2024 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

To find the points of intersection, we set the equations equal to each other:
\[2x + k = x^2 - 4x + 11\]
Rearranging into a standard quadratic form:
\[x^2 - 6x + (11 - k) = 0\]
For the line not to intersect the curve, the quadratic equation must have no real roots, so its discriminant must be less than zero (\(B^2 - 4AC < 0\)):
\[(-6)^2 - 4(1)(11 - k) < 0\]
\[36 - 44 + 4k < 0\]
\[-8 + 4k < 0\]
\[4k < 8 \implies k < 2\]

M1: Set up the discriminant of the formed quadratic equation to be less than zero.
A1: Correct final inequality \(k < 2\).

Let \(y = 3 - \sqrt{x-2}\).
To find the inverse, we rearrange to make \(x\) the subject of the equation:
\[\sqrt{x-2} = 3 - y\]
Squaring both sides:
\[x - 2 = (3 - y)^2\]
\[x = (3 - y)^2 + 2\]
Replacing \(y\) with \(x\) to express the inverse function:
\[\mathrm{f}^{-1}(x) = (3 - x)^2 + 2\]

M1: Attempt to make \(x\) the subject of the formula (at least squaring both sides correctly).
A1: Correct final expression for \(\mathrm{f}^{-1}(x)\) (accept \(x^2 - 6x + 11\)).

Using the midpoint formula for the \(x\)-coordinate:
\[\frac{k + 4}{2} = 1 \implies k + 4 = 2 \implies k = -2\]
Using the midpoint formula for the \(y\)-coordinate:
\[\frac{3 + (k + 5)}{2} = y\]
Substituting \(k = -2\) into the expression for \(y\):
\[y = \frac{3 + (-2 + 5)}{2} = \frac{6}{2} = 3\]
Thus, \(k = -2\) and \(y = 3\).

M1: Use the midpoint formula for the \(x\)-coordinate to solve for \(k\).
A1: Correct values of \(k = -2\) and \(y = 3\).

The perimeter of the sector is given by:
\[P = 2r + r\theta = 30 \implies r\theta = 30 - 2r\]
The area of the sector is given by:
\[A = \frac{1}{2}r^2\theta = 50 \implies r(r\theta) = 100\]
Substituting the expression for \(r\theta\) from the perimeter into the area equation:
\[r(30 - 2r) = 100 \implies 30r - 2r^2 = 100\]
\[2r^2 - 30r + 100 = 0 \implies r^2 - 15r + 50 = 0\]
Factoring the quadratic equation:
\[(r - 5)(r - 10) = 0 \implies r = 5 \text{ or } r = 10\]
If \(r = 5\), then \(\theta = \frac{30 - 2(5)}{5} = 4\), which contradicts \(\theta \le 2\).
If \(r = 10\), then \(\theta = \frac{30 - 2(10)}{10} = 1\), which satisfies \(\theta \le 2\).
Thus, \(r = 10\).

M1: Set up simultaneous equations for perimeter and area and obtain a quadratic equation in \(r\).
A1: Solve the quadratic to find \(r = 10\), rejecting the alternative value with appropriate justification.

Since the terms are in geometric progression, the common ratio \(r\) is constant:
\[\frac{k}{k+2} = \frac{k-1.2}{k}\]
Cross-multiplying gives:
\[k^2 = (k+2)(k-1.2)\]
\[k^2 = k^2 + 0.8k - 2.4\]
Subtracting \(k^2\) from both sides:
\[0.8k - 2.4 = 0 \implies 0.8k = 2.4 \implies k = 3\]

M1: Set up the correct ratio equation \(\frac{k}{k+2} = \frac{k-1.2}{k}\) and attempt to expand.
A1: Correctly solve for \(k = 3\).

First, find the \(x\)-coordinate of the point where the curve crosses the \(x\)-axis by setting \(y = 0\):
\[x^2 - \frac{8}{x} = 0 \implies x^3 = 8 \implies x = 2\]
Next, differentiate \(y = x^2 - 8x^{-1}\) with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 2x + 8x^{-2} = 2x + \frac{8}{x^2}\]
Substitute \(x = 2\) into the derivative to find the gradient:
\[\text{Gradient} = 2(2) + \frac{8}{2^2} = 4 + 2 = 6\]

M1: Set \(y = 0\) to find \(x = 2\), and find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) (at least one term correct).
A1: Obtain the correct gradient value of \(6\).

Rewrite the integrand as powers of \(x\):
\[\int_{1}^{4} \left(3x^{-1/2} - x\right) \mathrm{d}x\]
Integrate term by term:
\[\left[ 6x^{1/2} - \frac{1}{2}x^2 \right]_{1}^{4}\]
Substitute the upper limit of \(4\):
\[6(4)^{1/2} - \frac{1}{2}(4)^2 = 12 - 8 = 4\]
Substitute the lower limit of \(1\):
\[6(1)^{1/2} - \frac{1}{2}(1)^2 = 6 - 0.5 = 5.5\]
Subtract the lower limit value from the upper limit value:
\[4 - 5.5 = -1.5\]

M1: Attempt to integrate, obtaining at least one integrated term correct.
A1: Correctly substitute limits and find the final value of \(-1.5\) (or \(-\frac{3}{2}\)).

