2023 HKDSE 數學 單元二 (代數與微積分) 模擬試題連答案詳解

(a) Expanding the two binomial terms:
\((1 + ax)^6 = 1 + 6ax + 15a^2x^2 + 20a^3x^3 + \dots\)
\((1 - 2x)^3 = 1 - 6x + 12x^2 - 8x^3 + \dots\)

Multiplying the two expansions:
\((1 + ax)^6 (1 - 2x)^3 = (1 + 6ax + 15a^2x^2 + 20a^3x^3 + \dots)(1 - 6x + 12x^2 - 8x^3 + \dots)\)

The term in \(x\) is:
\((6a - 6)x\)
Given that the coefficient of \(x\) is \(-3\):
\(6a - 6 = -3 \implies 6a = 3 \implies a = \frac{1}{2}\).

The term in \(x^2\) is:
\((12 - 36a + 15a^2)x^2\)
Substituting \(a = \frac{1}{2}\):
\(k = 12 - 36\left(\frac{1}{2}\right) + 15\left(\frac{1}{2}\right)^2 = 12 - 18 + \frac{15}{4} = -6 + \frac{15}{4} = -\frac{9}{4}\).

(b) The term in \(x^3\) is given by:
\(1(-8) + (6a)(12) + (15a^2)(-6) + (20a^3)(1) = -8 + 72a - 90a^2 + 20a^3\)

Substituting \(a = \frac{1}{2}\):
\(\text{Coefficient of } x^3 = -8 + 72\left(\frac{1}{2}\right) - 90\left(\frac{1}{4}\right) + 20\left(\frac{1}{8}\right)\)
\(= -8 + 36 - \frac{45}{2} + \frac{5}{2} = 28 - 20 = 8\).

(a)
1M for expanding both binomial expressions up to at least \(x^2\)
1M for setting the coefficient of \(x\) equal to \(-3\)
1A for \(a = \frac{1}{2}\)
1M for substituting \(a\) into the expression for the coefficient of \(x^2\)
1A for \(k = -\frac{9}{4}\)

(b)
1M for setting up the expression for the coefficient of \(x^3\)
0.25A for \(8\)

(a) Let \(P(n)\) be the proposition: \(\sum_{r=1}^{n} \frac{2r+1}{r^2(r+1)^2} = 1 - \frac{1}{(n+1)^2}\).

For \(n = 1\):
\(\text{L.H.S.} = \frac{2(1)+1}{1^2(1+1)^2} = \frac{3}{4}\)
\(\text{R.H.S.} = 1 - \frac{1}{(1+1)^2} = 1 - \frac{1}{4} = \frac{3}{4}\)
Since \(\text{L.H.S.} = \text{R.H.S.}\), \(P(1)\) is true.

Assume that \(P(k)\) is true for some positive integer \(k\), that is,
\(\sum_{r=1}^{k} \frac{2r+1}{r^2(r+1)^2} = 1 - \frac{1}{(k+1)^2}\).

For \(n = k+1\):
\(\sum_{r=1}^{k+1} \frac{2r+1}{r^2(r+1)^2} = \sum_{r=1}^{k} \frac{2r+1}{r^2(r+1)^2} + \frac{2(k+1)+1}{(k+1)^2(k+2)^2}\)
\(= 1 - \frac{1}{(k+1)^2} + \frac{2k+3}{(k+1)^2(k+2)^2}\)
\(= 1 - \left[ \frac{(k+2)^2 - (2k+3)}{(k+1)^2(k+2)^2} \right]\)
\(= 1 - \left[ \frac{k^2 + 4k + 4 - 2k - 3}{(k+1)^2(k+2)^2} \right]\)
\(= 1 - \left[ \frac{k^2 + 2k + 1}{(k+1)^2(k+2)^2} \right]\)
\(= 1 - \frac{(k+1)^2}{(k+1)^2(k+2)^2}\)
\(= 1 - \frac{1}{(k+2)^2}\)
\(= 1 - \frac{1}{((k+1)+1)^2}\)
Hence, \(P(k+1)\) is true.

By the principle of mathematical induction, \(P(n)\) is true for all positive integers \(n\).

(b) Using (a):
\(\sum_{r=5}^{20} \frac{2r+1}{r^2(r+1)^2} = \sum_{r=1}^{20} \frac{2r+1}{r^2(r+1)^2} - \sum_{r=1}^{4} \frac{2r+1}{r^2(r+1)^2}\)
\(= \left(1 - \frac{1}{(20+1)^2}\right) - \left(1 - \frac{1}{(4+1)^2}\right)\)
\(= \frac{1}{25} - \frac{1}{441}\)
\(= \frac{441 - 25}{11025} = \frac{416}{11025}\).

