SACE Stage 2 Specialist Maths 2024 Suggested Answers

Questions are publicly available. Do comment if you spot any errors. Download the solutions here.

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

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

Furthermore, for any planar shape $S$ , we denote its area by $\mathcal A(S)$ . Finally, we use row vector notation for vectors:

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

Question 1. Polynomials

Solution. (a) Since $P$ has real coefficients, $3 - i$ is also a zero by the conjugate root theorem. Therefore, a real quadratic factor is

$(z - (3+i))(z - (3-i)) = (z-3)^2 - i^2 = z^2 - 6z + 10.$

(b) (i) By the factor theorem, $P(1) = 0$ :

$1 - 5 + a + b - 20 = 0 \quad \Rightarrow \quad a + b = 24.$

(b) (ii) By the remainder theorem, $P(-1) = -34$ :

$1 + 5 + a - b - 20 = -34 \quad \Rightarrow \quad a - b = -20.$

(b) (iii) Adding and subtracting the equations in (b) (i) and (b) (ii), $a = 2$ and $b = 22$ .

$P(z) = (z^2 - 6z + 10)(z^2 + pz - 2)$

for some $p$ , where we compared constant terms. Comparing the coefficients of $z^3$ , $p - 6 = -5$ so that $p = 1$ . Therefore,

$P(z) = (z^2 - 6z + 10)(z^2 + z - 2) = (z^2 - 6z + 10)(z - 1)(z + 2).$

Question 2. Planes and Lines

Solution. (a) The normal vectors $[1,2,-3]$ and $[1,1,2]$ of $P_1$ and $P_2$ are not scalar multiples of each other (comparing the first two components would require the scalar to be both $1$ and $2$ ). Therefore, the planes are not parallel.

(b) Applying $R_2 \to R_2 - R_1$ ,

$\left[ \begin{array}{ccc|c} 1 & 2 & -3 & 9 \\ 1 & 1 & 2 & 5 \end{array} \right] \to \left[ \begin{array}{ccc|c} 1 & 2 & -3 & 9 \\ 0 & -1 & 5 & -4 \end{array} \right].$

Setting $z = t$ , the second row yields $-y + 5t = -4$ , so that $y = 4 + 5t$ . The first row then yields

$x = 9 - 2(4 + 5t) + 3t = 1 - 7t.$

Therefore, the planes meet along the line $l_1$ with parametric equations

$x = 1 - 7t, y = 4 + 5t, z = t, \quad t \in \mathbb R.$

(c) Setting $t = 2$ , we obtain $x = 1 - 14 = -13$ , $y = 4 + 10 = 14$ and $z = 2$ . Hence, $A(-13, 14, 2)$ lies on $l_1$ .

(d) The plane $P_3$ contains the point $A(-13, 14, 2)$ on $l_1$ , the direction vector $[-7, 5, 1]$ of $l_1$ , and the vector

$\overrightarrow{AQ} = [-10,12,0] - [-13,14,2] = [3, -2, -2].$

Hence, a normal vector of $P_3$ is

$\begin{aligned} [-7, 5, 1] \times [3, -2, -2] &= [5 \cdot (-2) - 1 \cdot (-2),\ 1 \cdot 3 - (-7) \cdot (-2),\ -7 \cdot (-2) - 5 \cdot 3] \\ &= [-8, -11, -1]. \end{aligned}$

Taking the normal vector $[8, 11, 1]$ and the point $A$ ,

$P_3 : 8x + 11y + z = 8 \cdot (-13) + 11 \cdot 14 + 2 = -104 + 154 + 2 = 52.$

As a check, $Q$ satisfies

$8 \cdot (-10) + 11 \cdot 12 + 0 = -80 + 132 = 52.$

Question 3. Matrices and Mathematical Induction

Solution. (a) Denote $M = \begin{bmatrix} i & a \\ 0 & i \end{bmatrix}$ . Using $i^2 = -1$ ,

