2025 Edexcel IAL Mathematics (YMA01) 模擬試題連答案詳解

(a) First, find the gradient of \(l_1\):
\[m_1 = \frac{2 - (-2)}{3 - 1} = \frac{4}{2} = 2\]
Using the point-slope form with \(B(3, 2)\):
\[y - 2 = 2(x - 3)\]
\[y - 2 = 2x - 6\]
\[2x - y - 4 = 0\]

(b) Since \(l_2\) is perpendicular to \(l_1\), its gradient is:
\[m_2 = -\frac{1}{m_1} = -\frac{1}{2}\]
Using the point-slope form with \(C(2, 5)\):
\[y - 5 = -\frac{1}{2}(x - 2)\]
\[y - 5 = -\frac{1}{2}x + 1\]
\[y = -\frac{1}{2}x + 6\]

(c) To find the coordinates of \(D\), solve the equations of \(l_1\) and \(l_2\) simultaneously:
\[2x - 4 = -\frac{1}{2}x + 6\]
Multiply by 2:
\[4x - 8 = -x + 12\]
\[5x = 20 \implies x = 4\]
Substitute \(x = 4\) back into the equation of \(l_1\):
\[y = 2(4) - 4 = 4\]
Thus, the coordinates of \(D\) are \((4, 4)\).

(a)
- **M1**: For an attempt to find the gradient of \(l_1\).
- **M1**: For a correct method to find the equation of \(l_1\) using their gradient and one of the points.
- **A1**: \(2x - y - 4 = 0\) (or any non-zero integer multiple, e.g. \(4x - 2y - 8 = 0\)).

(b)
- **M1**: For using the perpendicular gradient rule \(m_2 = -\frac{1}{m_1}\) to find the gradient of \(l_2\).
- **A1**: \(y = -\frac{1}{2}x + 6\) or \(y = -0.5x + 6\).

(c)
- **M1**: For equating their \(l_1\) and \(l_2\) to find the \(x\)-coordinate.
- **A1**: \(x = 4\).
- **A1**: \(y = 4\) (or coordinates written as \((4, 4)\)).

(a) Differentiating \(y = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + 5\) term by term:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 2 \left(\frac{3}{2}x^{\frac{1}{2}}\right) - 8 \left(\frac{1}{2}x^{-\frac{1}{2}}\right)\]
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{\frac{1}{2}} - 4x^{-\frac{1}{2}}\]

(b) At the point \(P(4, 5)\), we have \(x = 4\).
Substitute \(x = 4\) into
\[\frac{\mathrm{d}y}{\mathrm{d}x}\]
to find the gradient of the tangent, \(m_T\):
\[m_T = 3(4)^{\frac{1}{2}} - 4(4)^{-\frac{1}{2}} = 3(2) - 4\left(\frac{1}{2}\right) = 6 - 2 = 4\]
Since the normal is perpendicular to the tangent, the gradient of the normal, \(m_N\), is:
\[m_N = -\frac{1}{m_T} = -\frac{1}{4}\]
Using the equation of a straight line with gradient \(-\frac{1}{4}\) passing through \(P(4, 5)\):
\[y - 5 = -\frac{1}{4}(x - 4)\]
\[4(y - 5) = -(x - 4)\]
\[4y - 20 = -x + 4\]
\[x + 4y - 24 = 0\]

(a)
- **M1**: For an attempt to differentiate: at least one power decreased by 1.
- **A1**: At least one term correct: \(3x^{\frac{1}{2}}\) or \(-4x^{-\frac{1}{2}}\).
- **A1**: Fully correct derivative: \(3x^{\frac{1}{2}} - 4x^{-\frac{1}{2}}\).

(b)
- **M1**: For substituting \(x = 4\) into their derivative to find the gradient of the tangent.
- **M1**: For finding the negative reciprocal of their tangent gradient to get the normal gradient.
- **M1**: For an attempt to find the equation of a line through \((4, 5)\) using their normal gradient.
- **A1**: \(x + 4y - 24 = 0\) (or any non-zero integer multiple).

(a) The curve crosses the \(x\)-axis when \(y = 0\):
\[9x - x^3 = 0\]
\[x(9 - x^2) = 0\]
\[x(3 - x)(3 + x) = 0\]
Since \(A\) is on the positive \(x\)-axis, we have \(x = 3\).
Thus, the coordinates of \(A\) are \((3, 0)\).

