2021 HKDSE Mathematics M2 (Algebra and Calculus) Answers & Marking Scheme

Let \(P(n)\) be the proposition \(\sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)} = \frac{n}{2n+1}\). For \(n=1\), \(\text{LHS} = \frac{1}{1 \times 3} = \frac{1}{3}\) and \(\text{RHS} = \frac{1}{2(1)+1} = \frac{1}{3}\). Since \(\text{LHS} = \text{RHS}\), \(P(1)\) is true. Assume that \(P(k)\) is true for some positive integer \(k\), i.e., \(\sum_{r=1}^{k} \frac{1}{(2r-1)(2r+1)} = \frac{k}{2k+1}\). For \(n=k+1\), \(\text{LHS} = \sum_{r=1}^{k} \frac{1}{(2r-1)(2r+1)} + \frac{1}{(2(k+1)-1)(2(k+1)+1)} = \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)} = \frac{k(2k+3) + 1}{(2k+1)(2k+3)} = \frac{2k^2+3k+1}{(2k+1)(2k+3)} = \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} = \frac{k+1}{2k+3} = \frac{k+1}{2(k+1)+1} = \text{RHS}\). Thus, \(P(k+1)\) is true. By the principle of mathematical induction, \(P(n)\) is true for all positive integers \(n\).

For testing \(n=1\) (1 mark); For assuming \(P(k)\) is true (0.5 mark); For using induction hypothesis to write down LHS for \(n=k+1\) (1.5 marks); For algebraic simplification to show \(P(k+1)\) is true (2.5 marks); For conclusion (0.75 mark).

(a) The coefficient of \(x\) is \(na = 12\) (1). The coefficient of \(x^2\) is \(\frac{n(n-1)}{2} a^2 = 60 \implies n(n-1)a^2 = 120\) (2). Dividing (2) by the square of (1) gives \(\frac{n(n-1)a^2}{n^2 a^2} = \frac{120}{144} \implies \frac{n-1}{n} = \frac{5}{6} \implies n = 6\). Substitute \(n = 6\) into (1) gives \(6a = 12 \implies a = 2\). (b) For \(a = 2\) and \(n = 6\), the expression is \((2-x)(1+2x)^6\). In the expansion of \((1+2x)^6\), the coefficient of \(x^2\) is \(\binom{6}{2}2^2 = 60\), and the coefficient of \(x^3\) is \(\binom{6}{3}2^3 = 160\). In the expansion of \((2-x)(1+2x)^6\), the coefficient of \(x^3\) is \(2 \times 160 - 1 \times 60 = 260\).

(a) For finding relations of coefficients (1 mark); for getting the ratio of equations (0.5 mark); for finding \(n = 6\) (1 mark); for finding \(a = 2\) (0.75 mark). (b) For finding coefficients of \(x^2\) and \(x^3\) in \((1+2x)^6\) (1 mark); for setting up the expression for the coefficient of \(x^3\) (1 mark); for the correct answer \(260\) (1 mark).

Using the sum-to-product formula, we have \(\cos 3\theta + \cos \theta = 2 \cos 2\theta \cos \theta\). The equation becomes \(2 \cos 2\theta \cos \theta = \cos 2\theta \implies \cos 2\theta (2 \cos \theta - 1) = 0\). Therefore, \(\cos 2\theta = 0\) or \(\cos \theta = \frac{1}{2}\). Since \(0 \le \theta \le \pi\): If \(\cos 2\theta = 0\), then \(2\theta = \frac{\pi}{2}\) or \(2\theta = \frac{3\pi}{2}\), which gives \(\theta = \frac{\pi}{4}\) or \(\theta = \frac{3\pi}{4}\). If \(\cos \theta = \frac{1}{2}\), then \(\theta = \frac{\pi}{3}\). Thus, the solutions are \(\theta = \frac{\pi}{4}, \frac{\pi}{3}, \frac{3\pi}{4}\).

For applying sum-to-product formula (1 mark); for simplifying to product form (1 mark); for factoring the equation (1.25 marks); for solving \(\cos 2\theta = 0\) (1 mark); for solving \(\cos \theta = 1/2\) (1 mark); for stating all correct solutions (1 mark).

