2025 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

To find the set of values of \(k\) for which the line and the curve do not intersect, we set their equations equal to each other: \(x^2 - 4x + 1 = kx - 3\). Rearranging this into a standard quadratic equation form gives: \(x^2 - (4 + k)x + 4 = 0\). For the line and the curve not to intersect, this quadratic equation must have no real roots, meaning its discriminant must be strictly less than zero (\(b^2 - 4ac < 0\)). Identifying the coefficients: \(a = 1\), \(b = -(4 + k)\), and \(c = 4\). Substituting these into the discriminant inequality: \((-(4 + k))^2 - 4(1)(4) < 0\), which simplifies to \((4 + k)^2 - 16 < 0\). Solving this inequality: \((4 + k)^2 < 16\), which gives \(-4 < 4 + k < 4\). Subtracting 4 from all parts of the inequality yields \(-8 < k < 0\). Therefore, the set of values of \(k\) is \(-8 < k < 0\).

M1: Attempts to equate the line and the curve and forms a three-term quadratic equation in \(x\) (condone minor sign errors). M1: Applies the condition \(b^2 - 4ac < 0\) to their quadratic coefficients to form an inequality in terms of \(k\). A1: Obtains the correct final inequality \(-8 < k < 0\) (or equivalent interval notation).

Let the first three terms of the geometric progression be:
\(u_1 = k + 6\)
\(u_2 = k\)
\(u_3 = k - 4\)

Since the terms form a geometric progression, there is a common ratio \(r\):
\(r = \frac{u_2}{u_1} = \frac{u_3}{u_2}\)

Therefore:
\(\frac{k}{k + 6} = \frac{k - 4}{k}\)

Multiply both sides by \(k(k + 6)\):
\(k^2 = (k - 4)(k + 6)\)
\(k^2 = k^2 + 2k - 24\)

Subtract \(k^2\) from both sides:
\(2k - 24 = 0\)
\(2k = 24\)
\(k = 12\)

Now, substitute \(k = 12\) back to find the first term \(a\) and the common ratio \(r\):
\(a = k + 6 = 12 + 6 = 18\)
\(r = \frac{k}{k + 6} = \frac{12}{18} = \frac{2}{3}\)

Since \(|r| < 1\), we can find the sum to infinity \(S_{\infty}\):
\(S_{\infty} = \frac{a}{1 - r}\)
\(S_{\infty} = \frac{18}{1 - \frac{2}{3}} = \frac{18}{\frac{1}{3}} = 54\)

**(a)**
* **M1**: Sets up a correct algebraic equation for the common ratio \(r\), e.g., \(\frac{k}{k+6} = \frac{k-4}{k}\).
* **A1**: Solves the equation correctly to find \(k = 12\).

**(b)**
* **M1**: Uses their value of \(k\) to find the first term \(a = 18\) and the common ratio \(r = \frac{2}{3}\).
* **M1**: Applies the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\) with their \(a\) and \(r\) (provided \(|r| < 1\)).
* **A1**: Obtains the correct sum to infinity \(54\).

To find the points of intersection, we equate the equations of the curve and the line:
\(2x^2 - 3x + 1 = kx - 1\)

Rearrange into a standard quadratic equation in the form \(ax^2 + bx + c = 0\):
\(2x^2 - 3x - kx + 2 = 0\)
\(2x^2 - (3 + k)x + 2 = 0\)

For the line and the curve to not intersect, the quadratic equation must have no real roots. Therefore, the discriminant \(\Delta\) must be less than zero:
\(\Delta = b^2 - 4ac < 0\)

Substitute the values \(a = 2\), \(b = -(3 + k)\), and \(c = 2\):
\((-(3 + k))^2 - 4(2)(2) < 0\)
\((3 + k)^2 - 16 < 0\)

Expand or solve directly:
\((3 + k)^2 < 16\)
\(-4 < 3 + k < 4\)

Subtract 3 from all parts:
\(-7 < k < 1\)

* **M1**: Equates the curve and line equations to form a single equation in \(x\).
* **A1**: Rearranges the equation correctly to get \(2x^2 - (3+k)x + 2 = 0\) (or equivalent with correct grouping of \(x\)-terms).
* **M1**: Attempts to use the discriminant condition \(b^2 - 4ac < 0\).
* **A1**: Finds the critical values \(k = -7\) and \(k = 1\) (or equivalent inequality boundary values).
* **A1**: States the correct final range \(-7 < k < 1\) (allow interval notation like \((-7, 1)\)).

