2022 HKDSE Mathematics M2 (Algebra and Calculus) Practice Paper with Answers

(a) \begin{aligned} f(2+h) - f(2) &= \frac{1}{\sqrt{3(2+h)+1}} - \frac{1}{\sqrt{3(2)+1}} \\ &= \frac{1}{\sqrt{3h+7}} - \frac{1}{\sqrt{7}} \\ &= \frac{\sqrt{7} - \sqrt{3h+7}}{\sqrt{7}\sqrt{3h+7}} \\ &= \frac{(\sqrt{7} - \sqrt{3h+7})(\sqrt{7} + \sqrt{3h+7})}{\sqrt{7}\sqrt{3h+7}(\sqrt{7} + \sqrt{3h+7})} \\ &= \frac{7 - (3h+7)}{\sqrt{7}\sqrt{3h+7}(\sqrt{3h+7}+\sqrt{7})} \\ &= \frac{-3h}{\sqrt{7}\sqrt{3h+7}(\sqrt{3h+7}+\sqrt{7})} \end{aligned}

(b) By first principles,
\begin{aligned} f'(2) &= \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} \\ &= \lim_{h \to 0} \frac{1}{h} \left( \frac{-3h}{\sqrt{7}\sqrt{3h+7}(\sqrt{3h+7}+\sqrt{7})} \right) \\ &= \lim_{h \to 0} \frac{-3}{\sqrt{7}\sqrt{3h+7}(\sqrt{3h+7}+\sqrt{7})} \\ &= \frac{-3}{\sqrt{7}\sqrt{7}(\sqrt{7}+\sqrt{7})} \\ &= \frac{-3}{7(2\sqrt{7})} \\ &= -\frac{3}{14\sqrt{7}} = -\frac{3\sqrt{7}}{98} \end{aligned}

(a)
- 1M for setting up difference of fractions and rationalizing numerator
- 1A for correctly completing the algebraic proof

(b)
- 1M for applying definition of derivative \(f'(2) = \lim_{h \to 0} \frac{f(2+h)-f(2)}{h}\)
- 1M for substituting expression from (a) and cancelling \(h\)
- 1.25A for correct exact value \(-\frac{3}{14\sqrt{7}}\) or \(-\frac{3\sqrt{7}}{98}\)

(a) \begin{aligned} \text{L.H.S.} &= \frac{1}{1-\sin\theta} + \frac{1}{1+\sin\theta} \\ &= \frac{(1+\sin\theta) + (1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)} \\ &= \frac{2}{1-\sin^2\theta} \\ &= \frac{2}{\cos^2\theta} \\ &= 2\sec^2\theta = \text{R.H.S.} \end{aligned}

(b) Using the result of (a),
\begin{aligned} 2\sec^2\theta &= 8\tan\theta \\ \sec^2\theta - 4\tan\theta &= 0 \\ (1 + \tan^2\theta) - 4\tan\theta &= 0 \\ \tan^2\theta - 4\tan\theta + 1 &= 0 \end{aligned}
Solving the quadratic equation for \(\tan\theta\):
\[ \tan\theta = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(1)}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3} \]
Since \(0 < \theta < \frac{\pi}{2}\):
For \(\tan\theta = 2 - \sqrt{3}\), \(\theta = \frac{\pi}{12}\).
For \(\tan\theta = 2 + \sqrt{3}\), \(\theta = \frac{5\pi}{12}\).

(a)
- 1M for combining fractions over common denominator \(1-\sin^2\theta\)
- 1A for using \(1-\sin^2\theta = \cos^2\theta\) to complete proof

(b)
- 1M for using (a) and substituting \(\sec^2\theta = 1+\tan^2\theta\)
- 1M for solving the quadratic in \(\tan\theta\)
- 2.25A for both correct angles \(\theta = \frac{\pi}{12}\) and \(\theta = \frac{5\pi}{12}\) (withhold 1M if extraneous roots not rejected or radians omitted)

(a) Let \(P(n)\) be the proposition \(\sum_{k=1}^n k(k+2) = \frac{n(n+1)(2n+7)}{6}\).
For \(n=1\),
\(\text{L.H.S.} = 1(1+2) = 3\)
\(\text{R.H.S.} = \frac{1(1+1)(2(1)+7)}{6} = \frac{1(2)(9)}{6} = 3\)
Since \(\text{L.H.S.} = \text{R.H.S.}\), \(P(1)\) is true.

Assume that \(P(m)\) is true for some positive integer \(m\), that is,
\[ \sum_{k=1}^m k(k+2) = \frac{m(m+1)(2m+7)}{6} \]

For \(n = m+1\),
\begin{aligned} \sum_{k=1}^{m+1} k(k+2) &= \sum_{k=1}^m k(k+2) + (m+1)((m+1)+2) \\ &= \frac{m(m+1)(2m+7)}{6} + (m+1)(m+3) \quad \text{(by induction hypothesis)} \\ &= \frac{m+1}{6} \left[ m(2m+7) + 6(m+3) \right] \\ &= \frac{m+1}{6} \left( 2m^2 + 7m + 6m + 18 \right) \\ &= \frac{m+1}{6} \left( 2m^2 + 13m + 18 \right) \\ &= \frac{m+1}{6} (m+2)(2m+9) \\ &= \frac{(m+1)((m+1)+1)(2(m+1)+7)}{6} \end{aligned}
Hence, \(P(m+1)\) is true.
By the principle of mathematical induction, \(P(n)\) is true for all positive integers \(n\).

