2025 Cambridge IAL Mathematics (9709) 模擬試題連答案詳解

First, we find the midpoint, \(M\), of the line segment \(AB\): \(M = \left(\frac{2+6}{2}, \frac{-3+5}{2}\right) = (4, 1)\). Next, we find the gradient of the line \(AB\): \(m_{AB} = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2\). Since the bisector is perpendicular to \(AB\), its gradient, \(m_{\perp}\), is: \(m_{\perp} = -\frac{1}{2}\). The equation of the perpendicular bisector is: \(y - 1 = -\frac{1}{2}(x - 4)\). Multiply both sides by 2: \(2y - 2 = -x + 4\). Rearranging into the form \(ax + by = c\) yields: \(x + 2y = 6\).

M1: For finding the correct midpoint of \(AB\) at \( (4, 1) \). M1: For finding the gradient of \(AB\), which is 2. A0.5: For stating the gradient of the perpendicular line is \(-\frac{1}{2}\). M1: For attempting to form the equation of a line using their midpoint and their perpendicular gradient. A1: For the correct simplified equation \(x + 2y = 6\).

Since the car is traveling at a constant speed, the acceleration is zero, meaning the driving force \(F\) is equal to the total resistance forces acting down the slope. The component of the car's weight acting down the slope is: \(W_{\text{slope}} = mg \sin\theta = 1200 \times 10 \times 0.05 = 600\text{ N}\). The total resistance down the slope is the sum of the weight component and the resistance force: \(F = 600 + 300 = 900\text{ N}\). The power \(P\) of the engine is given by the formula: \(P = Fv\) where \(v = 15\text{ m s}^{-1}\). Thus, \(P = 900 \times 15 = 13500\text{ W}\).

M1: For calculating the component of the weight down the slope using \(mg\sin\theta\). A0.5: For getting \(600\text{ N}\). M1: For setting up the equation of motion along the slope \(F = mg\sin\theta + R\) to find the driving force. M1: For using the formula \(P = Fv\). A1: For obtaining the correct power of \(13500\text{ W}\).

To find the total distance traveled, we must first find if and when the velocity changes sign in the interval \(0 \le t \le 3\). Set \(v = 0\): \(3t^2 - 12t + 9 = 0 \implies 3(t - 1)(t - 3) = 0\). Thus, the velocity changes sign at \(t = 1\). We integrate the velocity function to find the displacement function \(s(t)\) (taking \(s(0) = 0\)): \(s(t) = \int (3t^2 - 12t + 9) \, dt = t^3 - 6t^2 + 9t\). Evaluate the displacement at the key times: At \(t = 0\): \(s(0) = 0\). At \(t = 1\): \(s(1) = 1^3 - 6(1)^2 + 9(1) = 4\). At \(t = 3\): \(s(3) = 3^3 - 6(3)^2 + 9(3) = 0\). The distance traveled between \(t = 0\) and \(t = 1\) is \(|s(1) - s(0)| = |4 - 0| = 4\text{ m}\). The distance traveled between \(t = 1\) and \(t = 3\) is \(|s(3) - s(1)| = |0 - 4| = 4\text{ m}\). The total distance traveled is \(4 + 4 = 8\text{ m}\).

M1: For setting \(v = 0\) and solving for \(t\) to find the turning point at \(t = 1\). M1: For integrating \(v\) to obtain the form \(t^3 - 6t^2 + 9t\). A0.5: For the correct integrated expression. M1: For evaluating the displacement at the limits and realizing the need to sum the absolute values of the two segments of motion. A1: For obtaining the correct total distance of \(8\).

To find the points of intersection, we equate the two equations: \(x^2 - kx + 5 = 2x + 1\). Rearranging this into standard quadratic form: \(x^2 - (k + 2)x + 4 = 0\). The line and the curve do not intersect if this quadratic equation has no real roots. This occurs when the discriminant is less than zero (\(b^2 - 4ac < 0\)): \((- (k + 2))^2 - 4(1)(4) < 0 \implies k^2 + 4k + 4 - 16 < 0 \implies k^2 + 4k - 12 < 0\). We factorize the quadratic expression: \((k + 6)(k - 2) < 0\). The critical values are \(k = -6\) and \(k = 2\). Since we require the expression to be less than zero, the set of values of \(k\) is \(-6 < k < 2\).

M1: For equating the two equations and collecting terms on one side to form a quadratic in \(x\). A0.5: For finding the correct coefficients \(a = 1, b = -(k+2), c = 4\). M1: For calculating the discriminant and setting up the inequality \(b^2 - 4ac < 0\). M1: For finding the correct critical values \(k = -6\) and \(k = 2\). A1: For writing the correct inequality \(-6 < k < 2\).

The second term of a geometric progression is given by: \(ar = 12 \implies a = \frac{12}{r}\). The sum to infinity is given by: \(S_{\infty} = \frac{a}{1-r} = 50\). Substitute \(a = \frac{12}{r}\) into the sum to infinity equation: \(\frac{12/r}{1-r} = 50 \implies 12 = 50r(1-r) \implies 50r^2 - 50r + 12 = 0\). Divide the quadratic equation by 2: \(25r^2 - 25r + 6 = 0\). Factorizing the quadratic: \((5r - 2)(5r - 3) = 0\). This gives two possible values for \(r\): \(r = 0.4\) or \(r = 0.6\). If \(r = 0.4\), then \(a = \frac{12}{0.4} = 30\). If \(r = 0.6\), then \(a = \frac{12}{0.6} = 20\). Since we are given that the first term \(a > 25\), we choose \(a = 30\) and \(r = 0.4\).

M1: For using the formula for the \(n\)-th term of a GP to write \(ar = 12\). M1: For using the formula for the sum to infinity to write \(\frac{a}{1-r} = 50\). M1: For eliminating one variable to form a quadratic equation. A0.5: For obtaining a correct quadratic equation, such as \(25r^2 - 25r + 6 = 0\). A1: For solving and selecting the correct pair of values based on \(a > 25\) to get \(a = 30\) and \(r = 0.4\).

(a) The center of the circle is \(C(3, 2)\).

The line \(L\) is \(y = 2x - 1\), which has gradient \(m_1 = 2\).

The line perpendicular to \(L\) that passes through the center \(C(3,2)\) has gradient \(m_2 = -\frac{1}{2}\).

