2024 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

The vertex of the modulus function \(f(x) = |ax + b| + c\) occurs where the expression inside the absolute value is zero, which is \(ax + b = 0\). At this vertex, the function value is \(f(x) = c\). Since the vertex is given as \((3, -5)\), we have \(c = -5\) and \(a(3) + b = 0\), which gives \(b = -3a\). The function can therefore be written as \(f(x) = |ax - 3a| - 5\). Since the graph passes through the \(y\)-intercept at \((0, 4)\), we can substitute these coordinates into the equation: \(4 = |a(0) - 3a| - 5\), which simplifies to \(9 = |-3a|\). Given that \(a > 0\), this becomes \(3a = 9\), so \(a = 3\). Substituting \(a = 3\) into \(b = -3a\) yields \(b = -9\). Thus, the values are \(a = 3\), \(b = -9\), and \(c = -5\).

M1: Identify \(c = -5\) from the vertex coordinates. M1: Set up the equation \(3a + b = 0\) or equivalent. M1: Use the point \((0, 4)\) to set up an equation in terms of \(a\) and solve for \(a > 0\). A0.5: Correct values for all three constants \(a = 3\), \(b = -9\), and \(c = -5\).

Let \(Y = \ln y\) and \(X = x^2\). The graph of \(Y\) against \(X\) is a straight line. The gradient of this line is \(m = \frac{11 - 5}{4 - 1} = \frac{6}{3} = 2\), so \(k = 2\). Using the point-slope form with the point \((1, 5)\), the equation of the line is \(Y - 5 = 2(X - 1)\), which simplifies to \(Y = 2X + 3\). Substituting \(Y = \ln y\) and \(X = x^2\) back into the equation gives \(\ln y = 2x^2 + 3\). Taking the exponential of both sides, we get \(y = e^{2x^2 + 3} = e^3 e^{2x^2}\). Comparing this with \(y = A e^{kx^2}\), we find the exact values of the constants to be \(A = e^3\) and \(k = 2\).

M1: Calculate the gradient of the straight line to find \(k = 2\). M1: Formulate the linear equation and solve for the vertical intercept \(\ln A = 3\). A1.5: Correctly write the final equation as \(y = e^3 e^{2x^2}\) or state \(A = e^3\) and \(k = 2\).

For the curve \(y = a \cos(bx) + c\) where \(a > 0\), the maximum value is \(a + c\) and the minimum value is \(-a + c\). From the given points, the maximum is \(7\) and the minimum is \(-1\). This gives the simultaneous equations: \(a + c = 7\) and \(-a + c = -1\). Adding these equations gives \(2c = 6\), which means \(c = 3\). Substituting \(c = 3\) into the first equation gives \(a = 4\). The first minimum point occurs at \(x = \pi\). For a cosine graph starting at a maximum at \(x = 0\), the first minimum occurs at half a period, so \(bx = \pi\) when \(x = \pi\). This gives \(b\pi = \pi\), which simplifies to \(b = 1\). Therefore, \(a = 4\), \(b = 1\), and \(c = 3\).

M1: Set up simultaneous equations for \(a\) and \(c\) using the max and min values. A1: Solve to find \(c = 3\) and \(a = 4\). M1: Use the position of the first minimum to set up the equation \(b\pi = \pi\) or equivalent. A0.5: Obtain \(b = 1\).

Integrate the gradient function to find the equation of the curve: \( y = \int (3x^2 - 10x + 7) \mathrm{d}x = x^3 - 5x^2 + 7x + C \), where \( C \) is the constant of integration. Since \( (x-2) \) is a factor of \( y \), we have \( y = 0 \) when \( x = 2 \). Substituting \( x = 2 \) and \( y = 0 \) into the equation: \( 0 = (2)^3 - 5(2)^2 + 7(2) + C \). This simplifies to \( 0 = 8 - 20 + 14 + C \), which gives \( C = -2 \). Therefore, the equation of the curve is \( y = x^3 - 5x^2 + 7x - 2 \).

M1 for attempting to integrate the gradient function. A1 for correct integration to find \( y = x^3 - 5x^2 + 7x + C \). M1 for applying the factor theorem \( y(2) = 0 \). M1 for solving for \( C \). A1.5 for writing the final equation \( y = x^3 - 5x^2 + 7x - 2 \).

The volume of the box is \( V = x^2 y = 27 \), so \( y = \frac{27}{x^2} \). The total surface area of a closed box is \( A = 2x^2 + 4xy \). Substituting \( y \): \( A = 2x^2 + 4x \left(\frac{27}{x^2}\right) = 2x^2 + \frac{108}{x} \). To find the stationary point, differentiate with respect to \( x \): \( \frac{\mathrm{d}A}{\mathrm{d}x} = 4x - \frac{108}{x^2} \). Setting \( \frac{\mathrm{d}A}{\mathrm{d}x} = 0 \) gives \( 4x^3 = 108 \), which simplifies to \( x^3 = 27 \), so \( x = 3 \). To show this minimizes \( A \), find the second derivative: \( \frac{\mathrm{d}^2 A}{\mathrm{d}x^2} = 4 + \frac{216}{x^3} \). Substituting \( x = 3 \) gives \( \frac{\mathrm{d}^2 A}{\mathrm{d}x^2} = 4 + 8 = 12 \). Since \( \frac{\mathrm{d}^2 A}{\mathrm{d}x^2} > 0 \), the value \( x = 3 \) minimizes the surface area.