(b) \begin{aligned} \sum_{k=10}^{30} k(k+2) &= \sum_{k=1}^{30} k(k+2) - \sum_{k=1}^9 k(k+2) \\ &= \frac{30(31)(2(30)+7)}{6} - \frac{9(10)(2(9)+7)}{6} \\ &= 5(31)(67) - 15(25) \\ &= 10385 - 375 \\ &= 10010 \end{aligned}

(a)
- 1 for checking base case \(n=1\)
- 1M for stating induction assumption
- 1M for adding \((m+1)(m+3)\) to induction hypothesis
- 1 for completing algebra and conclusion

(b)
- 1M for expressing sum as \(\sum_{k=1}^{30} k(k+2) - \sum_{k=1}^9 k(k+2)\)
- 1.25A for correct final value \(10010\)

(a) By the binomial theorem,
\begin{aligned} \left(1 + \frac{x}{2}\right)^n &= 1 + C_1^n \left(\frac{x}{2}\right) + C_2^n \left(\frac{x}{2}\right)^2 + C_3^n \left(\frac{x}{2}\right)^3 + \cdots \\ &= 1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \frac{n(n-1)(n-2)}{48}x^3 + \cdots \end{aligned}

(b) \begin{aligned} (2-3x)\left(1 + \frac{x}{2}\right)^n &= (2-3x)\left( 1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \cdots \right) \\ &= 2 + nx + \frac{n(n-1)}{4}x^2 - 3x - \frac{3n}{2}x^2 + \cdots \\ &= 2 + (n-3)x + \left( \frac{n(n-1)}{4} - \frac{3n}{2} \right)x^2 + \cdots \end{aligned}
Given that the coefficient of \(x\) is \(4\):
\[ n - 3 = 4 \implies n = 7 \]

Now find \(p\), the coefficient of \(x^2\):
\[ p = \frac{7(7-1)}{4} - \frac{3(7)}{2} = \frac{42}{4} - \frac{21}{2} = \frac{21}{2} - \frac{21}{2} = 0 \]

(a)
- 1M for applying binomial expansion
- 1A for \(1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \frac{n(n-1)(n-2)}{48}x^3\)

(b)
- 1M for expanding product and collecting like terms
- 1A for finding \(n = 7\)
- 1M for substituting \(n = 7\) into coefficient of \(x^2\)
- 1.25A for \(p = 0\)

(a) Note that \(y = \frac{x(x-2)+4}{x-2} = x + \frac{4}{x-2}\).
As \(x \to 2^+\) or \(x \to 2^-\), \(y \to \pm \infty\), so the vertical asymptote is \(x = 2\).
As \(x \to \pm \infty\), \(\frac{4}{x-2} \to 0\), so the oblique asymptote is \(y = x\).

(b) \(\frac{dy}{dx} = 1 - \frac{4}{(x-2)^2}\).
Setting \(\frac{dy}{dx} = 0\):
\[ 1 - \frac{4}{(x-2)^2} = 0 \implies (x-2)^2 = 4 \implies x - 2 = \pm 2 \]
Thus \(x = 0\) or \(x = 4\).

Second derivative: \(\frac{d^2y}{dx^2} = \frac{8}{(x-2)^3}\).
- When \(x = 0\), \(y = \frac{4}{-2} = -2\) and \(\left.\frac{d^2y}{dx^2}\right|_{x=0} = \frac{8}{(-2)^3} = -1 < 0\).
Thus, \((0, -2)\) is the local maximum point.
- When \(x = 4\), \(y = 4 + \frac{4}{4-2} = 6\) and \(\left.\frac{d^2y}{dx^2}\right|_{x=4} = \frac{8}{2^3} = 1 > 0\).
Thus, \((4, 6)\) is the local minimum point.

(a)
- 1A for vertical asymptote \(x = 2\)
- 1M for rewriting \(y = x + \frac{4}{x-2}\) or polynomial division
- 1A for oblique asymptote \(y = x\)

(b)
- 1M for setting \(\frac{dy}{dx} = 0\) to solve for \(x\)
- 1M for testing nature of stationary points via first/second derivative
- 1.25A for local maximum point \((0, -2)\) and local minimum point \((4, 6)\)

(a) Let \(u = \sqrt{2x+3}\), then \(u^2 = 2x+3 \implies x = \frac{u^2-3}{2}\) and \(dx = u \, du\).
\begin{aligned} \int \frac{x}{\sqrt{2x+3}} \, dx &= \int \frac{\frac{u^2-3}{2}}{u} \cdot u \, du \\ &= \frac{1}{2} \int (u^2-3) \, du \\ &= \frac{1}{2} \left( \frac{u^3}{3} - 3u \right) + C \\ &= \frac{u^3}{6} - \frac{3u}{2} + C \\ &= \frac{(2x+3)^{3/2}}{6} - \frac{3\sqrt{2x+3}}{2} + C \\ &= \frac{\sqrt{2x+3}(x-3)}{3} + C \end{aligned}

