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

To find where the line and curve do not intersect, we set their equations equal:
\(kx - 3 = x^2 - 4x + 1\)

Rearranging into a standard quadratic equation:
\(x^2 - (4+k)x + 4 = 0\)

For there to be no points of intersection, the discriminant must be strictly less than zero (\(b^2 - 4ac < 0\)):
\((-(4+k))^2 - 4(1)(4) < 0\)

\((k+4)^2 - 16 < 0\)

\(k^2 + 8k + 16 - 16 < 0\)

\(k(k+8) < 0\)

Solving this inequality gives the range:
\(-8 < k < 0\)

M1: Equating the line and curve equations, and setting up the discriminant of the resulting quadratic inequality.
M1: Setting \(b^2 - 4ac < 0\) and solving the quadratic inequality.
A0.5: Correct final range \(-8 < k < 0\) (accept equivalent notation).

First, find the midpoint \(M\) of the line segment \(AB\):
\(M = \left(\frac{2+8}{2}, \frac{5+(-3)}{2}\right) = (5, 1)\)

Next, find the gradient \(m\) of the line \(AB\):
\(m = \frac{-3 - 5}{8 - 2} = \frac{-8}{6} = -\frac{4}{3}\)

The gradient of the perpendicular bisector, \(m'\), is the negative reciprocal of \(m\):
\(m' = -\frac{1}{m} = \frac{3}{4}\)

Now, find the equation of the line passing through \(M(5, 1)\) with gradient \(\frac{3}{4}\):
\(y - 1 = \frac{3}{4}(x - 5)\)

Multiply both sides by 4:

\(4(y - 1) = 3(x - 5)\)

\(4y - 4 = 3x - 15\)

Rearranging into the requested form \(ay + bx = c\) with \(a > 0\):

\(4y - 3x = -11\)

M1: Correctly finding both the midpoint of \(AB\) and the gradient of \(AB\).
M1: Finding the perpendicular gradient and attempting to find the equation of the line passing through the midpoint.
A0.5: Correct equation in the specified form: \(4y - 3x = -11\) (or \(4y - 3x + 11 = 0\) if rearranged, but must have integer coefficients with the coefficient of \(y\) positive).

The \(n\)-th term of an arithmetic progression is given by:
\(u_n = a + (n-1)d\)

For the 3rd term:
\(a + 2d = 11 \quad \text{--- (Equation 1)}\)

The sum of the first \(n\) terms is given by:
\(S_n = \frac{n}{2}[2a + (n-1)d]\)

For the sum of the first 8 terms:
\(S_8 = \frac{8}{2}[2a + 7d] = 100\)

\(4[2a + 7d] = 100\)

\(2a + 7d = 25 \quad \text{--- (Equation 2)}\)

Multiply Equation 1 by 2:
\(2a + 4d = 22\)

Subtract this from Equation 2:
\((2a + 7d) - (2a + 4d) = 25 - 22\)

\(3d = 3 \implies d = 1\)

Substitute \(d = 1\) back into Equation 1:
\(a + 2(1) = 11 \implies a = 9\)

M1: Writing two correct simultaneous equations in \(a\) and \(d\) using the term and sum formulae.
M1: Solving the simultaneous equations to find both \(a\) and \(d\).
A0.5: Correct values \(a = 9\) and \(d = 1\).

The perimeter \(P\) of a sector is given by:
\(P = 2r + r\theta\)

Given \(P = 30\):
\(2r + r\theta = 30 \implies r\theta = 30 - 2r\)

The area \(A\) of a sector is given by:
\(A = \frac{1}{2}r^2\theta\)

Given \(A = 50\):

\(\frac{1}{2}r^2\theta = 50 \implies r(r\theta) = 100\)

Substitute the expression for \(r\theta\) from the perimeter equation into the area equation:
\(r(30 - 2r) = 100\)

\(30r - 2r^2 = 100\)

\(2r^2 - 30r + 100 = 0\)

Divide by 2:
\(r^2 - 15r + 50 = 0\)

Factor the quadratic equation:
\((r - 5)(r - 10) = 0\)

Thus, the two possible values of \(r\) are \(5\) and \(10\).

M1: Stating correct formulas for the perimeter and area of a sector, and setting up the initial equations.
M1: Eliminating \(\theta\) to obtain a quadratic equation in \(r\) and attempting to solve it.
A0.5: Correctly identifying both possible values of \(r\) as \(5\) and \(10\).

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

To find the stationary point, set \(\frac{dy}{dx} = 0\):
\(3x^{\frac{1}{2}} - 9 = 0 \implies 3\sqrt{x} = 9 \implies \sqrt{x} = 3 \implies x = 9\)

Substitute \(x = 9\) back into the original curve equation to find the corresponding \(y\)-value:
\(y = 2(9)^{\frac{3}{2}} - 9(9) + 5 = 2(27) - 81 + 5 = 54 - 81 + 5 = -22\)

Thus, the coordinates of the stationary point are \((9, -22)\).

