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By binomial expansion, \(\left(1 + \frac{x}{k}\right)^n = 1 + \binom{n}{1}\left(\frac{x}{k}\right) + \binom{n}{2}\left(\frac{x}{k}\right)^2 + \dots = 1 + \frac{n}{k}x + \frac{n(n-1)}{2k^2}x^2 + \dots\). Comparing coefficients: \(\frac{n}{k} = 2\) --- (1), and \(\frac{n(n-1)}{2k^2} = \frac{3}{2}\) --- (2). From (1), we have \(k = \frac{n}{2}\). Substituting \(k = \frac{n}{2}\) into (2): \(\frac{n(n-1)}{2\left(\frac{n}{2}\right)^2} = \frac{3}{2}\), which simplifies to \(\frac{2(n-1)}{n} = \frac{3}{2}\), then \(4n - 4 = 3n\), so \(n = 4\). Since \(k = \frac{n}{2}\), we have \(k = 2\).

1M for binomial expansion up to \(x^2\) term. 1M for setting up both equations. 1M for substitution to eliminate one variable. 1A for \(n = 4\). 1A for \(k = 2\).

Let \(R\) be the event of drawing a red ball, \(X\) be the event of choosing Box \(X\) (i.e., rolling 1 or 2), and \(Y\) be the event of choosing Box \(Y\) (i.e., rolling 3, 4, 5, or 6). We have \(P(X) = \frac{2}{6} = \frac{1}{3}\) and \(P(Y) = \frac{4}{6} = \frac{2}{3}\). (a) By the law of total probability, \(P(R) = P(X)P(R|X) + P(Y)P(R|Y) = \frac{1}{3} \cdot \frac{3}{5} + \frac{2}{3} \cdot \frac{2}{6} = \frac{1}{5} + \frac{2}{9} = \frac{19}{45}\). (b) By Bayes' Theorem, the required probability is \(P(X|R) = \frac{P(X \cap R)}{P(R)} = \frac{P(X)P(R|X)}{P(R)} = \frac{\frac{1}{3} \cdot \frac{3}{5}}{\frac{19}{45}} = \frac{1/5}{19/45} = \frac{9}{19}\).

(a) 1M for summing the two conditional cases, 1A for \(\frac{19}{45}\). (b) 1M for the conditional probability definition, 1M for substitution of calculated values, 1A for \(\frac{9}{19}\).

We have \((1+ax)^n = 1 + nax + \frac{n(n-1)}{2}a^2 x^2 + \dots\) and \((1-3x)^2 = 1 - 6x + 9x^2\). In the expansion of \((1+ax)^n(1-3x)^2\), the coefficient of \(x\) is given by \(na(1) + 1(-6) = 4\), which simplifies to \(na = 10\) (Equation 1). The coefficient of \(x^2\) is given by \(\frac{n(n-1)}{2}a^2(1) + na(-6) + 1(9) = -11\), which simplifies to \(\frac{n(n-1)a^2}{2} - 6na + 9 = -11\) (Equation 2). Substituting Equation 1 into Equation 2, we get \(\frac{na(n-1)a}{2} - 6(10) + 9 = -11\), which gives \(\frac{10(n-1)a}{2} - 60 + 9 = -11\), so \(5a(n-1) - 51 = -11\), which simplifies to \(a(n-1) = 8\) (Equation 3). From Equation 1, we have \(a = \frac{10}{n}\). Substituting this into Equation 3 yields \(\frac{10}{n}(n-1) = 8\), which simplifies to \(10n - 10 = 8n\), so \(2n = 10\), giving \(n = 5\). Finally, substituting \(n = 5\) into Equation 1 gives \(5a = 10\), which yields \(a = 2\). Thus, \(a = 2\) and \(n = 5\).

For expanding both terms correctly up to \(x^2\) (can be implicit): 1M. For obtaining \(na = 10\): 1M. For obtaining \(\frac{n(n-1)}{2}a^2 - 6na + 9 = -11\): 1M. For solving the simultaneous equations (substituting \(na=10\) or equivalent): 1M. For finding \(n = 5\): 1A. For finding \(a = 2\): 1A.

