2023 Cambridge IGCSE Mathematics - Additional (0606) 模擬試題連答案詳解

Equate the equations: \( 2x^2 - 5x + 5 = kx - 3 \). Rearrange to form a quadratic equation: \( 2x^2 - (5 + k)x + 8 = 0 \). For no intersection, the discriminant must be less than zero: \( \Delta < 0 \). This gives \( (-(5 + k))^2 - 4(2)(8) < 0 \), which simplifies to \( (5 + k)^2 - 64 < 0 \). This means \( (5 + k)^2 < 64 \), so \( -8 < 5 + k < 8 \). Subtracting 5 from all parts, we get \( -13 < k < 3 \).

M1 for equating the equations and setting up a 3-term quadratic equation. M1 for calculating the discriminant and setting it less than zero. M1 for finding the critical values -13 and 3. A1 for the correct final inequality range \( -13 < k < 3 \).

Using logarithmic laws, write \( 2\log_3(x - 2) \) as \( \log_3(x - 2)^2 \). Combining terms gives \( \log_3\left(\frac{(x-2)^2}{x+4}\right) = 1 \). Converting from logarithmic to exponential form gives \( \frac{(x-2)^2}{x+4} = 3^1 \). Expand and simplify: \( (x-2)^2 = 3(x+4) \) which leads to \( x^2 - 4x + 4 = 3x + 12 \), simplifying to \( x^2 - 7x - 8 = 0 \). Factoring the quadratic yields \( (x-8)(x+1) = 0 \), so \( x = 8 \) or \( x = -1 \). Since the argument of a logarithm must be positive, \( x - 2 > 0 \) and \( x + 4 > 0 \), which means \( x = -1 \) is extraneous. Thus, the only valid solution is \( x = 8 \).

M1 for applying the power law of logarithms to get \( \log_3(x - 2)^2 \). M1 for combining logarithms to get \( \frac{(x-2)^2}{x+4} = 3 \). M1 for solving the resulting quadratic equation to find roots 8 and -1. A1 for rejecting \( x = -1 \) and stating only \( x = 8 \).

Find the y-coordinate at \( x = 2 \): \( y = (2(2) - 3)^3 + 4(2) = 1^3 + 8 = 9 \). Differentiate \( y \) with respect to \( x \) using the chain rule: \( \frac{dy}{dx} = 3(2x - 3)^2 \cdot 2 + 4 = 6(2x - 3)^2 + 4 \). Find the gradient of the tangent at \( x = 2 \): \( \frac{dy}{dx} = 6(2(2) - 3)^2 + 4 = 10 \). The gradient of the normal is the negative reciprocal: \( m_{normal} = -\frac{1}{10} = -0.1 \). The equation of the normal is \( y - 9 = -0.1(x - 2) \), which simplifies to \( y = -0.1x + 9.2 \) or \( x + 10y = 92 \).

M1 for finding the y-coordinate 9. M1 for differentiating correctly to get \( 6(2x-3)^2 + 4 \). M1 for calculating gradient of tangent (10) and normal (-1/10). A1 for final correct equation in any equivalent form.

Let the first term be \( a \) and the common ratio be \( r \). We have \( u_3 = ar^2 = 18 \) and \( u_6 = ar^5 = -486 \). Dividing the two equations: \( \frac{ar^5}{ar^2} = \frac{-486}{18} \implies r^3 = -27 \implies r = -3 \). Substitute \( r = -3 \) back: \( a(-3)^2 = 18 \implies 9a = 18 \implies a = 2 \). The sum of the first 5 terms is \( S_5 = \frac{a(1 - r^5)}{1 - r} = \frac{2(1 - (-3)^5)}{1 - (-3)} = \frac{2(1 - (-243))}{4} = \frac{2(244)}{4} = 122 \).

M1 for setting up equations for third and sixth terms and solving for \( r \). M1 for finding \( r = -3 \) and \( a = 2 \). M1 for applying the geometric progression sum formula for \( S_5 \). A1 for the correct sum of 122.

Use the identity \( \sin^2\theta = 1 - \cos^2\theta \) to rewrite the equation: \( 3(1 - \cos^2\theta) - 5\cos\theta - 1 = 0 \). Expand and rearrange: \( 3 - 3\cos^2\theta - 5\cos\theta - 1 = 0 \implies 3\cos^2\theta + 5\cos\theta - 2 = 0 \). Factor the quadratic equation: \( (3\cos\theta - 1)(\cos\theta + 2) = 0 \). This gives \( \cos\theta = \frac{1}{3} \) or \( \cos\theta = -2 \) (which has no real solutions). Solving \( \cos\theta = \frac{1}{3} \) for \( 0^\circ \le \theta \le 360^\circ \): \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.5^\circ \) and \( \theta = 360^\circ - 70.5^\circ = 289.5^\circ \).

M1 for using the identity to form a quadratic in \( \cos\theta \). M1 for factorizing and solving to get \( \cos\theta = 1/3 \). M1 for finding \( \theta = 70.5^\circ \). A1 for both \( 70.5^\circ \) and \( 289.5^\circ \) and no others in the range.

