2022 HKDSE 數學 單元一 (微積分與統計) 模擬試題連答案詳解

(a) A \(95\%\) confidence interval for the population mean \(\mu\) is
\[ \left( 8.5 - 1.96 \cdot \frac{1.2}{\sqrt{64}},\, 8.5 + 1.96 \cdot \frac{1.2}{\sqrt{64}} \right) = (8.5 - 0.294,\, 8.5 + 0.294) = (8.206,\, 8.794) \]

(b) The width of the \(95\%\) confidence interval is \(2(1.96)\left(\frac{1.2}{\sqrt{n}}\right)\).
We require:
\[ 2(1.96)\left(\frac{1.2}{\sqrt{n}}\right) \le 0.4 \]
\[ \frac{4.704}{\sqrt{n}} \le 0.4 \]
\[ \sqrt{n} \ge \frac{4.704}{0.4} = 11.76 \]
\[ n \ge 11.76^2 = 138.2976 \]
Since \(n\) must be an integer, the minimum value of \(n\) is \(139\).

(a)
\(8.5 \pm 1.96 \left(\frac{1.2}{\sqrt{64}}\right)\) : 1M
\((8.206, 8.794)\) (or \([8.206, 8.794]\)) : 1A

(b)
\(2(1.96)\left(\frac{1.2}{\sqrt{n}}\right) \le 0.4\) : 1M
\(n \ge 138.2976\) (or \(\sqrt{n} \ge 11.76\)) : 1M
\(n = 139\) : 1A

(a)
\[ \begin{aligned} (1 - 3x)^6 &= 1 + C_1^6(-3x) + C_2^6(-3x)^2 + \dots \\ &= 1 + 6(-3x) + 15(9x^2) + \dots \\ &= 1 - 18x + 135x^2 + \dots \end{aligned} \]

(b) (i)
\[ (2 + kx)(1 - 3x)^6 = (2 + kx)(1 - 18x + 135x^2 + \dots) \]
The term in \(x^2\) is
\[ 2(135x^2) + kx(-18x) = (270 - 18k)x^2 \]
Given that the coefficient of \(x^2\) is \(90\):
\[ 270 - 18k = 90 \]
\[ 18k = 180 \implies k = 10 \]

(ii) The term in \(x\) is
\[ 2(-18x) + kx(1) = (-36 + k)x \]
Substituting \(k = 10\):
\[ -36 + 10 = -26 \]
Thus, the coefficient of \(x\) is \(-26\).

(a)
For binomial expansion form \(1 + C_1^6(-3x) + C_2^6(-3x)^2\) : 1M
\(1 - 18x + 135x^2\) : 1A

(b)(i)
\(2(135) + k(-18) = 90\) : 1M
\(k = 10\) : 1A

(b)(ii)
Coefficient of \(x = 2(-18) + 10 = -26\) : 1A (f.t.)

(a) \(e^{-2kx} = 1 + (-2kx) + \frac{(-2kx)^2}{2!} + \dots = 1 - 2kx + 2k^2 x^2 + \dots\)

(b)(i) \((1 + 4x)^5 = 1 + \binom{5}{1}(4x) + \binom{5}{2}(4x)^2 + \dots = 1 + 20x + 160x^2 + \dots\)

\((1 + 4x)^5 e^{-2kx} = (1 + 20x + 160x^2 + \dots)(1 - 2kx + 2k^2x^2 + \dots)\)

Coefficient of \(x\):
\(20 - 2k = 14\)
\(2k = 6\)
\(k = 3\)

(ii) Coefficient of \(x^2\):
\(= 1(2k^2) + 20(-2k) + 160(1)\)
\(= 2(3)^2 - 40(3) + 160\)
\(= 18 - 120 + 160\)
\(= 58\)

(a)
\(e^{-2kx} = 1 + (-2kx) + \frac{(-2kx)^2}{2} + \dots\) (1M)
\(= 1 - 2kx + 2k^2x^2 + \dots\) (1A)

(b)(i)
\((1 + 4x)^5 = 1 + 20x + 160x^2 + \dots\) (1M for expanding \((1+4x)^5\))
\(20 - 2k = 14 \implies k = 3\) (1A)

(b)(ii)
\(\text{Coefficient of } x^2 = 2k^2 - 40k + 160\) (1M)
\(= 58\) (1A)

(a) Let \(u = \sqrt{2x+1}\).
Then \(u^2 = 2x+1 \implies x = \frac{u^2-1}{2}\) and \(\mathrm{d}x = u\,\mathrm{d}u\).

