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

Let the arithmetic progression have first term \(a\) and common difference \(d\). The sum of the first 10 terms is given by \(S_{10} = 5(2a + 9d) = 210\), which simplifies to \(2a + 9d = 42\). The 23rd, 8th, and 3rd terms of the AP are \(a + 22d\), \(a + 7d\), and \(a + 2d\) respectively. Since these form a GP, we have \((a + 7d)^2 = (a + 22d)(a + 2d)\). Expanding and simplifying yields \(a^2 + 14ad + 49d^2 = a^2 + 24ad + 44d^2\), which simplifies to \(5d^2 - 10ad = 0\). Since the progression is non-constant, \(d \neq 0\), which gives \(d = 2a\). Substituting \(d = 2a\) into \(2a + 9d = 42\) gives \(20a = 42\), so \(a = 2.1\) and \(d = 4.2\). The first term of the GP is \(g_1 = a + 22d = 2.1 + 22(4.2) = 94.5\). The second term is \(g_2 = a + 7d = 2.1 + 7(4.2) = 31.5\). The common ratio of the GP is \(r = \frac{31.5}{94.5} = \frac{1}{3}\). Since \(|r| < 1\), the sum to infinity exists and is given by \(S_{\infty} = \frac{g_1}{1 - r} = \frac{94.5}{1 - 1/3} = 141.75\).

M1: Set up the sum of AP equation. A1: Obtain \(2a + 9d = 42\). M1: Set up the GP term relation and simplify. A1: Obtain \(d = 2a\). M1: Find values for \(a\), \(d\), \(g_1\), and \(r\). A1: Correctly calculate the sum to infinity of the GP as 141.75.

Completing the square for \(\text{f}(x)\) gives \(\text{f}(x) = 2(x-3)^2 + 5\). (i) The vertex is at \(x = 3\). For \(\text{f}\) to be a one-to-one function on \(x \ge a\), the least value of \(a\) is 3. (ii) Let \(y = 2(x-3)^2 + 5\). Then \((x-3)^2 = \frac{y-5}{2}\). Since \(x \ge 3\), taking the positive root gives \(x = 3 + \sqrt{\frac{y-5}{2}}\), so \(\text{f}^{-1}(x) = 3 + \sqrt{\frac{x-5}{2}}\). The domain of \(\text{f}^{-1}\) is the range of \(\text{f}\), which is \(x \ge 5\). (iii) The equation \(\text{g}\text{f}(x) = 35\) becomes \(3(2(x-3)^2 + 5) - 1 = 35\), which simplifies to \(6(x-3)^2 + 14 = 35\), then \(6(x-3)^2 = 21\), so \((x-3)^2 = 3.5\). Since \(x \ge 3\), we take the positive square root to get \(x = 3 + \sqrt{3.5}\) (or approximately 4.87).

B1: Identify the least value of \(a\) is 3. M1: Complete the square and set \(y = \text{f}(x)\) to change the subject. A1: Obtain \(\text{f}^{-1}(x) = 3 + \sqrt{\frac{x-5}{2}}\). B1: State correct domain \(x \ge 5\). M1: Set up the composite equation \(\text{g}\text{f}(x) = 35\) and solve for \(x\). A1: Obtain the correct solution \(x = 3 + \sqrt{3.5}\) and reject the other root.

(i) Since \(P(2, 6)\) lies on the curve, substituting gives \(6 = \frac{a}{4} + 2b \implies a + 8b = 24\). Differentiating the curve equation gives \(\frac{dy}{dx} = -\frac{2a}{x^3} + b\). At the stationary point \(x = 2\), the gradient is zero, so \(-\frac{2a}{8} + b = 0 \implies a = 4b\). Solving these simultaneous equations yields \(4b + 8b = 24 \implies 12b = 24\), so \(b = 2\) and \(a = 8\). (ii) Find the second derivative: \(\frac{d^2y}{dx^2} = \frac{6a}{x^4}\). At \(x = 2\) and with \(a = 8\), \(\frac{d^2y}{dx^2} = \frac{48}{16} = 3 > 0\). Since the second derivative is positive, \(P\) is a local minimum. (iii) At \(x = 1\), the y-coordinate is \(y = \frac{8}{1} + 2(1) = 10\). The gradient of the tangent is \(\frac{dy}{dx} = -\frac{16}{1} + 2 = -14\). The gradient of the normal is the negative reciprocal, \(\frac{1}{14}\). The equation of the normal is \(y - 10 = \frac{1}{14}(x - 1)\), which simplifies to \(x - 14y + 139 = 0\).

