Cambridge IAS-Level · Thinka 原創模擬試題

2024 Cambridge IAS-Level Mathematics - Further (9231) 模擬試題連答案詳解

Thinka Nov 2024 (V3) Cambridge International A Level-Style Mock — Mathematics - Further (9231)

150 分240 分鐘2024

An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V3) Cambridge International A Level Mathematics - Further (9231) paper. Not affiliated with or reproduced from Cambridge.

卷一: Further Pure Mathematics 1

Answer all questions. Show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

8 題目 · 83 分

題目 1 · short_answer

10 分

The matrix $\mathbf{M}$ is given by
\[ \mathbf{M} = \begin{pmatrix} a & 2 \\ 3 & a - 1 \end{pmatrix} \]
where $a$ is a real constant.

(a) Find the two values of $a$ for which the transformation represented by $\mathbf{M}$ has a line of invariant points other than the origin, and find the equation of this line in each case. [4]

The matrix $\mathbf{A}$ is given by
\[ \mathbf{A} = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 0 & 3 & -1 \end{pmatrix} \]
The line $l$ has vector equation
\[ \mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \]
where $\lambda$ is a real parameter.

(b) Find a vector equation of the image of $l$ under the 3D linear transformation represented by $\mathbf{A}$. [6]

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解題

(a) A line of invariant points exists if there are non-trivial solutions to $\mathbf{M}\mathbf{x} = \mathbf{x}$, which can be written as $(\mathbf{M} - \mathbf{I})\mathbf{x} = \mathbf{0}$.
This requires the determinant of $\mathbf{M} - \mathbf{I}$ to be zero:
\[ \det(\mathbf{M} - \mathbf{I}) = \det \begin{pmatrix} a - 1 & 2 \\ 3 & a - 2 \end{pmatrix} = 0 \]
\[ (a - 1)(a - 2) - 6 = 0 \]
\[ a^2 - 3a - 4 = 0 \]
\[ (a - 4)(a + 1) = 0 \]
So the two values of $a$ are $a = 4$ and $a = -1$.

If $a = 4$, the system $(\mathbf{M} - \mathbf{I})\mathbf{x} = \mathbf{0}$ becomes:
\[ \begin{pmatrix} 3 & 2 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]
which gives the line of invariant points: $3x + 2y = 0$ (or $y = -\frac{3}{2}x$).

If $a = -1$, the system becomes:
\[ \begin{pmatrix} -2 & 2 \\ 3 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]
which gives the line of invariant points: $y = x$ (or $x - y = 0$).

(b) A general point $\mathbf{u}$ on the line $l$ has coordinates:
\[ \mathbf{u} = \begin{pmatrix} 1 + 2\lambda \\ -2 + \lambda \\ 3 - \lambda \end{pmatrix} \]
The image $\mathbf{u}'$ of this point under the transformation represented by $\mathbf{A}$ is given by:
\[ \mathbf{u}' = \mathbf{A}\mathbf{u} = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 0 & 3 & -1 \end{pmatrix} \begin{pmatrix} 1 + 2\lambda \\ -2 + \lambda \\ 3 - \lambda \end{pmatrix} \]
\[ \mathbf{u}' = \begin{pmatrix} 1(1+2\lambda) - 1(-2+\lambda) + 2(3-\lambda) \\ 2(1+2\lambda) + 1(-2+\lambda) + 1(3-\lambda) \\ 0(1+2\lambda) + 3(-2+\lambda) - 1(3-\lambda) \end{pmatrix} \]
\[ \mathbf{u}' = \begin{pmatrix} 1 + 2\lambda + 2 - \lambda + 6 - 2\lambda \\ 2 + 4\lambda - 2 + \lambda + 3 - \lambda \\ -6 + 3\lambda - 3 + \lambda \end{pmatrix} = \begin{pmatrix} 9 - \lambda \\ 3 + 4\lambda \\ -9 + 4\lambda \end{pmatrix} \]
Separating the constant and parameter parts, we obtain the vector equation of the image line:
\[ \mathbf{r} = \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} + t \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix} \]
(where $t$ is a real parameter, replacing $\lambda$).

Alternatively, we can find the image of two points on $l$:
For $\lambda = 0$, the point is $\mathbf{p}_1 = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$, with image:
\[ \mathbf{A}\mathbf{p}_1 = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 0 & 3 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} \]
For $\lambda = 1$, the point is $\mathbf{p}_2 = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$, with image:
\[ \mathbf{A}\mathbf{p}_2 = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 0 & 3 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 7 \\ -5 \end{pmatrix} \]
The direction vector of the image line is:
\[ \mathbf{A}\mathbf{p}_2 - \mathbf{A}\mathbf{p}_1 = \begin{pmatrix} 8 \\ 7 \\ -5 \end{pmatrix} - \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix} \]
Thus, the vector equation of the image line is:
\[ \mathbf{r} = \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} + t \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix} \]

評分準則

(a)
- **M1**: Sets up the condition for a line of invariant points, $\det(\mathbf{M} - \mathbf{I}) = 0$.
- **A1**: Correctly solves the quadratic equation to find $a = 4$ and $a = -1$.
- **A1**: Correctly finds the equation of the line for $a = 4$, which is $3x + 2y = 0$ (or any equivalent form).
- **A1**: Correctly finds the equation of the line for $a = -1$, which is $y = x$ (or any equivalent form).

(b)
- **M1**: Attempts to find the image of a general point on the line $l$, or the images of two distinct points on $l$.
- **A1**: Finds the image of one point on the line correctly, e.g., $\begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix}$.
- **A1**: Finds the image of a second point on the line correctly, e.g., $\begin{pmatrix} 8 \\ 7 \\ -5 \end{pmatrix}$ (or writes down the general image vector in terms of $\lambda$ correctly).
- **M1**: Finds a direction vector of the image line by subtracting two image points or by extracting the coefficient of $\lambda$.
- **A1**: Obtains a correct direction vector, e.g., $\begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix}$ (or any non-zero scalar multiple).
- **A1**: Writes the final answer in a correct vector equation format (including $\mathbf{r} = \dots$ and a parameter): $\mathbf{r} = \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} + t \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix}$.

題目 2 · essay

6 分

Prove by mathematical induction that $ 7^n + 4^n + 1 $ is divisible by 6 for all positive integers $ n $.

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解題

Let $ H(n) $ be the statement that $ 7^n + 4^n + 1 $ is divisible by 6. For $ n = 1 $: $ 7^1 + 4^1 + 1 = 12 $, which is divisible by 6, so the base case holds. Assume that the statement is true for $ n = k $, so $ 7^k + 4^k + 1 = 6m $ for some integer $ m $. For $ n = k+1 $, we have $ 7^{k+1} + 4^{k+1} + 1 = 7 \cdot 7^k + 4 \cdot 4^k + 1 $. Substituting $ 7^k = 6m - 4^k - 1 $ yields $ 7(6m - 4^k - 1) + 4 \cdot 4^k + 1 = 42m - 3 \cdot 4^k - 6 = 6(7m - 1) - 3 \cdot 4^k $. Since $ k \ge 1 $, $ 3 \cdot 4^k = 3 \cdot 2 \cdot 2 \cdot 4^{k-1} = 6(2 \cdot 4^{k-1}) $, which is a multiple of 6. Thus, the expression is equal to $ 6(7m - 1 - 2 \cdot 4^{k-1}) $, which is divisible by 6. Therefore, if the statement is true for $ n = k $, it is also true for $ n = k+1 $. Since the base case is true, by mathematical induction the statement is true for all positive integers $ n $.

評分準則

**B1**: Show that the base case $ n=1 $ is true: $ 7^1 + 4^1 + 1 = 12 $, which is divisible by 6. **M1**: State the inductive hypothesis clearly: assume $ 7^k + 4^k + 1 = 6m $ for some integer $ m $. **M1**: Write down the expression for $ n=k+1 $ and attempt to substitute the inductive hypothesis. **A1**: Obtain a correct simplified expression, such as $ 42m - 3 \cdot 4^k - 6 $ or equivalent. **A1**: Justify why the remaining term (e.g. $ 3 \cdot 4^k $) is divisible by 6. **A1**: Complete the proof with a correct concluding statement.

題目 3 · free_response

10 分

The cubic equation $x^3 - 3x^2 + 5x - 2 = 0$ has roots $\alpha$, $\beta$ and $\gamma$.

