2022 HKDSE Mathematics M2 (Algebra and Calculus) Answers & Marking Scheme

Let \(P(n)\) be the statement \(\sum_{r=1}^n r(r+3) = \frac{n(n+1)(n+5)}{3}\).

When \(n=1\),
\(\text{L.H.S.} = 1(1+3) = 4\)
\(\text{R.H.S.} = \frac{1(1+1)(1+5)}{3} = \frac{1(2)(6)}{3} = 4\)
Since \(\text{L.H.S.} = \text{R.H.S.}\), \(P(1)\) is true.

Assume that \(P(k)\) is true for some positive integer \(k\), i.e.,
\(\sum_{r=1}^k r(r+3) = \frac{k(k+1)(k+5)}{3}\)

When \(n=k+1\),
\(\sum_{r=1}^{k+1} r(r+3) = \sum_{r=1}^k r(r+3) + (k+1)(k+4)\)
\(= \frac{k(k+1)(k+5)}{3} + (k+1)(k+4)\)
\(= \frac{k+1}{3} [ k(k+5) + 3(k+4) ]\)
\(= \frac{k+1}{3} [ k^2 + 5k + 3k + 12 ]\)
\(= \frac{k+1}{3} [ k^2 + 8k + 12 ]\)
\(= \frac{(k+1)(k+2)(k+6)}{3}\)
So, \(P(k+1)\) is true.

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

- Prove the base case \(n=1\). (1.5 marks)
- State the induction hypothesis. (1 mark)
- Substitute the induction hypothesis into the expression for \(n=k+1\) and attempt to factorize/simplify. (2 marks)
- Correctly simplify to the required form. (1 mark)
- Correct conclusion with induction statement. (0.75 marks)

(a) The general term of the expansion of \((x^2 - \frac{2}{x})^n\) is:
\(T_{r+1} = \binom{n}{r} (x^2)^{n-r} (-\frac{2}{x})^r = \binom{n}{r} (-2)^r x^{2n - 3r}\)

The third term corresponds to \(r=2\).
Its coefficient is \(\binom{n}{2} (-2)^2 = 144 \implies 4 \frac{n(n-1)}{2} = 144 \implies n(n-1) = 72\).
Since \(n\) is a positive integer, we have \(n = 9\).

(b) For \(n = 9\), the expansion is \((x^2 - \frac{2}{x})^9 = \sum_{r=0}^9 \binom{9}{r} (-2)^r x^{18 - 3r}\).
We want to find the constant term in the expansion of \((1 + \frac{x^3}{4})(x^2 - \frac{2}{x})^9\).

Constant term \(= [1 \times (\text{coefficient of } x^0 \text{ in } (x^2 - \frac{2}{x})^9)] + [\frac{x^3}{4} \times (\text{coefficient of } x^{-3} \text{ in } (x^2 - \frac{2}{x})^9)]\)

- For \(x^0\): \(18 - 3r = 0 \implies r = 6\).
The term is \(\binom{9}{6} (-2)^6 = 84 \times 64 = 5376\).
- For \(x^{-3}\): \(18 - 3r = -3 \implies 3r = 21 \implies r = 7\).
The term is \(\frac{1}{4} \binom{9}{7} (-2)^7 = \frac{1}{4} \times 36 \times (-128) = -1152\).

Thus, the constant term is \(5376 + (-1152) = 4224\).

- Write down the general term of the expansion. (1 mark)
- Solve for \(n = 9\). (1.25 marks)
- Recognize the two cases for the constant term. (1.5 marks)
- Calculate the term for \(r=6\). (1 mark)
- Calculate the term for \(r=7\). (1 mark)
- Find the correct constant term \(4224\). (0.5 marks)

Using the sum-to-product formula, we have:
\(\sin 3\theta + \sin \theta = 2 \sin\left(\frac{3\theta + \theta}{2}\right) \cos\left(\frac{3\theta - \theta}{2}\right) = 2 \sin 2\theta \cos \theta\)

Thus, the equation becomes:
\(2 \sin 2\theta \cos \theta = \cos \theta\)
\(2 \sin 2\theta \cos \theta - \cos \theta = 0\)
\(\cos \theta (2 \sin 2\theta - 1) = 0\)

Therefore, we have either:
1) \(\cos \theta = 0\)
Since \(0 \le \theta \le \pi\), we get \(\theta = \frac{\pi}{2}\).

