SACE Stage 2 Specialist Maths 2025 Suggested Answers

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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:

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

and denote vector norms by $|\mathbf v| \equiv \| \mathbf v \|$ .

Question 1. Related Rates of Change

Solution. (a) Setting $V = \displaystyle \frac{200}3 \pi$ ,

$\displaystyle \frac{11\pi}2 h^2 - \frac{\pi}3 h^3 = \frac{200\pi}{3} \quad \Rightarrow \quad 2h^3-33h^2 +400 = 0.$

Solving for $0 \leq h \leq 5$ , we have $h = 4$ .

(b) Differentiating with respect to $h$ ,

$\begin{aligned} \frac{\mathrm dV}{\mathrm dh} &= \frac{\mathrm d}{\mathrm dh} \left( \frac{11\pi}2 h^2 - \frac{\pi}3 h^3 \right) \\ &= \frac{11\pi}2\frac{\mathrm d}{\mathrm dh} \cdot ( h^2 )- \frac{\pi}3 \cdot \frac{\mathrm d}{\mathrm dh} ( h^3 ) \\ &= \frac{11 \pi}{2} \cdot 2h - \frac{\pi}3 \cdot 3h^2 \\ &= \pi h ( 11 - h). \end{aligned}$

By the chain rule,

$\begin{aligned} \frac{\mathrm dV}{\mathrm dt} = \pi h ( 11 - h) \cdot \frac{\mathrm dh}{\mathrm dt} \quad \Rightarrow \quad \frac{\mathrm dh}{\mathrm dt} &= \frac 1{\pi h ( 11 - h)} \frac{\mathrm dV}{\mathrm dt}. \end{aligned}$

(c) By (a), $h = 4$ when $V = \displaystyle \frac{200}3 \pi$ . By the question, $\displaystyle \frac{\mathrm dV}{\mathrm dt} = 2$ . By (b),

$\displaystyle \frac{\mathrm dh}{\mathrm dt} = \frac 1{\pi \cdot 4 \cdot (11-4)} \cdot 2 = \frac 1{14\pi} \approx 0.0227\, \text{cm}/\text s.$

Question 2. Integral Calculus

Solution. (a) Integrating by parts and using the assumed integral,

$\begin{aligned} \int x^2 e^{2x}\, \mathrm dx &= \int x \cdot xe^{2x}\, \mathrm dx \\ &= \underbrace{\phantom. x \phantom.}_{u} \cdot \underbrace{\phantom. \left(\textstyle{\frac 12} xe^{2x} - \textstyle{\frac 14}e^{2x} \right)\phantom.}_{v} - \int \underbrace{\phantom.\left( \textstyle{\frac 12} xe^{2x} - \textstyle{\frac 14}e^{2x} \right) \phantom.}_{v} \cdot \underbrace{\phantom. 1\, \mathrm dx\phantom.}_{\mathrm du} \\ &= \textstyle{\frac 12} x^2e^{2x} - \textstyle{\frac 14}xe^{2x} - \left( \textstyle{\frac 12} \displaystyle \int xe^{2x}\, \mathrm dx - \textstyle{\frac 14} \displaystyle \int e^{2x}\, \mathrm dx \right) \\ &= \textstyle{\frac 12} x^2e^{2x} - \textstyle{\frac 14}xe^{2x} - \textstyle{\frac 12} \displaystyle \int xe^{2x}\, \mathrm dx + \textstyle{\frac 14} \displaystyle \int e^{2x}\, \mathrm dx \\ &= \textstyle{\frac 12} x^2e^{2x} - \textstyle{\frac 14}xe^{2x} - \textstyle{\frac 12} \cdot \left( \textstyle{\frac 12} xe^{2x} - \textstyle{\frac 14}e^{2x} \right) + \textstyle{\frac 14} \cdot \textstyle{\frac 12} e^{2x} + C \\ &= \textstyle{\frac 12} x^2e^{2x} - \textstyle{\frac 12}xe^{2x} + \textstyle{\frac 14}e^{2x} + C. \end{aligned}$

(b) Using the volume of revolution formula, the required volume $V$ is given by

$\begin{aligned} V &= \pi \int_{-1}^0 (xe^x)^2\, \mathrm dx \\ &= \pi \int_{-1}^0 x^2 e^{2x}\, \mathrm dx \\ &= \pi \left[ \textstyle{\frac 12} x^2e^{2x} - \textstyle{\frac 12}xe^{2x} + \textstyle{\frac 14}e^{2x} \right]_{-1}^0 \\ &= \textstyle{\frac 14} \pi ( 1-e^{-2} )\, \text{units}^3. \end{aligned}$