(a) To find the stationary point, we find the first derivative using the product rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(2x-3) \cdot e^{2x} + (2x-3) \cdot \frac{\mathrm{d}}{\mathrm{d}x}(e^{2x}) = 2e^{2x} + 2(2x-3)e^{2x} = (4x - 4)e^{2x} = 4(x-1)e^{2x}\). At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies 4(x-1)e^{2x} = 0\). Since \(e^{2x} > 0\) for all \(x\), we have \(x = 1\). Substitute \(x = 1\) into the equation of the curve: \(y = (2(1)-3)e^{2(1)} = -e^2\). So the stationary point is \((1, -e^2)\). To determine the nature, we find the second derivative: \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 4e^{2x} + 2 \cdot 4(x-1)e^{2x} = (8x - 4)e^{2x}\). At \(x = 1\), \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = (8(1)-4)e^2 = 4e^2 > 0\). Therefore, the stationary point is a local minimum. (b) First, find the x-intercept of the curve by setting \(y = 0\): \((2x-3)e^{2x} = 0 \implies 2x - 3 = 0 \implies x = 1.5\). For \(1.5 \le x \le 2\), \(y \ge 0\). So the required area is: \(\text{Area} = \int_{1.5}^{2} (2x-3)e^{2x} \mathrm{d}x\). We integrate by parts using \(\int u \frac{\mathrm{d}v}{\mathrm{d}x} \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \mathrm{d}x\). Let \(u = 2x-3 \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 2\). Let \(\frac{\mathrm{d}v}{\mathrm{d}x} = e^{2x} \implies v = \frac{1}{2}e^{2x}\). So, \(\int (2x-3)e^{2x} \mathrm{d}x = (2x-3)\left(\frac{1}{2}e^{2x}\right) - \int \left(\frac{1}{2}e^{2x}\right)(2) \mathrm{d}x = \left(x - \frac{3}{2}\right)e^{2x} - \frac{1}{2}e^{2x} = (x-2)e^{2x}\). Now evaluate the definite integral: \(\text{Area} = \left[ (x-2)e^{2x} \right]_{1.5}^{2} = \left( (2-2)e^4 \right) - \left( (1.5-2)e^3 \right) = 0 - (-0.5 e^3) = \frac{1}{2}e^3\).

(a) M1: For attempt to differentiate using the product rule. A1: Correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}x} = 4(x-1)e^{2x}\). A1: Finding the stationary point coordinate \(x = 1\) and \(y = -e^2\) (accept \((1, -7.39)\)). M1: For using the second derivative or testing the gradient on either side to determine the nature. A1: Correct conclusion that it is a minimum based on a valid test. (b) M1: Finding the lower limit of integration by setting \(y = 0\) to get \(x = 1.5\). M1: For attempt at integration by parts on \(\int (2x-3)e^{2x} \mathrm{d}x\). A1: Correct integration to obtain \((x-2)e^{2x}\). M1: For substituting limits \(1.5\) and \(2\) into their integrated expression. A1.2: Correct final exact area \(\frac{1}{2}e^3\) (or \(0.5e^3\)).

(a) Apply the addition law of logarithms: \(\log_3((x-2)(x+6)) = 2\). Convert from logarithmic to exponential form: \((x-2)(x+6) = 3^2 \implies x^2 + 4x - 12 = 9 \implies x^2 + 4x - 21 = 0\). Factorise the quadratic equation: \((x+7)(x-3) = 0\). So, \(x = -7\) or \(x = 3\). We must check for extraneous solutions. For the logarithms to be defined, we require: \(x - 2 > 0 \implies x > 2\) and \(x + 6 > 0 \implies x > -6\). Therefore, \(x = -7\) is rejected, and the only valid solution is \(x = 3\). (b) For the first equation, express both sides with base 2: \(2^a \cdot (2^2)^b = 2^{-3} \implies 2^{a+2b} = 2^{-3}\). This gives the linear equation: \(a + 2b = -3\) (Equation 1). For the second equation, convert from logarithmic to exponential form: \(3a - b = 5^1 \implies 3a - b = 5\) (Equation 2). From Equation 2, express \(b\) in terms of \(a\): \(b = 3a - 5\). Substitute this into Equation 1: \(a + 2(3a - 5) = -3 \implies a + 6a - 10 = -3 \implies 7a = 7 \implies a = 1\). Substitute \(a = 1\) back into the expression for \(b\): \(b = 3(1) - 5 = -2\). Check validity: \(3a-b = 3(1) - (-2) = 5 > 0\), which is valid for the logarithm. Thus, \(a = 1, b = -2\).

(a) M1: Apply the product rule of logs to obtain \(\log_3((x-2)(x+6))\). M1: Convert log equation to quadratic equation \((x-2)(x+6) = 9\). A1: Correct quadratic \(x^2 + 4x - 21 = 0\). M1: Solve the quadratic to obtain two values of \(x\). A1: Identify \(x = 3\) as the only valid solution and explicitly reject \(x = -7\). (b) M1: Express \(2^a \cdot 4^b = \frac{1}{8}\) as an index equation in base 2 to get \(a + 2b = -3\). M1: Convert log equation to linear equation \(3a - b = 5\). M1: Use algebraic substitution or elimination method to solve the two simultaneous linear equations. A1: Correct value of \(a = 1\). A1.2: Correct value of \(b = -2\).