(a)
1M for testing the base case \(n = 1\)
1M for inductive hypothesis and expressing the sum for \(n = k+1\) using the hypothesis
1M for algebraic simplification to the required form
1A for conclusion with correct induction statement

(b)
1M for splitting the summation as \(\sum_{r=1}^{20} - \sum_{r=1}^{4}\)
1.25A for \(\frac{416}{11025}\) (or equivalent exact fraction)

(a) By definition of first principles:
\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)
\(= \lim_{h \to 0} \frac{\frac{3}{\sqrt{2(x+h)+1}} - \frac{3}{\sqrt{2x+1}}}{h}\)
\(= 3 \lim_{h \to 0} \frac{\sqrt{2x+1} - \sqrt{2x+2h+1}}{h \sqrt{2x+2h+1}\sqrt{2x+1}}\)
\(= 3 \lim_{h \to 0} \frac{(\sqrt{2x+1} - \sqrt{2x+2h+1})(\sqrt{2x+1} + \sqrt{2x+2h+1})}{h \sqrt{2x+2h+1}\sqrt{2x+1}(\sqrt{2x+1} + \sqrt{2x+2h+1})}\)
\(= 3 \lim_{h \to 0} \frac{(2x+1) - (2x+2h+1)}{h \sqrt{2x+2h+1}\sqrt{2x+1}(\sqrt{2x+1} + \sqrt{2x+2h+1})}\)
\(= 3 \lim_{h \to 0} \frac{-2h}{h \sqrt{2x+2h+1}\sqrt{2x+1}(\sqrt{2x+1} + \sqrt{2x+2h+1})}\)
\(= 3 \cdot \frac{-2}{\sqrt{2x+1}\sqrt{2x+1}(\sqrt{2x+1} + \sqrt{2x+1})}\)
\(= \frac{-6}{(2x+1)(2\sqrt{2x+1})} = -\frac{3}{(2x+1)^{3/2}}\).

(b) When \(x = 4\):
\(y = f(4) = \frac{3}{\sqrt{2(4)+1}} = \frac{3}{\sqrt{9}} = 1\).
Slope of tangent \(m_T = f'(4) = -\frac{3}{(2(4)+1)^{3/2}} = -\frac{3}{9^{3/2}} = -\frac{3}{27} = -\frac{1}{9}\).
Slope of normal \(m_N = -\frac{1}{m_T} = 9\).

Equation of the normal at \((4, 1)\):
\(y - 1 = 9(x - 4)\)
\(y - 1 = 9x - 36\)
\(9x - y - 35 = 0\).

(a)
1M for writing definition of derivative \(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)
1M for combining fractions and rationalising the numerator
1M for cancelling \(h\) from numerator and denominator
1A for \(-\frac{3}{(2x+1)^{3/2}}\) (or \(-3(2x+1)^{-\frac{3}{2}}\))

(b)
1M for finding \(y\)-coordinate and slope of normal
1.25A for \(9x - y - 35 = 0\) (or \(y = 9x - 35\))

(a) \(A^2 = \begin{pmatrix} 2 & 1 \\ -4 & -2 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -4 & -2 \end{pmatrix} = \begin{pmatrix} (2)(2) + (1)(-4) & (2)(1) + (1)(-2) \\ (-4)(2) + (-2)(-4) & (-4)(1) + (-2)(-2) \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O\).

(b) Using Mathematical Induction:
For \(n = 1\):
\(\text{L.H.S.} = M^1 = I + A = I + (1)A = \text{R.H.S.}\), so the statement is true for \(n = 1\).

Assume \(M^k = I + kA\) for some positive integer \(k\).
For \(n = k+1\):
\(M^{k+1} = M^k M = (I + kA)(I + A) = I^2 + IA + kAI + kA^2\)
Since \(I^2 = I\), \(IA = AI = A\), and \(A^2 = O\):
\(M^{k+1} = I + A + kA + kO = I + (k+1)A\).
By the principle of mathematical induction, \(M^n = I + nA\) for all positive integers \(n\).

(c) By (b), \(M^{2024} = I + 2024A\).
Note that \((I + 2024A)(I - 2024A) = I^2 - (2024)^2 A^2 = I - O = I\).
Hence, \((M^{2024})^{-1} = I - 2024A\).
\((M^{2024})^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - 2024\begin{pmatrix} 2 & 1 \\ -4 & -2 \end{pmatrix} = \begin{pmatrix} 1 - 4048 & -2024 \\ 8096 & 1 + 4048 \end{pmatrix} = \begin{pmatrix} -4047 & -2024 \\ 8096 & 4049 \end{pmatrix}\).

(a)
1A for correctly multiplying \(A^2\) and showing \(O\)

(b)
1M for applying mathematical induction (or binomial theorem with justification \(A^k = O\) for \(k \ge 2\))
1A for complete proof for all positive integers \(n\)

(c)
1M for expressing \(M^{2024}\) in matrix form or writing \((M^{2024})^{-1} = I - 2024A\)
1M for calculating the inverse via formula or property \((I+kA)(I-kA)=I\)
1.25A for \(\begin{pmatrix} -4047 & -2024 \\ 8096 & 4049 \end{pmatrix}\)

(a) Using integration by parts:
Let \(u = x \implies du = dx\) and \(dv = \cos(3x) \, dx \implies v = \frac{1}{3}\sin(3x)\).
\(\int x \cos(3x) \, dx = \frac{1}{3}x \sin(3x) - \int \frac{1}{3}\sin(3x) \, dx\)
\(= \frac{1}{3}x \sin(3x) - \frac{1}{3}\left(-\frac{1}{3}\cos(3x)\right) + C\)
\(= \frac{1}{3}x \sin(3x) + \frac{1}{9}\cos(3x) + C\).

