MetadataPastPaper.sampleTitle

First, find the coordinates of the point where the curve \( C \) crosses the \( y \)-axis by setting \( x = 0 \). This gives \( y = \frac{3(0) - 1}{0^2 + 2} = -\frac{1}{2} \). So the point of contact is \( (0, -0.5) \). Next, find the gradient of the curve by differentiating \( y \) with respect to \( x \) using the quotient rule: \( \frac{dy}{dx} = \frac{3(x^2 + 2) - (3x - 1)(2x)}{(x^2 + 2)^2} \). Evaluating this derivative at \( x = 0 \) gives \( \frac{dy}{dx} = \frac{3(2) - 0}{2^2} = \frac{6}{4} = \frac{3}{2} \). The gradient of the normal is the negative reciprocal of the gradient of the tangent, which is \( m_n = -\frac{2}{3} \). Using the equation of a straight line, the equation of the normal is \( y - y_1 = m_n(x - x_1) \), which gives \( y - \left(-\frac{1}{2}\right) = -\frac{2}{3}(x - 0) \). Simplifying this equation gives \( y + \frac{1}{2} = -\frac{2}{3}x \), which can be rewritten as \( 6y + 3 = -4x \), or \( 4x + 6y + 3 = 0 \).

M1: For substituting \( x = 0 \) into the equation of the curve to find the \( y \)-coordinate. A1: Correctly obtaining \( y = -0.5 \) (or equivalent). M1: For applying the quotient rule correctly to differentiate \( y = \frac{3x - 1}{x^2 + 2} \). A1: Correctly finding the gradient of the tangent at \( x = 0 \) is \( \frac{3}{2} \). M1: For finding the gradient of the normal as the negative reciprocal of their tangent gradient. A0.5: Formulating the equation of the normal and simplifying to the required form \( 4x + 6y + 3 = 0 \) (or any non-zero integer multiple thereof).

Use the double-angle identity \( \cos 2\theta = 1 - 2\sin^2\theta \) to rewrite the equation in terms of \( \sin\theta \) only: \( 3(1 - 2\sin^2\theta) + 5\sin\theta - 4 = 0 \). Expanding and simplifying gives \( 3 - 6\sin^2\theta + 5\sin\theta - 4 = 0 \), which simplifies to \( 6\sin^2\theta - 5\sin\theta + 1 = 0 \). This quadratic equation can be factorised as \( (2\sin\theta - 1)(3\sin\theta - 1) = 0 \). This gives two possible values: \( \sin\theta = \frac{1}{2} \) or \( \sin\theta = \frac{1}{3} \). For \( \sin\theta = \frac{1}{2} \) within the range \( 0 \le \theta < 360^\circ \), the solutions are \( \theta = 30^\circ \) and \( \theta = 180^\circ - 30^\circ = 150^\circ \). For \( \sin\theta = \frac{1}{3} \), the principal value is \( \theta = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.5^\circ \), and the second value is \( \theta = 180^\circ - 19.5^\circ = 160.5^\circ \). Thus, the full set of solutions is \( \theta = 19.5^\circ, 30^\circ, 150^\circ, 160.5^\circ \).

M1: Substituting \( \cos 2\theta = 1 - 2\sin^2\theta \) into the equation. A0.5: Formulating a correct quadratic equation in terms of \( \sin\theta \): \( 6\sin^2\theta - 5\sin\theta + 1 = 0 \). M1: Solving the quadratic equation to obtain two values for \( \sin\theta \). A1.5: Obtaining \( \sin\theta = \frac{1}{2} \) and \( \sin\theta = \frac{1}{3} \). A0.75: Identifying the angles \( 30^\circ \) and \( 150^\circ \) from \( \sin\theta = \frac{1}{2} \). A0.75: Identifying the angles \( 19.5^\circ \) and \( 160.5^\circ \) from \( \sin\theta = \frac{1}{3} \), rounded to 1 decimal place.

The sum of the first \( n \) terms is \( S_n = 2n^2 + 5n \). We can find the first term \( a \) by evaluating \( S_1 \): \( a = S_1 = 2(1)^2 + 5(1) = 7 \). The sum of the first two terms is \( S_2 = 2(2)^2 + 5(2) = 8 + 10 = 18 \). Since \( S_2 = u_1 + u_2 = a + u_2 \), the second term is \( u_2 = S_2 - S_1 = 18 - 7 = 11 \). The common difference \( d \) of the arithmetic series is \( d = u_2 - a = 11 - 7 = 4 \). The general expression for the \( k \)-th term of an arithmetic series is \( u_k = a + (k - 1)d \). Substituting our values, we get \( u_k = 7 + (k - 1)4 = 4k + 3 \). Since the \( k \)-th term is 103, we can set up the equation \( 4k + 3 = 103 \). Solving for \( k \) gives \( 4k = 100 \), which simplifies to \( k = 25 \).

M1: Attempting to find \( S_1 \) and \( S_2 \) or using an alternative valid method to find terms of the series. A0.5: Correctly finding \( a = 7 \). M1: Finding the second term \( u_2 = 11 \) or establishing the common difference \( d = 4 \). A1: Correctly writing the general term of the series as \( u_k = 4k + 3 \). M1: Setting up the linear equation \( 4k + 3 = 103 \) to solve for \( k \). A1: Obtaining \( k = 25 \).