$\begin{aligned} M^2 &= \begin{bmatrix} i^2 & ia + ai \\ 0 & i^2 \end{bmatrix} = \begin{bmatrix} -1 & 2ai \\ 0 & -1 \end{bmatrix}, \\ M^3 &= M^2 M = \begin{bmatrix} -i & -a + 2ai^2 \\ 0 & -i \end{bmatrix} = \begin{bmatrix} -i & -3a \\ 0 & -i \end{bmatrix}. \end{aligned}$

(b) For the base case $n = 1$ , $\begin{bmatrix} i^1 & 1 \cdot a i^0 \\ 0 & i^1 \end{bmatrix} = \begin{bmatrix} i & a \\ 0 & i \end{bmatrix} = M$ , as required. Fix $k \in \mathbb N^+$ such that

$M^k = \begin{bmatrix} i^k & kai^{k-1} \\ 0 & i^k \end{bmatrix}.$

Then

$\begin{aligned} M^{k+1} = M^k M &= \begin{bmatrix} i^k & kai^{k-1} \\ 0 & i^k \end{bmatrix} \begin{bmatrix} i & a \\ 0 & i \end{bmatrix} \\ &= \begin{bmatrix} i^{k+1} & i^k a + kai^{k-1} \cdot i \\ 0 & i^{k+1} \end{bmatrix} \\ &= \begin{bmatrix} i^{k+1} & (k+1) a i^k \\ 0 & i^{k+1} \end{bmatrix}. \end{aligned}$

Since the identity holds for $n = 1$ and it holds for $n = k+1$ whenever it holds for $n = k$ , the identity holds for all $n \in \mathbb N^+$ by mathematical induction.

$M^{2027} = \begin{bmatrix} -i & -2027a \\ 0 & -i \end{bmatrix}.$

Therefore, by (a),

$M^3 - M^{2027} = \begin{bmatrix} -i & -3a \\ 0 & -i \end{bmatrix} - \begin{bmatrix} -i & -2027a \\ 0 & -i \end{bmatrix} = \begin{bmatrix} 0 & 2024a \\ 0 & 0 \end{bmatrix}.$

Question 4. Vector Geometry

Solution. (a) (i) Since $N$ is the midpoint of $OB$ , $\overrightarrow{AN} = \overrightarrow{ON} - \overrightarrow{OA} = \frac 12 \mathbf b - \mathbf a$ .

(a) (ii) Similarly, $\overrightarrow{BM} = \overrightarrow{OM} - \overrightarrow{OB} = \frac 12 \mathbf a - \mathbf b$ .

(b) Since $\overrightarrow{AN} \perp \overrightarrow{BM}$ , their dot product is zero:

$\begin{aligned} 0 &= \left( \tfrac 12 \mathbf b - \mathbf a \right) \cdot \left( \tfrac 12 \mathbf a - \mathbf b \right) \\ &= \tfrac 14 \mathbf a \cdot \mathbf b - \tfrac 12 |\mathbf b|^2 - \tfrac 12 |\mathbf a|^2 + \mathbf a \cdot \mathbf b \\ &= \tfrac 54 \mathbf a \cdot \mathbf b - \tfrac 12 (|\mathbf a|^2 + |\mathbf b|^2). \end{aligned}$

Rearranging, $|\mathbf a|^2 + |\mathbf b|^2 = \frac 52 \mathbf a \cdot \mathbf b$ .

$|\overrightarrow{AB}|^2 = |\mathbf a|^2 + |\mathbf b|^2 - 2|\mathbf a||\mathbf b| \cos \theta = |\mathbf a|^2 + |\mathbf b|^2 - 2 \mathbf a \cdot \mathbf b,$

since $\mathbf a \cdot \mathbf b = |\mathbf a||\mathbf b| \cos \theta$ by definition.

