2024 Cambridge IAS-Level Mathematics - Further (9231) Practice Paper with Answers

**(i)** Let the statement be \(P(n): \mathbf{M}^n = \begin{pmatrix} 2n+1 & -4n \\ n & 1-2n \end{pmatrix}\).

**Base case:** For \(n = 1\),
\(\mathbf{M}^1 = \begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\).
This is equal to \(\mathbf{M}\), so \(P(1)\) is true.

**Inductive step:** Assume that \(P(k)\) is true for some positive integer \(k\), i.e.,
\(\mathbf{M}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\).

We must show that \(P(k+1)\) is true:
\(\mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\)
\(= \begin{pmatrix} (2k+1)(3) + (-4k)(1) & (2k+1)(-4) + (-4k)(-1) \\ k(3) + (1-2k)(1) & k(-4) + (1-2k)(-1) \end{pmatrix}\)
\(= \begin{pmatrix} 6k+3-4k & -8k-4+4k \\ 3k+1-2k & -4k-1+2k \end{pmatrix}\)
\(= \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}\)
\(= \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix}\).

Since \(P(1)\) is true, and \(P(k) \implies P(k+1)\), by the principle of mathematical induction, \(P(n)\) is true for all positive integers \(n\).

**(ii)** Using the definition of \(\mathbf{A}\):
\(\mathbf{A} = \sum_{r=1}^N \mathbf{M}^r = \sum_{r=1}^N \begin{pmatrix} 2r+1 & -4r \\ r & 1-2r \end{pmatrix} = \begin{pmatrix} \sum_{r=1}^N (2r+1) & \sum_{r=1}^N (-4r) \\ \sum_{r=1}^N r & \sum_{r=1}^N (1-2r) \end{pmatrix}\).

We calculate each component sum using standard series formulae:
- \(\sum_{r=1}^N r = \frac{1}{2}N(N+1)\)
- \(\sum_{r=1}^N (2r+1) = 2\sum_{r=1}^N r + N = 2\left(\frac{1}{2}N(N+1)\right) + N = N^2 + 2N\)
- \(\sum_{r=1}^N (-4r) = -4 \sum_{r=1}^N r = -4 \left(\frac{1}{2}N(N+1)\right) = -2N(N+1)\)
- \(\sum_{r=1}^N (1-2r) = N - 2\sum_{r=1}^N r = N - N(N+1) = -N^2\).

Therefore:
\(\mathbf{A} = \begin{pmatrix} N^2+2N & -2N(N+1) \\ \frac{1}{2}N(N+1) & -N^2 \end{pmatrix}\).

**(i)**
- **M1**: Verifies base case \(n=1\).
- **M1**: Assumes statement for \(n=k\) and writes down \(\mathbf{M}^k \mathbf{M}\).
- **A1**: Correct expansion of the matrix multiplication.
- **A1**: Simplifies to the form in terms of \(k+1\).
- **A1**: Provides a complete and clear inductive conclusion. (6 marks)

**(ii)**
- **M1**: Recognizes that each element of the matrix sum can be summed individually.
- **M1**: Uses standard summation formulae for \(\sum r\).
- **A1**: Obtains correct expressions for diagonal elements.
- **A1**: Obtains correct expressions for non-diagonal elements. (4.7 marks)

From the given cubic equation \(2x^3 - 3x^2 + 4x - 5 = 0\), we have:
\(\sum \alpha = \frac{3}{2}\)
\(\sum \alpha\beta = 2\)
\(\alpha\beta\gamma = \frac{5}{2}\).

**(i)** We use the identity:
\(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\)
\(= \left(\frac{3}{2}\right)^2 - 2(2) = \frac{9}{4} - 4 = -\frac{7}{4}\).

**(ii)** Since \(\alpha, \beta, \gamma\) are roots of the cubic, they satisfy:
\(2x^3 - 3x^2 + 4x - 5 = 0\).
Summing over the three roots:
\(2\sum \alpha^3 - 3\sum \alpha^2 + 4\sum \alpha - 15 = 0\).

Substitute the values found in part (i) and the sum of roots:
\(2\sum \alpha^3 - 3\left(-\frac{7}{4}\right) + 4\left(\frac{3}{2}\right) - 15 = 0\)
\(2\sum \alpha^3 + \frac{21}{4} + 6 - 15 = 0\)
\(2\sum \alpha^3 = 9 - \frac{21}{4} = \frac{15}{4}\)
\(\sum \alpha^3 = \frac{15}{8}\).