From the first equation, \( \log_2(x - 2y) = 3 \), we can rewrite this in exponential form: \( x - 2y = 2^3 \), which simplifies to \( x - 2y = 8 \) (Equation 1). From the second equation, \( 3^x \cdot 9^y = 27^5 \), we can express all terms with base 3: \( 3^x \cdot (3^2)^y = (3^3)^5 \), which simplifies to \( 3^{x + 2y} = 3^{15} \). Equating the indices gives \( x + 2y = 15 \) (Equation 2). We now have a system of two linear equations: 1) \( x - 2y = 8 \) and 2) \( x + 2y = 15 \). Adding the two equations together: \( (x - 2y) + (x + 2y) = 8 + 15 \implies 2x = 23 \implies x = 11.5 \). Substituting \( x = 11.5 \) back into Equation 2: \( 11.5 + 2y = 15 \implies 2y = 3.5 \implies y = 1.75 \). Checking both values in \( x - 2y > 0 \) confirms that they are valid solutions.

M1: Converting the logarithmic equation into the linear equation \( x - 2y = 8 \). A0.5: Correctly simplified linear equation from log. M1: Expressing the second equation using a common base of 3 and equating powers. A1: Correctly obtaining the linear equation \( x + 2y = 15 \). M1: Attempting to solve the system of two simultaneous linear equations. A1: Correctly finding both \( x = 11.5 \) and \( y = 1.75 \) (or equivalent fractional values \( x = \frac{23}{2} \) and \( y = \frac{7}{4} \)).

(a) The total surface area of a closed cylinder is given by \(A = 2\pi r^2 + 2\pi r h = 150\pi\). This simplifies to \(2\pi r h = 150\pi - 2\pi r^2\), which gives \(h = \frac{150\pi - 2\pi r^2}{2\pi r} = \frac{75 - r^2}{r}\). The volume of the cylinder is \(V = \pi r^2 h = \pi r^2 \left(\frac{75 - r^2}{r}\right) = \pi r (75 - r^2) = 75\pi r - \pi r^3\). (b) Differentiating \(V\) with respect to \(r\) gives \(\frac{dV}{dr} = 75\pi - 3\pi r^2\). Setting \(\frac{dV}{dr} = 0\) for stationary points gives \(75\pi - 3\pi r^2 = 0 \implies 3r^2 = 75 \implies r^2 = 25\). Since \(r > 0\), we have \(r = 5\). Substituting \(r = 5\) back into the volume formula gives \(V = 75\pi(5) - \pi(5)^3 = 375\pi - 125\pi = 250\pi\) cm\(^3\). (c) The second derivative is \(\frac{d^2V}{dr^2} = -6\pi r\). At \(r = 5\), \(\frac{d^2V}{dr^2} = -30\pi < 0\). Since the second derivative is negative, the volume is a maximum.

(a) M1: Attempts to write an equation for the total surface area and sets equal to \(150\pi\). M1: Rearranges to express \(h\) in terms of \(r\). M1: Substitutes expression for \(h\) into the volume formula \(V = \pi r^2 h\). A1: Correctly simplifies to show \(V = 75\pi r - \pi r^3\) (no errors seen). (b) M1: Correctly differentiates \(V\) to find \(\frac{dV}{dr}\). M1: Sets derivative equal to zero. A1: Solves to find \(r = 5\) (accepts \(r = \pm 5\) but must reject negative for final A1). M1: Substitutes \(r = 5\) into the volume equation. A1: Correct volume of \(250\pi\). (c) M1: Differentiates again to find the second derivative \(\frac{d^2V}{dr^2}\). A1: Evaluates at \(r = 5\) and states that since it is less than 0, it is a maximum.

(a) Using the identity \ \cos 2\theta = 1 - 2\sin^2\theta\, we substitute this into the equation: \(3(1 - 2\sin^2\theta) + 8\sin\theta + 5 = 0\). Expanding and simplifying gives: \(3 - 6\sin^2\theta + 8\sin\theta + 5 = 0 \implies -6\sin^2\theta + 8\sin\theta + 8 = 0\). Multiplying the entire equation by \(-1\) gives the required form: \(6\sin^2\theta - 8\sin\theta - 8 = 0\). (b) We let \(y = \sin\theta\), so the equation is \(6y^2 - 8y - 8 = 0\). Dividing by 2: \(3y^2 - 4y - 4 = 0\). Factoring gives \((3y + 2)(y - 2) = 0\). Thus, \\sin\theta = -\frac{2}{3}\ or \(\sin\theta = 2\) (which has no real solutions). For \(\sin\theta = -\frac{2}{3}\), the basic acute angle is \(\sin^{-1}(2/3) \approx 41.81^\circ\). Since sine is negative, \(\theta\) lies in the third and fourth quadrants: \(\theta = 180^\circ + 41.81^\circ = 221.81^\circ \approx 222^\circ\) and \(\theta = 360^\circ - 41.81^\circ = 318.19^\circ \approx 318^\circ\). (c) Comparing the equations, let \(\theta = 2x - 10^\circ\). The range \(0^\circ \le x \le 180^\circ\) becomes \(-10^\circ \le 2x - 10^\circ \le 350^\circ\). In this range, our solutions for \(\theta\) are \(221.81^\circ\) and \(318.19^\circ\). For \(2x - 10^\circ = 221.81^\circ \implies 2x = 231.81^\circ \implies x \approx 115.9^\circ\). For \(2x - 10^\circ = 318.19^\circ \implies 2x = 328.19^\circ \implies x \approx 164.1^\circ\). To the nearest degree, \(x = 116^\circ\) or \(x = 164^\circ\).