M1: Substitute coordinates of \(P\) to get an equation in terms of \(a\) and \(b\). M1: Differentiate and set to zero at \(x=2\). A1: Correctly calculate \(a = 8\) and \(b = 2\). M1: Evaluate the second derivative at \(x = 2\). A1: Deduce that \(P\) is a local minimum. M1: Find the gradient of the normal at \(x = 1\) and construct the line equation. A1: Correctly output \(x - 14y + 139 = 0\).

(i) The center \(M\) of the circle is the midpoint of \(AB\): \(M = \left(\frac{-2+4}{2}, \frac{3+11}{2}\right) = (1, 7)\). The diameter length is \(AB = \sqrt{(4 - (-2))^2 + (11 - 3)^2} = \sqrt{6^2 + 8^2} = 10\). Thus, the radius is \(r = 5\). The equation of the circle \(C\) is \((x-1)^2 + (y-7)^2 = 25\). (ii) Using the perpendicular distance method, the distance from the center \((1, 7)\) to the line \(2x - y + k = 0\) must be equal to the radius \(5\). This gives \(\frac{|2(1) - 7 + k|}{\sqrt{2^2 + (-1)^2}} = 5 \implies \frac{|k-5|}{\sqrt{5}} = 5 \implies |k-5| = 5\sqrt{5}\). Thus, the exact possible values of \(k\) are \(k = 5 \pm 5\sqrt{5}\).

M1: Find the midpoint of \(AB\) to determine the center of the circle. A1: Correctly obtain the center \((1,7)\) and radius \(r = 5\). A1: State the correct circle equation. M1: Formulate the equation for tangent line condition (either using discriminant or perpendicular distance). A1: Set up the correct algebraic expression, e.g., \(|k-5| = 5\sqrt{5}\). A1: Deduce both correct values \(k = 5 \pm 5\sqrt{5}\).

(i) The perimeter of the sector is given by \(P = 2r + r\theta = 30\). Solving for \(\theta\) gives \(\theta = \frac{30}{r} - 2\). The area of the sector is given by \(A = \frac{1}{2}r^2\theta\). Substituting the expression for \(\theta\) yields \(A = \frac{1}{2}r^2\left(\frac{30}{r} - 2\right) = 15r - r^2\), as required. (ii) Setting \(A = 54\) gives \(15r - r^2 = 54 \implies r^2 - 15r + 54 = 0\). Factoring gives \((r-6)(r-9) = 0\), so \(r = 6\) or \(r = 9\). If \(r = 6\), then \(\theta = \frac{30}{6} - 2 = 3\) radians, which is rejected because \(\theta < 1.5\). If \(r = 9\), then \(\theta = \frac{30}{9} - 2 = \frac{4}{3} \approx 1.33\) radians, which satisfies the condition \(\theta < 1.5\). Therefore, \(r = 9\) and \(\theta = \frac{4}{3}\).

M1: Write down the perimeter equation and express \(\theta\) in terms of \(r\). A1: Substitute the expression for \(\theta\) into the area formula and obtain the given relation. M1: Set \(A = 54\) and write a quadratic equation in \(r\). A1: Solve the quadratic equation to find two potential values of \(r\) (6 and 9). M1: Evaluate \(\theta\) for both cases. A1: Correctly choose \(r = 9\), reject \(r = 6\), and state \(\theta = 4/3\).

To prove the identity: \(LHS = \frac{\sin^2 \theta + (1 - \cos \theta)^2}{\sin \theta(1 - \cos \theta)} = \frac{\sin^2 \theta + 1 - 2\cos \theta + \cos^2 \theta}{\sin \theta(1 - \cos \theta)}\). Since \(\sin^2 \theta + \cos^2 \theta = 1\), this becomes \(\frac{2 - 2\cos \theta}{\sin \theta(1 - \cos \theta)} = \frac{2(1 - \cos \theta)}{\sin \theta(1 - \cos \theta)} = \frac{2}{\sin \theta}\). (Identity proven). Now, substituting the identity into the given equation: \(\frac{2}{\sin \theta} = 4\tan \theta \implies \frac{2}{\sin \theta} = 4\frac{\sin \theta}{\cos \theta}\). Multiplying by \(\sin \theta \cos \theta\) (which is valid as \(\sin \theta \neq 0\) and \(\cos \theta \neq 0\) in the given domain): \(2\cos \theta = 4\sin^2 \theta \implies \cos \theta = 2(1 - \cos^2 \theta) \implies 2\cos^2 \theta + \cos \theta - 2 = 0\). Applying the quadratic formula for \(\cos \theta\): \(\cos \theta = \frac{-1 \pm \sqrt{1 - 4(2)(-2)}}{4} = \frac{-1 \pm \sqrt{17}}{4}\). Since \(\frac{-1 - \sqrt{17}}{4} < -1\), we reject this root. Thus, \(\cos \theta = \frac{\sqrt{17}-1}{4} \approx 0.7808\). Since \(0 < \theta < \pi\), we find \(\theta = \cos^{-1}(0.7808) \approx 0.675\) radians.