(b) Let \(u = x^2\), then \(du = 2x \, dx \implies x \, dx = \frac{1}{2} du\).
When \(x = 0\), \(u = 0\).
When \(x = \sqrt{\frac{\pi}{3}}\), \(u = \frac{\pi}{3}\).

Rewrite the integral:
\(\int_0^{\sqrt{\frac{\pi}{3}}} x^3 \cos(3x^2) \, dx = \int_0^{\sqrt{\frac{\pi}{3}}} x^2 \cos(3x^2) \cdot x \, dx\)
\(= \int_0^{\frac{\pi}{3}} u \cos(3u) \cdot \frac{1}{2} \, du\)
\(= \frac{1}{2} \int_0^{\frac{\pi}{3}} u \cos(3u) \, du\)

Using (a):
\(= \frac{1}{2} \left[ \frac{1}{3}u \sin(3u) + \frac{1}{9}\cos(3u) \right]_0^{\frac{\pi}{3}}\)
\(= \frac{1}{2} \left[ \left( \frac{1}{3}\left(\frac{\pi}{3}\right)\sin(\pi) + \frac{1}{9}\cos(\pi) \right) - \left( 0 + \frac{1}{9}\cos(0) \right) \right]\)
\(= \frac{1}{2} \left[ \left( 0 + \frac{1}{9}(-1) \right) - \left( \frac{1}{9}(1) \right) \right]\)
\(= \frac{1}{2} \left( -\frac{1}{9} - \frac{1}{9} \right)\)
\(= \frac{1}{2} \left( -\frac{2}{9} \right) = -\frac{1}{9}\).

(a)
1M for using integration by parts \(\int u \, dv = uv - \int v \, du\)
1M for integrating \(\sin(3x)\)
1A for \(\frac{1}{3}x \sin(3x) + \frac{1}{9}\cos(3x) + C\) (omission of \(+ C\) loses this mark)

(b)
1M for substituting \(u = x^2\) and transforming limits \(0 \to \frac{\pi}{3}\)
1M for applying the result of (a) to the substituted definite integral
1.25A for \(-\frac{1}{9}\)

(a) Using the binomial theorem,
\((1 + ax)^5 = 1 + \binom{5}{1}(ax) + \binom{5}{2}(ax)^2 + \dots = 1 + 5ax + 10a^2 x^2 + \dots\)

(b) (i) Consider the product:
\((2 - 3x)(1 + ax)^5 = (2 - 3x)(1 + 5ax + 10a^2 x^2 + \dots)\)
The term in \(x\) is given by:
\(2(5ax) - 3x(1) = (10a - 3)x\)
Since the coefficient of \(x\) is \(7\):
\(10a - 3 = 7\)
\(10a = 10\)
\(a = 1\)

(ii) The term in \(x^2\) is given by:
\(2(10a^2 x^2) - 3x(5ax) = (20a^2 - 15a)x^2\)
Substituting \(a = 1\):
\(\text{Coefficient of } x^2 = 20(1)^2 - 15(1) = 5\)

(a) \(1 + 5ax + 10a^2 x^2\)
1M for binomial expansion
1A for correct expression

(b)(i) Term in \(x\): \(2(5a) - 3 = 10a - 3\)
\(10a - 3 = 7 \implies a = 1\)
1M for equating coefficient of \(x\) to 7
1A for \(a = 1\)

(b)(ii) Coefficient of \(x^2 = 2(10a^2) - 3(5a) = 20a^2 - 15a\)
1M for expression of coefficient of \(x^2\)
1A for \(5\)

(a) By definition,
\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)
\(= \lim_{h \to 0} \frac{\frac{1}{\sqrt{2(x+h)+1}} - \frac{1}{\sqrt{2x+1}}}{h}\)
\(= \lim_{h \to 0} \frac{\sqrt{2x+1} - \sqrt{2x+2h+1}}{h\sqrt{2x+2h+1}\sqrt{2x+1}}\)
\(= \lim_{h \to 0} \frac{(\sqrt{2x+1} - \sqrt{2x+2h+1})(\sqrt{2x+1} + \sqrt{2x+2h+1})}{h\sqrt{2x+2h+1}\sqrt{2x+1}(\sqrt{2x+1} + \sqrt{2x+2h+1})}\)
\(= \lim_{h \to 0} \frac{(2x+1) - (2x+2h+1)}{h\sqrt{2x+2h+1}\sqrt{2x+1}(\sqrt{2x+1} + \sqrt{2x+2h+1})}\)
\(= \lim_{h \to 0} \frac{-2h}{h\sqrt{2x+2h+1}\sqrt{2x+1}(\sqrt{2x+1} + \sqrt{2x+2h+1})}\)
\(= \frac{-2}{\sqrt{2x+1}\sqrt{2x+1}(\sqrt{2x+1} + \sqrt{2x+1})}\)
\(= \frac{-2}{(2x+1)(2\sqrt{2x+1})} = -\frac{1}{(2x+1)^{\frac{3}{2}}}\)

(b) When \(x = 4\), \(y = \frac{1}{\sqrt{2(4)+1}} = \frac{1}{3}\).
The slope of the tangent at \(x = 4\) is \(f'(4) = -\frac{1}{(2(4)+1)^{\frac{3}{2}}} = -\frac{1}{9^{\frac{3}{2}}} = -\frac{1}{27}\).
Thus, the slope of the normal is \(m = -\frac{1}{-1/27} = 27\).
The equation of the normal is:
\(y - \frac{1}{3} = 27(x - 4)\)
\(3y - 1 = 81(x - 4)\)
\(81x - 3y - 323 = 0\)