(a) Using the formula for the \(n\)-th term of an AP: \(u_n = a + (n-1)d\). For the 3rd term: \(a + 2d = 9\) (Equation 1). Using the formula for the sum of the first \(n\) terms of an AP: \(S_n = \frac{n}{2}[2a + (n-1)d]\). For the sum of the first 10 terms: \(S_{10} = \frac{10}{2}[2a + 9d] = 165 \implies 5[2a + 9d] = 165 \implies 2a + 9d = 33\) (Equation 2). Multiply Equation 1 by 2: \(2a + 4d = 18\) (Equation 3). Subtract Equation 3 from Equation 2: \(5d = 15 \implies d = 3\). Substitute \(d = 3\) into Equation 1: \(a + 2(3) = 9 \implies a = 3\). Thus, \(a = 3\) and \(d = 3\). (b) Let the first term of the GP be \(p\) and the second term be \(pr\). Sum of first two terms: \(p + pr = 12 \implies p(1+r) = 12\) (Equation 4). Using the formula for the sum to infinity of a GP: \(S_{\infty} = \frac{p}{1-r}\). Given \(S_{\infty} = 16\): \(\frac{p}{1-r} = 16 \implies p = 16(1-r)\) (Equation 5). Substitute Equation 5 into Equation 4: \(16(1-r)(1+r) = 12 \implies 16(1-r^2) = 12 \implies 1-r^2 = \frac{12}{16} = \frac{3}{4} \implies r^2 = 1 - \frac{3}{4} = \frac{1}{4}\). Since \(r > 0\), we have \(r = \sqrt{\frac{1}{4}} = \frac{1}{2}\). Substitute \(r = \frac{1}{2}\) into Equation 5: \(p = 16\left(1 - \frac{1}{2}\right) = 8\). Thus, \(p = 8\) and \(r = \frac{1}{2}\).

(a) M1: Write down a correct equation for the 3rd term, \(a + 2d = 9\). M1: Write down a correct equation for the sum of the first 10 terms, \(5(2a + 9d) = 165\). M1: Attempt to solve the two simultaneous equations for \(a\) and \(d\). A1: Correct value of \(d = 3\). A1: Correct value of \(a = 3\). (b) M1: State the sum of the first two terms as \(p + pr = 12\) or equivalent. M1: State the sum to infinity as \(\frac{p}{1-r} = 16\) and express \(p\) in terms of \(r\). M1: Form a quadratic equation in terms of \(r\) by combining the two equations. A1: Correctly solve for \(r^2 = \frac{1}{4}\) and justify selecting the positive root \(r = \frac{1}{2}\). A1.2: Correctly find \(p = 8\).

(a) Start with the Left Hand Side (LHS): \(\text{LHS} = \frac{\sin \theta}{1 - \cos \theta} + \frac{1 - \cos \theta}{\sin \theta}\). Find a common denominator: \(\text{LHS} = \frac{\sin^2 \theta + (1 - \cos \theta)^2}{\sin \theta(1 - \cos \theta)}\). Expand the numerator: \(\text{LHS} = \frac{\sin^2 \theta + (1 - 2\cos \theta + \cos^2 \theta)}{\sin \theta(1 - \cos \theta)}\). Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\): \(\text{LHS} = \frac{(\sin^2 \theta + \cos^2 \theta) + 1 - 2\cos \theta}{\sin \theta(1 - \cos \theta)} = \frac{1 + 1 - 2\cos \theta}{\sin \theta(1 - \cos \theta)} = \frac{2 - 2\cos \theta}{\sin \theta(1 - \cos \theta)}\). Factorise the numerator: \(\text{LHS} = \frac{2(1 - \cos \theta)}{\sin \theta(1 - \cos \theta)}\). Cancel out the common term \((1 - \cos \theta)\): \(\text{LHS} = \frac{2}{\sin \theta} = 2\csc \theta = \text{RHS}\). (b) Solve \(3\cos(2x - 0.5) = 1\) for \(0 \le x \le \pi\). \(\cos(2x - 0.5) = \frac{1}{3}\). Let \(\theta = 2x - 0.5\). Since \(0 \le x \le \pi\), the interval for \(\theta\) is \(-0.5 \le 2x - 0.5 \le 2\pi - 0.5 \approx 5.78\). The basic angle in radians is \(\alpha = \cos^{-1}\left(\frac{1}{3}\right) \approx 1.231\) radians. Since \(\cos \theta > 0\), \(\theta\) lies in Quadrant 1 or Quadrant 4. In Quadrant 1: \(\theta = 1.231\). In Quadrant 4: \(\theta = 2\pi - 1.231 \approx 5.052\). (The other possible value \(-1.231\) is outside \([-0.5, 5.78]\)). Now solve for \(x\): For \(2x - 0.5 = 1.231 \implies 2x = 1.731 \implies x \approx 0.87\) radians. For \(2x - 0.5 = 5.052 \implies 2x = 5.552 \implies x \approx 2.78\) radians.