2) \(\sin 2\theta = \frac{1}{2}\)
Since \(0 \le \theta \le \pi\), we have \(0 \le 2\theta \le 2\pi\).
Thus, \(2\theta = \frac{\pi}{6}\) or \(2\theta = \frac{5\pi}{6}\).
So, \(\theta = \frac{\pi}{12}\) or \(\theta = \frac{5\pi}{12}\).

Combining the solutions, we have \(\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{\pi}{2}\).

- Apply sum-to-product formula correctly. (1.5 marks)
- Factorize the expression to get \(\cos \theta(2\sin 2\theta - 1) = 0\). (1.5 marks)
- Solve \(\cos \theta = 0\) to get \(\theta = \frac{\pi}{2}\). (1 mark)
- Solve \(\sin 2\theta = \frac{1}{2}\) to get \(\theta = \frac{\pi}{12}, \frac{5\pi}{12}\). (1.5 marks)
- Write down all correct solutions with no extra incorrect ones. (0.75 marks)

Let \(f(x) = \sqrt{2x+3}\).
By definition, the derivative from first principles is:
\(\frac{d}{dx}(\sqrt{2x+3}) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

\(= \lim_{h \to 0} \frac{\sqrt{2(x+h)+3} - \sqrt{2x+3}}{h}\)

\(= \lim_{h \to 0} \frac{\sqrt{2x+2h+3} - \sqrt{2x+3}}{h} \times \frac{\sqrt{2x+2h+3} + \sqrt{2x+3}}{\sqrt{2x+2h+3} + \sqrt{2x+3}}\)

\(= \lim_{h \to 0} \frac{(2x+2h+3) - (2x+3)}{h(\sqrt{2x+2h+3} + \sqrt{2x+3})}\)

\(= \lim_{h \to 0} \frac{2h}{h(\sqrt{2x+2h+3} + \sqrt{2x+3})}\)

\(= \lim_{h \to 0} \frac{2}{\sqrt{2x+2h+3} + \sqrt{2x+3}}\)

\(= \frac{2}{\sqrt{2x+3} + \sqrt{2x+3}}\)

\(= \frac{2}{2\sqrt{2x+3}}\)

\(= \frac{1}{\sqrt{2x+3}}\)

- Write down the correct definition of derivative from first principles. (1 mark)
- Set up the limit expression with rationalization. (2 marks)
- Simplify the numerator and cancel \(h\). (1.5 marks)
- Evaluate the limit correctly. (1.25 marks)
- Obtain final simplified answer. (0.5 marks)

To evaluate \(\int x^5 \cos(x^3) dx\), we can use integration by substitution.
Let \(u = x^3\), then \(du = 3x^2 dx \implies x^2 dx = \frac{1}{3} du\).

The integral becomes:
\(\int x^3 \cos(x^3) (x^2 dx) = \int u \cos u \left(\frac{1}{3} du\right) = \frac{1}{3} \int u \cos u du\)

Now, we use integration by parts for \(\int u \cos u du\):
Let \(w = u \implies dw = du\)
\(dv = \cos u du \implies v = \sin u\)

\(\int u \cos u du = u sin u - \int \sin u du\)
\(= u \sin u - (-\cos u)\)
\(= u \sin u + \cos u\)

Substituting \(u = x^3\) back, we get:
\(\int x^5 \cos(x^3) dx = \frac{1}{3} (x^3 \sin(x^3) + \cos(x^3)) + C\)
\(= \frac{1}{3} x^3 \sin(x^3) + \frac{1}{3} \cos(x^3) + C\)
where \(C\) is an arbitrary constant.