To determine the nature of this stationary point, find the second derivative \(\frac{d^2y}{dx^2}\):
\(\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^{\frac{1}{2}} - 9) = \frac{3}{2}x^{-\frac{1}{2}} = \frac{3}{2\sqrt{x}}\)

Evaluate the second derivative at \(x = 9\):
\(\frac{d^2y}{dx^2}\Big|_{x=9} = \frac{3}{2\sqrt{9}} = \frac{3}{6} = \frac{1}{2}\)

Since \(\frac{d^2y}{dx^2} > 0\) at \(x = 9\), the stationary point is a minimum.

M1: Finding \(\frac{dy}{dx}\), setting it to 0, and solving for \(x\).
M1: Substituting \(x\) to find \(y\), finding \(\frac{d^2y}{dx^2}\), and using its sign to determine the nature.
A0.5: Correctly stating the coordinates \((9, -22)\) and identifying the point as a minimum.

(a) Express \(f(x)\) in completed square form:
\(f(x) = 2(x^2 - 4x) + 5 = 2\left[(x-2)^2 - 4\right] + 5 = 2(x-2)^2 - 3\).
This represents a parabola with vertex at \((2, -3)\).
For \(f\) to be a one-to-one function (and therefore have an inverse), the domain must be restricted to one side of the vertex. Since the domain is \(x \ge k\), the smallest value of \(k\) is \(2\).

(b) Set \(y = 2(x-2)^2 - 3\) and rearrange to make \(x\) the subject:
\(y + 3 = 2(x-2)^2\)
\(\frac{y+3}{2} = (x-2)^2\)
Since \(x \ge 2\), we take the positive square root:
\(x - 2 = \sqrt{\frac{y+3}{2}}
\)x = 2 + \sqrt{\frac{y+3}{2}}
Therefore, \(f^{-1}(x) = 2 + \sqrt{\frac{x+3}{2}}\).
The domain of \(f^{-1}\) is the range of \(f\). Since \(f(x) \ge -3\) for \(x \ge 2\), the domain of \(f^{-1}(x)\) is \(x \ge -3\).

(c) We are given \(g(3) = 11\), so:
\(3a + b = 11\) (Equation 1)

We also have \(gf(2) = 5\).
Using \(f(2) = 2(2-2)^2 - 3 = -3\), we have:
\(g(-3) = 5 \implies -3a + b = 5\) (Equation 2)

Add Equation 1 and Equation 2:
\((3a + b) + (-3a + b) = 11 + 5\)
\(2b = 16 \implies b = 8\).

Substitute \(b = 8\) back into Equation 1:
\(3a + 8 = 11 \implies 3a = 3 \implies a = 1\).

(a)
B1: Attempt to complete the square or find vertex by differentiation (e.g., \(f'(x) = 4x - 8 = 0\)).
M1: Identify the x-coordinate of the turning point as the boundary of the domain.
A1: Correctly state \(k = 2\).

(b)
M1: Put \(y = 2(x-2)^2 - 3\) (or equivalent form) and attempt to isolate \((x-2)^2\).
A1: Correctly obtain \((x-2) = \pm\sqrt{\frac{y+3}{2}}\).
M1: Choose positive square root because \(x \ge 2\) and write inverse in terms of \(x\).
A1: Correctly state \(f^{-1}(x) = 2 + \sqrt{\frac{x+3}{2}}\) and state domain as \(x \ge -3\) (allow 0.5 marks for domain).

(c)
B1: Set up first equation \(3a + b = 11\).
M1: Evaluate \(f(2) = -3\).
A1: Set up second equation \(-3a + b = 5\).
M1: Attempt to solve the simultaneous equations for \(a\) and \(b\).
A1: Obtain \(b = 8\).
A1: Obtain \(a = 1\).

(a) Rearranging the equation of the circle by completing the square for both \(x\) and \(y\):
\(x^2 - 4x + y^2 - 6y = -8\)
\((x-2)^2 - 4 + (y-3)^2 - 9 = -8\)
\((x-2)^2 + (y-3)^2 = 5\)
This is in the form \((x-h)^2 + (y-k)^2 = r^2\).
Therefore, the center of the circle is \((2, 3)\) and the radius is \(\sqrt{5}\).

