2025 Cambridge IAL Mathematics - Further (9231) 模擬試題連答案詳解

Using partial fractions, the general term can be written as: \(\frac{2}{(2r+1)(2r+3)} = \frac{1}{2r+1} - \frac{1}{2r+3}\). Letting \(u_r = \frac{1}{2r+1}\), the sum becomes: \(\sum_{r=1}^{n} (u_r - u_{r+1}) = (u_1 - u_2) + (u_2 - u_3) + \dots + (u_n - u_{n+1}) = u_1 - u_{n+1} = \frac{1}{3} - \frac{1}{2n+3} = \frac{(2n+3) - 3}{3(2n+3)} = \frac{2n}{3(2n+3)}\).

M1: Expresses the general term as a difference of two fractions, e.g., \(\frac{1}{2r+1} - \frac{1}{2r+3}\). A1: Correctly writes down the sum showing cancellation of intermediate terms, leaving \(\frac{1}{3} - \frac{1}{2n+3}\). A1: Simplifies to a single fraction to obtain the final answer \(\frac{2n}{3(2n+3)}\).

First, find the sum of roots and sum of product of roots: \(\sum \alpha = 3\) and \(\sum \alpha\beta = 2\). We also have: \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 3^2 - 2(2) = 5\). Since \(\alpha, \beta, \gamma\) satisfy the original cubic equation, we have: \(\alpha^3 - 3\alpha^2 + 2\alpha - 5 = 0\), \(\beta^3 - 3\beta^2 + 2\beta - 5 = 0\), and \(\gamma^3 - 3\gamma^2 + 2\gamma - 5 = 0\). Summing these three equations yields: \(\sum \alpha^3 - 3\sum \alpha^2 + 2\sum \alpha - 15 = 0 \implies \sum \alpha^3 - 3(5) + 2(3) - 15 = 0 \implies \sum \alpha^3 - 15 + 6 - 15 = 0 \implies \sum \alpha^3 = 24\).

M1: Recalls or calculates \(\sum \alpha = 3\) and finds \(\sum \alpha^2 = 5\) using the identity \((\sum \alpha)^2 - 2\sum \alpha\beta\). M1: Sets up the relation \(\sum \alpha^3 - 3\sum \alpha^2 + 2\sum \alpha - 15 = 0\) by summing the cubic equations. A1: Obtains the correct value of 24.

Let \(H_n\) be the statement that \(8^n - 1\) is divisible by 7. For \(n = 1\): \(8^1 - 1 = 7\), which is divisible by 7, so \(H_1\) is true. Assume that \(H_k\) is true for some positive integer \(k\), so \(8^k - 1 = 7m\) for some integer \(m\). For \(n = k+1\): \(8^{k+1} - 1 = 8 \cdot 8^k - 1 = 8(7m + 1) - 1 = 56m + 8 - 1 = 56m + 7 = 7(8m + 1)\). Since \(8m + 1\) is an integer, \(8^{k+1} - 1\) is divisible by 7. Thus, \(H_{k+1}\) is true. Since \(H_1\) is true, and \(H_k \implies H_{k+1}\), by mathematical induction, \(8^n - 1\) is divisible by 7 for all positive integers \(n\).

B1: Verifies the base case \(n=1\) correctly. M1: Assumes true for \(n=k\) and correctly manipulates the algebraic expression for \(n=k+1\) to show divisibility by 7. A1: Completes the inductive step and gives a clear, logical concluding statement.