M1: Find a common denominator and combine the fractions. A1: Correctly use \(\sin^2 \theta + \cos^2 \theta = 1\). A1: Factorize and simplify to show the identity. M1: Rewrite the trigonometric equation using the identity and express \(\tan \theta\) as \(\sin \theta / \cos \theta\). M1: Set up a quadratic equation in \(\cos \theta\). A1: Solve for \(\cos \theta\) and identify the valid root. A1: Find the correct value of \(\theta \approx 0.675\) in radians.

(i) The area is given by \(A = \int_{1}^{2} (2x^2 + 3) dx = \left[ \frac{2}{3}x^3 + 3x \right]_{1}^{2}\). Evaluating at the limits: At \(x = 2\): \(\frac{16}{3} + 6 = \frac{34}{3}\). At \(x = 1\): \(\frac{2}{3} + 3 = \frac{11}{3}\). Thus, Area = \(\frac{34}{3} - \frac{11}{3} = \frac{23}{3}\). (ii) The volume is given by \(V = \pi \int_{1}^{2} y^2 dx = \pi \int_{1}^{2} (2x^2 + 3)^2 dx = \pi \int_{1}^{2} (4x^4 + 12x^2 + 9) dx\). Integrating term by term: \(V = \pi \left[ \frac{4}{5}x^5 + 4x^3 + 9x \right]_{1}^{2}\). Evaluating at the limits: At \(x = 2\): \(\frac{128}{5} + 32 + 18 = 25.6 + 50 = 75.6\). At \(x = 1\): \(\frac{4}{5} + 4 + 9 = 0.8 + 13 = 13.8\). Thus, Volume = \(\pi(75.6 - 13.8) = 61.8\pi\) (or \(\frac{309}{5}\pi\)).

M1: Perform integration on \(y\). A1: Obtain correct antiderivative \(\frac{2}{3}x^3 + 3x\). A1: Correctly calculate the area as \(23/3\). M1: Expand \(y^2\) correctly to \(4x^4 + 12x^2 + 9\). M1: Perform integration on \(y^2\) and insert limits. A1: Correctly calculate the volume as \(61.8\pi\).

To find points of intersection, set the line equation equal to the curve equation: \(kx - 2 = 2x^2 - 3x + k\). Rearranging into standard quadratic form gives: \(2x^2 - (3+k)x + (k+2) = 0\). For the line and the curve not to intersect, this quadratic equation must have no real roots. Therefore, the discriminant \(\Delta = b^2 - 4ac\) must be less than zero: \(\Delta = (-(3+k))^2 - 4(2)(k+2) < 0 \implies k^2 + 6k + 9 - 8k - 16 < 0 \implies k^2 - 2k - 7 < 0\). Finding the critical values by solving \(k^2 - 2k - 7 = 0\) using the quadratic formula: \(k = \frac{2 \pm \sqrt{4 - 4(1)(-7)}}{2} = \frac{2 \pm \sqrt{32}}{2} = 1 \pm 2\sqrt{2}\). For the quadratic expression to be negative, \(k\) must lie between these two roots. Hence, the set of values is \(1 - 2\sqrt{2} < k < 1 + 2\sqrt{2}\).

M1: Eliminate \(y\) to form a quadratic equation in \(x\). A1: Correctly find the quadratic equation \(2x^2 - (3+k)x + (k+2) = 0\). M1: Apply the discriminant condition \(b^2 - 4ac < 0\). A1: Obtain the quadratic inequality \(k^2 - 2k - 7 < 0\). M1: Solve the quadratic inequality to find critical values. A1: Deduce the correct final inequality range \(1 - 2\sqrt{2} < k < 1 + 2\sqrt{2}\).

