2023 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

(a) For the arithmetic progression:
\(T_n = a + (n-1)d\)
So, \(T_3 = a + 2d\) and \(T_{11} = a + 10d\).

For the geometric progression:
\(G_n = a r^{n-1}\)
So, \(G_3 = a r^2\) and \(G_5 = a r^4\).

From the given information:
1) \(a + 2d = a r^2 \implies 2d = a(r^2 - 1)\)
2) \(a + 10d = a r^4 \implies 10d = a(r^4 - 1)\)

Since \(10d = 5(2d)\), we can substitute the expression for \(2d\) into the equation for \(10d\):
\(a(r^4 - 1) = 5a(r^2 - 1)\)

Since \(a \neq 0\), we can divide both sides by \(a\):
\(r^4 - 1 = 5(r^2 - 1)\)
\(r^4 - 5r^2 + 4 = 0\)

Factorising this quadratic in \(r^2\):
\((r^2 - 1)(r^2 - 4) = 0\)
Since \(r > 1\), we have \(r^2 = 4 \implies r = 2\).

(b) Substituting \(r = 2\) into the formula for \(d\):
\(2d = a(2^2 - 1) \implies 2d = 3a \implies d = 1.5a\)

The sum of the first 10 terms of the AP is:
\(S_{10} = \frac{10}{2} [2a + 9d] = 5[2a + 9(1.5a)] = 5[2a + 13.5a] = 5[15.5a] = 77.5a\)

We are given that \(S_{10} = 310\):
\(77.5a = 310 \implies a = 4\).

(a)
- M1: Express \(T_3\), \(T_{11}\), \(G_3\), \(G_5\) correctly in terms of \(a\), \(d\), and \(r\).
- M1: Set up a system of equations to eliminate \(d\).
- A1: Correctly show the quartic equation \(r^4 - 5r^2 + 4 = 0\).
- M1: Solve the quadratic in \(r^2\).
- A1: Identify the correct value \(r = 2\) (rejecting \(r=1\) since \(r>1\)).

(b)
- M1: Substitute \(r=2\) to express \(d\) in terms of \(a\) and write down the sum of the first 10 terms.
- A1: Set up the correct linear equation for \(a\).
- A1: Obtain \(a = 4\).

(a) First, find the midpoint \(M\) of the line segment \(AB\):
\(M = \left( \frac{-2 + 6}{2}, \frac{5 + 1}{2} \right) = (2, 3)\)

Next, find the gradient \(m_1\) of the line \(L_1\):
\(m_1 = \frac{1 - 5}{6 - (-2)} = \frac{-4}{8} = -\frac{1}{2}\)

The gradient \(m_2\) of the perpendicular bisector \(L_2\) is the negative reciprocal of \(m_1\):
\(m_2 = -\frac{1}{-1/2} = 2\)

Now, find the equation of \(L_2\) using the gradient \(m_2 = 2\) and the midpoint \(M(2, 3)\):
\(y - 3 = 2(x - 2) \implies y = 2x - 1\)

(b) To find the intersection of \(L_2\) and the curve \(y^2 = 4x - 2\), substitute \(y = 2x - 1\) into the curve's equation:
\((2x - 1)^2 = 4x - 2\)
\(4x^2 - 4x + 1 = 4x - 2\)
\(4x^2 - 8x + 3 = 0\)

Factorising the quadratic equation:
\((2x - 1)(2x - 3) = 0\)
This gives \(x = 0.5\) or \(x = 1.5\).

Find the corresponding \(y\)-coordinates:
- For \(x = 0.5\): \(y = 2(0.5) - 1 = 0\). So \(P = (0.5, 0)\).
- For \(x = 1.5\): \(y = 2(1.5) - 1 = 2\). So \(Q = (1.5, 2)\).

Now, find the distance \(PQ\):
\(PQ = \sqrt{(1.5 - 0.5)^2 + (2 - 0)^2} = \sqrt{1^2 + 2^2} = \sqrt{5}\).

(a)
- M1: Correct method to find the midpoint of \(AB\).
- M1: Correct method to find the gradient of \(AB\).
- M1: Use of perpendicular gradient rule \(m_1 m_2 = -1\).
- A1: Correct equation of line \(L_2\) in the form \(y = 2x - 1\).

(b)
- M1: Substitute the linear equation into the curve equation to form a quadratic in \(x\).
- A1: Find the correct coordinates of both points of intersection: \((0.5, 0)\) and \((1.5, 2)\).
- M1: Apply the distance formula to find \(PQ\).
- A1: Obtain the exact value \(\sqrt{5}\).

