2022 AQA AS-Level Further Mathematics 7366 模擬試題連答案詳解

Let \(z = x + iy\). Then \(z - 2 = (x - 2) + iy\). The argument of \(z-2\) is given by \(\arg(z-2) = \arctan\left(\frac{y}{x-2}\right)\) for \(x > 2\). We are given that this is equal to \(\frac{\pi}{4}\), so \(\frac{y}{x-2} = \tan\left(\frac{\pi}{4}\right) = 1\). Thus, \(y = x - 2\) with the condition \(x > 2\) because the half-line originates at \((2,0)\) and goes into the first quadrant relative to that point.

B1: Correct Cartesian equation with the correct domain constraint.

For a matrix to be singular, its determinant must be equal to zero. Thus, \(\det(\mathbf{M}) = (k-2)(k+2) - (3)(4) = 0\). Simplifying this gives \(k^2 - 4 - 12 = 0\), which leads to \(k^2 - 16 = 0\). Solving for \(k\) gives \(k^2 = 16\), hence \(k = \pm 4\).

B1: Correct values of \(k\) identified.

From the relation between roots and coefficients of a cubic equation of the form \(ax^3 + bx^2 + cx + d = 0\), we have: \(\alpha + \beta + \gamma = -\frac{b}{a} = 3\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = 5\). Using the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\), we substitute the known values: \(\alpha^2 + \beta^2 + \gamma^2 = (3)^2 - 2(5) = 9 - 10 = -1\).

B1: Correct value of \(-1\) obtained.

For a matrix to be singular, its determinant must be equal to zero.

The determinant of \(\mathbf{M}\) is:
\(\det(\mathbf{M}) = k(k+3) - (2)(-1)\)
\(\det(\mathbf{M}) = k^2 + 3k + 2\)

Setting the determinant to zero to find the values of \(k\) for which the matrix is singular:
\(k^2 + 3k + 2 = 0\)

Factorising the quadratic equation:
\((k+1)(k+2) = 0\)

Thus, the possible values are:
\(k = -1\) or \(k = -2\)

Award 1 mark for the correct option (A).

**[1 Mark]**
- **B1**: Identifies \(k = -1\) or \(k = -2\) as the correct solutions.

**Step 1:** Base case
For \(n = 1\):
\(\text{LHS} = 1(1+3) = 4\)
\(\text{RHS} = \frac{1}{3}(1)(1+1)(1+5) = \frac{1}{3}(1)(2)(6) = 4\)
Since \(\text{LHS} = \text{RHS}\), the statement is true for \(n = 1\).

**Step 2:** Inductive step
Assume the statement is true for \(n = k\), where \(k \ge 1\) is an integer.
That is,
\(\sum_{r=1}^{k} r(r+3) = \frac{1}{3}k(k+1)(k+5)\)

We want to show that the statement is true for \(n = k+1\):
\(\sum_{r=1}^{k+1} r(r+3) = \frac{1}{3}(k+1)(k+2)(k+6)\)

Starting with the sum for \(n = k+1\):
\(\sum_{r=1}^{k+1} r(r+3) = \sum_{r=1}^{k} r(r+3) + (k+1)(k+1+3)\)
\(= \frac{1}{3}k(k+1)(k+5) + (k+1)(k+4)\)

Factor out \(\frac{1}{3}(k+1)\):
\(= \frac{1}{3}(k+1) \left[ k(k+5) + 3(k+4) \right]\)
\(= \frac{1}{3}(k+1) \left[ k^2 + 5k + 3k + 12 \right]\)
\(= \frac{1}{3}(k+1) \left[ k^2 + 8k + 12 \right]\)

Factorise the quadratic:
\(= \frac{1}{3}(k+1)(k+2)(k+6)\)

**Step 3:** Conclusion
Since the statement is true for \(n = 1\) and, if assumed true for \(n = k\), it is also true for \(n = k+1\), then by mathematical induction it is true for all positive integers \(n\).

**B1:** Verifies the base case \(n = 1\) is true by showing LHS = RHS = 4.
**M1:** Assumes statement is true for \(n = k\) and writes down the sum for \(n = k+1\) as \(\frac{1}{3}k(k+1)(k+5) + (k+1)(k+4)\).
**M1:** Factorises out common factor of \((k+1)\) or \(\frac{1}{3}(k+1)\).
**A1:** Obtains correct simplified quadratic \((k^2 + 8k + 12)\) or equivalent expression.
**A1:** Correctly factorises to show the result is \(\frac{1}{3}(k+1)(k+2)(k+6)\).
**C1:** Complete and clear explanation of mathematical induction with a correct concluding sentence.