M1 for expressing \( y \) in terms of \( x \). M1 for obtaining the area formula \( A = 2x^2 + \frac{108}{x} \). M1 for differentiating to find \( \frac{\mathrm{d}A}{\mathrm{d}x} \). A1 for setting to 0 and solving to get \( x = 3 \). M1 for finding the second derivative \( \frac{\mathrm{d}^2 A}{\mathrm{d}x^2} \) and verifying it is positive (minimum) [0.5 marks for the final verification].

First, find the derivative of \( y \): \( \frac{\mathrm{d}y}{\mathrm{d}x} = 2e^{2x} - 4e^x - 6 \). Set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) to locate the stationary point: \( 2e^{2x} - 4e^x - 6 = 0 \). Divide by 2 to get \( e^{2x} - 2e^x - 3 = 0 \). Let \( u = e^x \), giving the quadratic \( u^2 - 2u - 3 = 0 \). Factorise to find \( (u-3)(u+1) = 0 \), so \( u = 3 \) or \( u = -1 \). Since \( e^x \) must be positive, we discard \( u = -1 \), leaving \( e^x = 3 \), which gives \( x = \ln(3) \). Substitute \( x = \ln(3) \) back into the original curve equation: \( y = e^{2\ln(3)} - 4e^{\ln(3)} - 6\ln(3) = 9 - 12 - 6\ln(3) = -3 - 6\ln(3) \). Thus, the exact coordinates are \( (\ln(3), -3 - 6\ln(3)) \).

M1 for differentiating to find \( \frac{\mathrm{d}y}{\mathrm{d}x} \). M1 for setting the derivative to 0 and forming a quadratic in \( e^x \). A1 for solving the quadratic and obtaining \( e^x = 3 \). M1 for finding \( x = \ln(3) \). A1.5 for finding the exact \( y \)-coordinate and stating the coordinates as \( (\ln(3), -3 - 6\ln(3)) \).

Find the intersection points by setting the equations equal to each other: \( 4x - x^2 = 2x - 3 \). Rearranging gives the quadratic equation \( x^2 - 2x - 3 = 0 \). Factorise to get \( (x-3)(x+1) = 0 \), which gives the intersection points at \( x = -1 \) and \( x = 3 \). The area of the region is given by the integral of the upper curve minus the lower line: \( \int_{-1}^{3} ((4x - x^2) - (2x - 3)) \mathrm{d}x = \int_{-1}^{3} (-x^2 + 2x + 3) \mathrm{d}x \). Integrate each term: \( \left[ -\frac{x^3}{3} + x^2 + 3x \right]_{-1}^{3} \). Evaluate at the upper limit \( x = 3 \): \( -\frac{27}{3} + 9 + 9 = 9 \). Evaluate at the lower limit \( x = -1 \): \( -\frac{-1}{3} + 1 - 3 = -\frac{5}{3} \). Calculate the area: \( 9 - (-\frac{5}{3}) = \frac{32}{3} \approx 10.7 \).

M1 for setting the equations equal to find the intersection points. A1 for correct intersection x-values \( x = -1 \) and \( x = 3 \). M1 for setting up the correct integration expression \( \int_{-1}^{3} (-x^2 + 2x + 3) \mathrm{d}x \). M1 for integrating correctly. A1.5 for evaluating the limits and finding the final exact area \( \frac{32}{3} \) or 10.7.

Find the coordinates of \( P \): at \( x = 1.5 \), \( y = \frac{4}{2(1.5)-1} = 2 \), so \( P(1.5, 2) \). Differentiate the curve: \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(4(2x-1)^{-1}) = -4(2x-1)^{-2} \cdot 2 = -\frac{8}{(2x-1)^2} \). The gradient of the tangent at \( P \) is \( -\frac{8}{(2(1.5)-1)^2} = -2 \). The gradient of the normal is therefore \( m = 0.5 \). The equation of the normal is \( y - 2 = 0.5(x - 1.5) \), which simplifies to \( y = 0.5x + 1.25 \) or \( 4y = 2x + 5 \). To find the other intersection point, substitute the curve equation into the normal: \( 4\left(\frac{4}{2x-1}\right) = 2x + 5 \implies 16 = (2x+5)(2x-1) \implies 4x^2 + 8x - 21 = 0 \). Factorise the quadratic: \( (2x-3)(2x+7) = 0 \). Since \( 2x-3 = 0 \) gives \( x = 1.5 \) (the point \( P \)), the other root is \( 2x+7 = 0 \implies x = -3.5 \). Substitute \( x = -3.5 \) into the curve equation: \( y = \frac{4}{2(-3.5)-1} = -0.5 \). The coordinates of the intersection point are \( (-3.5, -0.5) \).