(a) Starting with the given equation:
\(2\sin\theta\tan\theta = 1 + \frac{1}{\cos\theta}\)

Substitute \(\tan\theta = \frac{\sin\theta}{\cos\theta}\):
\(2\sin\theta \left(\frac{\sin\theta}{\cos\theta}\right) = 1 + \frac{1}{\cos\theta}\)
\(\frac{2\sin^2\theta}{\cos\theta} = \frac{\cos\theta + 1}{\cos\theta}\)

Since \(\cos\theta \neq 0\), we can multiply both sides by \(\cos\theta\):
\(2\sin^2\theta = \cos\theta + 1\)

Substitute the identity \(\sin^2\theta = 1 - \cos^2\theta\):
\(2(1 - \cos^2\theta) = \cos\theta + 1\)
\(2 - 2\cos^2\theta = \cos\theta + 1\)
\(2\cos^2\theta + \cos\theta - 1 = 0\)

(b) Factorise the quadratic equation:
\((2\cos\theta - 1)(\cos\theta + 1) = 0\)

This gives two possible values for \(\cos\theta\):
1) \(\cos\theta = 0.5\)
For \(0^\circ \le \theta \le 360^\circ\):
\(\theta = 60^\circ\) or \(\theta = 360^\circ - 60^\circ = 300^\circ\).

2) \(\cos\theta = -1\)
For \(0^\circ \le \theta \le 360^\circ\):
\(\theta = 180^\circ\).

Thus, the solutions in the interval are \(\theta = 60^\circ, 180^\circ, 300^\circ\).

(a)
- M1: Use the identity \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and write the equation over a common denominator.
- M1: Correctly apply \(\sin^2\theta = 1 - \cos^2\theta\) to eliminate sine.
- A1.5: Simplify correctly to get \(2\cos^2\theta + \cos\theta - 1 = 0\) with clear steps shown.

(b)
- M1: Factorise the quadratic equation in \(\cos\theta\) to find the values of \(\cos\theta\).
- A1: Find \(\theta = 60^\circ\) and \(\theta = 300^\circ\) from \(\cos\theta = 0.5\).
- A1: Find \(\theta = 180^\circ\) from \(\cos\theta = -1\).
- A1: Ensure no extra incorrect solutions in range and all three solutions are found.

(a) To find \(f^{-1}(x)\), let \(y = 3x - 2\):
\(y + 2 = 3x \implies x = \frac{y + 2}{3}\)

So, \(f^{-1}(x) = \frac{x + 2}{3}\).
Since the domain of \(f\) is \(\mathbb{R}\), the range of \(f\) is \(\mathbb{R}\). Therefore, the domain of \(f^{-1}\) is \(x \in \mathbb{R}\).

(b) First find an expression for \(gf(x)\):
\(gf(x) = g(3x - 2) = \frac{6}{2(3x - 2) - 1} = \frac{6}{6x - 4 - 1} = \frac{6}{6x - 5}\)

We set this equal to 2:
\(\frac{6}{6x - 5} = 2\)
\(6 = 2(6x - 5)\)
\(6 = 12x - 10\)
\(12x = 16 \implies x = \frac{4}{3}\)

(c) For \(g(x) = \frac{6}{2x - 1}\) with domain \(x > \frac{1}{2}\):
As \(x \to \frac{1}{2}^+\), \(2x - 1 \to 0^+\), so \(g(x) \to \infty\).
As \(x \to \infty\), \(2x - 1 \to \infty\), so \(g(x) \to 0^+\).
Since \(g(x)\) is a continuous and strictly decreasing function on this domain, its range is \(g(x) > 0\).

(a)
- M1: Correct algebraic steps to change the subject of the formula.
- A1: Correct inverse function and its domain stated as \(x \in \mathbb{R}\).

(b)
- M1: Correctly substitute \(f(x)\) into \(g(x)\).
- M1: Form and solve a linear equation in \(x\).
- A0.5: Obtain \(x = \frac{4}{3}\).

(c)
- M1: Identify behaviour of function as \(x \to \frac{1}{2}^+\).
- M1: Identify behaviour of function as \(x \to \infty\).
- A1: State correct range: \(g(x) > 0\) (or \(y > 0\)).

