2023 Edexcel IGCSE Further Pure Mathematics 模擬試題連答案詳解

(a) The volume of a cylinder is \(V = \pi r^2 h = 128\pi\), which gives \(h = \frac{128}{r^2}\). The total surface area is \(S = 2\pi r^2 + 2\pi r h\). Substituting \(h\) gives \(S = 2\pi r^2 + 2\pi r \left(\frac{128}{r^2}\right) = 2\pi r^2 + \frac{256\pi}{r}\). (b) Differentiating \(S\) with respect to \(r\) gives \(\frac{\text{d}S}{\text{d}r} = 4\pi r - \frac{256\pi}{r^2}\). Setting \(\frac{\text{d}S}{\text{d}r} = 0\) gives \(4\pi r = \frac{256\pi}{r^2} \implies r^3 = 64 \implies r = 4\). Finding the second derivative gives \(\frac{\text{d}^2S}{\text{d}r^2} = 4\pi + \frac{512\pi}{r^3}\). For \(r = 4\), \(\frac{\text{d}^2S}{\text{d}r^2} = 4\pi + \frac{512\pi}{64} = 12\pi > 0\), which confirms a minimum. (c) Substituting \(r = 4\) into the expression for \(S\) gives \(S = 2\pi(4)^2 + \frac{256\pi}{4} = 32\pi + 64\pi = 96\pi\).

(a) M1: Uses \(V = 128\pi\) to express \(h\) in terms of \(r\). A1: Correctly substitutes \(h\) into \(S = 2\pi r^2 + 2\pi r h\) to show the given formula. (b) M1: Differentiates \(S\) to find \(\frac{\text{d}S}{\text{d}r}\). A1: Equates to 0 and solves to find \(r = 4\). M1: Finds \(\frac{\text{d}^2S}{\text{d}r^2}\) and evaluates at \(r = 4\). A1: Showing \(\frac{\text{d}^2S}{\text{d}r^2} > 0\) to justify minimum. (c) M1: Substitutes \(r = 4\) into \(S\). A1: Obtains \(96\pi\).

(a) Using \(\cos(A+B) = \cos A \cos B - \sin A \sin B\), we have \(\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta\). Substituting \(\cos 2\theta = 2\cos^2 \theta - 1\) and \(\sin 2\theta = 2\sin \theta \cos \theta\), we get \((2\cos^2 \theta - 1)\cos \theta - 2\sin^2 \theta \cos \theta = 2\cos^3 \theta - \cos \theta - 2(1 - \cos^2 \theta)\cos \theta = 4\cos^3 \theta - 3\cos \theta\). (b) The equation \(8\cos^3 \theta - 6\cos \theta - 1 = 0\) can be rewritten as \(2(4\cos^3 \theta - 3\cos \theta) = 1\), which simplifies using part (a) to \(2\cos 3\theta = 1 \implies \cos 3\theta = \frac{1}{2}\). Since \(0 \le \theta \le \pi\), we have \(0 \le 3\theta \le 3\pi\). This gives \(3\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}\). Solving for \(\theta\) gives \(\theta = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}\).

(a) M1: Uses \(\cos(2\theta+\theta)\) compound angle formula. M1: Applies double angle formulae for \(\cos 2\theta\) and \(\sin 2\theta\). M1: Uses \(\sin^2 \theta = 1 - \cos^2 \theta\). A1: Fully correct proof. (b) M1: Recognises the identity from part (a) to write \(2\cos 3\theta = 1\). A1: Finding \(\cos 3\theta = \frac{1}{2}\). M1: Finds at least two correct values for \(3\theta\) in the range. A1: Divides by 3 to find correct values of \(\theta\). A1: All three correct values: \(\frac{\pi}{9}\), \(\frac{5\pi}{9}\), \(\frac{7\pi}{9}\) and no others in the range.

From the given quadratic equation, the sum of roots is \(\alpha + \beta = -\frac{5}{2}\) and the product is \(\alpha\beta = \frac{4}{2} = 2\). (a) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\). (b) \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = \left(-\frac{5}{2}\right)^3 - 3(2)\left(-\frac{5}{2}\right) = -\frac{125}{8} + 15 = -\frac{5}{8}\). (c) The sum of the new roots is \(S = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2 \beta^2} = \frac{-5/8}{2^2} = -\frac{5}{32}\). The product of the new roots is \(P = \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} = \frac{1}{\alpha\beta} = \frac{1}{2}\). The new equation is \(x^2 - Sx + P = 0 \implies x^2 + \frac{5}{32}x + \frac{1}{2} = 0\). Multiplying by 32 to get integer coefficients gives \(32x^2 + 5x + 16 = 0\).

(a) B1: Correct sum \(\alpha+\beta = -2.5\) and product \(\alpha\beta = 2\). M1: Uses identity \(\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta\) to find \(\frac{9}{4}\). (b) M1: Uses identity \(\alpha^3+\beta^3 = (\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)\). A1: Correctly calculates \(-\frac{5}{8}\). (c) M1: Finds sum of new roots in terms of \(\alpha\) and \(\beta\). A1: Correctly calculates \(S = -\frac{5}{32}\). M1: Finds product of new roots \(P = \frac{1}{2}\). A1: Correct equation in the required form \(32x^2 + 5x + 16 = 0\).