\(\int \frac{x+2}{\sqrt{2x+1}}\,\mathrm{d}x = \int \frac{\frac{u^2-1}{2} + 2}{u} \cdot u\,\mathrm{d}u\)
\(= \int \frac{u^2 + 3}{2}\,\mathrm{d}u\)
\(= \frac{1}{2}\left(\frac{u^3}{3} + 3u\right) + C\)
\(= \frac{u^3}{6} + \frac{3u}{2} + C\)
\(= \frac{(2x+1)^{3/2}}{6} + \frac{3\sqrt{2x+1}}{2} + C\)
\(= \frac{(x+5)\sqrt{2x+1}}{3} + C\), where \(C\) is a constant.

(b) Since \(C\) passes through \((4, 10)\):
\(10 = \frac{(4+5)\sqrt{2(4)+1}}{3} + C\)
\(10 = \frac{9(3)}{3} + C\)
\(10 = 9 + C\)
\(C = 1\)

Thus, the equation of the curve \(C\) is \(y = \frac{(x+5)\sqrt{2x+1}}{3} + 1\).

(a)
Let \(u = \sqrt{2x+1}\) (or \(u = 2x+1\)), \(\mathrm{d}x = u\,\mathrm{d}u\) (1M)
\(\int \frac{\frac{u^2+3}{2}}{u} \cdot u\,\mathrm{d}u = \int \frac{u^2+3}{2}\,\mathrm{d}u\) (1M)
\(= \frac{u^3}{6} + \frac{3u}{2} + C\) (1M)
\(= \frac{(x+5)\sqrt{2x+1}}{3} + C\) (or equivalent) (1A)

(b)
Substituting \((4, 10)\) into the expression to find \(C\): \(10 = 9 + C \implies C = 1\) (1M)
\(y = \frac{(x+5)\sqrt{2x+1}}{3} + 1\) (1A)

(a)
Since the sum of all probabilities is 1,
\[ 0.2 + p + q = 1 \implies p + q = 0.8 \]
Given \(\mathrm{E}(X) = 1.1\),
\[ (-2)(0.2) + (1)(p) + (3)(q) = 1.1 \]
\[ -0.4 + p + 3q = 1.1 \implies p + 3q = 1.5 \]
Solving the system of equations:
\[ (p + 3q) - (p + q) = 1.5 - 0.8 \implies 2q = 0.7 \implies q = 0.35 \]
\[ p = 0.8 - 0.35 = 0.45 \]
Now find \(\mathrm{E}(X^2)\):
\[ \mathrm{E}(X^2) = (-2)^2(0.2) + (1)^2(0.45) + (3)^2(0.35) = 0.8 + 0.45 + 3.15 = 4.4 \]
\[ \mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 4.4 - (1.1)^2 = 4.4 - 1.21 = 3.19 \]

(b)
By the central limit theorem,
\[ \bar{X} \sim N\left(1.1, \frac{3.19}{100}\right) \quad \text{approximately, i.e. } N(1.1, 0.0319) \]
\[ \mathrm{P}(\bar{X} < 0.95) \approx \mathrm{P}\left(Z < \frac{0.95 - 1.1}{\sqrt{0.0319}}\right) \approx \mathrm{P}\left(Z < \frac{-0.15}{0.178606}\right) \approx \mathrm{P}(Z < -0.84) \]
\[ = 0.5 - 0.2995 = 0.2005 \]

(a)
\(0.2 + p + q = 1\) (1M)
\((-2)(0.2) + p + 3q = 1.1\) (1M)
\(p = 0.45\) and \(q = 0.35\) (1A for both correct)
\(\mathrm{Var}(X) = 3.19\) (1A)

