2023 HKDSE Mathematics M1 (Calculus and Statistics) Answers & Marking Scheme

(a) Since the sum of the probabilities is 1, we have:\n\(p + q + 0.3 + p = 1 \implies 2p + q = 0.7\) --- (1)\n\nSince \(E(X) = 4.1\), we have:\n\(1(p) + 3(q) + 5(0.3) + 7(p) = 4.1 \implies 8p + 3q = 2.6\) --- (2)\n\nFrom (1), we have \(q = 0.7 - 2p\).\nSubstituting this into (2):\n\(8p + 3(0.7 - 2p) = 2.6 \implies 2p = 0.5 \implies p = 0.25\)\n\nTherefore, \(q = 0.7 - 2(0.25) = 0.2\).\n\n(b) First, calculate \(E(X^2)\):\n\(E(X^2) = 1^2(0.25) + 3^2(0.2) + 5^2(0.3) + 7^2(0.25) = 0.25 + 1.8 + 7.5 + 12.25 = 21.8\)\n\nThen, the variance of \(X\) is:\n\(Var(X) = E(X^2) - [E(X)]^2 = 21.8 - 4.1^2 = 4.99\)\n\nHence, the required variance is:\n\(Var(3 - 2X) = (-2)^2 Var(X) = 4(4.99) = 19.96\)

(a) \n- For setting up \(2p + q = 0.7\) (1M)\n- For setting up \(8p + 3q = 2.6\) (1M)\n- For both \(p = 0.25\) and \(q = 0.2\) (1A)\n\n(b)\n- For finding \(E(X^2) = 21.8\) (1M)\n- For finding \(Var(X) = 4.99\) or using \(Var(3-2X) = 4Var(X)\) (1M)\n- For obtaining the correct answer \(19.96\) (1A)

(a) Since the number of accidents per week follows a Poisson distribution with mean \(\lambda = 4\), the population mean is \(\mu = 4\) and the population variance is \(\sigma^2 = 4\).
The mean of \(\bar{X}\) is:
\(E(\bar{X}) = \mu = 4\)
The variance of \(\bar{X}\) is:
\(Var(\bar{X}) = \frac{\sigma^2}{n} = \frac{4}{100} = 0.04\)

(b) Since the sample size \(n = 100\) is large (i.e., \(n \ge 30\)), by the Central Limit Theorem, \(\bar{X}\) is approximately normally distributed:
\(\bar{X} \sim \mathcal{N}(4, 0.04)\) approximately.

The required probability is:
\(P(3.7 \le \bar{X} \le 4.3) \approx P\left( \frac{3.7 - 4}{\sqrt{0.04}} \le Z \le \frac{4.3 - 4}{\sqrt{0.04}} \right)\)
\(= P\left( \frac{-0.3}{0.2} \le Z \le \frac{0.3}{0.2} \right)\)
\(= P(-1.5 \le Z \le 1.5)\)
\(= 2 \times P(0 \le Z \le 1.5)\)
\(= 2 \times 0.4332\)
\(= 0.8664\)

(a)
- Mean of \(\bar{X} = 4\) (1A)
- Variance of \(\bar{X} = 0.04\) (1A)

(b)
- Standardizing and applying Central Limit Theorem (1M)
- Expressing as \(P(-1.5 \le Z \le 1.5)\) or equivalent (1M)
- Correct answer 0.8664 (1A)

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(a) Let \(A\), \(B\), and \(C\) be the events that a component is from supplier \(A\), \(B\), and \(C\) respectively, and \(D\) be the event that a component is defective. \(P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) = (0.40)(0.02) + (0.35)(0.03) + (0.25)(0.05) = 0.008 + 0.0105 + 0.0125 = 0.031\). (b) The required probability is \(P(A \cup C | D) = \frac{P(A \cap D) + P(C \cap D)}{P(D)} = \frac{(0.40)(0.02) + (0.25)(0.05)}{0.031} = \frac{0.008 + 0.0125}{0.031} = \frac{41}{62} \approx 0.661\). Alternatively, \(P(B | D) = \frac{P(B \cap D)}{P(D)} = \frac{(0.35)(0.03)}{0.031} = \frac{21}{62}\), hence \(P(A \cup C | D) = 1 - P(B | D) = 1 - \frac{21}{62} = \frac{41}{62} \approx 0.661\).

Part (a): 1M for the total probability formula, 1M for correct substitution, 1A for the correct answer of 0.031 (or 31/1000). Part (b): 1M for Bayes' theorem formula, 1M for calculating the joint probabilities, 1M for substituting into the conditional probability, 1A for the correct answer of 41/62 (or approx. 0.661).

