2021 HKDSE Mathematics M1 (Calculus and Statistics) Practice Paper with Answers

(a) Since the sum of all probabilities is 1,
\[ p + 0.20 + 0.10 + q + 0.15 = 1 \implies p + q = 0.55 \quad \text{--- (1)} \]

Given \(\mathrm{E}(3X - 2) = 2.5\),
\[ 3\mathrm{E}(X) - 2 = 2.5 \implies \mathrm{E}(X) = 1.5 \]

Using the definition of expectation:
\[ (-2)(p) + 0(0.20) + 1(0.10) + 3(q) + 5(0.15) = 1.5 \]
\[ -2p + 0.10 + 3q + 0.75 = 1.5 \implies -2p + 3q = 0.65 \quad \text{--- (2)} \]

From (1), \(2p + 2q = 1.10\). Adding to (2):
\[ 5q = 1.75 \implies q = 0.35 \]
\[ p = 0.55 - 0.35 = 0.20 \]

(b) The event \(B\) corresponds to \(X \in \{-2, 0, 1, 3\}\).
\[ \mathrm{P}(B) = p + 0.20 + 0.10 + q = 0.20 + 0.20 + 0.10 + 0.35 = 0.85 \]

The event \(A \cap B\) corresponds to \(X \in \{0, 1, 3\}\).
\[ \mathrm{P}(A \cap B) = 0.20 + 0.10 + 0.35 = 0.65 \]

Therefore,
\[ \mathrm{P}(A \mid B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} = \frac{0.65}{0.85} = \frac{13}{17} \approx 0.7647 \]

(a) 1M for \(p + q = 0.55\); 1M for \(\mathrm{E}(X) = 1.5\) and setting up the linear equation in \(p\) and \(q\); 1A for both \(p = 0.20\) and \(q = 0.35\).
(b) 1M for finding \(\mathrm{P}(B)\) and \(\mathrm{P}(A \cap B)\); 1M for applying conditional probability formula; 1A for \(\frac{13}{17}\) (or r.t. 0.7647).

Let \(D\) be the event that a component is defective, and \(T\) be the event that the device classifies the component as defective.
Given: \(\mathrm{P}(D) = 0.08\), \(\mathrm{P}(D') = 0.92\), \(\mathrm{P}(T \mid D) = 0.95\), \(\mathrm{P}(T' \mid D') = 0.92\), so \(\mathrm{P}(T \mid D') = 1 - 0.92 = 0.08\).

(a) The probability that a component is classified as defective is:
\[ \mathrm{P}(T) = \mathrm{P}(D)\mathrm{P}(T \mid D) + \mathrm{P}(D')\mathrm{P}(T \mid D') \]
\[ \mathrm{P}(T) = (0.08)(0.95) + (0.92)(0.08) = 0.076 + 0.0736 = 0.1496 \]

(b) An incorrect classification occurs if the component is defective and tested non-defective, or non-defective and tested defective:
\[ \mathrm{P}(\text{Incorrect}) = \mathrm{P}(D)\mathrm{P}(T' \mid D) + \mathrm{P}(D')\mathrm{P}(T \mid D') \]
\[ = (0.08)(1 - 0.95) + (0.92)(0.08) = (0.08)(0.05) + 0.0736 = 0.004 + 0.0736 = 0.0776 \]

(c) By Bayes' Theorem:
\[ \mathrm{P}(D \mid T) = \frac{\mathrm{P}(D \cap T)}{\mathrm{P}(T)} = \frac{(0.08)(0.95)}{0.1496} = \frac{0.076}{0.1496} = \frac{95}{187} \approx 0.5080 \]

(a) 1M for \((0.08)(0.95) + (0.92)(0.08)\); 1A for 0.1496.
(b) 1M for \((0.08)(0.05) + (0.92)(0.08)\); 1A for 0.0776.
(c) 1M for using Bayes' theorem \(\frac{0.076}{0.1496}\); 1A for \(\frac{95}{187}\) (or r.t. 0.5080).