(b)
\(\bar{X} \sim N\left(1.1, \frac{3.19}{100}\right)\) (1M, can be absorbed in standardisation)
Standardisation \(\mathrm{P}\left(Z < \frac{0.95 - 1.1}{\sqrt{0.0319}}\right)\) (1M)
\(0.2005\) (r.t. 0.2005) (1A)

(a)
\[ \begin{aligned} \mathrm{P}(E) &= \mathrm{P}(E \cap F) + \mathrm{P}(E \cap F') \\ &= \mathrm{P}(F)\mathrm{P}(E \mid F) + \mathrm{P}(F')\mathrm{P}(E \mid F') \\ &= 0.4k + 0.25(1 - k) \\ &= 0.15k + 0.25 \end{aligned} \]

(b)(i)
\[ \mathrm{P}(F \mid E) = \frac{\mathrm{P}(E \cap F)}{\mathrm{P}(E)} = \frac{0.4k}{0.15k + 0.25} = \frac{8}{13} \]
\[ 13(0.4k) = 8(0.15k + 0.25) \]
\[ 5.2k = 1.2k + 2 \implies 4k = 2 \implies k = 0.5 \]

(ii)
When \(k = 0.5\),
\[ \mathrm{P}(E) = 0.15(0.5) + 0.25 = 0.325 \]
Since \(\mathrm{P}(E \mid F) = 0.4 \neq 0.325 = \mathrm{P}(E)\) (or \(\mathrm{P}(E \cap F) = 0.2 \neq \mathrm{P}(E)\mathrm{P}(F) = 0.1625\)),
\(E\) and \(F\) are not independent.

(iii)
\[ \mathrm{P}(E) + \mathrm{P}(G) = 0.325 + 0.75 = 1.075 > 1 \]
If \(E\) and \(G\) were mutually exclusive, then \(\mathrm{P}(E \cup G) = \mathrm{P}(E) + \mathrm{P}(G) = 1.075 > 1\), which contradicts the axiom that probability cannot exceed 1.
Thus, \(E\) and \(G\) cannot be mutually exclusive.

(a)
\(\mathrm{P}(E) = \mathrm{P}(E \cap F) + \mathrm{P}(E \cap F')\) (1M)
\(\mathrm{P}(E) = 0.15k + 0.25\) (1A)

(b)(i)
\(\frac{0.4k}{0.15k + 0.25} = \frac{8}{13}\) (1M)
\(k = 0.5\) (1A)

(ii)
Comparing \(\mathrm{P}(E \mid F)\) and \(\mathrm{P}(E)\) (or \(\mathrm{P}(E \cap F)\) and \(\mathrm{P}(E)\mathrm{P}(F)\)) with correct conclusion (1A)

(iii)
\(\mathrm{P}(E) + \mathrm{P}(G) = 1.075 > 1\) (1M)
Correct conclusion with explanation (1A f.t.)

(a)
Using the quotient rule:
\[ \begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{(x^2 + 7)(1) - (x + 3)(2x)}{(x^2 + 7)^2} \\ &= \frac{x^2 + 7 - 2x^2 - 6x}{(x^2 + 7)^2} \\ &= \frac{-x^2 - 6x + 7}{(x^2 + 7)^2} \\ &= \frac{-(x + 7)(x - 1)}{(x^2 + 7)^2} \end{aligned} \]

(b)
Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ -(x + 7)(x - 1) = 0 \implies x = 1 \text{ or } x = -7 \]
When \(x = 1\), \(y = \frac{1 + 3}{1^2 + 7} = \frac{4}{8} = \frac{1}{2}\).
When \(x = -7\), \(y = \frac{-7 + 3}{(-7)^2 + 7} = \frac{-4}{56} = -\frac{1}{14}\).

Testing the nature of the stationary points using the first derivative:
- For \(x < -7\), \(\frac{\mathrm{d}y}{\mathrm{d}x} < 0\).
- For \(-7 < x < 1\), \(\frac{\mathrm{d}y}{\mathrm{d}x} > 0\).
- For \(x > 1\), \(\frac{\mathrm{d}y}{\mathrm{d}x} < 0\).

Hence,
Maximum point is \(\left(1, \frac{1}{2}\right)\).
Minimum point is \(\left(-7, -\frac{1}{14}\right)\).