(a) Find the values of:
(i) $\alpha^2 + \beta^2 + \gamma^2$
(ii) $\alpha^3 + \beta^3 + \gamma^3$ [4]

(b) Find a cubic equation with integer coefficients whose roots are $\alpha^2 - 1$, $\beta^2 - 1$ and $\gamma^2 - 1$. [6]

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解題

(a) From the given equation, we have:
$\sum \alpha = 3$
$\sum \alpha\beta = 5$
$\alpha\beta\gamma = 2$

(i) Using the identity for the sum of squares:
$\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta$
$\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(5) = 9 - 10 = -1$

(ii) Since $\alpha$, $\beta$ and $\gamma$ are roots of the cubic equation, we have:
$\alpha^3 - 3\alpha^2 + 5\alpha - 2 = 0$
Summing this relation over the three roots yields:
$\sum \alpha^3 - 3\sum \alpha^2 + 5\sum \alpha - 6 = 0$
Substitute the known values:
$\sum \alpha^3 - 3(-1) + 5(3) - 6 = 0$
$\sum \alpha^3 + 3 + 15 - 6 = 0$
$\sum \alpha^3 + 12 = 0 \implies \alpha^3 + \beta^3 + \gamma^3 = -12$

Alternatively, using the identity:
$\sum \alpha^3 - 3\alpha\beta\gamma = (\sum \alpha)(\sum \alpha^2 - \sum \alpha\beta)$
$\sum \alpha^3 - 3(2) = 3(-1 - 5)$
$\sum \alpha^3 - 6 = -18 \implies \sum \alpha^3 = -12$.

(b) Let $y = x^2 - 1$, which means $x^2 = y + 1$.
Rearranging the original cubic equation to separate the terms with odd and even powers of $x$:
$x^3 + 5x = 3x^2 + 2$
$x(x^2 + 5) = 3x^2 + 2$
Squaring both sides:
$x^2(x^2 + 5)^2 = (3x^2 + 2)^2$
Substitute $x^2 = y + 1$ into this equation:
$(y + 1)(y + 1 + 5)^2 = (3(y + 1) + 2)^2$
$(y + 1)(y + 6)^2 = (3y + 5)^2$
Expand both sides:
$(y + 1)(y^2 + 12y + 36) = 9y^2 + 30y + 25$
$y^3 + 12y^2 + 36y + y^2 + 12y + 36 = 9y^2 + 30y + 25$
$y^3 + 13y^2 + 48y + 36 = 9y^2 + 30y + 25$
Simplifying by collecting all terms to one side:
$y^3 + 4y^2 + 18y + 11 = 0$

評分準則

Part (a):
- M1: For identifying the elementary symmetric sums: $\sum \alpha = 3$, $\sum \alpha\beta = 5$, and $\alpha\beta\gamma = 2$.
- A1: For obtaining $\alpha^2 + \beta^2 + \gamma^2 = -1$.
- M1: For applying a valid method to find $\sum \alpha^3$ (such as the cubic recurrence relation or the standard algebraic identity).
- A1: For obtaining $\alpha^3 + \beta^3 + \gamma^3 = -12$.

Part (b) (using substitution method):
- M1: For setting up the substitution $y = x^2 - 1$ (or equivalent) and isolating the odd/even powers of $x$.
- M1: For squaring both sides to produce an equation in terms of $x^2$ only.
- M1: For substituting $x^2 = y + 1$ into the squared equation.
- M1: For expanding both sides of the substituted equation correctly.
- A1: For combining terms to obtain the correct cubic expression $y^3 + 4y^2 + 18y + 11$.
- A1: For writing the final answer as a complete equation, i.e., $y^3 + 4y^2 + 18y + 11 = 0$ (accept any variable name instead of $y$).

Part (b) (alternative method using root relations):
- M1: For finding $\sum (\alpha^2 - 1) = \sum \alpha^2 - 3 = -4$.
- M1: For expressing $\sum (\alpha^2 - 1)(\beta^2 - 1) = \sum \alpha^2\beta^2 - 2\sum \alpha^2 + 3$.
- M1: For finding $\sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma\sum\alpha = 25 - 12 = 13$.
- M1: For calculating $\sum (\alpha^2 - 1)(\beta^2 - 1) = 13 - 2(-1) + 3 = 18$.
- A1: For calculating the constant term from the product: $(\alpha^2 - 1)(\beta^2 - 1)(\gamma^2 - 1) = 4 - 13 - 1 - 1 = -11$.
- A1: For writing the correct final equation $y^3 + 4y^2 + 18y + 11 = 0$.

題目 4 · free-text

8 分

Show that
\[ \frac{6}{(3r-2)(3r+1)(3r+4)} = \frac{1}{(3r-2)(3r+1)} - \frac{1}{(3r+1)(3r+4)}. \]

Hence find
\[ \sum_{r=1}^n \frac{1}{(3r-2)(3r+1)(3r+4)} \]
in terms of $ n $, simplifying your answer.

State the value of
\[ \sum_{r=1}^{\infty} \frac{1}{(3r-2)(3r+1)(3r+4)}. \]

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解題

To show the identity, we start with the right-hand side (RHS):
\[ \frac{1}{(3r-2)(3r+1)} - \frac{1}{(3r+1)(3r+4)} \]
Combining the fractions over a common denominator:
\[ = \frac{(3r+4) - (3r-2)}{(3r-2)(3r+1)(3r+4)} \]
\[ = \frac{6}{(3r-2)(3r+1)(3r+4)} \]
which matches the left-hand side (LHS).

Next, we use this identity to evaluate the sum:
\[ \sum_{r=1}^n \frac{1}{(3r-2)(3r+1)(3r+4)} = \frac{1}{6} \sum_{r=1}^n \left( \frac{1}{(3r-2)(3r+1)} - \frac{1}{(3r+1)(3r+4)} \right) \]
Let $ f(r) = \frac{1}{(3r-2)(3r+1)} $. The sum can be written as:
\[ \frac{1}{6} \sum_{r=1}^n \left( f(r) - f(r+1) \right) \]
Writing out the terms of this telescoping series:
\[ = \frac{1}{6} \left[ \left( f(1) - f(2) \right) + \left( f(2) - f(3) \right) + \dots + \left( f(n) - f(n+1) \right) \right] \]
\[ = \frac{1}{6} \left[ f(1) - f(n+1) \right] \]
We calculate the values for $ f(1) $ and $ f(n+1) $:
\[ f(1) = \frac{1}{(1)(4)} = \frac{1}{4} \]
\[ f(n+1) = \frac{1}{(3n+1)(3n+4)} \]
So the sum $ S_n $ is:
\[ S_n = \frac{1}{6} \left( \frac{1}{4} - \frac{1}{(3n+1)(3n+4)} \right) = \frac{1}{24} - \frac{1}{6(3n+1)(3n+4)} \]
We simplify this to a single algebraic fraction:
\[ S_n = \frac{(3n+1)(3n+4) - 4}{24(3n+1)(3n+4)} \]
\[ S_n = \frac{9n^2 + 15n + 4 - 4}{24(3n+1)(3n+4)} \]
\[ S_n = \frac{3n(3n+5)}{24(3n+1)(3n+4)} \]
\[ S_n = \frac{n(3n+5)}{8(3n+1)(3n+4)} \]

Finally, for the sum to infinity, we take the limit as $ n \to \infty $:
\[ S_{\infty} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{1}{24} - \frac{1}{6(3n+1)(3n+4)} \right) = \frac{1}{24} \]

評分準則

**M1**: For combining the RHS of the identity over a common denominator.
**A1**: For obtaining the correct LHS numerator of 6 and completing the verification.

**M1**: For expressing the series as a sum of differences using the identity.
**M1**: For showing the telescoping cancellation of terms, leaving only the first and last terms.
**A1**: For finding the correct unsimplified sum $ S_n = \frac{1}{6} \left( \frac{1}{4} - \frac{1}{(3n+1)(3n+4)} \right) $.
**A1**: For simplifying $ S_n $ correctly to the final form $ \frac{n(3n+5)}{8(3n+1)(3n+4)} $.