(b) Substitute \(y = 2x + k\) into the original circle equation:
\(x^2 + (2x + k)^2 - 4x - 6(2x + k) + 8 = 0\)
\(x^2 + 4x^2 + 4kx + k^2 - 4x - 12x - 6k + 8 = 0\)
\(5x^2 + (4k - 16)x + (k^2 - 6k + 8) = 0\)

Since the line is a tangent, the resulting quadratic equation must have exactly one real root, so its discriminant \(\Delta\) must be equal to zero:
\(\Delta = b^2 - 4ac = 0\)
\((4k - 16)^2 - 4(5)(k^2 - 6k + 8) = 0\)
\(16(k - 4)^2 - 20(k^2 - 6k + 8) = 0\)

Divide the entire equation by 4:
\(4(k^2 - 8k + 16) - 5(k^2 - 6k + 8) = 0\)
\(4k^2 - 32k + 64 - 5k^2 + 30k - 40 = 0\)
\(-k^2 - 2k + 24 = 0\)
\(k^2 + 2k - 24 = 0\) (as required)

Solving this quadratic equation:
\((k + 6)(k - 4) = 0\)
So, the possible values of \(k\) are \(k = 4\) and \(k = -6\).

(c) For \(k = 4\), substitute this value into the simplified quadratic equation in \(x\):
\(5x^2 + (4(4) - 16)x + (4^2 - 6(4) + 8) = 0\)
\(5x^2 + 0x + (16 - 24 + 8) = 0\)
\(5x^2 = 0 \implies x = 0\)

Substitute \(x = 0\) into the tangent equation \(y = 2x + 4\):
\(y = 2(0) + 4 = 4\).

Thus, the point of contact is \((0, 4)\).

(a)
M1: Attempt to complete the square for both \(x\) and \(y\).
A1: Identify center coordinates as \((2, 3)\).
A1: Obtain radius \(\sqrt{5}\) correctly.

(b)
M1: Substitute \(y = 2x + k\) into the circle equation.
A1: Simplify to a quadratic equation in \(x\): \(5x^2 + (4k - 16)x + (k^2 - 6k + 8) = 0\).
M1: Apply discriminant condition \(b^2 - 4ac = 0\) for tangency.
A1: Correctly expand and progress to a simplified quadratic in \(k\).
A1: Show clearly the derivation of \(k^2 + 2k - 24 = 0\).
A1: Solve to find both values: \(k = 4\) and \(k = -6\).

(c)
M1: Substitute \(k = 4\) back into the quadratic equation in terms of \(x\).
A1: Find \(x = 0\).
M1: Calculate the corresponding \(y\)-value using \(y = 2x + 4\).
A0.5: Correctly state point of contact as \((0, 4)\).

(a) To prove the identity, start with the Left Hand Side (LHS) and express over a common denominator:
\(\text{LHS} = \frac{\sin \theta}{1 - \cos \theta} + \frac{1 - \cos \theta}{\sin \theta}\)
\(= \frac{\sin^2 \theta + (1 - \cos \theta)^2}{\sin \theta (1 - \cos \theta)}\)

Expand the numerator:
\(\sin^2 \theta + (1 - 2\cos \theta + \cos^2 \theta) = (\sin^2 \theta + \cos^2 \theta) + 1 - 2\cos \theta\)

Since \(\sin^2 \theta + \cos^2 \theta = 1\):
\(= 1 + 1 - 2\cos \theta = 2 - 2\cos \theta = 2(1 - \cos \theta)\)

Now substitute this back into the fraction:
\(\text{LHS} = \frac{2(1 - \cos \theta)}{\sin \theta (1 - \cos \theta)}\)

Cancel the common factor of \((1 - \cos \theta)\) (where \(\cos \theta \ne 1\)):
\(\text{LHS} = \frac{2}{\sin \theta}\) (as required).

(b) Use the identity proven in part (a) to rewrite the equation:
\(\frac{2}{\sin \theta} = 3\tan \theta\)

Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\(\frac{2}{\sin \theta} = \frac{3\sin \theta}{\cos \theta}\)

Cross-multiply (noting that \(\sin \theta \ne 0\)):
\(2\cos \theta = 3\sin^2 \theta\)

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(2\cos \theta = 3(1 - \cos^2 \theta)\)
\(3\cos^2 \theta + 2\cos \theta - 3 = 0\)

Let \(u = \cos \theta\). Then we solve the quadratic equation \(3u^2 + 2u - 3 = 0\):
\(u = \frac{-2 \pm \sqrt{2^2 - 4(3)(-3)}}{2(3)}
\)u = \frac{-2 \pm \sqrt{4 + 36}}{6} = \frac{-2 \pm \sqrt{40}}{6} = \frac{-1 \pm \sqrt{10}}{3}

Calculating the two possible values of \(u\):
\(u_1 = \frac{-1 + \sqrt{10}}{3} \approx 0.7208\)
\(u_2 = \frac{-1 - \sqrt{10}}{3} \approx -1.387\)

Since \(\cos \theta\) must lie in the range \([-1, 1]\), there are no solutions for \(\cos \theta \approx -1.387\).