(a) Using the definition of conditional probability:\n\( P(B | A) = \frac{P(A \cap B)}{P(A)} \)\n\( 0.3 = \frac{P(A \cap B)}{0.4} \)\n\( P(A \cap B) = 0.12 \)\n\n(b) By De Morgan's Laws:\n\( P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \)\nSince \(P(A' \cap B') = 0.48\), we have:\n\( P(A \cup B) = 1 - 0.48 = 0.52 \)\nUsing the addition rule of probability:\n\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)\n\( 0.52 = 0.4 + P(B) - 0.12 \)\n\( 0.52 = 0.28 + P(B) \)\n\( P(B) = 0.24 \)\n\n(c) We check the definition of independent events:\n\( P(A) \times P(B) = 0.4 \times 0.24 = 0.096 \)\nSince \(P(A \cap B) = 0.12 \neq 0.096\) (or \(P(B | A) = 0.3 \neq P(B) = 0.24\)), the events \(A\) and \(B\) are not independent.

(a) \n1M: For using the definition of conditional probability: \(P(A \cap B) = P(B | A) \times P(A)\)\n1A: \(P(A \cap B) = 0.12\)\n\n(b)\n1M: For finding \(P(A \cup B) = 0.52\) or using \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)\n1A: \(P(B) = 0.24\)\n\n(c)\n1M: For calculating \(P(A) \times P(B)\) or comparing \(P(B | A)\) and \(P(B)\)\n1A: For stating that they are not independent with a correct justification.

(a) Using the binomial theorem:
\((1 - 2x)^3 = 1 - 6x + 12x^2 + \dots\)
\((1 + ax)^n = 1 + nax + \frac{n(n-1)}{2}a^2 x^2 + \dots\)
Thus,
\(f(x) = \left(1 - 6x + 12x^2 + \dots\right)\left(1 + nax + \frac{n(n-1)}{2}a^2 x^2 + \dots\right)\)
\(f(x) = 1 + (na - 6)x + \left(\frac{n(n-1)a^2}{2} - 6na + 12\right)x^2 + \dots\)

(b) Since the coefficient of \(x\) is \(2\), we have:
\(na - 6 = 2 \implies na = 8\) ... (1)

From the expansion in (a), we can find \(f''(0)\). Since \(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \dots\), the coefficient of \(x^2\) is \(\frac{f''(0)}{2}\).
Alternatively, differentiating \(f(x)\) twice:
\(f'(x) = (na - 6) + 2\left(\frac{n(n-1)a^2}{2} - 6na + 12\right)x + \dots\)
\(f''(x) = 2\left(\frac{n(n-1)a^2}{2} - 6na + 12\right) + \text{terms containing } x\)
Setting \(x = 0\):
\(f''(0) = n(n-1)a^2 - 12na + 24\)
We are given \(f''(0) = -24\), so:
\(n(n-1)a^2 - 12na + 24 = -24\)
\(n^2 a^2 - n a^2 - 12(8) + 24 = -24\) (substituting \(na = 8\))
\(64 - na^2 - 96 + 24 = -24\)
\(-8 - na^2 = -24 \implies na^2 = 16\) ... (2)

Dividing (2) by (1):
\(\frac{na^2}{na} = \frac{16}{8} \implies a = 2\)
Substituting \(a = 2\) back into (1):
\(n(2) = 8 \implies n = 4\)

**(a)**
For writing \(1 - 6x + 12x^2 + \dots\) **[1M]**
For writing \(1 + nax + \frac{n(n-1)a^2}{2}x^2 + \dots\) **[1M]**
For \(1 + (na - 6)x + \left(\frac{n(n-1)a^2}{2} - 6na + 12\right)x^2 + \dots\) **[1A]** (or equivalent)

**(b)**
For \(na = 8\) **[1M]**
For setting up the equation \(n(n-1)a^2 - 12na + 24 = -24\) (or setting coefficient of \(x^2\) equal to \(-12\)) **[1M]**
For obtaining \(na^2 = 16\) (or any single-variable equation in \(a\) or \(n\)) **[1M]**
For \(a = 2, n = 4\) **[1A]** (must get both correct)