- Choose appropriate substitution \(u = x^3\) and find \(du\). (1.5 marks)
- Express the integral entirely in terms of \(u\). (1 mark)
- Set up and apply integration by parts correctly. (2 marks)
- Substitute \(u = x^3\) back into the result. (1 mark)
- Include the constant of integration \(C\) and get the correct simplified expression. (0.75 marks)

To integrate \(\int_{0}^{1} x^2 \sqrt{1 - x^2} dx\), let \(x = \sin\theta\).
Then \(dx = \cos\theta d\theta\).

When \(x = 0\), \(\theta = 0\).
When \(x = 1\), \(\theta = \frac{\pi}{2}\).

Since \(0 \le \theta \le \frac{\pi}{2}\), \(\sqrt{1 - x^2} = \sqrt{1 - \sin^2\theta} = \cos\theta\).

The integral becomes:
\(\int_{0}^{\pi/2} \sin^2\theta \cdot \cos\theta \cdot \cos\theta d\theta = \int_{0}^{\pi/2} \sin^2\theta \cos^2\theta d\theta\)

Since \(\sin\theta \cos\theta = \frac{1}{2}\sin 2\theta\), we have:
\(\sin^2\theta \cos^2\theta = \frac{1}{4} \sin^2 2\theta = \frac{1}{4} \left( \frac{1 - cos 4\theta}{2} \right) = \frac{1}{8} (1 - \cos 4\theta)\)

Thus, the integral is:
\(\frac{1}{8} \int_{0}^{\pi/2} (1 - \cos 4\theta) d\theta = \frac{1}{8} \left[ \theta - \frac{\sin 4\theta}{4} \right]_{0}^{\pi/2}\)
\(= \frac{1}{8} \left[ \left(\frac{\pi}{2} - \frac{\sin 2\pi}{4}\right) - (0 - 0) \right]\)
\(= \frac{1}{8} \left[ \frac{\pi}{2} - 0 \right]\)
\(= \frac{\pi}{16}\)

- Use substitution \(x = \sin\theta\) and change limits of integration correctly. (1.5 marks)
- Transform integrand into \(\sin^2\theta \cos^2\theta\). (1 mark)
- Apply double-angle formulas to simplify integrand. (1.5 marks)
- Integrate term by term. (1 mark)
- Substitute limits and find the correct final answer \(\frac{\pi}{16}\). (1.25 marks)

(a)
\(A^2 = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 2(2)+1(-1) & 2(1)+1(0) \\ -1(2)+0(-1) & -1(1)+0(0)
\end{pmatrix} = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix}\)

\(A^2 - 2A + I = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix} - 2\begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
\(= \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix} - \begin{pmatrix} 4 & 2 \\ -2 & 0
\end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
\(= \begin{pmatrix} 3 - 4 + 1 & 2 - 2 + 0 \\ -2 - (-2) + 0 & -1 - 0 + 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0\)

(b)
From (a), we have \(A^2 = 2A - I\).

To find \(A^4\):
\(A^4 = (A^2)^2 = (2A - I)^2 = 4A^2 - 4AI + I^2 = 4A^2 - 4A + I\)
Substituting \(A^2 = 2A - I\):
\(A^4 = 4(2A - I) - 4A + I = 8A - 4I - 4A + I = 4A - 3I\)
Thus, \(A^4 = 4A - 3I\) (where \(\alpha = 4, \beta = -3\)).

To find \(A^{-1}\):
Since \(A^2 - 2A + I = 0\), we multiply by \(A^{-1}\):
\(A^{-1}(A^2 - 2A + I) = A^{-1}(0)\)
\(A - 2I + A^{-1} = 0\)
\(A^{-1} = -A + 2I\)
Thus, \(A^{-1} = -A + 2I\) (where \(\alpha = -1, \beta = 2\)).