M1 for finding the coordinates of \( P \) and differentiating the curve. A1 for finding the gradient of the normal (\( 0.5 \)) and the equation of the normal (\( 4y = 2x + 5 \)). M1 for setting up the simultaneous equation to find the intersection. M1 for solving the quadratic equation to find \( x = -3.5 \). A1.5 for calculating \( y = -0.5 \) and stating the coordinates as \( (-3.5, -0.5) \).

The volume \( V \) of a cylinder is \( V = \pi r^2 h \). Given \( h = 3r \), \( V = 3\pi r^3 \). Differentiating \( V \) with respect to \( r \): \( \frac{\mathrm{d}V}{\mathrm{d}r} = 9\pi r^2 \). Using the chain rule, \( \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r} \frac{\mathrm{d}r}{\mathrm{d}t} \implies 18\pi = 9\pi r^2 \frac{\mathrm{d}r}{\mathrm{d}t} \implies \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{2}{r^2} \). At \( r = 2 \), \( \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{2}{4} = 0.5 \text{ cm s}^{-1} \). The total surface area \( A \) is \( A = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r(3r) = 8\pi r^2 \). Differentiating \( A \) with respect to \( r \): \( \frac{\mathrm{d}A}{\mathrm{d}r} = 16\pi r \). The rate of change of the area is \( \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}r} \frac{\mathrm{d}r}{\mathrm{d}t} = (16\pi r) \left(\frac{2}{r^2}\right) = \frac{32\pi}{r} \). When \( r = 2 \), \( \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{32\pi}{2} = 16\pi \approx 50.3 \text{ cm}^2\text{s}^{-1} \).

M1 for expressing \( V \) and \( A \) in terms of \( r \) only. M1 for differentiating \( V \) to find \( \frac{\mathrm{d}V}{\mathrm{d}r} = 9\pi r^2 \). M1 for finding \( \frac{\mathrm{d}r}{\mathrm{d}t} = 0.5 \) when \( r = 2 \). M1 for differentiating \( A \) to find \( \frac{\mathrm{d}A}{\mathrm{d}r} = 16\pi r \). A1.5 for applying the chain rule correctly to obtain \( 16\pi \) (or 50.3) as the final answer.

Using the product rule, differentiate \( x \sin(2x) \): \( \frac{\mathrm{d}}{\mathrm{d}x}(x \sin(2x)) = (1)\sin(2x) + x(2\cos(2x)) = \sin(2x) + 2x\cos(2x) \). Integrating both sides gives \( \int (\sin(2x) + 2x\cos(2x)) \mathrm{d}x = x \sin(2x) \). This can be rearranged as: \( 2 \int x\cos(2x) \mathrm{d}x = x\sin(2x) - \int \sin(2x) \mathrm{d}x \). Thus, \( \int x\cos(2x) \mathrm{d}x = \frac{1}{2}x\sin(2x) - \frac{1}{2}\left(-\frac{1}{2}\cos(2x)\right) = \frac{1}{2}x\sin(2x) + \frac{1}{4}\cos(2x) \). Evaluating this from \( 0 \) to \( \frac{\pi}{4} \): at the upper limit \( x = \frac{\pi}{4} \), we have \( \frac{1}{2}\left(\frac{\pi}{4}\right)\sin(\frac{\pi}{2}) + \frac{1}{4}\cos(\frac{\pi}{2}) = \frac{\pi}{8}(1) + 0 = \frac{\pi}{8} \). At the lower limit \( x = 0 \), we have \( 0 + \frac{1}{4}\cos(0) = \frac{1}{4} \). Therefore, the value of the integral is \( \frac{\pi}{8} - \frac{1}{4} \).

M1 for correctly applying the product rule to differentiate \( x \sin(2x) \). A1 for obtaining \( \sin(2x) + 2x\cos(2x) \). M1 for expressing the integral of \( 2x\cos(2x) \) in terms of the derivative and \( \sin(2x) \). M1 for performing the integration of \( \sin(2x) \) correctly. A1.5 for evaluating the limits accurately and getting the exact value \( \frac{\pi}{8} - \frac{1}{4} \).

(a) Using the product rule:
\(\frac{\text{d}y}{\text{d}x} = \frac{\text{d}}{\text{d}x}(2x-1) \cdot \text{e}^{3x} + (2x-1) \cdot \frac{\text{d}}{\text{d}x}(\text{e}^{3x})\)
\(= 2\text{e}^{3x} + 3(2x-1)\text{e}^{3x}\)
\(= (6x - 1)\text{e}^{3x}\).

(b) At the stationary point, \(\frac{\text{d}y}{\text{d}x} = 0\).
\((6x - 1)\text{e}^{3x} = 0\).
Since \(\text{e}^{3x} \neq 0\) for all real \(x\), we have:
\(6x - 1 = 0 \implies x = \frac{1}{6}\).