(b) The region \(R\) is bounded by the curve and the positive \(x\)-axis, which corresponds to the interval from \(x = 0\) to \(x = 3\).
\[\text{Area} = \int_{0}^{3} (9x - x^3) \\, \mathrm{d}x\]
First, find the indefinite integral:
\[\int (9x - x^3) \\, \mathrm{d}x = \frac{9}{2}x^2 - \frac{1}{4}x^4 + c\]
Now apply the limits of integration:
\[\left[ \frac{9}{2}x^2 - \frac{1}{4}x^4 \right]_{0}^{3} = \left( \frac{9}{2}(3)^2 - \frac{1}{4}(3)^4 \right) - 0\]
\[= \left( \frac{9}{2}(9) - \frac{1}{4}(81) \right)\]
\[= \frac{81}{2} - \frac{81}{4} = \frac{81}{4} = 20.25\]

(a)
- **M1**: Sets \(9x - x^3 = 0\) and attempts to factorise/solve for \(x\).
- **A1**: \(A(3, 0)\) or \(x = 3, y = 0\). (Accept \(x = 3\) if coordinates are clearly implied).

(b)
- **M1**: For an attempt to integrate \(9x - x^3\) with at least one power increased by 1.
- **A1**: For at least one term correctly integrated: \(\frac{9}{2}x^2\) or \(-\frac{1}{4}x^4\).
- **A1**: For the completely correct integrated expression: \(\frac{9}{2}x^2 - \frac{1}{4}x^4\).
- **M1**: For substituting their limit \(3\) (from part a) and \(0\) into their integrated expression.
- **M1**: For performing the subtraction (upper limit value minus lower limit value).
- **A1**: \(\frac{81}{4}\) or \(20.25\) (or equivalent fraction).

(a) First find the gradient of \(l_1\):
\(m_1 = \frac{1 - 5}{6 - (-2)} = \frac{-4}{8} = -\frac{1}{2}\)

Now use the point-gradient formula with point \(B(6, 1)\):
\(y - 1 = -\frac{1}{2}(x - 6)\)
\(2(y - 1) = -(x - 6)\)
\(2y - 2 = -x + 6\)
\(x + 2y - 8 = 0\)

(b) Since \(l_2\) is perpendicular to \(l_1\), its gradient \(m_2\) is:
\(m_2 = -\frac{1}{m_1} = -\frac{1}{-1/2} = 2\)

Using the gradient \(m_2 = 2\) and the point \(C(4, 7)\):
\(y - 7 = 2(x - 4)\)
\(y - 7 = 2x - 8\)
\(y = 2x - 1\)

(c) To find the intersection point \(D\), substitute \(y = 2x - 1\) into the equation for \(l_1\):
\(x + 2(2x - 1) - 8 = 0\)
\(x + 4x - 2 - 8 = 0\)
\(5x - 10 = 0\)
\(5x = 10 \implies x = 2\)

Substitute \(x = 2\) back into the equation for \(l_2\) to find \(y\):
\(y = 2(2) - 1 = 3\)

So the coordinates of \(D\) are \((2, 3)\).

(a)
M1: Attempts to find the gradient of \(l_1\) using \\frac{y_2 - y_1}{x_2 - x_1}.
M1: Uses their gradient and one of the points \(A\) or \(B\) to form a linear equation.
A1: Correct equation in the form \(ax + by + c = 0\) (or any non-zero integer multiple, e.g., \(2x + 4y - 16 = 0\)).

(b)
M1: Uses the perpendicular gradient rule \(m_2 = -\frac{1}{m_1}\) with their gradient from part (a).
A1: Correct equation for \(l_2\), e.g., \(y = 2x - 1\) or \(2x - y - 1 = 0\).

(c)
M1: Formulates a method to solve the simultaneous equations, e.g., by substitution or elimination.
A1: Finds \(x = 2\) or \(y = 3\).
A1: Gives the correct coordinates of \(D\) as \((2, 3)\) or \(x = 2, y = 3\).