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

For definition of derivative (1 mark); for substituting \(f(x)\) (1 mark); for rationalizing numerator (1.25 marks); for simplifying numerator to \(2h\) (1 mark); for canceling \(h\) (1 mark); for taking the limit to get the final answer (1 mark).

(a) The volume of a sphere is \(V = \frac{4}{3}\pi r^3\). Differentiating both sides with respect to \(t\) gives \(\frac{\mathrm{d}V}{\mathrm{d}t} = 4\pi r^2 \frac{\mathrm{d}r}{\mathrm{d}t}\). Given \(\frac{\mathrm{d}V}{\mathrm{d}t} = 12\pi\), when \(r = 3\), we have \(12\pi = 4\pi (3)^2 \frac{\mathrm{d}r}{\mathrm{d}t} \implies 12\pi = 36\pi \frac{\mathrm{d}r}{\mathrm{d}t} \implies \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{1}{3} \text{ cm/s}\). (b) The surface area of a sphere is \(A = 4\pi r^2\). Differentiating both sides with respect to \(t\) gives \(\frac{\mathrm{d}A}{\mathrm{d}t} = 8\pi r \frac{\mathrm{d}r}{\mathrm{d}t}\). When \(r = 3\) and \(\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{1}{3}\), we have \(\frac{\mathrm{d}A}{\mathrm{d}t} = 8\pi (3) \left(\frac{1}{3}\right) = 8\pi \text{ cm}^2\text{s}^{-1}\).

(a) For differentiating \(V\) with respect to \(t\) (1.25 marks); for substituting values (1 mark); for correct answer with unit (1 mark). (b) For differentiating \(A\) with respect to \(t\) (1.25 marks); for substituting values (0.75 mark); for correct answer with unit (1 mark).

Using integration by parts: Let \(u = x^2\) and \(\mathrm{d}v = e^{2x} \, \mathrm{d}x\), then \(\mathrm{d}u = 2x \, \mathrm{d}x\) and \(v = \frac{1}{2} e^{2x}\). \(\int_{0}^{1} x^2 e^{2x} \, \mathrm{d}x = \left[ \frac{1}{2} x^2 e^{2x} \right]_{0}^{1} - \int_{0}^{1} x e^{2x} \, \mathrm{d}x = \frac{1}{2} e^2 - \int_{0}^{1} x e^{2x} \, \mathrm{d}x\). Apply integration by parts again to \(\int_{0}^{1} x e^{2x} \, \mathrm{d}x\): Let \(u_1 = x\) and \(\mathrm{d}v_1 = e^{2x} \, \mathrm{d}x\), then \(\mathrm{d}u_1 = \mathrm{d}x\) and \(v_1 = \frac{1}{2} e^{2x}\). \(\int_{0}^{1} x e^{2x} \, \mathrm{d}x = \left[ \frac{1}{2} x e^{2x} \right]_{0}^{1} - \int_{0}^{1} \frac{1}{2} e^{2x} \, \mathrm{d}x = \frac{1}{2} e^2 - \left[ \frac{1}{4} e^{2x} \right]_{0}^{1} = \frac{1}{2} e^2 - \left( \frac{1}{4} e^2 - \frac{1}{4} \right) = \frac{1}{4} e^2 + \frac{1}{4}\). Thus, \(\int_{0}^{1} x^2 e^{2x} \, \mathrm{d}x = \frac{1}{2} e^2 - \left( \frac{1}{4} e^2 + \frac{1}{4} \right) = \frac{e^2 - 1}{4}\).

For first integration by parts setup (1 mark); for evaluating the first boundary term and remaining integral expression (1.25 marks); for second integration by parts setup (1 mark); for evaluating the second boundary term (1 mark); for evaluating the final integral (1 mark); for correct simplified answer (1 mark).