(a) In the right-angled triangle \(OAC\):
- \(AC = OA \sin\theta = r \sin\theta\)
- \(OC = OA \cos\theta = r \cos\theta\)

Since \(OB = r\) is the radius of the circle, the length of \(BC\) is:
\(BC = OB - OC = r - r \cos\theta = r(1 - \cos\theta)\)

The length of the arc \(AB\) is:
\(\text{Arc } AB = r\theta\)

The perimeter \(P\) of the shaded region is the sum of arc \(AB\), line \(BC\), and line \(AC\):
\(P = r\theta + r(1 - \cos\theta) + r\sin\theta = r(\theta + \sin\theta + 1 - \cos\theta)\).

(b) The area of the sector \(OAB\) is:
\(\text{Area}_{\text{sector}} = \frac{1}{2} r^2 \theta\)

The area of the right-angled triangle \(OAC\) is:
\(\text{Area}_{\text{triangle}} = \frac{1}{2} \times OC \times AC = \frac{1}{2} (r\cos\theta)(r\sin\theta) = \frac{1}{2} r^2 \sin\theta\cos\theta\)

The area of the shaded region is:
\(\text{Area}_{\text{shaded}} = \text{Area}_{\text{sector}} - \text{Area}_{\text{triangle}} = \frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin\theta\cos\theta\)

The ratio of the shaded area to the sector area is:
\(\text{Ratio} = \frac{\frac{1}{2} r^2 (\theta - \sin\theta\cos\theta)}{\frac{1}{2} r^2 \theta} = 1 - \frac{\sin\theta\cos\theta}{\theta}\)

Substitute \(\theta = \frac{1}{3}\pi\):
\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\) and \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)
\(\sin\theta\cos\theta = \frac{\sqrt{3}}{4}\)

Thus, the ratio is:
\(1 - \frac{\frac{\sqrt{3}}{4}}{\frac{\pi}{3}} = 1 - \frac{3\sqrt{3}}{4\pi}\).
So, \(a = 3\) and \(b = 4\).

(a)
- M1: Correctly express \(AC\) as \(r \sin\theta\).
- M1: Correctly express \(OC\) as \(r \cos\theta\).
- A1: Find the correct expression for \(BC = r(1 - \cos\theta)\).
- M1: Add arc length \(r\theta\), \(AC\), and \(BC\) to find the perimeter.
- A0.5: Reach the given formula clearly.

(b)
- M1: Write down correct expressions for both the sector area and the triangle area.
- M1: Divide the shaded area expression by the sector area expression and substitute \(\theta = \frac{\pi}{3}\).
- A1: Obtain the correct ratio \(1 - \frac{3\sqrt{3}}{4\pi}\) (or identify \(a = 3\), \(b = 4\)).

(a) Given \(y = 4x + 9(x - 1)^{-1}\).
First, find \(\frac{dy}{dx}\) using the chain rule:
\(\frac{dy}{dx} = 4 - 9(x - 1)^{-2} = 4 - \frac{9}{(x - 1)^2}\)

At the stationary point, \(\frac{dy}{dx} = 0\):
\(4 - \frac{9}{(x - 1)^2} = 0 \implies 4 = \frac{9}{(x - 1)^2}\)
\((x - 1)^2 = \frac{9}{4}\)

Since \(x > 1\), we take the positive square root:
\(x - 1 = 1.5 \implies x = 2.5\)

Substitute \(x = 2.5\) back into the original curve equation:
\(y = 4(2.5) + \frac{9}{2.5 - 1} = 10 + \frac{9}{1.5} = 10 + 6 = 16\)

So the coordinates of the stationary point are \((2.5, 16)\).

To determine the nature, find the second derivative:
\(\frac{d^2y}{dx^2} = 18(x - 1)^{-3} = \frac{18}{(x - 1)^3}\)

Substitute \(x = 2.5\):
\(\frac{d^2y}{dx^2} = \frac{18}{(1.5)^3} = \frac{18}{3.375} = \frac{16}{3}\)
Since \(\frac{d^2y}{dx^2} > 0\), the stationary point \((2.5, 16)\) is a local minimum.

(b) We are given \(\frac{dx}{dt} = 0.2\).
We want to find \(\frac{dy}{dt}\) when \(x = 4\).

First, find \(\frac{dy}{dx}\) at \(x = 4\):
\(\frac{dy}{dx} = 4 - \frac{9}{(4 - 1)^2} = 4 - \frac{9}{9} = 3\)

Using the chain rule:
\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = 3 \times 0.2 = 0.6\) units per second.