Using the characteristic equation or the Cayley-Hamilton theorem, any \(2 \times 2\) matrix \(\mathbf{A}\) satisfies \(\mathbf{A}^2 - \text{Tr}(\mathbf{A})\mathbf{A} + \det(\mathbf{A})\mathbf{I} = \mathbf{0}\). Here, \(\text{Tr}(\mathbf{A}) = 2 + 3 = 5\) and \(\det(\mathbf{A}) = (2)(3) - (k)(-1) = 6 + k\). Thus, \(\mathbf{A}^2 - 5\mathbf{A} + (6+k)\mathbf{I} = \mathbf{0}\). Comparing this with the given equation \(\mathbf{A}^2 - 5\mathbf{A} + 7\mathbf{I} = \mathbf{0}\), we have \(6 + k = 7 \implies k = 1\). Alternatively, computing \(\mathbf{A}^2\) directly gives \(\begin{pmatrix} 4 - k & 5k \\ -5 & 9 - k \end{pmatrix}\), so \(\mathbf{A}^2 - 5\mathbf{A} + 7\mathbf{I} = \begin{pmatrix} 1-k & 0 \\ 0 & 1-k \end{pmatrix} = \mathbf{0}\), yielding \(k=1\).

M1: Attempts to compute \(\mathbf{A}^2\) or uses the Cayley-Hamilton/characteristic equation approach. A1: Correctly expresses the elements of the matrix equation (e.g., getting the diagonal elements as \(1-k\) or finding \(\det(\mathbf{A}) = 6+k\)). A1: Solves to find \(k = 1\).

We use the relationships \(x = r\cos\theta\), \(y = r\sin\theta\), and \(r^2 = x^2 + y^2\). Multiplying the polar equation \(r = 4\cos\theta - 2\sin\theta\) by \(r\) gives \(r^2 = 4r\cos\theta - 2r\sin\theta\). Substituting the Cartesian variables yields \(x^2 + y^2 = 4x - 2y\). Rearranging to the requested form gives \(x^2 + y^2 - 4x + 2y = 0\).

M1: Multiplies the polar equation by \(r\) to obtain \(r^2 = 4r\cos\theta - 2r\sin\theta\). M1: Uses the standard substitution formulas \(r^2 = x^2 + y^2\), \(r\cos\theta = x\), and \(r\sin\theta = y\). A1: Obtains the correct Cartesian equation in the specified form: \(x^2 + y^2 - 4x + 2y = 0\).

The Cartesian equation of the plane is \(2x - y + 2z - 12 = 0\). The formula for the shortest distance from point \((x_0, y_0, z_0)\) to plane \(Ax + By + Cz + D = 0\) is \(d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\). Substituting \(P(1, 2, 3)\) gives \(d = \frac{|2(1) - (2) + 2(3) - 12|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{|2 - 2 + 6 - 12|}{\sqrt{9}} = \frac{|-6|}{3} = 2\).

M1: Converts the vector plane equation to Cartesian form \(2x - y + 2z - 12 = 0\) (or equivalent). M1: Applies the perpendicular distance formula correctly using the point \((1, 2, 3)\). A1: Obtains the correct distance of 2.

Setting the denominator to zero gives the vertical asymptote \(x = 2\). For the oblique asymptote, polynomial division of \(3x^2 - 5x + 2\) by \(x - 2\) gives \((3x + 1)\) with a remainder of \(4\). Thus, \(y = 3x + 1 + \frac{4}{x - 2}\). As \(x \to \pm \infty\), \(\frac{4}{x-2} \to 0\), so the oblique asymptote is \(y = 3x + 1\).

B1: Identifies the vertical asymptote as \(x = 2\). M1: Attempts algebraic long division or equivalent to write \(y\) in the form \(ax + b + \frac{c}{x-2}\). A1: Obtains the correct oblique asymptote equation \(y = 3x + 1\).

The matrix for reflection in the line \(y = -x\) is \(\mathbf{R} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\). The matrix for a stretch of factor 3 parallel to the \(y\)-axis is \(\mathbf{S} = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}\). The combined transformation represents reflection followed by stretch, given by \(\mathbf{S}\mathbf{R} = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix}\).

B1: Writes down the correct reflection matrix \(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\) (or shows equivalent effect on base vectors). M1: Multiplies the stretch matrix and the reflection matrix in the correct order: \(\mathbf{S}\mathbf{R}\). A1: Obtains the correct final matrix \(\begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix}\).