Substituting \(x = \frac{1}{6}\) into the equation of the curve:
\(y = \left(2\left(\frac{1}{6}\right) - 1\right)\text{e}^{3(1/6)} = -\frac{2}{3}\text{e}^{1/2} = -\frac{2}{3}\sqrt{\text{e}}\).

So the stationary point is \(\left(\frac{1}{6}, -\frac{2}{3}\sqrt{\text{e}}\right)\).

To determine the nature, find the second derivative:
\(\frac{\text{d}^2y}{\text{d}x^2} = 6\text{e}^{3x} + 3(6x-1)\text{e}^{3x} = (18x + 3)\text{e}^{3x}\).
At \(x = \frac{1}{6}\):
\(\frac{\text{d}^2y}{\text{d}x^2} = \left(18\left(\frac{1}{6}\right) + 3\right)\text{e}^{1/2} = 6\sqrt{\text{e}}\).
Since \(6\sqrt{\text{e}} > 0\), the stationary point is a local minimum.

(c) The curve crosses the \(y\)-axis when \(x = 0\).
At \(x = 0\), \(y = (2(0) - 1)\text{e}^0 = -1\).
The gradient of the tangent at this point is \(\frac{\text{d}y}{\text{d}x}\big|_{x=0} = (6(0) - 1)\text{e}^0 = -1\).
Therefore, the gradient of the normal is \(m = -\frac{1}{-1} = 1\).

The equation of the normal is:
\(y - (-1) = 1(x - 0) \implies y = x - 1\).

(a)
M1: For applying the product rule with at least one term differentiated correctly.
A1: For \(2\text{e}^{3x} + 3(2x-1)\text{e}^{3x}\) or equivalent.
A1: For fully simplified derivative \((6x - 1)\text{e}^{3x}\).

(b)
M1: For setting their \(\frac{\text{d}y}{\text{d}x} = 0\) and solving to find \(x = 1/6\).
A1: For finding the exact coordinates \(\left(\frac{1}{6}, -\frac{2}{3}\sqrt{\text{e}}\right)\).
B1: For finding the second derivative and correctly showing it is positive at \(x = 1/6\), concluding it is a local minimum.

(c)
B1: For finding the correct intersection point \((0, -1)\) and the tangent gradient \(-1\).
M1: For using the negative reciprocal to find the gradient of the normal and setting up the linear equation.
A0.5: For the correct final equation \(y = x - 1\).

(a) LHS \(= \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta}\)
\(= \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta(1 + \cos \theta)}\)

Since \(\sin^2 \theta + \cos^2 \theta = 1\), this becomes:
\(= \frac{1 + 1 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2(1 + \cos \theta)}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2}{\sin \theta} = 2\text{cosec} \theta = \text{RHS}\).

(b) Using the identity from part (a) with \(\theta = 2x - 0.5\), the equation simplifies to:
\(2\text{cosec}(2x - 0.5) = 5\)
\(\text{cosec}(2x - 0.5) = 2.5\)
\(\sin(2x - 0.5) = 0.4\)

Given \(0 < x < \pi\), the range for the angle \(\theta = 2x - 0.5\) is:
\(0 < 2x < 2\pi \implies -0.5 < 2x - 0.5 < 2\pi - 0.5 \approx 5.783\) radians.

The basic angle in radians is:
\(\sin^{-1}(0.4) \approx 0.4115\) rad.

Since \(\sin(2x - 0.5)\) is positive, \(2x - 0.5\) must be in the first or second quadrant within the valid interval:

Case 1:
\(2x - 0.5 = 0.4115\)
\(2x = 0.9115 \implies x \approx 0.4557\)
Rounding to 2 decimal places gives \(x = 0.46\).

Case 2:
\(2x - 0.5 = \pi - 0.4115 \approx 2.7301\)
\(2x = 3.2301 \implies x \approx 1.6150\)
Rounding to 2 decimal places gives \(x = 1.62\).

(a)
M1: For writing fractions over a common denominator: \(\frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)}\).
M1: For expanding \((1 + \cos \theta)^2\) correctly and recognizing \(\sin^2 \theta + \cos^2 \theta = 1\).
A1: For obtaining \(\frac{2(1 + \cos \theta)}{\sin \theta(1 + \cos \theta)}\) through factoring.
A0.5: For simplifying to \(2\text{cosec} \theta\).

(b)
M1: For recognizing the connection to part (a) and establishing the equation \(\sin(2x - 0.5) = 0.4\).
B1: For finding the correct basic angle of approximately \(0.4115\) radians.
M1: For setting up two correct equations for the first and second quadrants within the interval: \(2x - 0.5 = 0.4115\) and \(2x - 0.5 = \pi - 0.4115\).
A1: For obtaining \(x = 0.46\).
A1: For obtaining \(x = 1.62\).