(a) Since \( N(2) = 500 + 10\ln 3 \), we have \( 500 + a\ln(2b+1) = 500 + 10\ln 3 \), which simplifies to \( a\ln(2b+1) = 10\ln 3 \) --- (1). Differentiating \( N(t) \) with respect to \( t \), we get \( \frac{dN}{dt} = \frac{ab}{bt+1} \). Given that the rate of change is \( \frac{10}{3} \) when \( t = 2 \), we have \( \frac{ab}{2b+1} = \frac{10}{3} \) --- (2). From (1), we have \( a = \frac{10\ln 3}{\ln(2b+1)} \). Substituting this into (2) gives \( \frac{10\ln 3}{\ln(2b+1)} \cdot \frac{b}{2b+1} = \frac{10}{3} \), which simplifies to \( \frac{b}{(2b+1)\ln(2b+1)} = \frac{1}{3\ln 3} \). By inspection, \( b = 1 \) is a solution. Substituting \( b = 1 \) into (1), we get \( a\ln 3 = 10\ln 3 \), so \( a = 10 \). (b) When \( t = 5 \), the rate of change of the number of bacteria is \( \frac{dN}{dt}\Big|_{t=5} = \frac{ab}{5b+1} = \frac{10(1)}{5(1)+1} = \frac{10}{6} = \frac{5}{3} \) per hour.

(a) Substituting \( t = 2 \) into \( N(t) \): \( a\ln(2b+1) = 10\ln 3 \) (1M). Finding derivative: \( \frac{dN}{dt} = \frac{ab}{bt+1} \) (1M). Setting up derivative equation at \( t=2 \): \( \frac{ab}{2b+1} = \frac{10}{3} \) (1M). Solving for \( a \) and \( b \) to get \( a = 10 \) and \( b = 1 \) (1A) (both correct). (b) Substituting \( a=10 \), \( b=1 \), and \( t=5 \) into \( \frac{dN}{dt} \) (1M). Correct answer: \( \frac{5}{3} \) (or equivalent fraction / decimal \( \approx 1.67 \)) (1A).

Let \(u = 1+3x^2\).\nThen \(du = 6x \, dx\), which means \(x \, dx = \frac{1}{6} du\).\nAlso, \(x^2 = \frac{u-1}{3}\).\nWhen \(x = 0\), \(u = 1\).\nWhen \(x = 1\), \(u = 4\).\n\n\(\int_{0}^{1} \frac{x^3}{\sqrt{1+3x^2}} \, dx = \int_{0}^{1} \frac{x^2}{\sqrt{1+3x^2}} (x \, dx)\)\n\(= \int_{1}^{4} \frac{\frac{u-1}{3}}{\sqrt{u}} \left( \frac{1}{6} du \right)\)\n\(= \frac{1}{18} \int_{1}^{4} \left( u^{1/2} - u^{-1/2} \right) du\)\n\(= \frac{1}{18} \left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right]_{1}^{4}\)\n\(= \frac{1}{18} \left\{ \left( \frac{2}{3}(4)^{3/2} - 2(4)^{1/2} \right) - \left( \frac{2}{3}(1)^{3/2} - 2(1)^{1/2} \right) \right\}\)\n\(= \frac{1}{18} \left\{ \left( \frac{16}{3} - 4 \right) - \left( \frac{2}{3} - 2 \right) \right\}\)\n\(= \frac{1}{18} \left( \frac{4}{3} - \left( -\frac{4}{3} \right) \right)\)\n\(= \frac{1}{18} \left( \frac{8}{3} \right)\)\n\(= \frac{4}{27}\)

- Letting \(u = 1+3x^2\) to find \(du = 6x\,dx\) (or equivalent substitution) (1M)\n- Changing limits of integration correctly to \(1\) and \(4\) (1M)\n- Expressing the integral as \(\frac{1}{18} \int_{1}^{4} (u^{1/2} - u^{-1/2}) \, du\) (1M)\n- Correct integration to obtain \(\frac{1}{18} \left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right]\) (1M)\n- Substituting limits \(u=4\) and \(u=1\) into the integrated expression (1M)\n- Final answer: \(\frac{4}{27}\) (1A)

(a) Given \(y = x e^{-x}\).
Using the product rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = e^{-x} - x e^{-x} = (1-x)e^{-x}\)
Using the product rule again:
\(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -e^{-x} - (1-x)e^{-x} = (x-2)e^{-x}\)