(a) The determinant of the coefficient matrix of \((E)\) is \(\det(M) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & a \end{vmatrix} = 1(2a - 9) - 1(a - 3) + 1(3 - 2) = a - 5\). For \((E)\) to have a unique solution, \(\det(M) \ne 0 \implies a \ne 5\). (b) When \(a = 5\), the augmented matrix is \(\left( \begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 5 & b \end{array} \right)\). Performing row operations: \(\sim \left( \begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 2 & 4 & b - 3 \end{array} \right) \sim \left( \begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & b - 5 \end{array} \right)\). For \((E)\) to be consistent, we must have \(b - 5 = 0 \implies b = 5\). In this case, the equations are \(y + 2z = 1\) and \(x + y + z = 3\). Let \(z = t\) where \(t \in \mathbb{R}\). Then \(y = 1 - 2t\) and \(x = 3 - (1-2t) - t = 2 + t\). Thus, the solution is \(x = 2 + t\), \(y = 1 - 2t\), \(z = t\) for any \(t \in \mathbb{R}\).

(a) For evaluating determinant (1 mark); for getting \(a \ne 5\) (1 mark). (b) For row operations on augmented matrix (2 marks); for finding \(b = 5\) (0.75 mark); for introducing parameter \(t\) and finding correct expressions for \(x, y, z\) (1.5 marks).

(a) Since \(\det(A) = 2(3) - 1(5) = 1 \ne 0\), we have \(A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\). (b) From \(A X A = B\), pre-multiplying by \(A^{-1}\) and post-multiplying by \(A^{-1}\) on both sides gives \(X = A^{-1} B A^{-1}\). First, \(A^{-1} B = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ -5 & -2 \end{pmatrix}\). Then, \(X = (A^{-1} B) A^{-1} = \begin{pmatrix} 3 & 1 \\ -5 & -2 \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3(3)+1(-5) & 3(-1)+1(2) \\ -5(3)+(-2)(-5) & -5(-1)+(-2)(2) \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ -5 & 1 \end{pmatrix}\).

(a) For finding determinant of \(A\) (1 mark); for correct \(A^{-1}\) (1 mark). (b) For showing \(X = A^{-1} B A^{-1}\) (2 marks); for calculating \(A^{-1} B\) (1 mark); for final correct matrix \(X\) (1.25 marks).

(a) Let \(A\) be the coefficient matrix of the system.
\(|A| = \begin{vmatrix} 1 & 2 & -1 \\ 2 & a+3 & 3 \\ 3 & 6 & a^2-4 \end{vmatrix} \).
Using row operations: \(R_2 \to R_2 - 2R_1\) and \(R_3 \to R_3 - 3R_1\),
\(|A| = \begin{vmatrix} 1 & 2 & -1 \\ 0 & a-1 & 5 \\ 0 & 0 & a^2-1 \end{vmatrix} = (a-1)(a^2-1) = (a-1)^2(a+1)\).
For the system to have a unique solution, \(|A| \neq 0 \implies a \neq 1\) and \(a \neq -1\).

(b)(i) When \(a = 1\), the augmented matrix is:
\(\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 2 & 4 & 3 & 4 \\ 3 & 6 & -3 & 3 \end{array}\right)\)
By performing \(R_2 \to R_2 - 2R_1\) and \(R_3 \to R_3 - 3R_1\):
\(\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 0 & 5 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right)\)
From Row 2, \(5z = 2 \implies z = \frac{2}{5}\).
Let \(y = t\) where \(t\) is any real number.
Then \(x + 2t - \frac{2}{5} = 1 \implies x = \frac{7}{5} - 2t\).
So the solution is \(x = \frac{7}{5} - 2t\), \(y = t\), \(z = \frac{2}{5}\) for \(t \in \mathbb{R}\).

(b)(ii) \(x^2 + y^2 = \left(\frac{7}{5} - 2t\right)^2 + t^2 = 5t^2 - \frac{28}{5}t + \frac{49}{25} = 5\left(t - \frac{14}{25}\right)^2 + \frac{49}{125}\).
Thus, the minimum value of \(x^2 + y^2\) is \(\frac{49}{125}\).

(c)(i) When \(a = -1\), the augmented matrix is:
\(\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 2 & 2 & 3 & 2 \\ 3 & 6 & -3 & -1 \end{array}\right)\)
Applying \(R_3 \to R_3 - 3R_1\):
\(\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & -2 & 5 & 0 \\ 0 & 0 & 0 & -4 \end{array}\right)\)
Since Row 3 implies \(0 = -4\), which is impossible, the system is inconsistent.