The equation of this perpendicular line is:
\(y - 2 = -\frac{1}{2}(x - 3) \implies 2y - 4 = -x + 3 \implies x + 2y = 7\).

To find the closest point (the point of intersection of these two lines), we solve simultaneously:
\(y = 2x - 1\)
\(x + 2y = 7\)

Substitute \(y\) into the second equation:
\(x + 2(2x - 1) = 7 \implies 5x - 2 = 7 \implies 5x = 9 \implies x = 1.8\) (or \(\frac{9}{5}\)).

Substitute \(x = 1.8\) back to find \(y\):
\(y = 2(1.8) - 1 = 2.6\) (or \(\frac{13}{5}\)).

Thus, the coordinates of the closest point, \(M\), are \((1.8, 2.6)\) or \(\left(\frac{9}{5}, \frac{13}{5}\right)\).

The perpendicular distance from \(C(3,2)\) to \(L\) is the distance between \(C(3,2)\) and \(M\left(\frac{9}{5}, \frac{13}{5}\right)\):
\(\text{Distance} = \sqrt{\left(3 - \frac{9}{5}\right)^2 + \left(2 - \frac{13}{5}\right)^2} = \sqrt{\left(\frac{6}{5}\right)^2 + \left(-\frac{3}{5}\right)^2} = \sqrt{\frac{36}{25} + \frac{9}{25}} = \sqrt{\frac{45}{25}} = \frac{3\sqrt{5}}{5} = \frac{3}{\sqrt{5}}\). (Shown)

(b) Let \(M\) be the midpoint of the chord \(PQ\), which is also the point on \(L\) closest to \(C\).

Thus, the distance \(CM = \frac{3}{\sqrt{5}}\).

The length of the chord \(PQ\) is \(2\sqrt{5}\), so the length of \(PM\) (half of the chord) is:
\(PM = \frac{1}{2} \times 2\sqrt{5} = \sqrt{5}\).

Since \(\triangle CMP\) is a right-angled triangle at \(M\), we apply Pythagoras' theorem:
\(CP^2 = CM^2 + PM^2\)

Since \(P\) lies on the circle, \(CP = r\):
\(r^2 = \left(\frac{3}{\sqrt{5}}\right)^2 + (\sqrt{5})^2 = \frac{9}{5} + 5 = 1.8 + 5 = 6.8 = \frac{34}{5}\).

Since \(r > 0\), the exact value of \(r\) is \(\sqrt{\frac{34}{5}}\) (or \(\frac{\sqrt{170}}{5}\)).

(a)
M1: For finding the gradient of the perpendicular line (\(-\frac{1}{2}\)) and attempting its equation.
A1: Correct equation of the perpendicular line, e.g., \(x + 2y = 7\).
M1: For attempting to solve the simultaneous equations to find the coordinates of \(M\).
A1: Correct coordinates \((1.8, 2.6)\) or equivalent, and showing the distance is \(\frac{3}{\sqrt{5}}\).

(b)
M1: Realising that the midpoint of the chord is \(M\) and that \(PM = \sqrt{5}\).
M1: Applying Pythagoras' theorem \(r^2 = CM^2 + PM^2\).
A1: Substituting correct values into the formula: \(r^2 = \frac{9}{5} + 5\).
A1: Correct exact value of \(r = \sqrt{\frac{34}{5}}\) (or equivalent exact expression).

(a) To find the stationary point, we differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = 4 \left(\frac{1}{2}x^{-1/2}\right) - 1 = 2x^{-1/2} - 1\).

At a stationary point, \(\frac{dy}{dx} = 0\):
\(2x^{-1/2} - 1 = 0 \implies \frac{2}{\sqrt{x}} = 1 \implies \sqrt{x} = 2 \implies x = 4\).

Substitute \(x = 4\) back into the curve equation:
\(y = 4(4)^{1/2} - 4 = 4(2) - 4 = 4\).

So the stationary point is \((4, 4)\).

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

Substitute \(x = 4\):
\(\frac{d^2y}{dx^2} = -(4)^{-3/2} = -\frac{1}{8}\).

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point \((4, 4)\) is a local maximum.

(b) To find the limits of the region enclosed by the curve and the positive \(x\)-axis, we set \(y = 0\):
\(4\sqrt{x} - x = 0 \implies \sqrt{x}(4 - \sqrt{x}) = 0\).

This gives \(x = 0\) and \(x = 16\).

The area \(A\) is given by:
\(A = \int_{0}^{16} (4x^{1/2} - x) \, dx\).

Integrating each term:
\(A = \left[ \frac{4x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{16} = \left[ \frac{8}{3}x^{3/2} - \frac{1}{2}x^2 \right]_{0}^{16}\).

Evaluate at the upper limit \(x = 16\):
\(\frac{8}{3}(16)^{3/2} - \frac{1}{2}(16)^2 = \frac{8}{3}(64) - 128 = \frac{512}{3} - 128 = \frac{128}{3}\).

Evaluate at the lower limit \(x = 0\):
\(0\).

So the exact area of the region is \(\frac{128}{3}\) (or \(42\frac{2}{3}\)).

(a)
M1: For differentiating to get \(\frac{dy}{dx} = 2x^{-1/2} - 1\) (at least one term correct).
A1: Finding \(x = 4\) and \(y = 4\).
M1: Finding the second derivative \(\frac{d^2y}{dx^2} = -x^{-3/2}\) and evaluating at \(x = 4\).
A1: Correctly identifying the stationary point \((4, 4)\) as a maximum with valid reasoning.

(b)
M1: For setting \(y = 0\) and finding the correct limits \(x = 0\) and \(x = 16\).
M1: For integrating to obtain \(\frac{8}{3}x^{3/2} - \frac{1}{2}x^2\) (allow one coefficient error).
A1: Correct integration.
A1: Substituting limits correctly to get \(\frac{128}{3}\) or \(42\frac{2}{3}\).

(a) Since the range of \(\cos(2x)\) is \([-1, 1]\) and \(q > 0\), the maximum value of \(f(x)\) occurs when \(\cos(2x) = -1\):
\(f_{\max} = p - q(-1) = p + q = 9\).

The minimum value of \(f(x)\) occurs when \(\cos(2x) = 1\):
\(f_{\min} = p - q(1) = p - q = 3\).