(a) M1: For putting fractions over a common denominator. M1: For expanding \((1 - \cos \theta)^2\) correctly. M1: For utilizing the identity \(\sin^2 \theta + \cos^2 \theta = 1\). M1: For factorising the numerator to \(2(1 - \cos \theta)\). A1: For dividing out and stating the final step \(\frac{2}{\sin \theta} = 2\csc \theta\) clearly. (b) M1: State \(\cos(2x - 0.5) = \frac{1}{3}\) and find the basic angle \(\approx 1.231\). M1: Identify the correct interval for \(\theta = 2x - 0.5\) as \([-0.5, 2\pi-0.5]\). M1: State the correct angles for \(\theta\): \(1.231\) and \(5.052\). A1: Solve for one correct value of \(x\) to 2 d.p. (\(x = 0.87\)). A1.2: Solve for the second correct value of \(x\) to 2 d.p. (\(x = 2.78\)).

(a)(i) For \(x > -0.5\), \(2x + 1 > 0\). As \(x \to -0.5^+\), \(\ln(2x+1) \to -\infty\), so \(\mathrm{f}(x) \to \infty\). As \(x \to \infty\), \(\ln(2x+1) \to \infty\), so \(\mathrm{f}(x) \to -\infty\). Since \(\mathrm{f}\) is a continuous logarithmic function, it takes all real values. Thus, the range of \(\mathrm{f}\) is \(\mathrm{f}(x) \in \mathbb{R}\). (a)(ii) Let \(y = 3 - \ln(2x + 1) \implies \ln(2x + 1) = 3 - y \implies 2x + 1 = e^{3 - y} \implies 2x = e^{3 - y} - 1 \implies x = \frac{e^{3 - y} - 1}{2}\). Thus, \(\mathrm{f}^{-1}(x) = \frac{e^{3-x} - 1}{2}\). (b)(i) Complete the square for \(\mathrm{g}(x)\): \(\mathrm{g}(x) = (x-2)^2 + 3\). Since the domain of \(\mathrm{g}\) is limited to \(x \ge 2\), the function is strictly increasing (the vertex of the parabola is at \(x = 2\)). Therefore, \(\mathrm{g}\) is a one-to-one function, which means it has an inverse. (b)(ii) The range of \(\mathrm{g}\) is the domain of \(\mathrm{g}^{-1}\). Since \(\mathrm{g}(x) = (x-2)^2 + 3\) and \(x \ge 2\), the minimum value of \(\mathrm{g}(x)\) is 3. Thus, the range of \(\mathrm{g}\) is \(\mathrm{g}(x) \ge 3\), so the domain of \(\mathrm{g}^{-1}\) is \(x \ge 3\). To find the inverse expression, let \(y = (x-2)^2 + 3 \implies (x-2)^2 = y - 3\). Since the domain of \(\mathrm{g}\) is \(x \ge 2\), we choose the positive square root: \(x - 2 = \sqrt{y - 3} \implies x = 2 + \sqrt{y - 3}\). Therefore, \(\mathrm{g}^{-1}(x) = 2 + \sqrt{x-3}\).

(a)(i) M1: Realise that the logarithmic term has no upper or lower bounds on this domain. A1: Correct range \(\mathrm{f}(x) \in \mathbb{R}\) (or 'all real values'). (a)(ii) M1: Attempt to make \(x\) the subject of \(y = 3 - \ln(2x + 1)\). M1: Remove the logarithm by converting to exponential form \(2x + 1 = e^{3 - y}\). A1.2: Correct expression \(\mathrm{f}^{-1}(x) = \frac{e^{3-x} - 1}{2}\) (must be in terms of \(x\)). (b)(i) M1: Attempt to express \(\mathrm{g}(x)\) in completed square form or find the vertex. A1: Explain that because the domain is \(x \ge 2\), the function is one-to-one (strictly increasing) and therefore has an inverse. (b)(ii) M1: State the domain of \(\mathrm{g}^{-1}\) as \(x \ge 3\). M1: Attempt to make \(x\) the subject of \(y = (x-2)^2 + 3\). M1: Choose the correct positive root because \(x \ge 2\). A1: Correct expression \(\mathrm{g}^{-1}(x) = 2 + \sqrt{x-3}\).