(a) M1: Uses the double angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\). M1: Substitutes and simplifies terms. A1: Correctly reaches the given equation. (b) M1: Solves the quadratic in \(\sin\theta\) (by factoring, formula or calculator). A1: Identifies \(\sin\theta = -2/3\) and states no solution for \(\sin\theta = 2\). M1: Finds one correct value for \(\theta\) in the range. A1: Finds \(222^\circ\) (accept 221.8). A1: Finds \(318^\circ\) (accept 318.2). (c) M1: Connects the two equations by setting \(\theta = 2x - 10^\circ\). A1: Correctly solves for one value of \(x\) (116 or 164). A1: Finds both correct values of \(x\) (116 and 164) to the nearest degree.

(a) Since the terms are in geometric progression, the common ratio is constant: \(\frac{k}{k+4} = \frac{2k-15}{k}\). Cross-multiplying gives \(k^2 = (k+4)(2k-15) \implies k^2 = 2k^2 - 15k + 8k - 60 \implies k^2 = 2k^2 - 7k - 60\). Rearranging gives the required quadratic: \(k^2 - 7k - 60 = 0\). (b) Factoring the quadratic: \((k-12)(k+5) = 0\). Since \(k > 0\), we must have \(k = 12\). (c) For \(k = 12\), the first three terms of \(G\) are: \(a_G = 16\), \(12\), and \(9\). The common ratio is \(r = \frac{12}{16} = 0.75\). The sum to infinity is \(S_\infty = \frac{a_G}{1-r} = \frac{16}{1 - 0.75} = \frac{16}{0.25} = 64\). (d) The 3rd term of \(G\) is 9, so the 3rd term of \(A\) is \(a + 2d = 9\). The 10th term of \(A\) is \(a + 9d = 37\). Subtracting these two equations: \(7d = 28 \implies d = 4\). Substituting back gives \(a + 2(4) = 9 \implies a = 1\). The sum of the first 20 terms is \(S_{20} = \frac{20}{2}[2a + 19d] = 10[2(1) + 19(4)] = 10[2 + 76] = 780\).

(a) M1: Uses the common ratio property of a GP to set up a fractional equation. M1: Expands parentheses and simplifies. A1: Obtains the correct quadratic equation without any errors. (b) M1: Solves the quadratic equation. A1: Chooses \(k = 12\) and rejects \(k = -5\). (c) M1: Evaluates the first term and common ratio of \(G\). M1: Uses the sum to infinity formula with their values of \(a\) and \(r\). A1: Correctly obtains 64. (d) M1: Sets up two simultaneous equations for \(a\) and \(d\). A1: Solves to find \(a = 1\) and \(d = 4\). A1: Applies sum formula of AP to obtain 780.

(a) The vector \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (7\mathbf{i} - 7\mathbf{j}) - (-\mathbf{i} - \mathbf{j}) = 8\mathbf{i} - 6\mathbf{j}\). The magnitude is \(|\overrightarrow{AB}| = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = 10\). (b) Since \(C\) is the midpoint of \(AB\), \(\overrightarrow{OC} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{1}{2}((-\mathbf{i} - \mathbf{j}) + (7\mathbf{i} - 7\mathbf{j})) = \frac{1}{2}(6\mathbf{i} - 8\mathbf{j}) = 3\mathbf{i} - 4\mathbf{j}\). (c) Given \(\overrightarrow{OD} = \lambda\mathbf{i} + 6\mathbf{j}\) is perpendicular to \(\overrightarrow{OC}\), their scalar product must be zero: \(\overrightarrow{OD} \cdot \overrightarrow{OC} = 0 \implies (\lambda\mathbf{i} + 6\mathbf{j}) \cdot (3\mathbf{i} - 4\mathbf{j}) = 0 \implies 3\lambda - 24 = 0 \implies 3\lambda = 24 \implies \lambda = 8\). (d) The magnitude of \(\overrightarrow{OC}\) is \(|\overrightarrow{OC}| = \sqrt{3^2 + (-4)^2} = 5\). The unit vector is \(\frac{\overrightarrow{OC}}{|\overrightarrow{OC}|} = \frac{3\mathbf{i} - 4\mathbf{j}}{5} = 0.6\mathbf{i} - 0.8\mathbf{j}\).