(a) The midpoint \(M\) of \(AB\) is \(\left(\frac{1+5}{2}, \frac{5-3}{2}\right) = (3, 1)\). The gradient of \(AB\) is \(m = \frac{-3-5}{5-1} = \frac{-8}{4} = -2\). The gradient of the perpendicular bisector is \(m_{\perp} = -\frac{1}{-2} = \frac{1}{2}\). The equation of the perpendicular bisector is \(y - 1 = \frac{1}{2}(x - 3) \implies 2y - 2 = x - 3 \implies 2y - x + 1 = 0\). (b) The line meets the \(x\)-axis when \(y = 0\), which gives \(2(0) - x + 1 = 0 \implies x = 1\). Thus, \(P\) is \((1, 0)\). (c) The area of triangle \(ABP\) with coordinates \(A(1, 5)\), \(B(5, -3)\), and \(P(1, 0)\) can be calculated using the formula \(\frac{1}{2} | 1(-3-0) + 5(0-5) + 1(5-(-3)) | = \frac{1}{2} | -3 - 25 + 8 | = \frac{1}{2} | -20 | = 10\).

(a) M1: Finds midpoint \(M(3, 1)\) of \(AB\). M1: Finds gradient of \(AB\) and negative reciprocal. M1: Finds equation of the perpendicular line. A1: Correct equation in required form, e.g. \(2y - x + 1 = 0\) or any integer multiple. (b) M1: Sets \(y = 0\) in their line equation. A1: Correct coordinates \(P(1, 0)\). (c) M1: Uses a valid method to find the area of the triangle (e.g. shoelace formula or base/height). A1: Correct distance calculations or correct substitution into formula. A1: Area = 10.

(a) The first, third, and eleventh terms of the arithmetic series are \(a\), \(a+2d\), and \(a+10d\) respectively. Since these terms are in a geometric series, we have \(\frac{a+2d}{a} = \frac{a+10d}{a+2d} \implies (a+2d)^2 = a(a+10d)\). Expanding both sides: \(a^2 + 4ad + 4d^2 = a^2 + 10ad \implies 4d^2 = 6ad\). Since \(d \neq 0\), we can divide by \(2d\) to get \(2d = 3a \implies d = \frac{3}{2}a\). (b) The common ratio \(r\) is given by \(\frac{a+2d}{a}\). Substituting \(d = \frac{3}{2}a\) gives \(r = \frac{a + 2(1.5a)}{a} = \frac{4a}{a} = 4\). (c) The sum of the first 20 terms of an arithmetic series is \(S_{20} = \frac{20}{2}[2a + 19d] = 10[2a + 19d] = 610 \implies 2a + 19d = 61\). Substituting \(d = 1.5a\) gives \(2a + 19(1.5a) = 61 \implies 30.5a = 61 \implies a = 2\). Since \(d = 1.5a\), we have \(d = 1.5(2) = 3\).

(a) M1: Sets up the geometric series relation \((a+2d)^2 = a(a+10d)\). M1: Expands and simplifies the equation. A1: Correctly deduces \(d = 1.5a\). (b) M1: Writes the ratio \(r = \frac{a+2d}{a}\) and substitutes \(d\). A1: Correctly shows \(r = 4\). (c) M1: Uses the sum of arithmetic series formula to write \(10[2a + 19d] = 610\). M1: Substitutes \(d = 1.5a\) to find a linear equation in one variable. A1: Finds \(a = 2\). A1: Finds \(d = 3\).

(a) Using the binomial expansion formula, \(\mathrm{f}(x) = (1 - 3x)^{-\frac{1}{3}} = 1 + \left(-\frac{1}{3}\right)(-3x) + \frac{(-\frac{1}{3})(-\frac{4}{3})}{2!}(-3x)^2 + \frac{(-\frac{1}{3})(-\frac{4}{3})(-\frac{7}{3})}{3!}(-3x)^3 + \dots = 1 + x + \frac{2}{9}(9x^2) + \frac{14}{81}(27x^3) + \dots = 1 + x + 2x^2 + \frac{14}{3}x^3\). (b) The expansion is valid for \(|-3x| < 1 \implies |x| < \frac{1}{3}\). (c) Substituting \(x = 0.01\) gives \((1 - 3(0.01))^{-\frac{1}{3}} = (0.97)^{-\frac{1}{3}} = \sqrt[3]{\frac{100}{97}}\). Substituting \(x = 0.01\) into the expansion gives \(1 + (0.01) + 2(0.01)^2 + \frac{14}{3}(0.01)^3 = 1 + 0.01 + 0.0002 + \frac{14}{3}(0.000001) \approx 1.0102 + 0.0000046667 = 1.01020467\).

(a) M1: Applies the binomial expansion formula with fractional exponent. M1: Obtains correct second term \(x\). M1: Obtains correct third term coefficient \(2\). A1: Obtains correct fourth term coefficient \(\frac{14}{3}\). (b) B1: Correct range \(|x| < \frac{1}{3}\) or equivalent. (c) M1: Shows that substituting \(x = 0.01\) into the function gives \(\sqrt[3]{\frac{100}{97}}\). M1: Substitutes \(x = 0.01\) into the polynomial. A1: Correctly calculates value to 8 decimal places: 1.01020467.