(b) The width of each subinterval is \(h = \frac{1-0}{4} = 0.25\).
Let \(f(x) = x e^{-x}\).
The values of \(f(x)\) at the grid points are:
\(f(0) = 0\)
\(f(0.25) = 0.25 e^{-0.25} \approx 0.194700\)
\(f(0.5) = 0.5 e^{-0.5} \approx 0.303265\)
\(f(0.75) = 0.75 e^{-0.75} \approx 0.354275\)
\(f(1) = e^{-1} \approx 0.367879\)
Using the trapezoidal rule, the estimated area is:
\(\text{Area} \approx \frac{0.25}{2} [f(0) + 2(f(0.25) + f(0.5) + f(0.75)) + f(1)]\)
\(\text{Area} \approx 0.125 [0 + 2(0.194700 + 0.303265 + 0.354275) + 0.367879]\)
\(\text{Area} \approx 0.125 [1.704480 + 0.367879]\)
\(\text{Area} \approx 0.259045 \approx 0.2590\) (correct to 4 decimal places)

(c) For \(0 \le x \le 1\), we have \(x-2 < 0\) and \(e^{-x} > 0\).
Therefore, \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = (x-2)e^{-x} < 0\) for all \(x \in [0, 1]\).
Since the second derivative is negative on the interval, the curve is concave downward.
Hence, the trapezoidal estimate is an under-estimate.

(a) \(\frac{\mathrm{d}y}{\mathrm{d}x} = (1-x)e^{-x}\) (or equivalent) (1M)
\(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = (x-2)e^{-x}\) (1A)

(b) Correct width \(h = 0.25\) and substitution into trapezoidal rule formula: (1M)
\(\text{Area} \approx \frac{0.25}{2} [f(0) + 2(f(0.25) + f(0.5) + f(0.75)) + f(1)]\)
Correct values of \(f(x)\) substituted: (1M)
\(\approx 0.125 [0 + 2(0.194700 + 0.303265 + 0.354275) + 0.367879]\)
\(\approx 0.2590\) (1A)

(c) Show that \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} < 0\) for all \(0 \le x \le 1\): (1M)
State that the estimate is an under-estimate (1A)

(a)(i)
Unbiased estimate of population mean \(\mu = \bar{x} = \frac{25200}{100} = 252\)
Unbiased estimate of population variance \(\sigma^2 = s^2 = \frac{1}{100-1}\sum (x_i - \bar{x})^2 = \frac{396}{99} = 4\)

(a)(ii)
Since \(n = 100\) is large, by the Central Limit Theorem, the 95% confidence interval for \(\mu\) is:
\(\bar{x} \pm 1.96 \frac{s}{\sqrt{n}}\)
\(= 252 \pm 1.96 \frac{\sqrt{4}}{\sqrt{100}}\)
\(= 252 \pm 1.96 \times 0.2\)
\(= 252 \pm 0.392\)
\(= (251.608, 252.392)\) (or \([251.608, 252.392]\))

(b)(i)
Let \(X\) be the weight of honey in a jar. \(X \sim N(252, 2^2)\).
The probability that a jar is underfilled is:
\(P(X < 248.5) = P\left(Z < \frac{248.5 - 252}{2}\right)\)
\(= P(Z < -1.75)\)
\(= 0.5 - P(0 \leq Z \leq 1.75)\)
\(= 0.5 - 0.4599\)
\(= 0.0401\)

(c)(i)
Let \(Y\) be the number of underfilled jars in a box of 12. Then \(Y \sim B(12, 0.0401)\).
The required probability is:
\(P(Y \geq 2) = 1 - P(Y = 0) - P(Y = 1)\)
\(= 1 - (1 - 0.0401)^{12} - 12(0.0401)(1 - 0.0401)^{11}\)
\(= 1 - 0.9599^{12} - 12(0.0401)(0.9599)^{11}\)
\(\approx 1 - 0.612089 - 0.306842\)
\(\approx 0.081069 \approx 0.0811\) (correct to 3 sig. fig.)

(c)(ii)
Let \(p = 0.081069\).
The probability that the 3rd box sent for inspection is the 8th box selected is:
\(P(\text{2 of the first 7 boxes sent for inspection}) \times P(\text{the 8th box sent for inspection})\)
\(= C^7_2 p^2 (1-p)^5 \times p\)
\(= 21 p^3 (1-p)^5\)
\(= 21 (0.081069)^3 (1 - 0.081069)^5\)
\(\approx 21 (0.00053279)(0.655998)\)
\(\approx 0.00734\)

(a)(i)
Unbiased estimate of \(\mu = 252\) (1A)
Unbiased estimate of \(\sigma^2 = \frac{396}{99} = 4\) (1A)