**M1**: For considering the limit of $ S_n $ as $ n \to \infty $.
**A1**: For the correct value $ \frac{1}{24} $.

題目 5 · free-text

8 分

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解題

評分準則

題目 6 · structured

13 分

The curve $C$ has polar equation $r = a(7 + 3\cos\theta)$, where $a$ is a positive constant and $-\pi < \theta \le \pi$. (i) Sketch the curve $C$, stating the polar coordinates of the points of intersection with the initial line and the line $\theta = \frac{\pi}{2}$. [3 marks] (ii) Find the polar coordinates of the points on $C$ where the tangent to $C$ is parallel to the initial line. [5 marks] (iii) Find, in terms of $a$ and $\pi$, the exact area of the region enclosed by $C$. [5 marks]

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解題

(i) To sketch $r = a(7 + 3\cos\theta)$, we note that for all $\theta$, $r > 0$ since the minimum value of $7 + 3\cos\theta$ is $4$. The curve is symmetric about the initial line because $\cos(-\theta) = \cos\theta$. Evaluating at key angles: when $\theta = 0$, $r = 10a$; when $\theta = \pi$, $r = 4a$; when $\theta = \pm\frac{\pi}{2}$, $r = 7a$. Thus, the curve is a smooth, closed, egg-like convex shape with no dimple or cusp, intersecting the initial line at $(10a, 0)$ and $(4a, \pi)$, and intersecting $\theta = \frac{\pi}{2}$ at $(7a, \frac{\pi}{2})$ and $(7a, -\frac{\pi}{2})$. (ii) The distance from the initial line is $y = r\sin\theta = a(7 + 3\cos\theta)\sin\theta = a(7\sin\theta + 3\sin\theta\cos\theta) = a(7\sin\theta + \frac{3}{2}\sin 2\theta)$. For the tangent to be parallel to the initial line, we set $\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0$. Differentiating: $\frac{\mathrm{d}y}{\mathrm{d}\theta} = a(7\cos\theta + 3\cos 2\theta) = a(7\cos\theta + 3(2\cos^2\theta - 1)) = a(6\cos^2\theta + 7\cos\theta - 3)$. Setting $6\cos^2\theta + 7\cos\theta - 3 = 0$ and factoring yields $(2\cos\theta + 3)(3\cos\theta - 1) = 0$. Since $\cos\theta$ must lie in the range $[-1, 1]$, we discard the root $\cos\theta = -1.5$. This leaves $3\cos\theta - 1 = 0 \implies \cos\theta = \frac{1}{3}$. The corresponding values of $\theta$ in $(-\pi, \pi]$ are $\theta = \pm\arccos(\frac{1}{3})$. Substituting $\cos\theta = \frac{1}{3}$ back into the polar equation gives $r = a(7 + 3(\frac{1}{3})) = 8a$. Thus, the polar coordinates of the points of contact are $(8a, \arccos(\frac{1}{3}))$ and $(8a, -\arccos(\frac{1}{3}))$. (iii) The area $A$ of the region enclosed by $C$ is given by $A = \frac{1}{2}\int_{-\pi}^{\pi} r^2 \mathrm{d}\theta = \frac{a^2}{2}\int_{-\pi}^{\pi} (7 + 3\cos\theta)^2 \mathrm{d}\theta = \frac{a^2}{2}\int_{-\pi}^{\pi} (49 + 42\cos\theta + 9\cos^2\theta) \mathrm{d}\theta$. Using the identity $\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)$, the integrand becomes: $49 + 42\cos\theta + \frac{9}{2}(1 + \cos 2\theta) = \frac{107}{2} + 42\cos\theta + \frac{9}{2}\cos 2\theta$. Integrating each term with respect to $\theta$: $A = \frac{a^2}{2} \left[ \frac{107}{2}\theta + 42\sin\theta + \frac{9}{4}\sin 2\theta \right]_{-\pi}^{\pi}$. Evaluating between the limits: $A = \frac{a^2}{2} \left[ \left( \frac{107\pi}{2} + 0 + 0 \right) - \left( -\frac{107\pi}{2} + 0 + 0 \right) \right] = \frac{a^2}{2} (107\pi) = \frac{107}{2}\pi a^2$.

評分準則

Part (i): [3 marks] B1: Correctly shaped closed curve (convex, symmetric about the initial line, with no dimple or loop). B1: Correct polar coordinates for intersections with the initial line: $(10a, 0)$ and $(4a, \pi)$. B1: Correct polar coordinates for intersections with the line $\theta = \pi/2$: $(7a, \pi/2)$ and $(7a, -\pi/2)$. Part (ii): [5 marks] M1: Expresses $y = r\sin\theta$ in terms of $\theta$. M1: Differentiates $y$ and equates $\frac{\mathrm{d}y}{\mathrm{d}\theta}$ to zero to obtain a quadratic equation in $\cos\theta$. A1: Correctly factors and solves the quadratic equation, obtaining $\cos\theta = 1/3$ and showing the rejection of $\cos\theta = -1.5$. M1: Substitutes $\cos\theta = 1/3$ into the polar equation to find $r = 8a$. A1: Correctly states both pairs of polar coordinates: $(8a, \pm\arccos(1/3))$ (allow equivalent exact expressions). Part (iii): [5 marks] M1: States and uses the polar area formula $\frac{1}{2}\int_{-\pi}^{\pi} r^2 \mathrm{d}\theta$ (or $\int_{0}^{\pi} r^2 \mathrm{d}\theta$). M1: Correctly expands $(7+3\cos\theta)^2$ and applies the identity $\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)$. A1: Obtains the correct integrated expression: $\frac{107}{2}\theta + 42\sin\theta + \frac{9}{4}\sin 2\theta$ (or equivalent). M1: Substitutes the limits correctly. A1: Obtains the final exact area $\frac{107}{2}\pi a^2$ (or $53.5\pi a^2$).

題目 7 · subjective

13 分

The curve $ C $ has equation
\[ y = \frac{x^2 - 3x + 6}{x - 1} \]

(i) Find the equations of the asymptotes of $ C $. [3]

(ii) Find the coordinates of the stationary points of $ C $. [4]

(iii) Sketch $ C $, showing the coordinates of the stationary points, the $ y $-intercept, and the equations of the asymptotes. [3]

(iv) By considering the intersection of $ C $ with the line $ y = k $, or otherwise, find the set of values of $ k $ for which the equation
\[ \frac{x^2 - 3x + 6}{x - 1} = k \]
has no real roots. [3]

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解題

(i) By algebraic division:
\[ y = \frac{x(x - 1) - 2(x - 1) + 4}{x - 1} = x - 2 + \frac{4}{x - 1} \]
As $ x \to \pm\infty $, \\frac{4}{x - 1} \\to 0 \), so the oblique asymptote is $ y = x - 2 $.
The denominator is zero when $ x = 1 $ and the numerator is non-zero, so the vertical asymptote is $ x = 1 $.

(ii) Differentiating $ y = x - 2 + 4(x - 1)^{-1} $ with respect to $ x $:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x - 1)^2} \]
Setting $ \frac{\mathrm{d}y}{\mathrm{d}x} = 0 $:
\[ 1 - \frac{4}{(x - 1)^2} = 0 \implies (x - 1)^2 = 4 \implies x - 1 = \pm 2 \]
This gives $ x = 3 $ or $ x = -1 $.
For $ x = 3 $: $ y = 3 - 2 + \frac{4}{3 - 1} = 3 $.
For $ x = -1 $: $ y = -1 - 2 + \frac{4}{-1 - 1} = -5 $.
Thus, the stationary points are $ (3, 3) $ and $ (-1, -5) $.

(iii) The sketch of $ C $ should feature:
- Two branches of the curve (one for $ x > 1 $ with a local minimum at $ (3, 3) $, and one for $ x < 1 $ with a local maximum at $ (-1, -5) $).
- The vertical asymptote $ x = 1 $ and oblique asymptote $ y = x - 2 $ drawn as dashed lines.
- The $ y $-intercept labeled at $ (0, -6) $.

(iv) Method 1 (Graphical/Range):
From the sketch, the local maximum value is $ -5 $ and the local minimum value is $ 3 $.
Therefore, the range of the function is $ y \le -5 $ and $ y \ge 3 $.
For the line $ y = k $ to have no intersection with the curve $ C $ (and thus no real roots to the equation), $ k $ must lie strictly between these two regions:
\[ -5 < k < 3 \]

Method 2 (Algebraic):
Rearranging the equation:
\[ x^2 - 3x + 6 = k(x - 1) \implies x^2 - (3 + k)x + (6 + k) = 0 \]
For no real roots, the discriminant $ \Delta < 0 $:
\[ (3 + k)^2 - 4(6 + k) < 0 \]
\[ k^2 + 6k + 9 - 24 - 4k < 0 \]
\[ k^2 + 2k - 15 < 0 \]
\[ (k + 5)(k - 3) < 0 \]
This yields $ -5 < k < 3 $.