(a) Differentiating \(y = 2x^2 - 8x^{1/2} + 5\) term by term with respect to \(x\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(2x^2) - \frac{\mathrm{d}}{\mathrm{d}x}(8x^{1/2}) + \frac{\mathrm{d}}{\mathrm{d}x}(5)\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 4x - 8 \left(\frac{1}{2}x^{-1/2}\right) + 0\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 4x - 4x^{-1/2}\)

(b) At the point where \(x = 4\):
Find the y-coordinate on \(C\):
\(y = 2(4)^2 - 8(4)^{1/2} + 5 = 2(16) - 8(2) + 5 = 32 - 16 + 5 = 21\)

Find the gradient of the tangent by substituting \(x = 4\) into \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\(m = 4(4) - 4(4)^{-1/2} = 16 - \frac{4}{\sqrt{4}} = 16 - 2 = 14\)

Using the equation of a straight line with \(m = 14\) and point \((4, 21)\):
\(y - 21 = 14(x - 4)\)
\(y - 21 = 14x - 56\)
\(y = 14x - 35\)

(a)
M1: Differentiates \(2x^2 \to ax\) where \(a\) is a constant.
M1: Differentiates \(-8x^{1/2} \to bx^{-1/2}\) where \(b\) is a constant.
A1: Fully correct derivative: \(4x - 4x^{-1/2}\) or equivalent.

(b)
M1: Substitutes \(x = 4\) into the original equation to find the corresponding y-coordinate.
M1: Substitutes \(x = 4\) into their derivative from part (a) to find the gradient of the tangent.
M1: Formulates the equation of the tangent using their point and gradient.
A1: Correct equation in the form \(y = mx + c\), which is \(y = 14x - 35\).

(a) To find where the curve intersects the x-axis, set \(y = 0\):
\(9x - x^3 = 0\)
\(x(9 - x^2) = 0\)
\(x(3 - x)(3 + x) = 0\)

The solutions are \(x = 0\), \(x = 3\), and \(x = -3\).
Since \(x_B > 0\), we have \(x_B = 3\).
Thus, the coordinates of \(B\) are \((3, 0)\).

(b) Integrating term by term:
\(\int (9x - x^3) \,\mathrm{d}x = \frac{9}{2}x^2 - \frac{1}{4}x^4 + C\)

(c) The region bounded by the curve and the x-axis for \(x \ge 0\) lies between \(x = 0\) and \(x = 3\).
Since \(y \ge 0\) on the interval \([0, 3]\), the area is given by the definite integral:
\(\text{Area} = \int_{0}^{3} (9x - x^3) \,\mathrm{d}x\)

Using the integration result from part (b):
\(\text{Area} = \left[ \frac{9}{2}x^2 - \frac{1}{4}x^4 \right]_{0}^{3}\)
\(\text{Area} = \left( \frac{9}{2}(3)^2 - \frac{1}{4}(3)^4 \right) - \left( 0 - 0 \right)\)
\(\text{Area} = \left( \frac{81}{2} - \frac{81}{4} \right) = \frac{81}{4} = 20.25\)

(a)
M1: Sets \(9x - x^3 = 0\) and attempts to factorise or solve for \(x\).
A1: Deduces \(x = 3\) and states the coordinates of \(B\) as \((3, 0)\) (accept \(x = 3\)).

(b)
M1: Integrates at least one term correctly (increasing power of \(x\) by 1).
A1: Correct integration: \(\frac{9}{2}x^2 - \frac{1}{4}x^4\) (constant of integration \(C\) not required).

(c)
M1: Identifies the limits of integration as \(0\) and their x-coordinate of \(B\) (which is \(3\)).
M1: Substitutes their limits into their integrated expression from part (b).
A1ft: Correct substitution, i.e., \(\left(\frac{9}{2}(3)^2 - \frac{1}{4}(3)^4\right) - 0\).
A1: Correct area of \(\frac{81}{4}\) or \(20.25\) (or equivalent fraction).

(a) Rewrite the curve equation as \( y = 3x^{\frac{3}{2}} - 8x^{-1} + 5 \). Differentiating with respect to x gives: \( \frac{\mathrm{d}y}{\mathrm{d}x} = 3 \times \frac{3}{2}x^{\frac{1}{2}} - 8 \times (-1)x^{-2} = \frac{9}{2}x^{\frac{1}{2}} + \frac{8}{x^2} \). (b) At the point P where \( x = 4 \), the gradient of the tangent is: \( m = \frac{9}{2}(4)^{\frac{1}{2}} + \frac{8}{4^2} = \frac{9}{2}(2) + \frac{8}{16} = 9 + \frac{1}{2} = \frac{19}{2} \). The equation of the tangent is given by: \( y - 27 = \frac{19}{2}(x - 4) \). Multiplying by 2: \( 2y - 54 = 19(x - 4) \) which simplifies to \( 2y - 54 = 19x - 76 \). Rearranging into the form \( ax + by + c = 0 \) yields: \( 19x - 2y - 22 = 0 \). (c) The tangent crosses the x-axis when \( y = 0 \). Substituting \( y = 0 \) into the equation of the tangent: \( 19x - 22 = 0 \Rightarrow x = \frac{22}{19} \). Therefore, the coordinates of Q are \( \left(\frac{22}{19}, 0\right) \).