(a) M1: Calculates \(\mathbf{b} - \mathbf{a}\) to find \(\overrightarrow{AB}\). A1: Correct vector \(8\mathbf{i} - 6\mathbf{j}\). A1: Correct magnitude of 10. (b) M1: Employs midpoint formula \(\frac{\mathbf{a}+\mathbf{b}}{2}\). A1: Correct \(\mathbf{i}\) component (3). A1: Correct \(\mathbf{j}\) component (-4). (c) M1: Set up the dot product equation \(\overrightarrow{OD} \cdot \overrightarrow{OC} = 0\). A1: Correctly expands to obtain \(3\lambda - 24 = 0\). A1: Solves to find \(\lambda = 8\). (d) M1: Attempts to divide \(\overrightarrow{OC}\) by its magnitude (5). A1: Correct unit vector \(0.6\mathbf{i} - 0.8\mathbf{j}\) (or in fraction form).

(a) Expand the equation of the curve: \(y = (x-2)(x^2 - 8x + 16) = x^3 - 8x^2 + 16x - 2x^2 + 16x - 32 = x^3 - 10x^2 + 32x - 32\). Differentiating with respect to \(x\): \(\frac{dy}{dx} = 3x^2 - 20x + 32\). Set the derivative to zero for stationary points: \(3x^2 - 20x + 32 = 0 \implies (3x - 8)(x - 4) = 0\). So \(x = 4\) or \(x = \frac{8}{3}\). When \(x = 4\), \(y = (4-2)(4-4)^2 = 0\). Stationary point is \((4, 0)\). When \(x = \frac{8}{3}\), \(y = (\frac{8}{3} - 2)(\frac{8}{3} - 4)^2 = (\frac{2}{3})(-\frac{4}{3})^2 = \frac{32}{27}\). Stationary point is \((\frac{8}{3}, \frac{32}{27})\). (b) When \(x = 0\), \(y = (0-2)(0-4)^2 = -32\). Thus the y-intercept is \((0, -32)\). When \(y = 0\), \(x = 2\) or \(x = 4\). The curve touches the x-axis at \((4,0)\) and crosses it at \((2,0)\). The sketch should show a cubic curve starting in quadrant 3, crossing the y-axis at \((0,-32)\), crossing the x-axis at \((2,0)\), reaching a maximum at \((\frac{8}{3}, \frac{32}{27})\), turning at a minimum at \((4,0)\), and increasing for \(x > 4\). (c) The region is bounded between \(x = 2\) and \(x = 4\). The area is: \(\int_{2}^{4} (x^3 - 10x^2 + 32x - 32) \, dx = \left[ \frac{1}{4}x^4 - \frac{10}{3}x^3 + 16x^2 - 32x \right]_{2}^{4}\). At \(x = 4\): \(\frac{256}{4} - \frac{640}{3} + 256 - 128 = 192 - \frac{640}{3} = -\frac{64}{3}\). At \(x = 2\): \(\frac{16}{4} - \frac{80}{3} + 64 - 64 = 4 - \frac{80}{3} = -\frac{68}{3}\). Area = \(-\frac{64}{3} - (-\frac{68}{3}) = \frac{4}{3}\).

(a) M1: Expands the expression of \(y\) correctly. M1: Differentiates to find \(\frac{dy}{dx}\). M1: Sets derivative to 0 and solves quadratic. A1: Obtains \((4, 0)\). A1: Obtains \((8/3, 32/27)\). (b) M1: Finds intercept on y-axis \((0, -32)\) and intercepts on x-axis \((2,0)\), \((4,0)\). A1: Correct shape of positive cubic curve. A1: Completely labelled sketch with all key coordinates shown. (c) M1: Integrates the expanded polynomial expression. A1: Substitutes limits 2 and 4 correctly. A1: Correctly evaluates the area to obtain \(4/3\).

(a) The vertical asymptote occurs where the denominator is zero: \(2x + 4 = 0 \implies x = -2\). The horizontal asymptote is found by taking the limit as \(x \to \pm\infty\): \(y = \frac{3}{2}\). (b) Intersection with y-axis (set \(x=0\)): \(y = \frac{-5}{4} = -1.25\). Point is \((0, -1.25)\). Intersection with x-axis (set \(y=0\)): \(3x - 5 = 0 \implies x = \frac{5}{3}\). Point is \((\frac{5}{3}, 0)\). (c) The sketch should show a rectangular hyperbola with branches in the top-right and bottom-left regions formed by the asymptotes \(x = -2\) and \(y = 1.5\). It must clearly show the intersections with the axes at \((0, -1.25)\) and \((\frac{5}{3}, 0)\). (d) To find the intersections, set the equations equal: \\frac{3x - 5}{2x + 4} = x - 3 \implies 3x - 5 = (x - 3)(2x + 4) = 2x^2 - 2x - 12\). Rearranging gives: \(2x^2 - 5x - 7 = 0 \implies (2x - 7)(x + 1) = 0\). The x-coordinates are \(x = -1\) and \(x = 3.5\). Substituting into \(y = x - 3\): for \(x = -1\), \(y = -4\); for \(x = 3.5\), \(y = 0.5\). The points are \((-1, -4)\) and \((3.5, 0.5)\).