First, we prove the algebraic identity by combining the terms on the right-hand side over a common denominator: \( \frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{r^2 + 2r + 1 - r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2} \). Next, we use the method of differences to find the sum: \( S_n = \sum_{r=1}^{n} \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right) \). Writing out the terms: for \(r=1\) we have \( \frac{1}{1^2} - \frac{1}{2^2} \), for \(r=2\) we have \( \frac{1}{2^2} - \frac{1}{3^2} \), up to \(r=n\) where we have \( \frac{1}{n^2} - \frac{1}{(n+1)^2} \). Summing these terms leads to a telescoping cancellation where only the first and last terms remain: \( S_n = 1 - \frac{1}{(n+1)^2} \). As \(n \to \infty\), \( \frac{1}{(n+1)^2} \to 0 \), which gives the sum to infinity: \( S_{\infty} = 1 \). We want the difference between \(S_{\infty}\) and \(S_n\) to be less than \(10^{-4}\): \( |S_{\infty} - S_n| < 10^{-4} \implies \frac{1}{(n+1)^2} < 10^{-4} \implies (n+1)^2 > 10^4 \implies n+1 > 100 \implies n > 99 \). Since \(n\) is an integer, the least value is \(n = 100\).

M1: For combining the right-hand side of the identity over a common denominator. M1: For writing out the summation terms showing clear cancellation. A1: For obtaining the correct simplified sum \(S_n = 1 - \frac{1}{(n+1)^2}\). B1: For stating the correct sum to infinity \(S_{\infty} = 1\). M1: For setting up the inequality \( \frac{1}{(n+1)^2} < 10^{-4} \). A1: For solving the inequality to find \(n > 99\). A1: For concluding that the least integer value is \(n = 100\).

Let \(y = x^2\), so \(x = \sqrt{y}\). Substituting this into the cubic equation gives: \( y\sqrt{y} - 3y + 5\sqrt{y} - 2 = 0 \). Rearranging to isolate the square roots on one side: \( \sqrt{y}(y + 5) = 3y + 2 \). Squaring both sides: \( y(y + 5)^2 = (3y + 2)^2 \implies y(y^2 + 10y + 25) = 9y^2 + 12y + 4 \implies y^3 + 10y^2 + 25y = 9y^2 + 12y + 4 \). Collecting all terms on one side yields: \( y^3 + y^2 + 13y - 4 = 0 \). Let the roots of this new equation be \(u = \alpha^2\), \(v = \beta^2\), and \(w = \gamma^2\). From the relations between roots and coefficients: \( u + v + w = -1 \) and \( uv + vw + wu = 13 \). We want to find \(\alpha^4 + \beta^4 + \gamma^4 = u^2 + v^2 + w^2\). Using the algebraic identity: \( u^2 + v^2 + w^2 = (u + v + w)^2 - 2(uv + vw + wu) \). Substituting the values: \( u^2 + v^2 + w^2 = (-1)^2 - 2(13) = 1 - 26 = -25 \). Thus, \(\alpha^4 + \beta^4 + \gamma^4 = -25\).

M1: For substituting \(y = x^2\) or using relations between roots. M1: For isolating terms involving \(\sqrt{y}\). A1: For squaring both sides correctly. M1: For expanding and gathering terms in standard form. A1: For obtaining the correct cubic equation \(y^3 + y^2 + 13y - 4 = 0\). M1: For using the sum of roots and sum of product of roots of the new equation in the identity for sum of squares. A1: For obtaining \(-25\).