Equate the line and the curve: (kx - 3 = x^2 - 5x + 1). Rearrange to form a quadratic equation: (x^2 - (k + 5)x + 4 = 0). For no intersection, the discriminant must be negative: (\Delta < 0). This gives ((-(k + 5))^2 - 4(1)(4) < 0). Expanding and simplifying: (k^2 + 10k + 25 - 16 < 0 \implies k^2 + 10k + 9 < 0). Factorising the quadratic: ((k + 9)(k + 1) < 0). The critical values are (k = -9) and (k = -1). Since we want the expression to be less than zero, the range of values is (-9 < k < -1).

M1: Equate line and curve equations and group terms. A1: Correct quadratic equation in (x) (e.g., (x^2 - (k+5)x + 4 = 0)). M1: Use of discriminant (\Delta < 0). A1: Correct critical values (k = -9) and (k = -1). A1: Correct final inequality (-9 < k < -1).

First, complete the square: (f(x) = 2(x^2 - 6x) + 13 = 2[(x - 3)^2 - 9] + 13 = 2(x - 3)^2 - 5). Now set (y = 2(x - 3)^2 - 5). Rearrange to make (x) the subject: (y + 5 = 2(x - 3)^2 \implies \frac{y + 5}{2} = (x - 3)^2). Taking the square root: since the domain of (f) is (x \le 3), we must have (x - 3 \le 0), so we take the negative square root: (x - 3 = -\sqrt{\frac{y + 5}{2}} \implies x = 3 - \sqrt{\frac{y + 5}{2}}). Thus, (f^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}}). The domain of (f^{-1}) is the range of (f). Since (x \le 3), (2(x-3)^2 \ge 0), which means (f(x) \ge -5). Therefore, the domain of (f^{-1}(x)) is (x \ge -5).

B1: Complete the square correctly to get (2(x-3)^2 - 5). M1: Set (y = 2(x-3)^2 - 5) and make ((x-3)^2) the subject. M1: Take the negative square root, justifying with (x \le 3). A1: Obtain the correct expression (f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}). B1: State the correct domain (x \ge -5).

Find the midpoint (M) of (AB): (M = \left(\frac{2 + 8}{2}, \frac{5 - 3}{2}\right) = (5, 1)). Find the gradient of (AB): (m = \frac{-3 - 5}{8 - 2} = \frac{-8}{6} = -\frac{4}{3}). The gradient of the perpendicular bisector is the negative reciprocal: (m_{\perp} = -\frac{1}{-4/3} = \frac{3}{4}). Use the point-gradient form with (M(5, 1)): (y - 1 = \frac{3}{4}(x - 5)). Multiply by 4 to clear fractions: (4(y - 1) = 3(x - 5) \implies 4y - 4 = 3x - 15). Rearranging into the form (ax + by = c): (3x - 4y = 11).

B1: Find the correct midpoint ((5, 1)). M1: Find the gradient of (AB) and show it is (-\frac{4}{3}). M1: Identify and use the perpendicular gradient (\frac{3}{4}). M1: Use the midpoint and perpendicular gradient to write a line equation. A1: Obtain the correct equation in the form (3x - 4y = 11) (or equivalent integer form).

The perimeter of the sector is given by (P = 2r + r\theta = 24). This can be rearranged to give (r\theta = 24 - 2r). The area of the sector is (A = \frac{1}{2}r^2\theta). Substitute (r\theta) into the area formula: (A = \frac{1}{2}r(r\theta) = \frac{1}{2}r(24 - 2r) = 12r - r^2). To find the maximum area, differentiate (A) with respect to (r): (\frac{\text{d}A}{\text{d}r} = 12 - 2r). For a stationary point, set (\frac{\text{d}A}{\text{d}r} = 0 \implies 12 - 2r = 0 \implies r = 6). Since (\frac{\text{d}^2A}{\text{d}r^2} = -2 < 0), this value of (r) gives the maximum area. The maximum area is (A = 12(6) - 6^2 = 72 - 36 = 36\text{ cm}^2).

M1: Express the perimeter as (2r + r\theta = 24) and express (\theta) or (r\theta) in terms of (r). A1: Correctly show that (A = 12r - r^2). M1: Differentiate (A) to find (\frac{\text{d}A}{\text{d}r} = 12 - 2r) and set to (0). A1: Find (r = 6). A1: Calculate the maximum area of (36\text{ cm}^2) (with second derivative check or vertex argument).

Use the identity (\sin^2 \theta = 1 - \cos^2 \theta): (3(1 - \cos^2 \theta) - 5 \cos \theta - 1 = 0 \implies 3 - 3\cos^2 \theta - 5\cos \theta - 1 = 0). Simplify to obtain the quadratic: (3\cos^2 \theta + 5\cos \theta - 2 = 0). Factorise the quadratic: ((3\cos \theta - 1)(\cos \theta + 2) = 0). This gives two possible equations: (\cos \theta = \frac{1}{3}) or (\cos \theta = -2). Since (\cos \theta = -2) has no solutions, we only solve (\cos \theta = \frac{1}{3}). The principal value is (\theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.528^\circ \approx 70.5^\circ). The other solution in the range (0^\circ \le \theta \le 360^\circ) is (\theta = 360^\circ - 70.528^\circ \approx 289.472^\circ \approx 289.5^\circ).