評分準則

(i)
M1: Attempt algebraic division or express the fraction in the form $ ax + b + \frac{c}{x-1} $.
A1: Correct oblique asymptote: $ y = x - 2 $.
A1: Correct vertical asymptote: $ x = 1 $.

(ii)
M1: Attempt differentiation of the function (using quotient rule or from the simplified form).
A1: Correct derivative, e.g., $ \frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x-1)^2} $ or $ \frac{x^2 - 2x - 3}{(x-1)^2} $.
M1: Set $ \frac{\mathrm{d}y}{\mathrm{d}x} = 0 $ and solve the resulting quadratic equation for $ x $.
A1: Correct stationary points: $ (3, 3) $ and $ (-1, -5) $.

(iii)
B1: Sketch both branches of the curve in the correct quadrants relative to the asymptotes.
B1: Draw and label both asymptotes with their correct equations.
B1: Label the stationary points $ (3, 3) $, $ (-1, -5) $ and the $ y $-intercept $ (0, -6) $.

(iv)
M1: Formulate a quadratic equation in $ x $ and find its discriminant, or utilize the stationary values from part (ii) to set up an inequality.
M1: Correctly solve the resulting quadratic inequality in $ k $ (e.g., $ k^2 + 2k - 15 < 0 $).
A1: Obtain the correct interval: $ -5 < k < 3 $ (or equivalent interval notation).

題目 8 · subjective

15 分

The lines $l_1$ and $l_2$ have vector equations:

$l_1: \mathbf{r} = \mathbf{i} + 4\mathbf{j} - \mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j})$

$l_2: \mathbf{r} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k} + \mu(\mathbf{j} + \mathbf{k})$

(a) Find the shortest distance between $l_1$ and $l_2$. [5]

(b) Find the Cartesian equation of the plane $\Pi_1$ which contains $l_1$ and is parallel to $l_2$. [3]

(c) Find the Cartesian equation of the plane $\Pi_2$ which contains $l_1$ and is perpendicular to $\Pi_1$. [4]

(d) Find the acute angle between the line $l_2$ and the plane $\Pi_2$, giving your answer to the nearest $0.1^\circ$. [3]

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解題

(a) To find the shortest distance between the skew lines, we first find a vector perpendicular to both direction vectors, $\mathbf{u} = 2\mathbf{i} + \mathbf{j}$ and $\mathbf{v} = \mathbf{j} + \mathbf{k}$:

$\mathbf{n} = \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \mathbf{i}(1-0) - \mathbf{j}(2-0) + \mathbf{k}(2-0) = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$

Now, we find its magnitude:

$|\mathbf{n}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3$

Let $A(1, 4, -1)$ be a point on $l_1$ and $B(2, 1, 3)$ be a point on $l_2$. The displacement vector $\mathbf{AB}$ is:

$\mathbf{AB} = (2-1)\mathbf{i} + (1-4)\mathbf{j} + (3 - (-1))\mathbf{k} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$

The shortest distance $d$ is the projection of $\mathbf{AB}$ onto the common perpendicular $\mathbf{n}$:

$d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|(1)(1) + (-3)(-2) + (4)(2)|}{3} = \frac{|1 + 6 + 8|}{3} = \frac{15}{3} = 5$

(b) The plane $\Pi_1$ contains $l_1$ and is parallel to $l_2$, so its normal is parallel to the common perpendicular $\mathbf{n}_1 = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$.

Since $\Pi_1$ contains $l_1$, it passes through the point $A(1, 4, -1)$.

The equation of the plane is:

$1(x-1) - 2(y-4) + 2(z+1) = 0$

$x - 2y + 2z - 1 + 8 + 2 = 0 \implies x - 2y + 2z = -9$

(c) The plane $\Pi_2$ contains $l_1$ and is perpendicular to $\Pi_1$. Thus, the normal to $\Pi_2$, which we call $\mathbf{n}_2$, is perpendicular to the direction vector of $l_1$ ($\mathbf{u} = 2\mathbf{i} + \mathbf{j}$) and the normal of $\Pi_1$ ($\mathbf{n}_1 = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$):

$\mathbf{n}_2 = \mathbf{u} \times \mathbf{n}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 0 \\ 1 & -2 & 2 \end{vmatrix} = \mathbf{i}(2-0) - \mathbf{j}(4-0) + \mathbf{k}(-4-1) = 2\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}$

Since $\Pi_2$ contains $l_1$, it also passes through $A(1, 4, -1)$. The equation of $\Pi_2$ is:

$2(x-1) - 4(y-4) - 5(z+1) = 0$

$2x - 4y - 5z - 2 + 16 - 5 = 0 \implies 2x - 4y - 5z = -9$

(d) The angle $\theta$ between the line $l_2$ (with direction $\mathbf{v} = \mathbf{j} + \mathbf{k}$) and the plane $\Pi_2$ (with normal $\mathbf{n}_2 = 2\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}$) satisfies:

$\sin\theta = \frac{|\mathbf{v} \cdot \mathbf{n}_2|}{|\mathbf{v}| |\mathbf{n}_2|}$

First, calculate the scalar product and magnitudes:

$\mathbf{v} \cdot \mathbf{n}_2 = (0)(2) + (1)(-4) + (1)(-5) = -9$

$|\mathbf{v}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$

$|\mathbf{n}_2| = \sqrt{2^2 + (-4)^2 + (-5)^2} = \sqrt{4 + 16 + 25} = \sqrt{45} = 3\sqrt{5}$

Thus:

$\sin\theta = \frac{|-9|}{\sqrt{2} \times 3\sqrt{5}} = \frac{9}{3\sqrt{10}} = \frac{3}{\sqrt{10}}$

$\theta = \arcsin\left(\frac{3}{\sqrt{10}}\right) \approx 71.565^\circ$

To the nearest $0.1^\circ$, $\theta = 71.6^\circ$.

評分準則

(a)
M1: For attempting the cross product of the direction vectors of $l_1$ and $l_2$.
A1: For obtaining correct normal vector $\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$ (or scalar multiple).
M1: For finding a vector connecting a point on each line, e.g., $\mathbf{AB} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$, and attempting the dot product with their normal vector.
M1: For using the correct projection formula $d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}$.
A1: For obtaining the correct distance of 5.

(b)
M1: For identifying that the normal vector of $\Pi_1$ is parallel to the common perpendicular found in part (a).
M1: For substituting a point on $l_1$ (e.g. $(1, 4, -1)$) into the equation $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$.
A1: For obtaining the correct Cartesian equation $x - 2y + 2z = -9$ (or equivalent).

(c)
M1: For realizing that the normal of $\Pi_2$ is perpendicular to the direction vector of $l_1$ and the normal of $\Pi_1$.
M1: For attempting the cross product of $2\mathbf{i} + \mathbf{j}$ and $\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$.
A1: For obtaining the correct normal vector $2\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}$ (or scalar multiple).
A1: For obtaining the correct Cartesian equation $2x - 4y - 5z = -9$ (or equivalent).

(d)
M1: For attempting to use the formula $\sin\theta = \frac{|\mathbf{v} \cdot \mathbf{n}_2|}{|\mathbf{v}| |\mathbf{n}_2|}$ with their direction vector of $l_2$ and normal of $\Pi_2$.
A1: For finding the correct value of $\sin\theta = \frac{3}{\sqrt{10}}$ (or equivalent).
A1: For obtaining the correct angle $71.6^\circ$ (allow $1.25$ radians if specified, otherwise degrees as standard).

卷二: Further Pure Mathematics 2

Answer all questions. Show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

21 題目 · 195 分

題目 1 · short_answer

4 分

Solve the equation $3\cosh x - 2\sinh x = 3$, giving your answers in exact logarithmic form.

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解題

Substitute the exponential definitions of $\cosh x$ and $\sinh x$:
$\cosh x = \frac{e^x + e^{-x}}{2}$, $\sinh x = \frac{e^x - e^{-x}}{2}$.

Substitute these into the given equation:
$3\left(\frac{e^x + e^{-x}}{2}\right) - 2\left(\frac{e^x - e^{-x}}{2}\right) = 3$.

Multiply by 2 to clear the denominators:
$3(e^x + e^{-x}) - 2(e^x - e^{-x}) = 6$.

Expand and collect terms:
$e^x + 5e^{-x} = 6$.

Multiply by $e^x$ to form a quadratic equation:
$(e^x)^2 - 6e^x + 5 = 0$.