(a) To find the largest value of \( k \) for which \( f(x) \) is one-to-one (and hence has an inverse), we find the vertex of the quadratic function.
Completing the square on \( f(x) \):
\( f(x) = 2(x^2 - 6x) + 13 \)
\( f(x) = 2[(x - 3)^2 - 9] + 13 \)
\( f(x) = 2(x - 3)^2 - 18 + 13 \)
\( f(x) = 2(x - 3)^2 - 5 \)
Alternatively, using differentiation:
\( f'(x) = 4x - 12 = 0 \implies x = 3 \)
The vertex is at \( x = 3 \). Since the domain is restricted to \( x \le k \), the function is decreasing and therefore one-to-one if \( k \le 3 \).
Thus, the largest value of \( k \) is \( 3 \).

(b) For \( k = 3 \), we set \( y = 2(x - 3)^2 - 5 \).
Rearranging to make \( x \) the subject:
\( y + 5 = 2(x - 3)^2 \)
\( \frac{y+5}{2} = (x - 3)^2 \)
Since \( x \le 3 \), \( x - 3 \le 0 \). Therefore, taking the negative square root:
\( x - 3 = -\sqrt{\frac{y+5}{2}} \)
\( x = 3 - \sqrt{\frac{y+5}{2}} \)
Replacing \( y \) with \( x \), we obtain:
\( f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}} \)

To find the domain of \( f^{-1} \), we find the range of \( f \) for \( x \le 3 \).
Since the vertex is at \( (3, -5) \) and the parabola opens upwards, the minimum value of \( f(x) \) is \( -5 \).
Thus, the range of \( f \) is \( f(x) \ge -5 \).
Therefore, the domain of \( f^{-1} \) is \( x \ge -5 \).

(a)
M1: For attempting to find the x-coordinate of the vertex of the quadratic expression (either by completing the square or differentiating and setting to 0).
A1: For obtaining the correct value of \( k = 3 \).

(b)
M1: For setting up the equation \( y = 2(x-3)^2 - 5 \) (or equivalent from their completed square form).
M1: For correctly isolating the squared term \( (x-3)^2 = \frac{y+5}{2} \).
M1: For identifying that the negative square root must be taken because \( x \le 3 \).
A1: For the correct expression \( f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}} \) (must be in terms of \( x \)).
B1: For stating the correct domain as \( x \ge -5 \) (or equivalent notation).

First, we find the length of the chord \(AB\).
We can split the triangle \(OAB\) into two congruent right-angled triangles by dropping a perpendicular from \(O\) to \(AB\). The perpendicular bisects the angle \(AOB\) and the chord \(AB\).
The half-angle is \( \frac{1}{2} \left(\frac{2}{3}\pi\right) = \frac{\pi}{3} \) radians.
Thus, \( \frac{1}{2}AB = 6 \sin\left(\frac{\pi}{3}\right) \).
Since \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\( \frac{1}{2}AB = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\text{ cm} \)
Therefore, the length of the chord \(AB = 6\sqrt{3}\text{ cm} \).
(Alternatively, using the cosine rule: \( AB^2 = 6^2 + 6^2 - 2(6)(6)\cos\left(\frac{2}{3}\pi\right) = 36 + 36 - 72\left(-\frac{1}{2}\right) = 108 \implies AB = \sqrt{108} = 6\sqrt{3}\text{ cm} \).)

The chord \(AB\) is the diameter of the semicircle. Therefore, the radius of the semicircle, \(R\), is:
\( R = \frac{1}{2} AB = 3\sqrt{3}\text{ cm} \).

Now, we find the length of the arc of the semicircle:
\( \text{Arc}_{\text{semicircle}} = \pi R = 3\sqrt{3}\pi\text{ cm} \).

Next, we find the length of the arc \(AB\) of the sector:
\( \text{Arc}_{\text{sector}} = r\theta = 6 \times \frac{2}{3}\pi = 4\pi\text{ cm} \).

The perimeter of the region is the sum of these two arc lengths:
\( P = \text{Arc}_{\text{semicircle}} + \text{Arc}_{\text{sector}} = 3\sqrt{3}\pi + 4\pi = \pi(4 + 3\sqrt{3})\text{ cm} \).

M1: For a valid method to find the length of the chord \(AB\) (using trigonometry on right-angled triangles or the cosine rule).
A1: For obtaining the exact chord length \( AB = 6\sqrt{3} \) (or \( \sqrt{108} \)).
B1: For finding the correct arc length of the sector: \( r\theta = 6 \times \frac{2}{3}\pi = 4\pi \).
M1: For finding the radius of the semicircle as \( R = \frac{1}{2}AB \) and writing down the formula for its arc length (\( \pi R \)).
A1: For obtaining the semicircle arc length as \( 3\sqrt{3}\pi \).
M1: For adding the two arc lengths to find the total perimeter.
A1: For the correct final exact perimeter \( \pi(4 + 3\sqrt{3}) \) or \( 4\pi + 3\sqrt{3}\pi \).