M1: Substitute (\sin^2 \theta = 1 - \cos^2 \theta) to form a quadratic in (\cos \theta). A1: Obtain the correct quadratic (3\cos^2 \theta + 5\cos \theta - 2 = 0). M1: Solve the quadratic to find (\cos \theta = \frac{1}{3}) and reject (\cos \theta = -2). A1: Find the first angle (\theta = 70.5^\circ). A1: Find the second angle (\theta = 289.5^\circ) and no other solutions.

Since the terms are in a geometric progression, the ratio between consecutive terms is constant: (\frac{x - 2}{x + 10} = \frac{x - 6}{x - 2}). Cross-multiply to solve for (x): ((x - 2)^2 = (x + 10)(x - 6) \implies x^2 - 4x + 4 = x^2 + 4x - 60). Simplify the equation: (-4x + 4 = 4x - 60 \implies 8x = 64 \implies x = 8). Now find the first term (a) and common ratio (r): (a = x + 10 = 8 + 10 = 18), second term is (x - 2 = 6), so (r = \frac{6}{18} = \frac{1}{3}). Since (|r| < 1), the sum to infinity exists: (S_{\infty} = \frac{a}{1 - r} = \frac{18}{1 - 1/3} = \frac{18}{2/3} = 27).

M1: Set up the correct ratio equation (\frac{x-2}{x+10} = \frac{x-6}{x-2}). M1: Expand and solve the linear equation to get (x = 8). A1: Find the first term (a = 18) and common ratio (r = \frac{1}{3}). M1: Use the sum to infinity formula (S_{\infty} = \frac{a}{1-r}) with their values of (a) and (r) (where (|r| < 1)). A1: Obtain the correct sum to infinity of (27).

First, express the curve's equation as (y = 4x + 9x^{-1}). Find the first derivative: (\frac{\text{d}y}{\text{d}x} = 4 - 9x^{-2} = 4 - \frac{9}{x^2}). Set (\frac{\text{d}y}{\text{d}x} = 0) to find stationary points: (4 - \frac{9}{x^2} = 0 \implies x^2 = \frac{9}{4} \implies x = \pm 1.5). Substitute (x) back into the curve equation to find the (y)-coordinates: For (x = 1.5), (y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12). Stationary point is ((1.5, 12)). For (x = -1.5), (y = 4(-1.5) + \frac{9}{-1.5} = -6 - 6 = -12). Stationary point is ((-1.5, -12)). To find their nature, find the second derivative: (\frac{\text{d}^2y}{\text{d}x^2} = 18x^{-3} = \frac{18}{x^3}). Evaluate at each point: At (x = 1.5), (\frac{\text{d}^2y}{\text{d}x^2} = \frac{18}{1.5^3} > 0), so ((1.5, 12)) is a minimum point. At (x = -1.5), (\frac{\text{d}^2y}{\text{d}x^2} = \frac{18}{(-1.5)^3} < 0), so ((-1.5, -12)) is a maximum point.

M1: Correctly differentiate to find (\frac{\text{d}y}{\text{d}x} = 4 - \frac{9}{x^2}). A1: Find stationary (x)-coordinates (x = \pm 1.5). A1: Find both correct stationary points ((1.5, 12)) and ((-1.5, -12)). M1: Correctly find the second derivative (\frac{\text{d}^2y}{\text{d}x^2} = \frac{18}{x^3}). A1: Correctly determine the nature of both points with clear mathematical justification.

(a) The 1st, 4th, and 10th terms of the arithmetic progression are given by:
\( T_1 = a \)
\( T_4 = a + 3d \)
\( T_{10} = a + 9d \)

Since these terms form a geometric progression, the common ratio is constant:
\( \frac{a + 3d}{a} = \frac{a + 9d}{a + 3d} \)

Cross-multiplying, we obtain:
\( (a + 3d)^2 = a(a + 9d) \)
\( a^2 + 6ad + 9d^2 = a^2 + 9ad \)

Subtracting \( a^2 \) from both sides and simplifying:
\( 9d^2 = 3ad \)

Since \( d \neq 0 \), we can divide both sides by \( 3d \):
\( 3d = a \)

Thus, \( a = 3d \).

(b) The sum of the first 5 terms of the arithmetic progression is given by:
\( S_5 = \frac{5}{2}(2a + 4d) = 45 \)

Simplify this expression:
\( 5(a + 2d) = 45 \implies a + 2d = 9 \)

Substituting \( a = 3d \) into this equation:
\( 3d + 2d = 9 \implies 5d = 9 \implies d = 1.8 \)

Therefore, \( a = 3(1.8) = 5.4 \).

The first term of the geometric progression is:
\( G_1 = a = 5.4 \)

The common ratio is:
\( r = \frac{a + 3d}{a} = \frac{5.4 + 3(1.8)}{5.4} = 2 \)

The sum of the first 6 terms of this geometric progression is:
\( S_6 = \frac{G_1(r^6 - 1)}{r - 1} = \frac{5.4(2^6 - 1)}{2 - 1} = 5.4(63) = 340.2 \).