Letting $u = e^x$, we have:
$u^2 - 6u + 5 = 0 \implies (u - 1)(u - 5) = 0$.

Thus $u = 1$ or $u = 5$.

Since $u = e^x$:
For $u = 1$, $e^x = 1 \implies x = 0$.
For $u = 5$, $e^x = 5 \implies x = \ln 5$.

評分準則

M1: For substituting the exponential definitions of $\cosh x$ and $\sinh x$ into the equation.
A1: For obtaining a correct quadratic equation in terms of $e^x$, such as $(e^x)^2 - 6e^x + 5 = 0$.
M1: For solving their quadratic equation to find two positive values for $e^x$.
A1: For both correct exact solutions $x = 0$ and $x = \ln 5$.

題目 2 · short_answer

4 分

Solve the equation $3\cosh x - 2\sinh x = 3$, giving your answers in exact logarithmic form.

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解題

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題目 3 · Differentiation and Curves

6 分

Use Leibniz's theorem to find the exact value of $\frac{\text{d}^8}{\text{d}x^8}\left(x^2 \cosh(2x)\right)$ at $x = 0$.

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解題

Let $y = x^2 \cosh(2x)$. We define $u = \cosh(2x)$ and $v = x^2$. By Leibniz's theorem, the $n$th derivative of a product is given by $y^{(n)} = \sum_{r=0}^n \binom{n}{r} u^{(n-r)} v^{(r)}$. The derivatives of $v = x^2$ are: $v' = 2x$, $v'' = 2$, and $v^{(r)} = 0$ for all $r \ge 3$. Applying this to the 8th derivative ($n = 8$), we get: $y^{(8)} = \binom{8}{0} u^{(8)} v + \binom{8}{1} u^{(7)} v' + \binom{8}{2} u^{(6)} v''$. Substituting the derivatives of $v$: $y^{(8)} = x^2 u^{(8)} + 8(u^{(7)})(2x) + 28(u^{(6)})(2) = x^2 u^{(8)} + 16x u^{(7)} + 56 u^{(6)}$. Evaluating at $x = 0$, we see that the first two terms vanish: $y^{(8)}(0) = 56 u^{(6)}(0)$. To find $u^{(6)}$, we differentiate $u = \cosh(2x)$ successively: $u' = 2\sinh(2x)$, $u'' = 4\cosh(2x)$, $u''' = 8\sinh(2x)$, and in general, the even derivatives of $u$ are $u^{(2k)} = 2^{2k}\cosh(2x)$. For $2k = 6$, we have $u^{(6)} = 2^6\cosh(2x) = 64\cosh(2x)$. At $x = 0$, this gives $u^{(6)}(0) = 64\cosh(0) = 64$. Therefore, $y^{(8)}(0) = 56 \times 64 = 3584$.

評分準則

M1: State or apply Leibniz's theorem for the 8th derivative of the product. A1: Obtain the simplified expression $y^{(8)} = x^2 u^{(8)} + 16x u^{(7)} + 56 u^{(6)}$ (or equivalent). M1: Substitute $x = 0$ to obtain $y^{(8)}(0) = 56 u^{(6)}(0)$. M1: Successively differentiate $u = \cosh(2x)$ to find the form of the 6th derivative. A1: Find $u^{(6)}(0) = 64$. A1: Compute the final correct value of 3584.

題目 4 · Differentiation and Curves

6 分

Use Leibniz's theorem to find the exact value of $\frac{\text{d}^8}{\text{d}x^8}\left(x^2 \cosh(2x)\right)$ at $x = 0$.

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解題

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題目 5 · structural

8 分

The curve $C$ has parametric equations
$$x = e^t \cos t, \quad y = e^t \sin t, \quad \text{for } 0 \le t \le \frac{1}{2}\pi.$$

Find the exact area of the surface of revolution generated when $C$ is rotated through $2\pi$ radians about the $x$-axis.

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解題

We use the formula for the surface area of revolution about the $x$-axis:
$$S = 2\pi \int_{a}^{b} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$$

First, find the derivatives of $x$ and $y$ with respect to $t$:
$$\frac{dx}{dt} = e^t \cos t - e^t \sin t = e^t(\cos t - \sin t)$$
$$\frac{dy}{dt} = e^t \sin t + e^t \cos t = e^t(\sin t + \cos t)$$

Next, calculate the term under the square root:
$$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = e^{2t}(\cos^2 t - 2\sin t\cos t + \sin^2 t) + e^{2t}(\sin^2 t + 2\sin t\cos t + \cos^2 t)$$
$$= e^{2t}(1 - 2\sin t\cos t + 1 + 2\sin t\cos t) = 2e^{2t}$$

Thus, the element of arc length is:
$$\frac{ds}{dt} = \sqrt{2e^{2t}} = \sqrt{2} e^t$$

Substitute $y$ and $\frac{ds}{dt}$ into the surface area formula:
$$S = 2\pi \int_{0}^{\pi/2} (e^t \sin t)(\sqrt{2} e^t) dt = 2\sqrt{2}\pi \int_{0}^{\pi/2} e^{2t} \sin t dt$$

To evaluate the integral $I = \int e^{2t} \sin t dt$, we use integration by parts:
Let $u = \sin t$ and $dv = e^{2t} dt$, so $du = \cos t dt$ and $v = \frac{1}{2}e^{2t}$.
$$I = \frac{1}{2}e^{2t} \sin t - \frac{1}{2} \int e^{2t} \cos t dt$$

Apply integration by parts again to the second integral:
Let $u = \cos t$ and $dv = e^{2t} dt$, so $du = -\sin t dt$ and $v = \frac{1}{2}e^{2t}$.
$$\int e^{2t} \cos t dt = \frac{1}{2}e^{2t} \cos t - \int \frac{1}{2}e^{2t} (-\sin t) dt = \frac{1}{2}e^{2t} \cos t + \frac{1}{2}I$$

Substitute this back into the expression for $I$:
$$I = \frac{1}{2}e^{2t} \sin t - \frac{1}{2}\left(\frac{1}{2}e^{2t} \cos t + \frac{1}{2}I\right)$$
$$I = \frac{1}{2}e^{2t} \sin t - \frac{1}{4}e^{2t} \cos t - \frac{1}{4}I$$
$$\frac{5}{4}I = \frac{1}{4}e^{2t}(2\sin t - \cos t) \implies I = \frac{1}{5}e^{2t}(2\sin t - \cos t)$$

Now, evaluate the limits from $0$ to $\frac{1}{2}\pi$:
$$\left[\frac{1}{5}e^{2t}(2\sin t - \cos t)\right]_{0}^{\pi/2} = \frac{1}{5}e^{\pi}(2\sin(\pi/2) - \cos(\pi/2)) - \frac{1}{5}e^{0}(2\sin(0) - \cos(0))$$
$$= \frac{1}{5}e^{\pi}(2(1) - 0) - \frac{1}{5}(1)(0 - 1)$$
$$= \frac{2}{5}e^{\pi} + \frac{1}{5} = \frac{2e^{\pi} + 1}{5}$$

Finally, multiply by the constant $2\sqrt{2}\pi$ to find the surface area $S$:
$$S = 2\sqrt{2}\pi \left(\frac{2e^{\pi} + 1}{5}\right) = \frac{2\sqrt{2}\pi}{5}(2e^{\pi} + 1)$$"

評分準則

**M1**: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ using the product rule.
**A1**: Obtain correct derivatives: $\frac{dx}{dt} = e^t(\cos t - \sin t)$ and $\frac{dy}{dt} = e^t(\sin t + \cos t)$.
**M1**: Substitute into $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$ and simplify.
**A1**: Obtain $\sqrt{2}e^t$ (or equivalent exact simplified form).
**M1**: Set up correct integral for surface area: $S = 2\pi \int_{0}^{\pi/2} (e^t \sin t)(\sqrt{2} e^t) dt$.
**M1**: Use integration by parts twice to integrate $e^{2t} \sin t$ (or state and apply the standard formula correctly).
**A1**: Obtain correct integrated term $\frac{1}{5}e^{2t}(2\sin t - \cos t)$ and evaluate limits correctly to get $\frac{2e^\pi + 1}{5}$.
**A1**: Obtain the final exact surface area $\frac{2\sqrt{2}\pi}{5}(2e^\pi + 1)$ (or any algebraically equivalent form).