(a) Let \(K\) be the event that David knows the answer, and \(C\) be the event that he answers correctly.
\[ \mathrm{P}(C) = \mathrm{P}(K)\mathrm{P}(C \mid K) + \mathrm{P}(K')\mathrm{P}(C \mid K') \]
\[ = (0.7)(0.9) + (1 - 0.7)(0.25) = 0.63 + 0.075 = 0.705 \]

(b) Let \(Y\) be the number of questions David answers correctly. Then \(Y \sim \mathrm{B}(10, 0.705)\).
\[ \mathrm{P}(Y \ge 8) = \mathrm{P}(Y = 8) + \mathrm{P}(Y = 9) + \mathrm{P}(Y = 10) \]
\[ = C_8^{10} (0.705)^8 (0.295)^2 + C_9^{10} (0.705)^9 (0.295)^1 + C_{10}^{10} (0.705)^{10} \]
\[ = 45 (0.0598818)(0.087025) + 10 (0.0422167)(0.295) + 1 (0.0297628) \]
\[ \approx 0.234493 + 0.124539 + 0.029763 = 0.388795 \approx 0.3888 \text{ (or } 0.3887 \text{)} \]
More precisely:
\(C_8^{10}(0.705)^8(0.295)^2 + C_9^{10}(0.705)^9(0.295) + (0.705)^{10} \approx 0.2344930 + 0.1245392 + 0.0297628 = 0.388795\).

(c) The probability that he knows the correct answer to all 10 questions and answers at least 8 correctly:
Probability he knows all 10 questions is \(0.7^{10}\).
Given he knows all 10 questions, the number of correct answers \(Z \sim \mathrm{B}(10, 0.9)\).
\[ \mathrm{P}(\text{Knows all 10 and answers } \ge 8 \text{ correctly}) = (0.7)^{10} \left[ C_8^{10} (0.9)^8 (0.1)^2 + C_9^{10} (0.9)^9 (0.1)^1 + (0.9)^{10} \right] \]
\[ = (0.7)^{10} [45(0.43046721)(0.01) + 10(0.387420489)(0.1) + 0.3486784401] \]
\[ = (0.7)^{10} [0.19371024 + 0.38742049 + 0.34867844] = (0.0282475249)(0.92980917) \approx 0.0262648 \]

Hence, the required conditional probability is:
\[ \frac{0.0262648}{0.388795} \approx 0.0676 \]

(a) 1M for \((0.7)(0.9) + (0.3)(0.25)\); 1A for 0.705.
(b) 1M for setting up binomial probability with \(n=10, p=0.705\); 1M for sum of terms for 8, 9, 10; 1A for r.t. 0.3888.
(c) 1M for evaluating \(0.7^{10} \times \mathrm{P}(Z \ge 8)\); 1M for conditional probability setup; 1A for r.t. 0.0676 (or follow through from (b)).

(a) (i) The sample proportion is:
\[ \hat{p} = \frac{28}{80} = 0.35 \]

(ii) For a 95% confidence interval, \(z_{0.025} = 1.96\).
\[ \hat{p} \pm 1.96 \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = 0.35 \pm 1.96 \sqrt{\frac{0.35 \times 0.65}{80}} \]
\[ = 0.35 \pm 1.96 \sqrt{0.00284375} = 0.35 \pm 1.96(0.0533268) = 0.35 \pm 0.10452 \]
\[ = (0.24548, 0.45452) \approx (0.2455, 0.4545) \]

(b) For an approximate 99% confidence interval, \(z_{0.005} = 2.575\).
Let \(N\) be the number of households to survey. The width of the confidence interval is:
\[ 2 \times 2.575 \sqrt{\frac{0.35(1 - 0.35)}{N}} \le 0.06 \]
\[ 5.15 \sqrt{\frac{0.2275}{N}} \le 0.06 \]
\[ \sqrt{N} \ge \frac{5.15 \times \sqrt{0.2275}}{0.06} \approx \frac{5.15 \times 0.4769696}{0.06} \approx 40.93989 \]
\[ N \ge (40.93989)^2 \approx 1676.07 \quad (\text{or with } 2.5758: N \ge 1677) \]
Using \(z = 2.575\):
\[ N \ge \frac{4 \times (2.575)^2 \times 0.2275}{(0.06)^2} = \frac{4 \times 6.630625 \times 0.2275}{0.0036} = \frac{6.03386875}{0.0036} \approx 1676.07 \]
Thus, the least number of households is 1677 (or 1675 if using \(z=2.575\) and rounding appropriately; using \(N \ge 1676.07\), least integer is 1677).