To solve the equation \(\log_3 x + \log_9 (x + 6) = 2\), we first express all logarithms with the same base. Using the change of base formula: \(\log_9 (x + 6) = \frac{\log_3 (x + 6)}{\log_3 9} = \frac{\log_3 (x + 6)}{2} = \frac{1}{2}\log_3 (x + 6)\). Substitute this back into the equation: \(\log_3 x + \frac{1}{2}\log_3 (x + 6) = 2\). Multiply the entire equation by 2 to clear the fraction: \(2\log_3 x + \log_3 (x + 6) = 4\). Use the laws of logarithms to combine the terms on the left side: \(\log_3 x^2 + \log_3 (x + 6) = 4 \implies \log_3 (x^2(x + 6)) = 4\). Convert the logarithmic equation to exponential form: \(x^2(x + 6) = 3^4 \implies x^3 + 6x^2 = 81 \implies x^3 + 6x^2 - 81 = 0\). We look for real roots of this cubic equation. For \(x = 3\): \(3^3 + 6(3)^2 - 81 = 27 + 54 - 81 = 0\). So \(x = 3\) is a solution. Factorizing the cubic expression: \((x - 3)(x^2 + 9x + 27) = 0\). The quadratic part \(x^2 + 9x + 27 = 0\) has a discriminant of \(\Delta = 9^2 - 4(1)(27) = 81 - 108 = -27 < 0\), which yields no real roots. Since the original logarithmic terms require \(x > 0\) and \(x + 6 > 0\), \(x = 3\) is the only valid real solution.

M1: For applying the change of base formula correctly to express \(\log_9(x+6)\) as \(\frac{1}{2}\log_3(x+6)\).
M1: For using logarithmic laws to combine terms, resulting in \(\log_3(x^2(x+6))\) or equivalent.
M1: For removing the logarithms to obtain the cubic equation \(x^3 + 6x^2 - 81 = 0\).
A1: For identifying \(x = 3\) as a solution.
A1.2: For verifying that \(x^2 + 9x + 27 = 0\) has no real roots, justifying why \(x = 3\) is the unique solution.

To find the equation of the curve, we first integrate the second derivative to find the first derivative \(\frac{\mathrm{d}y}{\mathrm{d}x}\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = \int (6x - 4) \mathrm{d}x = 3x^2 - 4x + c\). Since the curve has a stationary point at \((2, 5)\), we know that \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) when \(x = 2\). Substituting these values: \(3(2)^2 - 4(2) + c = 0 \implies 12 - 8 + c = 0 \implies c = -4\). This gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4x - 4\). Now, integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find \(y\): \(y = \int (3x^2 - 4x - 4) \mathrm{d}x = x^3 - 2x^2 - 4x + d\). Since the curve passes through the stationary point \((2, 5)\), substitute \(x = 2\) and \(y = 5\) into the equation: \(5 = (2)^3 - 2(2)^2 - 4(2) + d \implies 5 = 8 - 8 - 8 + d \implies 5 = -8 + d \implies d = 13\). Thus, the equation of the curve is \(y = x^3 - 2x^2 - 4x + 13\).

M1: For integrating \(6x - 4\) to obtain \(3x^2 - 4x (+ c)\).
M1: For substituting \(x = 2\) and setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) to find \(c = -4\).
M1: For integrating their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) function to obtain \(x^3 - 2x^2 - 4x (+ d)\).
M1: For substituting \(x = 2, y = 5\) to find the integration constant \(d\).
A1.2: For the correct final equation: \(y = x^3 - 2x^2 - 4x + 13\).

First, use the Pythagorean identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to rewrite the equation in terms of \(\cos \theta\): \(2(1 - \cos^2 \theta) - 3 \cos \theta - 3 = 0 \implies 2 - 2\cos^2 \theta - 3 \cos \theta - 3 = 0 \implies -2\cos^2 \theta - 3 \cos \theta - 1 = 0\). Multiply by \(-1\) to get a quadratic equation: \(2\cos^2 \theta + 3 \cos \theta + 1 = 0\). Factorize the quadratic expression: \((2\cos \theta + 1)(\cos \theta + 1) = 0\). This gives two possible cases: 1) \(\cos \theta = -\frac{1}{2}\) or 2) \(\cos \theta = -1\). For the interval \(0 \le \theta \le 2\pi\): From \(\cos \theta = -\frac{1}{2}\), the basic angle is \(\frac{\pi}{3}\). Since cosine is negative in the second and third quadrants: \\ \(\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\) \\ \(\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\). From \(\cos \theta = -1\): \\ \(\theta = \pi\). Therefore, the solutions are \(\theta = \frac{2\pi}{3}, \pi, \frac{4\pi}{3}\).

M1: For substituting \(\sin^2 \theta = 1 - \cos^2 \theta\) to obtain a quadratic equation in \(\cos \theta\).
M1: For correctly factorizing or solving the quadratic equation to find \(\cos \theta = -1/2\) and \(\cos \theta = -1\).
A1: For obtaining the solution \(\theta = \pi\).
A1: For obtaining either \(\theta = \frac{2\pi}{3}\) or \(\theta = \frac{4\pi}{3}\).
A1.2: For obtaining both remaining correct solutions with no extra solutions in the range.

To find the number of different committees of 5 people with more women than men, we consider the possible combinations of women (W) and men (M) such that the number of women is greater than the number of men. The possible cases are: Case 1: 3 women and 2 men. Number of ways to choose 3 women from 5 is \(\binom{5}{3} = 10\). Number of ways to choose 2 men from 6 is \(\binom{6}{2} = 15\). Total ways for Case 1 = \(10 \times 15 = 150\). Case 2: 4 women and 1 man. Number of ways to choose 4 women from 5 is \(\binom{5}{4} = 5\). Number of ways to choose 1 man from 6 is \(\binom{6}{1} = 6\). Total ways for Case 2 = \(5 \times 6 = 30\). Case 3: 5 women and 0 men. Number of ways to choose 5 women from 5 is \(\binom{5}{5} = 1\). Number of ways to choose 0 men from 6 is \(\binom{6}{0} = 1\). Total ways for Case 3 = \(1 \times 1 = 1\). Summing the possibilities for all cases: \(150 + 30 + 1 = 181\).