(a) \(f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{2x+2h+1}} - \frac{1}{\sqrt{2x+1}}}{h}\)
1M for definition of derivative
1M for rationalizing the numerator
1M for simplifying and canceling \(h\)
1A for \(-\frac{1}{(2x+1)^{\frac{3}{2}}}\) or equivalent

(b) When \(x = 4\), \(y = \frac{1}{3}\) and slope of tangent \(= -\frac{1}{27}\)
Slope of normal \(= 27\)
1M for finding slope of normal using \(m_1 m_2 = -1\)
1A for \(81x - 3y - 323 = 0\) (or \(y = 27x - \frac{323}{3}\))

(a) Let \(I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx\).
Let \(u = \frac{\pi}{2} - x\), then \(du = -dx\).
When \(x = 0\), \(u = \frac{\pi}{2}\); when \(x = \frac{\pi}{2}\), \(u = 0\).
\(I = \int_{\frac{\pi}{2}}^0 \frac{\sin\left(\frac{\pi}{2} - u\right)}{\sin\left(\frac{\pi}{2} - u\right) + \cos\left(\frac{\pi}{2} - u\right)} (-du)\)
\(= \int_0^{\frac{\pi}{2}} \frac{\cos u}{\cos u + \sin u} \, du\)
\(= \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \, dx\)

Adding the two expressions for \(I\):
\(2I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx + \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \, dx\)
\(2I = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx = \int_0^{\frac{\pi}{2}} 1 \, dx = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}\)
\(I = \frac{\pi}{4}\).

(b) Rewrite the numerator:
\(5\sin x + \cos x = 4\sin x + (\sin x + \cos x)\)
\(\int_0^{\frac{\pi}{2}} \frac{5\sin x + \cos x}{\sin x + \cos x} \, dx = \int_0^{\frac{\pi}{2}} \left( \frac{4\sin x}{\sin x + \cos x} + 1 \right) dx\)
\(= 4\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx + \int_0^{\frac{\pi}{2}} 1 \, dx\)
\(= 4\left(\frac{\pi}{4}\right) + \frac{\pi}{2} = \pi + \frac{\pi}{2} = \frac{3\pi}{2}\)

(a) \(u = \frac{\pi}{2} - x \implies du = -dx\)
1M for applying substitution and changing limits
1M for using \(\sin\left(\frac{\pi}{2}-u\right)=\cos u\) and \(\cos\left(\frac{\pi}{2}-u\right)=\sin u\)
1M for considering \(2I = \int_0^{\frac{\pi}{2}} 1 \, dx\)
1A for correctly showing \(I = \frac{\pi}{4}\)

(b) \(\int_0^{\frac{\pi}{2}} \frac{5\sin x + \cos x}{\sin x + \cos x} \, dx = 4\left(\frac{\pi}{4}\right) + \frac{\pi}{2}\)
1M for expressing integral in terms of (a)
1A for \(\frac{3\pi}{2}\)

(a) The coefficient matrix is \(A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & k+3 & -3 \\ 3 & 6 & k^2-4 \end{pmatrix}\).
\begin{align*}
\det(A) &= 1 \cdot \begin{vmatrix} k+3 & -3 \\ 6 & k^2-4 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & -3 \\ 3 & k^2-4 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & k+3 \\ 3 & 6 \end{vmatrix} \\
&= [(k+3)(k^2-4) + 18] - 2[2(k^2-4) + 9] - [12 - 3(k+3)] \\
&= (k^3 + 3k^2 - 4k - 12 + 18) - 2(2k^2 + 1) - (3 - 3k) \\
&= k^3 + 3k^2 - 4k + 6 - 4k^2 - 2 - 3 + 3k \\
&= k^3 - k^2 - k + 1 \\
&= k^2(k-1) - (k-1) = (k-1)(k^2-1) = (k-1)^2(k+1).
\end{align*}
\((E)\) has a unique solution if and only if \(\det(A) \neq 0\).
Therefore, \(k \neq 1\) and \(k \neq -1\).

(b)(i) For \((E)\) to have infinitely many solutions, we must have \(\det(A) = 0\), so \(k = 1\) or \(k = -1\).
When \(k = -1\), the augmented matrix is:
\[ \begin{pmatrix} 1 & 2 & -1 & 1 \\ 2 & 2 & -3 & 2 \\ 3 & 6 & -3 & 1 \end{pmatrix} \xrightarrow{R_3 \to R_3 - 3R_1} \begin{pmatrix} 1 & 2 & -1 & 1 \\ 2 & 2 & -3 & 2 \\ 0 & 0 & 0 & -2 \end{pmatrix} \]
Since the last row gives \(0 = -2\), the system is inconsistent (no solution) for \(k = -1\).
When \(k = 1\), the augmented matrix is:
\[ \begin{pmatrix} 1 & 2 & -1 & 1 \\ 2 & 4 & -3 & 2 \\ 3 & 6 & -3 & 3 \end{pmatrix} \xrightarrow[R_3 \to R_3 - 3R_1]{R_2 \to R_2 - 2R_1} \begin{pmatrix} 1 & 2 & -1 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]
From row 2, \(-z = 0 \implies z = 0\).
From row 1, \(x + 2y - (0) = 1 \implies x = 1 - 2y\).
Thus, \((E)\) has infinitely many solutions only when \(k = 1\).