(a) Let the first term of the arithmetic progression be \( a \) and the common difference be \( d \).
The 1st, 5th, and 15th terms of the AP are given by:
\( T_1 = a \)
\( T_5 = a + 4d \)
\( T_{15} = a + 14d \)

Since these three terms are the first three terms of a geometric progression, the common ratio between consecutive terms must be equal:
\( \frac{a + 4d}{a} = \frac{a + 14d}{a + 4d} \)

Cross-multiplying gives:
\( (a + 4d)^2 = a(a + 14d) \)
\( a^2 + 8ad + 16d^2 = a^2 + 14ad \)

Subtracting \( a^2 \) from both sides:
\( 8ad + 16d^2 = 14ad \)
\( 16d^2 = 6ad \)

Since \( d \ne 0 \), we can divide both sides by \( 2d \):
\( 8d = 3a \)
\( d = \frac{3}{8}a \) (as required).

(b) The sum of the first 20 terms of an AP is given by:
\( S_{20} = \frac{20}{2} [2a + 19d] = 10[2a + 19d] \)

We are given that \( S_{20} = 730 \), so:
\( 10[2a + 19d] = 730 \)
\( 2a + 19d = 73 \)

Substituting \( d = \frac{3}{8}a \) into this equation:
\( 2a + 19\left(\frac{3}{8}a\right) = 73 \)
\( 2a + \frac{57}{8}a = 73 \)
\( \frac{16a + 57a}{8} = 73 \)
\( \frac{73a}{8} = 73 \)
\( 73a = 584 \implies a = 8 \)

Now, substitute \( a = 8 \) back into the expression for \( d \):
\( d = \frac{3}{8}(8) = 3 \).

Thus, the values are \( a = 8 \) and \( d = 3 \).

(a)
B1: For expressing the 1st, 5th, and 15th terms of the AP correctly as \( a \), \( a+4d \), and \( a+14d \).
M1: For setting up the GP relation, e.g., \( (a+4d)^2 = a(a+14d) \).
A1: For expanding and simplifying correctly to \( 16d^2 = 6ad \) (or equivalent).
A1: For dividing by \( d \) (justified by \( d \ne 0 \)) and showing \( d = \frac{3}{8}a \) clearly.

(b)
M1: For setting up the sum equation \( 10(2a + 19d) = 730 \) or \( 2a + 19d = 73 \).
M1: For substituting \( d = \frac{3}{8}a \) (or \( a = \frac{8}{3}d \)) into their sum equation and attempting to solve for one of the variables.
A1: For obtaining \( a = 8 \) and \( d = 3 \).

(i) By the Factor Theorem, since \(p(x)\) is divisible by \(x-3\), we have \(p(3) = 0\):
\(2(3)^3 + a(3)^2 + b(3) - 15 = 0\)
\(54 + 9a + 3b - 15 = 0\)
\(3a + b = -13\) (Equation 1)

By the Remainder Theorem, since the remainder when divided by \(x+1\) is \(-24\), we have \(p(-1) = -24\):
\(2(-1)^3 + a(-1)^2 + b(-1) - 15 = -24\)
\(-2 + a - b - 15 = -24\)
\(a - b = -7\) (Equation 2)

Adding Equation 1 and Equation 2 gives:
\(4a = -20 \implies a = -5\)

Substituting \(a = -5\) back into Equation 2:
\(-5 - b = -7 \implies b = 2\).

(ii) Substituting the values of \(a\) and \(b\), we have \(p(x) = 2x^3 - 5x^2 + 2x - 15\).
Dividing \(p(x)\) by \(x-3\) by matching coefficients or polynomial division:
\(2x^3 - 5x^2 + 2x - 15 = (x-3)(2x^2 + x + 5)\).

To find the roots of \(p(x) = 0\):
\(x - 3 = 0 \implies x = 3\).

For the quadratic factor \(2x^2 + x + 5 = 0\), the discriminant is:
\(\Delta = 1^2 - 4(2)(5) = -39 < 0\).
Since the discriminant is negative, this quadratic factor has no real roots. Thus, the only real root of the equation is \(x = 3\).