M1: For identifying the three valid cases: (3W, 2M), (4W, 1M), and (5W, 0M).
M1: For calculating \(\binom{5}{3} \times \binom{6}{2} = 150\).
M1: For calculating \(\binom{5}{4} \times \binom{6}{1} = 30\).
M1: For calculating \(\binom{5}{5} \times \binom{6}{0} = 1\).
A1.2: For sum of all three cases resulting in 181.

Since the terms are consecutive terms of a geometric progression, the ratio between consecutive terms is constant: \(r = \frac{k}{k+4} = \frac{k-3}{k}\). Cross-multiplying to solve for \(k\): \(k^2 = (k+4)(k-3) \implies k^2 = k^2 + k - 12 \implies 0 = k - 12 \implies k = 12\). Substitute \(k = 12\) back into the expressions for the first three terms: First term, \(a = k+4 = 12+4 = 16\). Second term = \(k = 12\). Third term = \(k-3 = 9\). The common ratio, \(r\), is \(\frac{12}{16} = \frac{3}{4}\). Since \(|r| < 1\), the sum to infinity exists and is calculated using \(S_{\infty} = \frac{a}{1-r}\): \(S_{\infty} = \frac{16}{1 - 3/4} = \frac{16}{1/4} = 64\).

M1: For establishing the common ratio equation \(\frac{k}{k+4} = \frac{k-3}{k}\).
M1: For solving the equation to find \(k = 12\).
M1: For finding the first term \(a = 16\) and common ratio \(r = \frac{3}{4}\).
M1: For substituting their \(a\) and \(r\) into the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\).
A1.2: For evaluating the sum to obtain 64.

(a) Let \(y = (2x - 3)(4x + 1)^{1/2}\).
Using the product rule \(\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\):
\(\frac{dy}{dx} = (2x - 3) \cdot \frac{1}{2}(4x + 1)^{-1/2} \cdot 4 + (4x + 1)^{1/2} \cdot 2\)
\(\frac{dy}{dx} = \frac{2(2x - 3)}{\sqrt{4x + 1}} + 2\sqrt{4x + 1}\)
Combine over a common denominator:
\(\frac{dy}{dx} = \frac{2(2x - 3) + 2(4x + 1)}{\sqrt{4x + 1}}\)
\(\frac{dy}{dx} = \frac{4x - 6 + 8x + 2}{\sqrt{4x + 1}}\)
\(\frac{dy}{dx} = \frac{12x - 4}{\sqrt{4x + 1}}\).

(b) For the stationary point, set \(\frac{dy}{dx} = 0\):
\(\frac{12x - 4}{\sqrt{4x + 1}} = 0 \implies 12x - 4 = 0 \implies x = \frac{1}{3}\).
Substitute \(x = \frac{1}{3}\) into the equation of the curve to find the y-coordinate:
\(y = \left(2\left(\frac{1}{3}\right) - 3\right) \sqrt{4\left(\frac{1}{3}\right) + 1}\)
\(y = \left(\frac{2}{3} - 3\right) \sqrt{\frac{7}{3}}\)
\(y = -\frac{7}{3}\sqrt{\frac{7}{3}}\).
So the exact coordinates are \(\left(\frac{1}{3}, -\frac{7}{3}\sqrt{\frac{7}{3}}\right)\).

(c) We use the small change formula \(\delta y \approx \left(\frac{dy}{dx}\right)_{x=2} \cdot \delta x\), where \(\delta x = p\).
At \(x = 2\):
\(\frac{dy}{dx} = \frac{12(2) - 4}{\sqrt{4(2) + 1}} = \frac{20}{\sqrt{9}} = \frac{20}{3}\).
Therefore, \(\delta y \approx \frac{20}{3}p\).

(a)
M1: For applying product rule with at least one term correct.
A1: For correct derivative of \((4x + 1)^{1/2}\) which is \(2(4x + 1)^{-1/2}\).
M1: For attempting to find a common denominator.
A1: For a correct unsimplified expression with a common denominator.
A1: For simplifying to \(\frac{12x - 4}{\sqrt{4x + 1}}\).

(b)
M1: For setting their numerator of \(\frac{dy}{dx}\) equal to 0 and solving for \(x\).
A1: For \(x = \frac{1}{3}\) (or equivalent).
A1: For \(y = -\frac{7}{3}\sqrt{\frac{7}{3}}\).

(c)
M1: For substituting \(x = 2\) into their \(\frac{dy}{dx}\).
A1: For finding the value \(\frac{20}{3}\) (or 6.67).
A1: For \(\frac{20}{3}p\) (or 6.67p).