(a)(i) 1A for 0.35.
(a)(ii) 1M for \(0.35 \pm 1.96 \sqrt{\frac{0.35(0.65)}{80}}\); 1A for \((0.2455, 0.4545)\).
(b) 1M for \(2(2.575)\sqrt{\frac{0.35(0.65)}{N}} \le 0.06\); 1M for solving inequality for \(N\); 1A for 1677 (accept 1676 to 1678 depending on z-value used).

(a) Let \(u = 4 - x^2\), then \(\frac{\mathrm{d}u}{\mathrm{d}x} = -2x\), so \(x \, \mathrm{d}x = -\frac{1}{2} \, \mathrm{d}u\).
\[ \int x(4 - x^2)^{\frac{3}{2}} \, \mathrm{d}x = \int u^{\frac{3}{2}} \left(-\frac{1}{2}\right) \mathrm{d}u \]
\[ = -\frac{1}{2} \left( \frac{u^{\frac{5}{2}}}{\frac{5}{2}} \right) + C = -\frac{1}{5} u^{\frac{5}{2}} + C \]
\[ = -\frac{1}{5}(4 - x^2)^{\frac{5}{2}} + C \]

(b) The required area is:
\[ \text{Area} = \int_{0}^{2} x(4 - x^2)^{\frac{3}{2}} \, \mathrm{d}x \]
When \(x = 0\), \(u = 4\). When \(x = 2\), \(u = 0\).
\[ = \left[ -\frac{1}{5}(4 - x^2)^{\frac{5}{2}} \right]_{0}^{2} \]
\[ = \left( -\frac{1}{5}(0)^{\frac{5}{2}} \right) - \left( -\frac{1}{5}(4)^{\frac{5}{2}} \right) \]
\[ = 0 + \frac{1}{5}(32) = \frac{32}{5} \]

(a) 1M for substitution \(u = 4 - x^2\) and \(\mathrm{d}u = -2x\,\mathrm{d}x\); 1M for integrating \(u^{3/2}\); 1A for \(-\frac{1}{5}(4 - x^2)^{5/2} + C\).
(b) 1M for setting up \(\int_0^2 f(x)\,\mathrm{d}x\); 1M for substituting limits; 1A for \(\frac{32}{5}\) (or 6.4).

(a) By the Binomial Theorem:
\[ (1 - 2x)^5 = 1 + C_1^5(-2x) + C_2^5(-2x)^2 + C_3^5(-2x)^3 + \cdots \]
\[ = 1 + 5(-2x) + 10(4x^2) + 10(-8x^3) + \cdots \]
\[ = 1 - 10x + 40x^2 - 80x^3 + \cdots \]

(b) Consider the expansion:
\[ (2 + kx + 3x^2)(1 - 10x + 40x^2 - 80x^3 + \cdots) \]

The term in \(x^2\) comes from:
\[ 2(40x^2) + kx(-10x) + 3x^2(1) = (80 - 10k + 3)x^2 = (83 - 10k)x^2 \]

Given that the coefficient of \(x^2\) is \(-18\):
\[ 83 - 10k = -18 \]
\[ 10k = 101 \implies k = 10.1 \]

(a) 1M for binomial expansion formula with binomial coefficients; 1A for \(1 - 10x + 40x^2 - 80x^3\).
(b) 1M for selecting terms giving \(x^2\); 1M for forming equation \(83 - 10k = -18\); 1A for \(k = 10.1\) (or \(\frac{101}{10}\)).

(a) By the quotient rule:
\[ f'(x) = \frac{2(x^2 + 3) - (2x - 3)(2x)}{(x^2 + 3)^2} \]
\[ = \frac{2x^2 + 6 - (4x^2 - 6x)}{(x^2 + 3)^2} = \frac{-2x^2 + 6x + 6}{(x^2 + 3)^2} = \frac{-2(x^2 - 3x - 3)}{(x^2 + 3)^2} \]

(b) Set \(f'(x) = 0\):
\[ -2x^2 + 6x + 6 = 0 \implies x^2 - 3x - 3 = 0 \]
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-3)}}{2} = \frac{3 \pm \sqrt{21}}{2} \]

Since \(0 \le x \le 4\) and \(\sqrt{21} \approx 4.5826\):
\(x_1 = \frac{3 + \sqrt{21}}{2} \approx 3.7913 \in [0, 4]\),
\(x_2 = \frac{3 - \sqrt{21}}{2} \approx -0.7913 \notin [0, 4]\) (rejected).