Question 3. Complex Number Algebra

Solution. (a) (i) By Euler’s identity and the $2\pi$ -periodicity of cosine and sine,

$\mathrm{cis}\, 3\pi = \mathrm{cis}\, \pi = -1.$

By De Moivre’s theorem,

$\begin{aligned} \left(2\, \mathrm{cis}\, \frac{3\pi}{5} \right)^5 &= 2^5\, \mathrm{cis} \left( 5 \cdot \frac{3\pi}5 \right) \\ &= 32\, \mathrm{cis}\, 3\pi \\ &= 32 \cdot (-1) \\ &= -32. \end{aligned}$

(a) (ii) Let $z = r\, \mathrm{cis}\, \theta$ denote a solution to $z^5 = -32$ , where $r > 0$ and $-\pi < \theta \leq \pi$ . By De Moivre’s theorem,

$r^5\, \mathrm{cis}\, 5\theta = z^5 = -32 = 32\, \mathrm{cis}((2k+1)\pi).$

Comparing moduli, $r^5 = 32 \Rightarrow r = 2$ . Comparing arguments,

$\displaystyle \theta = \frac{2k+1}{5}\pi \in (-\pi, \pi].$

Solving for $k$ , we get

$-5 < 2k+1 \leq 5,$

so that $k=-2,-1,0,1,2$ . Therefore, we obtain the solutions

$\displaystyle -2, \quad 2\, \mathrm{cis}\, \frac{\pi}{5}, \quad 2\, \mathrm{cis}\, \frac{3\pi}{5}, \quad 2\, \mathrm{cis}\, \left(-\frac{3\pi}{5}\right), \quad 2\, \mathrm{cis}\, \left(-\frac{\pi}{5}\right)$

from $k=2,0,1,-2,-1$ respectively.

(b) (i) We take advantage of the geometric series:

$\begin{aligned} \frac{z^5 + 32}{z+2} &= \frac {-32}{-2} \cdot \frac{(-z/2)^5 - 1}{(-z/2)-1} \\ &= 16 \cdot (1 + (-z/2) + (-z/2)^2 + (-z/2)^3 + (-z/2)^4) \\ &= 16 \cdot \left( 1 -\frac z2 + \frac{z^2}4 - \frac{z^3}8 + \frac{z^4}{16} \right) \\ &= 16 - 8z + 4z^2 - 2z^3 + z^4, \end{aligned}$

as required, where the final result follows from the commutativity of addition.

(b) (ii) Using (a) (ii), since $z \neq -2$ , we obtain the solutions

$\displaystyle 2\, \mathrm{cis}\, \frac{\pi}{5}, \quad 2\, \mathrm{cis}\, \frac{3\pi}{5}, \quad 2\, \mathrm{cis}\, \left(-\frac{3\pi}{5}\right), \quad 2\, \mathrm{cis}\, \left(-\frac{\pi}{5}\right).$

$\displaystyle \mathcal A(\Delta AOD) = \frac 12 \cdot 2 \cdot 2 \sin \frac{2\pi}{5} = 2 \sin \frac{2\pi}5.$

$\displaystyle \mathcal A(\Delta BOC) = \frac 12 \cdot 2 \cdot 2 \sin \frac{4\pi}{5} = 2 \sin \frac{4\pi}5.$

By symmetry,

$\displaystyle \mathcal A(ABCD) = 6 \sin \frac{2\pi}5 + 2 \sin \frac{4\pi}5.$

Therefore, $(a,b) = (6,2)$ .

Question 4. Mathematical Induction

Solution. (a) By the product rule,

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx} ((2x+1)e^x ) \\ &= 2e^x + (2x+1) e^x \\ &= (2+2x+1)e^x \\ &= (2x+3)e^x.\end{aligned}$

(b) For the base case $n = 1$ , part (a) yields

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = (2x+3)e^x = (2x + (2 \cdot 1 + 1))e^x,$

as required. Fix $k \in \mathbb N^+$ such that

$\displaystyle \frac{\mathrm d^k y}{\mathrm dx^k} = (2x + (2k + 1))e^x.$

By the product rule,

$\begin{aligned} \frac{\mathrm d^{k+1} y}{\mathrm dx^{k+1}} &= \frac{\mathrm d}{\mathrm dx} ((2x + (2k + 1))e^x ) \\ &= 2e^x + (2x + (2k + 1)) e^x \\ &= (2+2x + (2k + 1))e^x \\ &= (2x + 2k+3)e^x \\ &= (2x + 2(k+1)+1)e^x. \end{aligned}$