The midpoint of \(AB\) is \(M\left(\frac{1+5}{2}, \frac{4-2}{2}\right) = (3, 1)\). The gradient of \(AB\) is \(m = \frac{-2-4}{5-1} = -\frac{3}{2}\). The gradient of the perpendicular bisector \(L_2\) is the negative reciprocal of \(m\), which is \(m' = \frac{2}{3}\). Using the point-slope form with \(M(3, 1)\), the equation is \(y - 1 = \frac{2}{3}(x - 3)\). Multiplying by 3 gives \(3(y - 1) = 2(x - 3)\), which simplifies to \(3y - 3 = 2x - 6\), and rearranging gives \(3y = 2x - 3\).

M1: For finding the correct midpoint of \(AB\) at \((3, 1)\). M1: For finding the gradient of \(AB\) as \(-\frac{3}{2}\) and using its negative reciprocal \(\frac{2}{3}\) as the gradient of the perpendicular line. A1.5: For writing the correct equation in the requested form \(3y = 2x - 3\) (or any equivalent integer multiple form such as \(6y = 4x - 6\)).

First, find the vector \(2\mathbf{a} + \mathbf{b}\): \(2\mathbf{a} + \mathbf{b} = 2(3\mathbf{i} - 4\mathbf{j}) + (k\mathbf{i} + 5\mathbf{j}) = (6+k)\mathbf{i} - 3\mathbf{j}\). For this vector to be parallel to \(\mathbf{i} - 2\mathbf{j}\), the ratio of the \(\mathbf{i}\)-component to the \(\mathbf{j}\)-component must be equal to that of \(\mathbf{i} - 2\mathbf{j}\): \(\frac{6+k}{1} = \frac{-3}{-2}\). Solving this gives \(6+k = 1.5\), which leads to \(k = -4.5\).

M1: Correct expression for \(2\mathbf{a} + \mathbf{b}\) in terms of \(k\) (e.g., \((6+k)\mathbf{i} - 3\mathbf{j}\)). M1: Setting up a ratio or scalar equation for parallel vectors (e.g., \(6+k = -0.5 \times (-3)\) or \(2(6+k) = 3\)). A1.5: Correct value of \(k = -4.5\) (or equivalent fraction \(-\frac{9}{2}\)).

To find the stationary point, we differentiate \(y\) with respect to \(x\) using the product rule: \(\frac{dy}{dx} = \frac{d}{dx}(x^2) \ln(x) + x^2 \frac{d}{dx}(\ln(x)) = 2x\ln(x) + x^2\left(\frac{1}{x}\right) = 2x\ln(x) + x\). For a stationary point, we set \(\frac{dy}{dx} = 0\), which gives \(x(2\ln(x) + 1) = 0\). Since \(x > 0\) is the domain of \(\ln(x)\), we have \(2\ln(x) + 1 = 0 \implies \ln(x) = -\frac{1}{2} \implies x = e^{-0.5}\).

M1.5: Correctly applying the product rule to obtain \(\frac{dy}{dx} = 2x\ln(x) + x\). M1: Setting \(\frac{dy}{dx} = 0\) and solving for \(\ln(x)\). A1: Finding the exact value \(x = e^{-0.5}\) (or \(\frac{1}{\sqrt{e}}\) or \(e^{-\frac{1}{2}}\)).

Using the change of base formula, we can rewrite \(\log_x(3)\) as \(\frac{1}{\log_3(x)}\). The equation becomes \(\log_3(x) + \frac{6}{\log_3(x)} = 5\). Let \(u = \log_3(x)\). Then \(u + \frac{6}{u} = 5\), which multiplies out to \(u^2 - 5u + 6 = 0\). Factoring the quadratic gives \((u - 2)(u - 3) = 0\), which means \(u = 2\) or \(u = 3\). Substituting back \(\log_3(x)\): if \(\log_3(x) = 2 \implies x = 3^2 = 9\); if \(\log_3(x) = 3 \implies x = 3^3 = 27\). Both values are valid solutions.

M1: Applying the change of base formula to express the equation in terms of \(\log_3(x)\). M1: Forming and solving a quadratic equation in terms of \(u\) (or \(\log_3(x)\)). A1.5: Obtaining both correct solutions \(x = 9\) and \(x = 27\) (award only 1 mark if only one correct solution is found or if left as \(u = 2, 3\)).

Using the trigonometric identity \(\tan^2 x = \sec^2 x - 1\), substitute this into the given equation: \(2(\sec^2 x - 1) - 5 \sec x - 1 = 0\). Simplifying the terms yields the quadratic equation: \(2\sec^2 x - 5 \sec x - 3 = 0\). Letting \(y = \sec x\), the equation becomes \(2y^2 - 5y - 3 = 0\), which factorizes to \((2y + 1)(y - 3) = 0\). This gives two possible solutions: \(\sec x = -0.5\) or \(\sec x = 3\). Recalling that \(\sec x = \frac{1}{\cos x}\), we analyze each case: 1) \(\cos x = -2\), which has no real solutions because the cosine function is bounded between -1 and 1. 2) \(\cos x = \frac{1}{3}\). For \(0 \le x \le 2\pi\), the principal angle is \(x = \cos^{-1}(1/3) \approx 1.23\) radians. The second angle in the fourth quadrant is \(x = 2\pi - \cos^{-1}(1/3) \approx 5.05\) radians.

