SACE Stage 2 Maths Methods 2024 Suggested Answers

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Remark 1. In SACE Stage 2 Mathematical Methods, $\Pr$ is used for probability; we write $\mathbb P$ instead.

Question 1. Normal Distribution

Solution. (a) (i) Since $X \sim \mathcal N(54, 6^2)$ , the required probability is

$\mathbb P(50 \leq X \leq 52) \approx 0.117.$

(a) (ii) Similarly,

$\mathbb P(X \geq 58) \approx 0.252.$

(b) Since $\mathbb P(X \geq k) = 0.10$ , we have $\mathbb P(X < k) = 0.90$ . Using $z$ -scores,

$\displaystyle \frac{k - 54}{6} = 1.2816 \quad \Rightarrow \quad k = 54 + 6 \cdot 1.2816 = 61.689 \approx 61.7.$

(c) Since the dashes are one standard deviation apart and the dotted line marks the mean, the three boxes (from left to right) are labelled $48$ , $54$ and $60$ .

(d) (i) The sample mean of $4$ independent observations has

$\displaystyle \mathbb E[\bar X_4] = \mu = 54, \quad \sigma_{\bar X_4} = \frac{\sigma}{\sqrt 4} = \frac 62 = 3.$

(d) (ii) The probability density function of $\bar X_4$ has the same centre $54$ , half the spread, and hence twice the maximum height. We sketch as follows.

Remark 2. Since the area under every probability density function is $1$ , halving the standard deviation must double the maximum height of the curve.

Question 2. Kinematics

Solution. (a) Substituting $t = 1$ ,

$v_A(1) = 3e^{1.5} - 3 \approx 10.4\ \text{m/s}.$

(b) (i) Integrating with respect to $t$ ,

$\displaystyle s_A(t) = \int (3e^{1.5t} - 3)\, \mathrm dt = \frac{3}{1.5} e^{1.5t} - 3t + C = 2e^{1.5t} - 3t + C.$

Since $s_A(0) = 0$ , we have $2 + C = 0$ so that $C = -2$ , hence $s_A(t) = 2e^{1.5t} - 3t - 2$ , as required.

(b) (ii) By (b) (i),

$s_A(2) = 2e^{3} - 6 - 2 = 2e^3 - 8 \approx 32.2\ \text{m}.$

(d) Solving $v_A(t) = v_B(t)$ using software yields $t = 0$ or $t \approx 1.58$ . By (c), the velocity of Car A was greater than that of Car B for $1.58 < t \leq 2$ .

(e) Integrating $v_B$ with $s_B(0) = 0$ ,

$\displaystyle s_B(2) = \int_0^2 (6.6t^2 + 8t)\, \mathrm dt = \left[ 2.2t^3 + 4t^2 \right]_0^2 = 17.6 + 16 = 33.6\ \text{m}.$

Since both cars only travel forwards, the distance each car travels equals its displacement. Since $33.6 > 32.2$ , the models suggest that Car B travelled the furthest.

Remark 3. Although Car A ends the $2$ seconds much faster, Car B was faster for most of the interval, namely $0 < t < 1.58$ . This is why Car B is narrowly ahead.

Question 3. Logarithmic Functions

Solution. (a) By the chain rule, $\displaystyle f'(x) = \frac{1}{x+3}$ .

(b) By (a), $\displaystyle f'(0) = \frac 13$ .

(c) Similarly, $\displaystyle g'(x) = \frac{b}{x+3}$ , so that $A = (0, \tfrac 13)$ and $B = (0, \tfrac b3)$ . Since $B$ lies above $A$ ,

$\displaystyle \frac b3 - \frac 13 = 2 \quad \Rightarrow \quad b - 1 = 6 \quad \Rightarrow \quad b = 7.$

(d) By (c), $g(x) = 7f(x)$ . Therefore, $g(x)$ is obtained from $f(x)$ by a dilation of factor $7$ parallel to the $y$ -axis (i.e. from the $x$ -axis).

Question 4. Confidence Intervals

Solution. (a) (i) The sample proportion is $\hat p = 23/100 = 0.23$ . Using a $95\%$ confidence level, $z \approx 1.96$ , so that the confidence interval is

$\displaystyle 0.23 - 1.96 \sqrt{\frac{0.23 \cdot 0.77}{100}} \leq p \leq 0.23 + 1.96 \sqrt{\frac{0.23 \cdot 0.77}{100}} \quad \Rightarrow \quad 0.148 \leq p \leq 0.312.$

(a) (ii) Yes.

(a) (iii) The value $p = 0.35$ does not lie in the confidence interval, since $0.35 > 0.312$ .

(b) (i) The confidence intervals labelled $18$ , $40$ and $47$ do not contain $0.35$ .