- Compute \(A^2\) correctly. (1.5 marks)
- Verify \(A^2 - 2A + I = 0\) by showing all matrix entries sum to 0. (1 mark)
- Use the relation \(A^2 = 2A - I\) to express \(A^4\) as \(4A - 3I\). (1.75 marks)
- Multiply the equation by \(A^{-1}\) (or equivalent method) to find \(A^{-1} = -A + 2I\). (2 marks)

(a)
\(\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 1 \end{vmatrix}\)
\(= \mathbf{i}(2(1) - (-1)(-1)) - \mathbf{j}(1(1) - (-1)(2)) + \mathbf{k}(1(-1) - 2(2))\)
\(= \mathbf{i}(2 - 1) - \mathbf{j}(1 + 2) + \mathbf{k}(-1 - 4)\)
\(= \mathbf{i} - 3\mathbf{j} - 5\mathbf{k}\)

(b)
The volume of the parallelepiped spanned by \(\mathbf{u}\), \(\mathbf{v}\) and \(\mathbf{w}\) is given by \(|(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}|\).

First, we compute the scalar triple product:
\((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = (\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}) \cdot (k\mathbf{i} + \mathbf{j} + 3\mathbf{k})\)
\(= (1)(k) + (-3)(1) + (-5)(3)\)
\(= k - 3 - 15\)
\(= k - 18\)

Given that the volume is 15:
\(|k - 18| = 15\)

Thus, we have:
\(k - 18 = 15 \implies k = 33\)
or
\(k - 18 = -15 \implies k = 3\)

So, the possible values of \(k\) are \(3\) and \(33\).

- Express cross product as determinant or expanded components correctly. (1 mark)
- Obtain correct cross product \(\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}\). (1.5 marks)
- Express the volume of the parallelepiped as the absolute value of the scalar triple product. (1 mark)
- Calculate the scalar triple product as \(k - 18\). (1 mark)
- Set up equation \(|k - 18| = 15\). (1 mark)
- Find both correct values \(k = 3\) and \(k = 33\). (0.75 marks)

(a) \( f'(x) = \frac{2x(x-1) - (x^2 + a)}{(x-1)^2} = \frac{x^2 - 2x - a}{(x-1)^2} \). Set \( f'(x) = 0 \), we have \( x^2 - 2x - a = 0 \). Since \( a > -1 \), \( \Delta = 4 - 4(-a) = 4(1+a) > 0 \), so there are two distinct roots: \( x = 1 \pm \sqrt{1+a} \). If \( x < 1 - \sqrt{1+a} \) or \( x > 1 + \sqrt{1+a} \), \( f'(x) > 0 \). If \( 1 - \sqrt{1+a} < x < 1 + \sqrt{1+a} \) (and \( x \neq 1 \)), \( f'(x) < 0 \). Thus, the local maximum point is \( (1 - \sqrt{1+a}, 2 - 2\sqrt{1+a}) \) and the local minimum point is \( (1 + \sqrt{1+a}, 2 + 2\sqrt{1+a}) \).
(b) The distance \( d \) between the two extreme points is given by: \( d^2 = [(1+\sqrt{1+a}) - (1-\sqrt{1+a})]^2 + [(2+2\sqrt{1+a}) - (2-2\sqrt{1+a})]^2 = (2\sqrt{1+a})^2 + (4\sqrt{1+a})^2 = 4(1+a) + 16(1+a) = 20(1+a) \). Given \( d = 2\sqrt{10} \), we have \( 20(1+a) = (2\sqrt{10})^2 = 40 \Rightarrow 1+a = 2 \Rightarrow a = 1 \).
(c) For \( a = 1 \), \( f(x) = \frac{x^2+1}{x-1} \).
(i) Since \( \lim_{x \to 1} f(x) = \pm\infty \), the vertical asymptote is \( x = 1 \). By division, \( f(x) = x + 1 + \frac{2}{x-1} \), so the oblique asymptote is \( y = x + 1 \).
(ii) \( f''(x) = \frac{d}{dx} \left( 1 - \frac{2}{(x-1)^2} \right) = \frac{4}{(x-1)^3} \). For concave upward, we need \( f''(x) > 0 \Rightarrow (x-1)^3 > 0 \Rightarrow x > 1 \).
(iii) Sketch of \( C \) showing the local minimum at \( (1+\sqrt{2}, 2+2\sqrt{2}) \), local maximum at \( (1-\sqrt{2}, 2-2\sqrt{2}) \), y-intercept at \( (0, -1) \), and asymptotes \( x = 1 \) and \( y = x + 1 \).