Evaluate \(f(x)\) at the critical point and endpoints:
- At \(x = 0\): \(f(0) = \frac{2(0) - 3}{0^2 + 3} = -1\).
- At \(x = 4\): \(f(4) = \frac{2(4) - 3}{4^2 + 3} = \frac{5}{19} \approx 0.2632\).
- At \(x = \frac{3 + \sqrt{21}}{2}\):
Since \(x^2 - 3x - 3 = 0 \implies x^2 + 3 = 3x + 6\),
\[ f(x) = \frac{2x - 3}{3x + 6} = \frac{2\left(\frac{3+\sqrt{21}}{2}\right) - 3}{3\left(\frac{3+\sqrt{21}}{2}\right) + 6} = \frac{\sqrt{21}}{\frac{21 + 3\sqrt{21}}{2}} = \frac{2\sqrt{21}}{3(7 + \sqrt{21})} = \frac{2\sqrt{21}(7 - \sqrt{21})}{3(49 - 21)} = \frac{2(7\sqrt{21} - 21)}{3(28)} = \frac{\sqrt{21} - 3}{6} \approx 0.2638 \]

Comparing values:
\(-1 < \frac{5}{19} < \frac{\sqrt{21} - 3}{6}\).

Therefore, the greatest value is \(\frac{\sqrt{21} - 3}{6}\) and the least value is \(-1\).

(a) 1M for applying quotient rule; 1A for \(f'(x) = \frac{-2x^2 + 6x + 6}{(x^2 + 3)^2}\).
(b) 1M for setting \(f'(x) = 0\) and solving for \(x = \frac{3 + \sqrt{21}}{2}\); 1M for evaluating \(f(x)\) at endpoints \(x = 0, 4\); 1M for evaluating \(f(x)\) at the stationary point; 1A for greatest value \(\frac{\sqrt{21}-3}{6}\) (or r.t. 0.2638); 1A for least value \(-1\).

(a) Given that the tangent at \(x = 0\) has slope 2, we have \(f'(0) = 2\).
\[ f'(0) = \frac{k e^0}{1 + e^0} = \frac{k}{1 + 1} = \frac{k}{2} \]
\[ \frac{k}{2} = 2 \implies k = 4 \]

(b) With \(k = 4\), \(f'(x) = \frac{4e^{2x}}{1 + e^{2x}}\).
\[ f(x) = \int \frac{4e^{2x}}{1 + e^{2x}} \, \mathrm{d}x \]
Let \(u = 1 + e^{2x}\), then \(\mathrm{d}u = 2e^{2x} \, \mathrm{d}x\), so \(4e^{2x} \, \mathrm{d}x = 2 \, \mathrm{d}u\).
\[ f(x) = \int \frac{2}{u} \, \mathrm{d}u = 2\ln|u| + C = 2\ln(1 + e^{2x}) + C \]

Since the curve passes through \(P(0, 3)\):
\[ f(0) = 2\ln(1 + 1) + C = 3 \implies 2\ln 2 + C = 3 \implies C = 3 - 2\ln 2 \]
Thus,
\[ f(x) = 2\ln(1 + e^{2x}) + 3 - 2\ln 2 \quad \left(\text{or } f(x) = 2\ln\left(\frac{1 + e^{2x}}{2}\right) + 3\right) \]

(c) The slope of the tangent at \(P(0, 3)\) is \(m = 2\).
Thus, the slope of the normal is \(m_{\text{normal}} = -\frac{1}{2}\).
The equation of the normal is:
\[ y - 3 = -\frac{1}{2}(x - 0) \implies 2y - 6 = -x \implies x + 2y - 6 = 0 \]

(a) 1M for substituting \(x=0\) into \(f'(x) = 2\); 1A for \(k = 4\).
(b) 1M for integration using substitution \(u = 1 + e^{2x}\); 1M for obtaining \(2\ln(1 + e^{2x}) + C\); 1M for using \(f(0) = 3\) to find \(C\); 1A for \(f(x) = 2\ln(1 + e^{2x}) + 3 - 2\ln 2\).
(c) 1A for \(x + 2y - 6 = 0\).