Let \(P(n)\) be the proposition that \(u_n = 2^{n+1} - 3^{n-1}\). For the base case \(n = 1\): the formula gives \(u_1 = 2^2 - 3^0 = 4 - 1 = 3\), which is correct. Assume \(P(k)\) is true for some positive integer \(k\), so \(u_k = 2^{k+1} - 3^{k-1}\). We want to prove \(P(k+1)\) is true, i.e., \(u_{k+1} = 2^{k+2} - 3^k\). Using the recurrence relation: \(u_{k+1} = 3u_k - 2^{k+1}\). Substituting the induction hypothesis: \(u_{k+1} = 3(2^{k+1} - 3^{k-1}) - 2^{k+1} = 3 \cdot 2^{k+1} - 3 \cdot 3^{k-1} - 2^{k+1}\). Grouping the powers of 2 gives: \(u_{k+1} = (3 - 1)2^{k+1} - 3^k = 2 \cdot 2^{k+1} - 3^k = 2^{k+2} - 3^k\). Thus, \(P(k+1)\) is true. By the principle of mathematical induction, the formula holds for all positive integers \(n\).

B1: For verifying the base case \(n = 1\) correctly. M1: For stating the inductive hypothesis clearly. M1: For substituting the inductive hypothesis into the recurrence relation for \(u_{k+1}\). A1: For expanding the expression correctly. M1: For grouping the terms involving powers of 2. A1: For showing the step-by-step algebra leads to \(2^{k+2} - 3^k\). B1: For a complete conclusion referencing the principle of mathematical induction.

(a) The system does not have a unique solution when the determinant of the coefficient matrix is zero: \( \det \begin{pmatrix} 1 & 2 & -1 \\ 2 & 5 & k \\ 3 & 7 & 1 \end{pmatrix} = 0 \). Expanding along the first row: \( 1(5 - 7k) - 2(2 - 3k) - 1(14 - 15) = 0 \implies 5 - 7k - 4 + 6k + 1 = 0 \implies 2 - k = 0 \implies k = 2 \). (b) Substituting \(k = 2\), we perform row operations on the augmented matrix: \( R_2 \leftarrow R_2 - 2R_1 \) gives \( y + 4z = -5 \). \( R_3 \leftarrow R_3 - 3R_1 \) gives \( y + 4z = -5 \). Since both operations yield the same consistent equation, the system is consistent. Setting the free parameter \(z = t\), we get \( y = -5 - 4t \). Substituting back into the first equation: \( x + 2(-5 - 4t) - t = 3 \implies x - 10 - 8t - t = 3 \implies x = 13 + 9t \). Thus, the general solution is \( x = 13 + 9t \), \( y = -5 - 4t \), \( z = t \).

M1: For setting the determinant of the coefficient matrix equal to 0. M1: For expanding the 3x3 determinant. A1: For obtaining \(k = 2\). M1: For substituting \(k=2\) and applying row reduction to the augmented matrix. A1: For showing consistency by demonstrating matching row equations or a row of zeros. M1: For parameterising \(z = t\) and solving for \(y\). A1: For finding the correct general expressions for \(x\) and \(y\) in terms of \(t\).

(a) The curve is a closed dimpled loop, symmetric about the initial line, with intercepts at \(r = 3a\) (when \(\theta=0\)), \(r=2a\) (when \(\theta=\pi/2\) and \(\theta=3\pi/2\)), and \(r=a\) (when \(\theta=\pi\)). It does not pass through the pole. (b) The area is given by \( A = \frac{1}{2} \int_{0}^{2
\pi} r^2 \, d\theta = \frac{1}{2} a^2 \int_{0}^{2\pi} (2 + \cos\theta)^2 \, d\theta \). Expanding: \( A = \frac{1}{2} a^2 \int_{0}^{2\pi} (4 + 4\cos\theta + \cos^2\theta) \, d\theta \). Using the identity \( \cos^2\theta = \frac{1 + \cos 2\theta}{2} \), we rewrite the integrand: \( A = \frac{1}{2} a^2 \int_{0}^{2\pi} \left( \frac{9}{2} + 4\cos\theta + \frac{1}{2}\cos 2\theta \right) \, d\theta \). Integrating term by term: \( A = \frac{1}{2} a^2 \left[ \frac{9}{2}\theta + 4\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{2\pi} \). Evaluating the limits gives: \( A = \frac{1}{2} a^2 \left( 9\pi - 0 \right) = \frac{9\pi a^2}{2} \).