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

Question 5. Vector Algebra

Solution. (a) Using cross product properties,

$\begin{aligned} (2\mathbf a + 3\mathbf b) \times (\mathbf a + 5\mathbf b) &= 2 \mathbf a \times \mathbf a + 10 \mathbf a \times \mathbf b + 3 \mathbf b \times \mathbf a + 15 \mathbf b \times \mathbf b \\ &= \mathbf 0 + 10 \mathbf a \times \mathbf b - 3 \mathbf a \times \mathbf b + \mathbf 0 \\ &= 7 \mathbf a \times \mathbf b. \end{aligned}$

(b) (i) Using vector addition,

$\begin{aligned} \overrightarrow{OE} &= \overrightarrow{OC} + \overrightarrow{CD} + \overrightarrow{DE} \\ &= \mathbf a + \mathbf a + 3\mathbf b \\ &= 2\mathbf a + 3\mathbf b. \end{aligned}$

(b) (ii) Similarly,

$\overrightarrow{OF} = \mathbf a + 5\mathbf b.$

By (a),

$\begin{aligned} \mathcal A(\Delta OEF) &= \frac 12 | \overrightarrow{OE} \times \overrightarrow{OF} | \\ &= \frac 12 | (2\mathbf a + 3\mathbf b) \times (\mathbf a + 5\mathbf b) | \\ &= \frac 12 | 7 \mathbf a \times \mathbf b | \\ &= \frac 72 | \mathbf a \times \mathbf b | . \end{aligned}$

(b) (iii) Similarly,

$\begin{aligned} \mathcal A(\Delta ODE) &= \frac 12 |\overrightarrow{OD} \times \overrightarrow{DE}| \\ &= \frac 12 |2\mathbf a \times 3\mathbf b| \\ &= \frac 12 \cdot 2 \cdot 3 \cdot |\mathbf a \times \mathbf b| \\ &= 3|\mathbf a \times \mathbf b|. \end{aligned}$

Therefore,

$\begin{aligned} \mathcal A(ODEF) &= \mathcal A(\Delta OEF) + \mathcal A(\Delta ODE) \\ &= \frac 72 | \mathbf a \times \mathbf b | + 3|\mathbf a \times \mathbf b| \\ &= \frac{13}{2} |\mathbf a \times \mathbf b|. \end{aligned}$

$\begin{aligned} \mathbf a \times \mathbf b &= [1,2,4] \times [3,1,-5] \\ &= [-10 - 4, -5-12, 1-6] \\ &= [-14, -17, -5] \\ &= -[14,17,5]. \end{aligned}$

Computing the vector norm,

$|\mathbf a \times \mathbf b| = \sqrt{14^2 + 17^2 + 5^2} = \sqrt{510}.$

By (b) (iii),

$\mathcal A(ODEF) = \frac{13}2 \sqrt{510} \approx 147\, \text{units}^2.$

Question 6. Complex Number Geometry

Solution. (a) We draw the ray as follows.

(b) (i) In Cartesian form, $P = k+ki$ .

(ii) In polar form, $P = k\sqrt 2\, \mathrm{cis}\, \frac{1}{4}\pi$ .

(d) We have

$|v - 2^2| = |1+3i| = \sqrt{10} < \sqrt{16} = 2^2,$

so that $v$ lies inside $C_2$ .

(e) Since $w$ has argument $\pi/4$ it lies on the ray drawn in (a). For any point $z$ on the ray, $z$ lies on $C_n$ if and only if $|z| = 2^{n+1/2}$ . Hence, we aim to find the smallest $n$ such that

$|w| < 2^{n+1/2}$

Since $|w| = 10$ , taking logarithms yields

$\displaystyle n > \frac{\ln 10}{\ln 2}-1/2 \approx 2.82.$

Therefore, $n = 3$ .