Solving these simultaneous equations:
Adding the equations gives \(2p = 12 \implies p = 6\).
Subtracting the equations gives \(2q = 6 \implies q = 3\).

Thus, \(p = 6\) and \(q = 3\).

(b) We solve the inequality \(f(x) \ge 7.5\) for \(0 \le x \le \pi\):
\(6 - 3 \cos(2x) \ge 7.5\)
\(-3 \cos(2x) \ge 1.5\)
\(\cos(2x) \le -0.5\).

Let \(\theta = 2x\). Since \(0 \le x \le \pi\), we have \(0 \le \theta \le 2\pi\).

We solve \(\cos \theta \le -0.5\).

The boundary equation is \(\cos \theta = -0.5\).

The basic angle (reference angle) is \(\theta_{\text{basic}} = \cos^{-1}(0.5) = \frac{\pi}{3}\).

Since \(\cos \theta\) is negative, \(\theta\) lies in the second and third quadrants:
\(\theta_1 =
\pi - \frac{\pi}{3} = \frac{2\pi}{3}\)
\(\theta_2 = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\).

The inequality \(\cos \theta \le -0.5\) is satisfied when:
\(\frac{2\pi}{3} \le \theta \le \frac{4\pi}{3}\).

Since \(\theta = 2x\):
\(\frac{2\pi}{3} \le 2x \le \frac{4\pi}{3} \implies \frac{\pi}{3} \le x \le \frac{2\pi}{3}\).

(a)
M1: Realising that the maximum is \(p+q\) and the minimum is \(p-q\).
A1: Writing the simultaneous equations \(p+q = 9\) and \(p-q = 3\).
A1: Correctly solving to find \(p = 6\) and \(q = 3\).

(b)
M1: Setting up the inequality \(6 - 3\cos(2x) \ge 7.5\) and simplifying to \(\cos(2x) \le -0.5\).
M1: Solving \(\cos(2x) = -0.5\) to find at least one correct boundary angle in terms of \(\pi\) (e.g. \(2x = \frac{2\pi}{3}\) or \(2x = \frac{4\pi}{3}\)).
A1: Finding both correct boundary values for \(x\): \(x = \frac{\pi}{3}\) and \(x = \frac{2\pi}{3}\).
M1: For selecting the correct interval between the boundary values.
A1: Correct final inequality: \\frac{\pi}{3} \le x \le \frac{2\pi}{3}\\ (accept interval notation \([\frac{\pi}{3}, \frac{2\pi}{3}]\)).

(a) To find the equation of the curve, we integrate \( f'(x) \):
\( f(x) = \int \left( 16(2x-5)^{-3} - 2 \right) dx \)
\( f(x) = \frac{16(2x-5)^{-2}}{-2 \times 2} - 2x + C \)
\( f(x) = -\frac{4}{(2x-5)^2} - 2x + C \)
Substitute the point \( (3, 5) \) to find the constant of integration \( C \):
\( 5 = -\frac{4}{(2(3)-5)^2} - 2(3) + C \)
\( 5 = -4 - 6 + C \implies C = 15 \)
So, the equation of the curve is:
\( f(x) = -\frac{4}{(2x-5)^2} - 2x + 15 \)

(b) To find the stationary point, set \( f'(x) = 0 \):
\( \frac{16}{(2x-5)^3} - 2 = 0 \implies (2x-5)^3 = 8 \implies 2x - 5 = 2 \implies x = 3.5 \)
Substitute \( x = 3.5 \) into \( f(x) \):
\( y = -\frac{4}{(2(3.5)-5)^2} - 2(3.5) + 15 = -1 - 7 + 15 = 7 \)
So the stationary point is \( (3.5, 7) \).
To determine its nature, find the second derivative:
\( f''(x) = \frac{d}{dx}\left(16(2x-5)^{-3} - 2\right) = -96(2x-5)^{-4} \)
At \( x = 3.5 \):
\( f''(3.5) = -96(2)^{-4} = -6 \)
Since \( f''(3.5) < 0 \), the stationary point is a maximum.

(c) Completing the square on \( g(x) \):
\( g(x) = 2(x^2 - 7x) + 31 \)
\( g(x) = 2\left(x - 3.5\right)^2 - 2(3.5)^2 + 31 \)
\( g(x) = 2(x - 3.5)^2 - 24.5 + 31 \)
\( g(x) = 2(x-3.5)^2 + 6.5 \)

(d) The quadratic function has its vertex at \( (3.5, 6.5) \). For \( g(x) \) to have an inverse, the domain must be restricted to a region where the function is one-to-one. Since the domain is defined for \( x \le k \), the largest possible value of \( k \) is the \( x \)-coordinate of the vertex:
\( k = 3.5 \)
To find the inverse, set \( y = 2(x-3.5)^2 + 6.5 \):
\( y - 6.5 = 2(x-3.5)^2 \implies \frac{y-6.5}{2} = (x-3.5)^2 \)
Since \( x \le 3.5 \), we take the negative square root:
\( x - 3.5 = -\sqrt{\frac{y-6.5}{2}} \implies x = 3.5 - \sqrt{\frac{y-6.5}{2}} \)
Hence, the inverse function is:
\( g^{-1}(x) = 3.5 - \sqrt{\frac{x-6.5}{2}} \) (or equivalent, such as \( 3.5 - \sqrt{0.5x - 3.25} \)) for \( x \ge 6.5 \).

(a)
- M1: Attempt to integrate \( f'(x) \) with at least one power increased by 1.
- A1: Correct integration of the fractional term: \( -4(2x-5)^{-2} \).
- A1: Correct integration of the constant term: \( -2x \).
- M1: Substitute \( (3, 5) \) into an expression with a constant of integration \( C \).
- A1: Correct value of \( C = 15 \) and correct final equation.

(b)
- M1: Set \( f'(x) = 0 \) and solve for \( x \).
- A1: Obtain \( x = 3.5 \) (or equivalent).
- A1: Obtain \( y = 7 \).
- M1: Evaluate the second derivative (or alternative method) at \( x = 3.5 \).
- A0.5: Show \( f''(3.5) = -6 < 0 \) and state that it is a maximum.

(c)
- B1: Correctly write in the form \( 2(x - 3.5)^2 + c \).
- B1: Correctly find \( c = 6.5 \).