(a) Let \(V\) be the volume of beverage in a bottle in mL, where \(V \sim N(\mu, \sigma^2)\).

$$P(V < 485) = 0.0668 \implies P\left(Z < \frac{485 - \mu}{\sigma}\right) = 0.0668$$
$$\frac{485 - \mu}{\sigma} = -1.50 \implies 485 - \mu = -1.5\sigma \quad \dots (1)$$

$$P(V \ge 510) = 0.1587 \implies P\left(Z \ge \frac{510 - \mu}{\sigma}\right) = 0.1587$$
$$\frac{510 - \mu}{\sigma} = 1.00 \implies 510 - \mu = 1.0\sigma \quad \dots (2)$$

Subtracting (1) from (2):
$$25 = 2.5\sigma \implies \sigma = 10$$
Substituting \(\sigma = 10\) into (2):
$$510 - \mu = 10 \implies \mu = 500$$

(b) Let \(p_O = 0.1587\) be the probability that a bottle is overfilled.
Probability that the 5th inspected bottle is the 2nd overfilled bottle:
$$= C_1^4 (p_O)^1 (1 - p_O)^3 \times p_O$$
$$= 4 (0.1587)^2 (1 - 0.1587)^3$$
$$= 4 (0.02518569)(0.5954622) \approx 0.059988 \approx 0.0600$$

(c) The probability that a bottle is standard is \(p_S = 1 - 0.0668 - 0.1587 = 0.7745\), and underfilled is \(p_U = 0.0668\).

(i) The event consists of two mutually exclusive cases:
- Case 1: 1 overfilled, 6 standard, 1 underfilled:
$$P(\text{Case 1}) = \frac{8!}{1!6!1!} (0.1587)^1 (0.7745)^6 (0.0668)^1 = 56(0.1587)(0.215579)(0.0668) \approx 0.12803$$
- Case 2: 1 overfilled, 7 standard, 0 underfilled:
$$P(\text{Case 2}) = \frac{8!}{1!7!0!} (0.1587)^1 (0.7745)^7 = 8(0.1587)(0.166966) \approx 0.21198$$
$$\text{Required probability} = 0.12803 + 0.21198 \approx 0.3400$$

(ii) Let \(A\) be the event of at least 6 standard bottles and \(B\) be the event of at least 1 overfilled bottle.
$$P(B) = 1 - (1 - p_O)^8 = 1 - (0.8413)^8 \approx 1 - 0.24838 = 0.75162$$
Event \(A \cap B\) includes:
- Exactly 1 overfilled and \(\ge 6\) standard: probability \(\approx 0.34001\)
- Exactly 2 overfilled and 6 standard (0 underfilled):
$$P(2\text{ overfilled}, 6\text{ standard}) = C_2^8 (0.1587)^2 (0.7745)^6 = 28 (0.0251857)(0.215579) \approx 0.15203$$
$$P(A \cap B) = 0.34001 + 0.15203 = 0.49204$$
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.49204}{0.75162} \approx 0.6546$$

(a)
1M for standardising and setting up equations using standard normal table
1M for \(\frac{485 - \mu}{\sigma} = -1.50\) and \(\frac{510 - \mu}{\sigma} = 1.00\)
1A for \(\sigma = 10\)
1A for \(\mu = 500\)

(b)
1M for \(C_1^4 (0.1587)(1-0.1587)^3\)
1M for multiplying by \(0.1587\)
1A for \(0.0600\) (r.t. 0.0600)

(c)(i)
1M for multinomial/binomial expression for 1 overfilled and 6 or 7 standard
1M for adding the two correct probabilities
1A for \(0.3400\) (r.t. 0.3400)

(c)(ii)
1M for finding \(P(\ge 1\text{ overfilled}) = 1 - (0.8413)^8\)
1M for finding \(P(\ge 6\text{ standard} \cap \ge 1\text{ overfilled})\)
0.5A for \(0.6546\) (r.t. 0.6546)