(c)
When \(x = 0\), \(y = \frac{0 + 3}{0^2 + 7} = \frac{3}{7}\).
The slope of the tangent at \(x = 0\) is
\[ \left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=0} = \frac{7}{(0 + 7)^2} = \frac{7}{49} = \frac{1}{7} \]
The equation of the tangent is
\[ y - \frac{3}{7} = \frac{1}{7}(x - 0) \implies x - 7y + 3 = 0 \]

(a)
Applying quotient rule (1M)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-x^2 - 6x + 7}{(x^2 + 7)^2}\) (or equivalent factored form) (1A)

(b)
Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) to find critical values \(x = 1, -7\) (1M)
Testing intervals or signs for maximum and minimum (1M)
Maximum point \(\left(1, \frac{1}{2}\right)\) and Minimum point \(\left(-7, -\frac{1}{14}\right)\) (1A for both)

(c)
Finding slope \(\frac{1}{7}\) at \(x = 0\) (1M)
\(x - 7y + 3 = 0\) (or \(y = \frac{1}{7}x + \frac{3}{7}\)) (1A)

(a)
\[ \begin{aligned} f'(x) &= 2e^{-2x} + (2x - k)(-2e^{-2x}) \\ &= 2(1 - 2x + k)e^{-2x} \\ &= 2(k + 1 - 2x)e^{-2x} \end{aligned} \]
\[ \begin{aligned} f''(x) &= -4e^{-2x} + 2(k + 1 - 2x)(-2e^{-2x}) \\ &= [-4 - 4(k + 1 - 2x)]e^{-2x} \\ &= 4(2x - k - 2)e^{-2x} \end{aligned} \]

(b)(i)
Since \(f(x)\) attains an extreme value at \(x = 2\), \(f'(2) = 0\).
\[ 2(k + 1 - 2(2))e^{-2(2)} = 0 \]
Since \(e^{-4} \neq 0\),
\[ k + 1 - 4 = 0 \implies k = 3 \]

(ii)
When \(k = 3\),
\[ f''(2) = 4(2(2) - 3 - 2)e^{-4} = 4(-1)e^{-4} = -4e^{-4} < 0 \]
By the second derivative test, the extreme value is a maximum value.

(iii)
Setting \(f''(x) = 0\) with \(k = 3\):
\[ 4(2x - 3 - 2)e^{-2x} = 0 \implies 2x - 5 = 0 \implies x = \frac{5}{2} \]
For \(x < \frac{5}{2}\), \(f''(x) < 0\), and for \(x > \frac{5}{2}\), \(f''(x) > 0\).
So the concavity changes at \(x = \frac{5}{2}\).
The corresponding \(y\)-coordinate is
\[ f\left(\frac{5}{2}\right) = \left(2\left(\frac{5}{2}\right) - 3\right)e^{-2(5/2)} = (5 - 3)e^{-5} = 2e^{-5} \]
Thus, the point of inflexion is \(\left(\frac{5}{2}, 2e^{-5}\right)\).

(a)
\(f'(x) = 2(k + 1 - 2x)e^{-2x}\) (1M for product rule, 1A for answer)
\(f''(x) = 4(2x - k - 2)e^{-2x}\) (1A)

(b)(i)
\(f'(2) = 0 \implies k = 3\) (1M)

(ii)
\(f''(2) = -4e^{-4} < 0\) with conclusion (1A)

(iii)
Setting \(f''(x) = 0 \implies x = \frac{5}{2}\) (1M)
Coordinates \(\left(\frac{5}{2}, 2e^{-5}\right)\) (1A)

(a)
Width of each sub-interval \( \Delta x = \frac{2-0}{4} = 0.5 \).