(a) A1: Correct vertical asymptote \(x = -2\). A1: Correct horizontal asymptote \(y = 1.5\) (or \(y = 3/2\)). (b) A1: Correct y-intercept \((0, -1.25)\). A1: Correct x-intercept \((5/3, 0)\). (c) M1: Draws a hyperbola with two branches. A1: Correct asymptotes drawn and labelled. A1: Intersects the axes in correct quadrants. A1: All four key features (2 asymptotes, 2 intercepts) correctly labelled with coordinates. (d) M1: Equates the curve and line, clears denominator to get a quadratic. A1: Solves to get \(x = -1\) and \(x = 3.5\). A1: Both points correctly found as \((-1, -4)\) and \((3.5, 0.5)\).

From the given equation, the sum and product of the roots are: \(\alpha + \beta = -\frac{5}{2}\) and \(\alpha\beta = -2\). (a) Using the identity: \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{5}{2}\right)^2 - 2(-2) = \frac{25}{4} + 4 = \frac{41}{4}\). (b) Using the identity: \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = \left(-\frac{5}{2}\right)^3 - 3(-2)\left(-\frac{5}{2}\right) = -\frac{125}{8} - 15 = -\frac{245}{8}\). (c) The sum of the new roots is: \(S = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2} = \frac{-245/8}{(-2)^2} = \frac{-245/8}{4} = -\frac{245}{32}\). The product of the new roots is: \(P = \frac{\alpha}{\beta^2} \cdot \frac{\beta}{\alpha^2} = \frac{\alpha\beta}{\alpha^2\beta^2} = \frac{1}{\alpha\beta} = \frac{1}{-2} = -\frac{1}{2}\). The new quadratic equation is: \(x^2 - Sx + P = 0 \implies x^2 - \left(-\frac{245}{32}\right)x + \left(-\frac{1}{2}\right) = 0 \implies x^2 + \frac{245}{32}x - \frac{1}{2} = 0\). Multiplying by 32 to get integer coefficients yields: \(32x^2 + 245x - 16 = 0\).

(a) M1: Identifies \(\alpha+\beta = -2.5\) and \(\alpha\beta = -2\). M1: Employs identity \(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\). A1: Evaluates to obtain \(41/4\) (or \(10.25\)). (b) M1: Employs algebraic identity for sum of cubes. M1: Substitutes their values of sum and product. A1: Correctly evaluates to obtain \(-245/8\) (or \(-30.625\)). (c) M1: Expresses the sum of new roots in terms of \(\alpha\) and \(\beta\). A1: Evaluates the new sum to be \(-245/32\). M1: Expresses and evaluates the product of new roots as \(-1/2\). M1: Constructs quadratic equation using their sum and product. A1: Correct equation with integer coefficients (e.g. \(32x^2 + 245x - 16 = 0\)).

We are given \(\frac{dV}{dt} = 12\).

The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\).

Differentiating with respect to \(r\):
\(\frac{dV}{dr} = 4\pi r^2\).

Using the chain rule, we can find \(\frac{dr}{dt}\):
\(\frac{dr}{dt} = \frac{dV}{dt} \div \frac{dV}{dr} = \frac{12}{4\pi r^2} = \frac{3}{\pi r^2}\).

The surface area of a sphere is given by \(A = 4\pi r^2\).

Differentiating with respect to \(r\):
\(\frac{dA}{dr} = 8\pi r\).

Using the chain rule to find \(\frac{dA}{dt}\):
\(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = 8\pi r \times \frac{3}{\pi r^2} = \frac{24}{r}\).

At the instant when \(r = 4\):
\(\frac{dA}{dt} = \frac{24}{4} = 6\text{ cm}^2/\text{s}\).

M1: Attempts to differentiate \(V = \frac{4}{3}\pi r^3\) to find \(\frac{dV}{dr}\).
M1: Uses the chain rule to express \(\frac{dr}{dt}\) in terms of \(r\).
M1: Differentiates \(A = 4\pi r^2\) to find \(\frac{dA}{dr}\).
M1: Formulates a correct chain rule expression for \(\frac{dA}{dt}\), e.g. \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\).
A1.2: Correctly calculates \(\frac{dA}{dt} = 6\).

Using the identity \(\cos 2\theta = 1 - 2\sin^2\theta\), we can rewrite the equation:
\(2(1 - 2\sin^2\theta) - 5\sin\theta - 3 = 0\)
\(2 - 4\sin^2\theta - 5\sin\theta - 3 = 0\)
\(-4\sin^2\theta - 5\sin\theta - 1 = 0\)

Multiply by \(-1\):
\(4\sin^2\theta + 5\sin\theta + 1 = 0\)

Factorize the quadratic equation:
\((4\sin\theta + 1)(\sin\theta + 1) = 0\)

This gives two cases:

Case 1: \(\sin\theta = -1\)
In the interval \(0^\circ \le \theta \le 360^\circ\), this gives \(\theta = 270^\circ\).

Case 2: \(\sin\theta = -0.25\)
The basic angle is \(\alpha = \arcsin(0.25) \approx 14.48^\circ\).
Since \(\sin\theta\) is negative, \(\theta\) lies in the 3rd and 4th quadrants:
\(\theta = 180^\circ + 14.48^\circ = 194.48^\circ \approx 194.5^\circ\) (to 1 d.p.)
\(\theta = 360^\circ - 14.48^\circ = 345.52^\circ \approx 345.5^\circ\) (to 1 d.p.).