(d) By (b), $2 \mathbf a \cdot \mathbf b = \frac 45 (|\mathbf a|^2 + |\mathbf b|^2)$ . Substituting into (c),

$|\overrightarrow{AB}|^2 = (|\mathbf a|^2 + |\mathbf b|^2) - \tfrac 45 (|\mathbf a|^2 + |\mathbf b|^2) = \tfrac 15 (|\mathbf a|^2 + |\mathbf b|^2).$

Remark 2. Equivalently, $|OA|^2 + |OB|^2 = 5|AB|^2$ : whenever two medians of a triangle are perpendicular, the squares of the two sides they bisect sum to five times the square of the third side.

Question 5. Inverse Functions and Volumes

Solution. (a) The curve starts at $(2, \ln 2) \approx (2, 0.693)$ and increases slowly, e.g. through $(6, \ln 4) \approx (6, 1.39)$ . We sketch it in blue below, together with $y = f^{-1}(x)$ in red for (b) (iii).

(b) (i) Let $y = \ln(\sqrt{x-2} + 2)$ . Then

$e^y = \sqrt{x-2} + 2 \quad \Rightarrow \quad \sqrt{x-2} = e^y - 2 \quad \Rightarrow \quad x = (e^y - 2)^2 + 2,$

where squaring is reversible since $e^y - 2 = \sqrt{x-2} \geq 0$ . Swapping $x$ and $y$ ,

$f^{-1}(x) = (e^x - 2)^2 + 2.$

(b) (ii) The domain of $f^{-1}$ is the range of $f$ . Since $f$ is increasing with $f(2) = \ln 2$ and $f(x) \to \infty$ as $x \to \infty$ , the domain of $f^{-1}$ is $[\ln 2, \infty)$ .

(b) (iii) The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ . It starts at $(\ln 2, 2)$ and increases steeply; see the red curve in the figure in (a).

$\displaystyle \pi \int_2^5 (\ln(\sqrt{y-2}+2))^2\, \mathrm dy$

only depends on the integrand and the limits, not on the name of the variable. Replacing $y$ by $x$ gives the volume formed when $y = f(x)$ is rotated about the $x$ -axis from $x = 2$ to $x = 5$ . Geometrically, by Figure 3, reflecting the region in $y = x$ swaps the two solids, so their volumes are equal.

$\displaystyle V = \pi \int_2^5 (\ln(\sqrt{x-2}+2))^2\, \mathrm dx \approx 12.4\ \text{units}^3.$

Question 6. Complex Roots

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

$r^6\, \mathrm{cis}\, 6\theta = 8\, \mathrm{cis}((2k+1)\pi),$

so that $r = 8^{1/6} = \sqrt 2$ and $\theta = \frac{(2k+1)\pi}{6}$ . Matching the positions in Figure 4,

$\begin{aligned} z_1 &= \sqrt 2\, \mathrm{cis}\, \tfrac{\pi}{6}, & z_2 &= \sqrt 2\, \mathrm{cis}\, \tfrac{\pi}{2}, & z_3 &= \sqrt 2\, \mathrm{cis}\, \tfrac{5\pi}{6}, \\ z_4 &= \sqrt 2\, \mathrm{cis} \left( -\tfrac{5\pi}{6} \right), & z_5 &= \sqrt 2\, \mathrm{cis} \left( -\tfrac{\pi}{2} \right), & z_6 &= \sqrt 2\, \mathrm{cis} \left( -\tfrac{\pi}{6} \right). \end{aligned}$

(a) (ii) Since $z_6$ is the conjugate of $z_1$ ,

$z_1 - z_6 = z_1 - \overline{z_1} = 2i\, \mathrm{Im}(z_1) = 2i \cdot \sqrt 2 \sin \tfrac{\pi}{6} = \sqrt 2\, i,$

so that $|z_1 - z_6| = \sqrt 2$ .

(b) (i) The triangle $Oz_1z_6$ has two sides of length $\sqrt 2$ with included angle $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{\pi}{3}$ . Hence,

$\mathcal A(\Delta O z_1 z_6) = \tfrac 12 \cdot \sqrt 2 \cdot \sqrt 2 \cdot \sin \tfrac{\pi}{3} = \tfrac{\sqrt 3}{2}.$

(b) (ii) By symmetry, the hexagon consists of six such triangles, so its area is $6 \cdot \frac{\sqrt 3}{2} = 3\sqrt 3$ .

