT}| = |\overrightarrow{BA}| \sin \theta$ . Since $\mathbf l$ points along $\overrightarrow{BT}$ , the angle between $\overrightarrow{BA}$ and $\mathbf l$ is $\theta$ . Therefore,

$\displaystyle |\overrightarrow{AT}| = |\overrightarrow{BA}| \sin \theta = \frac{|\overrightarrow{BA}| |\mathbf l| \sin \theta}{|\mathbf l|} = \frac{|\overrightarrow{BA} \times \mathbf l|}{|\mathbf l|}.$

(b) $\overrightarrow{BA} = [4, 0, 2] - [-1, -6, -2] = [5, 6, 4]$ .

$\displaystyle \overrightarrow{BA} \times \mathbf l = [5, 6, 4] \times [2, 2, -1] = [-6 - 8, 8 + 5, 10 - 12] = [-14, 13, -2],$

so that $|\overrightarrow{BA} \times \mathbf l| = \sqrt{196 + 169 + 4} = \sqrt{369} = 3\sqrt{41}$ .

$\displaystyle |\overrightarrow{AT}| = \frac{3\sqrt{41}}{3} = \sqrt{41}.$

$\displaystyle 2(-1 + 2\lambda) + 2(-6 + 2\lambda) - (-2 - \lambda) = 6 \quad \Rightarrow \quad 9\lambda - 12 = 6 \quad \Rightarrow \quad \lambda = 2.$

Setting $\lambda = 2$ , the intersection point is $T(3, -2, -4)$ .

(d) Since $\overrightarrow{BT}$ and $\overrightarrow{CV}$ are both normal to $P_1$ , they are parallel, and $T, V$ both lie on $P_1$ . Hence, the triangles $ATB$ and $AVC$ are similar, so that

$\displaystyle \frac{|AC|}{|AB|} = \frac{|AV|}{|AT|} = 5 \quad \Rightarrow \quad \overrightarrow{AC} = 5 \overrightarrow{AB},$

since $B$ and $C$ lie on the same side of $A$ . Therefore,

$\displaystyle \overrightarrow{OC} = [4, 0, 2] + 5[-5, -6, -4] = [-21, -30, -18].$

Since $C$ lies on $P_3$ , $k = 2(-21) + 2(-30) - (-18) = -84$ .

Remark 2. As a check, the distances from $P_1$ to $P_2$ and to $P_3$ are

$\displaystyle \frac{6 - (-12)}{3} = 6 \quad \text{and} \quad \frac{6 - (-84)}{3} = 30 = 5 \cdot 6$

respectively, as expected from the similar triangles.

Question 9. Complex Roots of Unity

Solution. (a) Let $z = r\, \mathrm{cis}\, \alpha$ with $r > 0$ and $-\pi < \alpha \leq \pi$ . By De Moivre’s theorem,

$\displaystyle r^5\, \mathrm{cis}\, 5\alpha = 1 = \mathrm{cis}(2k\pi),$

so that $r = 1$ and $\alpha = \frac{2k\pi}{5}$ for $k = 0, \pm 1, \pm 2$ . Matching the positions in Figure 11,

$\displaystyle z_1 = 1, \quad z_2 = \mathrm{cis}\, \tfrac{2\pi}{5}, \quad z_3 = \mathrm{cis}\, \tfrac{4\pi}{5}, \quad z_4 = \mathrm{cis} \left( -\tfrac{4\pi}{5} \right), \quad z_5 = \mathrm{cis} \left( -\tfrac{2\pi}{5} \right).$

(b) (i) Since $|z| = 1$ , $|z^n| = |z|^n = 1$ for every integer $n$ . By the triangle inequality,

$\displaystyle |f(z)| \leq |z^2| + |z| + 1 + \left| \frac 1z \right| + \left| \frac{1}{z^2} \right| = 1 + 1 + 1 + 1 + 1 = 5.$

(b) (ii) $z = 1$ , since $f(1) = 5$ .

$\displaystyle z^2 (z - 1) f(z) = (z - 1)(z^4 + z^3 + z^2 + z + 1) = z^5 - 1,$

where the final equality follows from the geometric series (or by expanding and cancelling).