(a)
- M1: Sets up the geometric progression ratio equation using \( a \), \( a+3d \), and \( a+9d \).
- A1: Expands and simplifies to obtain \( 9d^2 = 3ad \) or equivalent.
- A1: Correctly completes the proof to show \( a = 3d \) stating that \( d \neq 0 \).

(b)
- M1: Uses the sum of AP formula to set up \( \frac{5}{2}(2a + 4d) = 45 \).
- A1: Substitutes \( a = 3d \) and correctly solves for \( d = 1.8 \) and \( a = 5.4 \).
- M1: Finds the common ratio \( r = 2 \) and applies the sum of GP formula for \( n = 6 \).
- A1: Obtains the correct sum of \( 340.2 \).

(a) We complete the square for both \( x \) and \( y \) terms in the circle's equation:
\( x^2 - 4x + y^2 + 6y = 12 \)
\( (x - 2)^2 - 4 + (y + 3)^2 - 9 = 12 \)
\( (x - 2)^2 + (y + 3)^2 = 25 \)

Comparing this with the standard equation of a circle \( (x - h)^2 + (y - k)^2 = r^2 \):
- The coordinates of the centre, \( C \), are \( (2, -3) \).
- The radius of the circle is \( r = \sqrt{25} = 5 \).

(b) The coordinates of \( P \) are \( (5, 1) \) and the centre is \( C(2, -3) \).

The gradient of the radius \( CP \) is:
\( m_{CP} = \frac{1 - (-3)}{5 - 2} = \frac{4}{3} \)

Since the tangent is perpendicular to the radius at the point of contact:
\( m_{\text{tangent}} = -\frac{1}{m_{CP}} = -\frac{3}{4} \)

The equation of the tangent passing through \( P(5, 1) \) is:
\( y - 1 = -\frac{3}{4}(x - 5) \)

Multiply both sides by 4:
\( 4(y - 1) = -3(x - 5) \)
\( 4y - 4 = -3x + 15 \)
\( 3x + 4y = 19 \).

(a)
- M1: Attempt to complete the square for both \( x \) and \( y \) terms.
- A1: Obtain correct centre coordinates \( (2, -3) \).
- A1: Obtain correct radius \( 5 \).

(b)
- M1: Calculate the gradient of the radius \( CP \).
- M1: Use the perpendicular gradient property to find the gradient of the tangent as \( -\frac{3}{4} \).
- M1: Attempt to form the equation of the line passing through \( (5, 1) \) with their tangent gradient.
- A1: Obtain the correct equation in the specified integer form: \( 3x + 4y = 19 \).

(a) When \( x = 5 \):
\( y = (3(5) + 1)^{1/2} = \sqrt{16} = 4 \)
So the point of contact is \( P(5, 4) \).

To find the gradient of the tangent, we differentiate \( y = (3x + 1)^{1/2} \) using the chain rule:
\( \frac{dy}{dx} = \frac{1}{2}(3x + 1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x + 1}} \)

At \( x = 5 \):
\( \frac{dy}{dx} = \frac{3}{2\sqrt{16}} = \frac{3}{8} \)

The gradient of the normal is the negative reciprocal of the tangent gradient:
\( m_{\text{normal}} = -\frac{8}{3} \)

The equation of the normal at \( P(5, 4) \) is:
\( y - 4 = -\frac{8}{3}(x - 5) \)

Multiplying by 3 to clear the fraction:
\( 3(y - 4) = -8(x - 5) \)
\( 3y - 12 = -8x + 40 \)
\( 8x + 3y = 52 \).

(b) The volume \( V \) of the solid of revolution is given by:
\( V = \pi \int_{0}^{5} y^2 \, dx \)

Since \( y = (3x + 1)^{1/2} \), we have \( y^2 = 3x + 1 \).

Now, perform the integration:
\( V = \pi \int_{0}^{5} (3x + 1) \, dx \)
\( V = \pi \left[ \frac{3}{2}x^2 + x \right]_{0}^{5} \)

Evaluate the definite integral:
\( V = \pi \left( \left( \frac{3}{2}(25) + 5 \right) - 0 \right) \)
\( V = \pi \left( \frac{75}{2} + 5 \right) = \pi \left( 37.5 + 5 \right) = 42.5\pi \) (or \( \frac{85}{2}\pi \)).

(a)
- B1: Find correct y-coordinate \( y = 4 \) at \( x = 5 \).
- M1: Differentiate to find \( \frac{dy}{dx} \), showing the chain rule factor of 3.
- A1: Find correct gradient of the normal as \( -\frac{8}{3} \).
- A1: Obtain correct equation of the normal in the form \( 8x + 3y = 52 \).

(b)
- M1: Formulate the correct volume integral expression \( \pi \int_{0}^{5} (3x + 1) \, dx \).
- A1: Correct integration to obtain \( \frac{3}{2}x^2 + x \).
- M1: Correctly substitute limits \( 5 \) and \( 0 \).
- A1: Obtain final volume as \( 42.5\pi \) (or \( \frac{85}{2}\pi \)).