(b) \begin{aligned} y &= \int \frac{x+1}{\sqrt{2x+3}} \, dx \\ &= \int \frac{x}{\sqrt{2x+3}} \, dx + \int (2x+3)^{-1/2} \, dx \\ &= \left( \frac{(2x+3)^{3/2}}{6} - \frac{3\sqrt{2x+3}}{2} \right) + \sqrt{2x+3} + C_1 \\ &= \frac{(2x+3)^{3/2}}{6} - \frac{\sqrt{2x+3}}{2} + C_1 \\ &= \frac{(x-1)\sqrt{2x+3}}{3} + C_1 \end{aligned}
Since \(K\) passes through \((3, 4)\):
\[ 4 = \frac{(3-1)\sqrt{2(3)+3}}{3} + C_1 = \frac{2(3)}{3} + C_1 = 2 + C_1 \implies C_1 = 2 \]
Wait, let us re-verify: \(\frac{(2(3)+3)^{3/2}}{6} - \frac{\sqrt{2(3)+3}}{2} + C_1 = \frac{27}{6} - \frac{3}{2} + C_1 = \frac{9}{2} - \frac{3}{2} + C_1 = 3 + C_1 = 4 \implies C_1 = 1\).
Let us check \(\frac{(x-1)\sqrt{2x+3}}{3}\) at \(x=3\): \(\frac{(3-1)(3)}{3} = 2\). Then \(2 + 1 = 3 \neq 4\)? Wait!
\(\frac{(2x+3)^{3/2}}{6} - \frac{\sqrt{2x+3}}{2} = \sqrt{2x+3}\left(\frac{2x+3}{6} - \frac{1}{2}\right) = \sqrt{2x+3}\left(\frac{2x}{6}\right) = \frac{x\sqrt{2x+3}}{3}\).
At \(x=3\), \(\frac{3(3)}{3} = 3\). So \(3 + C_1 = 4 \implies C_1 = 1\).
Thus, \(y = \frac{x\sqrt{2x+3}}{3} + 1\) (or \(y = \frac{(2x+3)^{3/2}}{6} - \frac{\sqrt{2x+3}}{2} + 1\)).

(a)
- 1M for substitution \(u = \sqrt{2x+3}\) and \(dx = u\,du\)
- 1M for integrating polynomial in \(u\)
- 1A for \(\frac{(2x+3)^{3/2}}{6} - \frac{3\sqrt{2x+3}}{2} + C\) or equivalent

(b)
- 1M for using (a) to find general integral \(y = \frac{x\sqrt{2x+3}}{3} + C_1\)
- 1M for substituting \((3, 4)\) to find constant
- 1.25A for \(y = \frac{x\sqrt{2x+3}}{3} + 1\) or equivalent

(a) Using integration by parts with \(u = x\) and \(dv = e^{-x}dx\):
\begin{aligned} \int x e^{-x} \, dx &= -x e^{-x} - \int -e^{-x} \, dx \\ &= -x e^{-x} - e^{-x} + C \\ &= -(x+1)e^{-x} + C \end{aligned}

(b) Since \(x e^{-x} \ge 0\) for all \(x \in [0, k]\):
\begin{aligned} A &= \int_0^k x e^{-x} \, dx \\ &= \left[ -(x+1)e^{-x} \right]_0^k \\ &= -(k+1)e^{-k} - (-(0+1)e^0) \\ &= 1 - (k+1)e^{-k} \end{aligned}

(c) By the chain rule,
\[ \frac{dA}{dt} = \frac{dA}{dk} \cdot \frac{dk}{dt} \]
\(\frac{dA}{dk} = \frac{d}{dk}\left(1 - (k+1)e^{-k}\right) = -e^{-k} + (k+1)e^{-k} = k e^{-k}\).
\(\frac{dk}{dt} = \frac{d}{dt}(2\ln t) = \frac{2}{t}\).

When \(t = e\):
\(k = 2\ln e = 2\).
\[ \left.\frac{dA}{dt}\right|_{t=e} = \left( 2 e^{-2} \right) \left( \frac{2}{e} \right) = \frac{4}{e^3} \]

(a)
- 1M for integration by parts
- 1A for \(-(x+1)e^{-x} + C\)

(b)
- 1M for definite integral setup and evaluation
- 1A for \(A = 1 - (k+1)e^{-k}\)

(c)
- 1M for applying chain rule \(\frac{dA}{dt} = \frac{dA}{dk}\cdot\frac{dk}{dt}\)
- 1.25A for exact answer \(\frac{4}{e^3}\) (or \(4e^{-3}\))

(a) The coefficient determinant of \((E)\) is
\begin{aligned} \Delta &= \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda \\ 2 & 3 & \lambda^2-1 \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & \lambda-1 \\ 0 & 1 & \lambda^2-3 \end{vmatrix} \\ &= (\lambda^2-3) - (\lambda-1) \\ &= \lambda^2 - \lambda - 2 \\ &= (\lambda-2)(\lambda+1) \end{aligned}
\((E)\) has a unique solution if and only if \(\Delta \neq 0\).
Thus, the range of values of \(\lambda\) is \(\lambda \neq -1\) and \(\lambda \neq 2\).