題目 6 · 9 marks

9 分

The matrix $ A $ is given by
\[ A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 1 & -1 \\ 0 & 2 & 4 \end{pmatrix}. \]

(i) Find the characteristic equation of $ A $ and state how the Cayley-Hamilton Theorem is used to show that
\[ A^3 - 7A^2 + 16A - 12I = 0. \]
[3]

(ii) Hence, find $ A^{-1} $ in the form $ pA^2 + qA + rI $, where $ p, q, r $ are constants to be determined.
[2]

(iii) Express $ A^4 $ in the form $ aA^2 + bA + cI $, where $ a, b, c $ are constants to be determined.
[4]

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解題

**Part (i)**
To find the characteristic equation of $ A $, we set $ \det(A - \lambda I) = 0 $:
\[ \det(A - \lambda I) = \begin{vmatrix} 2-\lambda & 1 & 0 \\ 0 & 1-\lambda & -1 \\ 0 & 2 & 4-\lambda \end{vmatrix} = 0 \]
Expanding along the first column:
\[ (2-\lambda) \left[ (1-\lambda)(4-\lambda) - (-1)(2) \right] = 0 \]
\[ (2-\lambda) \left[ \lambda^2 - 5\lambda + 4 + 2 \right] = 0 \]
\[ (2-\lambda)(\lambda^2 - 5\lambda + 6) = 0 \]
\[ 2\lambda^2 - 10\lambda + 12 - \lambda^3 + 5\lambda^2 - 6\lambda = 0 \]
\[ -\lambda^3 + 7\lambda^2 - 16\lambda + 12 = 0 \]
Multiplying by $-1$, we obtain the characteristic equation:
\[ \lambda^3 - 7\lambda^2 + 16\lambda - 12 = 0 \]
By the Cayley-Hamilton Theorem, any square matrix satisfies its own characteristic equation. Replacing $ \lambda $ with $ A $ and the constant term $ 12 $ with $ 12I $ yields:
\[ A^3 - 7A^2 + 16A - 12I = 0 \]

**Part (ii)**
Since $ \det(A) = 12 \ne 0 $, the inverse matrix $ A^{-1} $ exists. Multiplying the relation from (i) by $ A^{-1} $:
\[ A^{-1}(A^3 - 7A^2 + 16A - 12I) = 0 \]
\[ A^2 - 7A + 16I - 12A^{-1} = 0 \]
Rearranging for $ A^{-1} $:
\[ 12A^{-1} = A^2 - 7A + 16I \]
\[ A^{-1} = \frac{1}{12}A^2 - \frac{7}{12}A + \frac{4}{3}I \]
Thus, the constants are $ p = \frac{1}{12} $, $ q = -\frac{7}{12} $, and $ r = \frac{4}{3} $.

**Part (iii)**
From the Cayley-Hamilton relation, we can write:
\[ A^3 = 7A^2 - 16A + 12I \]
Multiplying this entire equation by $ A $:
\[ A^4 = 7A^3 - 16A^2 + 12A \]
Now, substitute the expression for $ A^3 $ back into this equation:
\[ A^4 = 7(7A^2 - 16A + 12I) - 16A^2 + 12A \]
\[ A^4 = 49A^2 - 112A + 84I - 16A^2 + 12A \]
\[ A^4 = (49 - 16)A^2 + (-112 + 12)A + 84I \]
\[ A^4 = 33A^2 - 100A + 84I \]
Thus, the constants are $ a = 33 $, $ b = -100 $, and $ c = 84 $.

評分準則

**Part (i)**
* **M1**: For attempting to find the determinant $ \det(A - \lambda I) = 0 $.
* **A1**: For obtaining the correct characteristic equation $ \lambda^3 - 7\lambda^2 + 16\lambda - 12 = 0 $.
* **B1**: For clearly stating that by the Cayley-Hamilton Theorem, the matrix $ A $ satisfies its characteristic equation to show $ A^3 - 7A^2 + 16A - 12I = 0 $.

**Part (ii)**
* **M1**: For multiplying the equation by $ A^{-1} $ and attempting to isolate $ A^{-1} $.
* **A1**: For obtaining the correct expression $ A^{-1} = \frac{1}{12}A^2 - \frac{7}{12}A + \frac{4}{3}I $ (or equivalent values of $ p, q, r $).

**Part (iii)**
* **M1**: For multiplying the relation by $ A $ to get $ A^4 = 7A^3 - 16A^2 + 12A $.
* **A1**: For obtaining the correct expression for $ A^4 $ in terms of $ A^3 $.
* **M1**: For substituting $ A^3 = 7A^2 - 16A + 12I $ into the expression for $ A^4 $.
* **A1**: For obtaining the correct final expression $ A^4 = 33A^2 - 100A + 84I $ (or equivalent values of $ a, b, c $).

題目 7 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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解題

To find the complementary function, we solve the auxiliary equation: $m^2 + 4m + 4 = 0$, which gives $(m+2)^2 = 0$, so $m = -2$ (repeated root). Thus, the complementary function is $y_c = (A + Bx)e^{-2x}$. For the particular integral, we consider the two terms on the right-hand side separately. For the term $8e^{-2x}$, since $e^{-2x}$ and $xe^{-2x}$ are already in the complementary function, we try a particular integral of the form $y_{p1} = Cx^2 e^{-2x}$. Differentiating this gives $y_{p1}' = C(2x - 2x^2)e^{-2x}$ and $y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}$. Substituting these into the differential equation: $C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}$, which simplifies to $2Ce^{-2x} = 8e^{-2x}$, hence $C = 4$. For the term $25\cos x$, we try a particular integral of the form $y_{p2} = D\cos x + E\sin x$. Differentiating gives $y_{p2}' = -D\sin x + E\cos x$ and $y_{p2}'' = -D\cos x - E\sin x$. Substituting these into the differential equation: $(-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x$. Equating coefficients of $\cos x$ and $\sin x$ gives: $3D + 4E = 25$ and $-4D + 3E = 0$. Solving these simultaneous equations yields $D = 3$ and $E = 4$. Thus, the general solution is: $y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x$. Using the initial condition $y = 3$ when $x = 0$: $3 = A + 3$, which gives $A = 0$. Using the initial condition $\frac{dy}{dx} = 5$ when $x = 0$: First differentiate $y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x$ to get $\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x$. Substituting $x = 0$ and $\frac{dy}{dx} = 5$ gives $5 = B + 4$, so $B = 1$. Therefore, the particular solution is $y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x$.

評分準則

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function $y_c = (A+Bx)e^{-2x}$. M1: Correct form for the first part of the particular integral $y_{p1} = Cx^2 e^{-2x}$, differentiate and substitute. A1: Correct coefficient $C = 4$. M1: Correct form for the second part of the particular integral $y_{p2} = D\cos x + E\sin x$, differentiate and substitute. A1: Correct coefficients $D = 3$ and $E = 4$. A1: Correct general solution $y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x$. M1: Use initial condition $y(0)=3$ to find $A = 0$. M1: Differentiate and use initial condition $y'(0)=5$ to find $B = 1$. A1: Correct particular solution $y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x$.

題目 8 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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解題

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題目 9 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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解題

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題目 10 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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解題

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題目 11 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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解題

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題目 12 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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解題

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題目 13 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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題目 14 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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題目 15 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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題目 16 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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題目 17 · structured

10 分

Find the particular solution of the differential equation $\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 4y = 8e^{-2x} + 25\cos x$, given that $y = 3$ and $\frac{dy}{dx} = 5$ when $x = 0$.

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題目 18 · Structured

14 分

The function $ f $ is defined by $ f(x) = \frac{1}{2x+1} $ for $ x \ge 0 $.