(a)
- M1: Correct differentiation to find \(\frac{dy}{dx}\).
- M1: Set \(\frac{dy}{dx} = 0\) and solve for \(x\).
- A1: Find \(x = 2.5\) (and reject \(x = -0.5\) using \(x > 1\)).
- A1: Find \(y = 16\).
- M1: Use a valid second derivative check to determine the nature.
- A1: Correctly conclude that the point is a minimum.

(b)
- M1: Evaluate \(\frac{dy}{dx}\) at \(x = 4\).
- M1: Apply the connected rates of change formula.
- A0.5: Obtain \(\frac{dy}{dt} = 0.6\).

(a) When \(x = 4\):
\(y = \sqrt{2(4) + 1} = \sqrt{9} = 3\)
So the point of contact is \((4, 3)\).

Find \(\frac{dy}{dx}\) by using the chain rule on \(y = (2x + 1)^{1/2}\):
\(\frac{dy}{dx} = \frac{1}{2}(2x + 1)^{-1/2} \times 2 = \frac{1}{\sqrt{2x + 1}}\)

At \(x = 4\):
\(\frac{dy}{dx} = \frac{1}{\sqrt{9}} = \frac{1}{3}\)

The gradient of the tangent is \(\frac{1}{3}\). Therefore, the gradient of the normal, \(m\), is:
\(m = -3\)

The equation of the normal at \((4, 3)\) is:
\(y - 3 = -3(x - 4)\)
\(y = -3x + 15\)

(b) The volume of the solid of revolution \(V\) is given by:
\(V = \pi \int_{a}^{b} y^2 \, dx\)

Here, \(y^2 = 2x + 1\), and the limits are from \(x = 0\) to \(x = 4\):
\(V = \pi \int_{0}^{4} (2x + 1) \, dx\)

Integrate the expression:
\(V = \pi \left[ x^2 + x \right]_{0}^{4}\)

Substitute the limits:
\(V = \pi \left[ (4^2 + 4) - (0^2 + 0) \right] = \pi [16 + 4] = 20\pi\).

(a)
- M1: Find the \(y\)-coordinate at \(x = 4\).
- M1: Differentiate to find \(\frac{dy}{dx}\) using chain rule.
- M1: Use perpendicular gradient rule \(m_{\text{normal}} = -1 / m_{\text{tangent}}\).
- A0.5: Obtain the correct normal equation \(y = -3x + 15\).

(b)
- M1: State or apply the volume of revolution formula \(V = \pi \int y^2 \, dx\).
- M1: Integrate \(2x + 1\) correctly with respect to \(x\).
- M1: Apply the limits of integration from 0 to 4.
- A1: Find the correct exact volume \(20\pi\).

(a) To complete the square for \(2x^2 - 12x + 11\):
First, factor out 2 from the \(x^2\) and \(x\) terms:
\(2(x^2 - 6x) + 11\)

Now complete the square inside the bracket:
\(x^2 - 6x = (x - 3)^2 - 9\)

Substitute this back:
\(2[(x - 3)^2 - 9] + 11\)
\(= 2(x - 3)^2 - 18 + 11\)
\(= 2(x - 3)^2 - 7\)

So, \(a = 2\), \(b = 3\), and \(c = -7\).

(b) If the line \(y = kx - 7\) and the curve \(y = 2x^2 - 12x + 11\) do not intersect, then the equation:
\(kx - 7 = 2x^2 - 12x + 11\)
has no real solutions.

Rearranging to standard quadratic form:
\(2x^2 - 12x - kx + 11 + 7 = 0\)
\(2x^2 - (12 + k)x + 18 = 0\)

For no real solutions, the discriminant must be negative (\(\Delta < 0\)):
\(\Delta = [-(12 + k)]^2 - 4(2)(18) < 0\)
\((12 + k)^2 - 144 < 0\)
\((12 + k)^2 < 144\)

Taking square roots:
\(-12 < 12 + k < 12\)

Subtract 12 from all parts:
\(-24 < k < 0\)

So the set of values of \(k\) is the interval \(-24 < k < 0\).

(a)
- M1: Factorise out 2 (or equivalent correct step).
- M1: Complete the square on quadratic expression of the form \(x^2 - 6x\).
- A1: Find the correct expression \(2(x - 3)^2 - 7\).

(b)
- M1: Set up the simultaneous equation and rearrange into standard quadratic form.
- M1: Correctly find the discriminant in terms of \(k\).
- M1: Set the discriminant to be strictly less than 0 and initiate solving the inequality.
- A1.5: Correctly solve to find the range \(-24 < k < 0\).