(c) The solutions of $z^6 = k$ lie on a circle of radius $R = |k|^{1/6}$ , spaced $\frac{\pi}{3}$ apart. As in (b), the hexagon has area

$6 \cdot \tfrac 12 R^2 \sin \tfrac{\pi}{3} = \tfrac{3\sqrt 3}{2} R^2 = 9 \sqrt 3 \quad \Rightarrow \quad R^2 = 6 \quad \Rightarrow \quad |k| = R^6 = 216.$

Therefore, $k = -216$ works (and so does $k = 216$ ).

Question 7. Integration by Parts

Solution. (a) Integrating by parts with

$u = \ln(\cos x), \quad \mathrm dv = \sin x\, \mathrm dx, \quad \mathrm du = -\frac{\sin x}{\cos x}\, \mathrm dx, \quad v = -\cos x,$

we obtain

$\begin{aligned} \int \sin x \ln(\cos x)\, \mathrm dx &= -\cos x \ln(\cos x) - \int (-\cos x) \cdot \left( -\frac{\sin x}{\cos x} \right) \mathrm dx \\ &= -\cos x \ln(\cos x) - \int \sin x\, \mathrm dx \\ &= -\cos x \ln(\cos x) + \cos x + c. \end{aligned}$

(b) By (a),

$\begin{aligned} \int_0^{\pi/3} \sin x \ln(\cos x)\, \mathrm dx &= \left[ -\cos x \ln(\cos x) + \cos x \right]_0^{\pi/3} \\ &= \left( -\tfrac 12 \ln \tfrac 12 + \tfrac 12 \right) - (-1 \cdot \ln 1 + 1) \\ &= \tfrac 12 \ln 2 + \tfrac 12 - 1 \\ &= \tfrac 12 \ln 2 - \tfrac 12. \end{aligned}$

(c) For $x \geq 0$ , $f(|x|) = f(x)$ . For $x < 0$ , the graph of $y = f(|x|)$ is the reflection of the right half of $y = f(x)$ in the $y$ -axis. We sketch as follows.

(d) Since $f$ is odd,

$f(|x|) = f(-x) = -f(x),$

for $x < 0$ , and the two curves coincide for $x \geq 0$ . Hence, the bounded region lies in $-\frac{\pi}{3} \leq x \leq 0$ , where

$f(x) \geq 0 \geq f(|x|).$

Its area is

$\begin{aligned} \int_{-\pi/3}^0 (f(x) - f(|x|))\, \mathrm dx &= 2 \int_{-\pi/3}^0 f(x)\, \mathrm dx \\ &= -2 \int_0^{\pi/3} f(x)\, \mathrm dx \\ &= -2 \left( \tfrac 12 \ln 2 - \tfrac 12 \right) \\ &= 1 - \ln 2\ \text{units}^2, \end{aligned}$

where we used (b) and the substitution $x \mapsto -x$ together with $f(-x) = -f(x)$ .

Question 8. Parametric Curves

Solution. (a) The curve is a cycloid: it starts at $O(0,0)$ , rises to its maximum $(\pi, 2)$ at $t = \pi$ , and returns to $(2\pi, 0)$ at $t = 2\pi$ . We sketch as follows.

(b) (i) Differentiating $(x - \pi)^2 + (y - 1)^2 = 1$ implicitly with respect to $x$ ,

$\displaystyle 2(x - \pi) + 2(y-1) \frac{\mathrm dy}{\mathrm dx} = 0 \quad \Rightarrow \quad \frac{\mathrm dy}{\mathrm dx} = -\frac{x - \pi}{y - 1}.$

(b) (ii) By the chain rule,

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt} = \frac{\sin t}{1 - \cos t}.$

(c) The point $P(\pi, 2)$ lies on the circle since $0^2 + 1^2 = 1$ , and on the parametric curve with $t = \pi$ . By (b) (i), the circle has gradient $-\frac{0}{1} = 0$ at $P$ . By (b) (ii), the parametric curve has gradient

$\displaystyle \frac{\sin \pi}{1 - \cos \pi} = \frac 02 = 0$

at $P$ . Therefore, both curves have the common horizontal tangent $y = 2$ at $P$ .