(a) We start with the Left-Hand Side (LHS) of the identity:
\( \text{LHS} = \frac{\sin \theta \tan \theta}{1 - \cos \theta} \)

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

Use the Pythagorean identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\( \text{LHS} = \frac{1 - \cos^2 \theta}{\cos \theta (1 - \cos \theta)} \)

Factor the numerator as a difference of squares:
\( \text{LHS} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{\cos \theta (1 - \cos \theta)} \)

Divide out the non-zero common factor \( 1 - \cos \theta \):
\( \text{LHS} = \frac{1 + \cos \theta}{\cos \theta} = \frac{1}{\cos \theta} + 1 = 1 + \frac{1}{\cos \theta} = \text{RHS} \)

Thus, the identity is proven.

(b) Using the identity proven in part (a), we substitute \( \theta = 2x \). The equation becomes:
\( 1 + \frac{1}{\cos 2x} = 4 \)
\( \frac{1}{\cos 2x} = 3 \implies \cos 2x = \frac{1}{3} \)

Since \( 0^\circ \le x \le 180^\circ \), the range for \( 2x \) is \( 0^\circ \le 2x \le 360^\circ \).

Find the principal value for \( 2x \):
\( 2x = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53^\circ \)

Since cosine is positive in the first and fourth quadrants, the second solution within the range is:
\( 2x = 360^\circ - 70.53^\circ = 289.47^\circ \)

Solving for \( x \):
\( x = \frac{70.53^\circ}{2} \approx 35.3^\circ \) (to 1 decimal place)
\( x = \frac{289.47^\circ}{2} \approx 144.7^\circ \) (to 1 decimal place)

Thus, the solutions are \( x = 35.3^\circ \) and \( x = 144.7^\circ \).

(a)
- M1: Uses \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) to rewrite the expression as \( \frac{\sin^2 \theta}{\cos \theta(1 - \cos \theta)} \).
- M1: Replaces \( \sin^2 \theta \) with \( 1 - \cos^2 \theta \) and factors it as \( (1 - \cos \theta)(1 + \cos \theta) \).
- A1: Cancels the common factor \( 1 - \cos \theta \) and successfully completes the proof to obtain \( 1 + \frac{1}{\cos \theta} \).

(b)
- M1: Recognises the substitution to rewrite the equation as \( 1 + \frac{1}{\cos 2x} = 4 \).
- A1: Obtains the simplified trigonometric equation \( \cos 2x = \frac{1}{3} \).
- M1: Correctly calculates at least one angle for \( 2x \) and divides by 2.
- A1: Obtains both solutions \( x = 35.3^\circ \) and \( x = 144.7^\circ \), rounded to 1 decimal place (ignore extra solutions outside the interval).

Using the factor theorem, since \(3x - 2\) is a factor of \(p(x)\), we have \(p\left(\frac{2}{3}\right) = 0\).

Substitute \(x = \frac{2}{3}\) into the expression for \(p(x)\):

\[3\left(\frac{2}{3}\right)^3 + a\left(\frac{2}{3}\right)^2 - 7\left(\frac{2}{3}\right) + 6 = 0\]

\[3\left(\frac{8}{27}\right) + a\left(\frac{4}{9}\right) - \frac{14}{3} + 6 = 0\]

\[\frac{8}{9} + \frac{4a}{9} - \frac{42}{9} + \frac{54}{9} = 0\]

\[\frac{4a + 20}{9} = 0\]

\[4a + 20 = 0 \implies a = -5\]

M1: Substitute \(x = \frac{2}{3}\) into \(p(x)\) and equate to zero.
A1: Obtain a correct unsimplified equation in terms of \(a\), such as \(3\left(\frac{8}{27}\right) + \frac{4a}{9} - \frac{14}{3} + 6 = 0\).
A1: Obtain the correct value \(a = -5\).

Let \(u = e^x\). The equation becomes a quadratic in terms of \(u\):

\[u^2 - 5u + 6 = 0\]

Factorising the quadratic equation:

\[(u - 2)(u - 3) = 0\]

This gives:

\[u = 2 \quad \text{or} \quad u = 3\]

Substitute back \(u = e^x\):

\[e^x = 2 \implies x = \ln 2\]

\[e^x = 3 \implies x = \ln 3\]

So the solutions are \(x = \ln 2\) and \(x = \ln 3\).

M1: Substitute \(u = e^x\) and attempt to solve or factorise the resulting quadratic equation.
A1: Obtain \(e^x = 2\) and \(e^x = 3\).
A1: State both correct exact solutions: \(x = \ln 2\) and \(x = \ln 3\).

Using the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\), we substitute this into the equation:

\[3(1 + \tan^2 \theta) + 2 \tan \theta = 8\]

\[3 + 3 \tan^2 \theta + 2 \tan \theta = 8\]

\[3 \tan^2 \theta + 2 \tan \theta - 5 = 0\]

Factorising the quadratic in terms of \(\tan \theta\):

\[(3 \tan \theta + 5)(\tan \theta - 1) = 0\]

This yields two possible values for \(\tan \theta\):

1) \(\tan \theta = 1 \implies \theta = 45^\circ\) (which lies in the interval \(0^\circ < \theta < 180^\circ\))

2) \(\tan \theta = -\frac{5}{3} \implies \theta = 180^\circ - 59.04^\circ = 121.0^\circ\) (to 1 decimal place)

Thus, the solutions in the interval are \(\theta = 45^\circ\) and \(\theta = 121.0^\circ\).