(a) Using the quotient rule, we have \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}}{\mathrm{d}x}(e^{2x})(x-1) - e^{2x}\frac{\mathrm{d}}{\mathrm{d}x}(x-1)}{(x-1)^2} = \frac{2e^{2x}(x-1) - e^{2x}(1)}{(x-1)^2} = \frac{e^{2x}(2x-3)}{(x-1)^2}\). (b) To find the stationary points, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\). Since \(e^{2x} > 0\) and \(x > 1\), we have \(2x - 3 = 0\), which gives \(x = 1.5\). Using the first derivative test: For \(1 < x < 1.5\), \(2x-3 < 0\) so \(\frac{\mathrm{d}y}{\mathrm{d}x} < 0\). For \(x > 1.5\), \(2x-3 > 0\) so \(\frac{\mathrm{d}y}{\mathrm{d}x} > 0\). Therefore, \(C\) has a local minimum at \(x = 1.5\). When \(x = 1.5\), the \(y\)-coordinate is \(y = \frac{e^{2(1.5)}}{1.5-1} = \frac{e^3}{0.5} = 2e^3\). Thus, the coordinates of the local minimum point of \(C\) are \((1.5, 2e^3)\).

(a) For applying quotient rule (or product rule): 1M. For correct simplified expression: 1A. (b) For setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\): 1M. For finding \(x = 1.5\): 1A. For showing that this is a local minimum: 1M. For finding the \(y\)-coordinate and presenting the correct point \((1.5, 2e^3)\): 1A.

Using binomial expansion: \((1 + kx)^n = 1 + nkx + \frac{n(n-1)k^2}{2}x^2 + \dots\) and \((1 - 2x)^5 = 1 - 10x + 40x^2 - \dots\). Product expansion up to the \(x^2\) term: \((1 + kx)^n (1 - 2x)^5 = (1 + nkx + \frac{n(n-1)k^2}{2}x^2 + \dots)(1 - 10x + 40x^2 - \dots)\). The coefficient of \(x\) is \(nk - 10 = -2\), which gives \(nk = 8\) (Equation 1). The coefficient of \(x^2\) is \(40 - 10nk + \frac{n(n-1)k^2}{2} = -16\) (Equation 2). Substitute \(nk = 8\) into Equation 2: \(40 - 10(8) + \frac{n^2k^2 - nk^2}{2} = -16\), \(-40 + \frac{64 - 8k}{2} = -16\), \(-40 + 32 - 4k = -16\), \(-8 - 4k = -16\), \(4k = 8\), \(k = 2\). Substitute \(k = 2\) into Equation 1: \(n(2) = 8\), which gives \(n = 4\).

1M: Correct general expansion terms. 1M: Setting up coefficient equation of x. 1A: For nk = 8. 1M: Setting up coefficient equation of x^2. 1M: Substituting nk = 8 into the coefficient equation of x^2. 1A: For k = 2. 1A: For n = 4.

Let \(A\) be the event that the code is written by Alice, \(B\) be the event that the code is written by Bob, and \(E\) be the event that the code contains an error. Given: \(P(A) = 0.60\), \(P(B) = 0.40\), \(P(E|A) = 0.02\), \(P(E|B) = 0.05\). (a) By the law of total probability: \(P(E) = P(A)P(E|A) + P(B)P(E|B) = (0.60)(0.02) + (0.40)(0.05) = 0.012 + 0.020 = 0.032\). (b) By Bayes' theorem: \(P(B|E) = \frac{P(B \cap E)}{P(E)} = \frac{P(B)P(E|B)}{P(E)} = \frac{(0.40)(0.05)}{0.032} = \frac{0.020}{0.032} = 0.625\).

(a) 1M: Applying total probability formula. 1M: Correct substitution of values. 1A: Correct answer (0.032 or 4/125). (b) 1M: Applying Bayes' theorem formula. 1M: Correct numerator. 1M: Dividing by the answer in (a). 1A: Correct answer (0.625 or 5/8).