B1: For drawing a closed loop symmetric about the initial line that does not pass through the pole. B1: For indicating correct relative key coordinate distances on the axes. M1: For using the correct polar area integral formula. A1: For expanding \((2+\cos\theta)^2\) correctly. M1: For using the double-angle identity for \(\cos^2\theta\). M1: For performing integration of the terms correctly. A1: For obtaining the correct exact area of \(\frac{9\pi a^2}{2}\).

The direction vectors of \(l_1\) and \(l_2\) are \( \mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \) and \( \mathbf{d}_2 = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \). A normal vector \( \mathbf{n} \) perpendicular to both lines is found by the cross product: \( \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 1(2) - (-1)(-2) \\ (-1)(1) - 2(2) \\ 2(-2) - 1(1) \end{pmatrix} = \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix} \). We can simplify this normal direction to \( \mathbf{n}' = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \). Next, we find a vector \( \mathbf{v} \) between a point on \(l_1\), which is \(P_1(1, -1, 2)\), and a point on \(l_2\), which is \(P_2(2, 1, 3)\): \( \mathbf{v} = \mathbf{P_1 P_2} = \begin{pmatrix} 2 - 1 \\ 1 - (-1) \\ 3 - 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \). The shortest distance \(d\) is the projection of \( \mathbf{v} \) onto \( \mathbf{n}' \): \( d = \frac{|\mathbf{v} \cdot \mathbf{n}'|}{|\mathbf{n}'|} = \frac{|1(0) + 2(1) + 1(1)|}{\sqrt{0^2 + 1^2 + 1^2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \).

M1: For calculating the cross product of the two direction vectors. A1: For obtaining a correct normal vector. B1: For finding a vector connecting a point on \(l_1\) to a point on \(l_2\). M1: For using the projection formula for the shortest distance of skew lines. M1: For evaluating the dot product of the displacement vector and normal vector. A1: For finding the correct magnitude of the normal vector. A1: For obtaining the correct shortest distance as \(\frac{3\sqrt{2}}{2}\).

**(i)**
To find the intersection points, we equate the two equations:
\[ a(1 + \cos\theta) = a\sqrt{3}\sin\theta \implies 1 + \cos\theta = \sqrt{3}\sin\theta \]
Using the trigonometric identities \(1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)\) and \(\sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\), we obtain:
\[ 2\cos^2\left(\frac{\theta}{2}\right) = 2\sqrt{3}\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \]
This gives two possible cases:
1. \(\cos\left(\frac{\theta}{2}\right) = 0 \implies \frac{\theta}{2} = \frac{\pi}{2} \implies \theta = \pi\).
When \(\theta = \pi\), \(r = a(1 + \cos\pi) = 0\). This is the pole, \((0, \pi)\).
2. \(\cos\left(\frac{\theta}{2}\right) = \sqrt{3}\sin\left(\frac{\theta}{2}\right) \implies \tan\left(\frac{\theta}{2}\right) = \frac{1}{\sqrt{3}} \implies \frac{\theta}{2} = \frac{\pi}{6} \implies \theta = \frac{\pi}{3}\).
When \(\theta = \frac{\pi}{3}\), \(r = a\left(1 + \cos\frac{\pi}{3}\right) = \frac{3}{2}a\).

Thus, the points of intersection are \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\) and the pole \((0, \pi)\).

**(ii)**
- \(C_1\) is the upper half of a cardioid starting at \((2a, 0)\), passing through \((a, \frac{\pi}{2})\), and ending at the pole.
- \(C_2\) is a circle of diameter \(a\sqrt{3}\) lying in the direction \(\theta = \frac{\pi}{2}\), starting and ending at the pole.
- Both curves intersect at \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\) and at the pole.