M1: Substitute \(\sec^2 \theta = 1 + \tan^2 \theta\) and form a quadratic equation in \(\tan \theta\).
A1: Obtain \(\theta = 45^\circ\).
A1: Obtain \(\theta = 121.0^\circ\) (accept \(121^\circ\)), and no other values in the range.

To find the gradient, we differentiate \(y\) with respect to \(x\) using the quotient rule:

Let \(u = \ln(2x - 3) \implies \frac{du}{dx} = \frac{2}{2x - 3}\)

Let \(v = x \implies \frac{dv}{dx} = 1\)

Using the quotient rule formula \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\):

\[\frac{dy}{dx} = \frac{x \cdot \left(\frac{2}{2x - 3}\right) - \ln(2x - 3) \cdot 1}{x^2}\]

Substitute \(x = 2\) into the derivative:

\[\frac{dy}{dx} = \frac{2 \cdot \left(\frac{2}{2(2) - 3}\right) - \ln(2(2) - 3)}{2^2}\]

\[\frac{dy}{dx} = \frac{2 \cdot 2 - \ln(1)}{4}\]

Since \(\ln(1) = 0\):

\[\frac{dy}{dx} = \frac{4 - 0}{4} = 1\]

M1: Apply the quotient rule (or product rule) to differentiate, with at least one term differentiated correctly (e.g. \(\frac{d}{dx}(\ln(2x-3)) = \frac{2}{2x-3}\)).
A1: Obtain a correct unsimplified expression for \(\frac{dy}{dx}\).
A1: Obtain the correct exact gradient of \(1\).

Using the iterative formula, we compute each successive approximation:

- \(x_2 = \sqrt{5 + \frac{3}{2.4}} = \sqrt{5 + 1.25} = \sqrt{6.25} = 2.5000\)
- \(x_3 = \sqrt{5 + \frac{3}{2.5}} = \sqrt{5 + 1.2} = \sqrt{6.2} \approx 2.4900\)
- \(x_4 = \sqrt{5 + \frac{3}{2.48998}} \approx \sqrt{5 + 1.20483} = \sqrt{6.20483} \approx 2.4909\)

Comparing \(x_3 = 2.4900\) and \(x_4 = 2.4909\), both values round to \(2.49\) when rounded to 2 decimal places.

Thus, the root is \(2.49\).

M1: Attempt to calculate at least two iterations using the given formula, showing the substitution of \(x_1 = 2.4\).
A1: Obtain \(x_2 = 2.5000\), \(x_3 = 2.4900\), and \(x_4 = 2.4909\) (or \(2.4910\)) correctly rounded.
A1: State the root as \(2.49\) and show that it is correct to 2 decimal places.

Using the laws of logarithms: 1. Apply the power law to the first term: \(2 \ln(x - 1) = \ln(x - 1)^2\). 2. Apply the subtraction law to combine the logarithms on the left-hand side: \(\ln \left( \frac{(x - 1)^2}{x + 5} \right) = \ln(2)\). 3. Equate the arguments: \(\frac{(x - 1)^2}{x + 5} = 2\). 4. Expand and solve the resulting quadratic equation: \((x - 1)^2 = 2(x + 5) \implies x^2 - 2x + 1 = 2x + 10 \implies x^2 - 4x - 9 = 0\). 5. Use the quadratic formula to find \(x\): \(x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-9)}}{2} = \frac{4 \pm \sqrt{52}}{2} = 2 \pm \sqrt{13}\). 6. Evaluate the values: \(x_1 = 2 + \sqrt{13} \approx 5.61\) and \(x_2 = 2 - \sqrt{13} \approx -1.61\). Since \(\ln(x - 1)\) is only defined for \(x > 1\), the negative root \(x \approx -1.61\) is invalid. Therefore, the only solution is \(x \approx 5.61\).

M1: Apply the power law of logarithms to write \(2\ln(x-1) = \ln(x-1)^2\). M1: Apply the quotient/subtraction law to obtain a single logarithm equation and remove logarithms. A1: Obtain the correct quadratic equation \(x^2 - 4x - 9 = 0\). M1: Solve the quadratic equation and reject the invalid root \(x = 2 - \sqrt{13}\). A1: Obtain the correct final answer \(5.61\) correct to 3 significant figures.

We start by substituting the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) into the equation: \(3(1 + \tan^2 \theta) + 5 \tan \theta = 5\). Expand and simplify: \(3 + 3 \tan^2 \theta + 5 \tan \theta = 5 \implies 3 \tan^2 \theta + 5 \tan \theta - 2 = 0\). Factorise the quadratic equation in terms of \(\tan \theta\): \((3\tan \theta - 1)(\tan \theta + 2) = 0\). This gives two cases: Case 1: \(\tan \theta = \frac{1}{3} \implies \theta = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.4^\circ\). In the interval \(0^\circ \le \theta \le 360^\circ\), the other solution is \(\theta = 180^\circ + 18.4^\circ = 198.4^\circ\). Case 2: \(\tan \theta = -2\). The basic angle is \(\tan^{-1}(2) \approx 63.4^\circ\). Since tangent is negative in the second and fourth quadrants: \(\theta = 180^\circ - 63.4^\circ = 116.6^\circ\) and \(\theta = 360^\circ - 63.4^\circ = 296.6^\circ\). Combining all solutions, we get: \(\theta = 18.4^\circ, 116.6^\circ, 198.4^\circ, 296.6^\circ\).