**(iii)** Let \(y = x^2\), so \(x = y^{1/2}\).
Substitute \(x = y^{1/2}\) into the original equation:
\(2y^{3/2} - 3y + 4y^{1/2} - 5 = 0\)
\(y^{1/2}(2y + 4) = 3y + 5\).

Squaring both sides:
\(y(2y + 4)^2 = (3y + 5)^2\)
\(y(4y^2 + 16y + 16) = 9y^2 + 30y + 25\)
\(4y^3 + 16y^2 + 16y = 9y^2 + 30y + 25\)
\(4y^3 + 7y^2 - 14y - 25 = 0\).

**(i)**
- **M1**: Recalls and uses \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\).
- **M1**: Substitutes correct values of \(\sum \alpha\) and \(\sum \alpha\beta\).
- **A1**: Obtains \(-\frac{7}{4}\).

**(ii)**
- **M1**: Uses the original equation to set up a relation for \(\sum \alpha^3\).
- **A1**: Substitutes values correctly into \(2\sum \alpha^3 - 3\sum \alpha^2 + 4\sum \alpha - 15 = 0\).
- **M1**: Solves for \(\sum \alpha^3\) algebraically.
- **A1**: Obtains \(\frac{15}{8}\).

**(iii)**
- **M1**: Sets up a substitution \(y = x^2\) and isolates terms with fractional powers.
- **M1**: Squares both sides of the equation correctly.
- **A1**: Expands and collects terms.
- **A1**: Obtains \(4y^3 + 7y^2 - 14y - 25 = 0\) (or any integer multiple).

**(i)** Differentiating \(y\) with respect to \(x\) using the quotient rule:
\(\frac{dy}{dx} = \frac{(2x - 2)(x - 3) - (x^2 - 2x + a)(1)}{(x - 3)^2} = \frac{2x^2 - 8x + 6 - x^2 + 2x - a}{(x - 3)^2} = \frac{x^2 - 6x + (6 - a)}{(x - 3)^2}\).

For stationary points, \(\frac{dy}{dx} = 0\), which occurs when:
\(x^2 - 6x + (6 - a) = 0\).

For there to be two distinct stationary points, the discriminant \(\Delta\) of this quadratic must be strictly positive:
\(\Delta = (-6)^2 - 4(1)(6 - a) > 0\)
\(36 - 24 + 4a > 0\)
\(12 + 4a > 0 \implies a > -3\).

**(ii)** For \(a = 1\), \(y = \frac{x^2 - 2x + 1}{x - 3}\).

**Asymptotes:**
- Vertical asymptote is given by the denominator being zero: \(x = 3\).
- To find the oblique asymptote, write \(y\) in quotient-remainder form:
\(y = \frac{x^2 - 2x + 1}{x - 3} = \frac{(x+1)(x-3) + 4}{x - 3} = x + 1 + \frac{4}{x - 3}\).
As \(x \to \pm\infty\), \(y \to x + 1\). Thus, the oblique asymptote is \(y = x + 1\).

**Stationary points:**
Setting \(\frac{dy}{dx} = 0\) with \(a = 1\):
\(x^2 - 6x + 5 = 0 \implies (x - 1)(x - 5) = 0\).
Thus, \(x = 1\) or \(x = 5\).
- At \(x = 1\): \(y = \frac{1^2 - 2(1) + 1}{1 - 3} = 0\). Stationary point is \((1, 0)\).
- At \(x = 5\): \(y = \frac{5^2 - 2(5) + 1}{5 - 3} = \frac{16}{2} = 8\). Stationary point is \((5, 8)\).

**(iii)**
- The curve has a vertical asymptote at \(x = 3\) and an oblique asymptote at \(y = x + 1\).
- Intersections:
- \(x\)-axis: \(y = 0 \implies (x-1)^2 = 0 \implies (1, 0)\), which is a touch point.
- \(y\)-axis: \(x = 0 \implies y = -\frac{1}{3}\), giving \((0, -\frac{1}{3})\).
- Sketching:
- Branch 1: Lies to the left of \(x=3\). Passes through \((0, -1/3)\), touches the \(x\)-axis at the local maximum \((1, 0)\), and approaches the asymptotes \(y=x+1\) and \(x=3\).
- Branch 2: Lies to the right of \(x=3\). Has a local minimum at \((5, 8)\) and approaches the asymptotes \(x=3\) and \(y=x+1\).

**(i)**
- **M1**: Differentiates \(y\) correctly to find \(\frac{dy}{dx}\).
- **M1**: Expresses the condition for stationary points as a quadratic equation in \(x\) and sets its discriminant \(\Delta > 0\).
- **A1**: Shows clearly that this leads to \(a > -3\).