Part (i):
M1: Substitute \(x = 3\) to set \(p(3) = 0\) and obtain a linear equation.
A1: Obtain correct simplified equation \(3a + b = -13\).
M1: Substitute \(x = -1\) to set \(p(-1) = -24\) and obtain a second linear equation.
A1: Solve simultaneously to find \(a = -5\) and \(b = 2\).

Part (ii):
M1: Attempt polynomial division or matching coefficients to find the quadratic factor.
A1: Obtain correct factorised expression \((x-3)(2x^2 + x + 5)\).
A1: Compute the discriminant of \(2x^2+x+5\) to show there are no other real roots, and state \(x = 3\) as the only real root.

First, use the power law of logarithms to rewrite the first term:
\(2 \ln(2x - 3) = \ln(2x - 3)^2\)

Now, apply the subtraction law of logarithms:
\(\ln\frac{(2x - 3)^2}{x + 1} = \ln 2\)

Taking exponentials on both sides:
\(\frac{(2x - 3)^2}{x + 1} = 2\)
\(4x^2 - 12x + 9 = 2x + 2\)
\(4x^2 - 14x + 7 = 0\)

Solve the quadratic equation using the quadratic formula:
\(x = \frac{14 \pm \sqrt{(-14)^2 - 4(4)(7)}}{8} = \frac{14 \pm \sqrt{84}}{8} = \frac{7 \pm \sqrt{21}}{4}\)

Calculating the decimal values:
\(x_1 = \frac{7 + \sqrt{21}}{4} \approx 2.90\) (to 3 s.f.)
\(x_2 = \frac{7 - \sqrt{21}}{4} \approx 0.604\) (to 3 s.f.)

We must check the validity of the values in the original equation:
- For \(\ln(2x - 3)\) to be defined, we must have \(2x - 3 > 0 \implies x > 1.5\).
- Since \(2(0.604) - 3 = -1.792 < 0\), the value \(x_2 \approx 0.604\) is invalid.
- Since \(2(2.90) - 3 = 2.8 > 0\) and \(2.90 + 1 > 0\), the value \(x_1 \approx 2.90\) is valid.

Thus, the only solution is \(x = 2.90\).

M1: Apply power law of logarithms to write \(\ln(2x-3)^2\).
M1: Apply subtraction law of logarithms to obtain a single logarithm equation.
A1: Correctly remove logarithms to form the quadratic equation \(4x^2 - 14x + 7 = 0\).
M1: Solve the quadratic equation using a correct method.
A1: Identify two possible decimal roots: \(x \approx 2.90\) and \(x \approx 0.604\).
B1: State the condition \(x > 1.5\) or demonstrate that \(x = 0.604\) yields log of a negative number.
A1: State final answer \(x = 2.90\) and reject the other root.

(i) We expand \(R \cos(\theta + \alpha)\) using the compound angle identity:
\(R \cos(\theta + \alpha) = R \cos \theta \cos \alpha - R \sin \theta \sin \alpha\).
Comparing this with \(3 \cos \theta - 4 \sin \theta\):
\(R \cos \alpha = 3\)
\(R \sin \alpha = 4\)

Squaring and adding:
\(R^2 = 3^2 + 4^2 = 25 \implies R = 5\).
Dividing the equations:
\(\tan \alpha = \frac{4}{3} \implies \alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ\).

So, \(3 \cos \theta - 4 \sin \theta = 5 \cos(\theta + 53.13^\circ)\).

(ii) The equation is \(3 \cos 2x - 4 \sin 2x = 2\).
Using the identity from part (i) where \(\theta = 2x\):
\(5 \cos(2x + 53.13^\circ) = 2 \implies \cos(2x + 53.13^\circ) = 0.4\).

Let \(\phi = 2x + 53.13^\circ\).
Since \(0^\circ < x < 180^\circ\), the domain for \(\phi\) is:
\(0^\circ < 2x < 360^\circ \implies 53.13^\circ < 2x + 53.13^\circ < 413.13^\circ\).

Find solutions to \(\cos \phi = 0.4\) in this interval:
- Principal value: \(\phi_1 = \cos^{-1}(0.4) \approx 66.42^\circ\).
- Second value: \(\phi_2 = 360^\circ - 66.42^\circ = 293.58^\circ\).
- Third value: \(\phi_3 = 360^\circ + 66.42^\circ = 426.42^\circ\) (which is outside the range).