The perimeter \(P\) of a sector is given by:
\(P = 2r + r\theta\)
We are given that \(P = 30\), so:
\(2r + r\theta = 30 \implies r\theta = 30 - 2r \implies \theta = \frac{30 - 2r}{r}\)

The area \(A\) of a sector is given by:
\(A = \frac{1}{2}r^2\theta\)
We are given that \(A = 50\), so:
\(\frac{1}{2}r^2\theta = 50\)

Substitute \(\theta = \frac{30 - 2r}{r}\) into the area equation:
\(\frac{1}{2}r^2 \left(\frac{30 - 2r}{r}\right) = 50\)

Simplify the expression:
\(\frac{1}{2}r (30 - 2r) = 50\)
\(r(15 - r) = 50\)
\(15r - r^2 = 50\)

Rearrange into a standard quadratic form:
\(r^2 - 15r + 50 = 0\)

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

Thus, the two possible values of \(r\) are:
\(r = 5\) or \(r = 10\)

(Checking validity: if \(r = 5\), then \(\theta = 4\) rad, which is valid. If \(r = 10\), then \(\theta = 1\) rad, which is also valid.)

* **B1**: Formulates the two initial equations correctly: \(2r + r\theta = 30\) and \(\frac{1}{2}r^2\theta = 50\).
* **M1**: Eliminates \(\theta\) to create an equation in terms of \(r\) only.
* **A1**: Simplifies to obtain the quadratic equation \(r^2 - 15r + 50 = 0\) (or any equivalent 3-term quadratic form).
* **M1**: Solves their 3-term quadratic equation by factorisation, completing the square, or formula.
* **A1**: States the correct two values of \(r = 5\) and \(r = 10\).

We use the trigonometric identity \(\sin^2 x = 1 -
\cos^2 x\) to rewrite the equation in terms of \(\cos x\) only:
\(3(1 - \cos^2 x) - 5 \cos x - 1 = 0\)

Expand and simplify:
\(3 - 3 \cos^2 x - 5 \cos x - 1 = 0\)
\(-3 \cos^2 x - 5 \cos x + 2 = 0\)

Multiply by \(-1\) to get a positive leading coefficient:
\(3 \cos^2 x + 5 \cos x - 2 = 0\)

Let \(y = \cos x\). The equation becomes:
\(3y^2 + 5y - 2 = 0\)

Factorise the quadratic:
\((3y - 1)(y + 2) = 0\)

So, the solutions for \(y\) are:
\(y = \frac{1}{3}\) or \(y = -2\)

This gives:
1) \(\cos x = \frac{1}{3}\)
2) \(\cos x = -2\) (which has no real solutions since \(-1 \le \cos x \le 1\))

For \(\cos x = \frac{1}{3}\) in the interval \(0^\circ \le x \le 360^\circ\):
* The basic angle is \(x = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.529^\circ\), which rounds to \(70.5^\circ\) (to 1 decimal place).
* The second solution is in the fourth quadrant: \(x = 360^\circ - 70.529^\circ \approx 289.471^\circ\), which rounds to \(289.5^\circ\).

Therefore, the solutions are \(x = 70.5^\circ\) and \(x = 289.5^\circ\).

* **M1**: Uses the identity \(\sin^2 x = 1 - \cos^2 x\) to express the equation in terms of \(\cos x\) only.
* **A1**: Obtains the correct simplified quadratic equation \(3 \cos^2 x + 5 \cos x - 2 = 0\).
* **M1**: Solves the quadratic equation to find \(\text{cos } x =
\frac{1}{3}\) (accepts omitting/disregarding \(\cos x = -2\)).
* **A1**: Obtains one correct angle \(70.5^\circ\) (accept \(70.5\) or \(70.53\)).
* **A1**: Obtains the second correct angle \(289.5^\circ\) (accept \(289.5\) or \(289.47\)) and no extra solutions in range.

(a) LHS = \(\frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)} = \frac{\sin^2 \theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1+\cos\theta)}\). Since \(\sin^2\theta + \cos^2\theta = 1\), this becomes \(\frac{2 + 2\cos\theta}{\sin\theta(1+\cos\theta)} = \frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta} = \text{RHS}\). (b) Replacing \(\theta\) with \(2x\), the equation becomes \(\frac{2}{\sin 2x} = 4 \cos 2x\). Rearranging gives \(2 = 4 \sin 2x \cos 2x\), which simplifies to \(2 \sin 2x \cos 2x = 1\). Using the double angle identity \(\sin 2A = 2 \sin A \cos A\), we have \(\sin 4x = 1\). Given \(0 \le x \le \pi\), the range for \(4x\) is \(0 \le 4x \le 4\pi\). Solving \(\sin 4x = 1\) in this range gives \(4x = \frac{\pi}{2}, \frac{5\pi}{2}\). Dividing by 4, we find \(x = \frac{\pi}{8}\) or \(x = \frac{5\pi}{8}\).