(a) Since \(OP : PA = 2 : 1\), \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\). Since \(OQ : QB = 1 : 3\), \(\overrightarrow{OQ} = \frac{1}{4}\mathbf{b}\). Then \(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\), and \(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\). (b) Let \(\overrightarrow{OX} = \overrightarrow{OA} + \lambda \overrightarrow{AQ} = \mathbf{a} + \lambda(-\mathbf{a} + \frac{1}{4}\mathbf{b}) = (1-\lambda)\mathbf{a} + \frac{1}{4}\lambda \mathbf{b}\). Also, let \(\overrightarrow{OX} = \overrightarrow{OB} + \mu \overrightarrow{BP} = \mathbf{b} + \mu(\frac{2}{3}\mathbf{a} - \mathbf{b}) = \frac{2}{3}\mu \mathbf{a} + (1-\mu)\mathbf{b}\). Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): \(1-\lambda = \frac{2}{3}\mu\) and \(\frac{1}{4}\lambda = 1-\mu\). From the second equation, \(\mu = 1 - \frac{1}{4}\lambda\). Substituting into the first gives \(1-\lambda = \frac{2}{3}(1 - \frac{1}{4}\lambda) = \frac{2}{3} - \frac{1}{6}\lambda \implies \frac{1}{3} = \frac{5}{6}\lambda \implies \lambda = \frac{2}{5}\). Then \(\mu = \frac{9}{10}\). Substituting \(\lambda\) into \(\overrightarrow{OX}\) gives \(\overrightarrow{OX} = \frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\). (c) Since \(\overrightarrow{OX} = \overrightarrow{OB} + \mu \overrightarrow{BP}\) with \(\mu = \frac{9}{10}\), we have \(\overrightarrow{BX} = \frac{9}{10}\overrightarrow{BP}\), meaning \(PX = \frac{1}{10}BP\) and \(XB = \frac{9}{10}BP\). Thus, \(PX : XB = 1 : 9\).

(a) B1: Correct expression for \(\overrightarrow{AQ}\). B1: Correct expression for \(\overrightarrow{BP}\). (b) M1: Writes \(\overrightarrow{OX}\) as \(\overrightarrow{OA} + \lambda \overrightarrow{AQ}\). M1: Writes \(\overrightarrow{OX}\) as \(\overrightarrow{OB} + \mu \overrightarrow{BP}\). M1: Equates coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to form simultaneous equations. A1: Solves to find \(\lambda = 0.4\) or \(\mu = 0.9\). A1: Correct expression for \(\overrightarrow{OX} = 0.6\mathbf{a} + 0.1\mathbf{b}\). (c) M1: Recognises relationship between ratio and \(\mu\). A1: Correct ratio \(1 : 9\).

(a) Using the laws of logarithms, \(\log_3((x+4)(x-2)) = 3 \implies (x+4)(x-2) = 3^3 = 27\). Expanding and simplifying gives \(x^2 + 2x - 8 = 27 \implies x^2 + 2x - 35 = 0\). Factoring the quadratic gives \((x+7)(x-5) = 0 \implies x = -7\) or \(x = 5\). Since we must have \(x > 2\) for the original logarithms to be defined, we reject \(x = -7\). Hence, \(x = 5\). (b) From the first equation, \(2^x \times 2^{2y} = 2^5 \implies x + 2y = 5\). From the second equation, \(3x - y = 2^3 \implies 3x - y = 8\). Multiplying the second equation by 2 gives \(6x - 2y = 16\). Adding this to the first equation gives \(7x = 21 \implies x = 3\). Substituting \(x = 3\) into \(x+2y=5\) gives \(3 + 2y = 5 \implies y = 1\).

(a) M1: Uses product law for logarithms: \(\log_3((x+4)(x-2))\). M1: Converts log equation to quadratic: \(x^2+2x-8 = 27\). A1: Solves quadratic to find \(x = 5\) and \(x = -7\). A1: Rejects \(x = -7\) and gives \(x = 5\) as the only solution. (b) M1: Simplifies index equation to \(x + 2y = 5\). M1: Simplifies log equation to \(3x - y = 8\). M1: Solves simultaneous equations using substitution or elimination. A1: Correct value \(x = 3\). A1: Correct value \(y = 1\).

(a) The sum of the first \(n\) terms of an arithmetic series is given by \(S_n = \frac{n}{2}[2a + (n-1)d]\). For \(n = 10\): \(S_{10} = \frac{10}{2}(2a + 9d) = 5(2a + 9d) = 150\). Dividing by 5 gives: \(2a + 9d = 30\). (b) The 5th, 8th, and 14th terms of the arithmetic series are given by: \(u_5 = a + 4d\), \(u_8 = a + 7d\), and \(u_{14} = a + 13d\). Since these three terms form a geometric series, the common ratio is constant: \(\frac{a+7d}{a+4d} = \frac{a+13d}{a+7d}\). Multiplying out: \((a+7d)^2 = (a+4d)(a+13d)\) which expands to \(a^2 + 14ad + 49d^2 = a^2 + 17ad + 52d^2\). Simplifying this gives: \(3ad + 3d^2 = 0\), which factorizes to \(3d(a+d) = 0\). Since we are given that \(d \neq 0\), we must have: \(a+d = 0\), which leads to \(d = -a\). (c) Substitute \(d = -a\) into the equation from part (a): \(2(-d) + 9d = 30\) which simplifies to \(7d = 30\), so \(d = \frac{30}{7}\). Since \(a = -d\), we have \(a = -\frac{30}{7}\).

(a) M1: For attempting to use the sum formula of an arithmetic series with \(n=10\) and equating to 150. A1: For the simplified equation \(2a + 9d = 30\) or equivalent. (b) M1: For writing expressions for \(u_5\), \(u_8\), and \(u_{14}\) in terms of \(a\) and \(d\). M1: For setting up the geometric progression condition, e.g., \((a+7d)^2 = (a+4d)(a+13d)\). A1: For expanding and simplifying to \(3ad + 3d^2 = 0\) or equivalent. A1: For concluding \(d = -a\) with a valid reason that \(d \neq 0\). (c) M1: For substituting \(d = -a\) (or \(a = -d\)) into their equation from part (a). A1: For finding \(d = \frac{30}{7}\). A1: For finding \(a = -\frac{30}{7}\).