Therefore, the solutions are \(\theta = 194.5^\circ, 270^\circ, 345.5^\circ\).

M1: Uses the double angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\) to form a quadratic in \(\sin\theta\).
M1: Correctly factorizes or solves the quadratic equation \(4\sin^2\theta + 5\sin\theta + 1 = 0\) to obtain values for \(\sin\theta\).
A1: Identifies \(\theta = 270^\circ\).
M1: Uses \(\sin\theta = -0.25\) to find at least one correct quadrant angle.
A1.2: Both \(194.5^\circ\) and \(345.5^\circ\) correct (to 1 d.p.).

Let the first term of the geometric series be \(a\) and the common ratio be \(r\).
The sum to infinity is given by \(S_\infty = \frac{a}{1-r}\).
The second term is \(ar\).

According to the given condition:
\(\frac{a}{1-r} = 4.5 \times ar\)

Assuming \(a \ne 0\), we can divide both sides by \(a\):
\(\frac{1}{1-r} = \frac{9}{2}r\)

Multiply both sides by \(2(1-r)\):
\(2 = 9r(1-r)\)
\(2 = 9r - 9r^2\)

Rearranging into standard quadratic form:
\(9r^2 - 9r + 2 = 0\)

Factorizing the quadratic:
\((3r - 1)(3r - 2) = 0\)

This gives:
\(r = \frac{1}{3} or \)r = \frac{2}{3}\).

Both values are positive and satisfy \(|r| < 1\), so both are valid solutions.

M1: Sets up the equation \(\frac{a}{1-r} = 4.5ar\) using correct formulas for sum to infinity and the second term.
M1: Simplifies the equation by dividing by \(a\) to obtain an equation in terms of \(r\) only.
M1: Rearranges the equation to form a quadratic equation in standard form \(9r^2 - 9r + 2 = 0\).
M1: Solves the quadratic equation by factorization or formula.
A1.2: Obtains both \(r = \frac{1}{3} and \)r = \frac{2}{3}\).

For the quadratic equation to have real and distinct roots, its discriminant must be strictly greater than zero:
\(\Delta = b^2 - 4ac > 0\)

Here, \(a = 1\), \(b = -4p\), and \(c = 3p^2 + 4\).

Substitute these into the discriminant:
\((-4p)^2 - 4(1)(3p^2 + 4) > 0\)
\(16p^2 - 12p^2 - 16 > 0\)
\(4p^2 - 16 > 0\)

Divide by 4:
\(p^2 - 4 > 0\)
\(p^2 > 4\)

This inequality is satisfied when:
\(p > 2\) or \(p < -2\).

M1: Uses the condition for real and distinct roots, \(\Delta > 0\).
M1: Substitutes the coefficients into \(b^2 - 4ac\) to get an inequality in terms of \(p\).
A1: Simplifies the inequality to \(4p^2 - 16 > 0\) or equivalent.
M1: Solves the quadratic inequality \(p^2 - 4 > 0\) to find critical values \(2\) and \(-2\).
A1.2: Correctly expresses the final range as \(p < -2\) or \(p > 2\).

From the first equation:
\(\log_2 x - 2\log_4 y = 1\)

Using the change of base formula, \(\log_4 y = \frac{\log_2 y}{\log_2 4} = \frac{1}{2}\log_2 y\).
Substitute this back:
\(\log_2 x - 2\left(\frac{1}{2}\log_2 y\right) = 1\)
\(\log_2 x - \log_2 y = 1\)
\(\log_2\left(\frac{x}{y}\right) = 1\)
\(\frac{x}{y} = 2^1 \Rightarrow x = 2y\) (Equation 1)

From the second equation:
\(3^x \cdot 9^y = 3^9\)

Since \(9^y = (3^2)^y = 3^{2y}\):
\(3^x \cdot 3^{2y} = 3^9\)
\(3^{x + 2y} = 3^9\)

Equating exponents:
\(x + 2y = 9\) (Equation 2)

Substitute Equation 1 into Equation 2:
\(2y + 2y = 9\)
\(4y = 9 \Rightarrow y = 2.25\)

Using \(x = 2y\):
\(x = 2(2.25) = 4.5\).

Thus, the solution is \(x = 4.5, y = 2.25\).

M1: Applies the change of base formula to express \(\log_4 y\) in base 2.
M1: Simplifies the logarithmic equation to obtain \(x = 2y\).
M1: Uses index laws to rewrite the second equation as \(x + 2y = 9\).
M1: Solves the linear simultaneous equations.
A1.2: Obtains both \(x = 4.5\) and \(y = 2.25\).