(a) M1 for attempting to find a common denominator and expanding the numerator. A1 for correct expansion and simplifying using \(\sin^2 \theta + \cos^2 \theta = 1\). M1 for factoring out 2 in the numerator. A1 for fully proving the identity. (b) M1 for using the identity to write the equation as \(\frac{2}{\sin 2x} = 4 \cos 2x\). M1 for rearranging to \(4\sin 2x\cos 2x = 2\) or \(2\sin 2x\cos 2x = 1\). A1 for obtaining \(\sin 4x = 1\). M1 for identifying the correct values of \(4x\) within the range (at least one correct value). A1 for \(x = \frac{\pi}{8}\) (or 0.393). A1 for \(x = \frac{5\pi}{8}\) (or 1.96). A1 for having no other solutions in the interval.

(a) Rearranging the circle equation by completing the square: \((x-3)^2 - 9 + (y-4)^2 - 16 + 15 = 0 \implies (x-3)^2 + (y-4)^2 = 10\). Thus, the centre is \((3, 4)\) and the radius is \(\sqrt{10}\). (b) Substituting \(y = 3x + k\) into the circle equation: \(x^2 + (3x+k)^2 - 6x - 8(3x+k) + 15 = 0 \implies x^2 + 9x^2 + 6kx + k^2 - 6x - 24x - 8k + 15 = 0 \implies 10x^2 + (6k-30)x + (k^2-8k+15) = 0\). For a tangent, the discriminant must be zero: \((6k-30)^2 - 4(10)(k^2-8k+15) = 0 \implies 36k^2 - 360k + 900 - 40k^2 + 320k - 600 = 0 \implies -4k^2 - 40k + 300 = 0 \implies k^2 + 10k - 75 = 0\). Factoring gives \((k+15)(k-5) = 0\), so \(k = 5\) or \(k = -15\). (c) For the positive value \(k = 5\), substitute \(k = 5\) into the quadratic equation: \(10x^2 + (6(5)-30)x + (5^2-8(5)+15) = 0 \implies 10x^2 = 0 \implies x = 0\). Substitute \(x = 0\) into \(y = 3x + 5\) to get \(y = 5\). The point of contact is \((0, 5)\).

(a) M1 for attempting to complete the square. A1 for centre (3, 4). A1 for radius \(\sqrt{10}\). (b) M1 for substituting the line equation into the circle equation. A1 for a correct three-term quadratic in \(x\). M1 for setting the discriminant to zero. A1 for obtaining \(k^2 + 10k - 75 = 0\) or equivalent. A1 for \(k = 5\) and \(k = -15\). (c) M1 for substituting \(k=5\) into the quadratic. A1 for \(x = 0\). A1 for \(y = 5\), giving the coordinates \((0, 5)\).

(a) The 3rd term of the AP is \(a + 2d\) and the 3rd term of the GP is \(ar^2\). So \(a + 2d = ar^2 \implies 2d = a(r^2 - 1) \implies d = \frac{a(r^2 - 1)}{2}\). The 4th term of the AP is \(a + 3d\) and the 5th term of the GP is \(ar^4\). So \(a + 3d = ar^4\). Substituting \(d\): \(a + 3\left(\frac{a(r^2 - 1)}{2}\right) = ar^4\). Since \(a \ne 0\), we can divide by \(a\): \(1 + \frac{3(r^2 - 1)}{2} = r^4 \implies 2 + 3r^2 - 3 = 2r^4 \implies 2r^4 - 3r^2 + 1 = 0\). (b) Let \(u = r^2\), then \(2u^2 - 3u + 1 = 0 \implies (2u-1)(u-1) = 0 \implies r^2 = \frac{1}{2}\) or \(r^2 = 1\). Since the GP has a sum to infinity, the common ratio must satisfy \(|r| < 1\), which means \(r^2 \ne 1\). Therefore, \(r^2 = \frac{1}{2} \implies r = \pm \frac{1}{\sqrt{2}}\). (c) When \(r = \frac{1}{\sqrt{2}}\), \(r^2 = \frac{1}{2}\). Substituting this into the expression for \(d\): \(d = \frac{a(1/2 - 1)}{2} = -\frac{1}{4}a\). The sum to infinity is \(S_\infty = \frac{a}{1 - r} = \frac{a}{1 - 1/\sqrt{2}} = \frac{a\sqrt{2}}{\sqrt{2} - 1}\). Rationalising the denominator: \(S_\infty = \frac{a\sqrt{2}(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = a\sqrt{2}(\sqrt{2} + 1) = a(2 + \sqrt{2})\).