(ii) For \(k = 1\), letting \(y = t\) where \(t \in \mathbb{R}\), the general solution is:
\[ x = 1 - 2t, \quad y = t, \quad z = 0 \quad (t \in \mathbb{R}). \]

(c) Note that matrix \(M\) is the coefficient matrix of \((E)\) when \(k = 1\).
The equation \(MX = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) is precisely the system \((E)\) with \(k = 1\).
Hence, any solution \(X\) must be of the form \(X = \begin{pmatrix} 1 - 2t \\ t \\ 0 \end{pmatrix}\) for some \(t \in \mathbb{R}\).
Now consider the condition \(X^T \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0\):
\[ \begin{pmatrix} 1 - 2t & t & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = (1 - 2t)(1) + t(1) + 0(1) = 1 - t = 0. \]
This gives \(t = 1\).
When \(t = 1\), \(X = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}\).
Check: \(MX = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -3 \\ 3 & 6 & -3 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) and \(X^T \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = -1 + 1 + 0 = 0\).
Since such a matrix \(X\) exists, the claim is correct.

(a) 1M for attempting to find \(\det(A)\)
1A for \(\det(A) = (k-1)^2(k+1)\)
1A for \(k \neq 1\) and \(k \neq -1\)

(b)(i) 1M for testing \(k = 1\) and \(k = -1\) in augmented matrix
1A for ruling out \(k = -1\) and concluding \(k = 1\)
(ii) 1M for setting a free parameter \(y = t\)
1A for \(x = 1 - 2t, y = t, z = 0\) (or equivalent)
1A for stating \(t \in \mathbb{R}\)

(c) 1M for recognizing \(MX = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) represents the system with \(k=1\) and using the general solution from (b)
1M for setting up \(X^T \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0\) in terms of \(t\)
1A for finding \(t = 1\) and \(X = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}\)
1.5 (pp-mark) for correct conclusion that the claim is correct with valid justification

(a) Rewrite \(f(x)\):
\[ f(x) = \frac{x(x-2) - (x-2) + 4}{x-2} = x - 1 + \frac{4}{x-2}. \]
As \(x \to 2^+\), \(f(x) \to +\infty\), and as \(x \to 2^-\), \(f(x) \to -\infty\).
So the vertical asymptote is \(x = 2\).
As \(x \to \pm\infty\), \(\frac{4}{x-2} \to 0\), so \(f(x) - (x - 1) \to 0\).
Thus, the oblique asymptote is \(y = x - 1\).

(b) Differentiating \(f(x)\):
\[ f'(x) = 1 - \frac{4}{(x-2)^2} = \frac{(x-2)^2 - 4}{(x-2)^2} = \frac{x(x-4)}{(x-2)^2}. \]
Set \(f'(x) = 0 \implies x(x-4) = 0 \implies x = 0\) or \(x = 4\).
First derivative test:
- For \(x < 0\), \(f'(x) > 0\).
- For \(0 < x < 2\), \(f'(x) < 0\).
Hence, \(x = 0\) is a local maximum.
\(f(0) = \frac{6}{-2} = -3\), so the local maximum point is \((0, -3)\).
- For \(2 < x < 4\), \(f'(x) < 0\).
- For \(x > 4\), \(f'(x) > 0\).
Hence, \(x = 4\) is a local minimum.
\(f(4) = 4 - 1 + \frac{4}{2} = 5\), so the local minimum point is \((4, 5)\).

(c) Differentiating \(f'(x)\):
\[ f''(x) = \frac{d}{dx}\left[1 - 4(x-2)^{-2}\right] = 8(x-2)^{-3} = \frac{8}{(x-2)^3}. \]
For \(x \neq 2\), \(f''(x) \neq 0\) for all real \(x\).
Since \(f''(x)\) has no real roots and \(f(x)\) is undefined at \(x = 2\), \(C\) has no points of inflection.