(a) For \(\mathbf{M}\) to be singular, \(\det(\mathbf{M}) = 0\).
\(\det(\mathbf{M}) = a(a-1) - (2)(3) = a^2 - a - 6\)
Set \(\det(\mathbf{M}) = 0\):
\(a^2 - a - 6 = 0\)
\((a-3)(a+2) = 0\)
Thus, \(a = 3\) or \(a = -2\).

(b) When \(a = 4\), \(\mathbf{M} = \begin{pmatrix} 4 & 2 \\ 3 & 3 \end{pmatrix}\).
We have:
\(\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 3 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)
This gives:
\(x' = 4x + 2y\)
\(y' = 3x + 3y\)

Since \((x, y)\) lies on the line \(y = 2x - 1\), substitute this into the equations:
\(x' = 4x + 2(2x-1) = 8x - 2 \implies x = \frac{x'+2}{8}\)
\(y' = 3x + 3(2x-1) = 9x - 3\)

Substitute \(x\) into the expression for \(y'\):
\(y' = 9\left(\frac{x'+2}{8}\right) - 3\)
\(y' = \frac{9x' + 18 - 24}{8} = \frac{9}{8}x' - \frac{3}{4}\)

Thus, the equation of the image line is \(y = \frac{9}{8}x - \frac{3}{4}\).

**(a)**
**M1:** Sets \(\det(\mathbf{M}) = 0\) to form a quadratic equation in \(a\).
**A1:** Solves the quadratic to find \(a = 3\) and \(a = -2\).

**(b)**
**M1:** Writes down equations for \(x'\) and \(y'\) in terms of \(x\) and \(y\), and substitutes \(y = 2x - 1\).
**M1:** Expresses \(x\) in terms of \(x'\).
**A1:** Obtains a correct unsimplified equation linking \(x'\) and \(y'\), e.g., \(y' = 9\left(\frac{x'+2}{8}\right) - 3\).
**A1:** Correctly simplifies to find \(y = \frac{9}{8}x - \frac{3}{4}\).

(a)
Using the relationships between roots and coefficients:
(i) \(\alpha + \beta + \gamma = \frac{5}{2}\)
(ii) \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\)
(iii) \(\alpha\beta\gamma = \frac{3}{2}\)

(b)
Using the identity:
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)
\(\alpha^2 + \beta^2 + \gamma^2 = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\)

(c)
Let \(w = \frac{2}{x}\), so \(x = \frac{2}{w}\).
Substitute into the original equation:
\(2\left(\frac{2}{w}\right)^3 - 5\left(\frac{2}{w}\right)^2 + 4\left(\frac{2}{w}\right) - 3 = 0\)
\(\frac{16}{w^3} - \frac{20}{w^2} + \frac{8}{w} - 3 = 0\)

Multiply through by \(w^3\):
\(16 - 20w + 8w^2 - 3w^3 = 0\)
\(3w^3 - 8w^2 + 20w - 16 = 0\)

Using \(x\) as the variable, the equation with integer coefficients is:
\(3x^3 - 8x^2 + 20x - 16 = 0\)

**(a)**
**B1:** Correctly states \(\alpha + \beta + \gamma = \frac{5}{2}\).
**B1:** Correctly states \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\).
**B1:** Correctly states \(\alpha\beta\gamma = \frac{3}{2}\).

**(b)**
**M1:** Recalls or uses the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\).
**A1:** Obtains \(\frac{9}{4}\).

**(c)**
**M1:** Uses an appropriate substitution method such as \(x = \frac{2}{w}\).
**A1:** Substitutes into the cubic equation correctly.
**M1:** Multiplies through by \(w^3\) to clear denominators.
**A1:** Obtains \(3x^3 - 8x^2 + 20x - 16 = 0\) (or equivalent equation with integer coefficients).

(a) The equation \(\lvert z - (3 + 4\mathrm{i}) \rvert = 2\) represents a circle in the complex plane.
The centre of the circle is at the point representing \(3 + 4\mathrm{i}\) (coordinates \((3, 4)\)).
The radius of the circle is \(2\).
The sketch should show this circle in the first quadrant, not enclosing the origin.