(a) M1: Attempts to differentiate at least one term, e.g., \( x^{\frac{3}{2}} \to kx^{\frac{1}{2}} \) or \( x^{-1} \to kx^{-2} \). A1: One correct term: either \( \frac{9}{2}x^{\frac{1}{2}} \) or \( 8x^{-2} \). A1: Fully correct derivative: \( \frac{9}{2}x^{\frac{1}{2}} + \frac{8}{x^2} \). (b) M1: Substitutes \( x = 4 \) into their derivative to find the gradient of the tangent. M1: Uses their gradient and the coordinates \( (4, 27) \) to form a straight line equation. A1: Correct equation in the required integer form, e.g. \( 19x - 2y - 22 = 0 \) or equivalent. (c) M1: Sets \( y = 0 \) in their tangent equation and solves for x. A1: Correct coordinates \( \left(\frac{22}{19}, 0\right) \).

(a) The curve crosses the x-axis when \( y = 0 \), so \( 4x^{\frac{1}{2}} - x - 3 = 0 \). Let \( u = x^{\frac{1}{2}} \), then \( u^2 - 4u + 3 = 0 \). Factoring the quadratic gives: \( (u - 1)(u - 3) = 0 \), so \( u = 1 \) or \( u = 3 \). Since \( u = x^{\frac{1}{2}} \), we have \( x = 1 \) or \( x = 9 \). The coordinates are \( (1, 0) \) and \( (9, 0) \). (b) The area of the finite region is given by: \( \int_{1}^{9} (4x^{\frac{1}{2}} - x - 3) \mathrm{d}x \). Integrating term by term: \( \left[ 4 \left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right) - \frac{x^2}{2} - 3x \right]_{1}^{9} = \left[ \frac{8}{3}x^{\frac{3}{2}} - \frac{x^2}{2} - 3x \right]_{1}^{9} \). Evaluating at \( x = 9 \): \( \frac{8}{3}(27) - \frac{81}{2} - 27 = 72 - 40.5 - 27 = 4.5 = \frac{9}{2} \). Evaluating at \( x = 1 \): \( \frac{8}{3}(1) - \frac{1}{2} - 3 = \frac{16 - 3 - 18}{6} = -\frac{5}{6} \). Subtracting the lower limit value from the upper limit value: \( \frac{9}{2} - \left(-\frac{5}{6}\right) = \frac{27}{6} + \frac{5}{6} = \frac{32}{6} = \frac{16}{3} \).

(a) M1: Sets \( y = 0 \) and attempts to solve by substitution or factoring, reducing the equation to a quadratic in terms of \( x^{\frac{1}{2}} \). A1: Finds the roots \( x^{\frac{1}{2}} = 1 \) and \( x^{\frac{1}{2}} = 3 \). A1: Correctly identifies the coordinates as \( (1, 0) \) and \( (9, 0) \). (b) M1: Attempts to integrate the function \( 4x^{\frac{1}{2}} - x - 3 \) (at least one power increased by 1). A1: Any two terms integrated correctly. A1: Fully correct integrated expression: \( \frac{8}{3}x^{\frac{3}{2}} - \frac{x^2}{2} - 3x \). M1: Substitutes their limits 1 and 9 (or values from part a) into their integrated expression and subtracts. A1: Correct exact area of \( \frac{16}{3} \) (or equivalent).

(a) Using the identity \( \sin^2 x = 1 - \cos^2 x \), we substitute this into the equation: \( 6(1 - \cos^2 x) - \cos x - 4 = 0 \). Expanding and simplifying: \( 6 - 6 \cos^2 x - \cos x - 4 = 0 \Rightarrow -6 \cos^2 x - \cos x + 2 = 0 \). Multiplying by \(-1\) gives the required form: \( 6 \cos^2 x + \cos x - 2 = 0 \). (b) Solve the quadratic equation: \( 6 \cos^2 x + \cos x - 2 = 0 \). Factoring the quadratic: \( (3 \cos x + 2)(2 \cos x - 1) = 0 \). This gives \( \cos x = -\frac{2}{3} \) or \( \cos x = \frac{1}{2} \). For \( \cos x = \frac{1}{2} \), the solutions in the given range are \( x = 60^\circ \) and \( x = 360^\circ - 60^\circ = 300^\circ \). For \( \cos x = -\frac{2}{3} \), the principal value is \( x = \arccos\left(-\frac{2}{3}\right) \approx 131.81^\circ \approx 131.8^\circ \). The other solution in the range is \( x = 360^\circ - 131.81^\circ \approx 228.19^\circ \approx 228.2^\circ \). Thus, the complete set of solutions is \( x = 60^\circ, 131.8^\circ, 228.2^\circ, 300^\circ \).