**(ii)**
- **B1**: Identifies vertical asymptote \(x = 3\).
- **M1**: Uses algebraic division to find oblique asymptote.
- **A1**: Obtains oblique asymptote \(y = x + 1\).
- **A1**: Obtains correct stationary points \((1, 0)\) and \((5, 8)\).

**(iii)**
- **B1**: Sketches the two asymptotes with their equations labeled.
- **B1**: Correctly positions and shapes both branches of the curve.
- **B1**: Correctly labels the intercepts \((0, -1/3)\) and \((1, 0)\) and the stationary points.

**(i)** Expanding the right hand side:
\(2 \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right) = 2 \left( \frac{(r+1)^2 - r^2}{r^2(r+1)^2} \right)\)
\(= 2 \left( \frac{r^2 + 2r + 1 - r^2}{r^2(r+1)^2} \right)\)
\(= 2 \left( \frac{2r + 1}{r^2(r+1)^2} \right) = \frac{4r + 2}{r^2(r+1)^2}\). (Shown)

**(ii)** Let \(u_r = \frac{4r+2}{r^2(r+1)^2}\).
\(\sum_{r=1}^n u_r = 2 \sum_{r=1}^n \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right)\)
\(= 2 \left[ \left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \dots + \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) \right]\).

By telescoping cancellation, all intermediate terms cancel:
\(\sum_{r=1}^n u_r = 2 \left( 1 - \frac{1}{(n+1)^2} \right) = 2 - \frac{2}{(n+1)^2}\).

**(iii)** To find the sum to infinity, we take the limit as \(n \to \infty\):
\(\sum_{r=1}^\infty \frac{4r+2}{r^2(r+1)^2} = \lim_{n\to\infty} \left( 2 - \frac{2}{(n+1)^2} \right) = 2\).

**(iv)** Note that:
\(\sum_{r=N}^\infty u_r = \sum_{r=1}^\infty u_r - \sum_{r=1}^{N-1} u_r\)
\(= 2 - \left( 2 - \frac{2}{((N-1)+1)^2} \right) = \frac{2}{N^2}\).

We want:
\(\frac{2}{N^2} < 0.01 \implies N^2 > 200\).

Since \(14^2 = 196\) and \(15^2 = 225\), the smallest integer value of \(N\) is 15.

**(i)**
- **M1**: Finds a common denominator for the RHS expression.
- **A1**: Simplifies to show equivalence to LHS.

**(ii)**
- **M1**: Writes down the first few terms of the series to show the cancelling structure.
- **M1**: Identifies which terms remain after cancellation.
- **A1**: Correctly includes the factor of 2.
- **A1**: Obtains \(2 - \frac{2}{(n+1)^2}\).

**(iii)**
- **M1**: Applies limit as \(n \to \infty\) to the expression from (ii).
- **A1**: Obtains \(2\).

**(iv)**
- **M1**: Expresses the sum from \(N\) to infinity as \(\frac{2}{N^2}\).
- **M1**: Sets up and solves the inequality \(\frac{2}{N^2} < 0.01\).
- **A1**: Obtains \(N = 15\).

**(i)** The characteristic equation is given by \(\det(\mathbf{M} - \lambda \mathbf{I}) = 0\):
\[\begin{vmatrix} 1-\lambda & 1 & 2 \\ 0 & 2-\lambda & 1 \\ 0 & 3 & -\lambda \end{vmatrix} = 0\]

Expanding along the first column:
\((1-\lambda) \left[ (2-\lambda)(-\lambda) - 3(1) \right] = 0\)
\((1-\lambda) (\lambda^2 - 2\lambda - 3) = 0\)
\((1-\lambda)(\lambda - 3)(\lambda + 1) = 0\).

Thus, the eigenvalues are \(\lambda_1 = 1\), \(\lambda_2 = 3\), and \(\lambda_3 = -1\).

**(ii)** Corresponding eigenvectors:

- **For \(\lambda = 1\)**:
\(\mathbf{M} - \mathbf{I} = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 3 & -1 \end{pmatrix}\).
We solve \(\begin{pmatrix} 0 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 3 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\).
From the first and second rows: \(y + 2z = 0\) and \(y + z = 0 \implies y = 0\) and \(z = 0\).
The variable \(x\) is free. Choosing \(x = 1\), we get the eigenvector:
\(\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).