$\displaystyle \mathrm{cis}\, \tfrac{2\pi}{5}, \quad \mathrm{cis}\, \tfrac{4\pi}{5}, \quad \mathrm{cis} \left( -\tfrac{4\pi}{5} \right), \quad \mathrm{cis} \left( -\tfrac{2\pi}{5} \right).$

(d) (i) By De Moivre’s theorem, $z^n + z^{-n} = \mathrm{cis}(n\theta) + \mathrm{cis}(-n\theta) = 2\cos(n\theta)$ . Therefore,

$\displaystyle f(z) = (z^2 + z^{-2}) + (z + z^{-1}) + 1 = 2\cos 2\theta + 2\cos\theta + 1,$

which is real.

(d) (ii) Using the double angle identity $\cos 2\theta = 2\cos^2\theta - 1$ ,

$\displaystyle f(z) = 2(2\cos^2 \theta - 1) + 2\cos\theta + 1 = 4\cos^2 \theta + 2\cos\theta - 1.$

(d) (iii) Writing $c = \cos\theta \in [-1, 1]$ and completing the square,

$\displaystyle 4c^2 + 2c - 1 = 4 \left( c + \frac 14 \right)^2 - \frac 54 \geq -\frac 54,$

with equality when $c = -\frac 14 \in [-1, 1]$ . Therefore, the minimum value of $f(z)$ is $-\frac 54$ .

(d) (iv) The minimum occurs when $\cos\theta = -\frac 14$ , i.e.

$\displaystyle z = -\frac 14 \pm \frac{\sqrt{15}}{4} i.$

We mark these as follows.

Question 10. Parametric Curves in 3D

Solution. (a) Using the double angle identity $\cos^2 x = \frac 12 (1 + \cos 2x)$ ,

$\displaystyle \int \cos^2 x\, \mathrm dx = \int \frac 12 (1 + \cos 2x)\, \mathrm dx = \frac 12 x + \frac 14 \sin 2x + c.$

(b) The curve is a figure of eight through the origin, with $|x| \leq 4\sqrt 2 \approx 5.66$ and $|y| \leq \frac{\sqrt 2}{2} \approx 0.707$ . The turning points $\left( \pm 4, \pm \frac{\sqrt 2}{2} \right)$ occur at

$\displaystyle t = \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \frac{5\pi}{4}, \ \frac{7\pi}{4}.$

We sketch as follows.

$\displaystyle \mathbf v = \left[ 4\sqrt 2 \cos t,\ \sqrt 2 \cos 2t,\ 4 \right].$

$\displaystyle \begin{aligned} |\mathbf v|^2 &= 32c^2 + 2(2c^2 - 1)^2 + 16 \\ &= 32c^2 + 8c^4 - 8c^2 + 2 + 16 \\ &= 8c^4 + 24c^2 + 18 \\ &= 2(2c^2 + 3)^2. \end{aligned}$

Since $2c^2 + 3 > 0$ , the speed is

$\displaystyle |\mathbf v| = \sqrt 2 (2c^2 + 3) = 2\sqrt 2 \cos^2 t + 3\sqrt 2.$

$\displaystyle \begin{aligned} \int_{\pi/4}^{3\pi/4} (2\sqrt 2 \cos^2 t + 3\sqrt 2)\, \mathrm dt &= 2\sqrt 2 \left[ \frac t2 + \frac 14 \sin 2t \right]_{\pi/4}^{3\pi/4} + 3\sqrt 2 \cdot \frac{\pi}{2} \\ &= 2\sqrt 2 \left( \left( \frac{3\pi}{8} - \frac 14 \right) - \left( \frac{\pi}{8} + \frac 14 \right) \right) + \frac{3\sqrt 2 \pi}{2} \\ &= 2\sqrt 2 \left( \frac{\pi}{4} - \frac 12 \right) + \frac{3\sqrt 2 \pi}{2} \\ &= \sqrt 2 \left( \frac{\pi}{2} - 1 + \frac{3\pi}{2} \right) \\ &= \sqrt 2 (2\pi - 1). \end{aligned}$

Remark 3. The $z$ -component contributes a constant $4$ to the velocity, so the dragonfly climbs steadily while tracing the figure of eight from (b) when viewed from above.

Remark 4. The diagrams have been drawn and the solutions have been verified with the help of Generative AI.

—Joel Kindiak, 17 Oct 26, 0022H

KindiakMath

SACE Stage 2 Specialist Maths 2023 Suggested Answers

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SACE Stage 2 Specialist Maths 2023 Suggested Answers

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