(a) M1: Uses the identity \( \sin^2 x = 1 - \cos^2 x \) to rewrite the equation in terms of \( \cos x \). A1: Correctly simplifies to the given equation with no errors shown. (b) M1: Attempts to solve the quadratic equation to find values for \( \cos x \). A1: Identifies \( \cos x = \frac{1}{2} \) and \( \cos x = -\frac{2}{3} \). B1: Obtains \( 60^\circ \) and \( 300^\circ \) (both required for this mark). M1: Finds one correct angle for \( \cos x = -\frac{2}{3} \) and attempts to find the second angle using \( 360^\circ - \text{angle} \). A1: Obtains \( 131.8^\circ \) and \( 228.2^\circ \) (allow solutions rounded to 1 decimal place). Deduct 1 mark overall if any extra solutions in the range are present.

Part (a): We first rewrite the curve's equation by expressing the terms with fractional and negative indices as \( y = 2x^{\frac{3}{2}} - 8x^{-\frac{1}{2}} + 3 \). Differentiating with respect to x, we get \( \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \times \frac{3}{2}x^{\frac{1}{2}} - 8 \times \left(-\frac{1}{2}\right)x^{-\frac{3}{2}} \), which simplifies to \( \frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{\frac{1}{2}} + 4x^{-\frac{3}{2}} \). Part (b): First, find the y-coordinate of the point P where \( x = 4 \) by substituting into the curve's equation: \( y = 2(4)^{\frac{3}{2}} - \frac{8}{\sqrt{4}} + 3 = 2(8) - 4 + 3 = 15 \). So P is \( (4, 15) \). Next, find the gradient of the tangent at P by substituting \( x = 4 \) into the derivative: \( m = 3(4)^{\frac{1}{2}} + 4(4)^{-\frac{3}{2}} = 3(2) + 4\left(\frac{1}{8}\right) = 6 + \frac{1}{2} = \frac{13}{2} \). The equation of the tangent is \( y - 15 = \frac{13}{2}(x - 4) \). Multiplying by 2 gives \( 2(y - 15) = 13(x - 4) \), which expands to \( 2y - 30 = 13x - 52 \). Rearranging into the required form gives \( 2y - 13x + 22 = 0 \) (or \( 13x - 2y - 22 = 0 \)).

Part (a) [3 Marks]: M1 for attempting to differentiate at least one term (power decreased by 1, multiplied by original power). A1 for one correct and simplified term. A1 for the fully correct simplified derivative \( 3x^{\frac{1}{2}} + 4x^{-\frac{3}{2}} \). Part (b) [4.5 Marks]: M1 for substituting \( x = 4 \) into the original equation to find the y-coordinate (obtaining \( y = 15 \)). M1 for substituting \( x = 4 \) into their derivative to find the gradient (obtaining \( \frac{13}{2} \)). dM1 (dependent on the previous M mark) for attempting to form the tangent equation using their point and gradient. A1.5 for a correct integer equation in the form \( ay + bx + c = 0 \), such as \( 2y - 13x + 22 = 0 \) or \( 13x - 2y - 22 = 0 \).

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(a) Since \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), the equation becomes: \(3 \left(\frac{\sin \theta}{\cos \theta}\right) \sin \theta = 8\), which simplifies to \(\frac{3 \sin^2 \theta}{\cos \theta} = 8\), so \(3 \sin^2 \theta = 8 \cos \theta\). Using the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\), we have: \(3(1 - \cos^2 \theta) = 8 \cos \theta\), which gives \(3 - 3 \cos^2 \theta = 8 \cos \theta\). Rearranging this gives the required form: \(3 \cos^2 \theta + 8 \cos \theta - 3 = 0\). (b) Let \(y = \cos \theta\), then the quadratic equation is \(3y^2 + 8y - 3 = 0\), which factorises to \((3y - 1)(y + 3) = 0\). This gives \(\cos \theta = \frac{1}{3}\) or \(\cos \theta = -3\). Since \(-1 \le \cos \theta \le 1\), the equation \(\cos \theta = -3\) has no real solutions. For \(\cos \theta = \frac{1}{3}\), the first solution is \(\theta = \arccos(1/3) \approx 1.23\) radians. The second solution in the interval \(0 \le \theta < 2\pi\) is \(\theta = 2\pi - 1.230959... \approx 5.05\) radians.