(a) M1 for writing down equations for 3rd and 5th terms. M1 for expressing \(d\) in terms of \(a\) and \(r\). A1 for substituting \(d\) into the second relation. M1 for dividing by \(a\) and multiplying by 2. A1 for obtaining \(2r^4 - 3r^2 + 1 = 0\). (b) M1 for attempting to factorise/solve the quadratic in \(r^2\). A1 for identifying that \(r^2 = 1/2\) because \(|r| < 1\) for sum to infinity. A1 for \(r = \pm \frac{1}{\sqrt{2}}\). (c) B1 for \(d = -\frac{1}{4}a\). M1 for substituting \(r = \frac{1}{\sqrt{2}}\) into the sum to infinity formula and rationalising. A1 for fully simplifying to \(a(2 + \sqrt{2})\).

(a) Rewriting the curve equation as \(y = 16x^{-1} + x^2 - 3\). Differentiating once: \(\frac{\text{d}y}{\text{d}x} = -16x^{-2} + 2x = -\frac{16}{x^2} + 2x\). Differentiating again: \(\frac{\text{d}^2y}{\text{d}x^2} = 32x^{-3} + 2 = \frac{32}{x^3} + 2\). (b) For a stationary point, \(\frac{\text{d}y}{\text{d}x} = 0 \implies -\frac{16}{x^2} + 2x = 0 \implies 2x^3 = 16 \implies x^3 = 8 \implies x = 2\). When \(x = 2\), \(y = \frac{16}{2} + 2^2 - 3 = 8 + 4 - 3 = 9\). So the coordinates are \((2, 9)\). At \(x = 2\), \(\frac{\text{d}^2y}{\text{d}x^2} = \frac{32}{2^3} + 2 = 4 + 2 = 6\). Since \(\frac{\text{d}^2y}{\text{d}x^2} > 0\), the stationary point is a minimum. (c) We are given \(\frac{\text{d}y}{\text{d}t} = 3\). By the chain rule, \(\frac{\text{d}y}{\text{d}t} = \frac{\text{d}y}{\text{d}x} \times \frac{\text{d}x}{\text{d}t}\). At \(x = 4\), \(\frac{\text{d}y}{\text{d}x} = -\frac{16}{4^2} + 2(4) = -1 + 8 = 7\). Substituting these values into the chain rule formula: \(3 = 7 \times \frac{\text{d}x}{\text{d}t} \implies \frac{\text{d}x}{\text{d}t} = \frac{3}{7}\) units per second.

(a) M1 for differentiating \(16/x\) correctly. A1 for \(\frac{\text{d}y}{\text{d}x} = -\frac{16}{x^2} + 2x\). A1 for \(\frac{\text{d}^2y}{\text{d}x^2} = \frac{32}{x^3} + 2\). (b) M1 for setting \(\frac{\text{d}y}{\text{d}x} = 0\) and solving for \(x\). A1 for \((2, 9)\). M1 for substituting \(x = 2\) into their second derivative. A1 for concluding it is a minimum point based on a positive value of 6. (c) M1 for using the chain rule relation. M1 for substituting \(x=4\) into their \(\frac{\text{d}y}{\text{d}x}\). A1 for finding \(\frac{\text{d}y}{\text{d}x} = 7\). A1 for \(\frac{\text{d}x}{\text{d}t} = \frac{3}{7}\) (or 0.429).

(a) We find \(y\) by integrating \(\frac{\text{d}y}{\text{d}x}\): \(y = \int 6(3x+4)^{-1/2} \text{d}x = \frac{6(3x+4)^{1/2}}{(1/2) \times 3} + C = 4(3x+4)^{1/2} + C = 4\sqrt{3x+4} + C\). Substituting \((4, 10)\): \(10 = 4\sqrt{3(4)+4} + C \implies 10 = 4\sqrt{16} + C \implies 10 = 16 + C \implies C = -6\). Thus, the equation of the curve is \(y = 4\sqrt{3x+4} - 6\). (b) The area is given by the integral \(\int_{0}^{4} y \text{d}x = \int_{0}^{4} (4(3x+4)^{1/2} - 6) \text{d}x\). Performing the integration: \(\int (4(3x+4)^{1/2} - 6) \text{d}x = \frac{4(3x+4)^{3/2}}{(3/2) \times 3} - 6x + K = \frac{8}{9}(3x+4)^{3/2} - 6x + K\). Evaluating this between the limits \(0\) and \(4\): At \(x = 4\): \(\frac{8}{9}(3(4)+4)^{3/2} - 6(4) = \frac{8}{9}(16)^{3/2} - 24 = \frac{8}{9}(64) - 24 = \frac{512}{9} - \frac{216}{9} = \frac{296}{9}\). At \(x = 0\): \(\frac{8}{9}(3(0)+4)^{3/2} - 6(0) = \frac{8}{9}(4)^{3/2} = \frac{8}{9}(8) = \frac{64}{9}\). The area is \(\frac{296}{9} - \frac{64}{9} = \frac{232}{9}\) (or \(25\frac{7}{9}\) or approximately 25.8).