(c)(ii) If the third equation is replaced by \(3x + 6y - 3z = k\), the system becomes:
\(\begin{cases} x + 2y - z = 1 \\ 2x + 2y + 3z = 2 \\ 3x + 6y - 3z = k \end{cases}\)
Note that the LHS of the third equation is \(3(x + 2y - z)\).
Since \(x + 2y - z = 1\), we must have \(3(1) = k \implies k = 3\).
When \(k = 3\), the third equation becomes identical to three times the first, hence the system has infinitely many solutions and is consistent.

(a) 1M for setting up determinant equation, 1M for factoring determinant, 1A for range of \(a\).
(b)(i) 1M for row reduction, 1M for setting parameter \(t\), 1A for correct solutions.
(b)(ii) 1M for expressing \(x^2+y^2\) in terms of \(t\), 1A for completing square and finding minimum value.
(c)(i) 1M for substituting \(a=-1\) and row reduction, 1A for concluding contradiction \(0 = -4\).
(c)(ii) 1M for identifying relationship between first and modified third equation, 1A for finding \(k=3\).

(a) Let \(u = a+b-x\). Then \(dx = -du\).
When \(x = a\), \(u = b\). When \(x = b\), \(u = a\).
\(\int_a^b f(x) \, dx = \int_b^a f(a+b-u) (-du) = \int_a^b f(a+b-u) \, du = \int_a^b f(a+b-x) \, dx\).

(b)(i) Let \(a=0\) and \(b=\pi/2\) in (a). Then \(a+b-x = \frac{\pi}{2} - x\).
\(\int_0^{\pi/2} \frac{\sin^3 x}{\sin x + \cos x} \, dx = \int_0^{\pi/2} \frac{\sin^3(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x) + \cos(\frac{\pi}{2}-x)} \, dx\)
Since \(\sin(\frac{\pi}{2}-x) = \cos x\) and \(\cos(\frac{\pi}{2}-x) = \sin x\),
we have \(\int_0^{\pi/2} \frac{\sin^3 x}{\sin x + \cos x} \, dx = \int_0^{\pi/2} \frac{\cos^3 x}{\cos x + \sin x} \, dx = \int_0^{\pi/2} \frac{\cos^3 x}{\sin x + \cos x} \, dx\).

(b)(ii) Let \(I = \int_0^{\pi/2} \frac{\sin^3 x}{\sin x + \cos x} \, dx\).
\(2I = \int_0^{\pi/2} \frac{\sin^3 x}{\sin x + \cos x} \, dx + \int_0^{\pi/2} \frac{\cos^3 x}{\sin x + \cos x} \, dx\)
\(2I = \int_0^{\pi/2} \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} \, dx\)
Using \(A^3 + B^3 = (A+B)(A^2 - AB + B^2)\):
\(2I = \int_0^{\pi/2} \frac{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{\sin x + \cos x} \, dx = \int_0^{\pi/2} (1 - \sin x \cos x) \, dx\)
\(2I = \left[ x - \frac{1}{2}\sin^2 x \right]_0^{\pi/2} = \left(\frac{\pi}{2} - \frac{1}{2}\right) - 0 = \frac{\pi - 1}{2}\)
So \(I = \frac{\pi - 1}{4}\).

(c) Let \(t = \tan \frac{x}{2}\). Then \(dx = \frac{2}{1+t^2} \, dt\), \(\sin x = \frac{2t}{1+t^2}\), and \(\cos x = \frac{1-t^2}{1+t^2}\).
When \(x = 0\), \(t = 0\). When \(x = \pi/2\), \(t = 1\).
\(\int_0^{\pi/2} \frac{1}{\sin x + \cos x + 1} \, dx = \int_0^1 \frac{1}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} + 1} \left(\frac{2}{1+t^2}\right) \, dt\)
\(= \int_0^1 \frac{2}{2t + (1-t^2) + (1+t^2)} \, dt = \int_0^1 \frac{2}{2t + 2} \, dt = \int_0^1 \frac{1}{t+1} \, dt\)
\(= \left[ \ln|t+1| \right]_0^1 = \ln 2 - \ln 1 = \ln 2\).