M1: Substitute \(\sec^2 \theta = 1 + \tan^2 \theta\) into the equation. A1: Form the correct quadratic equation \(3\tan^2 \theta + 5\tan \theta - 2 = 0\). M1: Solve the quadratic equation to find \(\tan \theta = \frac{1}{3}\) and \(\tan \theta = -2\). A1: Find two correct angles. A1: Find all four correct angles: \(18.4^\circ, 116.6^\circ, 198.4^\circ, 296.6^\circ\), and no extra angles in the range.

To find the stationary point, we differentiate \(y\) with respect to \(x\) using the quotient rule. Let \(u = \ln x \implies u' = \frac{1}{x}\) and \(v = x^2 \implies v' = 2x\). Then, \(\frac{dy}{dx} = \frac{\frac{1}{x} \cdot x^2 - (\ln x)(2x)}{(x^2)^2} = \frac{x - 2x \ln x}{x^4} = \frac{1 - 2 \ln x}{x^3}\). For a stationary point, we set \(\frac{dy}{dx} = 0 \implies \frac{1 - 2 \ln x}{x^3} = 0 \implies 1 - 2 \ln x = 0 \implies \ln x = \frac{1}{2} \implies x = e^{1/2}\). Now substitute \(x = e^{1/2}\) back into the original equation to find the \(y\)-coordinate: \(y = \frac{\ln(e^{1/2})}{(e^{1/2})^2} = \frac{\frac{1}{2}}{e} = \frac{1}{2e}\). Thus, the exact coordinates of the stationary point are \(\left(e^{1/2}, \frac{1}{2e}\right)\).

M1: Attempt to differentiate using the quotient rule (or product rule) with correct structure. A1: Obtain the correct derivative \(\frac{dy}{dx} = \frac{1 - 2 \ln x}{x^3}\). M1: Set their \(\frac{dy}{dx} = 0\) and solve for \(\ln x\). A1: Obtain the exact x-coordinate \(x = e^{1/2}\). A1: Obtain the exact y-coordinate \(y = \frac{1}{2e}\).

We integrate each term individually: 1. \(\int 3\sin(2x) \, dx = -\frac{3}{2}\cos(2x)\). 2. \(\int 4\cos(3x) \, dx = \frac{4}{3}\sin(3x)\). Combining these, we get: \(\int_{0}^{\frac{\pi}{6}} (3\sin(2x) + 4\cos(3x)) \, dx = \left[ -\frac{3}{2}\cos(2x) + \frac{4}{3}\sin(3x) \right]_{0}^{\frac{\pi}{6}}\). Next, substitute the upper limit \(x = \frac{\pi}{6}\): \(-\frac{3}{2}\cos\left(\frac{\pi}{3}\right) + \frac{4}{3}\sin\left(\frac{\pi}{2}\right) = -\frac{3}{2}\left(\frac{1}{2}\right) + \frac{4}{3}(1) = -\frac{3}{4} + \frac{4}{3} = \frac{7}{12}\). Substitute the lower limit \(x = 0\): \(-\frac{3}{2}\cos(0) + \frac{4}{3}\sin(0) = -\frac{3}{2}(1) + 0 = -\frac{3}{2}\). Subtracting the lower limit value from the upper limit value: \(\frac{7}{12} - \left(-\frac{3}{2}\right) = \frac{7}{12} + \frac{18}{12} = \frac{25}{12}\). This completes the proof.

M1: Obtain integration results of the form \(a\cos(2x) + b\sin(3x)\) where \(a, b\) are non-zero constants. A1: Identify the correct integrated expression \(-\frac{3}{2}\cos(2x) + \frac{4}{3}\sin(3x)\). M1: Substitute limits \(\frac{\pi}{6}\) and \(0\) into an expression of the form \(a\cos(2x) + b\sin(3x)\). A1: Obtain either correct evaluation of the upper limit \(\frac{7}{12}\) or the lower limit \(-\frac{3}{2}\). A1: Show clear, correct algebraic steps to reach the final exact answer \(\frac{25}{12}\).