(a) M1 for attempting to integrate \(6(3x+4)^{-1/2}\). A1 for \(4\sqrt{3x+4}\) (ignoring constant). M1 for substituting \((4, 10)\) to find \(C\). A1 for \(y = 4\sqrt{3x+4} - 6\). (b) M1 for attempting to integrate their \(y\) expression. A1 for correct integration of the first term, \(\frac{8}{9}(3x+4)^{3/2}\). A1 for correct integration of the second term, \(-6x\). M1 for substituting \(x = 4\) into their integrated expression. M1 for substituting \(x = 0\) into their integrated expression (must show non-zero term calculation). A1 for obtaining both limit values correct (\(\frac{296}{9}\) and \(\frac{64}{9}\)). A1 for the final area \(\frac{232}{9}\) (or 25.8).

Part (i): Since \(2x - 1\) is a factor, \(p(0.5) = 0\). Substituting \(x = 0.5\) gives: \(2(0.5)^3 + a(0.5)^2 + b(0.5) - 6 = 0\) which simplifies to \(a + 2b = 23\) (Equation 1). Since dividing by \(x + 2\) gives a remainder of \(-30\), \(p(-2) = -30\). Substituting \(x = -2\) gives: \(2(-2)^3 + a(-2)^2 + b(-2) - 6 = -30\) which simplifies to \(2a - b = -4\) (Equation 2). Solving Equations 1 and 2 simultaneously: from Equation 2, \(b = 2a + 4\). Substitute this into Equation 1: \(a + 2(2a + 4) = 23 \implies 5a + 8 = 23 \implies a = 3\). Then \(b = 2(3) + 4 = 10\). Part (ii): With \(a = 3\) and \(b = 10\), the polynomial is \(p(x) = 2x^3 + 3x^2 + 10x - 6\). Since \(2x - 1\) is a factor, we can write \(p(x) = (2x - 1)(x^2 + kx + 6)\). Comparing the \(x^2\) term: \(2kx^2 - x^2 = 3x^2 \implies 2k - 1 = 3 \implies k = 2\). Thus, \(p(x) = (2x - 1)(x^2 + 2x + 6)\). For the quadratic factor \(x^2 + 2x + 6 = 0\), the discriminant is \(2^2 - 4(1)(6) = -20\). Since the discriminant is negative, this quadratic factor has no real roots. Therefore, \(x = 0.5\) is the only real root of the equation \(p(x) = 0\).

Part (i): [M1] Attempt to use the factor theorem with \(x = 0.5\) or the remainder theorem with \(x = -2\). [A1] Obtain a correct equation in \(a\) and \(b\) (e.g., \(a + 2b = 23\) or \(2a - b = -4\)). [M1] Solve the system of two linear equations. [A1] Obtain both correct values \(a = 3\) and \(b = 10\). Part (ii): [M1] Find the quadratic factor \(x^2 + 2x + 6\) by division or equating coefficients. [A1] Calculate the discriminant (or complete the square) of the quadratic factor to show it has no real roots, and state the conclusion.

Part (i): Using the laws of logarithms: \(\ln(2x + 1) - \ln(x^2) = \ln(3)\) which can be written as \(\ln\left(\frac{2x+1}{x^2}\right) = \ln(3)\). Thus, \(\frac{2x+1}{x^2} = 3 \implies 3x^2 - 2x - 1 = 0\). Solving this quadratic: \((3x+1)(x-1) = 0\) which gives \(x = -\frac{1}{3}\) or \(x = 1\). Since \(\ln(x)\) is only defined for \(x > 0\), the negative value \(x = -\frac{1}{3}\) must be rejected. The only valid solution is \(x = 1\). Part (ii): Taking the natural logarithm of both sides: \(\ln(3^{y+1}) = \ln(5^{2y-1}) \implies (y+1)\ln(3) = (2y-1)\ln(5)\). Expanding both sides: \(y\ln(3) + \ln(3) = 2y\ln(5) - \ln(5)\). Grouping terms containing \(y\): \(2y\ln(5) - y\ln(3) = \ln(3) + \ln(5) \implies y(2\ln(5) - \ln(3)) = \ln(15)\). Making \(y\) the subject: \(y = \frac{\ln(15)}{2\ln(5) - \ln(3)}\). Evaluating numerically: \(y \approx \frac{2.70805}{2(1.60944) - 1.09861} \approx \frac{2.70805}{2.12027} \approx 1.277\). Correct to 3 significant figures, \(y = 1.28\).