- M1: For utilizing the identity \(\tan^2 x = \sec^2 x - 1\) to rewrite the equation in terms of \(\sec x\) only. - A1: For obtaining the correct quadratic equation \(2\sec^2 x - 5 \sec x - 3 = 0\). - M1: For correctly factorizing or solving the quadratic equation to get \(\sec x = -0.5\) and \(\sec x = 3\). - A1: For identifying that \(\cos x = -2\) has no solutions and stating \(\cos x = \frac{1}{3}\). - A1.5: For finding the first angle \(x \approx 1.23\). - A1: For finding the second angle \(x \approx 5.05\) and no other incorrect solutions within the interval.

(a) The first, third, and eleventh terms of the arithmetic progression are given by \(T_1 = a\), \(T_3 = a + 2d\), and \(T_{11} = a + 10d\). Because these terms form a geometric progression, the common ratio remains constant: \(\frac{a + 2d}{a} = \frac{a + 10d}{a + 2d}\). Cross-multiplying gives \((a + 2d)^2 = a(a + 10d)\). Expanding both sides results in \(a^2 + 4ad + 4d^2 = a^2 + 10ad\). Simplifying this yields \(4d^2 = 6ad\). Since \(d \neq 0\), dividing by \(2d\) results in \(2d = 3a\), which simplifies to \(d = 1.5a\). (b) The sum of the first 8 terms of the arithmetic progression is given by \(S_8 = \frac{8}{2}[2a + 7d] = 200\), which simplifies to \(4[2a + 7d] = 200\), meaning \(2a + 7d = 50\). Substituting \(d = 1.5a\) into this equation yields \(2a + 7(1.5a) = 50\), which is \(12.5a = 50\). This gives \(a = 4\), and therefore \(d = 1.5 \times 4 = 6\). The first term of the geometric progression is \(G_1 = a = 4\), and its second term is \(G_2 = a + 2d = 4 + 12 = 16\). The common ratio of this geometric progression is \(r = \frac{16}{4} = 4\). Using the sum formula for a geometric progression, the sum of the first 6 terms is \(S_6 = \frac{4(4^6 - 1)}{4 - 1} = \frac{4(4096 - 1)}{3} = \frac{4 \times 4095}{3} = 5460\).

Part (a): - M1: For expressing the terms of the arithmetic progression and setting up the geometric ratio equation \((a + 2d)^2 = a(a + 10d)\). - M1: For expanding and simplifying to get \(4d^2 = 6ad\). - A1: For dividing by \(2d\) and clearly concluding the given result \(d = 1.5a\). Part (b): - M1: For utilizing the arithmetic sum formula \(S_8 = 4(2a + 7d) = 200\) and substituting \(d = 1.5a\). - A1: For solving to find \(a = 4\) and \(d = 6\). - M1: For finding the first term of the geometric progression as 4 and identifying the common ratio \(r = 4\). - A0.5: For using the geometric progression sum formula correctly to evaluate the final sum as 5460.

(a) To find the stationary point, first differentiate \(y = (2x - 3)e^{2x}\) with respect to \(x\) using the product rule: \(\frac{\text{d}y}{\text{d}x} = 2e^{2x} + 2(2x - 3)e^{2x} = 2e^{2x}(1 + 2x - 3) = 4(x - 1)e^{2x}\). Set \(\frac{\text{d}y}{\text{d}x} = 0\) to locate stationary points. Since \(e^{2x} \neq 0\) for all real \(x\), we have \(x - 1 = 0\), which gives \(x = 1\). Substitute \(x = 1\) into the original equation to obtain the corresponding y-coordinate: \(y = (2(1) - 3)e^{2(1)} = -e^2\). Thus, the exact coordinates of the stationary point are \((1, -e^2)\). (b) First find the x-intercept of the curve by setting \(y = 0\): \((2x - 3)e^{2x} = 0 \implies 2x - 3 = 0 \implies x = 1.5\). The region is therefore bounded between \(x = 1.5\) and \(x = 2\). Integrating the function from \(1.5\) to \(2\) using integration by parts, let \(u = 2x - 3\) so \(\text{d}u = 2\text{d}x\), and let \(\text{d}v = e^{2x}\text{d}x\) so \(v = 0.5e^{2x}\). This yields: \(\int (2x - 3)e^{2x}\text{d}x = 0.5(2x - 3)e^{2x} - \int e^{2x}\text{d}x = 0.5(2x - 3)e^{2x} - 0.5e^{2x} = (x - 2)e^{2x}\). Applying the limits: \(\text{Area} = [(x - 2)e^{2x}]_{1.5}^{2} = (2 - 2)e^4 - (1.5 - 2)e^3 = 0 - (-0.5)e^3 = 0.5e^3\).