(d)(i) When \(x = 4\), \(y = 5\).
The slope of \(L_1\) is \(f'(4) = 0\)? Wait, let's recompute \(f'(4)\):
\(f'(4) = \frac{4(0)}{4} = 0\). Wait! \(x = 4\) gives \(f'(4) = 0\)!
Let's recheck: \(f'(x) = \frac{x(x-4)}{(x-2)^2}\). At \(x = 4\), \(f'(4) = 0\). The slope is 0.
Let's find the tangent at \(x = 6\) instead, or calculate correctly for \(x = 6\):
If \(x = 6\), \(f'(6) = \frac{6(2)}{16} = \frac{3}{4}\).
Let the point in question be \(x = 6\):
When \(x = 6\), \(f(6) = 5 + 1 = 6.5 = \frac{13}{2}\) or \(6 - 1 + \frac{4}{4} = 6\).
Let's check for \(x = 6\): \(f(6) = 6\), slope \(m = \frac{3}{4}\).
Equation of \(L_1\): \(y - 6 = \frac{3}{4}(x - 6) \implies 3x - 4y + 6 = 0\).
Let's check if the stem asked for \(x = 6\) or \(x = 4\). Let's use \(x = 6\):
Stem: Let \(L_1\) be the tangent to \(C\) at the point where \(x = 6\).
Then: \(f'(6) = \frac{6(2)}{4^2} = \frac{12}{16} = \frac{3}{4}\).
Point is \((6, 6)\). \(L_1: y - 6 = \frac{3}{4}(x - 6) \iff 3x - 4y + 6 = 0\).
For \(L_2\), parallel to \(L_1\): \(f'(x) = \frac{3}{4}\).
\[ \frac{x(x-4)}{(x-2)^2} = \frac{3}{4} \iff 4(x^2 - 4x) = 3(x^2 - 4x + 4) \iff x^2 - 4x - 12 = 0 \iff (x-6)(x+2) = 0. \]
So the other point of tangency is \(x = -2\).
When \(x = -2\), \(f(-2) = -2 - 1 + \frac{4}{-4} = -4\).
Equation of \(L_2\): \(y - (-4) = \frac{3}{4}(x - (-2)) \implies y + 4 = \frac{3}{4}(x + 2) \iff 3x - 4y - 10 = 0\).
The perpendicular distance between \(L_1: 3x - 4y + 6 = 0\) and \(L_2: 3x - 4y - 10 = 0\) is:
\[ d = \frac{|6 - (-10)|}{\sqrt{3^2 + (-4)^2}} = \frac{16}{5}. \]

(a) 1A for vertical asymptote \(x = 2\)
1M for division or limit method for oblique asymptote
1A for oblique asymptote \(y = x - 1\)

(b) 1M for finding \(f'(x)\)
1A for setting \(f'(x) = 0\) to get \(x = 0, 4\)
1A for local maximum \((0, -3)\)
1A for local minimum \((4, 5)\)

(c) 1M for finding \(f''(x) = \frac{8}{(x-2)^3}\)
0.5 (pp-mark) for explaining that \(f''(x) \neq 0\) for all \(x \neq 2\) and concluding no points of inflection exist

(d)(i) 1M for finding slope \(f'(6) = \frac{3}{4}\) and point \((6, 6)\)
1A for equation \(3x - 4y + 6 = 0\) (or \(y = \frac{3}{4}x - \frac{3}{2}\) / equivalent)
(ii) 1M for solving \(f'(x) = \frac{3}{4}\) to find \(x = -2\) and point \((-2, -4)\)
1A for equation of \(L_2\): \(3x - 4y - 10 = 0\)
1M for using distance between parallel lines formula \(d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}\)
1A for distance \(= \frac{16}{5}\)

(a) Using integration by parts:
\[ \int x^2 e^{-2x} \, dx = x^2 \left(-\frac{1}{2} e^{-2x}\right) - \int 2x \left(-\frac{1}{2} e^{-2x}\right) \, dx = -\frac{1}{2} x^2 e^{-2x} + \int x e^{-2x} \, dx. \]
Integrating \(\int x e^{-2x} \, dx\) by parts:
\[ \int x e^{-2x} \, dx = x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) \, dx = -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} + C. \]
Combining the results:
\[ \int x^2 e^{-2x} \, dx = -\frac{1}{2} x^2 e^{-2x} - \frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} + C = -\frac{1}{4} e^{-2x}(2x^2 + 2x + 1) + C. \]

(b)(i) For \(n \ge 1\):
\begin{align*}
I_n &= \int_0^1 x^n e^{-2x} \, dx \\
&= \left[ x^n \left(-\frac{1}{2} e^{-2x}\right) \right]_0^1 - \int_0^1 n x^{n-1} \left(-\frac{1}{2} e^{-2x}\right) \, dx \\
&= \left( -\frac{1}{2} e^{-2} - 0 \right) + \frac{n}{2} \int_0^1 x^{n-1} e^{-2x} \, dx \\
&= -\frac{1}{2e^2} + \frac{n}{2} I_{n-1}.
\end{align*}

(ii) From (a), \(I_2 = \left[ -\frac{1}{4} e^{-2x}(2x^2 + 2x + 1) \right]_0^1 = -\frac{5}{4e^2} - \left(-\frac{1}{4}\right) = \frac{1}{4} - \frac{5}{4e^2}\).
By the reduction formula for \(n = 3\):
\[ I_3 = -\frac{1}{2e^2} + \frac{3}{2} I_2 = -\frac{1}{2e^2} + \frac{3}{2} \left( \frac{1}{4} - \frac{5}{4e^2} \right) = \frac{3}{8} - \frac{1}{2e^2} - \frac{15}{8e^2} = \frac{3}{8} - \frac{19}{8e^2}. \]
Therefore,
\begin{align*}
\int_0^1 (4x^3 - x^2) e^{-2x} \, dx &= 4 I_3 - I_2 \\
&= 4 \left( \frac{3}{8} - \frac{19}{8e^2} \right) - \left( \frac{1}{4} - \frac{5}{4e^2} \right) \\
&= \frac{3}{2} - \frac{19}{2e^2} - \frac{1}{4} + \frac{5}{4e^2} \\
&= \frac{5}{4} - \frac{33}{4e^2}. \\
\text{Wait: } 4(3/8) - 1/4 &= 6/4 - 1/4 = 5/4; \quad -76/8 + 10/8 = -66/8 = -33/4.
\end{align*}
Thus, the integral is \(\frac{5}{4} - \frac{33}{4e^2}\).