(a) 1M for substitution \(u=a+b-x\) and changing limits, 1A for completing proof.
(b)(i) 1M for applying (a) with \(a=0\), \(b=\pi/2\), 1A for correct simplification using co-function identities.
(b)(ii) 1M for writing \(2I\) and combining, 1M for algebraic simplification, 1A for finding \(I = \frac{\pi-1}{4}\).
(c) 1M for writing correct identities for \(dx\), \(\sin x\), \(\cos x\) in terms of \(t\), 1M for changing limits of integration, 1M for substituting and simplifying integrand to \(\frac{1}{t+1}\), 1M for integrating to get \(\ln|t+1|\), 2A for final answer \(\ln 2\).

(a) \(\vec{AB} = (0-2)\mathbf{i} + (3-1)\mathbf{j} + (2-0)\mathbf{k} = -2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).
\(\vec{AC} = (1-2)\mathbf{i} + (0-1)\mathbf{j} + (4-0)\mathbf{k} = -\mathbf{i} - \mathbf{j} + 4\mathbf{k}\).
\(\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 2 & 2 \\ -1 & -1 & 4 \end{vmatrix}\)
\(= (8 - (-2))\mathbf{i} - (-8 - (-2))\mathbf{j} + (2 - (-2))\mathbf{k}\)
\(= 10\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}\).

Area of \(\triangle ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{10^2 + 6^2 + 4^2} = \frac{1}{2} \sqrt{152} = \sqrt{38}\).

(b)(i) \(\vec{AD} = (k-2)\mathbf{i} + (2-1)\mathbf{j} + (-1-0)\mathbf{k} = (k-2)\mathbf{i} + \mathbf{j} - \mathbf{k}\).
The volume of tetrahedron \(ABCD\) is:
\(V = \frac{1}{6} |(\vec{AB} \times \vec{AC}) \cdot \vec{AD}|\)
\(= \frac{1}{6} |(10\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}) \cdot ((k-2)\mathbf{i} + \mathbf{j} - \mathbf{k})|\)
\(= \frac{1}{6} |10(k-2) + 6(1) + 4(-1)|\)
\(= \frac{1}{6} |10k - 20 + 6 - 4| = \frac{1}{6} |10k - 18| = \frac{1}{3} |5k - 9|\).

(b)(ii) Given \(V = 5\):
\(\frac{1}{3} |5k - 9| = 5 \implies |5k - 9| = 15\).
If \(5k - 9 = 15 \implies 5k = 24 \implies k = \frac{24}{5}\).
If \(5k - 9 = -15 \implies 5k = -6 \implies k = -\frac{6}{5}\).

(c)(i) A normal vector to the plane \(ABC\) is \(\mathbf{n} = 10\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}\).
A vector making an obtuse angle with the positive z-axis must have a negative z-component.
Thus, we choose \(-\mathbf{n} = -10\mathbf{i} - 6\mathbf{j} - 4\mathbf{k}\).
The unit normal vector \(\hat{\mathbf{n}}\) is:
\(\hat{\mathbf{n}} = -\frac{10\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}}{\sqrt{152}} = -\frac{10\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}}{2\sqrt{38}} = -\frac{5}{\sqrt{38}}\mathbf{i} - \frac{3}{\sqrt{38}}\mathbf{j} - \frac{2}{\sqrt{38}}\mathbf{k}\).

(c)(ii) When \(k = -\frac{6}{5}\), \(\vec{AD} = -\frac{16}{5}\mathbf{i} + \mathbf{j} - \mathbf{k}\).
The shortest distance from \(D\) to the plane \(ABC\) is:
\(d = |\vec{AD} \cdot \hat{\mathbf{n}}|\)
\(= \left| \left(-\frac{16}{5}\mathbf{i} + \mathbf{j} - \mathbf{k}\right) \cdot \left(-\frac{5}{\sqrt{38}}\mathbf{i} - \frac{3}{\sqrt{38}}\mathbf{j} - \frac{2}{\sqrt{38}}\mathbf{k}\right) \right|\)
\(= \left| \frac{16}{\sqrt{38}} - \frac{3}{\sqrt{38}} + \frac{2}{\sqrt{38}} \right| = \frac{15}{\sqrt{38}} = \frac{15\sqrt{38}}{38}\).