(a) When \(x = 3\), the \(y\)-coordinate of \(P\) is: \(y = 2(3) + \frac{9}{3} = 6 + 3 = 9\). So the coordinates of \(P\) are \((3, 9)\). We differentiate \(y = 2x + 9x^{-1}\) to find the gradient function: \(\frac{dy}{dx} = 2 - 9x^{-2} = 2 - \frac{9}{x^2}\). At \(x = 3\), the gradient of the tangent is: \(\frac{dy}{dx} = 2 - \frac{9}{3^2} = 2 - 1 = 1\). Therefore, the gradient of the normal is \(-\frac{1}{1} = -1\). The equation of the normal at \(P(3, 9)\) is: \(y - 9 = -1(x - 3)\) which simplifies to \(y - 9 = -x + 3\), leading to \(x + y - 12 = 0\). (b) To find the coordinates of \(Q\), we solve the equation of the normal and the curve simultaneously: \(y = 12 - x\) and \(y = 2x + \frac{9}{x}\). Equating the two expressions: \(12 - x = 2x + \frac{9}{x}\), which simplifies to \(3x - 12 + \frac{9}{x} = 0\). Multiplying through by \(x\) (since \(x > 0\)): \(3x^2 - 12x + 9 = 0\), which simplifies to \(x^2 - 4x + 3 = 0\). Factorizing the quadratic equation: \((x - 3)(x - 1) = 0\). Since \(x = 3\) corresponds to point \(P\), the point \(Q\) must have \(x = 1\). Substituting \(x = 1\) into the normal equation: \(y = 12 - 1 = 11\). Thus, the coordinates of \(Q\) are \((1, 11)\).

(a) B1: For finding the correct \(y\)-coordinate of \(P\) as 9. M1: For differentiating the curve equation to find \(\frac{dy}{dx} = 2 - \frac{9}{x^2}\). M1: For substituting \(x = 3\) into their derivative to find the tangent gradient, and calculating the normal gradient as the negative reciprocal. A1: For a correct normal gradient of \(-1\). A1: For the equation of the normal in the required form \(x + y - 12 = 0\) (or any integer multiple). (b) M1: For equating their normal equation and the curve equation to form an equation in \(x\). A1: For obtaining the simplified quadratic equation \(x^2 - 4x + 3 = 0\) or equivalent. M1: For solving the quadratic to find \(x = 1\) as the alternative solution. A1: For finding the correct coordinates of \(Q\) as \((1, 11)\).

(a) Using the compound angle formula: \(\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta\). We substitute the double angle formulae: \(\cos 2\theta = 2\cos^2\theta - 1\) and \(\sin 2\theta = 2\sin\theta\cos\theta\). This gives: \(\cos 3\theta = (2\cos^2\theta - 1)\cos\theta - (2\sin\theta\cos\theta)\sin\theta = 2\cos^3\theta - \cos\theta - 2\sin^2\theta \cos\theta\). Using the identity \(\sin^2\theta = 1 - \cos^2\theta\): \(\cos 3\theta = 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta = 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta = 4\cos^3\theta - 3\cos\theta\). (b) Substitute the identity from part (a) into the given equation: \(\cos 3\theta + 2\cos\theta = 0\) leads to \((4\cos^3\theta - 3\cos\theta) + 2\cos\theta = 0\), which simplifies to \(4\cos^3\theta - \cos\theta = 0\). Factorizing: \(\cos\theta(4\cos^2\theta - 1) = 0\). This gives two cases: Case 1: \(\cos\theta = 0\), which gives \(\theta = \frac{\pi}{2}\). Case 2: \(\cos^2\theta = \frac{1}{4}\), which gives \(\cos\theta = \pm\frac{1}{2}\). For \(\cos\theta = \frac{1}{2}\), \(\theta = \frac{\pi}{3}\). For \(\cos\theta = -\frac{1}{2}\), \(\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\). Therefore, the solutions in the interval \(0 \le \theta \le \pi\) are \(\theta = \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}\).

(a) M1: For applying the compound angle formula to \(\cos(2\theta + \theta)\). M1: For substituting the double angle formulae for both \(\cos 2\theta\) and \(\sin 2\theta\). M1: For substituting \(\sin^2\theta = 1 - \cos^2\theta\) to obtain an expression purely in terms of \(\cos\theta\). A1: For obtaining the final expression \(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\) with no errors in working. (b) M1: For substituting the identity from part (a) and simplifying to \(4\cos^3\theta - \cos\theta = 0\). M1: For factorizing to find \(\cos\theta = 0\) and \(\cos\theta = \pm\frac{1}{2}\). A1: For finding \(\theta = \frac{\pi}{2}\). B1: For finding \(\theta = \frac{\pi}{3}\). B1: For finding \(\theta = \frac{2\pi}{3}\).

(a) Using the quotient rule with \(u = 2x - 3\) and \(v = x + 2\), we have \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = 1\). Thus, \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(x + 2) - 1(2x - 3)}{(x + 2)^2} = \frac{2x + 4 - 2x + 3}{(x + 2)^2} = \frac{7}{(x + 2)^2}\). (b) The curve crosses the \(y\)-axis when \(x = 0\). Substituting \(x = 0\) into the equation of the curve gives \(y = \frac{2(0) - 3}{0 + 2} = -\frac{3}{2}\). The gradient of the tangent at \(x = 0\) is \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{7}{(0 + 2)^2} = \frac{7}{4}\). The gradient of the normal is the negative reciprocal of the tangent gradient, which is \(-\frac{4}{7}\). The equation of the normal is: \(y - \left(-\frac{3}{2}\right) = -\frac{4}{7}(x - 0) \implies y + \frac{3}{2} = -\frac{4}{7}x\). Multiplying by 14 to clear denominators: \(14y + 21 = -8x \implies 8x + 14y + 21 = 0\).