(a)(ii)
95% confidence interval = \(252 \pm 1.96 \frac{\sqrt{4}}{\sqrt{100}}\) (1M)
\(= 252 \pm 0.392\)
\(= (251.608, 252.392)\) (1A+1A)

(b)(i)
\(P(X < 248.5) = P\left(Z < \frac{248.5 - 252}{2}\right)\) (1M)
\(= P(Z < -1.75)\) (1M)
\(= 0.0401\) (1A)

(c)(i)
\(P(Y \geq 2) = 1 - P(Y = 0) - P(Y = 1)\) (1M)
\(= 1 - 0.9599^{12} - 12(0.0401)(0.9599)^{11}\) (1M)
\(\approx 0.0811\) (1A)

(c)(ii)
Required probability = \(C^7_2 p^2 (1-p)^5 \times p\) (1M)
\(\approx 0.00734\) (1A)

(a)(i)
Unbiased estimate of population mean \(\mu = \bar{x} = \frac{25200}{100} = 252\)
Unbiased estimate of population variance \(\sigma^2 = s^2 = \frac{1}{100-1}\sum (x_i - \bar{x})^2 = \frac{396}{99} = 4\)

(a)(ii)
Since \(n = 100\) is large, by the Central Limit Theorem, the 95% confidence interval for \(\mu\) is:
\(\bar{x} \pm 1.96 \frac{s}{\sqrt{n}}\)
\(= 252 \pm 1.96 \frac{\sqrt{4}}{\sqrt{100}}\)
\(= 252 \pm 1.96 \times 0.2\)
\(= 252 \pm 0.392\)
\(= (251.608, 252.392)\) (or \([251.608, 252.392]\))

(b)(i)
Let \(X\) be the weight of honey in a jar. \(X \sim N(252, 2^2)\).
The probability that a jar is underfilled is:
\(P(X < 248.5) = P\left(Z < \frac{248.5 - 252}{2}\right)\)
\(= P(Z < -1.75)\)
\(= 0.5 - P(0 \leq Z \leq 1.75)\)
\(= 0.5 - 0.4599\)
\(= 0.0401\)

(c)(i)
Let \(Y\) be the number of underfilled jars in a box of 12. Then \(Y \sim B(12, 0.0401)\).
The required probability is:
\(P(Y \geq 2) = 1 - P(Y = 0) - P(Y = 1)\)
\(= 1 - (1 - 0.0401)^{12} - 12(0.0401)(1 - 0.0401)^{11}\)
\(= 1 - 0.9599^{12} - 12(0.0401)(0.9599)^{11}\)
\(\approx 1 - 0.612089 - 0.306842\)
\(\approx 0.081069 \approx 0.0811\) (correct to 3 sig. fig.)

(c)(ii)
Let \(p = 0.081069\).
The probability that the 3rd box sent for inspection is the 8th box selected is:
\(P(\text{2 of the first 7 boxes sent for inspection}) \times P(\text{the 8th box sent for inspection})\)
\(= C^7_2 p^2 (1-p)^5 \times p\)
\(= 21 p^3 (1-p)^5\)
\(= 21 (0.081069)^3 (1 - 0.081069)^5\)
\(\approx 21 (0.00053279)(0.655998)\)
\(\approx 0.00734\)

(a)(i)
Unbiased estimate of \(\mu = 252\) (1A)
Unbiased estimate of \(\sigma^2 = \frac{396}{99} = 4\) (1A)

(a)(ii)
95% confidence interval = \(252 \pm 1.96 \frac{\sqrt{4}}{\sqrt{100}}\) (1M)
\(= 252 \pm 0.392\)
\(= (251.608, 252.392)\) (1A+1A)

(b)(i)
\(P(X < 248.5) = P\left(Z < \frac{248.5 - 252}{2}\right)\) (1M)
\(= P(Z < -1.75)\) (1M)
\(= 0.0401\) (1A)

(c)(i)
\(P(Y \geq 2) = 1 - P(Y = 0) - P(Y = 1)\) (1M)
\(= 1 - 0.9599^{12} - 12(0.0401)(0.9599)^{11}\) (1M)
\(\approx 0.0811\) (1A)

(c)(ii)
Required probability = \(C^7_2 p^2 (1-p)^5 \times p\) (1M)
\(\approx 0.00734\) (1A)