(d)
- B1: State \( k = 3.5 \) (accept \( \frac{7}{2} \)).
- M1: Make \( x \) the subject of their completed-square equation.
- A1: Correctly identify the negative square root to get \( g^{-1}(x) = 3.5 - \sqrt{\frac{x-6.5}{2}} \).

(a) When \( x = 2 \):
\( y = \frac{4}{(2(2)-3)^{1.5}} = \frac{4}{1^{1.5}} = 4 \)
So the point of contact is \( (2, 4) \).
Now differentiate \( y = 4(2x-3)^{-1.5} \) with respect to \( x \):
\( \frac{dy}{dx} = 4 \times (-1.5)(2x-3)^{-2.5} \times 2 \)
\( \frac{dy}{dx} = -12(2x-3)^{-2.5} \)
Substitute \( x = 2 \) into the derivative to find the gradient of the tangent:
\( m = -12(2(2)-3)^{-2.5} = -12 \)
Equation of the tangent:
\( y - 4 = -12(x-2) \)
\( y - 4 = -12x + 24 \implies 12x + y = 28 \)

(b) Find the intercepts of the tangent \( 12x + y = 28 \):
For \( A \) on the \( x \)-axis, set \( y = 0 \):
\( 12x = 28 \implies x = \frac{7}{3} \implies A = \left(\frac{7}{3}, 0\right) \)
For \( B \) on the \( y \)-axis, set \( x = 0 \):
\( y = 28 \implies B = (0, 28) \)
The area of the right-angled triangle \( OAB \) is:
\( \text{Area} = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times \frac{7}{3} \times 28 = \frac{98}{3} \) (or \( 32\frac{2}{3} \))

(c) The formula for the volume of revolution about the \( x \)-axis is:
\( V = \pi \int_{a}^{b} y^2 dx \)
Here, \( y^2 = \left( \frac{4}{(2x-3)^{1.5}} \right)^2 = \frac{16}{(2x-3)^3} = 16(2x-3)^{-3} \).
\( V = \pi \int_{2}^{4} 16(2x-3)^{-3} dx \)
Integrate \( 16(2x-3)^{-3} \):
\( \int 16(2x-3)^{-3} dx = \frac{16(2x-3)^{-2}}{-2 \times 2} = -4(2x-3)^{-2} \)
Now evaluate between the limits \( 2 \) and \( 4 \):
\( V = \pi \left[ -\frac{4}{(2x-3)^2} \right]_{2}^{4} \)
At the upper limit \( x = 4 \):
\( -\frac{4}{(8-3)^2} = -\frac{4}{25} \)
At the lower limit \( x = 2 \):
\( -\frac{4}{(4-3)^2} = -4 \)
So the volume is:
\( V = \pi \left( -\frac{4}{25} - (-4) \right) = \pi \left( 4 - \frac{4}{25} \right) = \pi \left( \frac{96}{25} \right) = \frac{96}{25}\pi \) (or \( 3.84\pi \))

(a)
- B1: Find the correct \( y \)-coordinate at \( x = 2 \) as \( 4 \).
- M1: Attempt to differentiate \( y = 4(2x-3)^{-1.5} \) using the chain rule.
- A1: Obtain correct derivative \( \frac{dy}{dx} = -12(2x-3)^{-2.5} \).
- M1: Substitute \( x = 2 \) to find gradient and formulate the equation of the tangent.
- A0.5: Obtain \( 12x + y = 28 \) (or equivalent form).

(b)
- M1: Identify \( x \) and \( y \) intercepts of their tangent line.
- A1: Correct coordinates \( A\left(\frac{7}{3}, 0\right) \) and \( B(0, 28) \) (or correct from their tangent).
- A1: Calculate the area of the triangle to get \( \frac{98}{3} \) (or exact equivalent).

(c)
- M1: Set up the volume integral \( V = \pi \int y^2 dx \).
- A1: Obtain correct integrand \( 16(2x-3)^{-3} \).
- M1: Attempt to integrate the power of \( (2x-3) \).
- A1: Correctly integrated expression: \( -4(2x-3)^{-2} \).
- M1: Show correct substitution of limits \( 4 \) and \( 2 \) into their integrated expression.
- A1: Correct numerical evaluation of the terms: \( -\frac{4}{25} \) and \( -4 \).
- A1: Obtain the final exact answer \( \frac{96}{25}\pi \) (or \( 3.84\pi \)).

Let \(\frac{2x^2 - 7x + 13}{(x - 1)(x - 2)^2} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2}\). Multiplying both sides by the denominator gives \(2x^2 - 7x + 13 = A(x - 2)^2 + B(x - 1)(x - 2) + C(x - 1)\). Substituting \(x = 1\) gives \(2(1)^2 - 7(1) + 13 = A(1 - 2)^2\), which simplifies to \(8 = A\), so \(A = 8\). Substituting \(x = 2\) gives \(2(2)^2 - 7(2) + 13 = C(2 - 1)\), which simplifies to \(7 = C\), so \(C = 7\). To find \(B\), compare the coefficients of \(x^2\): \(2 = A + B\). Since \(A = 8\), we have \(2 = 8 + B\), which gives \(B = -6\). Therefore, the partial fractions are \(\frac{8}{x - 1} - \frac{6}{x - 2} + \frac{7}{(x - 2)^2}\).

M1: State a correct form of partial fractions with constants \(A\), \(B\), and \(C\). A1: Obtain \(A = 8\). A1: Obtain \(C = 7\). M1: Use a valid method to find \(B\), such as equating coefficients or substituting a third value of \(x\). A0.5: Obtain \(B = -6\) and state the final correct expression.