(a) M1 for applying the quotient rule correctly with at least one term correct in the numerator. A1 for correct simplified numerator of 7. A1 for correct denominator \((x+2)^2\). (b) M1 for setting \(x=0\) and finding \(y = -\frac{3}{2}\). M1 for finding the gradient of the tangent at \(x=0\), which is \(\frac{7}{4}\). M1 for using the negative reciprocal to find the gradient of the normal, \(-\frac{4}{7}\). M1 for substituting their point and normal gradient into the equation of a straight line. A1 for correct equation in the required form \(8x + 14y + 21 = 0\) (or any integer multiple).

(a) The volume \(V\) of the solid of revolution is given by \(V = \pi \int_{1}^{k} y^2 \mathrm{d}x\). Substituting \(y = \sqrt{3x + 1}\): \(V = \pi \int_{1}^{k} (3x + 1) \mathrm{d}x = \pi \left[ \frac{3}{2}x^2 + x \right]_{1}^{k}\). Evaluating at the limits: \(V = \pi \left( \left(\frac{3}{2}k^2 + k\right) - \left(\frac{3}{2}(1)^2 + 1\right) \right) = \pi \left( \frac{3}{2}k^2 + k - \frac{5}{2} \right)\). We are given \(V = 14\pi\), so: \(\pi \left( \frac{3}{2}k^2 + k - \frac{5}{2} \right) = 14\pi \implies \frac{3}{2}k^2 + k - \frac{5}{2} = 14\). Multiplying both sides by 2: \(3k^2 + 2k - 5 = 28 \implies 3k^2 + 2k - 33 = 0\) (as required). (b) Factoring the quadratic equation: \(3k^2 + 2k - 33 = 0 \implies (3k + 11)(k - 3) = 0\). This yields \(k = -\frac{11}{3}\) or \(k = 3\). Since we are given \(k > 1\), we reject the negative root, so \(k = 3\).

(a) M1 for writing the volume formula \(V = \pi \int y^2 \mathrm{d}x\). M1 for substituting \(y^2 = 3x+1\) and integrating to obtain \(\frac{3}{2}x^2 + x\). M1 for substituting the limits \(1\) and \(k\) correctly. A1 for equating their volume expression to \(14\pi\) and simplifying to remove \(\pi\). A1 for completing the algebra to show the given quadratic equation \(3k^2 + 2k - 33 = 0\). (b) M1 for attempting to solve the quadratic equation (by factoring or formula). A1 for finding both roots \(k = 3\) and \(k = -\frac{11}{3}\). A1 for selecting \(k = 3\) and explaining that \(k > 1\).

(a) Using the trigonometric identity \(\cos 2\theta = 1 - 2\sin^2\theta\): \(5(1 - 2\sin^2\theta) + 9\sin\theta = 7 \implies 5 - 10\sin^2\theta + 9\sin\theta = 7\). Rearranging terms to set the equation to zero: \(10\sin^2\theta - 9\sin\theta + 2 = 0\) (as required). (b) Let \(x = \sin\theta\). The quadratic equation is \(10x^2 - 9x + 2 = 0\). Factoring gives: \((2x - 1)(5x - 2) = 0 \implies x = \frac{1}{2}\) or \(x = \frac{2}{5}\). Case 1: \(\sin\theta = \frac{1}{2}\). In the interval \(0 \le \theta < 360^\circ\), the solutions are \(\theta = 30^\circ\) and \(\theta = 180^\circ - 30^\circ = 150^\circ\). Case 2: \(\sin\theta = 0.4\). In the interval \(0 \le \theta < 360^\circ\), the reference angle is \(\sin^{-1}(0.4) \approx 23.578^\circ\). The solutions are \(\theta \approx 23.6^\circ\) and \(\theta \approx 180^\circ - 23.578^\circ = 156.4^\circ\). Thus, the complete set of solutions is \(\theta = 23.6^\circ, 30^\circ, 150^\circ, 156.4^\circ\).

(a) M1 for using the identity \(\cos 2\theta = 1 - 2\sin^2\theta\). M1 for expanding the brackets. A1 for rearranging terms correctly to obtain the given quadratic in \(\sin\theta\). (b) M1 for factoring or using the quadratic formula on the quadratic in \(\sin\theta\) to find values of \(\sin\theta\). A1 for \(\sin\theta = \frac{1}{2}\) and \(\sin\theta = 0.4\) (or equivalent). B1 for \(\theta = 30^\circ, 150^\circ\). M1 for finding at least one angle for \(\sin\theta = 0.4\) to 1 decimal place. A1 for \(\theta = 23.6^\circ\) and \(156.4^\circ\).

From the equation \(2x^2 - 5x + 4 = 0\), we have \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = \frac{4}{2} = 2\). (a) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\). (b) \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{5}{8}\). (c) For the new roots \(\frac{1}{\alpha^2}\) and \(\frac{1}{\beta^2}\): Sum of roots \(S = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{9/4}{4} = \frac{9}{16}\). Product of roots \(P = \frac{1}{\alpha^2\beta^2} = \frac{1}{(\alpha\beta)^2} = \frac{1}{4}\). The quadratic equation is \(x^2 - Sx + P = 0 \implies x^2 - \frac{9}{16}x + \frac{1}{4} = 0\). Multiplying by 16 to get integer coefficients: \(16x^2 - 9x + 4 = 0\).