(a) Let \(X\) be the number of customer inquiries arriving at the customer service desk in an hour. Then \(X \sim \text{Po}(3.2)\). (i) \(P(X = 2) = \frac{e^{-3.2} \cdot 3.2^2}{2!} = 5.12 e^{-3.2} \approx 0.208702484 \approx 0.2087\). (ii) \(P(X \ge 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) = 1 - e^{-3.2} - 3.2 e^{-3.2} - 5.12 e^{-3.2} = 1 - 9.32 e^{-3.2} \approx 1 - 0.379903741 = 0.620096259 \approx 0.6201\). (b) Let \(Y\) be the number of working hours in a day (out of 8 hours) with at least 3 inquiries. Then \(Y \sim \text{B}(8, p)\), where \(p = 1 - 9.32 e^{-3.2} \approx 0.620096259\). The required probability is \(P(Y \ge 6) = P(Y = 6) + P(Y = 7) + P(Y = 8) = \binom{8}{6} p^6 (1-p)^2 + \binom{8}{7} p^7 (1-p) + p^8 \approx 28(0.620096259)^6 (0.379903741)^2 + 8(0.620096259)^7 (0.379903741) + (0.620096259)^8 \approx 0.229712 + 0.107128 + 0.021857 = 0.358697 \approx 0.3587\). (c) Given that exactly 4 inquiries arrived, let \(C\) be the number of complaints among these 4 inquiries. Since each inquiry is classified as a complaint with probability 0.25 independently, \(C\) follows a binomial distribution \(C \sim \text{B}(4, 0.25)\). The required probability is \(P(C \ge 2 \mid X = 4) = 1 - P(C = 0) - P(C = 1) = 1 - (0.75)^4 - \binom{4}{1} (0.25)^1 (0.75)^3 = 1 - 0.31640625 - 0.421875 = 0.26171875 \approx 0.2617\) (or \(\frac{67}{256}\)).

(a)(i) 1M for Poisson formula, 1A for 0.2087 (accept 5.12 e^{-3.2}). (a)(ii) 1M for 1 - P(0) - P(1) - P(2), 1M for substituting Poisson values, 1A for 0.6201 (accept 0.620). (b) 1M for identifying Binomial distribution framework, 1M for substituting p = 0.6201 into the sum of Binomial terms, 1A for 0.3587 (accept 0.359). (c) 1M for recognizing the conditional distribution as Binomial(4, 0.25), 1M for using 1 - P(0) - P(1), 1M for substituting Binomial parameters correctly, 1A for 0.2617 (accept 67/256).

(a) Using the product rule to differentiate \(C(t)\) with respect to \(t\):
\(C'(t) = A e^{-0.5t} - 0.5A(t+1)e^{-0.5t} = 0.5A(1-t)e^{-0.5t}\)

Since \(A > 0\) and \(e^{-0.5t} > 0\) for all \(t\):
- For \(0 \le t < 1\), \(C'(t) > 0\), so \(C(t)\) is increasing for \(0 \le t \le 1\).
- For \(t > 1\), \(C'(t) < 0\), so \(C(t)\) is decreasing for \(t \ge 1\).

At \(t = 1\), \(C(1) = A(1+1)e^{-0.5} = 2Ae^{-0.5}\).
Since \(C'(t)\) changes from positive to negative at \(t=1\), the local maximum is at \((1, 2Ae^{-0.5})\).

(b) Differentiating \(C'(t)\) to find the second derivative:
\(C''(t) = 0.5A[-e^{-0.5t} + (1-t)(-0.5)e^{-0.5t}] = 0.25A(t-3)e^{-0.5t}\)

Setting \(C''(t) = 0 \implies t = 3\).
- For \(0 \le t < 3\), \(C''(t) < 0\) (concave downward).
- For \(t > 3\), \(C''(t) > 0\) (concave upward).
Since the concavity changes at \(t = 3\), \(t = 3\) is a point of inflection.
At \(t = 3\), \(C(3) = 4Ae^{-1.5}\).
The coordinates of the point of inflection are \((3, 4Ae^{-1.5})\).

(c) The graph starts at the \(y\)-intercept \((0, A)\), rises to a local maximum at \((1, 2Ae^{-0.5})\), and then decreases, passing through the point of inflection at \((3, 4Ae^{-1.5})\) and asymptotically approaching the \(t\)-axis.

(d)(i) Differentiating \(D(t)\):
\(D'(t) = B[2t e^{-0.5t} - 0.5 t^2 e^{-0.5t}] = Bt(2-0.5t)e^{-0.5t}\)

At \(t = 1\) (where \(C(t)\) reaches its maximum):
\(D'(1) = B(1)(2 - 0.5)e^{-0.5} = 1.5Be^{-0.5}\)
Setting \(1.5Be^{-0.5} = 1.5e^{-0.5} \implies B = 1\).