(b) (ii) Only the confidence interval labelled $40$ lies entirely above $0.35$ .

Remark 4. Here $3/50 = 6\%$ of the confidence intervals miss the population proportion, which is close to the $5\%$ that we expect from $95\%$ confidence intervals.

$\mathbb P(X = 5) \approx 0.180.$

$\mathbb P(X \geq 1) = 1 - \mathbb P(X = 0) = 1 - 0.95^{100} \approx 0.994.$

(d) Let $Y \sim \mathrm B(n, 0.05)$ denote the number of $n$ independent $95\%$ confidence intervals that will not contain the population proportion. We require

$\begin{aligned} \mathbb P(Y \geq 1) = 1 - 0.95^n &> 0.5 \\ 0.95^n &< 0.5 \\ n \ln 0.95 &< \ln 0.5 \\ n &> \frac{\ln 0.5}{\ln 0.95} \approx 13.5, \end{aligned}$

where the inequality flips since $\ln 0.95 < 0$ . Therefore, the minimum number is $n = 14$ . As a check, $0.95^{13} \approx 0.513$ while $0.95^{14} \approx 0.488$ .

Question 5. Differentiation from First Principles

Solution. (a) For $x \neq 2$ and small $h \neq 0$ ,

$\begin{aligned} f(x+h) - f(x) &= \frac{x+h}{x+h-2} - \frac{x}{x-2} \\ &= \frac{(x+h)(x-2) - x(x+h-2)}{(x+h-2)(x-2)} \\ &= \frac{(x^2 - 2x + hx - 2h) - (x^2 + hx - 2x)}{(x+h-2)(x-2)} \\ &= \frac{-2h}{(x+h-2)(x-2)}. \end{aligned}$

Therefore,

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{-2}{(x+h-2)(x-2)} = -\frac{2}{(x-2)^2}.$

(b) By (a), the slope of the tangent at $x = 6$ is $\displaystyle f'(6) = -\frac{2}{16} = -\frac 18$ .

Question 6. Continuous Random Variables

Solution. (a) Since $f$ is a probability density function,

$\displaystyle 1 = \int_{400}^{440} \frac kx\, \mathrm dx = k \left[ \ln x \right]_{400}^{440} = k \ln \left( \frac{440}{400} \right) = k \ln 1.1 \quad \Rightarrow \quad k = \frac{1}{\ln 1.1}.$

(b) (i) Similarly,

$\displaystyle \int_{400}^{410} \frac{1}{(\ln 1.1) x}\, \mathrm dx = \frac{\ln(410/400)}{\ln 1.1} = \frac{\ln 1.025}{\ln 1.1} \approx 0.259.$

(b) (ii) The probability that a randomly chosen bag contains between $400$ and $410$ grams of sand (i.e. at most $10$ grams of excess sand) is approximately $0.259$ .

$\displaystyle \mu_X = \int_{400}^{440} x \cdot \frac{1}{(\ln 1.1)x}\, \mathrm dx = \frac{1}{\ln 1.1} \int_{400}^{440} 1\, \mathrm dx = \frac{440 - 400}{\ln 1.1} = \frac{40}{\ln 1.1}\ \text{grams}.$

(d) (i) Using a $99\%$ confidence level, $z \approx 2.5758$ , so that the confidence interval is

$\displaystyle 417 - 2.5758 \cdot \frac{11.5}{\sqrt{25}} \leq \mu \leq 417 + 2.5758 \cdot \frac{11.5}{\sqrt{25}} \quad \Rightarrow \quad 411 \leq \mu \leq 423.$

(d) (ii) No. By (c), the mean net weight before the alteration is $40/\ln 1.1 \approx 419.7$ grams, which lies inside the confidence interval. Hence, with $99\%$ confidence, the confidence interval does not rule out that the mean net weight is unchanged.

Question 7. Tangents and Conjectures

Solution. (a) Writing $f(x) = x^{-2}$ , we have $f'(x) = -2x^{-3}$ . Hence, $f(1) = 1$ and $f'(1) = -2$ , so that the tangent at $x = 1$ is

$y - 1 = -2(x - 1) \quad \Rightarrow \quad y = -2x + 3.$

(b) Setting $y = 0$ , $x = \frac 32$ . The $x$ -intercept is $(\frac 32, 0)$ .

(c) The $x$ -intercept of the tangent to $g(x)$ at $x = 1$ is $\displaystyle \left( \frac{k+3}{2}, 0 \right)$ . (This also agrees with (b), where $k = 0$ .)