(a) By drawing a sketch of the curve $ y = f(x) $ and using appropriate rectangles, show that for any positive integer $ n $,
\[ \int_{1}^{n+1} \frac{1}{2x+1} \, dx < \sum_{r=1}^n \frac{1}{2r+1} < \int_{0}^{n} \frac{1}{2x+1} \, dx \] [4]

(b) Hence show that
\[ \frac{1}{2} \ln \left( \frac{2n+3}{3} \right) < \sum_{r=1}^n \frac{1}{2r+1} < \frac{1}{2} \ln(2n+1) \] [4]

(c) By considering the function $ g(x) = \frac{1}{(2x+1)^2} $ for $ x \ge 1 $ and using a similar method, find a lower bound and an upper bound for the infinite sum
\[ S = \sum_{r=1}^{\infty} \frac{1}{(2r+1)^2} \]
by considering the first term of the sum separately. Show all your working and express your bounds as rational numbers in their simplest form. [6]

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解題

(a) Since $ f(x) = \frac{1}{2x+1} $ is a strictly decreasing function for $ x \ge 0 $:
- For any interval $ [r-1, r] $ where $ r \ge 1 $, the minimum value of $ f(x) $ occurs at the right endpoint, so $ f(r) < f(x) $ for all $ x \in [r-1, r) $. Taking the integral, we get:
\[ f(r) < \int_{r-1}^r f(x) \, dx \]
Summing this inequality from $ r = 1 $ to $ n $:
\[ \sum_{r=1}^n f(r) < \sum_{r=1}^n \int_{r-1}^r f(x) \, dx = \int_0^n f(x) \, dx \]
Substituting $ f(r) = \frac{1}{2r+1} $:
\[ \sum_{r=1}^n \frac{1}{2r+1} < \int_0^n \frac{1}{2x+1} \, dx \]

- Similarly, for any interval $ [r, r+1] $ where $ r \ge 1 $, the maximum value of $ f(x) $ occurs at the left endpoint, so $ f(x) < f(r) $ for all $ x \in (r, r+1] $. Taking the integral, we get:
\[ \int_r^{r+1} f(x) \, dx < f(r) \]
Summing this inequality from $ r = 1 $ to $ n $:
\[ \int_1^{n+1} f(x) \, dx = \sum_{r=1}^n \int_r^{r+1} f(x) \, dx < \sum_{r=1}^n f(r) \]
Substituting $ f(r) = \frac{1}{2r+1} $:
\[ \int_1^{n+1} \frac{1}{2x+1} \, dx < \sum_{r=1}^n \frac{1}{2r+1} \]
Combining both results gives the required inequality:
\[ \int_{1}^{n+1} \frac{1}{2x+1} \, dx < \sum_{r=1}^n \frac{1}{2r+1} < \int_{0}^{n} \frac{1}{2x+1} \, dx \]

(b) First, evaluate the lower bound integral:
\[ \int_1^{n+1} \frac{1}{2x+1} \, dx = \left[ \frac{1}{2} \ln(2x+1) \right]_1^{n+1} = \frac{1}{2} \ln(2n+3) - \frac{1}{2} \ln(3) = \frac{1}{2} \ln \left( \frac{2n+3}{3} \right) \]
Next, evaluate the upper bound integral:
\[ \int_0^n \frac{1}{2x+1} \, dx = \left[ \frac{1}{2} \ln(2x+1) \right]_0^n = \frac{1}{2} \ln(2n+1) - \frac{1}{2} \ln(1) = \frac{1}{2} \ln(2n+1) \]
Substituting these evaluated integrals back into the inequality from part (a) yields:
\[ \frac{1}{2} \ln \left( \frac{2n+3}{3} \right) < \sum_{r=1}^n \frac{1}{2r+1} < \frac{1}{2} \ln(2n+1) \]

(c) The function $ g(x) = \frac{1}{(2x+1)^2} $ is also strictly decreasing for $ x \ge 1 $. Using the same rectangular method for $ r \ge 2 $:
\[ \int_2^{\infty} g(x) \, dx < \sum_{r=2}^{\infty} g(r) < \int_1^{\infty} g(x) \, dx \]
We find the general integral:
\[ \int_k^{\infty} \frac{1}{(2x+1)^2} \, dx = \left[ -\frac{1}{2(2x+1)} \right]_k^{\infty} = 0 - \left( -\frac{1}{2(2k+1)} \right) = \frac{1}{2(2k+1)} \]
For the lower bound (with $ k = 2 $):
\[ \int_2^{\infty} \frac{1}{(2x+1)^2} \, dx = \frac{1}{2(5)} = \frac{1}{10} \]
For the upper bound (with $ k = 1 $):
\[ \int_1^{\infty} \frac{1}{(2x+1)^2} \, dx = \frac{1}{2(3)} = \frac{1}{6} \]
Thus, we have:
\[ \frac{1}{10} < \sum_{r=2}^{\infty} \frac{1}{(2r+1)^2} < \frac{1}{6} \]
Now, we add the first term of the sum (for $ r = 1 $), which is $ T_1 = \frac{1}{(2(1)+1)^2} = \frac{1}{9} $:
\[ \frac{1}{9} + \frac{1}{10} < S < \frac{1}{9} + \frac{1}{6} \]
\[ \frac{19}{90} < S < \frac{5}{18} \]
Hence, the lower bound is $ \frac{19}{90} $ and the upper bound is $ \frac{5}{18} $.

評分準則

(a)
- M1: Draw/sketch the curve y = 1/(2x+1) for x >= 0 with rectangles of width 1.
- A1: Explain or show that the sum of rectangle areas for r=1 to n on intervals [r-1, r] lies under the curve, hence establishing the upper bound.
- A1: Explain or show that the sum of rectangle areas for r=1 to n on intervals [r, r+1] lies over the curve, hence establishing the lower bound.
- A1: Combine inequalities clearly to obtain the final stated inequality.

(b)
- M1: Integrate 1/(2x+1) to obtain k ln(2x+1) with constant k = 1/2.
- A1: Show evaluation of lower bound integral at n+1 and 1 to obtain 1/2 ln((2n+3)/3).
- A1: Show evaluation of upper bound integral at n and 0 to obtain 1/2 ln(2n+1).
- A1: Conclude the final combined logarithmic inequality.

(c)
- M1: State or use the inequality: integral from 2 to infinity of g(x) dx < sum from r=2 to infinity of g(r) < integral from 1 to infinity of g(x) dx.
- M1: Integrate 1/(2x+1)^2 to obtain -1/(2(2x+1)) with correct constant.
- A1: Show that the lower integral evaluates to 1/10.
- A1: Show that the upper integral evaluates to 1/6.
- M1: Add the first term T_1 = 1/9 to both bounds.
- A1: Obtain the correct lower bound 19/90 and upper bound 5/18 as simplified rational numbers.

題目 19 · free_response

10 分

(a) Find the general solution of the differential equation
$$\cos x \frac{\mathrm{d}y}{\mathrm{d}x} + 2y\sin x = \cos^3 x \sin x$$
for $-\frac{\pi}{2} < x < \frac{\pi}{2}$, giving $y$ in terms of $x$.

(b) Given also that $y = 0$ when $x = \frac{\pi}{3}$:
(i) Find the particular solution of the differential equation, giving $y$ in terms of $x$.
(ii) Determine the maximum value of $y$ in the interval $-\frac{\pi}{2} < x < \frac{\pi}{2}$, justifying your answer.

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解題

**(a)**

First, we divide the given differential equation by $\cos x$ to express it in standard linear form:
$$\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\tan x = \cos^2 x \sin x$$

The integrating factor $I(x)$ is:
$$I(x) = e^{\int 2\tan x \,\mathrm{d}x} = e^{2\ln|\sec x|} = e^{\ln(\sec^2 x)} = \sec^2 x$$

Multiplying both sides of the differential equation by $\sec^2 x$ gives:
$$\sec^2 x \frac{\mathrm{d}y}{\mathrm{d}x} + 2y\sec^2 x \tan x = \cos^2 x \sin x \sec^2 x$$
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(y\sec^2 x\right) = \sin x$$

Integrating both sides with respect to $x$:
$$y\sec^2 x = \int \sin x \,\mathrm{d}x = -\cos x + C$$

Multiplying by $\cos^2 x$ to solve for $y$:
$$y = C\cos^2 x - \cos^3 x$$

**(b)(i)**

We use the boundary condition $y = 0$ when $x = \frac{\pi}{3}$:
$$0 = C\cos^2\left(\frac{\pi}{3}\right) - \cos^3\left(\frac{\pi}{3}\right)$$

Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$:
$$0 = C\left(\frac{1}{4}\right) - \frac{1}{8} \implies C = \frac{1}{2}$$

The particular solution is:
$$y = \frac{1}{2}\cos^2 x - \cos^3 x$$

**(b)(ii)**

Let $u = \cos x$. For $-\frac{\pi}{2} < x < \frac{\pi}{2}$, we have $0 < u \le 1$.

Substitute $u$ into the expression for $y$:
$$y = \frac{1}{2}u^2 - u^3$$

Differentiating $y$ with respect to $u$:
$$\frac{\mathrm{d}y}{\mathrm{d}u} = u - 3u^2 = u(1-3u)$$

Setting $\frac{\mathrm{d}y}{\mathrm{d}u} = 0$, and since $u > 0$, we get:
$$1 - 3u = 0 \implies u = \frac{1}{3}$$

To confirm the nature of this stationary point, find the second derivative:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}u^2} = 1 - 6u$$

At $u = \frac{1}{3}$:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}u^2} = 1 - 6\left(\frac{1}{3}\right) = -1 < 0$$

Since the second derivative is negative, $u = \frac{1}{3}$ corresponds to a local maximum.