- **For \(\lambda = 3\)**:
\(\mathbf{M} - 3\mathbf{I} = \begin{pmatrix} -2 & 1 & 2 \\ 0 & -1 & 1 \\ 0 & 3 & -3 \end{pmatrix}\).
From the second row: \(-y + z = 0 \implies y = z\).
From the first row: \(-2x + y + 2z = 0 \implies 2x = 3y \implies x = 1.5y\).
Choosing \(y = 2\), we get \(x = 3, z = 2\). An eigenvector is:
\(\mathbf{e}_2 = \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}\).

- **For \(\lambda = -1\)**:
\(\mathbf{M} - (-1)\mathbf{I} = \mathbf{M} + \mathbf{I} = \begin{pmatrix} 2 & 1 & 2 \\ 0 & 3 & 1 \\ 0 & 3 & 1 \end{pmatrix}\).
From the second row: \(3y + z = 0 \implies z = -3y\).
From the first row: \(2x + y + 2z = 0 \implies 2x + y - 6y = 0 \implies 2x = 5y \implies x = 2.5y\).
Choosing \(y = 2\), we get \(x = 5, z = -6\). An eigenvector is:
\(\mathbf{e}_3 = \begin{pmatrix} 5 \\ 2 \\ -6 \end{pmatrix}\).

**(iii)**
By diagonalization, \(\mathbf{M} = \mathbf{P} \mathbf{D}_{0} \mathbf{P}^{-1}\), where columns of \(\mathbf{P}\) are the eigenvectors, and \(\mathbf{D}_{0}\) contains the eigenvalues.
Then \(\mathbf{M}^4 = \mathbf{P} \mathbf{D} \mathbf{P}^{-1}\) where \(\mathbf{D} = \mathbf{D}_{0}^4\).
Thus,
\(\mathbf{P} = \begin{pmatrix} 1 & 3 & 5 \\ 0 & 2 & 2 \\ 0 & 2 & -6 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 1^4 & 0 & 0 \\ 0 & 3^4 & 0 \\ 0 & 0 & (-1)^4 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) (or any permutation of columns, with matching order of eigenvalues).

**(i)**
- **M1**: Sets up \(\det(\mathbf{M} - \lambda \mathbf{I}) = 0\).
- **M1**: Expands determinant correctly to obtain the characteristic equation.
- **A1**: Solves quadratic part of the characteristic equation.
- **A1**: Obtains eigenvalues \(1, 3, -1\).

**(ii)**
- **M1**: Solves \((\mathbf{M} - \lambda \mathbf{I})\mathbf{v} = 0\) for one eigenvalue.
- **A1**: Obtains eigenvector for \(\lambda = 1\).
- **A1**: Obtains eigenvector for \(\lambda = 3\).
- **A1**: Obtains eigenvector for \(\lambda = -1\).
- **B1**: Writes down three distinct correct eigenvectors.

**(iii)**
- **B1**: States a correct matrix \(\mathbf{P}\) whose columns are their eigenvectors.
- **B1**: States a correct diagonal matrix \(\mathbf{D}\) with elements \(1^4, 3^4, (-1)^4\) in matching order.

**(i)** First, find a vector perpendicular to both lines by taking the cross product of their direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\):
\(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\)
\(\mathbf{n} = \begin{pmatrix} (-1)(1) - (3)(1) \\ (3)(1) - (2)(1) \\ (2)(1) - (-1)(1) \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix}\).

Let \(A = (1, 2, -1)\) be a point on \(l_1\) and \(B = (2, 4, 1)\) be a point on \(l_2\).
The vector connecting \(A\) and \(B\) is:
\(\mathbf{AB} = \begin{pmatrix} 2 - 1 \\ 4 - 2 \\ 1 - (-1) \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\).

The shortest distance \(d\) is the projection of \(\mathbf{AB}\) onto the common perpendicular \(\mathbf{n}\):
\(d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\)
\(\mathbf{AB} \cdot \mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix} = -4 + 2 + 6 = 4\).
\(|\mathbf{n}| = \sqrt{(-4)^2 + 1^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26}\).

So,
\(d = \frac{4}{\sqrt{26}} = \frac{2\sqrt{26}}{13}\).

**(ii)**
The plane \(\Pi\) contains \(l_1\) and is parallel to \(l_2\). Therefore, its normal vector is the same common perpendicular vector \(\mathbf{n} = \begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix}\).

The vector equation of the plane is:
\(\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}\), where \(\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) is a point on \(l_1\).
\(\mathbf{r} \cdot \begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix} = -4 + 2 - 3 = -5\).

This gives the Cartesian equation:
\-4x + y + 3z = -5
or \(4x - y - 3z = 5\).