For \(\cos \theta \approx 0.7208\):
\(\theta = \cos^{-1}(0.7208) \approx 43.88^\circ\)

The other solution in the interval \([0^\circ, 360^\circ]\) is:
\(\theta = 360^\circ - 43.88^\circ = 316.12^\circ\)

Correct to 1 decimal place, the solutions are:
\(\theta = 43.9^\circ\) and \(\theta = 316.1^\circ\).

(a)
M1: Express LHS as a single fraction with a common denominator.
A1: Correctly expand the numerator to get \(\sin^2\theta + 1 - 2\cos\theta + \cos^2\theta\).
M1: Use identity \(\sin^2\theta + \cos^2\theta = 1\) to simplify the numerator to \(2 - 2\cos\theta\).
A1.5: Factorize numerator and cancel \(1 - \cos\theta\) from numerator and denominator to get \(\frac{2}{\sin\theta}\).

(b)
M1: Substitute the identity into the given equation.
M1: Write \(\tan\theta\) as \(\frac{\sin\theta}{\cos\theta}\).
M1: Rearrange to form a equation with only \(\sin\theta\) and \(\cos\theta\).
M1: Use \(\sin^2\theta = 1 - \cos^2\theta\) to form a quadratic equation in \(\cos\theta\).
A1: Obtain correct quadratic: \(3\cos^2\theta + 2\cos\theta - 3 = 0\).
M1: Solve quadratic equation to find two values for \(\cos\theta\) and reject the value outside \([-1, 1]\).
A1: Obtain first angle \(\theta = 43.9^\circ\) (accept 43.9 or 44).
A1: Obtain second angle \(\theta = 316.1^\circ\) (accept 316.1 or 316).

(a) The first, third, and fourth terms of the arithmetic progression are:
\(T_1 = a\)
\(T_3 = a + 2d\)
\(T_4 = a + 3d\)

Since these three terms form a geometric progression, the ratio between consecutive terms is constant:
\(\frac{a+3d}{a+2d} = \frac{a+2d}{a}\)

Cross-multiplying gives:
\(a(a+3d) = (a+2d)^2\)
\(a^2 + 3ad = a^2 + 4ad + 4d^2\)

Subtract \(a^2 + 3ad\) from both sides:
\(ad + 4d^2 = 0\)
\(d(a + 4d) = 0\)

Since \(d \ne 0\), we must have:
\(a + 4d = 0 \implies a = -4d\)

The common ratio \(r\) of the geometric progression is given by:
\(r = \frac{T_3}{T_1} = \frac{a + 2d}{a}\)

Substitute \(a = -4d\) into the expression for \(r\):
\(r = \frac{-4d + 2d}{-4d} = \frac{-2d}{-4d} = \frac{1}{2}\) (as required).

(b) The sum of the first 10 terms of the arithmetic progression is given by:
\(S_{10} = \frac{10}{2} [2a + 9d] = 5[2a + 9d] = 35\)
\(2a + 9d = 7\)

Substitute \(a = -4d\) into this equation:
\(2(-4d) + 9d = 7\)
\(-8d + 9d = 7 \implies d = 7\)

Using \(a = -4d\):
\(a = -4(7) = -28\).

(c) The first three terms of the geometric progression are \(T_1 = a\), \(T_3 = a+2d\), \(T_4 = a+3d\).
Hence, the first term of the GP is \(a_1 = a = -28\).
The common ratio is \(r = \frac{1}{2}\).

The sum to infinity of the geometric progression is:
\(S_\infty = \frac{a_1}{1 - r} = \frac{-28}{1 - \frac{1}{2}} = \frac{-28}{\frac{1}{2}} = -56\).

(a)
B1: State correct terms for AP: \(T_1 = a\), \(T_3 = a+2d\), \(T_4 = a+3d\).
M1: Formulate equation using the geometric progression ratio condition: \(a(a+3d) = (a+2d)^2\).
A1: Expand and simplify to obtain \(ad + 4d^2 = 0\).
M1: Use condition \(d \ne 0\) to deduce \(a = -4d\).
A1.5: Substitute \(a = -4d\) into \(r = \frac{a+2d}{a}\) and simplify to obtain \(r = \frac{1}{2}\).

(b)
M1: Use AP sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) with \(n = 10\) and equate to 35.
A1: Obtain simplified linear equation: \(2a + 9d = 7\).
M1: Substitute \(a = -4d\) to solve for \(d\) and \(a\).
A1: Obtain both \(d = 7\) and \(a = -28\).