(a) B1 for identifying \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\). M1 for using the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\). A1 for \(\frac{9}{4}\). (b) M1 for using the identity \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\) or \((\alpha+\beta)(\alpha^2 - \alpha\beta + \beta^2)\). A1 for \(\frac{5}{8}\). (c) M1 for finding the sum of the new roots \(S = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}\). M1 for finding the product of the new roots \(P = \frac{1}{(\alpha\beta)^2}\). A1 for forming the equation \(16x^2 - 9x + 4 = 0\) (must have integer coefficients and be equal to 0).

(a) The vertical asymptote occurs when the denominator is zero: \(2x + 1 = 0 \implies x = -0.5\). As \(x \to \pm\infty\), \(y \to \frac{3}{2}\), so the horizontal asymptote is \(y = 1.5\). (b) When \(x = 0\), \(y = \frac{3(0) - 5}{2(0) + 1} = -5\), so the curve crosses the \(y\)-axis at \((0, -5)\). When \(y = 0\), \(3x - 5 = 0 \implies x = \frac{5}{3}\), so the curve crosses the \(x\)-axis at \(\left(\frac{5}{3}, 0\right)\). (c) To find the points of intersection, set \(y = \frac{3x - 5}{2x + 1}\) equal to \(y = x - 2\): \(\frac{3x - 5}{2x + 1} = x - 2 \implies 3x - 5 = (x - 2)(2x + 1) \implies 3x - 5 = 2x^2 - 3x - 2\). Rearranging into standard quadratic form: \(2x^2 - 6x + 3 = 0\). Using the quadratic formula: \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(3)}}{2(2)} = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}\).

(a) B1 for vertical asymptote \(x = -0.5\) (or equivalent). B1 for horizontal asymptote \(y = 1.5\) (or equivalent). (b) B1 for \((0, -5)\). B1 for \(\left(\frac{5}{3}, 0\right)\). (c) M1 for setting the two equations equal and clearing the fraction. M1 for expanding and simplifying to a standard quadratic equation. A1 for reaching \(2x^2 - 6x + 3 = 0\). M1 for attempting to solve using the quadratic formula. A1 for \(x = \frac{3 \pm \sqrt{3}}{2}\) (or equivalent exact simplified form).

(a) The formula for the sum of the first \(n\) terms of an arithmetic series is \(S_n = \frac{n}{2}(2a + (n-1)d)\). For \(n = 5\): \(S_5 = \frac{5}{2}(2a + 4d) = 5(a + 2d) = 65 \implies a + 2d = 13\). The formula for the \(n\)-th term is \(u_n = a + (n-1)d\). For \(n = 10\): \(u_{10} = a + 9d = 41\). We now have a system of two simultaneous linear equations: 1) \(a + 2d = 13\) and 2) \(a + 9d = 41\). Subtracting equation (1) from equation (2): \(7d = 28 \implies d = 4\). Substituting \(d = 4\) into equation (1): \(a + 2(4) = 13 \implies a + 8 = 13 \implies a = 5\). (b) The sum of the first 20 terms is \(S_{20} = \frac{20}{2}(2a + 19d) = 10(2(5) + 19(4)) = 10(10 + 76) = 10(86) = 860\).

(a) M1 for using the sum formula with \(n=5\) to write \(5(a+2d)=65\) or equivalent. M1 for using the term formula with \(n=10\) to write \(a+9d=41\). M1 for attempting to solve the simultaneous equations to eliminate one variable. A1 for \(d = 4\). A1 for \(a = 5\). (b) M1 for using the sum formula with \(n=20\), using their values of \(a\) and \(d\). A1 for correct substitution. A1 for calculating \(S_{20} = 860\).

(a) Since \(OP : PA = 2 : 1\), \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\). Since \(Q\) is the midpoint of \(OB\), \(\overrightarrow{OQ} = \frac{1}{2}\mathbf{b}\). Thus: \(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{1}{2}\mathbf{b}\). \(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\). (b) Expressing \(\overrightarrow{OX}\) in two different ways: Method 1: \(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + \lambda \overrightarrow{AQ} = \mathbf{a} + \lambda\left(-\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = (1 - \lambda)\mathbf{a} + \frac{1}{2}\lambda \mathbf{b}\). Method 2: \(\overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + \mu \overrightarrow{BP} = \mathbf{b} + \mu\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\mu \mathbf{a} + (1 - \mu)\mathbf{b}\). Since \(a\) and \(b\) are non-parallel vectors, we equate the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): 1) \(1 - \lambda = \frac{2}{3}\mu \implies 3\lambda + 2\mu = 3\). 2) \(\frac{1}{2}\lambda = 1 - \mu \implies \lambda + 2\mu = 2\). Subtracting (2) from (1): \(2\lambda = 1 \implies \lambda = \frac{1}{2}\). Substituting \(\lambda = \frac{1}{2}\) into (2): \(\frac{1}{2} + 2\mu = 2 \implies 2\mu = \frac{3}{2} \implies \mu = \frac{3}{4}\). (c) Since \(\overrightarrow{AX} = \frac{1}{2}\overrightarrow{AQ}\), the point \(X\) is the midpoint of the line segment \(AQ\). Therefore, the ratio \(AX : XQ = 1 : 1\).