(d) By symmetry of the curve about $x = \pi$ , the arc $PA$ has the same length as the arc $OP$ , and the length of each arc is greater than the length of the corresponding chord. By the triangle inequality in triangle $OPA$ ,

$\text{length of curve} = \text{arc } OP + \text{arc } PA > |OP| + |PA| > |OA| = 2\pi.$

Since the circle has radius $1$ , its circumference is $2\pi$ . Hence, the length of the parametric curve is greater than the circumference of the circle.

(e) (i) Since $x'(t) = 1 - \cos t$ and $y'(t) = \sin t$ ,

$(x'(t))^2 + (y'(t))^2 = 1 - 2\cos t + \cos^2 t + \sin^2 t = 2 - 2\cos t = 4 \sin^2 \tfrac t2,$

using the double angle identity $\cos t = 1 - 2\sin^2 \frac t2$ . For $0 \leq t \leq 2\pi$ , $\sin \frac t2 \geq 0$ , so that the length of the curve is

$\displaystyle \int_0^{2\pi} \sqrt{(x'(t))^2 + (y'(t))^2}\, \mathrm dt = \int_0^{2\pi} 2 \sin \left( \frac t2 \right) \mathrm dt.$

(e) (ii) By (e) (i),

$\displaystyle \int_0^{2\pi} 2 \sin \left( \frac t2 \right) \mathrm dt = \left[ -4 \cos \frac t2 \right]_0^{2\pi} = -4(-1) + 4 = 8\ \text{units}.$

Remark 3. This confirms (d): $8 > 2\pi \approx 6.28$ . In fact, the bound in (d) is $2\sqrt{\pi^2 + 4} \approx 7.45$ , since $|OP| = |PA| = \sqrt{\pi^2 + 4}$ .

Question 9. Differential Equations

Solution. (a) (i) By the method of separable variables,

$\begin{aligned} \int y^{-2}\, \mathrm dy &= \int -\tfrac{1}{10} e^{-0.5t}\, \mathrm dt \\ -\frac 1y &= \tfrac{1}{5} e^{-0.5t} + C \\ \frac 1y &= c - 0.2 e^{-0.5t} \\ y &= \frac{1}{c - 0.2 e^{-0.5t}}, \end{aligned}$

where we set $c := -C$ .

(a) (ii) Setting $t = 0$ and $y = 10$ ,

$\displaystyle 10 = \frac{1}{c - 0.2} \quad \Rightarrow \quad c = 0.3.$

As $t \to \infty$ , $e^{-0.5t} \to 0$ , so that

$\displaystyle y = \frac{1}{0.3 - 0.2e^{-0.5t}} \to \frac{1}{0.3} = \frac{10}{3}.$

(b) The solution curve starts at $(0, 10)$ , decreases, and approaches $y = \frac{10}{3}$ from above. We sketch as follows.

(c) Differentiating $\frac{\mathrm dy}{\mathrm dt} = -\frac{1}{10} e^{-0.5t} y^2$ with respect to $t$ using the product rule and the chain rule,

$\begin{aligned} \frac{\mathrm d^2 y}{\mathrm dt^2} &= -\frac 1{10} \left( -\frac 12 e^{-0.5t} y^2 + e^{-0.5t} \cdot 2y \frac{\mathrm dy}{\mathrm dt} \right) \\ &= -\frac 12 \cdot \left( -\frac 1{10} e^{-0.5t} y^2 \right) - \frac 15 e^{-0.5t} y \frac{\mathrm dy}{\mathrm dt} \\ &= -\frac 12 \frac{\mathrm dy}{\mathrm dt} - \frac 15 e^{-0.5t} y \frac{\mathrm dy}{\mathrm dt} \\ &= -\frac{\mathrm dy}{\mathrm dt} \left( \frac 12 + \frac 15 e^{-0.5t} y \right). \end{aligned}$