(a) M1: Substitutes \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) into the given equation. M1: Substitutes the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to form an equation in \(\cos \theta\) only. A1*: Obtains the correct quadratic equation with no errors shown in their working. (b) M1: Attempts to solve the quadratic equation to find at least one value for \(\cos \theta\). A1: Correctly identifies \(\cos \theta = \frac{1}{3}\) and states or implies that \(\cos \theta = -3\) has no solutions. M1: Uses \(\arccos(1/3)\) to find one value of \(\theta\) in radians. A1: \(\theta \approx 1.23\). M1: Subtracts their first angle from \(2\pi\) to find the second solution. A1: \(\theta \approx 5.05\).

(a) The sum of the first two terms is given by \(a + ar = a(1 + r) = 15\). The sum to infinity of the series is given by \(\frac{a}{1 - r} = 27\), which rearranges to \(a = 27(1 - r)\). Substituting this expression for \(a\) into the first equation yields \(27(1 - r)(1 + r) = 15\). This simplifies to \(27(1 - r^2) = 15\), which gives \(1 - r^2 = \frac{15}{27} = \frac{5}{9}\). Thus, \(r^2 = 1 - \frac{5}{9} = \frac{4}{9}\). (b) Since \(r > 0\), we take the positive square root to find \(r = \frac{2}{3}\). Substituting this into the equation for \(a\) gives \(a = 27\left(1 - \frac{2}{3}\right) = 27 \times \frac{1}{3} = 9\). (c) The sum of the first 5 terms is given by \(S_5 = \frac{a(1 - r^5)}{1 - r} = \frac{9\left(1 - (2/3)^5\right)}{1 - 2/3} = 27\left(1 - \frac{32}{243}\right) = 27 \times \frac{211}{243} = \frac{211}{9} \approx 23.44\).

(a) M1: Writes a correct equation for the sum of the first two terms. M1: Writes a correct equation for the sum to infinity. M1: Eliminates \(a\) between the two equations to obtain an equation in \(r\) only. A1*: Correctly simplifies to obtain \(r^2 = \frac{4}{9}\) with no errors shown. (b) M1: Identifies \(r = \frac{2}{3}\) and attempts to find \(a\) using a correct equation. A1: \(a = 9\). (c) M1: Correctly substitutes their values of \(a\) and \(r\) into the sum formula for \(S_5\). A1: \(23.44\) (must be rounded to 2 decimal places).

(a) Writing the terms in index form, we obtain \(\int \left(3x^{1/2} - 4x^{-2} - 2\right) \mathrm{d}x\). Integrating term by term: \(\int 3x^{1/2} \mathrm{d}x = \frac{3x^{3/2}}{3/2} = 2x^{3/2}\). \(\int -4x^{-2} \mathrm{d}x = \frac{-4x^{-1}}{-1} = 4x^{-1} = \frac{4}{x}\). \(\int -2 \mathrm{d}x = -2x\). Adding the constant of integration, we get \(2x^{3/2} + \frac{4}{x} - 2x + C\). (b) Using the result from part (a), the definite integral is evaluated as follows: \(\left[ 2x^{3/2} + \frac{4}{x} - 2x \right]_{1}^{4}\). At \(x = 4\), the value is \(2(4)^{3/2} + \frac{4}{4} - 2(4) = 2(8) + 1 - 8 = 9\). At \(x = 1\), the value is \(2(1)^{3/2} + \frac{4}{1} - 2(1) = 2 + 4 - 2 = 4\). Subtracting the lower limit value from the upper limit value gives \(9 - 4 = 5\).

(a) M1: Expresses \(\sqrt{x}\) as \(x^{1/2}\) or \(\frac{1}{x^2}\) as \(x^{-2}\). M1: Integrates at least one term of the form \(x^n \to \frac{x^{n+1}}{n+1}\). A1: Correct simplified term \(2x^{3/2}\). M1: Correct integration of the second term to get \(\frac{4}{x}\). A1: Fully correct expression with simplified terms and the constant of integration \(+ C\). (b) M1: Substitutes the limits of 4 and 1 into their integrated expression. M1: Evaluates the expression at both limits and subtracts the lower limit value from the upper limit value. A1: Correct exact value of \(5\).