(a) Start with LHS: \(\frac{\sin 3\theta}{\sin \theta} - \frac{\cos 3\theta}{\cos \theta} = \frac{\sin 3\theta \cos \theta - \cos 3\theta \sin \theta}{\sin \theta \cos \theta}\). Using the compound angle identity \(\sin(A - B) = \sin A \cos B - \cos A \sin B\), the numerator becomes: \(\sin 3\theta \cos \theta - \cos 3\theta \sin \theta = \sin(3\theta - \theta) = \sin 2\theta\). Using the double angle identity \(\sin 2\theta = 2\sin\theta\cos\theta\), we get: \(\frac{\sin 2\theta}{\sin\theta\cos\theta} = \frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2\). (b) The equation simplifies to: \(2 = 4\cos^2\theta - 1\), which gives \(4\cos^2\theta = 3\), so \\cos^2\theta = \frac{3}{4}\) and \(\cos\theta = \pm\frac{\sqrt{3}}{2}\). For \(0^\circ \le \theta \le 180^\circ\), the solutions are: if \(\cos\theta = \frac{\sqrt{3}}{2}\), then \(\theta = 30^\circ\); if \(\cos\theta = -\frac{\sqrt{3}}{2}\), then \(\theta = 150^\circ\). Both values are within the range and do not make \(\sin\theta = 0\) or \(\cos\theta = 0\).

(a) M1: Attempts to write as a single fraction with a common denominator. M1: Uses the sine compound angle formula correctly on the numerator. M1: Uses the double angle formula for \(\sin 2\theta\). A1: Completes the proof to reach 2. (b) M1: Equates the simplified LHS to the RHS to get \(2 = 4\cos^2\theta - 1\). M1: Rearranges to make \(\cos^2\theta\) or \(\cos\theta\) the subject. A1: Obtains \(\cos\theta = \pm\frac{\sqrt{3}}{2}\). M1: Finds one correct principal value for \(\theta\). A1: Obtains both \(\theta = 30^\circ\) and \(\theta = 150^\circ\) (and no extra values in range).

(a) Applying the product rule to \(y = e^x (2x + 1)^{\frac{1}{2}}\): \(\frac{\text{d}y}{\text{d}x} = e^x (2x+1)^{\frac{1}{2}} + e^x \cdot \frac{1}{2}(2x+1)^{-\frac{1}{2}} \cdot 2 = e^x (2x+1)^{\frac{1}{2}} + e^x (2x+1)^{-\frac{1}{2}}\). Factoring out \(e^x (2x+1)^{-\frac{1}{2}}\): \(\frac{\text{d}y}{\text{d}x} = e^x (2x+1)^{-\frac{1}{2}} ((2x+1) + 1) = \frac{2e^x(x+1)}{\sqrt{2x+1}}\). (b) The volume of revolution is \(V = \pi \int_{0}^{1} y^2 \text{d}x = \pi \int_{0}^{1} (2x + 1)e^{2x} \text{d}x\). Using integration by parts with \(u = 2x+1\) and \(\text{d}v = e^{2x} \text{d}x\), we get \(\int (2x+1)e^{2x} \text{d}x = (2x+1)\frac{e^{2x}}{2} - \int e^{2x} \text{d}x = \frac{2x+1}{2}e^{2x} - \frac{1}{2}e^{2x} = xe^{2x}\). Evaluating between the limits 0 and 1: \(V = \pi [xe^{2x}]_{0}^{1} = \pi (1e^2 - 0) = \pi e^2\).

(a) M1: Applies the product rule or quotient rule to differentiate \(y\). A1: Correctly differentiates \((2x+1)^{\frac{1}{2}}\) to get \(\frac{1}{\sqrt{2x+1}}\). M1: Attempts to put terms over a common denominator. A1: Obtains the correct given expression showing all steps. (b) M1: Uses the volume formula \(V = \pi \int y^2 \text{d}x\) with limits 0 and 1. A1: Correctly writes the integrand as \((2x + 1)e^{2x}\). M1: Applies integration by parts with correct choice of \(u\) and \(\text{d}v\). A1: Obtains \(xe^{2x}\) as the integral. M1: Substitutes limits 1 and 0 into their integrated expression. A1: Final answer of \(\pi e^2\) (must be exact).

(a) The terms are \(u_1 = a\), \(u_3 = a + 2d\), and \(u_7 = a + 6d\). Since these form a GP, \(\frac{a + 2d}{a} = \frac{a + 6d}{a + 2d}\), so \((a + 2d)^2 = a(a + 6d)\). This simplifies to \(a^2 + 4ad + 4d^2 = a^2 + 6ad\), which yields \(4d^2 - 2ad = 0\). Since \(d \neq 0\), we divide by \(2d\) to get \(2d - a = 0 \implies d = \frac{1}{2}a\). The common ratio is \(r = \frac{a+2d}{a} = \frac{a+a}{a} = 2\). (b)(i) The sum of the first 10 terms of \(A\) is \(S_{10} = 5(2a + 9d) = 130\). Substituting \(d = \frac{1}{2}a\): \(5(2a + 4.5a) = 130 \implies 32.5a = 130 \implies a = 4\). (ii) The first term of \(G\) is \(a = 4\) and the common ratio is \(r = 2\). The sum of the first 8 terms of \(G\) is \(S_8 = \frac{4(2^8 - 1)}{2 - 1} = 4(255) = 1020\).

(a) M1: Writes expressions for terms of \(A\) in terms of \(a\) and \(d\). M1: Sets up the geometric progression condition. A1: Correctly simplifies and solves to show \(d = \frac{1}{2}a\). A1: Deduces that \(r = 2\). (b)(i) M1: Uses the sum formula for an AP with \(n=10\) set to 130. M1: Substitutes \(d = \frac{1}{2}a\). A1: Solves to find \(a = 4\). (b)(ii) M1: Uses the geometric series sum formula with \(n=8\). A1: Obtains 1020.