(a) Let \(X\) be the number of urgent tickets received in an hour, \(X \sim \text{Po}(1.6)\).
$$P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$$
$$= 1 - e^{-1.6} - 1.6 e^{-1.6} = 1 - 2.6 e^{-1.6} \approx 1 - 0.524907 \approx 0.4751$$

(b) Let \(Y\) be the total number of support tickets received in 2 hours.
Since the sum of two independent Poisson random variables is Poisson, the mean per hour is \(1.6 + 2.4 = 4.0\).
For a 2-hour period, the mean is \(\lambda = 4.0 \times 2 = 8.0\), so \(Y \sim \text{Po}(8.0)\).
$$P(Y = 6) = \frac{e^{-8.0} 8^6}{6!} = \frac{262144 e^{-8}}{720} \approx 0.1221$$

(c) Let \(W\) be the total number of support tickets in 1 hour, \(W \sim \text{Po}(4.0)\). Let \(X\) be urgent and \(R\) be regular tickets.
$$P(X \ge 2 \mid W = 3) = \frac{P(X = 2, R = 1) + P(X = 3, R = 0)}{P(W = 3)}$$
$$P(X = 2, R = 1) = \left(\frac{e^{-1.6} 1.6^2}{2!}\right)\left(\frac{e^{-2.4} 2.4^1}{1!}\right) = \frac{1.28 \times 2.4 \times e^{-4}}{1} = 3.072 e^{-4}$$
$$P(X = 3, R = 0) = \left(\frac{e^{-1.6} 1.6^3}{3!}\right)\left(\frac{e^{-2.4} 2.4^0}{0!}\right) = \frac{4.096}{6} e^{-4} = \frac{2.048}{3} e^{-4} \approx 0.68267 e^{-4}$$
$$P(W = 3) = \frac{e^{-4} 4^3}{3!} = \frac{64}{6} e^{-4} = \frac{32}{3} e^{-4}$$
$$\text{Required probability} = \frac{3.072 + 2.048/3}{32/3} = \frac{9.216 + 2.048}{32} = \frac{11.264}{32} = 0.3520$$

(d) The condition is \(W < 3\), so \(W \in \{0, 1, 2\}\).
$$P(W < 3) = P(W = 0) + P(W = 1) + P(W = 2) = e^{-4}\left(1 + 4 + \frac{4^2}{2}\right) = 13 e^{-4}$$
We want \(P(X = 0 \cap W < 3) = P(X = 0, R = 0) + P(X = 0, R = 1) + P(X = 0, R = 2)\):
$$= e^{-1.6} \cdot e^{-2.4} \left(1 + 2.4 + \frac{2.4^2}{2}\right) = e^{-4} (1 + 2.4 + 2.88) = 6.28 e^{-4}$$
$$\text{Required probability} = \frac{P(X = 0 \cap W < 3)}{P(W < 3)} = \frac{6.28 e^{-4}}{13 e^{-4}} = \frac{6.28}{13} = \frac{157}{325} \approx 0.4831$$

(a)
1M for \(1 - P(X = 0) - P(X = 1)\)
1M for substituting Poisson terms with \(\lambda = 1.6\)
1A for \(0.4751\) (r.t. 0.4751)

(b)
1M for Poisson with parameter \(\lambda = 8.0\)
1M for evaluating \(\frac{e^{-8} 8^6}{6!}\)
1A for \(0.1221\) (r.t. 0.1221)

(c)
1M for conditional probability formula \(\frac{P(X \ge 2 \cap W = 3)}{P(W = 3)}\)
1M for calculating joint probabilities \(P(X=2, R=1)\) and \(P(X=3, R=0)\)
0.5M for calculating \(P(W=3)\)
1A for \(0.3520\) (accept \(\frac{44}{125}\))

(d)
1M for finding \(P(W < 3) = 13 e^{-4}\)
1M for finding \(P(X = 0 \cap W < 3) = 6.28 e^{-4}\)
1A for \(0.4831\) (accept \(\frac{157}{325}\), r.t. 0.4831)

(a) Using the quotient rule:
$$f'(x) = \frac{\sqrt{x+4}(1) - x\left(\frac{1}{2\sqrt{x+4}}\right)}{x+4} = \frac{2(x+4) - x}{2(x+4)^{3/2}} = \frac{x+8}{2(x+4)^{3/2}}$$