(b) The distance from the origin \((0,0)\) to the centre of the circle \((3,4)\) is:
\(d = \sqrt{3^2 + 4^2} = 5\)

(i) The minimum value of \(\lvert z \rvert\) is the distance from the origin to the closest point on the circle:
\(\text{Minimum} = d - r = 5 - 2 = 3\)

(ii) The maximum value of \(\lvert z \rvert\) is the distance from the origin to the furthest point on the circle:
\(\text{Maximum} = d + r = 5 + 2 = 7\)

(c) The complex number \(z\) for which \(\lvert z \rvert\) is a minimum lies on the line segment connecting the origin to the centre of the circle \(3 + 4\mathrm{i}\).
The distance of this point from the origin is \(3\).
We can find this by scaling the vector from the origin to the centre of the circle by the ratio of the minimum distance to the distance of the centre:
\(z = 3 \times \frac{3 + 4\mathrm{i}}{5} = \frac{9}{5} + \frac{12}{5}\mathrm{i} = 1.8 + 2.4\mathrm{i}\)

(a)
M1: For drawing a circle with the centre in the first quadrant.
A1: For a fully correct diagram: circle with centre in the first quadrant, not enclosing the origin, clearly indicating centre \((3,4)\) or \(3+4\mathrm{i}\) and radius \(2\).

(b)
M1: For calculating the distance from the origin to the centre of the circle as \(5\).
A1: For both correct minimum value of \(3\) and maximum value of \(7\).

(c)
M1: For identifying that the point closest to the origin lies on the line from the origin to the centre, or writing a vector/complex equation like \(z = 3 \times \frac{3+4\mathrm{i}}{5}\).
M1: For a valid method to scale the vector/complex number to a magnitude of \(3\).
A1: For the correct complex number \(z = 1.8 + 2.4\mathrm{i}\) (or \(\frac{9}{5} + \frac{12}{5}\mathrm{i}\)).

(a) Comparing the given equation \(2x^3 - 5x^2 + 4x - 3 = 0\) with \(ax^3 + bx^2 + cx + d = 0\), we have \(a=2\), \(b=-5\), \(c=4\), \(d=-3\):
(i) \(\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}\)
(ii) \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{4}{2} = 2\)
(iii) \(\alpha\beta\gamma = -\frac{d}{a} = -\frac{-3}{2} = \frac{3}{2}\)

(b) Using the identity for sum of squares:
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)
\(\alpha^2 + \beta^2 + \gamma^2 = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\)

(c) Method 1: Since \(\alpha\), \(\beta\), and \(\gamma\) are roots of \(2x^3 - 5x^2 + 4x - 3 = 0\):
\(2\alpha^3 - 5\alpha^2 + 4\alpha - 3 = 0\)
\(2\beta^3 - 5\beta^2 + 4\beta - 3 = 0\)
\(2\gamma^3 - 5\gamma^2 + 4\gamma - 3 = 0\)

Summing these equations:
\(2(\alpha^3 + \beta^3 + \gamma^3) - 5(\alpha^2 + \beta^2 + \gamma^2) + 4(\alpha + \beta + \gamma) - 9 = 0\)

Substitute the values from (a) and (b):
\(2(\alpha^3 + \beta^3 + \gamma^3) - 5\left(\frac{9}{4}\right) + 4\left(\frac{5}{2}\right) - 9 = 0\)
\(2(\alpha^3 + \beta^3 + \gamma^3) - \frac{45}{4} + 10 - 9 = 0\)
\(2(\alpha^3 + \beta^3 + \gamma^3) - \frac{41}{4} = 0\)
\(2(\alpha^3 + \beta^3 + \gamma^3) = \frac{41}{4} \implies \alpha^3 + \beta^3 + \gamma^3 = \frac{41}{8}\)

Method 2: Using the identity:
\(\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - (\alpha\beta + \beta\gamma + \gamma\alpha))\)
\(\alpha^3 + \beta^3 + \gamma^3 - 3\left(\frac{3}{2}\right) = \left(\frac{5}{2}\right)\left(\frac{9}{4} - 2\right)\)
\(\alpha^3 + \beta^3 + \gamma^3 - \frac{9}{2} = \frac{5}{2} \left(\frac{1}{4}\right) = \frac{5}{8}\)
\(\alpha^3 + \beta^3 + \gamma^3 = \frac{5}{8} + \frac{36}{8} = \frac{41}{8}\)

(a)
B1: For correctly identifying at least two of the roots relationships: \(\alpha+\beta+\gamma = \frac{5}{2}\), \(\alpha\beta+\beta\gamma+\gamma\alpha = 2\), \(\alpha\beta\gamma = \frac{3}{2}\).
B1: For all three relationships correct.