(a) Using the product rule and chain rule: \(\frac{\mathrm{d}P}{\mathrm{d}t} = 12 \left( \frac{\mathrm{d}}{\mathrm{d}t}(t^2) e^{-0.5t} + t^2 \frac{\mathrm{d}}{\mathrm{d}t}(e^{-0.5t}) \right) = 12 \left( 2t e^{-0.5t} + t^2 (-0.5 e^{-0.5t}) \right) = 12 e^{-0.5t} (2t - 0.5t^2) = 6t e^{-0.5t} (4 - t)\). (b) Set \(\frac{\mathrm{d}P}{\mathrm{d}t} = 0\): \(6t e^{-0.5t} (4 - t) = 0\). Since \(t > 0\) for critical values within the range and \(e^{-0.5t} > 0\), we get \(t = 4\). For \(0 < t < 4\), \(\frac{\mathrm{d}P}{\mathrm{d}t} > 0\); for \(t > 4\), \(\frac{\mathrm{d}P}{\mathrm{d}t} < 0\). Hence, \(P\) attains its maximum at \(t = 4\). The maximum profit is \(P(4) = 12(4)^2 e^{-0.5(4)} + 5 = 12(16) e^{-2} + 5 = 192 e^{-2} + 5\) million dollars.

(a) 1M: Applying product rule. 1M: Correct derivative of e^{-0.5t}. 1A: Correct derivative expression. (b) 1M: Set derivative to zero and solve for t. 1A: t = 4. 1M: Performing first derivative test to verify maximum. 1A: Maximum profit = 192 e^{-2} + 5.

We integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the equation of \(C\): \(y = \int x \sqrt{2x^2 + 9} \mathrm{d}x\). Let \(u = 2x^2 + 9\). Then \(\mathrm{d}u = 4x \mathrm{d}x\), which gives \(x \mathrm{d}x = \frac{1}{4} \mathrm{d}u\). Substituting these into the integral: \(y = \int \sqrt{u} \cdot \frac{1}{4} \mathrm{d}u = \frac{1}{4} \int u^{1/2} \mathrm{d}u = \frac{1}{4} \left( \frac{u^{3/2}}{3/2} \right) + C = \frac{1}{6} u^{3/2} + C = \frac{1}{6} (2x^2 + 9)^{3/2} + C\), where \(C\) is a constant. Since the curve passes through \((0, 4)\), substitute \(x = 0\) and \(y = 4\): \(4 = \frac{1}{6} (2(0)^2 + 9)^{3/2} + C\), \(4 = \frac{1}{6} (27) + C\), \(4 = 4.5 + C\), \(C = -0.5\). Thus, the equation of the curve \(C\) is \(y = \frac{1}{6} (2x^2 + 9)^{3/2} - \frac{1}{2}\).

1M: Expressing equation as indefinite integral. 1M: Using integration by substitution method. 1A: Correct rewritten integral in terms of u. 1M: Integrating u^{1/2}. 1A: Correct expression with C. 1M: Substituting (0, 4) to solve for C. 1A: Correct final equation of curve.

(a) Let \(R(t) = \frac{80 \ln(t+1)}{(t+1)^2}\) for \(t \ge 0\).
\(R'(t) = 80 \cdot \frac{\frac{1}{t+1}(t+1)^2 - \ln(t+1) \cdot 2(t+1)}{(t+1)^4} = \frac{80[1 - 2\ln(t+1)]}{(t+1)^3}\).
Set \(R'(t) = 0\), we have \(1 - 2\ln(t+1) = 0 \Rightarrow \ln(t+1) = 0.5 \Rightarrow t = e^{0.5} - 1\).
For \(0 \le t < e^{0.5} - 1\), \(R'(t) > 0\); for \(t > e^{0.5} - 1\), \(R'(t) < 0\).
Therefore, \(R(t)\) is maximum when \(t = e^{0.5} - 1\) weeks.