(a) Finding \( f'(x) \): 1M. Setting \( f'(x) = 0 \) to find \( x \)-coordinates: 1M. Correct y-coordinates: 1A. Differentiating local min/max points: 1A.
(b) Setting up distance equation: 1M. Simplifying to \( 20(1+a) = 40 \): 1M. Finding \( a = 1 \): 1A.
(c)(i) Correct equations of both asymptotes: 2A (1A for each).
(c)(ii) Finding second derivative: 1M. Correct range \( x > 1 \): 1A.
(c)(iii) Correctly sketching the curve with key features: 1A.

(a) \( \det(M) = 1(1-10) - p(p-2) + 1(5p-1) = -9 - p^2 + 2p + 5p - 1 = -p^2 + 7p - 10 \). For \( M \) to be singular, \( \det(M) = 0 \Rightarrow p^2 - 7p + 10 = 0 \Rightarrow (p-2)(p-5) = 0 \Rightarrow p = 2 \text{ or } p = 5 \).
(b) When \( p = 3 \), \( \det(M) = -(3)^2 + 7(3) - 10 = 2 \). Co-factor matrix of \( M \) is: \( C = \begin{pmatrix} -9 & -1 & 14 \\ 2 & 0 & -2 \\ 5 & 1 & -8 \end{pmatrix} \). Therefore, the adjoint matrix of \( M \) is \( C^T = \begin{pmatrix} -9 & 2 & 5 \\ -1 & 0 & 1 \\ 14 & -2 & -8 \end{pmatrix} \). Hence, \( M^{-1} = \frac{1}{2} \begin{pmatrix} -9 & 2 & 5 \\ -1 & 0 & 1 \\ 14 & -2 & -8 \end{pmatrix} = \begin{pmatrix} -4.5 & 1 & 2.5 \\ -0.5 & 0 & 0.5 \\ 7 & -1 & -4 \end{pmatrix} \).
(c)(i) When \( p = 2 \), write the augmented matrix: \( \begin{pmatrix} 1 & 2 & 1 & | & 4 \\ 2 & 1 & 2 & | & q \\ 1 & 5 & 1 & | & q + 2 \end{pmatrix} \). Perform row operations:
\( R_2 \to R_2 - 2R_1 \) and \( R_3 \to R_3 - R_1 \): \( \begin{pmatrix} 1 & 2 & 1 & | & 4 \\ 0 & -3 & 0 & | & q - 8 \\ 0 & 3 & 0 & | & q - 2 \end{pmatrix} \).
\( R_3 \to R_3 + R_2 \): \( \begin{pmatrix} 1 & 2 & 1 & | & 4 \\ 0 & -3 & 0 & | & q - 8 \\ 0 & 0 & 0 & | & 2q - 10 \end{pmatrix} \). For the system to be consistent, we must have \( 2q - 10 = 0 \Rightarrow q = 5 \).
(ii) Substituting \( q = 5 \) back into the augmented matrix: \( \begin{pmatrix} 1 & 2 & 1 & | & 4 \\ 0 & -3 & 0 & | & -3 \\ 0 & 0 & 0 & | & 0 \end{pmatrix} \). From row 2, \( -3y = -3 \Rightarrow y = 1 \). From row 1, \( x + 2(1) + z = 4 \Rightarrow x + z = 2 \). Let \( z = t \), then \( x = 2 - t \), where \( t \in \mathbb{R} \). Thus, the solution is \( \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2-t \\ 1 \\ t \end{pmatrix} \) (or \( x = 2 - t, y = 1, z = t \) for any real number \( t \)).