(a) The sum of the first two terms is given by:
\(a + ar = 15 \implies a(1+r) = 15 \quad [1]\)

The sum to infinity is given by:
\(\frac{a}{1-r} = 27 \implies a = 27(1-r) \quad [2]\)

Substitute \([2]\) into \([1]\):
\(27(1-r)(1+r) = 15\)

Using the difference of two squares:
\(27(1-r^2) = 15\)

\(27 - 27r^2 = 15 \implies 27r^2 = 12\) (as required).

(b) From part (a):
\(r^2 = \frac{12}{27} = \frac{4}{9} \implies r = \pm \frac{2}{3}\)

If \(r = \frac{2}{3}\):
\(a = 27\left(1 - \frac{2}{3}\right) = 9\)

If \(r = -\frac{2}{3}\):
\(a = 27\left(1 - \left(-\frac{2}{3}\right)\right) = 45\)

So the possible pairs are \(r = \frac{2}{3}, a = 9\) and \(r = -\frac{2}{3}, a = 45\).

(c) If all terms are positive, we must have \(r > 0\) and \(a > 0\), so we select \(r = \frac{2}{3}\) and \(a = 9\).

Using the sum formula for a geometric series for \(n = 4\):
\(S_4 = \frac{a(1-r^4)}{1-r}\)

\(S_4 = \frac{9\left(1 - \left(\frac{2}{3}\right)^4\right)}{1 - \frac{2}{3}} = \frac{9\left(1 - \frac{16}{81}\right)}{\frac{1}{3}} = 27 \times \frac{65}{81} = \frac{65}{3}\

(a)
- M1: Attempts to use \(a(1+r) = 15\) or \(a+ar = 15\).
- M1: Attempts to use \(\frac{a}{1-r} = 27\) to find an expression for \(a\).
- A1*: Combines both equations correctly to show \(27r^2 = 12\) with no errors.

(b)
- M1: Solves \(27r^2 = 12\) to find at least one value for \(r\).
- A1: Obtains \(r = \pm \frac{2}{3}\) (both positive and negative values required).
- A1: Obtains both corresponding values: \(a = 9\) when \(r = \frac{2}{3}\), and \(a = 45\) when \(r = -\frac{2}{3}\).

(c)
- M1: Selects \(r = \frac{2}{3}\) and \(a = 9\) and substitutes these into the formula for \(S_4\) (or manually adds the first 4 terms: \(9 + 6 + 4 + \frac{8}{3}\)).
- A1.33: Correctly computes the final answer as \(\frac{65}{3}\) (or \(21\frac{2}{3}\)).

(a) Given equation:
\(\log_3(x + 5) - \log_3(2x - 1) = 2 - \log_3(x - 1)\)

Rearranging to group log terms:
\(\log_3(x + 5) + \log_3(x - 1) - \log_3(2x - 1) = 2\)

Applying the laws of logarithms:
\(\log_3\left( \frac{(x+5)(x-1)}{2x-1} \right) = 2\)

Converting from logarithmic form to exponential form:
\(\frac{(x+5)(x-1)}{2x-1} = 3^2\)

\(\frac{x^2 + 4x - 5}{2x-1} = 9\)

\(x^2 + 4x - 5 = 9(2x - 1)\)

\(x^2 + 4x - 5 = 18x - 9\)

\(x^2 - 14x + 4 = 0\) (as required).

(b) Using the quadratic formula for \(x^2 - 14x + 4 = 0\):
\(x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(4)}}{2(1)}\)

\(x = \frac{14 \pm \sqrt{196 - 16}}{2} = \frac{14 \pm \sqrt{180}}{2}\)

Simplifying the radical:
\(\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}\)

So,
\(x = \frac{14 \pm 6\sqrt{5}}{2} = 7 \pm 3\sqrt{5}\)

Since \(x > 1\) is given (necessary for \(\log_3(x-1)\) to be defined):
\(7 - 3\sqrt{5} \approx 7 - 3(2.236) = 0.292 < 1\).

Therefore, we must reject \(x = 7 - 3\sqrt{5}\) as it leads to the logarithm of a negative number.

Thus, the only valid solution is:
\(x = 7 + 3\sqrt{5}\).