Using the trapezoidal rule,
\(\begin{aligned} \int_{0}^{2} f(x) \, dx &\approx \frac{0.5}{2} \left[ f(0) + f(2) + 2\left(f(0.5) + f(1) + f(1.5)\right) \right] \\ &= 0.25 \left[ (0+2)e^0 + (2+2)e^{-2} + 2\left(2.5e^{-0.5} + 3e^{-1} + 3.5e^{-1.5}\right) \right] \\ &= 0.25 \left[ 2 + 4e^{-2} + 5e^{-0.5} + 6e^{-1} + 7e^{-1.5} \right] \\ &\approx 0.25 \left[ 2 + 0.54134113 + 3.03265330 + 2.20727665 + 1.56191111 \right] \\ &= 0.25(9.34318219) \\ &\approx 2.3358 \end{aligned}\)

(b)
(i)
\(\begin{aligned} \frac{d}{dx} \left( (x+3)e^{-x} \right) &= (1)e^{-x} + (x+3)(-e^{-x}) \\ &= e^{-x} - (x+3)e^{-x} \\ &= -(x+2)e^{-x} \end{aligned}\)

(ii)
\(\begin{aligned} \int f(x) \, dx &= \int (x+2)e^{-x} \, dx \\ &= - \int \frac{d}{dx} \left( (x+3)e^{-x} \right) dx \\ &= -(x+3)e^{-x} + C \end{aligned}\)
where \( C \) is an arbitrary constant.

(c)
(i)
Since \( f(x) = (x+2)e^{-x} > 0 \) for all \( 0 \le x \le 2 \),
\(\begin{aligned} \text{Area of } R &= \int_{0}^{2} (x+2)e^{-x} \, dx \\ &= \left[ -(x+3)e^{-x} \right]_{0}^{2} \\ &= -5e^{-2} - (-3e^0) \\ &= 3 - 5e^{-2} \end{aligned}\)

(ii)
\( f'(x) = -(x+1)e^{-x} \)
\(\begin{aligned} \frac{d^2}{dx^2} f(x) &= \frac{d}{dx} \left( -(x+1)e^{-x} \right) \\ &= -e^{-x} + (x+1)e^{-x} \\ &= x e^{-x} \end{aligned}\)
For \( 0 < x < 2 \), \( x > 0 \) and \( e^{-x} > 0 \), so \( \frac{d^2}{dx^2} f(x) > 0 \).
Thus, the curve is concave upwards on \( [0, 2] \).
Therefore, the estimate obtained in (a) is an over-estimate.

(a)
1M: for applying trapezoidal rule
1M: for correct values of \( f(x) \)
1A: for \( 2.3358 \) (accept r.t. 2.3358)

(b)(i)
1A: for \( -(x+2)e^{-x} \)

(b)(ii)
1M: for using the result in (b)(i)
1A: for \( -(x+3)e^{-x} + C \) (must include constant of integration)

(c)(i)
1M: for substituting limits into the antiderivative
1A: for \( 3 - 5e^{-2} \) (or \( 3 - \frac{5}{e^2} \))

(c)(ii)
1M: for finding \( \frac{d^2}{dx^2} f(x) = x e^{-x} \)
1M: for showing \( \frac{d^2}{dx^2} f(x) > 0 \) for \( 0 < x < 2 \)
1A (f.t.): for concluding that it is an over-estimate

(a) \( f(x) = kx e^{-2x} \)
\( f'(x) = k e^{-2x} + kx(-2e^{-2x}) = k(1-2x)e^{-2x} \)
\( f''(x) = k(-2)e^{-2x} + k(1-2x)(-2e^{-2x}) = k(-2 - 2 + 4x)e^{-2x} = 4k(x-1)e^{-2x} \)

(b) (i) Setting \( f'(x) = 0 \):
\( k(1-2x)e^{-2x} = 0 \implies x = \frac{1}{2} \)
Since \( f''\left(\frac{1}{2}\right) = 4k\left(\frac{1}{2}-1\right)e^{-1} = -2ke^{-1} < 0 \), \( f(x) \) attains its maximum at \( x = \frac{1}{2} \).
\( f\left(\frac{1}{2}\right) = \frac{k(1/2)}{e^{2(1/2)}} = \frac{k}{2e} \)
Thus, the coordinates of the maximum point are \( \left(\frac{1}{2}, \frac{k}{2e}\right) \).