(b) (i) If \((E)\) has infinitely many solutions, \(\Delta = 0 \implies \lambda = 2\) or \(\lambda = -1\).
The augmented matrix of \((E)\) is:
\[ \begin{pmatrix} 1 & 1 & 1 & 2 \\ 1 & 2 & \lambda & 3 \\ 2 & 3 & \lambda^2-1 & \mu \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 & 2 \\ 0 & 1 & \lambda-1 & 1 \\ 0 & 1 & \lambda^2-3 & \mu-4 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 & 2 \\ 0 & 1 & \lambda-1 & 1 \\ 0 & 0 & \lambda^2-\lambda-2 & \mu-5 \end{pmatrix} \]
For infinitely many solutions, we must have \(\mu - 5 = 0 \implies \mu = 5\).
Thus, \((\lambda = 2, \mu = 5)\) or \((\lambda = -1, \mu = 5)\).

(ii) Case 1: \(\lambda = 2, \mu = 5\).
The system becomes:
\[ \begin{cases} x + y + z = 2 \\ y + z = 1 \end{cases} \]
Let \(z = t\), then \(y = 1 - t\) and \(x = 2 - (1-t) - t = 1\).
Solution set: \(\{(1, 1-t, t) : t \in \mathbb{R}\}\).

Case 2: \(\lambda = -1, \mu = 5\).
The system becomes:
\[ \begin{cases} x + y + z = 2 \\ y - 2z = 1 \end{cases} \]
Let \(z = t\), then \(y = 1 + 2t\) and \(x = 2 - (1+2t) - t = 1 - 3t\).
Solution set: \(\{(1-3t, 1+2t, t) : t \in \mathbb{R}\}\).

(a)
- 1M for evaluating coefficient determinant \(\Delta\)
- 1A for \(\lambda \neq -1\) and \(\lambda \neq 2\)

(b)(i)
- 1M for row reducing augmented matrix
- 1A for \(\mu = 5\) with \(\lambda = 2\) or \(\lambda = -1\)

(b)(ii)
- 1M for parameterizing solutions
- 1.25A for both solution sets \(\{(1, 1-t, t) : t \in \mathbb{R}\}\) and \(\{(1-3t, 1+2t, t) : t \in \mathbb{R}\}\)

(a) Since \( \lim_{x \to 2^+} f(x) = \infty \) and \( \lim_{x \to 2^-} f(x) = -\infty \), the vertical asymptote is \( x = 2 \).
Rewrite \( f(x) \) by polynomial division:
\[ f(x) = \frac{x^2 - 4x + 8}{x - 2} = \frac{(x - 2)^2 + 4}{x - 2} = (x - 2) + \frac{4}{x - 2} \]
As \( x \to \pm\infty \), \( \frac{4}{x - 2} \to 0 \), so \( \lim_{x \to \pm\infty} [f(x) - (x - 2)] = 0 \).
Thus, the oblique asymptote is \( y = x - 2 \).

(b) Differentiating \( f(x) = x - 2 + 4(x - 2)^{-1} \):
\[ f'(x) = 1 - \frac{4}{(x - 2)^2} = \frac{(x - 2)^2 - 4}{(x - 2)^2} = \frac{x^2 - 4x}{(x - 2)^2} = \frac{x(x - 4)}{(x - 2)^2} \]
Setting \( f'(x) = 0 \) gives \( x = 0 \) or \( x = 4 \).

First derivative test:
For \( x < 0 \), \( f'(x) > 0 \);
For \( 0 < x < 2 \), \( f'(x) < 0 \);
For \( 2 < x < 4 \), \( f'(x) < 0 \);
For \( x > 4 \), \( f'(x) > 0 \).

When \( x = 0 \), \( y = \frac{8}{-2} = -4 \).
When \( x = 4 \), \( y = \frac{16 - 16 + 8}{2} = 4 \).
Therefore, the local maximum point is \( (0, -4) \) and the local minimum point is \( (4, 4) \).

(c) Intercepts:
\( y \)-intercept: \( (0, -4) \).
\( x \)-intercept: \( x^2 - 4x + 8 = 0 \implies \Delta = (-4)^2 - 4(1)(8) = -16 < 0 \), so there is no \( x \)-intercept.

(d) The region \( R \) lies above the oblique asymptote \( L: y = x - 2 \) for \( x \in [3, 6] \).
Let \( u \) be the coordinate along the axis of rotation \( L \) and \( r(x) \) be the perpendicular distance from a point \( (x, f(x)) \) on \( G \) to the line \( x - y - 2 = 0 \).
The distance from \( (x, f(x)) \) to the line \( x - y - 2 = 0 \) is:
\[ r(x) = \frac{|x - f(x) - 2|}{\sqrt{1^2 + (-1)^2}} = \frac{f(x) - (x - 2)}{\sqrt{2}} = \frac{4}{\sqrt{2}(x - 2)} = \frac{2\sqrt{2}}{x - 2} \]
The projection of \( (x, x - 2) \) onto \( L \) has coordinate increment \( \mathrm{d}u = \sqrt{(\mathrm{d}x)^2 + (\mathrm{d}(x-2))^2} = \sqrt{1 + 1^2}\,\mathrm{d}x = \sqrt{2}\,\mathrm{d}x \).
Therefore, the volume is:
\[ V = \pi \int_{x=3}^{6} [r(x)]^2 \,\mathrm{d}u = \pi \int_{3}^{6} \left(\frac{2\sqrt{2}}{x - 2}\right)^2 \sqrt{2}\,\mathrm{d}x = \pi \int_{3}^{6} \frac{8}{(x - 2)^2} \sqrt{2}\,\mathrm{d}x \]
\[ V = 8\sqrt{2}\pi \left[ -\frac{1}{x - 2} \right]_3^6 = 8\sqrt{2}\pi \left( -\frac{1}{4} - (-1) \right) = 8\sqrt{2}\pi \left( \frac{3}{4} \right) = 6\sqrt{2}\pi \] (Wait: \( 8\sqrt{2}\pi \times \frac{3}{4} = 6\sqrt{2}\pi \)).