First, express \(w\) in polar form. The modulus of \(w\) is \(|w| = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = 8\). The argument of \(w\) is \(\theta = \pi - \arctan\left(\frac{4\sqrt{3}}{4}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\). Thus, \(w = 8\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}\). The equation \(z^3 = 8\mathrm{e}^{\mathrm{i}\left(\frac{2\pi}{3} + 2k\pi\right)}\) for integers \(k\) gives roots of the form \(z = 8^{1/3}\mathrm{e}^{\mathrm{i}\left(\frac{2\pi/3 + 2k\pi}{3}\right)}\). The modulus of each root is \(r = 8^{1/3} = 2\). Taking \(k = 0\), the argument is \(\frac{2\pi}{9}\), giving the root \(2\mathrm{e}^{\mathrm{i}\frac{2\pi}{9}}\). Taking \(k = 1\), the argument is \(\frac{2\pi/3 + 2\pi}{3} = \frac{8\pi}{9}\), giving the root \(2\mathrm{e}^{\mathrm{i}\frac{8\pi}{9}}\). Taking \(k = -1\), the argument is \(\frac{2\pi/3 - 2\pi}{3} = -\frac{4\pi}{9}\), giving the root \(2\mathrm{e}^{-\mathrm{i}\frac{4\pi}{9}}\).

M1: Find the modulus of \(w\) as 8 and find the argument of \(w\) as \(\frac{2\pi}{3}\). A1: State the polar form of \(w\) correctly. M1: Use De Moivre's theorem or fractional index to find the modulus \(r = 2\) and set up the arguments. A1: Find two correct roots. A0.5: Find the third correct root in the specified range.

Using the factor theorem, since \(p(x)\) is divisible by \(x - 2\), we have \(p(2) = 0\), which gives \(3(2)^3 + a(2)^2 + b(2) - 10 = 0\). Simplifying, we get \(24 + 4a + 2b - 10 = 0\), so \(2a + b = -7\). Using the remainder theorem, since the remainder when divided by \(x + 1\) is \(-12\), we have \(p(-1) = -12\), which gives \(3(-1)^3 + a(-1)^2 + b(-1) - 10 = -12\). Simplifying, we get \(-3 + a - b - 10 = -12\), so \(a - b = 1\). Adding the two equations \(2a + b = -7\) and \(a - b = 1\) yields \(3a = -6\), so \(a = -2\). Substituting \(a = -2\) into \(a - b = 1\) gives \(b = -3\). The polynomial is therefore \(p(x) = 3x^3 - 2x^2 - 3x - 10\). Since \(x - 2\) is a factor, we can perform polynomial division or algebraic matching to get \(p(x) = (x - 2)(3x^2 + 4x + 5)\). For the quadratic factor \(3x^2 + 4x + 5\), the discriminant is \(4^2 - 4(3)(5) = 16 - 60 = -44 < 0\). Since the discriminant is negative and the leading coefficient is positive, \(3x^2 + 4x + 5 > 0\) for all real \(x\). Thus, \(p(x) > 0\) if and only if \(x - 2 > 0\), which gives \(x > 2\).

M1: Use the factor theorem to set up an equation in terms of \(a\) and \(b\). M1: Use the remainder theorem to set up a second equation in terms of \(a\) and \(b\). A1: Solve the simultaneous equations to find \(a = -2\) and \(b = -3\). M1: Factorise the polynomial to obtain the quadratic factor \(3x^2 + 4x + 5\). A0.5: Show that the quadratic factor is always positive and conclude that the solution is \(x > 2\).

The inequality \(|z - (3 + 4\mathrm{i})| \le 2\) represents a closed disc of radius \(R = 2\) centered at the point \(C\) representing \(3 + 4\mathrm{i}\). The distance from the origin to the center \(C\) is \(|OC| = \sqrt{3^2 + 4^2} = 5\). (i) The greatest value of \(|z|\) is \(|OC| + R = 5 + 2 = 7\). The least value of \(|z|\) is \(|OC| - R = 5 - 2 = 3\). (ii) The greatest value of \(\arg z\) occurs at the upper tangent line from the origin to the circle. Let \(\theta\) be the argument of the center \(C\), so \(\theta = \arctan\left(\frac{4}{3}\right) \approx 0.9273\) radians. Let \(\alpha\) be the angle between the line \(OC\) and the tangent line. In the right-angled triangle formed by the origin, the tangent point, and the center \(C\), we have \(\sin\alpha = \frac{\text{radius}}{|OC|} = \frac{2}{5} = 0.4\), which gives \(\alpha = \arcsin(0.4) \approx 0.4115\) radians. The greatest value of \(\arg z\) is \(\theta + \alpha = 0.9273 + 0.4115 = 1.3388 \approx 1.34\) radians.

M1: Calculate the distance from the origin to the center of the circle as 5. A1: State the greatest value of \(|z|\) is 7 and the least value is 3. M1: Express the maximum argument as \(\theta + \alpha\), where \(\theta = \arg(C)\) and \(\sin\alpha = 0.4\). A1: Calculate \(\theta = 0.927\) and \(\alpha = 0.412\) (or equivalent in degrees then converted). A0.5: Obtain the final answer of 1.34 radians.

Since both sides of the inequality are non-negative, we can square both sides: \((2x - 5)^2 < 9(x + 1)^2\). Expanding both sides gives \(4x^2 - 20x + 25 < 9(x^2 + 2x + 1)\), which simplifies to \(4x^2 - 20x + 25 < 9x^2 + 18x + 9\). Rearranging all terms to one side yields \(0 < 5x^2 + 38x - 16\). Factorising the quadratic expression, we get \((5x - 2)(x + 8) > 0\). The critical values are \(x = 0.4\) and \(x = -8\). Since we require the expression to be greater than zero, the solution set is \(x < -8\) or \(x > 0.4\).

M1: Square both sides and attempt to expand. A1: Obtain the correct expanded inequality \(4x^2 - 20x + 25 < 9x^2 + 18x + 9\). M1: Rearrange to form a three-term quadratic and find critical values by factorising or using the formula. A1: Identify the critical values as \(-8\) and \(0.4\). A0.5: State the correct final range \(x < -8\) or \(x > 0.4\).

(a) We equate the components of the two lines to find the point of intersection:

- \( x \)-components: \( 1 + \lambda = 4 + 2\mu \implies \lambda - 2\mu = 3 \) (Equation 1)
- \( y \)-components: \( 2 - \lambda = 4 + 3\mu \implies \lambda + 3\mu = -2 \) (Equation 2)

Subtracting Equation 2 from Equation 1:
\[ (\lambda - 2\mu) - (\lambda + 3\mu) = 3 - (-2) \]
\[ -5\mu = 5 \implies \mu = -1 \]

Substituting \( \mu = -1 \) back into Equation 1:
\[ \lambda - 2(-1) = 3 \implies \lambda = 1 \]

Now, we equate the \( z \)-components:
\[ -1 + 2\lambda = 3 + a\mu \]

Substituting \( \lambda = 1 \) and \( \mu = -1 \):
\[ -1 + 2(1) = 3 + a(-1) \]
\[ 1 = 3 - a \implies a = 2 \]

(b) The direction vectors of the two lines are \( \mathbf{d}_1 = \mathbf{i} - \mathbf{j} + 2\mathbf{k} \) and \( \mathbf{d}_2 = 2\mathbf{i} + 3\mathbf{j} + 2\mathbf{k} \).