(b)
M1: For using the identity \(\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)\).
A1: For the correct value \(\frac{9}{4}\).

(c)
M1: For setting up the sum of the cubic equations for each root: \(2S_3 - 5S_2 + 4S_1 - 9 = 0\), or using the algebraic identity for \(\alpha^3+\beta^3+\gamma^3\).
M1: For substituting their values of \(S_2\), \(S_1\) (or \(\alpha+\beta+\gamma\), etc.) into their expression.
A1: For completing the algebraic steps to show that \(\alpha^3+\beta^3+\gamma^3 = \frac{41}{8}\) with no errors.

(a) Using the standard integration result \(\int \frac{1}{x^2 + a^2} \, \mathrm{d}x = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + c\) with \(a = 2\):
\(\int_{2}^{b} \frac{k}{x^2 + 4} \, \mathrm{d}x = \left[ \frac{k}{2} \arctan\left(\frac{x}{2}\right) \right]_{2}^{b}\)
\(= \frac{k}{2} \arctan\left(\frac{b}{2}\right) - \frac{k}{2} \arctan\left(\frac{2}{2}\right)\)
Since \(\arctan(1) = \frac{\pi}{4}\), we have:
\(\int_{2}^{b} \frac{k}{x^2 + 4} \, \mathrm{d}x = \frac{k}{2} \arctan\left(\frac{b}{2}\right) - \frac{k}{2} \left(\frac{\pi}{4}\right) = \frac{k}{2} \arctan\left(\frac{b}{2}\right) - \frac{k\pi}{8}\) (as required).

(b) The value of the improper integral is defined as the limit as \(b \to \infty\):
\(\int_{2}^{\infty} \frac{k}{x^2 + 4} \, \mathrm{d}x = \lim_{b \to \infty} \left( \frac{k}{2} \arctan\left(\frac{b}{2}\right) - \frac{k\pi}{8} \right)\)
As \(b \to \infty\), \(\frac{b}{2} \to \infty\), so \(\arctan\left(\frac{b}{2}\right) \to \frac{\pi}{2}\).
Therefore:
\(\lim_{b \to \infty} \left( \frac{k}{2} \arctan\left(\frac{b}{2}\right) - \frac{k\pi}{8} \right) = \frac{k}{2} \left(\frac{\pi}{2}\right) - \frac{k\pi}{8} = \frac{k\pi}{4} - \frac{k\pi}{8} = \frac{k\pi}{8}\)
We are given that the improper integral is equal to \(3\pi\):
\(\frac{k\pi}{8} = 3\pi\)
Dividing both sides by \(\pi\) and multiplying by \(8\):
\(k = 24\)

(a)
M1: For applying the standard integration result for \(\int \frac{1}{x^2+a^2} \, \mathrm{d}x\) to obtain a term of the form \(A \arctan\left(\frac{x}{2}\right)\).
A1: For the correct indefinite integral \(\frac{k}{2} \arctan\left(\frac{x}{2}\right)\).
A1: For substituting the limits \(2\) and \(b\) and evaluating \(\arctan(1) = \frac{\pi}{4}\) to obtain the given expression \(\frac{k}{2} \arctan\left(\frac{b}{2}\right) - \frac{k\pi}{8}\).

(b)
M1: For expressing the improper integral as a limit as \(b \to \infty\).
M1: For evaluating the limit \(\lim_{b \to \infty} \arctan\left(\frac{b}{2}\right) = \frac{\pi}{2}\).
A1: For obtaining the simplified limit expression \(\frac{k\pi}{8}\).
A1: For setting \(\frac{k\pi}{8} = 3\pi\) and solving to find \(k = 24\).