(i) The perimeter of the sector is given by:
\[ P = 2r + r\theta = 24 \implies r\theta = 24 - 2r \]

The area of the sector is given by:
\[ A = \frac{1}{2}r^2\theta = 32 \implies r^2\theta = 64 \]

Substitute \(r\theta = 24 - 2r\) into the area equation:
\[ r(r\theta) = 64 \implies r(24 - 2r) = 64 \]
\[ 24r - 2r^2 = 64 \]

Dividing the entire equation by 2:
\[ 12r - r^2 = 32 \implies r^2 - 12r + 32 = 0 \]
(as required).

(ii) Solving the quadratic equation:
\[ r^2 - 12r + 32 = 0 \implies (r - 4)(r - 8) = 0 \]
So \(r = 4\) or \(r = 8\).

If \(r = 4\):
\[ \theta = \frac{24 - 2(4)}{4} = \frac{16}{4} = 4\text{ radians} \]

If \(r = 8\):
\[ \theta = \frac{24 - 2(8)}{8} = \frac{8}{8} = 1\text{ radian} \]

Since we are given that \(0 < \theta < \pi\) (and \π \approx 3.142\), the value \θ = 4 is rejected because \(4 > \pi\).

Therefore, the only valid solution is \(r = 8\) and \(\theta = 1\).

(i)
M1: For writing down a correct expression for the perimeter, \(2r + r\theta = 24\).
M1: For writing down a correct expression for the area, \(\frac{1}{2}r^2\theta = 32\).
M1: For substituting \(r\theta\) from the perimeter equation into the area equation to eliminate \(\theta\).
A1: For completing the algebra correctly to arrive at the given quadratic equation: \(r^2 - 12r + 32 = 0\).

(ii)
M1: For solving the quadratic equation to find two values of \(r\) (\(r = 4\) and \(r = 8\)).
A1: For finding the corresponding values of \(\theta\) (\(\theta = 4\) and \(\theta = 1\)).
A1: For selecting \(r = 8, \theta = 1\) and stating a valid reason for rejecting \(\theta = 4\) (since \(4 > \pi\)).

(i) First, find the midpoint \(M\) of the line segment \(AB\):
\[ M = \left(\frac{6 + (-2)}{2}, \frac{1 + 5}{2}\right) = (2, 3) \]

Next, find the gradient of the line segment \(AB\):
\[ m_{AB} = \frac{5 - 1}{-2 - 6} = \frac{4}{-8} = -\frac{1}{2} \]

The gradient of the perpendicular bisector is the negative reciprocal:
\[ m = -\frac{1}{m_{AB}} = 2 \]

Now, find the equation of the perpendicular bisector passing through \(M(2, 3)\) with gradient \(m = 2\):
\[ y - 3 = 2(x - 2) \implies y - 3 = 2x - 4 \implies y = 2x - 1 \]

(ii) To find the coordinates of \(C\), solve \(y = 2x - 1\) and \(3x + y = 14\) simultaneously:
Substitute \(y = 2x - 1\) into \(3x + y = 14\):
\[ 3x + (2x - 1) = 14 \]
\[ 5x - 1 = 14 \implies 5x = 15 \implies x = 3 \]

Now find \(y\):
\[ y = 2(3) - 1 = 5 \]

Thus, the coordinates of \(C\) are \((3, 5)\).

(iii) Any circle passing through \(A\) and \(B\) must have its centre on the perpendicular bisector of \(AB\). Since the centre of this circle also lies on \(L\), the centre must be the intersection point of the perpendicular bisector and \(L\), which is \(C(3, 5)\).

The radius \(r\) of this circle is the distance from the centre \(C(3, 5)\) to either point \(A(6, 1)\) or \(B(-2, 5)\):
\[ r^2 = (6 - 3)^2 + (1 - 5)^2 = 3^2 + (-4)^2 = 9 + 16 = 25 \]

The equation of the circle is:
\[ (x - 3)^2 + (y - 5)^2 = 25 \]

(i)
M1: For calculating the midpoint \(M(2, 3)\) of \(AB\).
M1: For finding the gradient of \(AB\) and taking the negative reciprocal to get \(m = 2\).
M1: For using their midpoint and perpendicular gradient to form a linear equation.
A1: For the correct final equation, e.g., \(y = 2x - 1\) or \(2x - y = 1\).

(ii)
M1: For solving their perpendicular bisector equation simultaneously with \(3x + y = 14\).
A1: For obtaining the correct coordinates \(C(3, 5)\).