(c)
M1: Identify the first term of the GP as \(a = -28\).
M1: Apply GP sum to infinity formula \(S_\infty = \frac{a}{1-r}\) with \(r = \frac{1}{2}\).
A1: Obtain \(S_\infty = -56\).

(a) The equation of the curve is \(y = 4x + 9(x-1)^{-1}\).
Using the chain rule to differentiate with respect to \(x\):
\(\frac{\text{d}y}{\text{d}x} = 4 - 9(x-1)^{-2} = 4 - \frac{9}{(x-1)^2}\)

Differentiating again to find the second derivative:
\(\frac{\text{d}^2y}{\text{d}x^2} = (-2)(-9)(x-1)^{-3} = 18(x-1)^{-3} = \frac{18}{(x-1)^3}\).

(b) At a stationary point, \(\frac{\text{d}y}{\text{d}x} = 0\):
\(4 - \frac{9}{(x-1)^2} = 0\)
\(\frac{9}{(x-1)^2} = 4\)
\((x-1)^2 = \frac{9}{4}\)

Taking square roots, and since \(x > 1\), we choose the positive root:
\(x - 1 = \frac{3}{2} = 1.5\)
\(x = 2.5\)

To find the \(y\)-coordinate of the stationary point:
\(y = 4(2.5) + \frac{9}{2.5-1} = 10 + \frac{9}{1.5} = 10 + 6 = 16\).
So the stationary point is at \((2.5, 16)\).

To determine the nature, substitute \(x = 2.5\) into \(\frac{\text{d}^2y}{\text{d}x^2}\):
\(\frac{\text{d}^2y}{\text{d}x^2} = \frac{18}{(2.5-1)^3} = \frac{18}{1.5^3} = \frac{18}{3.375} = \frac{16}{3} \approx 5.33\).
Since \(\frac{\text{d}^2y}{\text{d}x^2} > 0\), the stationary point \((2.5, 16)\) is a minimum point.

(c) We are given \(\frac{\text{d}x}{\text{d}t} = 0.2\) units per second.
We want to find \(\frac{\text{d}y}{\text{d}t}\) when \(x = 3\).

By the chain rule:
\(\frac{\text{d}y}{\text{d}t} = \frac{\text{d}y}{\text{d}x} \times \frac{\text{d}x}{\text{d}t}\)

First, evaluate \(\frac{\text{d}y}{\text{d}x}\) at \(x = 3\):
\(\frac{\text{d}y}{\text{d}x} = 4 - \frac{9}{(3-1)^2} = 4 - \frac{9}{4} = 4 - 2.25 = 1.75\).

Now, calculate \(\frac{\text{d}y}{\text{d}t}\):
\(\frac{\text{d}y}{\text{d}t} = 1.75 \times 0.2 = 0.35\) units per second.

(a)
M1: Attempt to differentiate terms, applying power rule to \((x-1)^{-1}\).
A1: Obtain correct first derivative: \(4 - \frac{9}{(x-1)^2}\).
A1: Obtain correct second derivative: \(\frac{18}{(x-1)^3}\).

(b)
M1: Set first derivative to zero and attempt to solve for \(x\).
A1: Correctly solve to find \(x = 2.5\) (ignore root \(x = -0.5\) due to domain restriction).
M1: Substitute \(x = 2.5\) into original curve equation to find \(y\).
A1: Obtain coordinates \((2.5, 16)\).
A1: Substitute \(x = 2.5\) into the second derivative and correctly identify the nature as a Minimum based on positive value.

(c)
M1: Apply the rate of change formula/chain rule: \(\frac{\text{d}y}{\text{d}t} = \frac{\text{d}y}{\text{d}x} \times \frac{\text{d}x}{\text{d}t}\).
M1: Evaluate \(\frac{\text{d}y}{\text{d}x}\) at \(x = 3\).
A1: Obtain \(\frac{\text{d}y}{\text{d}x} = 1.75\).
A1.5: Multiply correctly to obtain final rate of change \(0.35\).

To solve \( 4^{x-1} = 3^{x+2} \), we take the natural logarithm of both sides: \( \ln(4^{x-1}) = \ln(3^{x+2}) \). Using the power law of logarithms, this becomes: \( (x-1)\ln 4 = (x+2)\ln 3 \). Expanding the brackets: \( x\ln 4 - \ln 4 = x\ln 3 + 2\ln 3 \). Grouping the terms in \( x \) on one side: \( x\ln 4 - x\ln 3 = 2\ln 3 + \ln 4 \). Factoring out \( x \): \( x(\ln 4 - \ln 3) = \ln 9 + \ln 4 \). This simplifies to: \( x\ln\left(\frac{4}{3}\right) = \ln 36 \). Solving for \( x \): \( x = \frac{\ln 36}{\ln(1.333...)} \approx \frac{3.5835}{0.28768} \approx 12.456 \). Correct to 3 significant figures, \( x = 12.5 \).