**(iii)**
The boundary of the region common to both curves is defined by:
- \(C_2\) for \(0 \le \theta \le \frac{\pi}{3}\)
- \(C_1\) for \(\frac{\pi}{3} \le \theta \le \pi\)

The total area \(A\) is given by:
\[ A = \frac{1}{2}\int_{0}^{\frac{\pi}{3}} r_2^2 \, d\theta + \frac{1}{2}\int_{\frac{\pi}{3}}^{\pi} r_1^2 \, d\theta \]

Evaluating the first integral \(I_1\):
\[ I_1 = \frac{1}{2}\int_{0}^{\frac{\pi}{3}} 3a^2\sin^2\theta \, d\theta = \frac{3a^2}{4}\int_{0}^{\frac{\pi}{3}} (1 - \cos 2\theta) \, d\theta \]
\[ I_1 = \frac{3a^2}{4}\left[ \theta - \frac{1}{2}\sin 2\theta \right]_{0}^{\frac{\pi}{3}} = \frac{3a^2}{4}\left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) = a^2\left( \frac{\pi}{4} - \frac{3\sqrt{3}}{16} \right) \]

Evaluating the second integral \(I_2\):
\[ I_2 = \frac{1}{2}\int_{\frac{\pi}{3}}^{\pi} a^2(1 + 2\cos\theta + \cos^2\theta) \, d\theta = \frac{a^2}{2}\int_{\frac{\pi}{3}}^{\pi} \left( \frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta \right) \, d\theta \]
\[ I_2 = \frac{a^2}{2}\left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{\frac{\pi}{3}}^{\pi} \]
At the upper limit \(\theta = \pi\):
\[ \frac{3\pi}{2} \]
At the lower limit \(\theta = \frac{\pi}{3}\):
\[ \frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8} = \frac{\pi}{2} + \frac{9\sqrt{3}}{8} \]
\[ I_2 = \frac{a^2}{2}\left( \frac{3\pi}{2} - \frac{\pi}{2} - \frac{9\sqrt{3}}{8} \right) = \frac{a^2}{2}\left( \pi - \frac{9\sqrt{3}}{8} \right) = a^2\left( \frac{\pi}{2} - \frac{9\sqrt{3}}{16} \right) \]

Summing \(I_1\) and \(I_2\):
\[ A = a^2\left( \frac{\pi}{4} - \frac{3\sqrt{3}}{16} + \frac{\pi}{2} - \frac{9\sqrt{3}}{16} \right) = a^2\left( \frac{3\pi}{4} - \frac{12\sqrt{3}}{16} \right) = \frac{3}{4}a^2(\pi - \sqrt{3}) \]

**(i)**
* **M1**: For equating the two expressions and attempting to solve for \(\theta\).
* **A1**: For obtaining the angle \(\theta = \frac{\pi}{3}\) (or equivalent).
* **A1**: For both correct coordinates: \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\) and \((0, \pi)\) (or pole).

**(ii)**
* **B1**: For a correctly shaped half-cardioid (upper part only, starting on the positive initial line and terminating at the pole).
* **B1**: For a correctly drawn circle passing through the pole, meeting the cardioid at \(\theta = \frac{\pi}{3}\).

**(iii)**
* **M1**: For expressing the common area as the sum of two integrals with appropriate limits (0 to \(\frac{\pi}{3}\) for \(C_2\) and \(\frac{\pi}{3}\) to \(\pi\) for \(C_1\)).
* **M1**: For using the identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\) and integrating both expressions.
* **A1**: For obtaining correct integrated expressions and evaluating limits to find \(I_1 = a^2\left( \frac{\pi}{4} - \frac{3\sqrt{3}}{16} \right)\) and \(I_2 = a^2\left( \frac{\pi}{2} - \frac{9\sqrt{3}}{16} \right)\) (or equivalent intermediate expressions).
* **A1**: For simplifying to the final correct answer: \(\frac{3}{4}a^2(\pi - \sqrt{3})\).