(a) 1M for finding vectors \(\vec{AB}\) and \(\vec{AC}\), 1M for computing cross product, 1A for cross product, 1A for area.
(b)(i) 1M for finding vector \(\vec{AD}\), 1M for scalar triple product formula, 1A for volume in terms of \(k\).
(b)(ii) 1M for setting up absolute value equation, 1A for finding both values of \(k\).
(c)(i) 1M for choosing opposite direction of normal vector, 1A for dividing by magnitude to get correct unit normal.
(c)(ii) 1M for using projection/dot product formula, 1A for correct distance.

(a) Since \(\lim_{x \to 4^+} \frac{x^2-3x}{x-4} = +\infty\) and \(\lim_{x \to 4^-} \frac{x^2-3x}{x-4} = -\infty\),
the vertical asymptote is \(x = 4\).
By long division:
\(f(x) = x + 1 + \frac{4}{x-4}\).
As \(x \to \pm\infty\), \(\frac{4}{x-4} \to 0\).
So the oblique asymptote of \(C\) is \(y = x + 1\).

(b) \(f'(x) = \frac{d}{dx} \left(x + 1 + \frac{4}{x-4}\right) = 1 - \frac{4}{(x-4)^2} = \frac{(x-4)^2 - 4}{(x-4)^2} = \frac{x^2 - 8x + 12}{(x-4)^2} = \frac{(x-2)(x-6)}{(x-4)^2}\).
For stationary points, set \(f'(x) = 0 \implies (x-2)(x-6) = 0 \implies x = 2\) or \(x = 6\).
We can test the nature using the first derivative table:
For \(x < 2\), \(f'(x) > 0\).
For \(2 < x < 4\), \(f'(x) < 0\).
Thus, \(x = 2\) is a local maximum.
At \(x = 2\), \(y = f(2) = \frac{2^2 - 3(2)}{2-4} = 1\).
So the local maximum point is \((2, 1)\).

For \(4 < x < 6\), \(f'(x) < 0\).
For \(x > 6\), \(f'(x) > 0\).
Thus, \(x = 6\) is a local minimum.
At \(x = 6\), \(y = f(6) = \frac{6^2 - 3(6)}{6-4} = 9\).
So the local minimum point is \((6, 9)\).

(c) \(f''(x) = \frac{d}{dx} \left(1 - 4(x-4)^{-2}\right) = 8(x-4)^{-3} = \frac{8}{(x-4)^3}\).
For \(C\) to be concave upward, we need \(f''(x) > 0\).
Since the numerator \(8 > 0\), we must have \((x-4)^3 > 0 \implies x > 4\).
So \(C\) is concave upward for \(x > 4\).

(d) Sketching \(C\):
1. Draw the vertical asymptote \(x = 4\) and the oblique asymptote \(y = x + 1\).
2. Plot the local maximum \((2, 1)\) and local minimum \((6, 9)\).
3. Note the intercepts: \(y = 0 \implies x(x-3) = 0 \implies x = 0\) or \(x = 3\).
4. Draw the two branches of the curve: the left branch passes through \((0,0)\) and \((3,0)\), peaks at \((2,1)\), and approaches the asymptotes; the right branch lies in the region \(x > 4\), has a minimum at \((6,9)\), and approaches the asymptotes.

(a) 1A for vertical asymptote, 2A for oblique asymptote (1M for long division/limits calculation).
(b) 1M for finding \(f'(x)\), 1M for setting \(f'(x)=0\) to find critical values, 1A for local maximum, 1A for local minimum.
(c) 1M for finding \(f''(x)\), 1A for setting \(f''(x)>0\) and solving to get \(x > 4\).
(d) 1A for sketching correct asymptotes, 1A for plotting stationary points with coordinates, 1A for drawing left branch with correct shape and intercepts, 1A for drawing right branch with correct shape.