(b) (i) Let \(u = t+1\), then \(du = dt\).
\(\int \frac{\ln(t+1)}{(t+1)^2} dt = \int \frac{\ln u}{u^2} du = \int \ln u d(-u^{-1}) = -\frac{\ln u}{u} - \int -u^{-1} \cdot \frac{1}{u} du = -\frac{\ln u}{u} + \int u^{-2} du = -\frac{\ln u + 1}{u} + C = -\frac{\ln(t+1) + 1}{t+1} + C\).

(ii) \(X(t) = \int \frac{dX}{dt} dt = -80 \cdot \frac{\ln(t+1) + 1}{t+1} + C\).
Using \(X(0) = 100\):
\(100 = -80 \cdot \frac{\ln(1) + 1}{1} + C \Rightarrow 100 = -80 + C \Rightarrow C = 180\).
So \(X(t) = 180 - \frac{80(\ln(t+1) + 1)}{t+1}\).
When \(t = 5\):
\(X(5) = 180 - \frac{80(\ln(6) + 1)}{6} = \frac{500 - 40\ln(6)}{3} \approx 142.78\) grams.
The mass of chemical \(X\) after 5 weeks is \(142.78\) grams.

(a) 1M for quotient rule, 1A for finding \(R'(t)\), 1M for solving \(R'(t) = 0\), 1A for identifying maximum with justification.
(b)(i) 1M for integration by parts, 1M for intermediate integration, 1A for final answer with C.
(b)(ii) 1M for integrating dX/dt, 1M for using initial condition to find C, 1A for correct X(t), 1A for correct mass at t=5 (accept exact value or rounded correct to 2 decimal places).

(a) Since the initial concentration is \(4\text{ ppm}\),
\(C(0) = 4 \implies \frac{b}{3} = 4 \implies b = 12\).

Next, we have \(C(t) = \frac{at + 12}{t^2 + 3}\).
By the quotient rule:
\(C'(t) = \frac{a(t^2 + 3) - 2t(at + 12)}{(t^2 + 3)^2} = \frac{-at^2 - 24t + 3a}{(t^2 + 3)^2}\).
Since the rate of change at \(t = 1\) is \(0\), we have \(C'(1) = 0\):
\(\frac{-a(1)^2 - 24(1) + 3a}{(1^2 + 3)^2} = 0\)
\(2a - 24 = 0 \implies a = 12\).

(b) Substituting \(a = 12\) into \(C'(t)\):
\(C'(t) = \frac{-12t^2 - 24t + 36}{(t^2 + 3)^2} = \frac{-12(t-1)(t+3)}{(t^2 + 3)^2}\).
For \(t > 1\), we have \(t-1 > 0\), \(t+3 > 0\), and \((t^2 + 3)^2 > 0\).
Therefore, \(C'(t) < 0\) for all \(t > 1\).
Thus, the concentration of the pollutant is decreasing for \(t > 1\).

(c) When \(t = 3\):
\(C'(3) = \frac{-12(3-1)(3+3)}{(3^2 + 3)^2} = \frac{-12(2)(6)}{144} = -1\text{ ppm/hour}\).

(d) (i) Differentiating \(C'(t) = \frac{-12t^2 - 24t + 36}{(t^2 + 3)^2}\) with respect to \(t\) again:
Let \(u = -12t^2 - 24t + 36 \implies u' = -24t - 24\)
Let \(v = (t^2 + 3)^2 \implies v' = 4t(t^2 + 3)\)
Using the quotient rule:
\(C''(t) = \frac{(-24t-24)(t^2+3)^2 - (-12t^2-24t+36)(4t(t^2+3))}{(t^2+3)^4}\)
\(C''(t) = \frac{(-24t-24)(t^2+3) - 4t(-12t^2-24t+36)}{(t^2+3)^3}\)
\(C''(t) = \frac{(-24t^3 - 24t^2 - 72t - 72) + (48t^3 + 96t^2 - 144t)}{(t^2+3)^3}\)
\(C''(t) = \frac{24t^3 + 72t^2 - 216t - 72}{(t^2+3)^3} = \frac{24(t^3 + 3t^2 - 9t - 3)}{(t^2+3)^3}\).