(iii)
M1: For identifying that \(C\) is the centre of the circle and calculating the value of \(r^2\) (or \(r = 5\)).
A1: For the correct equation of the circle: \((x - 3)^2 + (y - 5)^2 = 25\).

First, square both sides of the inequality to eliminate the absolute value signs: \((3x - 2)^2 < 4(x + 4)^2\). Expanding both sides: \(9x^2 - 12x + 4 < 4(x^2 + 8x + 16)\), which simplifies to \(9x^2 - 12x + 4 < 4x^2 + 32x + 64\). Rearranging into a standard quadratic inequality gives: \(5x^2 - 44x - 60 < 0\). Factoring the quadratic expression: \((5x + 6)(x - 10) < 0\). The critical values are \(x = -1.2\) and \(x = 10\). Since we require the expression to be less than zero, the solution lies between these critical values: \(-1.2 < x < 10\).

M1: For squaring both sides or setting up two linear equations to find critical values. A1: For obtaining a correct quadratic equation, e.g., \(5x^2 - 44x - 60 = 0\) (or equivalent). M1: For solving their quadratic or system of equations to find two critical values. A1: For correct critical values \(10\) and \(-1.2\) (or \(-\frac{6}{5}\)). M1: For choosing the correct inside region for '<'. A1.14: For the final correct inequality \(-1.2 < x < 10\).

Use the laws of logarithms to simplify the left-hand side: \(\ln(5x + 3) - \ln(x^2) = \ln(4)\), which becomes \(\ln\left(\frac{5x + 3}{x^2}\right) = \ln(4)\). Take exponentials of both sides: \(\frac{5x + 3}{x^2} = 4\), which leads to \(5x + 3 = 4x^2\). Rearrange to form a quadratic equation: \(4x^2 - 5x - 3 = 0\). Solve the quadratic equation using the quadratic formula: \(x = \frac{5 \pm \sqrt{(-5)^2 - 4(4)(-3)}}{2(4)} = \frac{5 \pm \sqrt{73}}{8}\). This gives two potential solutions: \(x \approx 1.693\) or \(x \approx -0.443\). Since the original equation contains the term \(2\ln(x)\), \(x\) must be positive. Thus, the negative solution is rejected. Therefore, the only valid solution is \(x \approx 1.69\) (to 3 significant figures).

M1: Use the power law of logarithms to write \(2\ln(x)\) as \(\ln(x^2)\). M1: Use subtraction law of logarithms to combine terms into a single logarithm. A1: Obtain the equation \(\frac{5x+3}{x^2} = 4\) or equivalent. M1: Solve the resulting quadratic equation \(4x^2 - 5x - 3 = 0\). A1: Obtain the two roots \(x = \frac{5 \pm \sqrt{73}}{8}\). A1.14: State the final answer \(1.69\) and explain or show that the negative root is rejected because \(\ln(x)\) is undefined for \(x < 0\).

We can express the left-hand side in the form \(R\cos(2\theta - \alpha)\) where \(R\cos(2\theta - \alpha) = R\cos(2\theta)\cos(\alpha) + R\sin(2\theta)\sin(\alpha)\). Comparing coefficients with \(3\cos(2\theta) + \sin(2\theta)\) gives \(R\cos\alpha = 3\) and \(R\sin\alpha = 1\). This yields \(R = \sqrt{3^2 + 1^2} = \sqrt{10}\) and \(\tan\alpha = \frac{1}{3} \implies \alpha \approx 18.43^\circ\). So the equation becomes \(\sqrt{10}\cos(2\theta - 18.43^\circ) = 2\), which simplifies to \(\cos(2\theta - 18.43^\circ) = \frac{2}{\sqrt{10}}\). The principal value is \(2\theta - 18.43^\circ = \cos^{-1}\left(\frac{2}{\sqrt{10}}\right) \approx 50.77^\circ\). Since \(0^\circ \le \theta \le 180^\circ\), the range for \(2\theta - 18.43^\circ\) is \(-18.43^\circ \le 2\theta - 18.43^\circ \le 341.57^\circ\). The possible values in this range are: 1) \(2\theta - 18.43^\circ = 50.77^\circ \implies 2\theta = 69.20^\circ \implies \theta \approx 34.6^\circ\). 2) \(2\theta - 18.43^\circ = 360^\circ - 50.77^\circ = 309.23^\circ \implies 2\theta = 327.66^\circ \implies \theta \approx 163.8^\circ\). So the solutions are \(\theta = 34.6^\circ\) and \(\theta = 163.8^\circ\).