(a) Finding determinant: 1M. Setting determinant to 0: 1M. Correct values of \( p \): 1A.
(b) Finding \( \det(M) = 2 \): 1M. Finding cofactor matrix (or adjoint matrix): 2M (1M for partial correctness). Finding \( M^{-1} \) correctly: 1A.
(c)(i) Performing correct row operations on augmented matrix: 1M. Showing the row relation of consistency: 1M. Finding \( q = 5 \): 1A.
(c)(ii) Correct value for \( y \): 1A. Expressing \( x \) in terms of parameter \( t \) (or similar): 1M. Correct general solution: 1A.

(a) Let \( u = a - x \), then \( dx = -du \).
When \( x = 0 \), \( u = a \). When \( x = a \), \( u = 0 \).
Therefore, \( \int_{0}^{a} f(x) \, dx = \int_{a}^{0} f(a-u) (-du) = \int_{0}^{a} f(a-u) \, du = \int_{0}^{a} f(a-x) \, dx \).
(b) Let \( I = \int_{0}^{\pi/4} \ln(1 + \tan x) \, dx \). By (a), we have:
\( I = \int_{0}^{\pi/4} \ln\left(1 + \tan\left(\frac{\pi}{4}-x\right)\right) \, dx \).
Since \( \tan\left(\frac{\pi}{4}-x\right) = \frac{\tan(\pi/4) - \tan x}{1 + \tan(\pi/4)\tan x} = \frac{1 - \tan x}{1 + \tan x} \),
we have \( 1 + \tan\left(\frac{\pi}{4}-x\right) = 1 + \frac{1 - \tan x}{1 + \tan x} = \frac{2}{1 + \tan x} \).
Thus, \( I = \int_{0}^{\pi/4} \ln\left(\frac{2}{1+\tan x}\right) \, dx = \int_{0}^{\pi/4} [\ln 2 - \ln(1 + \tan x)] \, dx \)
\( I = \int_{0}^{\pi/4} \ln 2 \, dx - \int_{0}^{\pi/4} \ln(1 + \tan x) \, dx = [x \ln 2]_{0}^{\pi/4} - I \).
\( 2I = \frac{\pi}{4} \ln 2 \Rightarrow I = \frac{\pi}{8} \ln 2 \).
(c) Let \( J = \int_{0}^{\pi/4} \frac{x \sec^2 x}{1 + \tan x} \, dx \).
Using integration by parts: let \( U = x \) and \( dV = \frac{\sec^2 x}{1 + \tan x} \, dx \).
Then \( dU = dx \) and \( V = \ln(1 + \tan x) \) (since \( 1 + \tan x > 0 \) on \( [0, \pi/4] \)).
\( J = [x \ln(1 + \tan x)]_{0}^{\pi/4} - \int_{0}^{\pi/4} \ln(1 + \tan x) \, dx \).
Since \( [x \ln(1 + \tan x)]_{0}^{\pi/4} = \frac{\pi}{4} \ln(1 + \tan(\pi/4)) - 0 = \frac{\pi}{4} \ln 2 \), and from (b), \( \int_{0}^{\pi/4} \ln(1 + \tan x) \, dx = \frac{\pi}{8} \ln 2 \),
we obtain \( J = \frac{\pi}{4} \ln 2 - \frac{\pi}{8} \ln 2 = \frac{\pi}{8} \ln 2 \).