(ii) Let \(R(t) = -C'(t)\) be the rate of decrease. To maximize \(R(t)\), we need \(R'(t) = -C''(t) = 0\).
When \(t = 3\):
\(C''(3) = \frac{24(3^3 + 3(3^2) - 9(3) - 3)}{(3^2+3)^3} = \frac{24(24)}{12^3} = \frac{576}{1728} = \frac{1}{3} \neq 0\).
Since \(R'(3) = -C''(3) = -\frac{1}{3} \neq 0\), \(t = 3\) is not a stationary point of \(R(t)\).
Hence, the rate of decrease is not the fastest when \(t = 3\). The researcher's claim is incorrect (disagree).

(a)
- 1M for substituting \(t=0\) into \(C(t) = 4\)
- 1A for \(b=12\)
- 1M for finding \(C'(t)\) using quotient rule and setting \(C'(1) = 0\)
- 1A for \(a=12\)

(b)
- 1M for factorizing \(C'(t) = \frac{-12(t-1)(t+3)}{(t^2+3)^2}\)
- 1A for concluding \(C'(t) < 0\) for \(t > 1\) with clear explanation

(c)
- 1M for substituting \(t=3\) into \(C'(t)\)
- 1A for \(-1\text{ ppm/hour}\) (accept \(-1\))

(d)(i)
- 1M for differentiating \(C'(t)\) using quotient rule
- 1A for \(C''(t) = \frac{24(t^3 + 3t^2 - 9t - 3)}{(t^2+3)^3}\) (or equivalent simplified form)

(d)(ii)
- 1M for substituting \(t=3\) into \(C''(t)\) to find \(C''(3) = \frac{1}{3}\)
- 1A for explaining \(C''(3) \neq 0\) hence \(R'(3) \neq 0\) and concluding disagree

(a) Using the quotient rule, we have:
\( S'(t) = \frac{\frac{d}{dt}[a \ln(t+1) + b] \cdot (t+1) - [a \ln(t+1) + b] \cdot \frac{d}{dt}(t+1)}{(t+1)^2} \)
\( S'(t) = \frac{\frac{a}{t+1}(t+1) - (a \ln(t+1) + b)}{(t+1)^2} \)
\( S'(t) = \frac{a - b - a \ln(t+1)}{(t+1)^2} \)
Set \( S'(t) = 0 \):
\( a - b - a \ln(t+1) = 0 \implies \ln(t+1) = 1 - \frac{b}{a} \)
\( t+1 = e^{1 - \frac{b}{a}} \implies t = e^{1 - \frac{b}{a}} - 1 \)
Since \( a > 0 \):
For \( t < e^{1 - \frac{b}{a}} - 1 \), \( \ln(t+1) < 1 - \frac{b}{a} \implies S'(t) > 0 \).
For \( t > e^{1 - \frac{b}{a}} - 1 \), \( \ln(t+1) > 1 - \frac{b}{a} \implies S'(t) < 0 \).
Therefore, the maximum value of \( S(t) \) occurs at \( t = e^{1 - \frac{b}{a}} - 1 \).

(b)(i) Since the maximum occurs at \( t = 1 \):
\( e^{1 - \frac{b}{a}} - 1 = 1 \implies e^{1 - \frac{b}{a}} = 2 \implies 1 - \frac{b}{a} = \ln 2 \implies b = a(1 - \ln 2) \) --- (1)
Also, the maximum value is 12:
\( S(1) = 12 \implies \frac{a \ln 2 + b}{2} = 12 \implies a \ln 2 + b = 24 \) --- (2)
Substitute (1) into (2):
\( a \ln 2 + a(1 - \ln 2) = 24 \implies a = 24 \)
Substitute \( a = 24 \) back into (1):
\( b = 24(1 - \ln 2) = 24 - 24 \ln 2 \approx 7.36 \)
(ii) At \( t = 0 \):
\( S(0) = \frac{a \ln(1) + b}{1} = b = 24 - 24 \ln 2 \approx 7.36 \) thousand units per month.