M1: Attempt to express the LHS in the form \(R\cos(2\theta - \alpha)\) or \(R\sin(2\theta + \alpha)\). A1: Find \(R = \sqrt{10}\) and \(\alpha \approx 18.43^\circ\) (or equivalent). M1: Set up the equation \(\cos(2\theta - 18.43^\circ) = \frac{2}{\sqrt{10}}\). A1: Obtain one correct value of \(2\theta - 18.43^\circ\) (e.g. \(50.77^\circ\) or \(309.23^\circ\)). A1: Solve to find \(\theta = 34.6^\circ\). A1.14: Solve to find \(\theta = 163.8^\circ\) and no other values in the range.

Use the quotient rule to find \(\frac{dy}{dx}\). Let \(u = \mathrm{e}^{2x}\) and \(v = 2x - 1\). Then \(\frac{du}{dx} = 2\mathrm{e}^{2x}\) and \(\frac{dv}{dx} = 2\). Applying the quotient rule: \(\frac{dy}{dx} = \frac{(2x - 1)(2\mathrm{e}^{2x}) - 2\mathrm{e}^{2x}}{(2x - 1)^2}\). Factor out \(2\mathrm{e}^{2x}\) in the numerator: \(\frac{dy}{dx} = \frac{2\mathrm{e}^{2x}(2x - 1 - 1)}{(2x - 1)^2} = \frac{4\mathrm{e}^{2x}(x - 1)}{(2x - 1)^2}\). At a stationary point, \(\frac{dy}{dx} = 0\), which gives \(\frac{4\mathrm{e}^{2x}(x - 1)}{(2x - 1)^2} = 0\). Since \(4\mathrm{e}^{2x} \ne 0\) for all real \(x\), we must have \(x - 1 = 0 \implies x = 1\). Substitute \(x = 1\) back into the original curve equation: \(y = \frac{\mathrm{e}^{2(1)}}{2(1) - 1} = \mathrm{e}^2\). Thus, the exact coordinates of the stationary point are \((1, \mathrm{e}^2)\).

M1: Apply quotient rule or product rule to differentiate \(y = \frac{\mathrm{e}^{2x}}{2x - 1}\). A1: Obtain a correct expression for \(\frac{dy}{dx}\), e.g., \(\frac{2\mathrm{e}^{2x}(2x - 1) - 2\mathrm{e}^{2x}}{(2x - 1)^2}\). M1: Set \(\frac{dy}{dx} = 0\) and attempt to solve for \(x\). A1: Obtain \(x = 1\). M1: Substitute \(x = 1\) into the original equation to find the y-coordinate. A2.14: Obtain the exact y-coordinate \(\mathrm{e}^2\) and state the coordinate pair as \((1, \mathrm{e}^2)\).

(i) Integrating, we get \(\int_{0}^{2} \frac{6}{3x + 2} \mathrm{d}x = \left[ 6 \cdot \frac{1}{3} \ln|3x + 2| \right]_{0}^{2} = \left[ 2 \ln|3x + 2| \right]_{0}^{2}\). Evaluating at the limits: At \(x = 2\), \(2 \ln(8)\). At \(x = 0\), \(2 \ln(2)\). Subtracting: \(2 \ln(8) - 2 \ln(2) = 2(\ln 8 - \ln 2) = 2 \ln(4) = \ln(4^2) = \ln(16)\). So \(a = 16\). (ii) With 4 intervals from \(x = 0\) to \(x = 2\), the width of each interval is \(h = \frac{2 - 0}{4} = 0.5\). The x-values are \(x_0 = 0\, x_1 = 0.5\, x_2 = 1.0\, x_3 = 1.5\, x_4 = 2.0\). Evaluating \(f(x) = \frac{6}{3x + 2}\) at these points: \(y_0 = 3\, y_1 \approx 1.714286\, y_2 = 1.2\, y_3 \approx 0.923077\, y_4 = 0.75\). Using the trapezium rule formula: \(I \approx \frac{0.5}{2} \left[ 3 + 0.75 + 2(1.714286 + 1.2 + 0.923077) \right] = 0.25 [ 3.75 + 7.674726 ] = 2.85618\). Correct to 3 decimal places, the approximation is \(2.856\).