Let \( \theta \) be the angle between the two direction vectors:
\[ \cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|} \]

Calculate the scalar product:
\[ \mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(2) + (-1)(3) + (2)(2) = 2 - 3 + 4 = 3 \]

Calculate the magnitudes:
\[ |\mathbf{d}_1| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6} \]
\[ |\mathbf{d}_2| = \sqrt{2^2 + 3^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \]

Thus:
\[ \cos\theta = \frac{3}{\sqrt{6}\sqrt{17}} = \frac{3}{\sqrt{102}} \approx 0.29704 \]
\[ \theta = \arccos(0.29704) \approx 72.7^{\circ} \]

(a)
- M1: Equate components to form at least two equations in \( \lambda \) and \( \mu \).
- A1: Solve the simultaneous equations to find correct values \( \lambda = 1 \) and \( \mu = -1 \).
- M1: Equate the third components using their values of \( \lambda \) and \( \mu \).
- A1.5: Obtain \( a = 2 \).

(b)
- M1: Use correct formula for the scalar product of the two direction vectors.
- A1: Obtain correct scalar product \( 3 \).
- A1: Calculate correct magnitudes \( \sqrt{6} \) and \( \sqrt{17} \).
- A1: Obtain final angle \( 72.7^{\circ} \) (accept \( 72.7 \)).

(a) We expand \( R\cos(2\theta + \alpha) \) using the compound angle identity:
\[ R\cos(2\theta + \alpha) = R\cos 2\theta \cos\alpha - R\sin 2\theta \sin\alpha \]

Comparing this with \( 3\cos 2\theta - 4\sin 2\theta \), we get:
\[ R\cos\alpha = 3 \]
\[ R\sin\alpha = 4 \]

We find \( R \):
\[ R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \]

We find \( \alpha \):
\[ \tan\alpha = \frac{4}{3} \implies \alpha = \arctan\left(\frac{4}{3}\right) \approx 0.927295 \approx 0.9273 \text{ radians} \]

So the expression is \( 5\cos(2\theta + 0.9273) \).

(b) The equation becomes:
\[ 5\cos(2\theta + 0.9273) = 2 \]
\[ \cos(2\theta + 0.9273) = 0.4 \]

Given \( 0 < \theta < \pi \), we have:
\[ 0 < 2\theta < 2\pi \implies 0.9273 < 2\theta + 0.9273 < 2\pi + 0.9273 \approx 7.2105 \]

The principal value is:
\[ 2\theta + 0.9273 = \arccos(0.4) \approx 1.1593 \]

Since \( 1.1593 \) lies in the interval \( (0.9273, 7.2105) \):
\[ 2\theta = 1.1593 - 0.9273 = 0.2320 \implies \theta \approx 0.116 \text{ radians} \]

The second value in the interval is:
\[ 2\theta + 0.9273 = 2\pi - 1.1593 \approx 6.2832 - 1.1593 = 5.1239 \]
\[ 2\theta = 5.1239 - 0.9273 = 4.1966 \implies \theta \approx 2.10 \text{ radians} \]

(a)
- B1: State \( R = 5 \).
- M1: Use \( \tan\alpha = \frac{4}{3} \) or equivalent.
- A1.5: Obtain \( \alpha = 0.9273 \) (or 0.927).

(b)
- B1: Rewrite the equation as \( \cos(2\theta + 0.9273) = 0.4 \).
- M1: Find at least one value of \( 2\theta + 0.9273 \) in the interval \( (0.9273, 7.2105) \).
- A1: Obtain \( \theta = 0.116 \).
- M1: Find the second valid angle value for \( 2\theta + 0.9273 \).
- A1: Obtain \( \theta = 2.10 \) (accept 2.1).

(a) Let the foot of the perpendicular be \( F \). Since \( F \) lies on the line \( l \), its position vector is given by:
\[ \mathbf{f} = (2 + t)\mathbf{i} + (-1 + 2t)\mathbf{j} + (1 + 3t)\mathbf{k} \]

We find the vector \( \vec{PF} = \mathbf{f} - \mathbf{p} \):
\[ \vec{PF} = \begin{pmatrix} 2 + t \\ -1 + 2t \\ 1 + 3t \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} = \begin{pmatrix} t \\ 2t - 1 \\ 3t - 4 \end{pmatrix} \]

The direction vector of the line \( l \) is \( \mathbf{d} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \).

Since \( PF \) is perpendicular to the line \( l \), the scalar product of \( \vec{PF} \) and \( \mathbf{d} \) must be zero:
\[ \vec{PF} \cdot \mathbf{d} = 0 \]
\[ t(1) + (2t - 1)(2) + (3t - 4)(3) = 0 \]
\[ t + 4t - 2 + 9t - 12 = 0 \]
\[ 14t - 14 = 0 \implies t = 1 \]

Substituting \( t = 1 \) back into the equation of \( l \) to find the position vector of \( F \):
\[ \mathbf{f} = (2 + 1)\mathbf{i} + (-1 + 2(1))\mathbf{j} + (1 + 3(1))\mathbf{k} = 3\mathbf{i} + \mathbf{j} + 4\mathbf{k} \]

(b) The perpendicular distance from \( P \) to \( l \) is the magnitude of \( \vec{PF} \).

Using \( t = 1 \), we have:
\[ \vec{PF} = \begin{pmatrix} 1 \\ 2(1) - 1 \\ 3(1) - 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -1
\end{pmatrix} \]

Calculate the magnitude:
\[ |\vec{PF}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \]

(a)
- M1: Express a general vector from \( P \) to the line \( l \) in terms of \( t \).
- A1: Obtain correct vector \( \vec{PF} = \begin{pmatrix} t \\ 2t-1 \\ 3t-4 \end{pmatrix} \) or equivalent.
- M1: Set up the scalar product of \( \vec{PF} \) and the direction vector \( \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) equal to zero.
- A1: Solve for \( t = 1 \).
- M1: Substitute their \( t \) into the equation of \( l \) to get coordinates/vector of \( F \).
- A0.5: Obtain \( 3\mathbf{i} + \mathbf{j} + 4\mathbf{k} \) (or equivalent column vector format).