Part (a): - M1: For utilizing the product rule to differentiate \(y\). - A1: For obtaining the correct derivative \(\frac{\text{d}y}{\text{d}x} = 4(x - 1)e^{2x}\). - A1: For finding \(x = 1\) and substituting it back to get the coordinates \((1, -e^2)\). Part (b): - M1: For identifying the boundary limits of integration by finding the x-intercept at \(x = 1.5\). - M1: For using integration by parts to integrate the function. - A1: For achieving the correct anti-derivative \((x - 2)e^{2x}\). - A0.5: For substituting limits correctly to yield the exact area value of \(0.5e^3\) or \(\frac{1}{2}e^3\).

(a) The perimeter of the sector is given by the formula \(P = 2r + r\theta\). Since the perimeter is 30 cm, we have \(2r + r\theta = 30\), which can be rearranged to give \(r\theta = 30 - 2r\), or \(\theta = \frac{30 - 2r}{r}\). The area of the sector is given by \(A = \frac{1}{2}r^2\theta\). Since the area is \(50\text{ cm}^2\), we have \(\frac{1}{2}r^2\theta = 50\), which simplifies to \(r^2\theta = 100\). Substituting the expression for \(\theta\) into this equation: \(r^2 \left(\frac{30 - 2r}{r}\right) = 100\), which simplifies to \(r(30 - 2r) = 100\). Expanding and rearranging the terms gives \(30r - 2r^2 = 100 \implies 2r^2 - 30r + 100 = 0\). Dividing the entire equation by 2 yields \(r^2 - 15r + 50 = 0\), as required. (b) To find the values of \(r\), we solve the quadratic equation \(r^2 - 15r + 50 = 0\). Factorizing the quadratic equation gives \((r - 5)(r - 10) = 0\). This gives two possible values for the radius: \(r = 5\) cm or \(r = 10\) cm. We then substitute these values back into \(\theta = \frac{30 - 2r}{r}\) to find the corresponding values of \(\theta\): 1) If \(r = 5\), then \(\theta = \frac{30 - 10}{5} = 4\) radians. 2) If \(r = 10\), then \(\theta = \frac{30 - 20}{10} = 1\) radian.

Part (a): - B1: For setting up the perimeter equation \(2r + r\theta = 30\). - B1: For setting up the area equation \(0.5r^2\theta = 50\). - M1: For eliminating \(\theta\) to form an equation in terms of \(r\) only. - A0.5: For simplifying correctly to obtain the given quadratic equation \(r^2 - 15r + 50 = 0\) with no algebraic errors. Part (b): - M1: For attempting to solve the quadratic equation \(r^2 - 15r + 50 = 0\). - A1: For finding the two correct values of the radius, \(r = 5\) and \(r = 10\). - A1: For finding the correct corresponding values of the angle, \(\theta = 4\) and \(\theta = 1\) (with units where appropriate).

**(a)**
Using the product rule with \(u = 3x-2\) and \(v = (x+1)^4\):

\(\frac{du}{dx} = 3\)

\(\frac{dv}{dx} = 4(x+1)^3\)

\(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)

\(\frac{dy}{dx} = (3x-2) \cdot 4(x+1)^3 + 3(x+1)^4\)

Factorising out \((x+1)^3\):

\(\frac{dy}{dx} = (x+1)^3 [4(3x-2) + 3(x+1)]\)

\(\frac{dy}{dx} = (x+1)^3 [12x - 8 + 3x + 3]\)

\(\frac{dy}{dx} = (15x-5)(x+1)^3\)

So \(A = 15\) and \(B = -5\).

**(b)**
For stationary points, set \(\frac{dy}{dx} = 0\):

\((15x-5)(x+1)^3 = 0\)

This gives \(x = \frac{1}{3}\) or \(x = -1\).
Since we require \(x > -1\), the stationary point of interest has \(x\)-coordinate \(x = \frac{1}{3}\).

To determine the nature, we find the second derivative:

\(\frac{d^2y}{dx^2} = 15(x+1)^3 + 3(15x-5)(x+1)^2\)

At \(x = \frac{1}{3}\):

\(\frac{d^2y}{dx^2} = 15\left(\frac{4}{3}\right)^3 + 3(0)\left(\frac{4}{3}\right)^2 = 15 \cdot \frac{64}{27} = \frac{320}{9} > 0\)

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point at \(x = \frac{1}{3}\) is a **local minimum**.