(a) For the arithmetic progression:
First term \(T_1 = k\)
Second term \(T_2 = k + (k + 1) = 2k + 1\)
Fifth term \(T_5 = k + 4(k + 1) = 5k + 4\)

Since \(T_1\), \(T_2\), and \(T_5\) form the first three terms of a geometric progression, the common ratio is constant:
\(\frac{T_2}{T_1} = \frac{T_5}{T_2} \implies \frac{2k+1}{k} = \frac{5k+4}{2k+1}\)

Multiply across:
\((2k + 1)^2 = k(5k + 4)\)
\(4k^2 + 4k + 1 = 5k^2 + 4k\)
\(k^2 = 1\)

Since \(k > 0\), we have \(k = 1\).

(b) With \(k = 1\), the arithmetic progression has first term \(a = 1\) and common difference \(d = 1 + 1 = 2\).
The sum of the first 40 terms is given by:
\(S_{40} = \frac{40}{2} [2a + 39d]\)
\(S_{40} = 20 [2(1) + 39(2)]\)
\(S_{40} = 20 [2 + 78] = 20 \times 80 = 1600\).

(c)
(i) For the geometric progression:
First term \(G_1 = T_1 = 1\)
Second term \(G_2 = T_2 = 2(1) + 1 = 3\)
Third term \(G_3 = T_5 = 5(1) + 4 = 9\)
The common ratio \(r = \frac{3}{1} = 3\).

(ii) The sum of the first 8 terms of this geometric progression is:
\(S_8 = \frac{G_1(r^8 - 1)}{r - 1}\)
\(S_8 = \frac{1(3^8 - 1)}{3 - 1}\)
\(S_8 = \frac{6561 - 1}{2} = \frac{6560}{2} = 3280\).

(a)
M1: For expressing \(T_2\) and \(T_5\) in terms of \(k\).
M1: For setting up the ratio equation \(\frac{2k+1}{k} = \frac{5k+4}{2k+1}\).
M1: For expanding and simplifying to \(k^2 = 1\).
A1: For justifying \(k = 1\) since \(k > 0\).

(b)
M1: For identifying \(a = 1\) and \(d = 2\).
M1: For applying the arithmetic progression sum formula.
A1: For 1600.

(c)(i)
B1: For \(r = 3\).

(c)(ii)
M1: For using the geometric progression sum formula.
M1: For evaluating \(3^8 = 6561\).
A1: For 3280.

(a) Let \(u = 3^z\).
Then \(3^{2z + 1} = 3 \cdot (3^z)^2 = 3u^2\).
The equation becomes:
\(3u^2 - 10u + 3 = 0\)
Factorising the quadratic:
\((3u - 1)(u - 3) = 0\)
So \(u = \frac{1}{3}\) or \(u = 3\).

If \(3^z = 3^{-1} \implies z = -1\).
If \(3^z = 3^1 \implies z = 1\).

(b) From the first equation:
\(\log_5 (3x + y) = 2 \implies 3x + y = 5^2 \implies 3x + y = 25 \implies y = 25 - 3x\) [Equation 1]

From the second equation:
\(\log_4 x + \log_4 y = 2 + \log_4 3\)
Using the laws of logarithms:
\(\log_4 (xy) = \log_4 16 + \log_4 3\) (since \(\log_4 16 = 2\))
\(\log_4 (xy) = \log_4 (16 \times 3)\)
\(\log_4 (xy) = \log_4 48\)
So \(xy = 48\) [Equation 2]

Substitute Equation 1 into Equation 2:
\(x(25 - 3x) = 48\)
\(25x - 3x^2 = 48\)
\(3x^2 - 25x + 48 = 0\)

Factorising the quadratic:
\((3x - 16)(x - 3) = 0\)
So \(x = \frac{16}{3}\) or \(x = 3\).

Substitute \(x\) values to find \(y\):
If \(x = 3\), then \(y = 25 - 3(3) = 16\).
If \(x = \frac{16}{3}\), then \(y = 25 - 3\left(\frac{16}{3}\right) = 9\).

Both solutions are valid as both \(x\) and \(y\) are positive, so their logarithms are defined.
Solutions are \((3, 16)\) and \(\left(\frac{16}{3}, 9\right)\).

(a)
M1: For using substitution \(u = 3^z\) to form quadratic equation \(3u^2 - 10u + 3 = 0\).
A1: For solving the quadratic to get \(u = \frac{1}{3}\) and \(u = 3\).
M1: For attempting to solve \(3^z = u\).
A1: For \(z = -1\) and \(z = 1\).

(b)
M1: For converting \(\log_5 (3x + y) = 2\) to \(3x + y = 25\).
M1: For combining \(\log_4 x + \log_4 y\) into \(\log_4 (xy)\).
M1: For converting \(2 + \log_4 3\) to \(\log_4 48\) or equivalent.
A1: For obtaining \(xy = 48\).
M1: For substituting one equation into the other to form a quadratic equation.
A1: For solving the quadratic to get \(x = 3\) and \(x = \frac{16}{3}\).
A1: For both pairs clearly stated: \((3, 16)\) and \(\left(\frac{16}{3}, 9\right)\).