(a) \( 1\text{A} \) for vertical asymptote \( x = 2 \)
\( 1\text{M} \) for finding the form \( (x - 2) + \frac{4}{x - 2} \)
\( 1\text{A} \) for oblique asymptote \( y = x - 2 \)

(b) \( 1\text{M} \) for differentiating \( f(x) \)
\( 1\text{M} \) for testing signs of \( f'(x) \)
\( 1\text{A} \) for local maximum point \( (0, -4) \)
\( 1\text{A} \) for local minimum point \( (4, 4) \)

(c) \( 1\text{A} \) for correct shape of both branches
\( 1\text{A} \) for showing asymptotes and turning points correctly

(d) \( 1\text{M} \) for setting up distance \( r(x) \) to line \( y = x - 2 \)
\( 1\text{M} \) for setting up the integral \( \pi \int [r(x)]^2 \sqrt{1 + (m)^2}\,\mathrm{d}x \)
\( 1\text{A} \) for obtaining \( 6\sqrt{2}\pi \)

(a) Let \( I = \int_0^a x h(x) \,\mathrm{d}x \).
Using substitution \( u = a - x \), \( \mathrm{d}x = -\mathrm{d}u \):
When \( x = 0 \), \( u = a \); when \( x = a \), \( u = 0 \).
\[ I = \int_a^0 (a - u) h(a - u) (-\mathrm{d}u) = \int_0^a (a - u) h(u) \,\mathrm{d}u \]
\[ I = a \int_0^a h(u) \,\mathrm{d}u - \int_0^a u h(u) \,\mathrm{d}u = a \int_0^a h(x) \,\mathrm{d}x - I \]
\[ 2I = a \int_0^a h(x) \,\mathrm{d}x \implies I = \frac{a}{2} \int_0^a h(x) \,\mathrm{d}x \]

(b) (i)
\[ \frac{1}{1 + \sin x} = \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} = \frac{1 - \sin x}{1 - \sin^2 x} = \frac{1 - \sin x}{\cos^2 x} \]

(ii) Let \( h(x) = \frac{\sin x}{1 + \sin x} \).
\( h(\pi - x) = \frac{\sin(\pi - x)}{1 + \sin(\pi - x)} = \frac{\sin x}{1 + \sin x} = h(x) \).
By (a):
\[ \int_0^{\pi} \frac{x \sin x}{1 + \sin x} \,\mathrm{d}x = \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{1 + \sin x} \,\mathrm{d}x \]
Notice that \( \frac{\sin x}{1 + \sin x} = \frac{1 + \sin x - 1}{1 + \sin x} = 1 - \frac{1}{1 + \sin x} \).
Using (b)(i):
\[ \int \frac{1}{1 + \sin x} \,\mathrm{d}x = \int (\sec^2 x - \sec x \tan x) \,\mathrm{d}x = \tan x - \sec x \]
\[ \int_0^{\pi} \frac{\sin x}{1 + \sin x} \,\mathrm{d}x = \int_0^{\pi} 1 \,\mathrm{d}x - \left[ \tan x - \sec x \right]_0^{\pi} \]
\( = [\pi - 0] - [(\tan\pi - \sec\pi) - (\tan 0 - \sec 0)] = \pi - [ (0 - (-1)) - (0 - 1) ] = \pi - 2 \).
Therefore:
\[ \int_0^{\pi} \frac{x \sin x}{1 + \sin x} \,\mathrm{d}x = \frac{\pi}{2}(\pi - 2) = \frac{\pi^2}{2} - \pi \]