(a) B1 for \(\overrightarrow{AQ} = -\mathbf{a} + \frac{1}{2}\mathbf{b}\). B1 for \(\overrightarrow{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\). (b) M1 for finding an expression for \(\overrightarrow{OX}\) using \(\lambda\) or \(\mu\). M1 for finding a second expression for \(\overrightarrow{OX}\) using the other parameter. M1 for equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to obtain two equations. A1 for \(\lambda = \frac{1}{2}\). A1 for \(\mu = \frac{3}{4}\). (c) B1 for stating the ratio \(1 : 1\) (or equivalent).

(a) Let \(u = 3^x\). Then \(3^{2x+1} = 3 \cdot (3^x)^2 = 3u^2\). The equation becomes: \(3u^2 - 10u + 3 = 0\). Factoring this quadratic equation: \((3u - 1)(u - 3) = 0 \implies u = \frac{1}{3} \text{or} u = 3\). If \(u = \frac{1}{3}\), then \(3^x = 3^{-1} \implies x = -1\). If \(u = 3\), then \(3^x = 3^1 \implies x = 1\). So the solutions are \(x = 1\) or \(x = -1\). (b) From the first equation: \(\log_2 y - \log_2 x = 3 \implies \log_2\left(\frac{y}{x}\right) = 3 \implies \frac{y}{x} = 2^3 = 8 \implies y = 8x\). From the second equation: \(2^y = 16^{x+1} \implies 2^y = (2^4)^{x+1} = 2^{4x+4} \implies y = 4x + 4\). Equating the two expressions for \(y\): \(8x = 4x + 4 \implies 4x = 4 \implies x = 1\). Substituting \(x = 1\) back to find \(y\): \(y = 8(1) = 8\). Since both \(x=1\) and \(y=8\) are positive, they are valid within the domains of the log functions.

(a) M1 for substituting \(u = 3^x\) and writing the quadratic \(3u^2 - 10u + 3 = 0\). M1 for attempting to solve the quadratic equation. A1 for finding \(u = \frac{1}{3}\) and \(u = 3\). M1 for equating \(3^x\) to their values of \(u\) and attempting to solve for \(x\). A1 for \(x = 1\) and \(x = -1\). (b) M1 for using log division rule to get \(\log_2(y/x) = 3\) and then \(y = 8x\). M1 for expressing \(16^{x+1}\) as a power of 2 and equating indices to get \(y = 4x + 4\). M1 for solving the simultaneous equations. A1 for \(x = 1, y = 8\).

(a) Using the compound angle formula for cosine:
\(\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta\)

Substitute the double angle formulas \(\cos 2\theta = 2\cos^2\theta - 1\) and \(\sin 2\theta = 2\sin\theta\cos\theta\):
\(\cos(2\theta + \theta) = (2\cos^2\theta - 1)\cos\theta - (2\sin\theta\cos\theta)\sin\theta\)
\(\cos 3\theta = 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta\)

Substitute \(\sin^2\theta = 1 - \cos^2\theta\):
\(\cos 3\theta = 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta\)
\(\cos 3\theta = 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta\)

\(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\) (as required).

(b) The given equation is \(8\cos^3\theta - 6\cos\theta = 1\).
Factor out 2 on the left-hand side:
\(2(4\cos^3\theta - 3\cos\theta) = 1\)

Using the identity from part (a):
\(2\cos 3\theta = 1 \implies \cos 3\theta = \frac{1}{2}\)

Given \(0^\circ \le \theta \le 180^\circ\), the range for \(3\theta\) is:
\(0^\circ \le 3\theta \le 540^\circ\)

Solving \(\cos 3\theta = \frac{1}{2}\) in this interval:
\(3\theta = 60^\circ, 300^\circ, 420^\circ\)

Dividing by 3:
\(\theta = 20^\circ, 100^\circ, 140^\circ\)

Part (a) [4 marks]:
- M1: For attempting to write \(\cos 3\theta\) as \(\cos(2\theta + \theta)\) and applying the compound angle formula.
- M1: For substituting correct double angle formulas for \(\cos 2\theta\) and \(\sin 2\theta\).
- M1: For using \(\sin^2\theta = 1 - \cos^2\theta\) to write the expression entirely in terms of \(\cos\theta\).
- A1: Correct algebraic simplification leading to the given identity with no errors.

Part (b) [5 marks]:
- M1: For identifying that the left hand side of the equation can be written as \(2\cos 3\theta\).
- A1: For obtaining \(\cos 3\theta = \frac{1}{2}\).
- M1: For finding at least two correct values for \(3\theta\) in the interval \([0^\circ, 540^\circ]\).
- A1: For at least two correct values of \(\theta\).
- A1: For all three correct solutions \(\theta = 20^\circ, 100^\circ, 140^\circ\) and no extras in the range.

(a) To find the stationary point, we differentiate \(y = \sqrt{x} e^{-x} = x^{1/2} e^{-x}\) using the product rule:
\(\frac{dy}{dx} = \frac{1}{2}x^{-1/2} e^{-x} - x^{1/2} e^{-x} = e^{-x}\left(\frac{1}{2\sqrt{x}} - \sqrt{x}\right)\)

At the stationary point, \(\frac{dy}{dx} = 0\):
\(e^{-x}\left(\frac{1 - 2x}{2\sqrt{x}}\right) = 0\)

Since \(e^{-x} \ne 0\) and \(x > 0\) for the derivative to be defined, we must have:
\(1 - 2x = 0 \implies x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) back into the curve equation to find the \(y\)-coordinate:
\(y = \sqrt{\frac{1}{2}} e^{-1/2} = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{e}} = \frac{1}{\sqrt{2e}}\)

So the stationary point coordinates are \(\left(\frac{1}{2}, \frac{1}{\sqrt{2e}}\right)\).