**(c)**
From part (a), we know that:

\(\frac{d}{dx} [(3x-2)(x+1)^4] = (15x-5)(x+1)^3 = 5(3x-1)(x+1)^3\)

Therefore:

\(\int (3x-1)(x+1)^3 dx = \frac{1}{5}(3x-2)(x+1)^4 + c\)

Evaluating the definite integral from \(0\) to \(1\):

\(\int_{0}^{1} (3x-1)(x+1)^3 dx = \left[ \frac{1}{5}(3x-2)(x+1)^4 \right]_0^1\)

\(= \left( \frac{1}{5}(3(1)-2)(1+1)^4 \right) - \left( \frac{1}{5}(3(0)-2)(0+1)^4 \right)\)

\(= \left( \frac{1}{5}(1)(16) \right) - \left( \frac{1}{5}(-2)(1) \right)\)

\(= \frac{16}{5} + \frac{2}{5} = \frac{18}{5} = 3.6\)

**(a)**
* **M1**: For an attempt to apply the product rule to get \(k(3x-2)(x+1)^3 + 3(x+1)^4\), where \(k=4\) or \(1\).
* **B1**: For correctly differentiating \((x+1)^4\) to obtain \(4(x+1)^3\).
* **A1**: For obtaining the correct unsimplified derivative: \(4(3x-2)(x+1)^3 + 3(x+1)^4\).
* **A1**: For fully simplifying to \((15x-5)(x+1)^3\) or identifying \(A = 15, B = -5\).

**(b)**
* **M1**: For setting their \(\frac{dy}{dx} = 0\) and solving to find \(x = 1/3\).
* **M1**: For a valid method to determine the nature of the stationary point (e.g., evaluating \(\frac{d^2y}{dx^2}\) or checking signs of \(\frac{dy}{dx}\) on either side of \(1/3\)).
* **A1**: For correctly concluding that it is a **minimum** with fully correct supporting working.

**(c)**
* **M1**: For recognising the relationship with part (a) and writing \(\int (3x-1)(x+1)^3 dx = \frac{1}{5}(3x-2)(x+1)^4\).
* **A1**: For substituting limits correctly and obtaining the final exact value \(\frac{18}{5}\) or \(3.6\).

**(a)**
Since \(OP : PA = 2 : 1\), \(OP = \frac{2}{3}OA\), so \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
Then:

\(\overrightarrow{PB} = \overrightarrow{OB} - \overrightarrow{OP} = \mathbf{b} - \frac{2}{3}\mathbf{a}\)

**(b)**
Since \(Q\) is the midpoint of \(AB\):

\(\overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AQ} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB}\)

\(\overrightarrow{OQ} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\)

**(c)**
First expression for \(\overrightarrow{OX}\):

\(\overrightarrow{OX} = \mu \overrightarrow{OQ} = \mu\left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = \frac{1}{2}\mu\mathbf{a} + \frac{1}{2}\mu\mathbf{b}\)

Second expression for \(\overrightarrow{OX}\):

\(\overrightarrow{OX} = \overrightarrow{OP} + \overrightarrow{PX} = \frac{2}{3}\mathbf{a} + \lambda \overrightarrow{PB}\)

\(\overrightarrow{OX} = \frac{2}{3}\mathbf{a} + \lambda\left(\mathbf{b} - \frac{2}{3}\mathbf{a}\right) = \left(\frac{2}{3} - \frac{2}{3}\lambda\right)\mathbf{a} + \lambda\mathbf{b}\)

Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
From \(\mathbf{b}\): \(\frac{1}{2}\mu = \lambda\)
From \(\mathbf{a}\): \(\frac{1}{2}\mu = \frac{2}{3} - \frac{2}{3}\lambda\)

Substitute \(\frac{1}{2}\mu = \lambda\) into the second equation:

\(\lambda = \frac{2}{3} - \frac{2}{3}\lambda\)

\(\frac{5}{3}\lambda = \frac{2}{3} \implies \lambda = \frac{2}{5}\)

Then:

\(\mu = 2\lambda = \frac{4}{5}\)

**(d)**
Since \(\overrightarrow{PX} = \lambda \overrightarrow{PB} = \frac{2}{5}\overrightarrow{PB}\), the point \(X\) lies on \(PB\) such that the ratio of the length \(PX\) to the entire length \(PB\) is \(2 : 5\).
Therefore, \(XB\) corresponds to \(\frac{3}{5}\) of \(PB\), so the ratio \(PX : XB\) is \(2 : 3\).

**(a)**
* **B1**: For correct expression \(\mathbf{b} - \frac{2}{3}\mathbf{a}\) (or any equivalent form).

**(b)**
* **M1**: For writing \(\overrightarrow{OQ} = \mathbf{a} + \frac{1}{2}(\mathbf{b}-\mathbf{a})\) or equivalent vector addition.
* **A1**: For correct simplified expression \(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

**(c)**
* **B1**: For correct first expression of \(\overrightarrow{OX} = \frac{1}{2}\mu\mathbf{a} + \frac{1}{2}\mu\mathbf{b}\).
* **M1**: For attempting the second expression \(\overrightarrow{OX} = \overrightarrow{OP} + \lambda \overrightarrow{PB}\) to get \(\left(\frac{2}{3} - \frac{2}{3}\lambda\right)\mathbf{a} + \lambda\mathbf{b}\).
* **M1**: For equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up two simultaneous equations.
* **A1**: For finding correct values \(\lambda = \frac{2}{5}\) and \(\mu = \frac{4}{5}\).

**(d)**
* **B1**: For correct ratio \(2 : 3\) or equivalent.