Question 7. Functions

Solution. (a) (i) Using the Pythagorean identity,

$\begin{aligned} \sin^3 x &= \sin x \cdot \sin^2 x \\ &= \sin x \cdot (1-\cos^2 x) \\ &= \sin x - \sin x \cos^2 x. \end{aligned}$

(a) (ii) Using linearity and the reverse chain rule,

$\begin{aligned} \int \sin^3 x\, \mathrm dx &= \int \sin x\, \mathrm dx + \int(-\sin x) (\cos x)^2\, \mathrm dx \\ &= -\cos x + \frac{(\cos x)^3}{3} + C \\ &= -\cos x + \textstyle{\frac 13} \cos^3 x + C. \end{aligned}$

Substituting the limits of the definite integral,

$\begin{aligned} \int_0^{\pi/3} \sin^3 x\, \mathrm dx &= \left[ -\cos x + \textstyle{\frac 13} \cos^3 x \right]_0^{\pi/3} \\ &= \left( -\cos(\pi/3) + \textstyle{\frac 13} \cos^3(\pi/3) \right) - \left( -1 + \textstyle{\frac 13} \cdot 1 \right) \\ &= \left( - \frac 12 + \frac 13 \cdot \left( \frac 12 \right)^3 \right) + \frac 23 \\ &= \frac 5{24}. \end{aligned}$

(b) We sketch as follows.

(d) The area of the white space, by (a) (ii), is $5/24$ . By symmetry, the total area of the white space is $8 \times 5/24 = 5/3$ . Since the square has an area of $(2\pi/3)^2 = 4\pi^2 /9$ , the required area is what is left over:

$\displaystyle \left( \frac{4\pi^2}9 - \frac 53\right)\, \text{units}^2.$

Question 8. Vector Geometry

Solution. (a) (i) Substituting $x = d_1/a, y = 0, z = 0$ , we get

$\displaystyle a \cdot \frac{d_1}a + b \cdot 0 + c \cdot 0 = d_1,$

as required.

(a) (ii) Similarly, the point $(d_2/a, 0, 0)$ lies on the plane

$ax + by + cz - d_2 = 0.$

By (a) (i), the distance between the planes is the distance from $(d_1/a,0,0)$ to the second plane:

$\displaystyle \frac{ |a \cdot d_1/a + b\cdot 0 + c \cdot 0 - d_2| }{\sqrt{a^2+b^2+c^2}} = \frac{ |d_1 - d_2| }{\sqrt{a^2+b^2+c^2}}.$

(b) (i) Particularising (a) (ii) to $(a,b,c,d_1,d_2) = (2,-1,1,20,-10)$ , we obtain a distance of

$\displaystyle \frac{ |20 - (-10)| }{\sqrt{2^2+(-1)^2+1^2}} = \frac{30}{\sqrt 6} = 5\sqrt{6} \approx 12.2\, \text{units}.$

(b) (ii) Since the normal vector of $P_1$ is given by $[2,-1,1]$ , the normal is given by the line equation

$[x,y,z] = [10,2,2] + t[2,-1,1],\quad t \in \mathbb R.$

(b) (iii) Writing in parametric form,

$\overrightarrow{OB} = [10+2t, 2-t, 2 + t]$

for some $t$ . When the normal intersects $P_2$ at $B$ ,

$2(10+2t) - (2-t) + (2+t) = -10 \quad \Rightarrow \quad t = -5.$

Therefore,

$\overrightarrow{OB} = [10+2\cdot(-5), 2-(-5), 2 + (-5)] = [0, 7, -3],$

so that $B(0,7, -3)$ .

$P_3 : 2x - y + z = d_3$

for some $d_3$ to be determined. Just like in (a), $(d_3/2, 0, 0)$ lies on $P_3$ . Denote the distance between $P_1$ and $P_3$ by $\delta$ . Since $P_1$ lies between $P_2$ and $P_3$ , by (b) (i) the distance between $P_2$ and $P_3$ is $30/\sqrt 6 + \delta$ . By the question, $30/\sqrt 6 + \delta = 3\delta$ , so that $\delta = 15/\sqrt 6$ . Therefore, by (a) (ii),

$\displaystyle \frac{|20 - d_3|}{\sqrt 6} = \frac{15}{\sqrt 6} \quad \Rightarrow \quad d_3 \in \{5, 35\}.$

Likewise, the distance between $P_2$ and $P_3$ is $3\delta = 45/\sqrt 6$ . Therefore, by (a) (ii),

$\displaystyle \frac{| {-10 - d_3} |}{\sqrt 6} = \frac{45}{\sqrt 6} \quad \Rightarrow \quad d_3 \in \{-55, 35\}.$

Therefore, $d_3 \in \{5,35\} \cap \{-55,35\} = \{35\}$ , i.e. $d_3 = 35$ . Hence,

$P_3 : 2x - y + z = 35.$

(d) (i) Let $Q$ denote the foot of perpendicular from $A$ to the line $BC$ , so that $Q = A$ if $A$ lies on $BC$ . By Pythagoras’ theorem (trivially if $Q \in \{A, B\}$ ),

$\begin{aligned} |BA|^2 = |BQ|^2 + |QA|^2 \geq |BQ|^2. \end{aligned}$

Taking square roots, $|QB| \leq |AB|$ . Similarly, $|QC| \leq |AC|$ . Since $B, Q, C$ are collinear,

$|BC| \leq |QB| + |QC| \leq |AB| + |AC|.$

Remark 2. This is simply the triangle inequality.