To prove the statement by mathematical induction:

**Base case:**
For \( n = 1 \):
\( \text{LHS} = \sum_{r=1}^{1} r(r!) = 1(1!) = 1 \)
\( \text{RHS} = (1+1)! - 1 = 2! - 1 = 2 - 1 = 1 \)
Since \( \text{LHS} = \text{RHS} \), the statement is true for \( n = 1 \).

**Inductive step:**
Assume the statement is true for \( n = k \), where \( k \ge 1 \) is an integer. That is:
\[ \sum_{r=1}^{k} r(r!) = (k+1)! - 1 \]

Now, consider the statement for \( n = k + 1 \):
\[ \sum_{r=1}^{k+1} r(r!) = \left( \sum_{r=1}^{k} r(r!) \right) + (k+1)(k+1)! \]

Substitute the inductive assumption:
\[ \sum_{r=1}^{k+1} r(r!) = (k+1)! - 1 + (k+1)(k+1)! \]

Factorise \( (k+1)! \) from the first and third terms:
\[ = (k+1)! \left( 1 + (k+1) \right) - 1 \]
\[ = (k+1)! (k+2) - 1 \]
\[ = (k+2)! - 1 \]
\[ = ((k+1)+1)! - 1 \]

**Conclusion:**
Since the statement is true for \( n = 1 \), and if it is true for \( n = k \) then it is true for \( n = k+1 \), the statement is true by mathematical induction for all positive integers \( n \).

**M1**: Verifies the base case \( n = 1 \) correctly showing LHS = RHS = 1.
**M1**: Clearly states the inductive hypothesis (assumes the result is true for \( n = k \)).
**M1**: Expresses the sum up to \( k+1 \) as the sum up to \( k \) plus the \( (k+1) \)-th term.
**A1**: Substitutes the inductive hypothesis correctly to get \( (k+1)! - 1 + (k+1)(k+1)! \).
**M1**: Factorises \( (k+1)! \) to obtain \( (k+1)!(k+2) - 1 \).
**A1**: Simplifies to the required form \( (k+2)! - 1 \).
**A1**: Provides a complete and clear concluding statement linking all parts of the induction proof.

**(a) Find the value of \( k \):**
Let the roots in arithmetic progression be represented as:
\( \alpha = a - d \), \( \beta = a \), \( \gamma = a + d \)

Using the relation for the sum of the roots of a cubic equation:
\( \alpha + \beta + \gamma = -\frac{b}{a_{3}} \)
\( (a - d) + a + (a + d) = -\frac{-12}{4} \)
\( 3a = 3 \implies a = 1 \)

Since \( a = 1 \) is one of the roots (specifically \( \beta = 1 \)), it must satisfy the cubic equation:
\( 4(1)^3 - 12(1)^2 + k(1) - 6 = 0 \)
\( 4 - 12 + k - 6 = 0 \)
\( k - 14 = 0 \implies k = 14 \)

**(b) Solve the equation completely:**
Substitute \( k = 14 \) back into the cubic equation:
\( 4x^3 - 12x^2 + 14x - 6 = 0 \)
Divide the entire equation by 2:
\( 2x^3 - 6x^2 + 7x - 3 = 0 \)

Since \( x = 1 \) is a root, \( (x - 1) \) is a factor. Perform polynomial division:
\( 2x^3 - 6x^2 + 7x - 3 = (x - 1)(2x^2 - 4x + 3) = 0 \)

Now, solve the quadratic equation \( 2x^2 - 4x + 3 = 0 \) using the quadratic formula:
\( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(3)}}{2(2)} \)
\( x = \frac{4 \pm \sqrt{16 - 24}}{4} \)
\( x = \frac{4 \pm \sqrt{-8}}{4} \)
\( x = \frac{4 \pm 2\sqrt{2}i}{4} \)
\( x = 1 \pm \frac{\sqrt{2}}{2}i \)

Therefore, the three roots are:
\( x = 1 \), \( x = 1 + \frac{\sqrt{2}}{2}i \), and \( x = 1 - \frac{\sqrt{2}}{2}i \).