(c)(i) \( I(T) = \int_0^T \frac{24 \ln(t+1) + b}{t+1} dt \)
Let \( u = \ln(t+1) \). Then \( du = \frac{1}{t+1} dt \).
When \( t = 0 \), \( u = 0 \).
When \( t = T \), \( u = \ln(T+1) \).
\( I(T) = \int_0^{\ln(T+1)} (24u + b) du = [12u^2 + bu]_0^{\ln(T+1)} = 12(\ln(T+1))^2 + b \ln(T+1) \)
Thus, \( I(T) = 12(\ln(T+1))^2 + (24 - 24 \ln 2)\ln(T+1) \).
(ii) Substitute \( T = 5 \) into the expression:
\( I(5) = 12(\ln 6)^2 + (24 - 24 \ln 2)\ln 6 \)
Since \( \ln 6 \approx 1.791759 \) and \( 24 - 24 \ln 2 \approx 7.364444 \):
\( I(5) \approx 12(1.791759)^2 + 7.364444(1.791759) \approx 38.5249 + 13.1953 = 51.7202 \) thousand units.
Since \( 51.7202 > 50 \), the total sales will exceed 50 thousand units. Thus, the claim is agreed.

(a)
1M: For quotient rule differentiation.
1A: For \( S'(t) = \frac{a - b - a\ln(t+1)}{(t+1)^2} \).
1M: For setting \( S'(t) = 0 \) and solving for \( t \).
1A: For explaining the first derivative test to show it is a maximum.

(b)(i)
1M: For using (a) to set up \( e^{1-b/a} - 1 = 1 \) or equivalent.
1M: For using \( S(1) = 12 \) to set up equation.
1A: For \( a = 24 \) and \( b = 24 - 24\ln 2 \) (or \( b \approx 7.36 \)).
(b)(ii)
1A: For finding \( S(0) \approx 7.36 \) (accept 7.36 to 7.4).

(c)(i)
1M: For substitution \( u = \ln(t+1) \) and \( du = \frac{1}{t+1} dt \).
1M: For integration to get \( 12u^2 + bu \).
1A: For \( I(T) = 12(\ln(T+1))^2 + (24 - 24 \ln 2)\ln(T+1) \).
(c)(ii)
1M: For substituting \( T = 5 \) and calculating \( I(5) \).
1A: For finding \( I(5) \approx 51.7 > 50 \) and concluding "Agree".

(a) (i) Let \(u = 4-t\). Then \(dt = -du\).
When \(t = 0\), \(u = 4\); when \(t = 4\), \(u = 0\).
\\begin{aligned}
\\int_0^4 t \\sqrt{4-t} dt &= \\int_4^0 (4-u) \\sqrt{u} (-du) \\\\
&= \\int_0^4 (4u^{1/2} - u^{3/2}) du \\\\
&= \\left[ \\frac{8}{3}u^{3/2} - \\frac{2}{5}u^{5/2} \\right]_0^4 \\\\
&= \\frac{8}{3}(4)^{3/2} - \\frac{2}{5}(4)^{5/2} \\\\
&= \\frac{64}{3} - \\frac{64}{5} \\\\
&= \\frac{128}{15}
\\end{aligned}
(ii) At \(t=4\), the amount of pollutant is:
\(A(4) = A(0) + \int_0^4 t \sqrt{4-t} dt = 20 + \frac{128}{15} = \frac{428}{15} \approx 28.5333\text{ kg}\).