Part (i): [M1] Combine terms into a single logarithm using laws of logarithms. [A1] Solve the resulting quadratic equation to obtain \(x = 1\) and \(x = -1/3\). [A1] Identify and state that \(x = 1\) is the only valid solution, with the negative root rejected. Part (ii): [M1] Apply logarithms on both sides and use the power rule. [M1] Collect terms of \(y\) and make \(y\) the subject. [A1] Obtain correct answer \(y = 1.28\).

Part (i): Consider the denominator of the Left Hand Side (LHS): \(\cot\theta + \tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}\). Substituting this back into the LHS: \(\text{LHS} = \frac{2\left(\frac{\cos\theta}{\sin\theta}\right)}{\frac{1}{\sin\theta\cos\theta}} = 2\left(\frac{\cos\theta}{\sin\theta}\right) \times (\sin\theta\cos\theta) = 2\cos^2\theta\). Using the double-angle identity for cosine: \(\cos 2\theta = 2\cos^2\theta - 1 \implies 2\cos^2\theta = \cos 2\theta + 1\). Thus, \(\text{LHS} \equiv \cos 2\theta + 1\) (Proved). Part (ii): Using the identity from part (i), the equation becomes: \(\cos 2\theta + 1 = 1.5 \implies \cos 2\theta = 0.5\). Since the domain is \(0^\circ < \theta < 180^\circ\), the domain for \(2\theta\) is \(0^\circ < 2\theta < 360^\circ\). Solving \(\cos 2\theta = 0.5\) in this range: \(2\theta = 60^\circ\) or \(2\theta = 300^\circ\). Dividing by 2 gives: \(\theta = 30^\circ\) or \(\theta = 150^\circ\).

Part (i): [M1] Write \(\tan\theta\) and \(\cot\theta\) in terms of \(\sin\theta\) and \(\cos\theta\). [A1] Simplify the expression to \(2\cos^2\theta\). [A1] Complete the proof by linking \(2\cos^2\theta\) to \(\cos 2\theta + 1\). Part (ii): [M1] Rewrite the equation as \(\cos 2\theta = 0.5\). [A1] Obtain at least one correct value for \(\theta\) (either \(30^\circ\) or \(150^\circ\)). [A1] Obtain both correct values for \(\theta\) and no extras in the range.

Part (i): Let \(u = \ln(3x - 1)\) and \(v = x^2\). The derivative of \(u\) with respect to \(x\) is \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{3}{3x-1}\). The derivative of \(v\) with respect to \(x\) is \(\frac{\mathrm{d}v}{\mathrm{d}x} = 2x\). Using the quotient rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{x^2\left(\frac{3}{3x-1}\right) - 2x\ln(3x-1)}{x^4}\). Dividing both numerator and denominator by \(x\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{3x}{3x-1} - 2\ln(3x-1)}{x^3}\). Part (ii): At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\). Since \(x > \frac{1}{3}\), the denominator \(x^3 \neq 0\). Setting the numerator equal to 0: \(\frac{3x}{3x-1} - 2\ln(3x-1) = 0 \implies \frac{3x}{3x-1} = 2\ln(3x-1)\). Multiplying both sides by \(3x-1\): \(3x = 2(3x-1)\ln(3x-1)\). Dividing both sides by 3: \(x = \frac{2(3x-1)}{3}\ln(3x-1)\) (Proved).

Part (i): [M1] Correctly apply the quotient or product rule template. [A1] Correctly differentiate \(\ln(3x-1)\) to get \(\frac{3}{3x-1}\). [A1] Simplify the derivative to obtain a correct single-fraction expression. Part (ii): [M1] Set their derivative (or numerator) equal to 0. [M1] Rearrange the terms to group the logarithmic and algebraic factors. [A1] Correctly obtain the given iterative relation with no errors in algebra.