**M1**: Sets up roots as \( a-d, a, a+d \) (or equivalent) and uses sum of roots formula \( 3a = 3 \).
**A1**: Correctly deduces that one root is \( x = 1 \).
**M1**: Substitutes \( x = 1 \) into the cubic equation to set up an equation in terms of \( k \).
**A1**: Obtains \( k = 14 \).
**M1**: Divides the cubic polynomial by \( (x - 1) \) to find the quadratic factor \( 2x^2 - 4x + 3 \) (or \( 4x^2 - 8x + 6 \)).
**M1**: Solves the resulting quadratic equation using a valid method.
**A1**: Obtains the two complex roots \( 1 \pm \frac{\sqrt{2}}{2}i \) (or equivalent form like \( 1 \pm \frac{1}{\sqrt{2}}i \)) and lists all three roots.

The volume \( V \) of a solid of revolution generated by rotating a curve about the \( x \)-axis is given by:

\[ V = \pi \int_{a}^{b} y^2 \, dx \]

First, find \( y^2 \):
\[ y^2 = \left( \frac{2}{\sqrt{3x^2 + 4}} \right)^2 = \frac{4}{3x^2 + 4} \]

Now, substitute the limits \( x = 0 \) and \( x = 2 \) into the volume formula:
\[ V = \pi \int_{0}^{2} \frac{4}{3x^2 + 4} \, dx \]

Rearrange the integrand to match a standard form:
\[ V = \pi \int_{0}^{2} \frac{4}{3(x^2 + \frac{4}{3})} \, dx = \frac{4\pi}{3} \int_{0}^{2} \frac{1}{x^2 + \left(\frac{2}{\sqrt{3}}\right)^2} \, dx \]

Using the standard integration formula \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) \) where \( a = \frac{2}{\sqrt{3}} \):
\[ V = \frac{4\pi}{3} \left[ \frac{\sqrt{3}}{2} \arctan\left(\frac{\sqrt{3}x}{2}\right) \right]_{0}^{2} \]

Simplify the constant factor outside the brackets:
\[ V = \frac{2\sqrt{3}\pi}{3} \left[ \arctan\left(\frac{\sqrt{3}x}{2}\right) \right]_{0}^{2} \]

Now evaluate at the limits:
At \( x = 2 \):
\[ \arctan\left(\frac{2\sqrt{3}}{2}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3} \]

At \( x = 0 \):
\[ \arctan(0) = 0 \]

Therefore:
\[ V = \frac{2\sqrt{3}\pi}{3} \left( \frac{\pi}{3} - 0 \right) = \frac{2\sqrt{3}\pi^2}{9} \]

**M1**: States or applies the correct volume of revolution formula \( V = \pi \int y^2 \, dx \).
**A1**: Correctly squares the function to get \( y^2 = \frac{4}{3x^2 + 4} \).
**M1**: Identifies the integral as being of the form involving \( \arctan \) and identifies \( a = \frac{2}{\sqrt{3}} \) (or equivalent substitution method).
**A1**: Correctly integrates to get \( \frac{2\sqrt{3}}{3}\arctan\left(\frac{\sqrt{3}x}{2}\right) \) (ignoring constants/limits for this mark).
**M1**: Demonstrates correct substitution of limits \( x = 2 \) and \( x = 0 \) into their integrated expression.
**A1**: Evaluates \( \arctan(\sqrt{3}) = \frac{\pi}{3} \) correctly.
**A1**: Obtains the final exact volume \( \frac{2\sqrt{3}\pi^2}{9} \) (or equivalent exact simplified form).

(a) Let a point on the line \(y = mx\) be represented by the vector \(\begin{pmatrix} x \\ mx \end{pmatrix}\).

Under the transformation \(T\), this point is mapped to:
\(\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 3 & 8 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} 3x + 2mx \\ 3x + 8mx \end{pmatrix}\).

For the line \(y = mx\) to be invariant, any image point \(\begin{pmatrix} x' \\ y' \end{pmatrix}\) must also lie on the line \(y = mx\), which means \(y' = mx'\).

Substituting the expressions for \(x'\) and \(y'\):
\(3x + 8mx = m(3x + 2mx)\).

Since this must hold for any point on the line (excluding the origin where \(x=0\)), we can divide by \(x \neq 0\):
\(3 + 8m = m(3 + 2m)\)
\(3 + 8m = 3m + 2m^2\)
\(2m^2 - 5m - 3 = 0\).

Solving this quadratic equation for \(m\):
\((2m + 1)(m - 3) = 0\)
\(m = -\frac{1}{2}\) or \(m = 3\).