Now we check the boundary points and the stationary point:
- As $u \to 0^+$, $y \to 0$.
- At $u = 1$, $y = \frac{1}{2}(1)^2 - (1)^3 = -\frac{1}{2}$.
- At $u = \frac{1}{3}$:
$$y = \frac{1}{2}\left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^3 = \frac{1}{18} - \frac{1}{27} = \frac{1}{54}$$

Thus, the maximum value of $y$ in the interval is $\frac{1}{54}$.

評分準則

**(a)**
* **M1**: For dividing by $\cos x$ to obtain the standard linear form $\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\tan x = \cos^2 x \sin x$.
* **M1**: For a correct attempt to find the integrating factor, evaluating $\int 2\tan x \,\mathrm{d}x$.
* **A1**: For obtaining the correct integrating factor $\sec^2 x$ (or $\frac{1}{\cos^2 x}$).
* **M1**: For writing the left-hand side as $\frac{\mathrm{d}}{\mathrm{d}x}(y \times \text{IF})$ and integrating the right-hand side correctly.
* **A1**: For the correct integration, leading to $y\sec^2 x = -\cos x + C$.
* **A1**: For expressing $y$ explicitly in terms of $x$ to get the general solution $y = C\cos^2 x - \cos^3 x$.

**(b)(i)**
* **M1**: For substituting $x = \frac{\pi}{3}$ and $y = 0$ into their general solution to find $C$.
* **A1**: For finding $C = \frac{1}{2}$ and stating the correct particular solution $y = \frac{1}{2}\cos^2 x - \cos^3 x$.

**(b)(ii)**
* **M1**: For a valid method to find the maximum value, e.g., substituting $u = \cos x$ and differentiating $y$ with respect to $u$, or differentiating $y$ directly with respect to $x$ and finding critical values.
* **A1**: For showing that the maximum occurs at $\cos x = \frac{1}{3}$ and finding the maximum value of $y = \frac{1}{54}$ with justification (such as checking boundaries or second derivative).

題目 20 · free_response

10 分

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題目 21 · structured

14 分

Answer all questions. Show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

(a) Let $z = \cos \theta + \text{i} \sin \theta$.

(i) Show that $z^n - z^{-n} = 2\text{i} \sin n\theta$. [1]

(ii) By expanding $(z - z^{-1})^6$, show that
\[\sin^6 \theta = -\frac{1}{32} \cos 6\theta + \frac{3}{16} \cos 4\theta - \frac{15}{32} \cos 2\theta + \frac{5}{16}.\] [5]

(b) Let $I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \\, \text{d}x$, where $n \ge 0$.

(i) Prove that, for $n \ge 2$,
\[I_n = n \left(\frac{\pi}{2}\right)^{n-1} - n(n-1) I_{n-2}.\] [4]

(ii) Find the exact value of $I_4$. [4]

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解題

(a)(i) Using de Moivre's theorem or Euler's formula:
$z^n = \cos n\theta + \text{i} \sin n\theta$
$z^{-n} = \cos(-n\theta) + \text{i} \sin(-n\theta) = \cos n\theta - \text{i} \sin n\theta$.

Subtracting the second equation from the first gives:
$z^n - z^{-n} = 2\text{i} \sin n\theta$.

(a)(ii) Using the binomial expansion:
$(z - z^{-1})^6 = z^6 - 6z^4 + 15z^2 - 20 + 15z^{-2} - 6z^{-4} + z^{-6}$
$= (z^6 + z^{-6}) - 6(z^4 + z^{-4}) + 15(z^2 + z^{-2}) - 20$.

Since $z^k + z^{-k} = 2 \cos k\theta$, we can substitute these into the expansion:
$(z - z^{-1})^6 = 2 \cos 6\theta - 12 \cos 4\theta + 30 \cos 2\theta - 20$.

From part (a)(i), we know that $z - z^{-1} = 2\text{i} \sin \theta$. Therefore:
$(z - z^{-1})^6 = (2\text{i} \sin \theta)^6 = 64\text{i}^6 \sin^6 \theta = -64 \sin^6 \theta$.

Equating the two expressions for $(z - z^{-1})^6$:
$-64 \sin^6 \theta = 2 \cos 6\theta - 12 \cos 4\theta + 30 \cos 2\theta - 20$.

Dividing both sides by $-64$ yields:
$\sin^6 \theta = -\frac{1}{32} \cos 6\theta + \frac{3}{16} \cos 4\theta - \frac{15}{32} \cos 2\theta + \frac{5}{16}$.

(b)(i) Given $I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \\, \text{d}x$.
Using integration by parts with $u = x^n$ and $\text{d}v = \sin x \\, \text{d}x$, we have:
$\text{d}u = n x^{n-1} \\, \text{d}x$ and $v = -\cos x$.

Applying the integration by parts formula:
$I_n = \left[-x^n \cos x\right]_0^{\frac{\pi}{2}} + n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \\, \text{d}x$.

For $n \ge 2$, the boundary term evaluates to:
$\left[-(\frac{\pi}{2})^n \cos(\frac{\pi}{2})\right] - \left[-0^n \cos 0\right] = 0 - 0 = 0$.

Thus, $I_n = n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \\, \text{d}x$.

We apply integration by parts a second time to this integral, with $u = x^{n-1}$ and $\text{d}v = \cos x \\, \text{d}x$:
$\text{d}u = (n-1) x^{n-2} \\, \text{d}x$ and $v = \sin x$.

This gives:
$\int_0^{\frac{\pi}{2}} x^{n-1} \cos x \\, \text{d}x = \left[x^{n-1} \sin x\right]_0^{\frac{\pi}{2}} - (n-1) \int_0^{\frac{\pi}{2}} x^{n-2} \sin x \\, \text{d}x$
$= \left(\frac{\pi}{2}\right)^{n-1} \sin(\frac{\pi}{2}) - 0 - (n-1) I_{n-2}$
$= \left(\frac{\pi}{2}\right)^{n-1} - (n-1) I_{n-2}$.

Substituting this back into the expression for $I_n$:
$I_n = n \left[ \left(\frac{\pi}{2}\right)^{n-1} - (n-1) I_{n-2} \right] = n \left(\frac{\pi}{2}\right)^{n-1} - n(n-1) I_{n-2}$.

(b)(ii) First, compute $I_0$:
$I_0 = \int_0^{\frac{\pi}{2}} \sin x \\, \text{d}x = [-\cos x]_0^{\frac{\pi}{2}} = -\cos(\frac{\pi}{2}) + \cos 0 = 1$.

Using the reduction formula with $n = 2$:
$I_2 = 2 \left(\frac{\pi}{2}\right)^1 - 2(1) I_0 = \pi - 2(1) = \pi - 2$.

Using the reduction formula with $n = 4$:
$I_4 = 4 \left(\frac{\pi}{2}\right)^3 - 4(3) I_2 = 4 \left(\frac{\pi^3}{8}\right) - 12 (\pi - 2) = \frac{\pi^3}{2} - 12\pi + 24$.

評分準則

(a)(i)
B1: For showing the correct derivation of $z^n - z^{-n} = 2\text{i} \sin n\theta$.

(a)(ii)
M1: For expanding $(z - z^{-1})^6$ using the binomial theorem.
M1: For pairing the terms and substituting $z^k + z^{-k} = 2\cos k\theta$.
M1: For identifying that $(z-z^{-1})^6 = -64 \sin^6 \theta$.
A1: For the correct unsimplified equation linking $\sin^6 \theta$ to multiple angles of cosine.
A1: For dividing by $-64$ and obtaining the given identity in the exact form requested.

(b)(i)
M1: For integrating by parts once and correctly evaluating the boundary term to 0.
A1: For obtaining the correct expression $I_n = n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \\, \text{d}x$.
M1: For integrating by parts a second time and evaluating the boundary term correctly.
A1: For completing the steps to establish the given reduction formula.

(b)(ii)
B1: For finding the correct value of $I_0 = 1$.
M1: For applying the reduction formula to find $I_2$.
A1: For obtaining the correct exact value of $I_2 = \pi - 2$.
A1: For applying the reduction formula a second time and obtaining $I_4 = \frac{\pi^3}{2} - 12\pi + 24$.

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