Solving for \(x\):
1) \(2x + 53.13^\circ = 66.42^\circ \implies 2x = 13.29^\circ \implies x \approx 6.6^\circ\).
2) \(2x + 53.13^\circ = 293.58^\circ \implies 2x = 240.45^\circ \implies x \approx 120.2^\circ\).

Thus, the solutions in the interval are \(x = 6.6^\circ\) and \(x = 120.2^\circ\).

Part (i):
B1: State \(R = 5\).
M1: Attempt \(\tan \alpha = 4/3\) or equivalent to find \(\alpha\).
A1: Obtain \(\alpha = 53.13^\circ\).

Part (ii):
M1: Substitute into the equation to get \(5 \cos(2x + 53.13^\circ) = 2\) and obtain \(\cos^{-1}(0.4)\).
A1: Obtain at least one correct value for \(2x + 53.13^\circ\) (e.g. \(66.42^\circ\) or \(293.58^\circ\)).
M1: Carry out subtraction and division to solve for \(x\).
A1: Obtain both \(x = 6.6^\circ\) and \(x = 120.2^\circ\), and no other solutions in range.

(i) We use the quotient rule with \(u = e^{2x}\) and \(v = 2x + 1\):
\(\frac{du}{dx} = 2e^{2x}\)
\(\frac{dv}{dx} = 2\)

Applying the formula \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\):
\(\frac{dy}{dx} = \frac{(2x + 1)(2e^{2x}) - e^{2x}(2)}{(2x + 1)^2}\)
\(\frac{dy}{dx} = \frac{e^{2x}(4x + 2 - 2)}{(2x + 1)^2} = \frac{4x e^{2x}}{(2x + 1)^2}\).

(ii) At the stationary point, \(\frac{dy}{dx} = 0\):
\(\frac{4x e^{2x}}{(2x + 1)^2} = 0 \implies 4x e^{2x} = 0\).
Since \(e^{2x} > 0\) for all real \(x\), we must have \(x = 0\).
Substituting \(x = 0\) into the original equation:
\(y = \frac{e^{2(0)}}{2(0) + 1} = \frac{1}{1} = 1\).

So, the exact coordinates of the stationary point are \((0, 1)\).

(iii) To determine the nature of the stationary point, we can check the sign of the first derivative \(\frac{dy}{dx}\) slightly to the left and right of \(x = 0\):
- For \(x = -0.1\): \(\frac{dy}{dx} = \frac{4(-0.1)e^{-0.2}}{(0.8)^2} < 0\).
- For \(x = 0.1\): \(\frac{dy}{dx} = \frac{4(0.1)e^{0.2}}{(1.2)^2} > 0\).

Since the gradient changes from negative to positive, the stationary point \((0, 1)\) is a minimum point.

Part (i):
M1: Apply the quotient rule correctly.
A1: Correct unsimplified derivative.
A1: Simplify to obtain \(\frac{4x e^{2x}}{(2x + 1)^2}\).

Part (ii):
M1: Set derivative to 0 and solve for \(x\).
A1: Obtain exact coordinates \((0, 1)\).

Part (iii):
M1: Evaluate first derivative signs on either side of \(x=0\) (or calculate second derivative at \(x=0\)).
A1: Correctly conclude that the point is a minimum.

(i) To integrate \(\cos^2(3x)\), we use the double-angle identity:
\(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\)
Substituting \(\theta = 3x\):
\(\cos^2(3x) = \frac{1 + \cos(6x)}{2}\)

Now we integrate:
\(\int_{0}^{\pi/6} \cos^2(3x) \, dx = \int_{0}^{\pi/6} \left(\frac{1}{2} + \frac{1}{2}\cos(6x)\right) \, dx\)
\(= \left[ \frac{1}{2}x + \frac{1}{12}\sin(6x) \right]_{0}^{\pi/6}\)
\(= \left( \frac{1}{2}\left(\frac{\pi}{6}\right) + \frac{1}{12}\sin(\pi) \right) - (0 + 0)\)
\(= \frac{\pi}{12} + 0 = \frac{\pi}{12}\).

(ii) To integrate \(\frac{6}{2x + 1}\), we use the standard integration formula for logarithmic forms:
\(\int_{1}^{4} \frac{6}{2x + 1} \, dx = \left[ 3 \ln|2x + 1| \right]_{1}^{4}\)
\(= 3 \ln|2(4) + 1| - 3 \ln|2(1) + 1|\)
\(= 3 \ln 9 - 3 \ln 3\)
\(= 3 \ln\left(\frac{9}{3}\right) = 3 \ln 3\) (or \(\ln 27\)).