(ii) With \(A = 4\) and \(B = 1\):
\(T(t) = 4(t+1)e^{-0.5t} + t^2 e^{-0.5t} = (t^2 + 4t + 4)e^{-0.5t} = (t+2)^2 e^{-0.5t}\)

Differentiating \(T(t)\):
\(T'(t) = 2(t+2)e^{-0.5t} - 0.5(t+2)^2 e^{-0.5t} = (t+2)(1-0.5t)e^{-0.5t}\)

Setting \(T'(t) = 0\):
Since \(t \ge 0\), \(t+2 > 0\) and \(e^{-0.5t} > 0\).
Thus, \(1 - 0.5t = 0 \implies t = 2\).

Checking the sign of \(T'(t)\):
- For \(0 \le t < 2\), \(1-0.5t > 0 \implies T'(t) > 0\).
- For \(t > 2\), \(1-0.5t < 0 \implies T'(t) < 0\).
Since \(T'(t)\) changes from positive to negative at \(t = 2\), \(T(t)\) attains its absolute maximum at \(t = 2\).
Therefore, the researcher's claim is correct.

**(a)**
- For finding \(C'(t) = 0.5A(1-t)e^{-0.5t}\) (or equivalent) using product rule. [1M]
- For correct intervals of increase: \(0 \le t \le 1\), and decrease: \(t \ge 1\). [1A]
- For correct coordinates of the local maximum: \((1, 2Ae^{-0.5})\). [1A]

**(b)**
- For finding \(C''(t) = 0.25A(t-3)e^{-0.5t}\). [1M]
- For setting \(C''(t) = 0\) to get \(t=3\), and verifying the change in concavity around \(t=3\). [1M]
- For correct coordinates of the point of inflection: \((3, 4Ae^{-1.5})\). [1A]

**(c)**
- For a curve starting from \((0,A)\), rising to a maximum, then decreasing asymptotically to the horizontal axis. [1A]
- For clearly labeling the \(y\)-intercept \((0, A)\), local maximum \((1, 2Ae^{-0.5})\), and point of inflection \((3, 4Ae^{-1.5})\) on the sketch. [1A]

**(d)**
**(i)**
- For finding \(D'(t) = Bt(2-0.5t)e^{-0.5t}\). [1M]
- For substituting \(t = 1\) and solving \(1.5Be^{-0.5} = 1.5e^{-0.5}\) to get \(B = 1\). [1A]

**(ii)**
- For setting up \(T(t) = (t+2)^2 e^{-0.5t}\) and obtaining \(T'(t) = (t+2)(1-0.5t)e^{-0.5t}\). [1M]
- For setting \(T'(t) = 0\) to find \(t = 2\), and justifying that it yields a maximum. [1M]
- For concluding that the researcher's claim is correct. [1A]

(a) Let \(u = 3 + e^{-0.1 t^2}\). Then we have:
\[ du = e^{-0.1 t^2} \cdot (-0.2 t) dt = -0.2 t e^{-0.1 t^2} dt \]
This gives:
\[ t e^{-0.1 t^2} dt = -5 du \]
Substituting these into the integral:
\[ \int \frac{t e^{-0.1 t^2}}{(3 + e^{-0.1 t^2})^2} dt = \int \frac{-5}{u^2} du = -5 \int u^{-2} du = \frac{5}{u} + C = \frac{5}{3 + e^{-0.1 t^2}} + C \]
where \(C\) is an arbitrary constant.

(b) (i) We have:
\[ P(t) = \int \frac{dP}{dt} dt = \int \frac{160 t e^{-0.1 t^2}}{(3 + e^{-0.1 t^2})^2} dt = 160 \int \frac{t e^{-0.1 t^2}}{(3 + e^{-0.1 t^2})^2} dt \]
Using the result from (a):
\[ P(t) = 160 \left( \frac{5}{3 + e^{-0.1 t^2}} \right) + C_1 = \frac{800}{3 + e^{-0.1 t^2}} + C_1 \]
where \(C_1\) is a constant.
Using the initial condition \(P(0) = 150\):
\[ 150 = \frac{800}{3 + e^0} + C_1 \implies 150 = \frac{800}{4} + C_1 \implies 150 = 200 + C_1 \implies C_1 = -50 \]
Thus, we have:
\[ P(t) = \frac{800}{3 + e^{-0.1 t^2}} - 50 \]

(ii) As \(t \to \infty\), \(e^{-0.1 t^2} \to 0\).
Therefore:
\[ \lim_{t \to \infty} P(t) = \frac{800}{3 + 0} - 50 = \frac{800}{3} - 50 = \frac{650}{3} \approx 216.67 \]
So the amount of pollutant after a very long time is \(\frac{650}{3}\text{ mg}\) (or approximately \(216.67\text{ mg}\)).