(a) Since \(OP : PA = 2 : 1\), \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\). Since \(OQ : QB = 1 : 3\), \(\overrightarrow{OQ} = \frac{1}{4}\mathbf{b}\). Therefore, \(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\) and \(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\). (b) Since \(X\) lies on \(AQ\), \(\overrightarrow{OX} = \mathbf{a} + \lambda \overrightarrow{AQ} = (1 - \lambda)\mathbf{a} + \frac{1}{4}\lambda\mathbf{b}\). Since \(X\) also lies on \(BP\), \(\overrightarrow{OX} = \mathbf{b} + \mu \overrightarrow{BP} = \frac{2}{3}\mu\mathbf{a} + (1 - \mu)\mathbf{b}\). Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): \(1 - \lambda = \frac{2}{3}\mu\) and \(\frac{1}{4}\lambda = 1 - \mu\). Solving these simultaneous equations gives \(\lambda = \frac{2}{5}\) and \(\mu = \frac{9}{10}\). Substituting \(\lambda = \frac{2}{5}\) back gives \(\overrightarrow{OX} = \frac{3}{5}\mathbf{a} + \frac{1}{10}\mathbf{b}\). (c) Since \(\overrightarrow{AX} = \lambda \overrightarrow{AQ} = \frac{2}{5}\overrightarrow{AQ}\), then \(\overrightarrow{XQ} = \frac{3}{5}\overrightarrow{AQ}\). Thus, the ratio \(AX : XQ = 2 : 3\).

(a) M1: Finds \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\) in terms of vectors. A1: Obtains correct expressions for \(\overrightarrow{AQ}\) and \(\overrightarrow{BP}\). (b) M1: Expresses \(\overrightarrow{OX}\) in two different ways using scalar parameters. M1: Equates coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up simultaneous equations. A1: Solves to find a parameter correctly. A1: Obtains the correct vector \(\overrightarrow{OX}\). (c) M1: Relates the ratio to the parameter \(\lambda\). A1: Deduces the ratio \(2 : 3\).

(a) The midpoint \(M\) of \(AB\) is \((\frac{1+5}{2}, \frac{3+11}{2}) = (3, 7)\). The gradient of \(AB\) is \(m = \frac{11 - 3}{5 - 1} = 2\). The gradient of the perpendicular bisector is \(m_{\perp} = -\frac{1}{2}\). The equation is \(y - 7 = -\frac{1}{2}(x - 3) \implies 2y - 14 = -x + 3 \implies 2y + x - 17 = 0\). (b) Setting \(x = 0\) in the equation of the line: \(2y - 17 = 0 \implies y = 8.5\), so \(C = (0, 8.5)\). (c) Using the coordinates \(A(1, 3)\), \(B(5, 11)\), and \(C(0, 8.5)\) in the area formula: \(\text{Area} = \frac{1}{2} |1(11 - 8.5) + 5(8.5 - 3) + 0(3 - 11)| = \frac{1}{2} |2.5 + 27.5| = 15\).

(a) M1: Finds the midpoint of \(AB\). M1: Finds the gradient of \(AB\) and uses it to find the perpendicular gradient. M1: Writes down the equation of the line using their midpoint and perpendicular gradient. A1: Obtains \(2y + x - 17 = 0\) or integer multiple. (b) B1: Correctly substitutes \(x=0\) to find \(C(0, 8.5)\). (c) M1: Applies a valid method to find the area of a triangle given three vertices. M1: Substitutes coordinates correctly. A1: Obtains 15.

(a) The vertical asymptote is when the denominator is zero: \(x = -1\). The horizontal asymptote is \(y = \lim_{x \to \pm\infty} f(x) = 2\). (b) For \(y = 0\), \(2x - 3 = 0 \implies x = 1.5\), giving \((1.5, 0)\). For \(x = 0\), \(y = -3\), giving \((0, -3)\). (c) The sketch should show a rectangular hyperbola with branches in the upper-left and lower-right regions defined by the asymptotes \(x = -1\) and \(y = 2\), passing through the axes at \((1.5, 0)\) and \((0, -3)\). (d) The line is \(y = x - 3\). To find the intersections, set \(\frac{2x - 3}{x + 1} = x - 3 \implies 2x - 3 = x^2 - 2x - 3 \implies x^2 - 4x = 0\), giving \(x = 0\) and \(x = 4\). From the sketch, the curve lies below or on the line when \(-1 < x \le 0\) or \(x \ge 4\).

(a) B1: Correct vertical asymptote \(x = -1\). B1: Correct horizontal asymptote \(y = 2\). (b) B1: Both coordinates \((1.5, 0)\) and \((0, -3)\) correct. (c) B1: Correct shape with two branches. B1: Both asymptotes drawn and labeled. B1: Axes intercepts correctly shown. (d) M1: Solves \(\frac{2x - 3}{x + 1} = x - 3\). A1: Finds intersection coordinates \(x = 0\) and \(x = 4\). M1: Identifies intervals graphically or algebraically. A1: Obtains \(-1 < x \le 0\) or \(x \ge 4\).