(a) Correct substitution \( u = a - x \) and change of limits: 1M. Completing the proof: 1A.
(b) Applying (a): 1M. Using the compound angle formula for tangent correctly: 1M. Simplifying to \( \ln(2/(1+\tan x)) \): 1M. Using properties of logarithms: 1M. Completing the proof: 1A.
(c) Applying integration by parts (choosing \( U, dV \) correctly): 1M. Finding \( V = \ln(1+\tan x) \): 1A. Evaluating the boundary term \( [x \ln(1+\tan x)]_0^{\pi/4} = \frac{\pi}{4} \ln 2 \): 1M. Substituting result from (b): 1M. Correct answer: 1A.

(a) \( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 1 \end{vmatrix} = \mathbf{i}(2-1) - \mathbf{j}(1-(-2)) + \mathbf{k}(-1-4) = \mathbf{i} - 3\mathbf{j} - 5\mathbf{k} \).
(b) The volume of the tetrahedron is given by \( V = \frac{1}{6} |(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}| \).
We have \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (1)(1) + (-3)(1) + (-5)(z) = -2 - 5z \).
Thus, \( \frac{1}{6} |-2 - 5z| = \frac{7}{6} \Rightarrow |-2 - 5z| = 7 \).
Case 1: \( -2 - 5z = 7 \Rightarrow 5z = -9 \Rightarrow z = -1.8 \).
Case 2: \( -2 - 5z = -7 \Rightarrow 5z = 5 \Rightarrow z = 1 \).
Therefore, the possible values of \( z \) are \( 1 \) and \( -1.8 \).
(c) When \( z = 1 \), \( \mathbf{c} = \mathbf{i} + \mathbf{j} + \mathbf{k} \).
(i) The normal vector to plane \( OAB \) is \( \mathbf{n} = \mathbf{a} \times \mathbf{b} = \mathbf{i} - 3\mathbf{j} - 5\mathbf{k} \).
The projection of \( \mathbf{c} \) onto \( \mathbf{n} \) is:
\( \mathbf{p}_{n} = \left( \frac{\mathbf{c} \cdot \mathbf{n}}{|\mathbf{n}|^2} \right) \mathbf{n} = \frac{1(1) + 1(-3) + 1(-5)}{1^2 + (-3)^2 + (-5)^2} (\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}) = \frac{-7}{35}(\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}) = -\frac{1}{5}(\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}) \).
The projection of \( \overrightarrow{OC} \) onto plane \( OAB \) is:
\( \mathbf{c} - \mathbf{p}_{n} = (\mathbf{i} + \mathbf{j} + \mathbf{k}) - \left(-\frac{1}{5}\mathbf{i} + \frac{3}{5}\mathbf{j} + \mathbf{k}\right) = \frac{6}{5}\mathbf{i} + \frac{2}{5}\mathbf{j} \).
(ii) The shortest distance from \( C \) to plane \( OAB \) is the magnitude of the projection of \( \mathbf{c} \) onto the normal vector \( \mathbf{n} \):
\( d = |\mathbf{p}_{n}| = \left| -\frac{1}{5}(\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}) \right| = \frac{1}{5}\sqrt{1^2 + (-3)^2 + (-5)^2} = \frac{\sqrt{35}}{5} \).

(a) Applying cross product correctly: 1M. Correct answer: 1A.
(b) Setting up formula \( V = \frac{1}{6} |(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}| \): 1M. Finding \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = -2 - 5z \): 1A. Considering two cases: 1M. Correct values of \( z \): 1A.
(c)(i) Identifying normal vector: 1M. Setting up formula for projection on plane: 1M. Calculating projection of vector onto normal vector: 1A. Correct projection onto plane: 1A.
(c)(ii) Correct formula for shortest distance: 1M. Calculating the magnitude correctly: 1M. Correct distance: 1A.