(d) We prove the conjecture. Writing $g(x) = (x^2 + k)^{-1}$ , by the chain rule,

$\displaystyle g'(x) = -(x^2 + k)^{-2} \cdot 2x = -\frac{2x}{(x^2 + k)^2}.$

Since $k \geq 0$ , $1 + k \neq 0$ , so that

$\displaystyle g(1) = \frac{1}{1+k}, \quad g'(1) = -\frac{2}{(1+k)^2}.$

Therefore, the tangent to $g(x)$ at $x = 1$ has equation

$\displaystyle y - \frac{1}{1+k} = -\frac{2}{(1+k)^2} (x - 1).$

Setting $y = 0$ and multiplying both sides by $-(1+k)^2/2$ ,

$\displaystyle x - 1 = \frac{1+k}{2} \quad \Rightarrow \quad x = \frac{k+3}{2}.$

Hence, the $x$ -intercept is $\displaystyle \left( \frac{k+3}{2}, 0 \right)$ for every non-negative constant $k$ , which proves the conjecture.

Question 8. Integration Applications

Solution. (a) (i) At $9$ :00 am ( $9$ hours after midnight), vehicles were entering Park-o-Rama at a rate of $50.6$ vehicles per hour.

(a) (ii) (1) The graph of $y = L(t)$ passes through $P_2$ and $P_5$ , and its maximum occurs at $P_5$ . Therefore, vehicles were leaving at the maximum rate $17$ hours after midnight.

(a) (ii) (2) By Table 2, $\displaystyle \int_0^{24} E(t)\, \mathrm dt = 471$ vehicles entered during the $24$ -hour period.

(a) (ii) (3) By Table 2, the number of vehicles that left during the final $7$ hours is

$\displaystyle \int_{17}^{24} L(t)\, \mathrm dt = \int_0^{24} L(t)\, \mathrm dt - \int_0^{17} L(t)\, \mathrm dt = 406 - 298 = 108.$

(a) (iii) The number of vehicles in Park-o-Rama at time $t$ is $\displaystyle N(t) = \int_0^t E(s)\, \mathrm ds - \int_0^t L(s)\, \mathrm ds$ , which is maximised when $E(t) = L(t)$ with $E$ dropping below $L$ , i.e. at $P_4$ . Indeed, $N(15) = 429 - 227 = 202$ . Therefore, the maximum occurred $15$ hours after midnight.

(b) (i) EVs weigh $2.25 \times 1.36 = 3.06$ tonnes, and the proportion of non-EVs is $1 - 0.07 = 0.93$ .

(b) (ii) By (b) (i),

$\mu_X = 2.25 \cdot 0.93 + 3.06 \cdot 0.07 = 2.3067 \approx 2.31\ \text{tonnes}.$

(b) (iii) By (a) (iii), there are at most $n = 202$ vehicles in Park-o-Rama at any point during the day. Let $T$ denote the total weight of these $202$ vehicles. By the central limit theorem,

$T \sim \mathcal N(202 \cdot 2.31, 202 \cdot 0.207^2) = \mathcal N(466.62, 8.6555) \quad \text{approximately}.$

Hence,

$\mathbb P(T > 475) \approx 0.00220 < 0.005.$

At any other time, there are fewer vehicles in Park-o-Rama, so the probability that their total weight exceeds $475$ tonnes is even smaller. Therefore, the engineers would not be concerned.

Question 9. Exponential Functions

Solution. (a) For $P$ , $f'(x) = e^x - 8e^{2x} = e^x(1 - 8e^x) = 0$ yields $e^x = \frac 18$ , so that

$\displaystyle y = \frac 18 - 4 \cdot \frac{1}{64} = \frac{1}{16}.$

For $Q$ , $g'(x) = 3e^{3x} - 8e^{2x} = e^{2x}(3e^x - 8) = 0$ yields $e^x = \frac 83$ , so that

$\displaystyle y = \left( \frac 83 \right)^3 - 4 \left( \frac 83 \right)^2 = \left( \frac 83 \right)^2 \left( \frac 83 - 4 \right) = -\frac{256}{27}.$

Therefore, the $y$ -coordinates of $P$ and $Q$ are $\frac{1}{16}$ and $-\frac{256}{27}$ respectively.

(b) (i) Differentiating, the stationary points satisfy

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} = ae^{ax} - 8e^{2x} &= 0 \\ ae^{ax} &= 8e^{2x} \\ e^{(a-2)x} &= \frac 8a \\ (a-2)x &= \ln \left( \frac 8a \right) \\ x &= \frac{1}{a-2} \ln \left( \frac 8a \right), \end{aligned}$

where we used $8/a > 0$ (since $a > 0$ ) and $a \neq 2$ . Since this equation has exactly one solution, this is the only stationary point.