Part (i): Integrating \(\cos(2x - \frac{\pi}{6})\): \(\int_0^{\pi/6} \cos(2x - \frac{\pi}{6}) \mathrm{d}x = \left[ \frac{1}{2}\sin\left(2x - \frac{\pi}{6}\right) \right]_0^{\pi/6}\). Substitute upper limit \(x = \pi/6\): \(\frac{1}{2}\sin\left(2(\pi/6) - \frac{\pi}{6}\right) = \frac{1}{2}\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}(0.5) = 0.25\). Substitute lower limit \(x = 0\): \(\frac{1}{2}\sin\left(-\frac{\pi}{6}\right) = \frac{1}{2}(-0.5) = -0.25\). Subtracting: \(0.25 - (-0.25) = 0.5\). Part (ii): The interval width is \(h = \frac{3-0}{3} = 1\). The boundary points are \(x_0=0, x_1=1, x_2=2, x_3=3\). Calculate the corresponding \(y\)-values using \(y = \frac{6}{x+2}\): \(y_0 = \frac{6}{2} = 3\), \(y_1 = \frac{6}{3} = 2\), \(y_2 = \frac{6}{4} = 1.5\), \(y_3 = \frac{6}{5} = 1.2\). Using the trapezium rule formula: \(\text{Area} \approx \frac{1}{2}h[y_0 + y_3 + 2(y_1 + y_2)] = \frac{1}{2}(1)[3 + 1.2 + 2(2 + 1.5)] = 0.5[4.2 + 7.0] = 0.5[11.2] = 5.6\).

Part (i): [M1] Integrate to obtain \(k\sin(2x - \frac{\pi}{6})\) where \(k = 0.5\). [M1] Show correct substitution of both limits into their integrated expression. [A1] Obtain exact answer \(0.5\). Part (ii): [M1] Calculate the correct step width \(h=1\) and find all correct \(y\)-values. [M1] Substitute their values correctly into the trapezium rule formula. [A1] Obtain the correct numerical approximation of \(5.6\).

(i)
We can write \(3\sin 2x - 4\cos 2x = R\sin(2x - \alpha) = R\sin 2x \cos \alpha - R\cos 2x \sin \alpha\).
Comparing coefficients:
\(R\cos \alpha = 3\)
\(R\sin \alpha = 4\)

To find \(R\):
\(R = \sqrt{3^2 + 4^2} = 5\)

To find \(\alpha\):
\(\tan \alpha = \frac{4}{3} \implies \alpha = \arctan\left(\frac{4}{3}\right) \approx 0.927295... \approx 0.927\) radians.

Thus, the expression is \(5\sin(2x - 0.927)\).

(ii)
We solve the inequality:
\(5\sin(2x - 0.927) > 2.5\)
\(\sin(2x - 0.927) > 0.5\)

Since \(0 < x < \pi\), we have \(-0.927 < 2x - 0.927 < 2\pi - 0.927 \approx 5.356\).
Within this range, \(\sin \theta > 0.5\) occurs when:
\(\frac{\pi}{6} < 2x - 0.9273 < \frac{5\pi}{6}\)

Calculating the boundaries:
\(2x - 0.9273 > \frac{\pi}{6} \implies 2x > 0.5236 + 0.9273 = 1.4509 \implies x > 0.725\) (to 3 s.f.)
\(2x - 0.9273 < \frac{5\pi}{6} \implies 2x < 2.6180 + 0.9273 = 3.5453 \implies x < 1.77\) (to 3 s.f.)

Thus, the set of values is \(0.725 < x < 1.77\).

(iii)
We integrate \((3\sin 2x - 4\cos 2x)^2 = 25\sin^2(2x - \alpha)\):
Using the identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\), we have:
\(25\sin^2(2x - \alpha) = \frac{25}{2}(1 - \cos(4x - 2\alpha))\)

Now integrate from \(0\) to \(\frac{\pi}{2}\):
\(\int_0^{\pi/2} \frac{25}{2}(1 - \cos(4x - 2\alpha)) \mathrm{d}x = \frac{25}{2} \left[ x - \frac{1}{4}\sin(4x - 2\alpha) \right]_0^{\pi/2}\)

Substitute the limits:
Upper limit \(x = \frac{\pi}{2}\):
\(\frac{\pi}{2} - \frac{1}{4}\sin(2\pi - 2\alpha) = \frac{\pi}{2} + \frac{1}{4}\sin 2\alpha\)

Lower limit \(x = 0\):
\(0 - \frac{1}{4}\sin(-2\alpha) = \frac{1}{4}\sin 2\alpha\)

Subtracting lower limit from upper limit:
\(\left(\frac{\pi}{2} + \frac{1}{4}\sin 2\alpha\right) - \left(\frac{1}{4}\sin 2\alpha\right) = \frac{\pi}{2}\)

Multiply by the constant factor \(\frac{25}{2}\):
\(\frac{25}{2} \times \frac{\pi}{2} = \frac{25\pi}{4}\)

(i)
M1: Attempt to solve \(R\cos\alpha = 3\), \(R\sin\alpha = 4\) to find \(R\).
A1: Obtain \(R = 5\).
A1: Obtain \(\alpha = 0.927\) (or \(0.9273\)).