(d) (ii) By (d) (i), equality holds if and only if $|QA| = 0$ and $Q$ lies between $B$ and $C$ , i.e. $A$ lies on the line segment $BC$ . In that case, $C$ lies on the line $AB$ , which is the normal from (b) (ii). Since $C$ also lies on $P_3$ , it is the intersection of the normal with $P_3$ . The normal has equation

$\displaystyle [x,y,z] = [10,2,2] + t[2,-1,1] = [10+2t, 2-t, 2+t],\quad t \in \mathbb R$

while $P_3 : 2x - y +z = 35$ . Intersecting the two objects,

$2(10+2t) - (2-t) + (2+t) = 35 \quad \Rightarrow \quad t = 5/2.$

Therefore,

$\overrightarrow{OC} = [10,2,2] + \frac 52 [2,-1,1] = [15, -1/2, 9/2].$

Question 9. Parametric Equations

Solution. (a) Using the double angle identity,

$\sqrt{1+ \cos 2x} = \sqrt{2 \cos^2 x} = \sqrt 2 |{\cos x}| = \sqrt 2 \cos x,$

for $-\pi/2 \leq x \leq \pi/2$ . Therefore,

$\displaystyle \int \sqrt{1 + \cos 2x}\, \mathrm dx = \sqrt 2 \int \cos x\, \mathrm dx = \sqrt 2 \sin x + C.$

(b) We sketch as follows

$\begin{aligned} x'(t) &= a((2t)' + (\sin 2t)') \\ &= a(2 + \cos 2t \cdot 2) \\ &= 2a + 2a \cos 2t. \end{aligned}$

Using the double angle identity again,

$\begin{aligned} y'(t) &= (-2a \cos^2 t)' \\ &= -2a \cdot ((\cos t)^2)' \\ &= -2a \cdot 2\cos t \cdot (-\sin t) \\ &= 2a \cdot 2 \sin t \cos t \\ &= 2a \sin 2t. \end{aligned}$

$\begin{aligned} \mathbf v \cdot \mathbf v &= (2a)^2 \cdot ((1+\cos 2t)^2 + \sin^2 2t) \\ &= 4a^2 \cdot (1 + 2 \cos 2t + \cos^2 2t + \sin^2 2t) \\ &= 4a^2 \cdot (2 + 2 \cos 2t) \\ &= 8a^2 \cdot (1+ \cos 2t). \end{aligned}$

Taking square roots,

$\sqrt{\mathbf v \cdot \mathbf v} = 2a\sqrt 2 \cdot \sqrt{1+\cos 2t}.$

By linearity, the required arc length is given by

$\begin{aligned} I = \int_{-\pi/2}^{\pi/2} 2a\sqrt 2 \cdot \sqrt{1+\cos 2t}\, \mathrm dt &= 2a\sqrt 2 \int_{-\pi/2}^{\pi/2} \sqrt{1+\cos 2t}\, \mathrm dt. \end{aligned}$

(d) (i) By (a) and (c) (ii),

$\begin{aligned} I &= 2a\sqrt 2 \cdot \left[ \sqrt 2 \sin t \right]_{-\pi/2}^{\pi/2} \\ &= 2a\sqrt 2 \cdot \sqrt 2 \cdot (1-(-1)) \\ &= 8a. \end{aligned}$

Since $a = 1$ and the string length is $4$ units, we have $8\cdot 1 = 4k \Rightarrow k = 2$ .

(d) (ii) The arc length is given by $I = 8$ .