(b) (ii) (1) Denote the stationary point by $x_0$ . By (b) (i), $e^{(a-2)x_0} = 8/a$ , so that

$\displaystyle e^{2x_0} = \left( e^{(a-2)x_0} \right)^{\frac{2}{a-2}} = \left( \frac 8a \right)^{\frac{2}{a-2}}, \quad e^{ax_0} = e^{(a-2)x_0} \cdot e^{2x_0} = \frac 8a \left( \frac 8a \right)^{\frac{2}{a-2}}.$

Therefore,

$\begin{aligned} y &= e^{ax_0} - 4e^{2x_0} \\ &= \left( \frac 8a \right)^{\frac{2}{a-2}} \left( \frac 8a - 4 \right) \\ &= 4 \left( \frac 8a \right)^{\frac{2}{a-2}} \left( \frac{2-a}{a} \right). \end{aligned}$

(b) (ii) (2) Since $8/a > 0$ , $4 (8/a)^{\frac{2}{a-2}} > 0$ . Hence, the $y$ -coordinate is positive if and only if $\frac{2-a}{a} > 0$ . Since $a > 0$ , this holds if and only if $2 - a > 0$ , i.e. $0 < a < 2$ .

Remark 5. As a check, setting $a = 1$ and $a = 3$ in (b) (ii) (1) recovers the $y$ -coordinates $\frac{1}{16} > 0$ and $-\frac{256}{27} < 0$ from (a).

Question 10. Calculus of Exponential and Trigonometric Functions

Solution. (a) At $P$ , $f(x) = f'(x)$ :

$e^{-\cos x} = e^{-\cos x} \sin x \quad \Rightarrow \quad e^{-\cos x} (1 - \sin x) = 0.$

Since $e^{-\cos x} > 0$ , $\sin x = 1$ , so that for $0 \leq x \leq 2\pi$ , the $x$ -coordinate of $P$ is $x = \frac{\pi}{2}$ .

(b) (i) By the product rule and the chain rule,

$\begin{aligned} f''(x) &= \frac{\mathrm d}{\mathrm dx} \left( e^{-\cos x} \sin x \right) \\ &= e^{-\cos x} \sin x \cdot \sin x + e^{-\cos x} \cos x \\ &= e^{-\cos x} (\sin^2 x + \cos x) \\ &= e^{-\cos x} (-\cos^2 x + \cos x + 1). \end{aligned}$

(b) (ii) Setting $f''(x) = f(x)$ and dividing by $e^{-\cos x} > 0$ ,

$-\cos^2 x + \cos x + 1 = 1 \quad \Rightarrow \quad \cos x (1 - \cos x) = 0.$

Therefore, $\cos x = 0$ or $\cos x = 1$ , so that $x = 0, \frac{\pi}{2}, \frac{3\pi}{2}, 2\pi$ .

(c) By (a), $f'(\frac{\pi}{2}) = f(\frac{\pi}{2})$ . By (b) (ii), $f''(\frac{\pi}{2}) = f(\frac{\pi}{2})$ . Hence, $f'(\frac{\pi}{2}) = f''(\frac{\pi}{2})$ , so that the graphs of $y = f'(x)$ and $y = f''(x)$ intersect at $x = \frac{\pi}{2}$ . This is the intersection labelled $k$ in Figure 10, so that $k = \frac{\pi}{2}$ .

(d) Let $c$ denote the other intersection point. On region $A$ , $f'(x) \geq f''(x)$ (e.g. $f'(\pi) = 0 > -e = f''(\pi)$ ), while on region $B$ , $f''(x) \geq f'(x)$ (e.g. $f''(\frac{3\pi}{2}) = 1 > -1 = f'(\frac{3\pi}{2})$ ). Therefore,

$\begin{aligned} A - B &= \int_{\pi/2}^{c} (f'(x) - f''(x))\, \mathrm dx - \int_c^{2\pi} (f''(x) - f'(x))\, \mathrm dx \\ &= \int_{\pi/2}^{2\pi} (f'(x) - f''(x))\, \mathrm dx \\ &= \left[ f(x) - f'(x) \right]_{\pi/2}^{2\pi} \\ &= (e^{-1} - 0) - (1 - 1) \\ &= \frac 1e. \end{aligned}$

Therefore, the area of region $A$ exceeds the area of region $B$ by exactly $\frac 1e$ units $^2$ .

Remark 6. We never needed the value of $c \approx 3.72$ : it cancels when the two integrals are combined. Furthermore, the two graphs in Figure 9 touch at $P$ because both their values and their gradients agree there: $f(\frac{\pi}{2}) = f'(\frac{\pi}{2}) = 1$ and, by (c), $f'(\frac{\pi}{2}) = f''(\frac{\pi}{2})$ .

Remark 7. The diagrams have been drawn and the solutions have been verified with the help of Generative AI.

—Joel Kindiak, 16 Oct 26, 0138H

KindiakMath

SACE Stage 2 Maths Methods 2024 Suggested Answers

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SACE Stage 2 Maths Methods 2024 Suggested Answers

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