(c) Using substitution \( t = x - \pi \), when \( x = 0 \), \( t = -\pi \); when \( x = 2\pi \), \( t = \pi \).
\[ \int_0^{2\pi} \frac{|x - \pi| \sin^2 x}{1 + \sin x} \,\mathrm{d}x = \int_{-\pi}^{\pi} \frac{|t| \sin^2(t + \pi)}{1 + \sin(t + \pi)} \,\mathrm{d}t = \int_{-\pi}^{\pi} \frac{|t| \sin^2 t}{1 - \sin t} \,\mathrm{d}t \]
Since the integrand is not purely even due to \( 1 - \sin t \), split into \( [-\pi, 0] \) and \( [0, \pi] \):
For \( t \in [-\pi, 0] \), let \( t = -v \):
\[ \int_{-\pi}^0 \frac{-t \sin^2 t}{1 - \sin t} \,\mathrm{d}t = \int_{\pi}^0 \frac{v \sin^2(-v)}{1 - \sin(-v)} (-\mathrm{d}v) = \int_0^{\pi} \frac{v \sin^2 v}{1 + \sin v} \,\mathrm{d}v \]
Thus, the total integral is:
\[ \int_0^{\pi} \frac{t \sin^2 t}{1 - \sin t} \,\mathrm{d}t + \int_0^{\pi} \frac{t \sin^2 t}{1 + \sin t} \,\mathrm{d}t = \int_0^{\pi} t \sin^2 t \left( \frac{1}{1 - \sin t} + \frac{1}{1 + \sin t} \right) \,\mathrm{d}t \]
\[ = \int_0^{\pi} t \sin^2 t \left( \frac{2}{1 - \sin^2 t} \right) \,\mathrm{d}t = 2 \int_0^{\pi} t \tan^2 t \,\mathrm{d}t \]
Let \( h(t) = \tan^2 t \). Since \( \tan^2(\pi - t) = (-\tan t)^2 = \tan^2 t \), by (a):
\[ 2 \int_0^{\pi} t \tan^2 t \,\mathrm{d}t = 2 \left( \frac{\pi}{2} \int_0^{\pi} \tan^2 t \,\mathrm{d}t \right) = \pi \int_0^{\pi} (\sec^2 t - 1) \,\mathrm{d}t \]
Since \( \tan t \) has a discontinuity at \( t = \frac{\pi}{2} \), we evaluate by symmetry \( 2 \int_0^{\pi/2} \tan^2 t \,\mathrm{d}t \) or directly:
\[ = 2\pi \lim_{\epsilon \to 0^+} [\tan t - t]_0^{\pi/2 - \epsilon} \] which is improper, but notice \( \frac{\sin^2 t}{1+\sin t} = \frac{1 - \cos^2 t}{1+\sin t} = 1 - \sin t \).
Wait! \( \frac{\sin^2 t}{1 + \sin t} = 1 - \sin t \) is NOT correct because \( 1 - \sin^2 t = \cos^2 t \). However, \( \frac{\sin^2 t}{1 + \sin t} = \frac{\sin^2 t - 1 + 1}{1 + \sin t} = \sin t - 1 + \frac{1}{1 + \sin t} \).
Let's evaluate \( \int_0^{\pi} \frac{t \sin^2 t}{1 + \sin t} \,\mathrm{d}t \):
By (a), since \( \frac{\sin^2(\pi-t)}{1+\sin(\pi-t)} = \frac{\sin^2 t}{1+\sin t} \):
\[ \int_0^{\pi} \frac{t \sin^2 t}{1 + \sin t} \,\mathrm{d}t = \frac{\pi}{2} \int_0^{\pi} \left( \sin t - 1 + \frac{1}{1 + \sin t} \right) \,\mathrm{d}t \]
\[ = \frac{\pi}{2} \left( [-\cos t - t]_0^{\pi} + (\pi - 2) \right) = \frac{\pi}{2} ( (1 - \pi) - (-1 - 0) + \pi - 2 ) = \frac{\pi}{2} (2 - \pi + \pi - 2) = 0 \]?
Wait! \( \int_0^\pi (\sin t - 1) \mathrm{d}t = [-\cos t - t]_0^\pi = (1 - \pi) - (-1) = 2 - \pi \).
And \( \int_0^\pi \frac{1}{1+\sin t} \mathrm{d}t = \int_0^\pi (\sec^2 t - \sec t \tan t) \mathrm{d}t = [\tan t - \sec t]_0^\pi = (0 - (-1)) - (0 - 1) = 2 \).
So \( \int_0^{\pi} \left(\sin t - 1 + \frac{1}{1+\sin t}\right) \mathrm{d}t = (2 - \pi) + 2 = 4 - \pi \).
Hence, \( \int_0^{\pi} \frac{t \sin^2 t}{1 + \sin t} \,\mathrm{d}t = \frac{\pi}{2}(4 - \pi) \).
Similarly, for the other half:
\[ \int_{-\pi}^0 \frac{|t| \sin^2 t}{1 - \sin t} \,\mathrm{d}t = \int_0^\pi \frac{v \sin^2 v}{1 + \sin v} \,\mathrm{d}v = \frac{\pi}{2}(4 - \pi) \]
Therefore, the required integral is:
\[ 2 \times \frac{\pi}{2}(4 - \pi) = 4\pi - \pi^2 \].

(a) \( 1\text{M} \) for using substitution \( u = a - x \)
\( 1\text{M} \) for combining integrals to get \( 2I = a \int_0^a h(x)\,\mathrm{d}x \)

(b)(i) \( 1\text{A} \) for complete algebraic proof
(b)(ii) \( 1\text{M} \) for applying property (a)
\( 1\text{M} \) for rewriting \( \frac{\sin x}{1+\sin x} = 1 - \frac{1}{1+\sin x} \)
\( 1\text{M} \) for integrating \( \sec^2 x - \sec x\tan x \)
\( 1\text{A} \) for \( \frac{\pi^2}{2} - \pi \)

(c) \( 1\text{M} \) for splitting integral at \( x = \pi \) or using substitution \( t = x - \pi \)
\( 1\text{M} \) for transforming \( \int_{-\pi}^0 \) into \( \int_0^\pi \)
\( 1\text{M} \) for decomposing \( \frac{\sin^2 t}{1+\sin t} = \sin t - 1 + \frac{1}{1+\sin t} \)
\( 1\text{M} \) for applying property (a)
\( 1\text{A} \) for obtaining \( 4\pi - \pi^2 \)

(a) (i) Characteristic polynomial of \( M \):
\[ \det(M - \lambda I) = \det\begin{pmatrix} 3 - \lambda & -1 \\ 2 & -\lambda \end{pmatrix} = -\lambda(3 - \lambda) - (-2) = \lambda^2 - 3\lambda + 2 = (\lambda - 1)(\lambda - 2) = 0 \]
Eigenvalues are \( \lambda_1 = 1, \lambda_2 = 2 \).
For \( \lambda_1 = 1 \):
\[ \begin{pmatrix} 2 & -1 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 2u_1 - u_2 = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \]
For \( \lambda_2 = 2 \):
\[ \begin{pmatrix} 1 & -1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies v_1 - v_2 = 0 \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \]
Thus, \( P = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \) and \( D = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \).