(c) Using the quotient rule on \( \frac{dP}{dt} = \frac{160 t e^{-0.1 t^2}}{(3 + e^{-0.1 t^2})^2} \):
Let \(u(t) = 160 t e^{-0.1 t^2}\) and \(v(t) = (3 + e^{-0.1 t^2})^2\).
Then:
\[ u'(t) = 160 e^{-0.1 t^2} + 160 t (-0.2 t e^{-0.1 t^2}) = 160 e^{-0.1 t^2} (1 - 0.2 t^2) \]
\[ v'(t) = 2(3 + e^{-0.1 t^2})(-0.2 t e^{-0.1 t^2}) = -0.4 t e^{-0.1 t^2}(3 + e^{-0.1 t^2}) \]
Applying the quotient rule:
\[ \frac{d^2P}{dt^2} = \frac{u' v - u v'}{v^2} = \frac{160 e^{-0.1 t^2} (1 - 0.2 t^2) (3 + e^{-0.1 t^2})^2 - 160 t e^{-0.1 t^2} [-0.4 t e^{-0.1 t^2} (3 + e^{-0.1 t^2})]}{(3 + e^{-0.1 t^2})^4} \]
Simplifying the fraction by dividing numerator and denominator by \(3 + e^{-0.1 t^2}\):
\[ \frac{d^2P}{dt^2} = \frac{160 e^{-0.1 t^2} \left[ (1 - 0.2 t^2)(3 + e^{-0.1 t^2}) + 0.4 t^2 e^{-0.1 t^2} \right]}{(3 + e^{-0.1 t^2})^3} \]
At \(t = 5\):
\( e^{-0.1(5^2)} = e^{-2.5} \)
\( 1 - 0.2(5^2) = -4 \)
\[ \left.\frac{d^2P}{dt^2}\right|_{t=5} = \frac{160 e^{-2.5} \left[ -4(3 + e^{-2.5}) + 10 e^{-2.5} \right]}{(3 + e^{-2.5})^3} = \frac{160 e^{-2.5} (-12 + 6 e^{-2.5})}{(3 + e^{-2.5})^3} \]
Using the values:
\( e^{-2.5} \approx 0.082085 \)
Numerator \( \approx 160(0.082085)(-12 + 0.492510) \approx 13.1336(-11.50749) \approx -151.1348 \)
Denominator \( \approx (3 + 0.082085)^3 \approx 29.2778 \)
\[ \left.\frac{d^2P}{dt^2}\right|_{t=5} \approx \frac{-151.1348}{29.2778} \approx -5.16 \]
Since \( \left.\frac{d^2P}{dt^2}\right|_{t=5} < 0 \), the rate of increase of the amount of pollutant is decreasing at \(t = 5\).

Part (a)
- For substitution: let \(u = 3 + e^{-0.1 t^2}\) [1M]
- For finding \(du = -0.2 t e^{-0.1 t^2} dt\) or equivalent [1M]
- For substituting to get \( \int -5 u^{-2} du \) [1M]
- For the correct answer: \( \frac{5}{3 + e^{-0.1 t^2}} + C \) (accept missing \(C\)) [1A]

Part (b)
(i)
- For integrating using (a)'s result: \( P(t) = \frac{800}{3 + e^{-0.1 t^2}} + C_1 \) [1M]
- For substituting \(t = 0, P = 150\) to find \(C_1\) [1M]
- For the correct expression of \(P(t)\): \( P(t) = \frac{800}{3 + e^{-0.1 t^2}} - 50 \) [1A]
(ii)
- For finding the limit: \( P \to \frac{650}{3}\text{ mg} \) (or approximately \(216.67\text{ mg}\)) [1A]

Part (c)
- For attempting to differentiate \(\frac{dP}{dt}\) using quotient or product rule [1M]
- For finding the correct expression of \(\frac{d^2P}{dt^2}\) [1A]
- For substituting \(t = 5\) into \(\frac{d^2P}{dt^2}\) [1M]
- For finding \(\left.\frac{d^2P}{dt^2}\right|_{t=5} \approx -5.16\) (accept range -5.16 to -5.17) [1A]
- For concluding that the rate of increase is decreasing since the second derivative is negative [1A]