(i) M1: Integrate to obtain \(k \ln(3x + 2)\). A1: Obtain the correct integral \(2 \ln(3x + 2)\). M1: Substitute limits 2 and 0 correctly and use log laws to simplify. A1: Obtain \(\ln 16\). (ii) M1: Find correct interval width \(h = 0.5\) and the intermediate x-values. A1: Calculate all correct y-values (to at least 3 decimal places). M1: Apply the trapezium rule formula correctly with their values. A0.14: Obtain \(2.856\).

(i) Let \(f(x) = x^3 - 5x + 1\). Evaluating at the endpoints: \(f(2) = 2^3 - 5(2) + 1 = -1\) and \(f(2.5) = 2.5^3 - 5(2.5) + 1 = 4.125\). Since \(f(2) < 0\) and \(f(2.5) > 0\), the change of sign in a continuous function indicates that a root exists between \(x = 2\) and \(x = 2.5\). (ii) Rearranging \(x^3 - 5x + 1 = 0\) gives \(x^3 = 5x - 1\). Since \(x \ne 0\) in our interval, dividing by \(x\) yields \(x^2 = 5 - \frac{1}{x}\). Since \(x > 0\) in this interval, taking the square root gives \(x = \sqrt{5 - \frac{1}{x}}\). (iii) Using the formula \(x_{n+1} = \sqrt{5 - \frac{1}{x_n}}\) with \(x_1 = 2\): \(x_2 = \sqrt{5 - 0.5} = \sqrt{4.5} \approx 2.12132\); \(x_3 = \sqrt{5 - \frac{1}{2.12132}} \approx 2.12805\); \(x_4 = \sqrt{5 - \frac{1}{2.12805}} \approx 2.12840\); \(x_5 = \sqrt{5 - \frac{1}{2.12840}} \approx 2.12842\). The values converge to \(2.128\) correct to 3 decimal places.

(i) M1: Evaluate \(f(2)\) and \(f(2.5)\). A1: Obtain \(f(2) = -1\) and \(f(2.5) = 4.125\), and write a conclusion referring to the change of sign. (ii) M1: For clear algebraic steps showing division of \(x^3 = 5x - 1\) by \(x\) or equivalent. A1: Complete the proof to get \(x = \sqrt{5 - \frac{1}{x}}\). (iii) M1: Apply the iterative formula at least twice. A1: Calculate \(x_2 \approx 2.12132\) and \(x_3 \approx 2.12805\). A0.14: Obtain the final answer \(2.128\), supported by iterations converging to this value.

(i) Use double-angle formulas: \(1 - \cos 2x = 2\sin^2 x\) and \(\sin 2x = 2\sin x \cos x\). Substituting these into the left-hand side: \(\frac{1 - \cos 2x}{\sin 2x} = \frac{2\sin^2 x}{2\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x\). Thus, the identity is proven. (ii) The given equation is \(\frac{2(1 - \cos 2x)}{\sin 2x} = 3\sec^2 x - 5\). Using the identity from part (i) gives \(2\tan x = 3\sec^2 x - 5\). Since \(\sec^2 x = 1 + \tan^2 x\), substitute this to form a quadratic in \(\tan x\): \(2\tan x = 3(1 + \tan^2 x) - 5\), which simplifies to \(3\tan^2 x - 2\tan x - 2 = 0\). Let \(t = \tan x\). Using the quadratic formula on \(3t^2 - 2t - 2 = 0\): \(t = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-2)}}{6} = \frac{1 \pm \sqrt{7}}{3}\). This yields \(\tan x \approx 1.21525\) or \(\tan x \approx -0.54858\). For \(0 < x < \pi\): If \(\tan x = 1.21525 \implies x \approx 0.882\) radians. If \(\tan x = -0.54858 \implies x = \pi + (-0.5011) \approx 2.64\) radians. So the solutions are \(x = 0.882\) and \(x = 2.64\).

(i) M1: Use the double angle formula for \(\cos 2x\) to express \(1 - \cos 2x\) in terms of \(\sin x\). M1: Use \(\sin 2x = 2\sin x \cos x\). A1: Complete proof cleanly to show \(\tan x\). (ii) M1: Recognise the LHS is \(2\tan x\) and substitute \(\sec^2 x = 1 + \tan^2 x\) to obtain a quadratic in \(\tan x\). A1: Obtain the correct quadratic \(3\tan^2 x - 2\tan x - 2 = 0\). M1: Solve this quadratic to find two values of \(\tan x\). A1: Obtain one correct value of \(x\) (e.g. \(0.882\)). A1.14: Obtain the second correct value of \(x\) (e.g. \(2.64\)) and no others in the range.