(ii) \( \det(P) = 1(1) - 1(2) = -1 \).
\[ P^{-1} = \frac{1}{-1} \begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} \]
For any positive integer \( n \):
\[ M^n = P D^n P^{-1} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 \\ 0 & 2^n \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} \]
\[ = \begin{pmatrix} 1 & 2^n \\ 2 & 2^n \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} -1 + 2\cdot 2^n & 1 - 2^n \\ -2 + 2\cdot 2^n & 2 - 2^n \end{pmatrix} = \begin{pmatrix} 2^{n+1}-1 & 1-2^n \\ 2^{n+1}-2 & 2-2^n \end{pmatrix} \]

(b) (i) The coefficient matrix is \( A = \begin{pmatrix} 3 & -1 & k \\ 2 & 0 & 2 \\ 1 & 1 & k - 2 \end{pmatrix} \).
\[ \det(A) = 3(0 - 2) - (-1)(2(k-2) - 2) + k(2 - 0) \]
\[ = -6 + (2k - 6) + 2k = 4k - 12 \]
Wait, let's recalculate:
\( \det(A) = 3(0 - 2) - (-1)(2k - 4 - 2) + k(2 - 0) = -6 + (2k - 6) + 2k = 4k - 12 \).
Setting \( \det(A) = 0 \implies 4k - 12 = 0 \implies k = 3 \).

(ii) When \( k = 3 \), the augmented matrix is:
\[ \begin{pmatrix} 3 & -1 & 3 & | & 2 \\ 2 & 0 & 2 & | & 4 \\ 1 & 1 & 1 & | & 1 \end{pmatrix} \]
\( R_2 \leftarrow \frac{1}{2}R_2 \):
\[ \begin{pmatrix} 1 & 0 & 1 & | & 2 \\ 3 & -1 & 3 & | & 2 \\ 1 & 1 & 1 & | & 1 \end{pmatrix} \]
\( R_2 \leftarrow R_2 - 3R_1 \), \( R_3 \leftarrow R_3 - R_1 \):
\[ \begin{pmatrix} 1 & 0 & 1 & | & 2 \\ 0 & -1 & 0 & | & -4 \\ 0 & 1 & 0 & | & -1 \end{pmatrix} \]
\( R_3 \leftarrow R_3 + R_2 \):
\[ \begin{pmatrix} 1 & 0 & 1 & | & 2 \\ 0 & -1 & 0 & | & -4 \\ 0 & 0 & 0 & | & -5 \end{pmatrix} \]
Since the last row gives \( 0 = -5 \), which is a contradiction, \( (E) \) has no solution when \( k = 3 \).

(a)(i) \( 1\text{M} \) for finding characteristic equation and eigenvalues
\( 1\text{M} \) for finding eigenvectors
\( 1\text{A} \) for correct \( P \) and \( D \)

(a)(ii) \( 1\text{M} \) for finding \( P^{-1} \)
\( 1\text{M} \) for using \( M^n = P D^n P^{-1} \)
\( 1\text{A} \) for correct matrix multiplication to arrive at the result

(b)(i) \( 1\text{M} \) for setting \( \det(A) = 0 \)
\( 1\text{A} \) for \( k = 3 \)

(b)(ii) \( 1\text{M} \) for setting up augmented matrix with \( k = 3 \)
\( 1\text{M} \) for applying Gaussian elimination
\( 1\text{A} \) for concluding no solution due to inconsistency

(a) (i)
\[ \vec{AB} = \vec{OB} - \vec{OA} = (4 - 2)\mathbf{i} + (3 - 1)\mathbf{j} + (3 - 2)\mathbf{k} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} \]
\[ \vec{AC} = \vec{OC} - \vec{OA} = (1 - 2)\mathbf{i} + (4 - 1)\mathbf{j} + (5 - 2)\mathbf{k} = -\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} \]
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ -1 & 3 & 3 \end{vmatrix} = (6 - 3)\mathbf{i} - (6 - (-1))\mathbf{j} + (6 - (-2))\mathbf{k} = 3\mathbf{i} - 7\mathbf{j} + 8\mathbf{k} \]
Wait, let's recalculate:
\( \mathbf{i}(2\times 3 - 1\times 3) - \mathbf{j}(2\times 3 - 1\times(-1)) + \mathbf{k}(2\times 3 - 2\times(-1)) = 3\mathbf{i} - 7\mathbf{j} + 8\mathbf{k} \).
Let's check dot products: \( (3)(2) + (-7)(2) + 8(1) = 6 - 14 + 8 = 0 \). Correct.
\( (3)(-1) + (-7)(3) + 8(3) = -3 - 21 + 24 = 0 \). Correct.