(ii) For the graph to be concave downward, \( f''(x) < 0 \):
\( 4k(x-1)e^{-2x} < 0 \)
Since \( k > 0 \) and \( e^{-2x} > 0 \) for all \( x \), we have:
\( x - 1 < 0 \implies x < 1 \)

(c) (i) \( \frac{\mathrm{d}}{\mathrm{d}x}\left( (2x+1)e^{-2x} \right) = 2e^{-2x} + (2x+1)(-2e^{-2x}) = -4xe^{-2x} \)
Thus, \( xe^{-2x} = -\frac{1}{4}\frac{\mathrm{d}}{\mathrm{d}x}\left( (2x+1)e^{-2x} \right) \)
\( \int f(x) \,\mathrm{d}x = k \int xe^{-2x} \,\mathrm{d}x = -\frac{k}{4}(2x+1)e^{-2x} + C \), where \( C \) is a constant.

(ii) (1) Since \( f(0.5) = 4e^{-1} \):
\( \frac{k(0.5)}{e^{2(0.5)}} = 4e^{-1} \implies 0.5k e^{-1} = 4e^{-1} \implies k = 8 \)

(2) The total profit in the first 2 years is:
\( P(2) - P(0) = \int_0^2 P'(t) \,\mathrm{d}t \)
\( = \int_0^2 \left( 8te^{-2t} + \frac{2}{t+1} \right) \mathrm{d}t \)
\( = \left[ -2(2t+1)e^{-2t} + 2\ln(t+1) \right]_0^2 \)
\( = \left( -2(5)e^{-4} + 2\ln 3 \right) - \left( -2(1)e^0 + 2\ln 1 \right) \)
\( = 2 - 10e^{-4} + 2\ln 3 \)
\( \approx 2 - 0.183156 + 2.197225 \approx 4.0141 \) million dollars.

(a)
\( f'(x) = k(1-2x)e^{-2x} \) : 1M
For using product rule to differentiate \( f'(x) \) : 1M
\( f''(x) = 4k(x-1)e^{-2x} \) : 1A

(b) (i)
\( x = \frac{1}{2} \) and \( y = \frac{k}{2e} \) with justification of maximum : 1M+1A
(ii)
\( x < 1 \) : 1A

(c) (i)
\( \frac{\mathrm{d}}{\mathrm{d}x}\left( (2x+1)e^{-2x} \right) = -4xe^{-2x} \) : 1M
\( \int f(x)\,\mathrm{d}x = -\frac{k}{4}(2x+1)e^{-2x} + C \) : 1A
(ii) (1)
\( k = 8 \) : 1A
(2)
Setting up \( \int_0^2 P'(t)\,\mathrm{d}t \) : 1M
Using results from (c)(i) and integrating \( \frac{2}{t+1} \) : 1M
\( 4.0141 \) (accept \( 2 - 10e^{-4} + 2\ln 3 \)) : 1A

(a) Let \(f(x) = x e^{2x}\).
With \(n = 4\) sub-intervals on \([0, 2]\), the width of each sub-interval is \(\Delta x = \frac{2 - 0}{4} = 0.5\).
Using the trapezoidal rule:
\[ \begin{aligned} \int_{0}^{2} x e^{2x} \, dx &\approx \frac{0.5}{2} \left[ f(0) + f(2) + 2\left( f(0.5) + f(1) + f(1.5) \right) \right] \\ &= 0.25 \left[ 0(e^0) + 2e^4 + 2\left( 0.5e^1 + 1e^2 + 1.5e^3 \right) \right] \\ &= 0.25 \left( 2e^4 + e + 2e^2 + 3e^3 \right) \\ &\approx 0.25 (109.196300 + 2.718282 + 14.778112 + 60.256611) \\ &\approx 46.737326 \\ &\approx 46.7373 \end{aligned} \]

(b) (i)
\[ \begin{aligned} \frac{d}{dx}\left( (2x - 1)e^{2x} \right) &= \frac{d}{dx}(2x - 1) \cdot e^{2x} + (2x - 1) \cdot \frac{d}{dx}(e^{2x}) \\ &= 2e^{2x} + (2x - 1)(2e^{2x}) \\ &= 2e^{2x} + 4x e^{2x} - 2e^{2x} \\ &= 4x e^{2x} \end{aligned} \]

(ii) From (b)(i), \(\frac{d}{dx}\left( (2x - 1)e^{2x} \right) = 4x e^{2x}\).
Integrating both sides with respect to \(x\):
\[ 4 \int x e^{2x} \, dx = (2x - 1)e^{2x} + C_1 \]
\[ \int x e^{2x} \, dx = \frac{1}{4}(2x - 1)e^{2x} + C \]
where \(C\) is an arbitrary constant.