Thus, the equations of the two invariant lines are \(y = -\frac{1}{2}x\) and \(y = 3x\).

(b)
- An **invariant line** is a line where every point on the line maps to a point that is also on the same line (the line as a whole is mapped to itself).
- A **line of invariant points** is a line where every individual point on the line is mapped to itself (i.e. \(T(x,y) = (x,y)\) for all points on the line).

**Part (a)**
* **M1**: Sets up the matrix multiplication and obtains expressions for \(x'\) and \(y'\) in terms of \(x\) and \(m\).
* **M1**: Applies the condition for an invariant line, \(y' = mx'\), to form an equation in \(m\) and \(x\).
* **A1**: Correctly simplifies to the quadratic equation \(2m^2 - 5m - 3 = 0\).
* **M1**: Solves their quadratic equation by factoring or using the formula.
* **A1**: Obtains both correct equations: \(y = -\frac{1}{2}x\) and \(y = 3x\).

**Part (b)**
* **B1**: Explains clearly that for an invariant line, points on the line map to other points on the same line.
* **B1**: Explains clearly that for a line of invariant points, each individual point maps to itself.

Let the proposition \(P(n)\) be defined as:
\(\sum_{r=1}^n (3r-2)(3r+1) = n(3n^2 + 3n - 2)\).

**Base case:**
For \(n = 1\):
LHS \(= (3(1)-2)(3(1)+1) = 1 \times 4 = 4\).
RHS \(= 1(3(1)^2 + 3(1) - 2) = 1(3 + 3 - 2) = 4\).
Since LHS = RHS, the proposition \(P(1)\) is true.

**Inductive step:**
Assume that \(P(k)\) is true for some positive integer \(k\).
That is, \(\sum_{r=1}^k (3r-2)(3r+1) = k(3k^2 + 3k - 2)\).

We must show that \(P(k+1)\) is also true, i.e.,
\(\sum_{r=1}^{k+1} (3r-2)(3r+1) = (k+1)(3(k+1)^2 + 3(k+1) - 2)\).

First, expand the target RHS for \(n = k+1\):
\((k+1)(3(k^2 + 2k + 1) + 3k + 3 - 2) = (k+1)(3k^2 + 6k + 3 + 3k + 1) = (k+1)(3k^2 + 9k + 4)\).

Now, express the sum for \(n = k+1\) as:
\(\sum_{r=1}^{k+1} (3r-2)(3r+1) = \sum_{r=1}^k (3r-2)(3r+1) + (3(k+1)-2)(3(k+1)+1)\).

Substitute the inductive hypothesis:
\(= k(3k^2 + 3k - 2) + (3k+1)(3k+4)\).

Expand both parts:
\(= 3k^3 + 3k^2 - 2k + (9k^2 + 12k + 3k + 4)\)

Simplify the expression:
\(= 3k^3 + 12k^2 + 13k + 4\).

Now let us expand the target expression to show they are identical:
\((k+1)(3k^2 + 9k + 4) = 3k^3 + 9k^2 + 4k + 3k^2 + 9k + 4 = 3k^3 + 12k^2 + 13k + 4\).

Since the two expressions are identical, the statement holds for \(n = k+1\).

**Conclusion:**
Since \(P(1)\) is true, and if \(P(k)\) is true then \(P(k+1)\) is true, then by mathematical induction, the proposition is true for all positive integers \(n\).

* **B1**: Verifies the base case \(n = 1\) by showing LHS = RHS = 4.
* **M1**: States the inductive hypothesis clearly, assuming the statement is true for \(n = k\).
* **M1**: Writes down the sum for \(n = k+1\) as the sum to \(k\) terms plus the \((k+1)\)-th term: \(k(3k^2 + 3k - 2) + (3k+1)(3k+4)\).
* **A1**: Correctly expands the algebraic expression to obtain \(3k^3 + 12k^2 + 13k + 4\).
* **M1**: Expands the target RHS for \(n = k+1\), which is \((k+1)(3(k+1)^2 + 3(k+1) - 2)\).
* **A1**: Correctly shows that both expressions simplify to the same cubic expression, \(3k^3 + 12k^2 + 13k + 4\).
* **B1**: Provides a complete and correct concluding statement of induction, mentioning that \(P(1)\) is true, \(P(k) \implies P(k+1)\), and concluding it holds for all integers \(n \ge 1\).