Question 10. Differential Equations

Solution. (a) By the chain rule,

$\displaystyle \frac{\mathrm dA}{\mathrm dX} \cdot \frac{\mathrm dX}{\mathrm dt} = \frac{\mathrm dA}{\mathrm dt} = -\frac 14 \frac{\mathrm dX}{\mathrm dt}.$

Therefore,

$\displaystyle \frac{\mathrm dA}{\mathrm dX} = -\frac 14 \quad \Rightarrow \quad A = \int -\frac 14\, \mathrm dX = -\frac 14 X + C.$

Since $A(0) = 2$ , we have $2 = 0 + C$ so that $C = 2$ , hence

$A = -\frac 14 X + 2.$

(b) Substituting the expressions for $A$ and $B$ ,

$\begin{aligned} \frac{\mathrm dX}{\mathrm dt} &= \frac 13 \left( -\frac 14 X + 2 \right) \left( -\frac 34X + 3 \right) \\ &= \frac 13 \cdot \frac 14 \cdot \frac 34 \cdot (-X + 8) \cdot (-X + 4) \\ &= \frac 1{16} \cdot (-1) \cdot (X-8) \cdot (-1) \cdot (X-4) \\ &= \frac 1{16} (X-8)(X-4). \end{aligned}$

$\displaystyle \frac 1{(X-8)(X-4)} = \frac A{X-8} + \frac B{X-4}.$

Clearing denominators,

$1 = A(X-4) + B(X-8).$

Setting $X = 8$ yields $A = 1/4$ , and setting $X = 4$ yields $B = -1/4$ , so that

$\displaystyle \frac 1{(X-8)(X-4)} = \frac {1/4}{X-8} + \frac {-1/4}{X-4} = \frac 14 \left( \frac 1{X-8} - \frac 1{X-4} \right).$

(d) (i) By the method of separable variables,

$\displaystyle \int \frac {1}{(X-8)(X-4)}\, \mathrm dX = \int \frac 1{16}\, \mathrm dt = \frac 1{16}t + C_1.$

To evaluate the left-hand side, we use (c) and linearity:

$\begin{aligned} \int \frac {1}{(X-8)(X-4)}\, \mathrm dX &= \frac 14 \int \left( \frac 1{X-8} - \frac 1{X-4} \right)\, \mathrm dX \\ &= \frac 14 \left( \ln|X-8| - \ln |X-4| \right) +C_2\\ &= \frac 14 \ln \left| \frac{X-8}{X-4} \right| + C_2. \end{aligned}$

Abbreviating $C := C_1 - C_2$ ,

$\begin{aligned} \frac 14 \ln \left| \frac{X-8}{X-4} \right| &= \frac 1{16}t + C \\ \left| \frac{X-8}{X-4} \right| &= e^{\frac 1{4}t + 4C} \\ \frac{X-8}{X-4} &= \pm e^{4C} e^{\frac 14 t} =: Ae^{\frac 14 t} \\ X(1- Ae^{\frac 14 t}) &= 4(2 - Ae^{\frac 14 t}) \\ X &= \frac{ 4(2 - Ae^{\frac 14 t}) }{ 1 - Ae^{\frac 14 t} }. \end{aligned}$

Using $X(1) = 1$ ,

$\displaystyle \frac{1-8}{1-4} = Ae^{\frac 14} \quad \Rightarrow \quad A = \frac{7}{3}e^{-\frac 14}.$

Therefore,

$\begin{aligned} X &= \frac{ 4\left(2 - \frac{7}{3}e^{-\frac 14} \cdot e^{\frac 14 t}\right) }{ 1 - \frac{7}{3}e^{-\frac 14} \cdot e^{\frac 14 t} } \\ &= \frac{ 4\left(6 - 7 e^{\frac 14 (t-1)}\right) }{ 3 - 7\cdot e^{\frac 14 (t-1)} } \\ &= \frac{ 24 - 28 e^{\frac 14 (t-1)} }{ 3 - 7e^{\frac 14 (t-1)} }. \end{aligned}$

(d) (ii) Writing $24 = 12 + 12$ ,

$\begin{aligned} X &= \frac{ 12 - 28 e^{\frac 14 (t-1)} + 12 }{ 3 - 7e^{\frac 14 (t-1)} }\\ &= \frac{ 4(3 - 7 e^{\frac 14 (t-1)}) + 12 }{ 3 - 7e^{\frac 14 (t-1)} } \\ &= 4 + \frac{ 12 }{ 3 - 7e^{\frac 14 (t-1)} }. \end{aligned}$

(d) (iii) We note that as $t \to \infty$ , we have

$\displaystyle X \to 4 + 0 = 4.$

(e) We draw as follows.

Remark 3. The diagrams have been drawn and the solutions have been verified with the help of Generative AI.

—Joel Kindiak, 15 Oct 26, 1234H

KindiakMath

SACE Stage 2 Specialist Maths 2025 Suggested Answers

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SACE Stage 2 Specialist Maths 2025 Suggested Answers

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