(ii)
M1: Set \(\sin(2x - \alpha) = 0.5\) and find at least one boundary value.
A1: Obtain one correct boundary value (either \(x \approx 0.725\) or \(x \approx 1.77\)).
A1: Obtain the correct inequality \(0.725 < x < 1.77\).

(iii)
M1: Use the double-angle formula to rewrite the integrand as \(\frac{25}{2}(1 - \cos(4x - 2\alpha))\).
M1: Integrate to obtain \(k\left[x - \frac{1}{4}\sin(4x - 2\alpha)\right]\).
A1: Correct substitution of limits \(0\) and \(\frac{\pi}{2}\) and simplification of the sine terms.
A1: Obtain exact value of \(\frac{25\pi}{4}\).

(i)
To find the stationary points, we first differentiate \(y\) with respect to \(x\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{2x} - 6\mathrm{e}^x + 3\)

Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(2\mathrm{e}^{2x} - 6\mathrm{e}^x + 3 = 0\)

Let \(u = \mathrm{e}^x\). The equation becomes quadratic in \(u\):
\(2u^2 - 6u + 3 = 0\)

Using the quadratic formula:
\(u = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(3)}}{2(2)}\)
\(u = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm \sqrt{12}}{4} = \frac{3 \pm \sqrt{3}}{2}\)

Since both values \(\frac{3 + \sqrt{3}}{2} \approx 2.366\) and \(\frac{3 - \sqrt{3}}{2} \approx 0.634\) are positive, we can take the natural logarithm of both:
\(x = \ln\left(\frac{3 + \sqrt{3}}{2}\right)\) and \(x = \ln\left(\frac{3 - \sqrt{3}}{2}\right)\)

(ii)
Let \(f(x) = \mathrm{e}^{2x} - 6\mathrm{e}^x + 3x\).
Evaluate \(f(x)\) at the boundaries of the interval:
\(f(1.5) = \mathrm{e}^{3} - 6\mathrm{e}^{1.5} + 3(1.5) \approx 20.0855 - 6(4.4817) + 4.5 = -2.3047\)
\(f(2.0) = \mathrm{e}^{4} - 6\mathrm{e}^{2} + 3(2.0) \approx 54.5982 - 6(7.3891) + 6.0 = 16.2636\)

Since \(f(1.5) < 0\) and \(f(2.0) > 0\), the sign change indicates that there is at least one root \(a\) in the interval \(1.5 < a < 2.0\).

(iii)
Using the iterative formula \(x_{n+1} = \frac{1}{2}\ln(6\mathrm{e}^{x_n} - 3x_n)\) with \(x_1 = 1.7\):
\(x_2 = \frac{1}{2}\ln(6\mathrm{e}^{1.7} - 3(1.7)) = \frac{1}{2}\ln(32.8437 - 5.1) \approx 1.6614\)
\(x_3 = \frac{1}{2}\ln(6\mathrm{e}^{1.6614} - 3(1.6614)) \approx 1.6409\)
\(x_4 = \frac{1}{2}\ln(6\mathrm{e}^{1.6409} - 3(1.6409)) \approx 1.6299\)
\(x_5 = \frac{1}{2}\ln(6\mathrm{e}^{1.6299} - 3(1.6299)) \approx 1.6240\)
\(x_6 = \frac{1}{2}\ln(6\mathrm{e}^{1.6240} - 3(1.6240)) \approx 1.6209\)
\(x_7 = \frac{1}{2}\ln(6\mathrm{e}^{1.6209} - 3(1.6209)) \approx 1.6192\)
\(x_8 = \frac{1}{2}\ln(6\mathrm{e}^{1.6192} - 3(1.6192)) \approx 1.6183\)

Thus, rounding to 2 decimal places, we have \(a \approx 1.62\).

(i)
M1: Differentiate to obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{2x} - 6\mathrm{e}^x + 3\).
M1: Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solve the quadratic equation in \(\mathrm{e}^x\).
A1: Obtain \(\mathrm{e}^x = \frac{3 \pm \sqrt{3}}{2}\).
A1: Take logarithms to obtain both exact values \(x = \ln\left(\frac{3 \pm \sqrt{3}}{2}\right)\).

(ii)
M1: Substitute both 1.5 and 2.0 into the function \(f(x)\).
A1: Calculate correct values (approx. \(-2.30\) and \(16.26\)) and conclude that the change of sign indicates a root in the interval.

(iii)
M1: Correctly calculate the first iteration value \(x_2\).
A1: Show iterations \(x_2 \approx 1.6614\), \(x_3 \approx 1.6409\), and \(x_4 \approx 1.6299\).
A1: Show further convergence with values leading to 1.62 (e.g. \(x_5\) or \(x_6\)).
A1: Conclude with final answer \(a = 1.62\) after checking or sufficient convergence.