Part (i):
M1: Use double-angle identity to write \(\cos^2(3x)\) in terms of \(\cos(6x)\).
A1: Integrate to obtain \(\frac{1}{2}x + \frac{1}{12}\sin(6x)\).
M1: Substitute the limits correctly.
A1: State exact answer \(\frac{\pi}{12}\).

Part (ii):
M1: Integrate to obtain a logarithmic form of the type \(k \ln(2x+1)\).
A1: Obtain correct term \(3 \ln(2x+1)\).
A1: Correctly substitute limits and apply logarithm rules to simplify to \(3 \ln 3\) or \(\ln 27\).

(i) Let \(f(x) = x^3 - 5x - 3\).
\(f(2) = 2^3 - 5(2) - 3 = 8 - 10 - 3 = -5 < 0\)
\(f(3) = 3^3 - 5(3) - 3 = 27 - 15 - 3 = 9 > 0\)
Since there is a sign change and \(f(x)\) is a continuous polynomial function, the interval \(2 < \alpha < 3\) must contain a root.

(ii) Rearranging \(x^3 - 5x - 3 = 0\):
\(x^3 = 5x + 3\)
Since \(x \neq 0\) in the given interval, we can divide both sides by \(x\):
\(x^2 = 5 + \frac{3}{x}\)
Taking the positive square root of both sides (since the root lies in \(2 < x < 3\)) gives:
\(x = \sqrt{5 + \frac{3}{x}}\).

(iii) Iterating with \(x_1 = 2.5\):
\(x_2 = \sqrt{5 + \frac{3}{2.5}} = \sqrt{6.2} \approx 2.48998\)
\(x_3 = \sqrt{5 + \frac{3}{2.48998}} \approx 2.49095\)
\(x_4 = \sqrt{5 + \frac{3}{2.49095}} \approx 2.49086\)
\(x_5 = \sqrt{5 + \frac{3}{2.49086}} \approx 2.49086\)

Since the values of the iterations have stabilized to \(2.49086\), the root correct to 3 decimal places is \(\alpha = 2.491\).

Part (i):
M1: Evaluate function at 2 and 3.
A1: Confirm change of sign with clear values (-5 and 9) and write conclusion.

Part (ii):
M1: Re-arrange cubic equation to get \(x^2\) or divide by \(x\).
A1: Complete the algebra to show \(x = \sqrt{5 + \frac{3}{x}}\).

Part (iii):
M1: Calculate at least two iterations correctly to 5 d.p.
A1: Show correct sequence of values \(x_2 \approx 2.48998\), \(x_3 \approx 2.49095\), and \(x_4 \approx 2.49086\).
A1: Give final answer \(\alpha = 2.491\) and show that iterations support this accuracy.

(i) Squaring both sides of the inequality \(|3x - 2| < |2x + 5|\):
\((3x - 2)^2 < (2x + 5)^2\)
\(9x^2 - 12x + 4 < 4x^2 + 20x + 25\)
\(5x^2 - 32x - 21 < 0\)

Finding critical values by solving the corresponding quadratic equation:
\(5x^2 - 32x - 21 = 0 \implies (5x + 3)(x - 7) = 0\)
So, the critical values are \(x = -0.6\) and \(x = 7\).

Since the quadratic inequality is \(< 0\), the solution is:
\(-0.6 < x < 7\).

(ii) Comparing the second inequality with the first, we substitute \(x = e^{-y}\).
From part (i), we have:
\(-0.6 < e^{-y} < 7\)

Since the exponential function is always positive, \(e^{-y} > 0 > -0.6\) is always true.
Thus, we only need to solve:
\(e^{-y} < 7\)

Taking natural logarithms on both sides:
\(-y < \ln 7\)
Multiplying by \(-1\) reverses the inequality:
\(y > -\ln 7\).

Part (i):
M1: Square both sides or set up equations for critical values.
A1: Obtain correct quadratic inequality \(5x^2 - 32x - 21 < 0\) (or critical values \(-0.6\) and \(7\)).
M1: Correct method to solve quadratic inequality.
A1: Obtain the correct range \(-0.6 < x < 7\).

Part (ii):
M1: Recognize the substitution \(x = e^{-y}\) and set up the corresponding range.
B1: Realise that \(e^{-y} > -0.6\) is always satisfied, leaving only \(e^{-y} < 7\).
A1: Correctly apply logarithms to obtain the final answer \(y > -\ln 7\) (or equivalent).