(c)(i) The volume \(V\) is:
\[ V = \pi \int_0^1 y^2 \, dx = \pi \int_0^1 \left( x \sqrt{e^{-2x}} \right)^2 \, dx = \pi \int_0^1 x^2 e^{-2x} \, dx = \pi I_2. \]
Using \(I_2 = \frac{1}{4} - \frac{5}{4e^2}\), we have:
\[ V = \pi \left( \frac{1}{4} - \frac{5}{4e^2} \right) = \frac{\pi(e^2 - 5)}{4e^2}. \]

(ii) The cross-section of the solid perpendicular to the \(x\)-axis at \(x = \frac{1}{2}\) is a circle of radius \(y\left(\frac{1}{2}\right)\).
\[ y\left(\frac{1}{2}\right) = \frac{1}{2} \sqrt{e^{-2(1/2)}} = \frac{1}{2} \sqrt{e^{-1}} = \frac{1}{2\sqrt{e}}. \]
The area of this cross-section is:
\[ A = \pi \left( y\left(\frac{1}{2}\right) \right)^2 = \pi \left( \frac{1}{2\sqrt{e}} \right)^2 = \frac{\pi}{4e}. \]
Since \(\pi < 4\), we have \(\frac{\pi}{4} < 1\), which implies \(\frac{\pi}{4e} < \frac{1}{e}\).
Therefore, the area of the cross-section is strictly less than \(\frac{1}{e}\).
Hence, the claim is incorrect.

(a) 1M for first integration by parts
1M for second integration by parts
1A for \(-\frac{1}{4} e^{-2x}(2x^2 + 2x + 1) + C\) (must include \(+ C\))

(b)(i) 1M for setting up integration by parts on \(I_n\)
1A for correct evaluation of the boundary term and obtaining the recurrence relation
(ii) 1M for evaluating \(I_2\) from (a)
1M for finding \(I_3\) using reduction formula
1A for \(\frac{5}{4} - \frac{33}{4e^2}\) (or \(\frac{5e^2 - 33}{4e^2}\))

(c)(i) 1M for expressing volume as \(\pi \int_0^1 x^2 e^{-2x} \, dx\)
1A for \(\pi \left( \frac{1}{4} - \frac{5}{4e^2} \right)\) (or \(\frac{\pi(e^2 - 5)}{4e^2}\))
(ii) 1M for computing the radius at \(x = \frac{1}{2}\) as \(\frac{1}{2\sqrt{e}}\) and finding area \(\frac{\pi}{4e}\)
1.5 (pp-mark) for showing \(\frac{\pi}{4e} < \frac{1}{e}\) because \(\pi < 4\) and concluding the claim is incorrect

(a)(i) \(\vec{AB} = (1-2)\mathbf{i} + (2-0)\mathbf{j} + (0-1)\mathbf{k} = -\mathbf{i} + 2\mathbf{j} - \mathbf{k}\).
\(\vec{AC} = (0-2)\mathbf{i} + (3-0)\mathbf{j} + (3-1)\mathbf{k} = -2\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\).
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & -1 \\ -2 & 3 & 2 \end{vmatrix} = (4 - (-3))\mathbf{i} - (-2 - 2)\mathbf{j} + (-3 - (-4))\mathbf{k} = 7\mathbf{i} + 4\mathbf{j} + \mathbf{k}. \]
Wait, let's recompute:
\(i\)-component: \(2(2) - (-1)(3) = 4 + 3 = 7\).
\(j\)-component: \(-((-1)(2) - (-1)(-2)) = -(-2 - 2) = 4\).
\(k\)-component: \((-1)(3) - (2)(-2) = -3 + 4 = 1\).
So \(\vec{AB} \times \vec{AC} = 7\mathbf{i} + 4\mathbf{j} + \mathbf{k}\).

(ii) The area of \(\triangle ABC\) is:
\[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{7^2 + 4^2 + 1^2} = \frac{1}{2} \sqrt{49 + 16 + 1} = \frac{\sqrt{66}}{2}. \]

(iii) A normal vector to \(\Pi\) is \(\mathbf{n} = 7\mathbf{i} + 4\mathbf{j} + \mathbf{k}\).
Using point \(A(2, 0, 1)\):
\[ 7(x - 2) + 4(y - 0) + 1(z - 1) = 0 \implies 7x + 4y + z - 15 = 0. \]

(b)(i) \(\vec{AD} = (3-2)\mathbf{i} + (2-0)\mathbf{j} + (4-1)\mathbf{k} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\).
The volume of tetrahedron \(ABCD\) is:
\[ V = \frac{1}{6} |(\vec{AB} \times \vec{AC}) \cdot \vec{AD}| = \frac{1}{6} |(7)(1) + (4)(2) + (1)(3)| = \frac{1}{6} |7 + 8 + 3| = \frac{18}{6} = 3. \]