**(i)**
- **M1**: Realizes the need to find the cross product of the direction vectors of \(l_1\) and \(l_2\).
- **A1**: Obtains correct perpendicular direction vector \(\begin{pmatrix}-4\\1\\3\end{pmatrix}\) (or any scalar multiple).
- **M1**: Selects points on both lines and finds vector \(\mathbf{AB}\).
- **M1**: Uses projection formula \(d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\).
- **A1**: Obtains correct numerator \(4\).
- **A1**: Obtains correct distance \(\frac{4}{\sqrt{26}}\) or \(\frac{2\sqrt{26}}{13}\).

**(ii)**
- **M1**: Uses normal vector \(\mathbf{n}\) found in part (i) as the normal to the plane.
- **M1**: Substitutes a point on \(l_1\) into the plane equation \(\mathbf{r} \cdot \mathbf{n} = d_0\).
- **A1**: Evaluates the constant term correctly as \(-5\) or \(5\).
- **A1**: Obtains correct Cartesian equation \(4x - y - 3z = 5\) (or equivalent form).

**(i)** The curve is a cardioid (specifically the top half since \(0 \le \theta \le \pi\)).
At \(\theta = 0\), \(r = 4\).
At \(\theta = \frac{\pi}{2}\), \(r = 2\).
At \(\theta = \pi\), \(r = 0\).
The sketch starts at \((4,0)\) on the initial line, curves smoothly through \((2,\pi/2)\), and finishes at the pole \((0,\pi)\).

**(ii)** The area \(A\) of the region is:
\(A = \int_{0}^{\pi} \frac{1}{2} r^2 d\theta = \int_{0}^{\pi} \frac{1}{2} \left(2(1 + \cos \theta)\right)^2 d\theta\)
\(= 2 \int_{0}^{\pi} (1 + 2\cos \theta + \cos^2 \theta) d\theta\).

Using the identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\):
\(A = 2 \int_{0}^{\pi} \left(1 + 2\cos \theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right) d\theta\)
\(= 2 \int_{0}^{\pi} \left(\frac{3}{2} + 2\cos \theta + \frac{1}{2}\cos 2\theta\right) d\theta\)
\(= 2 \left[ \frac{3}{2}\theta + 2\sin \theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi}\)
\(= 2 \left[ \left(\frac{3}{2}\pi + 0 + 0\right) - (0) \right] = 3\pi\).

**(iii)**
The Cartesian coordinate \(x\) is given by:
\(x = r\cos \theta = 2(1 + \cos \theta)\cos \theta = 2\cos \theta + 2\cos^2 \theta\).

A tangent is perpendicular to the initial line when \(\frac{dx}{d\theta} = 0\):
\(\frac{dx}{d\theta} = -2\sin \theta - 4\cos \theta \sin \theta = -2\sin \theta(1 + 2\cos \theta)\).

For \(\frac{dx}{d\theta} = 0\) within \(0 < \theta < \pi\):
Since \(\sin \theta \neq 0\) in this open interval, we have:
\(1 + 2\cos \theta = 0 \implies \cos \theta = -\frac{1}{2}\).

For \(0 < \theta < \pi\), this gives:
\(\theta = \frac{2\pi}{3}\).

Substitute this back into the polar equation to find \(r\):
\(r = 2\left(1 + \cos\frac{2\pi}{3}\right) = 2\left(1 - \frac{1}{2}\right) = 1\).

Thus, the polar coordinates of the point are \((1, \frac{2\pi}{3})\).

**(i)**
- **B1**: Smooth curve starting on positive initial line, passing through a point on the positive y-axis, and ending at the origin.
- **B1**: Correct orientation with key points \((4,0)\), \((2,\pi/2)\) and pole labeled or clearly indicated.

**(ii)**
- **M1**: Uses the correct integration formula \(\int \frac{1}{2} r^2 d\theta\) with correct limits \(0\) and \(\pi\).
- **M1**: Expands \((1+\cos\theta)^2\) correctly.
- **M1**: Uses the double angle identity \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\) to integrate.
- **A1**: Correctly integrates each term.
- **A1**: Substitutes limits and simplifies to obtain \(3\pi\).

**(iii)**
- **M1**: Writes \(x = r\cos\theta\) in terms of \(\theta\).
- **M1**: Differentiates \(x\) with respect to \(\theta\) and sets equal to 0.
- **A1**: Solves \(\frac{dx}{d\theta} = 0\) to find \(\theta = \frac{2\pi}{3}\).
- **A1**: Finds the corresponding value of \(r = 1\) and states the coordinates as \((1, \frac{2\pi}{3})\).