Differentiating \(f'(x)\):
$$f''(x) = \frac{2(x+4)^{3/2}(1) - (x+8)\left(3(x+4)^{1/2}\right)}{4(x+4)^3} = \frac{2(x+4) - 3(x+8)}{4(x+4)^{5/2}} = -\frac{x+16}{4(x+4)^{5/2}}$$

(b) (i) For \(n = 4\) sub-intervals on \([0, 4]\), the step size is \(h = \frac{4 - 0}{4} = 1\).
The partition points are \(x_0 = 0, x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4\).
- \(f(0) = 0\)
- \(f(1) = \frac{1}{\sqrt{5}} \approx 0.4472136\)
- \(f(2) = \frac{2}{\sqrt{6}} \approx 0.8164966\)
- \(f(3) = \frac{3}{\sqrt{7}} \approx 1.1338934\)
- \(f(4) = \frac{4}{\sqrt{8}} = \sqrt{2} \approx 1.4142136\)

$$I \approx \frac{1}{2}\left[f(0) + f(4) + 2(f(1) + f(2) + f(3))\right]$$
$$= \frac{1}{2}\left[0 + 1.4142136 + 2(0.4472136 + 0.8164966 + 1.1338934)\right]$$
$$= \frac{1}{2}\left[1.4142136 + 2(2.3976036)\right] = \frac{1}{2}(6.2094208) \approx 3.1047$$

(ii) For all \(x \in [0, 4]\), \(x + 16 > 0\) and \((x+4)^{5/2} > 0\), so \(f''(x) = -\frac{x+16}{4(x+4)^{5/2}} < 0\).
Since the curve \(y = f(x)\) is concave downward on \([0, 4]\), the trapezoids lie below the curve. Therefore, the estimate is an under-estimate.

(c) Let \(u = \sqrt{x+4}\), then \(u^2 = x + 4 \implies x = u^2 - 4\) and \(dx = 2u \, du\).
When \(x = 0\), \(u = 2\); when \(x = 4\), \(u = \sqrt{8} = 2\sqrt{2}\).
$$\int_0^4 \frac{x}{\sqrt{x+4}} \, dx = \int_2^{2\sqrt{2}} \frac{u^2 - 4}{u} (2u \, du) = 2 \int_2^{2\sqrt{2}} (u^2 - 4) \, du$$
$$= 2 \left[\frac{u^3}{3} - 4u\right]_2^{2\sqrt{2}}$$
$$= 2 \left[\left(\frac{(2\sqrt{2})^3}{3} - 4(2\sqrt{2})\right) - \left(\frac{2^3}{3} - 4(2)\right)\right]$$
$$= 2 \left[\left(\frac{16\sqrt{2}}{3} - 8\sqrt{2}\right) - \left(\frac{8}{3} - 8\right)\right] = 2 \left[-\frac{8\sqrt{2}}{3} + \frac{16}{3}\right] = \frac{32 - 16\sqrt{2}}{3}$$

(a)
1M for applying quotient/product rule for \(f'(x)\)
1A for \(f'(x) = \frac{x+8}{2(x+4)^{3/2}}\) (or equivalent)
1M for differentiating \(f'(x)\)
1A for \(f''(x) = -\frac{x+16}{4(x+4)^{5/2}}\) (or equivalent)

(b)(i)
1M for correct trapezoidal rule formula with \(h = 1\)
1M for substituting correct values of \(f(x_i)\)
1.5A for \(3.1047\) (r.t. 3.1047)

(b)(ii)
1M for stating \(f''(x) < 0\) on \([0, 4]\)
1A for concluding it is an under-estimate

(c)
1M for using substitution \(u = \sqrt{x+4}\) and converting limits/differential correctly
1M for integrating \(2(u^2 - 4)\) to get \(2\left(\frac{u^3}{3} - 4u\right)\)
1A for exact value \(\frac{32 - 16\sqrt{2}}{3}\)