(c) Using the antiderivative found in (b)(ii):
\[ \begin{aligned} \int_{0}^{2} x e^{2x} \, dx &= \left[ \frac{1}{4}(2x - 1)e^{2x} \right]_{0}^{2} \\ &= \frac{1}{4}(2(2) - 1)e^{2(2)} - \frac{1}{4}(2(0) - 1)e^{2(0)} \\ &= \frac{3e^4}{4} - \left(-\frac{1}{4}\right) \\ &= \frac{3e^4 + 1}{4} \end{aligned} \]

(d) (i) Let \(y = x e^{2x}\).
\[ \frac{dy}{dx} = e^{2x} + 2x e^{2x} = (2x + 1)e^{2x} \]
\[ \frac{d^2 y}{dx^2} = 2e^{2x} + 2(2x + 1)e^{2x} = (4x + 4)e^{2x} = 4(x + 1)e^{2x} \]

(ii) For all \(x \in [0, 2]\), \(x + 1 > 0\) and \(e^{2x} > 0\), so
\[ \frac{d^2}{dx^2}\left( x e^{2x} \right) = 4(x + 1)e^{2x} > 0 \]
Since the curve \(y = x e^{2x}\) is concave upwards on \([0, 2]\), the line segments connecting the consecutive points lie above the curve.
Therefore, the estimate in (a) is an over-estimate.

(e) For \(0 \le x \le 2\), \((x + 1)e^{2x} \ge x e^{2x}\).
\[ \begin{aligned} S &= \int_{0}^{2} \left( (x + 1)e^{2x} - x e^{2x} \right) \, dx \\ &= \int_{0}^{2} e^{2x} \, dx \\ &= \left[ \frac{1}{2}e^{2x} \right]_{0}^{2} \\ &= \frac{1}{2}e^{4} - \frac{1}{2}e^{0} \\ &= \frac{e^4 - 1}{2} \end{aligned} \]

(a)
- 1M: Correct application of the trapezoidal rule formula with \(h = 0.5\)
- 1M: Correct substitutions of \(f(0), f(0.5), f(1), f(1.5), f(2)\)
- 1A: Correct estimate \(\approx 46.7373\) (r.t. 46.7373)

(b)
(i)
- 1M: Product rule differentiation
- 1A: \(4x e^{2x}\)
(ii)
- 1A: \(\frac{1}{4}(2x - 1)e^{2x} + C\) (must include constant of integration \(C\))

(c)
- 1M: Applying Fundamental Theorem of Calculus with limits \(0\) and \(2\)
- 1A: \(\frac{3e^4 + 1}{4}\) (exact value)

(d)
(i)
- 1M: Differentiating first derivative using product rule
- 1A: \(4(x + 1)e^{2x}\)
(ii)
- 1A: Showing \(\frac{d^2 y}{dx^2} > 0\) for \(x \in [0, 2]\) and concluding it is an over-estimate (f.t.)

(e)
- 1M: Setting up the definite integral for area \(\int_0^2 ((x+1)e^{2x} - xe^{2x}) \, dx\)
- 1A: \(\frac{e^4 - 1}{2}\) (exact value)

(a)
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t}\left[(pt + q)e^{-0.2t}\right] &= p e^{-0.2t} + (pt + q)(-0.2e^{-0.2t}) \\
&= (-0.2pt + p - 0.2q)e^{-0.2t}
\end{align*}
Setting \((-0.2pt + p - 0.2q)e^{-0.2t} = (5t + 10)e^{-0.2t}\):
Comparing coefficients:
\begin{align*}
-0.2p &= 5 \implies p = -25 \\
p - 0.2q &= 10 \implies -25 - 0.2q = 10 \implies q = -175
\end{align*}
Therefore,
\[ \int (5t + 10)e^{-0.2t}\,\mathrm{d}t = (-25t - 175)e^{-0.2t} + C = -25(t + 7)e^{-0.2t} + C, \]
where \(C\) is an arbitrary constant.