M1: Take logs of both sides and apply the power law to obtain a linear equation in \( x \).
M1: Group terms and make \( x \) the subject of the equation.
A1: Obtain \( 12.5 \) (or a value that rounds to 12.5, accept 12.46).

First, find the derivative \( \frac{dy}{dx} \) using the product rule \( \frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx} \). Let \( u = 2x - 1 \) and \( v = e^{3x} \). Then \( \frac{du}{dx} = 2 \) and \( \frac{dv}{dx} = 3e^{3x} \). Applying the rule: \( \frac{dy}{dx} = (2x - 1)(3e^{3x}) + e^{3x}(2) \). Simplifying the expression: \( \frac{dy}{dx} = e^{3x}(6x - 3 + 2) = (6x - 1)e^{3x} \). The curve crosses the \( y \)-axis where \( x = 0 \). Substituting \( x = 0 \) into the derivative: \( \frac{dy}{dx} = (6(0) - 1)e^{0} = -1(1) = -1 \).

M1: Apply the product rule to differentiate \( (2x - 1)e^{3x} \).
A1: Obtain correct derivative \( \frac{dy}{dx} = (6x - 1)e^{3x} \) (or equivalent unsimplified form).
A1: Substitute \( x = 0 \) to obtain the gradient of \( -1 \).

We use the trigonometric identity \( \sec^2 \theta = 1 + \tan^2 \theta \) to rewrite the equation: \( 1 + \tan^2 \theta + 2\tan\theta = 4 \). Rearranging into standard quadratic form: \( \tan^2 \theta + 2\tan\theta - 3 = 0 \). Factoring the quadratic expression: \( (\tan\theta - 1)(\tan\theta + 3) = 0 \). This gives two possible equations: \( \tan\theta = 1 \) or \( \tan\theta = -3 \). For the interval \( 0^\circ \le \theta \le 180^\circ \): If \( \tan\theta = 1 \), then \( \theta = 45^\circ \). If \( \tan\theta = -3 \), then \( \theta = 180^\circ - \tan^{-1}(3) \approx 180^\circ - 71.57^\circ = 108.4^\circ \). Thus, the solutions are \( \theta = 45^\circ \) and \( \theta = 108.4^\circ \).

M1: Substitute \( \sec^2 \theta = 1 + \tan^2 \theta \) to obtain a quadratic equation in \( \tan\theta \).
M1: Solve the quadratic to find \( \tan\theta = 1 \) and \( \tan\theta = -3 \), and find at least one correct value for \( \theta \).
A1: Obtain both correct values \( 45^\circ \) and \( 108.4^\circ \) (and no other values in the range).

We integrate the function using the standard result \( \int \frac{1}{ax+b} \, dx = \frac{1}{a} \ln|ax+b| \): \( \int_{1}^{3} \frac{6}{2x-1} \, dx = \left[ \frac{6}{2} \ln(2x-1) \right]_1^3 = \left[ 3\ln(2x-1) \right]_1^3 \). Evaluating this at the limits: Upper limit \( x=3 \): \( 3\ln(2(3)-1) = 3\ln 5 \). Lower limit \( x=1 \): \( 3\ln(2(1)-1) = 3\ln 1 = 0 \). Subtracting the lower limit value from the upper limit value: \( 3\ln 5 - 0 = 3\ln 5 \). Using the power law of logarithms: \( 3\ln 5 = \ln(5^3) = \ln 125 \). Since the required form is \( \ln k \), the value of \( k \) is 125, making the final answer \( \ln 125 \).

M1: Integrate to obtain an expression of the form \( A\ln(2x-1) \) where \( A \neq 6 \).
A1: Obtain the correct integrated expression \( 3\ln(2x-1) \).
A1: Substitute limits correctly and apply the power law of logarithms to write the exact final answer as \( \ln 125 \) (or identify \( k = 125 \)).

(i) Using the identity \(\cos(2\theta) = 1 - 2\sin^2\theta\):
\[3(1 - 2\sin^2\theta) + 8\sin\theta = 5\]
\[3 - 6\sin^2\theta + 8\sin\theta = 5\]
Rearranging all terms to one side gives:
\[6\sin^2\theta - 8\sin\theta + 2 = 0\]

(ii) The equation simplifies to \(3\sin^2\theta - 4\sin\theta + 1 = 0\) by dividing by 2.
Factorising this quadratic equation in \(\sin\theta\):
\[(3\sin\theta - 1)(\sin\theta - 1) = 0\]
This gives:
\[\sin\theta = \frac{1}{3} \quad \text{or} \quad \sin\theta = 1\]
For \(\sin\theta = 1/3\):
\[\theta = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ \approx 19.5^\circ\]
\[\theta = 180^\circ - 19.47^\circ \approx 160.5^\circ\]
For \(\sin\theta = 1\):
\[\theta = 90.0^\circ\]
So the solutions are \(\theta = 19.5^\circ, 90.0^\circ, 160.5^\circ\).