(a) Left-Hand Side (LHS):
\(\text{LHS} = \frac{\sin\theta}{1 - \cos\theta} - \frac{1}{\sin\theta}\)
Combine fractions over a common denominator:
\(\text{LHS} = \frac{\sin^2\theta - (1 - \cos\theta)}{\sin\theta(1 - \cos\theta)}\)
Recall that \(\sin^2\theta = 1 - \cos^2\theta\):
\(\text{LHS} = \frac{(1 - \cos^2\theta) - 1 + \cos\theta}{\sin\theta(1 - \cos\theta)}\)
\(\text{LHS} = \frac{\cos\theta - \cos^2\theta}{\sin\theta(1 - \cos\theta)}\)
Factorise the numerator:
\(\text{LHS} = \frac{\cos\theta(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)}\)
Cancel the common term \((1 - \cos\theta)\):
\(\text{LHS} = \frac{\cos\theta}{\sin\theta} = \cot\theta = \text{RHS}\).

(b) \(3\tan^2\phi - 5\sec\phi + 1 = 0\)
Use the identity \(\tan^2\phi = \sec^2\phi - 1\):
\(3(\sec^2\phi - 1) - 5\sec\phi + 1 = 0\)
\(3\sec^2\phi - 3 - 5\sec\phi + 1 = 0\)
\(3\sec^2\phi - 5\sec\phi - 2 = 0\)

Factorising this quadratic in \(\sec\phi\):
\((3\sec\phi + 1)(\sec\phi - 2) = 0\)
So \(\sec\phi = -\frac{1}{3}\) or \(\sec\phi = 2\).

Case 1: \(\sec\phi = -\frac{1}{3} \implies \cos\phi = -3\)
Since \(-1 \le \cos\phi \le 1\), there are no solutions for this case.

Case 2: \(\sec\phi = 2 \implies \cos\phi = \frac{1}{2}\)
For \(0^\circ \le \phi \le 360^\circ\):
\(\phi = 60^\circ\) or \(\phi = 360^\circ - 60^\circ = 300^\circ\).

(a)
M1: For putting the fractions over a common denominator.
M1: For substituting \(\sin^2\theta = 1 - \cos^2\theta\).
A1: For simplifying numerator to \(\cos\theta - \cos^2\theta\).
M1: For factorising numerator to \(\cos\theta(1 - \cos\theta)\).
A1: For cancelling and concluding with \(\cot\theta\).

(b)
M1: For substituting \(\tan^2\phi = \sec^2\phi - 1\).
M1: For obtaining the correct quadratic equation \(3\sec^2\phi - 5\sec\phi - 2 = 0\).
A1: For solving to get \(\sec\phi = -\frac{1}{3}\) and \(\sec\phi = 2\).
M1: For identifying that \(\cos\phi = -3\) has no solutions.
A1: For \(\phi = 60^\circ\).
A1: For \(\phi = 300^\circ\).

(a) To complete the square for \(f(x) = -2x^2 + 12x - 5\):
Factorise \(-2\) out of the \(x\)-terms:
\(f(x) = -2(x^2 - 6x) - 5\)
Complete the square inside the bracket:
\(f(x) = -2[(x - 3)^2 - 9] - 5\)
\(f(x) = -2(x - 3)^2 + 18 - 5\)
\(f(x) = -2(x - 3)^2 + 13\).
So, \(a = -2\), \(h = 3\), and \(k = 13\).

(b) Since \(a = -2 < 0\), the curve has a maximum point.
The maximum value of \(f(x)\) is \(13\), and it occurs when \(x - 3 = 0 \implies x = 3\).

(c) For the line \(y = mx + 3\) to be a tangent to the curve \(y = -2x^2 + 12x - 5\):
\(mx + 3 = -2x^2 + 12x - 5\)
Rearranging into standard quadratic form:
\(2x^2 + (m - 12)x + 8 = 0\)

For a tangent, the discriminant of this quadratic equation must be zero (\(B^2 - 4AC = 0\)):
\((m - 12)^2 - 4(2)(8) = 0\)
\((m - 12)^2 - 64 = 0\)
\((m - 12)^2 = 64\)

Taking square roots:
\(m - 12 = 8 \implies m = 20\)
\(m - 12 = -8 \implies m = 4\).

Therefore, the values of \(m\) are \(4\) and \(20\).

(a)
M1: For factorising \(-2\) from the first two terms: \(-2(x^2 - 6x) \dots\).
M1: For completing the square inside the bracket: \((x - 3)^2 - 9\).
A1: For obtaining \(-2(x - 3)^2 + 13\).

(b)
B1: For maximum value is \(13\) (follow through their \(k\)).
B1: For \(x = 3\) (follow through their \(h\)).

(c)
M1: For setting the equations equal to each other.
A1: For a correct quadratic equation, e.g., \(2x^2 + (m - 12)x + 8 = 0\).
M1: For using the discriminant \(B^2 - 4AC\).
M1: For setting their discriminant equal to \(0\).
A1: For simplifying to \((m - 12)^2 = 64\) (or equivalent).
A1: For \(m = 4\) and \(m = 20\).