(a) Let \(g(t) = \frac{dV}{dt} = t(16 - t^2)^{1/2}\).
$$g'(t) = (1)(16 - t^2)^{1/2} + t \left(\frac{1}{2}(16 - t^2)^{-1/2}(-2t)\right) = \sqrt{16 - t^2} - \frac{t^2}{\sqrt{16 - t^2}} = \frac{16 - 2t^2}{\sqrt{16 - t^2}}$$
Setting \(g'(t) = 0\):
$$16 - 2t^2 = 0 \implies t^2 = 8 \implies t = 2\sqrt{2} \quad (\text{since } 0 \le t \le 4)$$
Testing the nature of the stationary point:
- For \(0 \le t < 2\sqrt{2}\), \(g'(t) > 0\)
- For \(2\sqrt{2} < t \le 4\), \(g'(t) < 0\)
Hence, \(g(t)\) attains its maximum at \(t = 2\sqrt{2}\). Thus, \(T = 2\sqrt{2}\).

(b) The volume at \(t = T = 2\sqrt{2}\) is:
$$V = \int_0^{2\sqrt{2}} t\sqrt{16 - t^2} \, dt$$
Let \(u = 16 - t^2\), so \(du = -2t \, dt \implies t \, dt = -\frac{1}{2} du\).
When \(t = 0\), \(u = 16\); when \(t = 2\sqrt{2}\), \(u = 16 - 8 = 8\).
$$V = \int_{16}^8 u^{1/2} \left(-\frac{1}{2} du\right) = \frac{1}{2} \int_8^{16} u^{1/2} \, du = \frac{1}{2} \left[\frac{2}{3} u^{3/2}\right]_8^{16} = \frac{1}{3}\left(16^{3/2} - 8^{3/2}\right)$$
$$= \frac{1}{3}(64 - 8\sqrt{8}) = \frac{1}{3}(64 - 16\sqrt{2}) = \frac{64 - 16\sqrt{2}}{3}$$

(c) (i) Let \(r\) be the radius of the water surface when the depth is \(h\).
By similar triangles, \(\frac{r}{h} = \frac{2}{4} = \frac{1}{2} \implies r = \frac{h}{2}\).
The volume of water is:
$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi}{12} h^3$$
Differentiating with respect to \(t\):
$$\frac{dV}{dt} = \frac{\pi}{12} (3h^2) \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$$
Thus, \(Q = \frac{\pi}{4}\).

(ii) When \(t = T = 2\sqrt{2}\):
$$\left.\frac{dV}{dt}\right|_{t=2\sqrt{2}} = (2\sqrt{2})\sqrt{16 - 8} = (2\sqrt{2})(2\sqrt{2}) = 8$$
From (b), \(V = \frac{\pi}{12} h^3 = \frac{64 - 16\sqrt{2}}{3}\):
$$h^3 = \frac{4(64 - 16\sqrt{2})}{\pi} = \frac{256 - 64\sqrt{2}}{\pi} \implies h = \left(\frac{256 - 64\sqrt{2}}{\pi}\right)^{1/3} \approx 3.7489$$
Using \(\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}\):
$$8 = \frac{\pi}{4} h^2 \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{32}{\pi h^2} = \frac{32}{\pi \left(\frac{256 - 64\sqrt{2}}{\pi}\right)^{2/3}} \approx 0.7237\text{ m/hour}$$

(a)
1M for differentiating \(\frac{dV}{dt}\)
1M for setting \(\frac{d}{dt}\left(\frac{dV}{dt}\right) = 0\)
1M for testing maximum (first derivative test or second derivative test)
1A for \(T = 2\sqrt{2}\) (accept \(\sqrt{8}\))

(b)
1M for setting up integral \(\int_0^{2\sqrt{2}} t\sqrt{16-t^2} \, dt\)
1M for substitution method (e.g., \(u = 16 - t^2\))
0.5M for changing limits or back substituting
1A for exact value \(\frac{64 - 16\sqrt{2}}{3}\)

(c)(i)
1M for expressing \(r = \frac{h}{2}\) and \(V = \frac{\pi}{12}h^3\)
1M for differentiating \(V\) with respect to \(t\)
0.5A for \(Q = \frac{\pi}{4}\)

(c)(ii)
1M for substituting \(\left.\frac{dV}{dt}\right|_{t=T} = 8\) and finding \(h\)
1.5A for \(\frac{dh}{dt} \approx 0.7237\) (r.t. 0.7237, or exact expression)