(b) (i)
\begin{align*}
\frac{\mathrm{d}P}{\mathrm{d}t} &= 5e^{-0.2t} + (5t + 10)(-0.2e^{-0.2t}) \\
&= (5 - t - 2)e^{-0.2t} \\
&= (3 - t)e^{-0.2t}
\end{align*}

(ii) Set \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = 0\):
Since \(e^{-0.2t} > 0\) for all \(t\),
\[ 3 - t = 0 \implies t = 3 \]
Testing the nature of the stationary point:
- For \(0 \le t < 3\), \(\dfrac{\mathrm{d}P}{\mathrm{d}t} > 0\).
- For \(t > 3\), \(\dfrac{\mathrm{d}P}{\mathrm{d}t} < 0\).

Thus, \(P(t)\) attains its maximum value at \(t = 3\).
The maximum value of \(P(t)\) is:
\[ P(3) = (5(3) + 10)e^{-0.2(3)} = 25e^{-0.6} \approx 13.7203\text{ kg/day} \]

(c) Since \(A'(t) = P(t)\),
\[ A(t) = \int (5t + 10)e^{-0.2t}\,\mathrm{d}t = -25(t + 7)e^{-0.2t} + C_1 \]
Given \(A(0) = 50\):
\[ 50 = -25(0 + 7)e^0 + C_1 \implies 50 = -175 + C_1 \implies C_1 = 225 \]
Thus,
\[ A(t) = 225 - 25(t + 7)e^{-0.2t} \]

(d)
For Claim 1:
Since \(\lim_{t \to \infty} 25(t + 7)e^{-0.2t} = 0\),
\[ \lim_{t \to \infty} A(t) = 225 - 0 = 225 \]
Since \(A'(t) = P(t) > 0\) for all \(t \ge 0\), \(A(t)\) is strictly increasing for \(t \ge 0\), which implies \(A(t) < 225 < 230\) for all \(t \ge 0\).
Thus, the total amount of pollutant will never exceed \(230\text{ kg}\). Claim 1 is incorrect.

For Claim 2:
At \(t = 6\),
\[ A(6) = 225 - 25(6 + 7)e^{-0.2(6)} = 225 - 325e^{-1.2} \approx 225 - 325(0.301194) \approx 127.1119\text{ kg} \]
Since \(A(6) \approx 127.1119 < 150\) and \(A(t)\) is strictly increasing, \(A(t) = 150\) occurs at some \(t > 6\).
Therefore, the time taken for the amount of pollutant to reach \(150\text{ kg}\) is greater than \(6\) days. Claim 2 is incorrect.

(a)
- \(\frac{\mathrm{d}}{\mathrm{d}t}\left[(pt+q)e^{-0.2t}\right] = (-0.2pt + p - 0.2q)e^{-0.2t}\): 1M
- Equating coefficients to obtain \(p = -25\) and \(q = -175\): 1M
- \(-25(t+7)e^{-0.2t} + C\): 1A

(b)
(i) \(\frac{\mathrm{d}P}{\mathrm{d}t} = (3-t)e^{-0.2t}\): 1A
(ii)
- Setting \(\frac{\mathrm{d}P}{\mathrm{d}t} = 0\) and solving \(t = 3\): 1M
- Justification of maximum using first derivative test or second derivative test: 1M
- Maximum value \(25e^{-0.6}\) (or r.t. 13.72): 1A

(c)
- Using \(A(t) = \int P(t)\,\mathrm{d}t\): 1M
- Substituting \(t = 0, A = 50\) to find constant \(C_1 = 225\): 1M
- \(A(t) = 225 - 25(t+7)e^{-0.2t}\): 1A

(d)
- Considering \(\lim_{t \to \infty} A(t) = 225\) and concluding Claim 1 is incorrect: 1M + 1A (f.t.)
- Calculating \(A(6) \approx 127.1119 < 150\) and concluding Claim 2 is incorrect: 1M + 1A (f.t.)