(b)
- M1: Find vector \( \vec{PF} \) using their \( F \) and \( P \).
- M1: Apply distance formula to find magnitude of their vector \( \vec{PF} \).
- A1: Obtain exact value \( \sqrt{3} \).

(a) To prove the identity, we start with the Left Hand Side (LHS) and combine the terms over a common denominator:
\[ \text{LHS} = \frac{\sin 3\theta \cos \theta - \cos 3\theta \sin \theta}{\sin \theta \cos \theta} \]

We apply the compound angle formula \( \sin(A - B) = \sin A \cos B - \cos A \sin B \) to the numerator, with \( A = 3\theta \) and \( B = \theta \):
\[ \sin 3\theta \cos \theta - \cos 3\theta \sin \theta = \sin(3\theta - \theta) = \sin 2\theta \]

So the expression becomes:
\[ \text{LHS} = \frac{\sin 2\theta}{\sin \theta \cos \theta} \]

Now, we apply the double angle identity \( \sin 2\theta = 2\sin\theta\cos\theta \) to the numerator:
\[ \text{LHS} = \frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} \]

For \( \sin\theta \ne 0 \) and \( \cos\theta \ne 0 \), we can cancel the terms in the numerator and denominator:
\[ \text{LHS} = 2 \]

Hence, the identity is proved.

(b) Using the identity proved in part (a), we substitute \( 2 \) for the left-hand side of the equation:
\[ 2 = 4\cos^2\theta - 1 \]
\[ 4\cos^2\theta = 3 \]
\[ \cos^2\theta = 0.75 \]
\[ \cos\theta = \pm \frac{\sqrt{3}}{2} \]

We solve for \( \theta \) in the interval \( 0^{\circ} < \theta < 180^{\circ} \):
- For \( \cos\theta = \frac{\sqrt{3}}{2} \):
\[ \theta = 30^{\circ} \]
- For \( \cos\theta = -\frac{\sqrt{3}}{2} \):
\[ \theta = 180^{\circ} - 30^{\circ} = 150^{\circ} \]

Both solutions are valid as they are within the specified interval.

(a)
- M1: Express LHS as a single fraction with denominator \( \sin\theta\cos\theta \).
- M1: Apply compound angle identity \( \sin(3\theta - \theta) \) correctly to the numerator.
- A1: Obtain \( \frac{\sin 2\theta}{\sin\theta\cos\theta} \).
- M1: Apply double angle identity \( \sin 2\theta = 2\sin\theta\cos\theta \) correctly.
- A0.5: Simplify to obtain \( 2 \) and state conclusion.

(b)
- B1: Use the identity from (a) to write the equation as \( 2 = 4\cos^2\theta - 1 \).
- M1: Solve for \( \cos\theta \) to obtain \( \cos\theta = \pm\frac{\sqrt{3}}{2} \) (or equivalent decimal).
- A1: Obtain \( \theta = 30^{\circ} \).
- A1: Obtain \( \theta = 150^{\circ} \).

(i) We separate the variables in the differential equation:
$$\int \frac{4+y}{y^2} \text{d}y = \int \frac{2 \ln x}{x} \text{d}x$$

Integrating the left-hand side:
$$\int \left( \frac{4}{y^2} + \frac{1}{y} \right) \text{d}y = -\frac{4}{y} + \ln y$$

Integrating the right-hand side, we use the substitution \(u = \ln x\), so \(\text{d}u = \frac{1}{x} \text{d}x\):
$$\int 2u \text{d}u = u^2 + C = (\ln x)^2 + C$$

Equating the two sides gives:
$$-\frac{4}{y} + \ln y = (\ln x)^2 + C$$

Using the initial conditions \(x = 1\) and \(y = 4\):
$$-\frac{4}{4} + \ln 4 = (\ln 1)^2 + C \implies -1 + \ln 4 = C$$

Substituting \(C\) back into the equation:
$$-\frac{4}{y} + \ln y = (\ln x)^2 - 1 + \ln 4$$

Rearranging to group the logarithmic terms:
$$\ln y - \ln 4 + 1 - \frac{4}{y} = (\ln x)^2$$
$$\ln(y/4) + 1 - \frac{4}{y} = (\ln x)^2$$

(ii) When \(x = e\), we substitute \(x = e\) and use the fact that \(\ln e = 1\):
$$\ln(y/4) + 1 - \frac{4}{y} = (\ln e)^2 = 1$$
$$\ln(y/4) - \frac{4}{y} = 0$$
$$\ln(y/4) = \frac{4}{y}$$

Taking the exponential of both sides:
$$\frac{y}{4} = e^{4/y} \implies y = 4e^{4/y}$$

(iii) Using the iterative formula \(y_{n+1} = 4e^{4/y_n}\) with \(y_1 = 7.0\):

\(y_1 = 7.0\)
\(y_2 = 4 e^{4/7.0} = 7.0832\)
\(y_3 = 4 e^{4/7.08317} = 7.0358\)
\(y_4 = 4 e^{4/7.03584} = 7.0627\)
\(y_5 = 4 e^{4/7.06265} = 7.0474\)
\(y_6 = 4 e^{4/7.04743} = 7.0561\)
\(y_7 = 4 e^{4/7.05606} = 7.0512\)
\(y_8 = 4 e^{4/7.05116} = 7.0539\)
\(y_9 = 4 e^{4/7.05394} = 7.0524\)

Thus, the value of \(y\) is \(7.05\) correct to 3 significant figures.

(i)
M1: Attempt to separate variables and integrate both sides.
A1: Obtain correct LHS term \(-\frac{4}{y} + \ln y\).
M1: Attempt integration of RHS using substitution or inspection.
A1: Obtain correct RHS term \((\ln x)^2\).
M1: Substitute \(x = 1, y = 4\) to find the constant \(C\).
A1: Obtain \(C = \ln 4 - 1\) (or equivalent).
A1: Obtain the given relation \(\ln(y/4) + 1 - 4/y = (\ln x)^2\) with clear intermediate steps.