(d) Since $\frac{\mathrm dy}{\mathrm dt} \neq 0$ (as $y > 0$ ), by (c),

$-\left( \frac 12 + \frac 15 e^{-0.5t} y \right) = -1 \quad \Rightarrow \quad e^{-0.5t} y = \frac 52.$

Write $u = e^{-0.5t}$ . By (a) (ii),

$\displaystyle y = \frac{1}{0.3 - 0.2u},$

so that

$\displaystyle \frac{u}{0.3 - 0.2u} = \frac 52 \quad \Rightarrow \quad u = 0.75 - 0.5u \quad \Rightarrow \quad u = \frac 12.$

Therefore, $e^{-0.5t} = \frac 12$ , so that $t = 2 \ln 2$ .

Question 10. Complex Roots and Regions

Solution. (a) (i) By De Moivre’s theorem, $z_2 = \sqrt{0.36}\, \mathrm{cis}\, \frac{\theta}{2} = 0.6\, \mathrm{cis}\, \frac{\theta}{2}$ satisfies $z_2^2 = 0.36\, \mathrm{cis}\, \theta = z_1$ . Since $0 < \frac{\theta}{2} < \frac{\pi}{4}$ , $z_2$ is in the first quadrant.

(a) (ii) The point $z_2$ has modulus $0.6$ and half the argument of $z_1$ , shown in blue in the figure below.

(b) (i) The set $S$ is the closed disc of radius $0.1$ centred at $1$ , shaded in blue below.

(b) (ii) The set $T$ is the closed half-plane on and to the right of the line $\mathrm{Re}(z) = 0.9$ , shaded in red below.

$\begin{aligned} (0.36)^{\frac 1n} \geq 0.9 \quad &\Leftrightarrow \quad \frac 1n \ln 0.36 \geq \ln 0.9 \\ &\Leftrightarrow \quad n \geq \frac{\ln 0.36}{\ln 0.9} \approx 9.70, \end{aligned}$

where we multiplied by $n/\ln 0.9 < 0$ . Therefore, the smallest such integer is $n = 10$ .

(c) (ii) By De Moivre’s theorem, $w = (0.36)^{\frac{1}{10}}\, \mathrm{cis}\, \frac{\theta}{10}$ , which is in the first quadrant since $0 < \frac{\theta}{10} < \frac{\pi}{20}$ .

$w = (0.36)^{0.1} \cos \tfrac{\pi}{50} + i\, (0.36)^{0.1} \sin \tfrac{\pi}{50} \approx 0.901 + 0.0567i.$

(c) (iii) (2) Yes. Since $\mathrm{Re}(w) \approx 0.901 \geq 0.9$ , $w$ is in $T$ . (This is guaranteed by (c) (i), since $\mathrm{Re}(w) = |w| \cos \frac{\pi}{50}$ with $|w| = 0.903$ .)

$|w - 1| \approx |{-0.0989} + 0.0567i| = \sqrt{0.0989^2 + 0.0567^2} \approx 0.114 > 0.1,$

so that $w$ is not within $S$ .

Remark 4. In particular, $S$ is a proper subset of $T$ , and $w$ lies in $T \backslash S$ : it is close enough to the real axis to satisfy $\mathrm{Re}(w) \geq 0.9$ , but its imaginary part is too large for it to lie within $0.1$ of $1$ .

Remark 5. The diagrams have been drawn and the solutions have been verified with the help of Generative AI.

—Joel Kindiak, 16 Oct 26, 2344H

KindiakMath

SACE Stage 2 Specialist Maths 2024 Suggested Answers

Leave a Reply Cancel reply

SACE Stage 2 Specialist Maths 2024 Suggested Answers

Share this:

Leave a Reply Cancel reply