(ii) Area of \( \triangle ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{3^2 + (-7)^2 + 8^2} = \frac{1}{2}\sqrt{9 + 49 + 64} = \frac{1}{2}\sqrt{122} \).
A normal vector to \( \Pi \) is \( \mathbf{n} = 3\mathbf{i} - 7\mathbf{j} + 8\mathbf{k} \).
The equation of \( \Pi \) is:
\[ 3(x - 2) - 7(y - 1) + 8(z - 2) = 0 \implies 3x - 7y + 8z - 15 = 0 \]

(b) (i) \( \vec{AD} = \vec{OD} - \vec{OA} = (3 - 2)\mathbf{i} + (-1 - 1)\mathbf{j} + (8 - 2)\mathbf{k} = \mathbf{i} - 2\mathbf{j} + 6\mathbf{k} \).
\[ \vec{AD} \cdot (\vec{AB} \times \vec{AC}) = (1)(3) + (-2)(-7) + (6)(8) = 3 + 14 + 48 = 65 \]
Volume of tetrahedron \( ABCD = \frac{1}{6} |\vec{AD} \cdot (\vec{AB} \times \vec{AC})| = \frac{65}{6} \).

(ii) The line passing through \( D(3, -1, 8) \) perpendicular to \( \Pi \) has parametric equations:
\[ x = 3 + 3t, \quad y = -1 - 7t, \quad z = 8 + 8t \]
Substitute into plane equation:
\[ 3(3 + 3t) - 7(-1 - 7t) + 8(8 + 8t) - 15 = 0 \]
\[ 9 + 9t + 7 + 49t + 64 + 64t - 15 = 0 \]
\[ 122t + 65 = 0 \implies t = -\frac{65}{122} \]
Thus, the coordinates of \( H \) are:
\[ x = 3 + 3\left(-\frac{65}{122}\right) = \frac{366 - 195}{122} = \frac{171}{122} \]
\[ y = -1 - 7\left(-\frac{65}{122}\right) = \frac{-122 + 455}{122} = \frac{333}{122} \]
\[ z = 8 + 8\left(-\frac{65}{122}\right) = \frac{976 - 520}{122} = \frac{456}{122} = \frac{228}{61} \]
So \( H = \left( \frac{171}{122}, \frac{333}{122}, \frac{228}{61} \right) \).

(iii) \( \vec{OE} = \frac{2\vec{OA} + 1\vec{OB}}{3} = \frac{2(2, 1, 2) + (4, 3, 3)}{3} = \left( \frac{8}{3}, \frac{5}{3}, \frac{7}{3} \right) \).
\[ \vec{DE} = \vec{OE} - \vec{OD} = \left( \frac{8}{3} - 3 \right)\mathbf{i} + \left( \frac{5}{3} - (-1) \right)\mathbf{j} + \left( \frac{7}{3} - 8 \right)\mathbf{k} = -\frac{1}{3}\mathbf{i} + \frac{8}{3}\mathbf{j} - \frac{17}{3}\mathbf{k} \]
Let \( \theta \) be the angle between \( DE \) and the plane \( \Pi \).
\[ \sin\theta = \frac{|\vec{DE} \cdot \mathbf{n}|}{|\vec{DE}| |\mathbf{n}|} \]
\[ \vec{DE} \cdot \mathbf{n} = \left(-\frac{1}{3}\right)(3) + \left(\frac{8}{3}\right)(-7) + \left(-\frac{17}{3}\right)(8) = -1 - \frac{56}{3} - \frac{136}{3} = -1 - 64 = -65 \]
\[ |\vec{DE}| = \frac{1}{3}\sqrt{(-1)^2 + 8^2 + (-17)^2} = \frac{1}{3}\sqrt{1 + 64 + 289} = \frac{1}{3}\sqrt{354} \]
\[ |\mathbf{n}| = \sqrt{122} \]
\[ \sin\theta = \frac{65}{\frac{1}{3}\sqrt{354}\sqrt{122}} = \frac{195}{\sqrt{43188}} \approx \frac{195}{207.817} \approx 0.938325 \]
\[ \theta = \arcsin(0.938325) \approx 69.8^\circ \]

(a)(i) \( 1\text{M} \) for finding \( \vec{AB} \) and \( \vec{AC} \)
\( 1\text{A} \) for \( 3\mathbf{i} - 7\mathbf{j} + 8\mathbf{k} \)

(a)(ii) \( 1\text{M} \) for formula of area \( \frac{1}{2}|\vec{AB}\times\vec{AC}| \)
\( 1\text{A} \) for area \( \frac{\sqrt{122}}{2} \)
\( 1\text{A} \) for plane equation \( 3x - 7y + 8z - 15 = 0 \)

(b)(i) \( 1\text{M} \) for volume formula \( \frac{1}{6}|\vec{AD}\cdot(\vec{AB}\times\vec{AC})| \)
\( 1\text{A} \) for \( \frac{65}{6} \)

(b)(ii) \( 1\text{M} \) for setting up parametric line for \( DH \)
\( 1\text{A} \) for coordinates \( \left(\frac{171}{122}, \frac{333}{122}, \frac{228}{61}\right) \)

(b)(iii) \( 1\text{M} \) for finding coordinates of \( E \) and vector \( \vec{DE} \)
\( 1\text{M} \) for using \( \sin\theta = \frac{|\vec{DE}\cdot\mathbf{n}|}{|\vec{DE}||\mathbf{n}|} \)
\( 1\text{A} \) for \( 69.8^\circ \)