(ii)
M1: Substitute \(x = e\) and apply \(\ln e = 1\).
A1: Show steps clearly to arrive at \(y = 4e^{4/y}\).

(iii)
M1: Calculate at least two iterations to 5 s.f. correctly.
A1: State the sequence of values \(7.0832\), \(7.0358\), \(7.0627\), etc.
A1: State final answer \(7.05\) correct to 3 s.f.

(i) To find the stationary point, we differentiate \(y = \frac{\ln x}{x^3}\) using the quotient rule:
$$\frac{\text{d}y}{\text{d}x} = \frac{\frac{1}{x} \cdot x^3 - \ln x \cdot 3x^2}{(x^3)^2} = \frac{x^2 - 3x^2 \ln x}{x^6} = \frac{1 - 3\ln x}{x^4}$$

Setting the derivative to 0 for a stationary point:
$$1 - 3\ln x = 0 \implies \ln x = \frac{1}{3} \implies x = e^{1/3}$$

Substituting \(x = e^{1/3}\) back into the equation of \(C\):
$$y = \frac{\ln(e^{1/3})}{(e^{1/3})^3} = \frac{1/3}{e} = \frac{1}{3e}$$

So the exact coordinates of the stationary point are \(\left(e^{1/3}, \frac{1}{3e}\right)\).

(ii) The area of the region \(R\) is given by:
$$\text{Area} = \int_{1}^{e} \frac{\ln x}{x^3} \text{d}x$$

We use integration by parts with \(u = \ln x\) and \(\text{d}v = x^{-3} \text{d}x\), which gives \(\text{d}u = \frac{1}{x} \text{d}x\) and \(v = -\frac{1}{2x^2}\):
$$\int \frac{\ln x}{x^3} \text{d}x = -\frac{\ln x}{2x^2} - \int -\frac{1}{2x^3} \text{d}x = -\frac{\ln x}{2x^2} - \frac{1}{4x^2}$$

Evaluating this from \(1\) to \(e\):
$$\left[ -\frac{\ln x}{2x^2} - \frac{1}{4x^2} \right]_{1}^{e} = \left( -\frac{\ln e}{2e^2} - \frac{1}{4e^2} \right) - \left( -\frac{\ln 1}{2(1)^2} - \frac{1}{4(1)^2} \right)$$
$$= \left( -\frac{1}{2e^2} - \frac{1}{4e^2} \right) - \left( 0 - \frac{1}{4} \right)$$
$$= -\frac{3}{4e^2} + \frac{1}{4} = \frac{1}{4} - \frac{3}{4e^2}$$

(iii) The volume \(V\) of the solid of revolution is given by:
$$V = \pi \int_{1}^{e} y^2 \text{d}x = \pi \int_{1}^{e} \frac{(\ln x)^2}{x^6} \text{d}x$$

We use integration by parts with \(u = (\ln x)^2\) and \(\text{d}v = x^{-6} \text{d}x\), which gives \(\text{d}u = \frac{2\ln x}{x} \text{d}x\) and \(v = -\frac{1}{5x^5}\):
$$\int \frac{(\ln x)^2}{x^6} \text{d}x = -\frac{(\ln x)^2}{5x^5} + \frac{2}{5} \int \frac{\ln x}{x^6} \text{d}x$$

Now we evaluate \(\int \frac{\ln x}{x^6} \text{d}x\) by parts again, with \(U = \ln x\) and \(\text{d}V = x^{-6} \text{d}x\), giving \(\text{d}U = \frac{1}{x}\text{d}x\) and \(V = -\frac{1}{5x^5}\):
$$\int \frac{\ln x}{x^6} \text{d}x = -\frac{\ln x}{5x^5} - \int -\frac{1}{5x^6} \text{d}x = -\frac{\ln x}{5x^5} - \frac{1}{25x^5}$$

Substituting this back into the main integral:
$$\int \frac{(\ln x)^2}{x^6} \text{d}x = -\frac{(\ln x)^2}{5x^5} - \frac{2\ln x}{25x^5} - \frac{2}{125x^5}$$

Evaluating this from \(1\) to \(e\):
$$\text{At } x = e: \quad -\frac{1}{5e^5} - \frac{2}{25e^5} - \frac{2}{125e^5} = \frac{-25 - 10 - 2}{125e^5} = -\frac{37}{125e^5}$$
$$\text{At } x = 1: \quad 0 - 0 - \frac{2}{125} = -\frac{2}{125}$$

Subtracting the lower limit from the upper limit:
$$\int_{1}^{e} \frac{(\ln x)^2}{x^6} \text{d}x = -\frac{37}{125} e^{-5} - \left(-\frac{2}{125}\right) = \frac{2}{125} - \frac{37}{125} e^{-5}$$

Multiplying by \(\pi\):
$$V = \pi \left( \frac{2}{125} - \frac{37}{125} e^{-5} \right)$$

Here, \(a = \frac{2}{125}\) and \(b = \frac{37}{125}\).

(i)
M1: Attempt to differentiate \(y\) using quotient rule (or product rule).
A1: Obtain correct derivative \(\frac{1-3\ln x}{x^4}\).
M1: Set derivative to 0 and solve for \(x\).
A1: Obtain correct coordinates \(\left(e^{1/3}, \frac{1}{3e}\right)\) in exact form.

(ii)
M1: Express area as integral and attempt integration by parts with \(u = \ln x, \text{d}v = x^{-3}\text{d}x\).
A1: Obtain correct term \(-\frac{\ln x}{2x^2} - \frac{1}{4x^2}\).
M1: Apply limits \(1\) and \(e\) correctly.
A1: Obtain exact area \(\frac{1}{4} - \frac{3}{4e^2}\) (or equivalent).

(iii)
M1: Express volume as \(\pi \int y^2 \text{d}x\) and attempt first integration by parts.
A1: Obtain correct first-stage integration by parts expression.
M1: Perform the second integration by parts successfully and substitute limits.
A1: Obtain final answer \(\pi\left(\frac{2}{125} - \frac{37}{125}e^{-5}\right)\) (accept \(a = \frac{2}{125}, b = \frac{37}{125}\)).