(iii) Using double angle formulae, \(\sin 2\phi = 2\sin\phi\cos\phi\) and \(\cos 2\phi = 2\cos^2\phi - 1\):
\[\frac{\sin 2\phi}{1 + \cos 2\phi} = \frac{2\sin\phi\cos\phi}{1 + (2\cos^2\phi - 1)} = \frac{2\sin\phi\cos\phi}{2\cos^2\phi} = \frac{\sin\phi}{\cos\phi} = \tan\phi\]

(iv) Let \(\phi = 15^\circ\). Then \(2\phi = 30^\circ\).
Using the result from (iii):
\[\tan 15^\circ = \frac{\sin 30^\circ}{1 + \cos 30^\circ} = \frac{1/2}{1 + \sqrt{3}/2} = \frac{1}{2 + \sqrt{3}}\]
Rationalising the denominator:
\[\tan 15^\circ = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = 2 - \sqrt{3}\]

(i)
- M1: For substituting \(\cos 2\theta = 1 - 2\sin^2\theta\) into the given equation.
- M1: For expanding and collecting terms correctly.
- A1: For obtaining the given quadratic equation cleanly.

(ii)
- M1: For factorising or using the quadratic formula on the equation to find two values for \(\sin\theta\).
- A1: For obtaining \(\sin\theta = 1/3\) and \(\sin\theta = 1\).
- B1: For obtaining \(\theta = 90.0^\circ\).
- A1: For obtaining \(\theta = 19.5^\circ\) (accept 19.47).
- A1: For obtaining \(\theta = 160.5^\circ\) (accept 160.53) and no other solutions in range.

(iii)
- M1: For substituting correct double angle identities for \(\sin 2\phi\) and \(\cos 2\phi\).
- A1: For completing the simplification to obtain \(\tan\phi\) convincingly.

(iv)
- M1: For evaluating expression with \(\phi = 15^\circ\) using exact values of \(\sin 30^\circ\) and \(\cos 30^\circ\).
- M1: For rationalising the denominator of the fraction.
- A0.6: For obtaining \(2 - \sqrt{3}\) as the final exact answer.

(i) Let \(u = 3^x\). Then \(3^{2x+1} = 3^1 \cdot (3^x)^2 = 3u^2\).
The equation becomes:
\[3u^2 - 10u + 3 = 0\]
Factorising the quadratic equation:
\[(3u - 1)(u - 3) = 0\]
This gives \(u = \frac{1}{3}\) or \(u = 3\).
Now we solve for \(x\):
For \(3^x = 1/3 = 3^{-1} \implies x = -1\).
For \(3^x = 3 = 3^1 \implies x = 1\).
So, the solutions are \(x = -1\) and \(x = 1\).

(ii) Take the natural logarithm of both sides of \(y = A e^{kx}\):
\[\ln y = \ln(A e^{kx}) = \ln A + kx\]
This is a linear relation of the form \(Y = mX + c\) where \(Y = \ln y\), \(X = x\), gradient \(m = k\), and vertical intercept \(c = \ln A\).
Calculate the gradient \(k\):
\[k = \frac{8.2 - 2.5}{4 - 1} = \frac{5.7}{3} = 1.9\]
Using the point \((1, 2.5)\) to find \(\ln A\):
\[2.5 = 1.9(1) + \ln A \implies \ln A = 0.6\]
Finding \(A\):
\[A = e^{0.6} \approx 1.822\]
Therefore, correct to 3 significant figures, \(A = 1.82\) and \(k = 1.90\).

(iii) Combine the logarithms:
\[\ln\left(\frac{5z - 2}{z}\right) = \ln(2^2) = \ln 4\]
Since \(\ln\) is a one-to-one function, we can equate the arguments:
\[\frac{5z - 2}{z} = 4\]
\[5z - 2 = 4z \implies z = 2\]

(i)
- M1: For attempting to write the equation as a quadratic in \(3^x\).
- A1: For obtaining the correct quadratic equation \(3u^2 - 10u + 3 = 0\) (or equivalent).
- M1: For correctly factorising or solving their quadratic equation to obtain two positive values for \(3^x\).
- A1: For finding \(3^x = 1/3\) and \(3^x = 3\).
- A1: For obtaining the final correct answers \(x = -1\) and \(x = 1\).