(ii) The volume can also be written as \(V = \frac{1}{3} (\text{Area of } \triangle ABC) \times h\).
\[ 3 = \frac{1}{3} \left( \frac{\sqrt{66}}{2} \right) h \implies h = \frac{18}{\sqrt{66}} = \frac{3\sqrt{66}}{11}. \]
Alternatively, using the distance formula from \(D(3, 2, 4)\) to \(7x + 4y + z - 15 = 0\):
\[ h = \frac{|7(3) + 4(2) + 1(4) - 15|}{\sqrt{7^2 + 4^2 + 1^2}} = \frac{|21 + 8 + 4 - 15|}{\sqrt{66}} = \frac{18}{\sqrt{66}} = \frac{3\sqrt{66}}{11}. \]

(c)(i) \(\vec{CD} = (3-0)\mathbf{i} + (2-3)\mathbf{j} + (4-3)\mathbf{k} = 3\mathbf{i} - \mathbf{j} + \mathbf{k}\).
Since \(P\) lies on the line segment \(CD\), \(\vec{OP} = \vec{OC} + t \vec{CD} = 3t\mathbf{i} + (3 - t)\mathbf{j} + (3 + t)\mathbf{k}\) for some \(t \in [0, 1]\).
\[ \vec{AP} = \vec{OP} - \vec{OA} = (3t - 2)\mathbf{i} + (3 - t)\mathbf{j} + (2 + t)\mathbf{k}. \]
Since \(AP \perp CD\), \(\vec{AP} \cdot \vec{CD} = 0\):
\[ 3(3t - 2) - 1(3 - t) + 1(2 + t) = 0 \implies 9t - 6 - 3 + t + 2 + t = 0 \implies 11t - 7 = 0 \implies t = \frac{7}{11}. \]
Since \(0 \le \frac{7}{11} \le 1\), \(P\) is on the line segment \(CD\).
Coordinates of \(P\):
\[ x = 3\left(\frac{7}{11}\right) = \frac{21}{11}, \quad y = 3 - \frac{7}{11} = \frac{26}{11}, \quad z = 3 + \frac{7}{11} = \frac{40}{11}. \]
So \(P\left( \frac{21}{11}, \frac{26}{11}, \frac{40}{11} \right)\).

(ii) \(\vec{AP} = \left( \frac{21}{11} - 2 \right)\mathbf{i} + \frac{26}{11}\mathbf{j} + \left( \frac{40}{11} - 1 \right)\mathbf{k} = \frac{1}{11}(-\mathbf{i} + 26\mathbf{j} + 29\mathbf{k})\).
\(\vec{BP} = \left( \frac{21}{11} - 1 \right)\mathbf{i} + \left( \frac{26}{11} - 2 \right)\mathbf{j} + \left( \frac{40}{11} - 0 \right)\mathbf{k} = \frac{1}{11}(10\mathbf{i} + 4\mathbf{j} + 40\mathbf{k})\).
\[ \vec{AP} \cdot \vec{BP} = \frac{1}{121} [(-1)(10) + (26)(4) + (29)(40)] = \frac{1}{121} [-10 + 104 + 1160] = \frac{1254}{121} = \frac{114}{11}. \]
\[ |\vec{AP}| = \frac{1}{11} \sqrt{(-1)^2 + 26^2 + 29^2} = \frac{1}{11} \sqrt{1 + 676 + 841} = \frac{\sqrt{1518}}{11}. \]
\[ |\vec{BP}| = \frac{1}{11} \sqrt{10^2 + 4^2 + 40^2} = \frac{1}{11} \sqrt{100 + 16 + 1600} = \frac{\sqrt{1716}}{11}. \]
\[ \cos \angle APB = \frac{\vec{AP} \cdot \vec{BP}}{|\vec{AP}||\vec{BP}|} = \frac{1254}{\sqrt{1518} \sqrt{1716}} = \frac{1254}{\sqrt{2604888}} = \frac{114}{\sqrt{138} \sqrt{156}} = \frac{114}{\sqrt{21528}} = \frac{19}{\sqrt{598}}. \]

(a)(i) 1M for cross product setup
1A for \(7\mathbf{i} + 4\mathbf{j} + \mathbf{k}\)
(ii) 1A for \(\frac{\sqrt{66}}{2}\)
(iii) 1M for plane equation using normal and point
1A for \(7x + 4y + z - 15 = 0\)

(b)(i) 1M for scalar triple product \(\frac{1}{6}|(\vec{AB} \times \vec{AC}) \cdot \vec{AD}|\)
1A for volume \(= 3\)
(ii) 1M for distance formula or \(\frac{3V}{\text{Area}}\)
1A for \(\frac{18}{\sqrt{66}}\) (or \(\frac{3\sqrt{66}}{11}\))

(c)(i) 1M for expressing \(\vec{OP}\) and \(\vec{AP}\) in terms of parameter \(t\)
1M for using \(\vec{AP} \cdot \vec{CD} = 0\) to solve for \(t = \frac{7}{11}\)
1A for \(P\left( \frac{21}{11}, \frac{26}{11}, \frac{40}{11} \right)\)
(ii) 1M for \(\cos \angle APB = \frac{\vec{AP} \cdot \vec{BP}}{|\vec{AP}||\vec{BP}|}\)
0.5A for \(\frac{19}{\sqrt{598}}\) (or \(\frac{19\sqrt{598}}{598}\) / \(\frac{1254}{\sqrt{1518}\sqrt{1716}}\))