(b) The volume \(V\) of the solid generated by rotating the region about the \(x\)-axis is given by:
\(V = \pi \int_{0}^{\ln 2} y^2 \, dx\)

Substitute \(y^2 = x e^{-2x}\):
\(V = \pi \int_{0}^{\ln 2} x e^{-2x} \, dx\)

Using integration by parts on \(\int x e^{-2x} \, dx\):
Let \(u = x\) and \(dv = e^{-2x} \, dx\)
Then \(du = dx\) and \(v = -\frac{1}{2}e^{-2x}\)

\(\int x e^{-2x} \, dx = -\frac{1}{2}x e^{-2x} - \int -\frac{1}{2} e^{-2x} \, dx = -\frac{1}{2}x e^{-2x} - \frac{1}{4} e^{-2x}\)

Now evaluate the definite integral from \(0\) to \(\ln 2\):
At the upper limit \(x = \ln 2\):
\(\left[-\frac{1}{2}(\ln 2) e^{-2\ln 2} - \frac{1}{4} e^{-2\ln 2}\right] = -\frac{1}{2}(\ln 2)\left(\frac{1}{4}\right) - \frac{1}{4}\left(\frac{1}{4}\right) = -\frac{1}{8}\ln 2 - \frac{1}{16}\)

At the lower limit \(x = 0\):
\(\left[0 - \frac{1}{4} e^0\right] = -\frac{1}{4}\)

Subtracting the lower limit from the upper limit:
\(V = \pi \left[\left(-\frac{1}{8}\ln 2 - \frac{1}{16}\right) - \left(-\frac{1}{4}\right)\right]\)
\(V = \pi \left(\frac{3}{16} - \frac{1}{8}\ln 2\right)\)

Here \(a = \frac{3}{16}\) and \(b = \frac{1}{8}\).

Part (a) [4 marks]:
- M1: For attempting to differentiate \(y = \sqrt{x} e^{-x}\) using the product rule.
- A1: Correct expression for \(\frac{dy}{dx}\).
- M1: Setting \(\frac{dy}{dx} = 0\) and solving for \(x\).
- A1: Correct coordinates \(x = \frac{1}{2}\) and \(y = \frac{1}{\sqrt{2e}}\) (or equivalent exact form).

Part (b) [5 marks]:
- M1: Setting up the correct integral formula for volume: \(\pi \int_{0}^{\ln 2} x e^{-2x} \, dx\).
- M1: Applying integration by parts correctly to obtain \(-\frac{1}{2}x e^{-2x} - \frac{1}{4} e^{-2x}\).
- A1: Correct integrated expression with limits.
- M1: Substituting limits \(\ln 2\) and \(0\) correctly, showing step-by-step evaluation.
- A1: Simplifying to the exact form \(\pi \left(\frac{3}{16} - \frac{1}{8}\ln 2\right)\) (accept any equivalent form that clearly identifies \(a\) and \(b\) as rational numbers).

(a) For the equation \(2x^2 - 5x + 4 = 0\):
\(\alpha + \beta = \frac{5}{2}\)
\(\alpha\beta = \frac{4}{2} = 2\)

First, calculate \(\alpha^2 + \beta^2\):
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\) (as required).

Next, calculate \(\alpha^3 + \beta^3\):
\(\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = \left(\frac{5}{2}\right)\left(\frac{9}{4} - 2\right) = \left(\frac{5}{2}\right)\left(\frac{1}{4}\right) = \frac{5}{8}\).

(b) Let the roots of the new quadratic equation be \(\gamma = \frac{\alpha}{\beta^2}\) and \(\delta = \frac{\beta}{\alpha^2}\).

The sum of the roots \(S\) is:
\(S = \gamma + \delta = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2}\)

Since \(\alpha^2\beta^2 = (\alpha\beta)^2 = 2^2 = 4\):
\(S = \frac{5/8}{4} = \frac{5}{32}\)

The product of the roots \(P\) is:
\(P = \gamma\delta = \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} = \frac{\alpha\beta}{\alpha^2\beta^2} = \frac{1}{\alpha\beta} = \frac{1}{2}\)

The quadratic equation with roots \(\gamma\) and \(\delta\) is given by:
\(x^2 - Sx + P = 0\)
\(x^2 - \frac{5}{32}x + \frac{1}{2} = 0\)

Multiply the entire equation by 32 to get integer coefficients:
\(32x^2 - 5x + 16 = 0\)

Part (a) [4 marks]:
- B1: State \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = 2\).
- M1: For utilizing \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\) to show the value \(\frac{9}{4}\).
- M1: For utilizing a correct expansion for \(\alpha^3 + \beta^3\) (e.g., \((\alpha+\beta)(\alpha^2 - \alpha\beta + \beta^2)\) or \((\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)\)).
- A1: For obtaining the correct value of \(\frac{5}{8}\).

Part (b) [5 marks]:
- M1: Write the sum of roots as \(S = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2}\).
- A1: Correct evaluation of the sum: \(S = \frac{5}{32}\).
- M1: Write the product of roots as \(P = \frac{1}{\alpha\beta}\) and evaluate to get \(P = \frac{1}{2}\).
- M1: Use of \(x^2 - Sx + P = 0\) with their calculated sum and product.
- A1: Clear final equation with integer coefficients: \(32x^2 - 5x + 16 = 0\) (or any integer multiple thereof, e.g., \(-32x^2 + 5x - 16 = 0\)).