(ii)
- M1: For taking logarithms on both sides of \(y = A e^{kx}\) and using logarithm laws correctly.
- A1: For identifying that gradient is \(k\) and intercept is \(\ln A\).
- M1: For calculating the gradient \(k = 1.9\).
- M1: For solving \(\ln A = 2.5 - 1.9\) to find \(A\).
- A1.1: For obtaining both \(A = 1.82\) and \(k = 1.90\) correct to 3 s.f.

(iii)
- M1: For applying logarithm laws to combine terms: \(\ln((5z-2)/z)\) and \(\ln 4\).
- M1: For removing logarithms to obtain a linear equation in \(z\).
- A0.5: For obtaining \(z = 2\) (accept after verifying \(5z-2 > 0\) and \(z > 0\)).

(i) Integrate each term with respect to \(x\):
\[\int 3e^{2x} \, dx = \frac{3}{2}e^{2x}\]
\[\int \frac{2}{2x+1} \, dx = \ln|2x+1|\]
Now evaluate the definite integral from 0 to 2:
\[\left[ \frac{3}{2}e^{2x} + \ln|2x+1| \right]_{0}^{2} = \left( \frac{3}{2}e^4 + \ln 5 \right) - \left( \frac{3}{2}e^0 + \ln 1 \right)\]
Since \(e^0 = 1\) and \(\ln 1 = 0\), we get:
\[= \frac{3}{2}e^4 + \ln 5 - \frac{3}{2}\]
So \(a = 1.5\), \(b = 5\), and \(c = -1.5\).

(ii) Let \(f(x) = \sin^2(x/2)\). The interval is \([0, \pi]\) with 4 intervals, so the width of each interval is \(h = \frac{\pi - 0}{4} = \frac{\pi}{4}\).
The values of \(x_i\) and corresponding \(y_i\) are:
- \(x_0 = 0\): \(y_0 = \sin^2 0 = 0\)
- \(x_1 = \frac{\pi}{4}\): \(y_1 = \sin^2(\pi/8) \approx 0.146447\)
- \(x_2 = \frac{\pi}{2}\): \(y_2 = \sin^2(\pi/4) = 0.5\)
- \(x_3 = \frac{3\pi}{4}\): \(y_3 = \sin^2(3
\pi/8) \approx 0.853553\)
- \(x_4 = \pi\): \(y_4 = \sin^2(\pi/2) = 1\)
Using the formula for the trapezium rule:
\[\int_{0}^{\pi} f(x) \, dx \approx \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]\]
\[\approx \frac{\pi}{8} [0 + 1 + 2(0.146447 + 0.5 + 0.853553)]\]
Note that \(y_1 + y_3 = \sin^2(\pi/8) + \cos^2(\pi/8) = 1\).
\[\approx \frac{\pi}{8} [1 + 2(1.5)] = \frac{\pi}{8} [4] = \frac{\pi}{2} \approx 1.571\]

(iii) Using the double-angle formula \(\cos 2\theta = 1 - 2\sin^2\theta\), we can let \(\theta = x/2\):
\[\cos x = 1 - 2\sin^2(x/2) \implies \sin^2(x/2) = \frac{1}{2}(1 - \cos x)\]
Now integrate this expression:
\[\int_{0}^{\pi} \frac{1}{2}(1 - \cos x) \, dx = \left[ \frac{1}{2}x - \frac{1}{2}\sin x \right]_{0}^{\pi}\]
\[= \left( \frac{\pi}{2} - \frac{1}{2}\sin\pi \right) - \left( 0 - \frac{1}{2}\sin 0 \right) = \frac{\pi}{2}\]

(i)
- M1: For integrating \(3e^{2x}\) to obtain \(k e^{2x}\).
- A1: For obtaining the correct term \(\frac{3}{2}e^{2x}\).
- M1: For integrating \(\frac{2}{2x+1}\) to obtain \(m \ln|2x+1|\).
- A1: For obtaining the correct term \(\ln(2x+1)\).
- A1: For substituting limits 0 and 2 correctly and getting the exact expression \(\frac{3}{2}e^4 + \ln 5 - \frac{3}{2}\).

(ii)
- B1: For correct interval width \(h = \pi/4\) (or equivalent).
- M1: For evaluating at least 4 of the correct \(y\)-values (either exact or decimal format).
- M1: For correctly applying the trapezium rule formula with their values.
- A1: For calculating the sum inside the brackets correctly (equals 4).
- A1: For obtaining final approximation \(1.571\) (correct to 3 d.p.).

(iii)
- M1: For substituting \(\sin^2(x/2) = \frac{1}{2}(1 - \cos x)\).
- A1: For integrating to obtain \(\frac{1}{2}x - \frac{1}{2}\sin x\).
- A0.6: For completing the integration, substituting limits, and showing the result is \(\frac{\pi}{2}\).