(i) Using the double angle identity \(\cos(2A) = 2\cos^2(A) - 1\), we substitute \(A = 2x\) to obtain: \(\cos(4x) = 2\cos^2(2x) - 1\). Rearranging for \(\cos^2(2x)\): \(2\cos^2(2x) = 1 + \cos(4x) \implies \cos^2(2x) = \frac{1}{2}(1 + \cos(4x))\). (ii) The area \(A\) is given by the definite integral: \(A = \int_{0}^{\frac{\pi}{6}} 4\cos^2(2x) dx\). Using the identity from part (i), we can rewrite the integrand: \(4\cos^2(2x) = 2(1 + \cos(4x))\). Now integrate: \(A = \int_{0}^{\frac{\pi}{6}} (2 + 2\cos(4x)) dx = \left[ 2x + \frac{1}{2}\sin(4x) \right]_{0}^{\frac{\pi}{6}}\). Substituting the limits: At \(x = \frac{\pi}{6}\), we get \(2\left(\frac{\pi}{6}\right) + \frac{1}{2}\sin\left(\frac{4\pi}{6}\right) = \frac{\pi}{3} + \frac{1}{2}\sin\left(\frac{2\pi}{3}\right) = \frac{\pi}{3} + \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{4}\). At \(x = 0\), the value is 0. So the exact area is \(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\). (iii) For three intervals between \(0\) and \(\frac{\pi}{6}\), the interval width \(h\) is \(h = \frac{\pi}{18}\). The \(x\)-values are \(x_0 = 0\), \(x_1 = \frac{\pi}{18}\), \(x_2 = \frac{\pi}{9}\), and \(x_3 = \frac{\pi}{6}\). We calculate the corresponding \(y\)-values \(y = 4\cos^2(2x)\): \(y_0 = 4\cos^2(0) = 4\), \(y_1 = 4\cos^2\left(\frac{\pi}{9}\right) \approx 3.53209\), \(y_2 = 4\cos^2\left(\frac{2\pi}{9}\right) \approx 2.34730\), and \(y_3 = 4\cos^2\left(\frac{\pi}{3}\right) = 1\). Applying the trapezium rule: \(\int_{0}^{\frac{\pi}{6}} 4\cos^2(2x) dx \approx \frac{h}{2} [y_0 + y_3 + 2(y_1 + y_2)] = \frac{\pi}{36} [4 + 1 + 2(3.53209 + 2.34730)] = \frac{\pi}{36} [5 + 2(5.87939)] = \frac{\pi}{36} [16.75878] \approx 1.46246\). So the approximation is \(1.462\) (correct to 3 decimal places).

(i) B1: For correctly applying the double-angle formula to obtain the given expression. (ii) M1: For expressing the integrand in terms of \(\cos(4x)\) using part (i). A1: For correct integration, obtaining \(2x + \frac{1}{2}\sin(4x)\). M1: For substituting the limits \(0\) and \(\frac{\pi}{6}\) correctly. A1: For obtaining the exact area \(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\). (iii) B1: For calculating the correct interval width \(h = \frac{\pi}{18}\) (or equivalent decimal value \(\approx 0.1745\)). M1: For calculating the intermediate \(y\)-values and applying the trapezium rule formula correctly. A1: For obtaining the final answer \(1.462\) correct to 3 decimal places.

(i) Let \(f(x) = x^2 + \ln x - 3\). Evaluating at the endpoints: \(f(1.5) = 1.5^2 + \ln(1.5) - 3 = 2.25 + 0.40547 - 3 = -0.34453\) and \(f(1.7) = 1.7^2 + \ln(1.7) - 3 = 2.89 + 0.53063 - 3 = 0.42063\). Since \(f(1.5) < 0\) and \(f(1.7) > 0\), there is a sign change. Since the function is continuous for \(x > 0\), there must be a root \(\alpha\) in the interval \(1.5 < x < 1.7\). (ii) From \(x^2 + \ln x - 3 = 0\), we rearrange: \(x^2 = 3 - \ln x\). Since \(x > 0\) in our domain, we take the positive square root to get \(x = \sqrt{3 - \ln x}\). (iii) Applying the iterative formula with \(x_1 = 1.6\): \(x_1 = 1.6\), \(x_2 = \sqrt{3 - \ln(1.6)} \approx 1.59164\), \(x_3 = \sqrt{3 - \ln(1.59164)} \approx 1.59220\), \(x_4 = \sqrt{3 - \ln(1.59220)} \approx 1.59216\), \(x_5 = \sqrt{3 - \ln(1.59216)} \approx 1.59217\). Since successive iterations round to \(1.592\), we find \(\alpha = 1.592\) correct to 3 decimal places. (iv) At \(x = 1\), \(y = 1^2 + \ln(1) - 3 = -2\). The derivative is \(\frac{dy}{dx} = 2x + \frac{1}{x}\). At \(x = 1\), the gradient is \(m = 2(1) + \frac{1}{1} = 3\). The equation of the tangent is \(y - (-2) = 3(x - 1) \implies y + 2 = 3x - 3 \implies y = 3x - 5\).

(i) M1: For evaluating \(f(1.5)\) and \(f(1.7)\) with at least one correct calculation to 2 s.f. A1: For obtaining both values correct with opposite signs and stating a correct conclusion. (ii) B1: For showing the algebraic steps clearly to obtain the rearranged form. (iii) M1: For calculating at least two successive iterations correctly to 5 decimal places. A1: For showing iterations that clearly converge, with \(x_2 \approx 1.59164\) and \(x_3 \approx 1.59220\). A1: For stating \(\alpha = 1.592\) correct to 3 decimal places. (iv) M1: For finding the \(y\)-coordinate \(-2\) and finding the derivative \(\frac{dy}{dx} = 2x + \frac{1}{x}\) to calculate the gradient \(m = 3\). A1: For the correct equation of the tangent \(y = 3x - 5\) (or any equivalent form).