(b) (i) Let \(u = t-3\), so \(du = dt\).
\\begin{aligned}
\\int \\frac{\\ln(t-3)}{(t-3)^2} dt &= \\int \\frac{\\ln u}{u^2} du \\\\
&= -\\frac{\\ln u}{u} - \\int -\\frac{1}{u^2} du \quad \\text{(by parts, let } w = \\ln u, dv = u^{-2}du\\text{)} \\\\
&= -\\frac{\\ln u}{u} - \\frac{1}{u} + C \\\\
&= -\\frac{\\ln(t-3)}{t-3} - \\frac{1}{t-3} + C
\\end{aligned}
(ii) For \(t > 4\):
\(A(t) = k \left( -\frac{\ln(t-3)}{t-3} - \frac{1}{t-3} \right) + C_2\).
Since \(A(t)\) is continuous at \(t=4\):
\(\lim_{t \to 4^+} A(t) = A(4)\)
\(k \left( -\frac{\ln(1)}{1} - \frac{1}{1} \right) + C_2 = \frac{428}{15}\)
\(-k + C_2 = \frac{428}{15} \implies C_2 = \frac{428}{15} + k\).
As \(t \to \infty\), \(A(t) \to 40\).
Since \(\lim_{t \to \infty} \frac{\ln(t-3)}{t-3} = 0\) and \(\lim_{t \to \infty} \frac{1}{t-3} = 0\),
\(C_2 = 40\).
Therefore,
\(\frac{428}{15} + k = 40 \implies k = 40 - \frac{428}{15} = \frac{172}{15}\).

(c) (i) Let \(H(t) = \frac{172}{15} \frac{\ln(t-3)}{(t-3)^2}\).
With \(n=2\) subintervals, \(h = \frac{10-6}{2} = 2\).
The values of \(H(t)\) at \(t=6\), \(8\), \(10\) are:
\(H(6) = \frac{172}{15} \frac{\ln 3}{9} \approx 1.399558\)
\(H(8) = \frac{172}{15} \frac{\ln 5}{25} \approx 0.738194\)
\(H(10) = \frac{172}{15} \frac{\ln 7}{49} \approx 0.454925\)
The trapezoidal estimate is:
\(I \approx \frac{2}{2} [H(6) + 2H(8) + H(10)] \approx 1.399558 + 2(0.738194) + 0.454925 \approx 3.330871 \approx 3.33\text{ kg}\).
(ii) Let \(y = H(t) = \frac{172}{15} \frac{\ln(t-3)}{(t-3)^2}\).
\(\frac{dy}{dt} = \frac{172}{15} \frac{1 - 2\ln(t-3)}{(t-3)^3}\)
\(\frac{d^2y}{dt^2} = \frac{172}{15} \frac{6\ln(t-3) - 5}{(t-3)^4}\)
For \(6 \le t \le 10\), we have \(t-3 \ge 3 > e^{5/6} \approx 2.30\).
Thus, \(6\ln(t-3) - 5 \ge 6\ln 3 - 5 \approx 1.59 > 0\).
Since \(\frac{d^2 y}{dt^2} > 0\) on \([6, 10]\), the curve is concave upwards.
Therefore, the estimate in (c)(i) is an overestimate.

(a) (i) 1M for substitution \(u=4-t\), 1M for integration to get \(\frac{8}{3}u^{3/2} - \frac{2}{5}u^{5/2}\), 1A for \(\frac{128}{15}\) (or 8.53).
(a) (ii) 1A for \(20 + \frac{128}{15} = \frac{428}{15}\) (or 28.5).
(b) (i) 1M for substitution \(u=t-3\), 1M for integration by parts, 1A for \(-\frac{\ln(t-3)}{t-3} - \frac{1}{t-3} + C\).
(b) (ii) 1M for integrating \(H(t)\) to get \(A(t)\) with constant, 1M for continuity condition, 1M for limit at infinity, 1A for showing \(k = \frac{172}{15}\) correctly.
(c) (i) 1M for using trapezoidal rule with 2 subintervals, 1A for \(3.33\).
(c) (ii) 1A for showing \(\frac{d^2 H}{dt^2} > 0\) on \([6, 10]\) and concluding it is an overestimate.