(a)
- M1: Rearranges the equation to collect at least two logarithmic terms on one side.
- M1: Correctly applies the addition law of logs: \(\log_3(x+5) + \log_3(x-1) = \log_3((x+5)(x-1))\).
- M1: Correctly applies the subtraction law of logs to write as a single logarithm: \(\log_3\left(\frac{(x+5)(x-1)}{2x-1}\right)\).
- M1: Eliminates logs by rewriting in index form: \(\frac{(x+5)(x-1)}{2x-1} = 3^2\) or \(9\).
- A1*: Expands and simplifies correctly to obtain the given quadratic \(x^2 - 14x + 4 = 0\) with no algebraic errors.

(b)
- M1: Uses a valid method to solve the quadratic equation \(x^2 - 14x + 4 = 0\) (e.g., quadratic formula or completing the square).
- A1: Obtains the two roots \(x = 7 \pm 3\sqrt{5}\).
- B1.33: Rejects \(x = 7 - 3\sqrt{5}\) with a valid reason (e.g., \(7 - 3\sqrt{5} < 1\), which makes \(\log_3(x-1)\) undefined) and states the final solution is \(x = 7 + 3\sqrt{5}\).

(a) Using the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\):
\(6(1 - \sin^2 \theta) - 5\sin \theta = 5\)

\(6 - 6\sin^2 \theta - 5\sin \theta = 5\)

Rearranging all terms to the right-hand side:
\(0 = 6\sin^2 \theta + 5\sin \theta - 1\)

So,
\(6\sin^2 \theta + 5\sin \theta - 1 = 0\) (as required).

(b) Factorising the quadratic equation:
\((6\sin \theta - 1)(\sin \theta + 1) = 0\)

This gives:
\(\sin \theta = \frac{1}{6}\) or \(\sin \theta = -1\)

Case 1: \(\sin \theta = \frac{1}{6}\)
\(\theta = \arcsin\left(\frac{1}{6}\right) \approx 9.594^\circ \approx 9.6^\circ\)

The second solution in the interval \([-180^\circ, 180^\circ]\) is:
\(\theta = 180^\circ - 9.594^\circ = 170.406^\circ \approx 170.4^\circ\)

Case 2: \(\sin \theta = -1\)
In the interval \([-180^\circ, 180^\circ]\):
\(\theta = -90^\circ\)

Combining all solutions, we get:
\(\theta = -90^\circ, 9.6^\circ, 170.4^\circ\)

(a)
- M1: Uses the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to rewrite the equation in terms of \(\sin \theta\) only.
- A1*: Simplifies correctly to show the given quadratic \(6\sin^2 \theta + 5\sin \theta - 1 = 0\).

(b)
- M1: Solves the quadratic equation to find \(\sin \theta = \frac{1}{6}\) and \(\sin \theta = -1\).
- A1: Correctly identifies both values for \(\sin \theta\).
- M1: Attempts to find at least one angle in the correct range from their values of \(\sin \theta\).
- A1: Obtains \(\theta = -90^\circ\).
- A1: Obtains \(\theta = 9.6^\circ\) (or 9.59°).
- A1.33: Obtains \(\theta = 170.4^\circ\) (or 170.41°) and no other extraneous solutions in the range.

(a) The sum of the first two terms of a geometric series is given by:
\[S_2 = a + ar = a(1+r)\]
So, \(a(1+r) = 15\) [Equation 1]

The sum to infinity is given by:
\[S_\infty = \frac{a}{1-r} = 27\]
So, \(a = 27(1-r)\) [Equation 2]

Substitute [Equation 2] into [Equation 1]:
\[27(1-r)(1+r) = 15\]
\[27(1-r^2) = 15\]
\[27 - 27r^2 = 15\]
\[27r^2 = 12\]
which is the required equation.

(b) From part (a):
\[r^2 = \frac{12}{27} = \frac{4}{9}\]
Taking the square root of both sides gives:
\[r = \pm \frac{2}{3}\]

If \(r = \frac{2}{3}\), substitute into [Equation 2]:
\[a = 27\left(1 - \frac{2}{3}\right) = 27\left(\frac{1}{3}\right) = 9\]

If \(r = -\frac{2}{3}\), substitute into [Equation 2]:
\[a = 27\left(1 - \left(-\frac{2}{3}\right)\right) = 27\left(\frac{5}{3}\right) = 45\]

Therefore, the two possible pairs of values are:
\(r = \frac{2}{3}, a = 9\) and \(r = -\frac{2}{3}, a = 45\).

Part (a):
- M1: Uses the formulas for \(S_2\) and \(S_\infty\) to write down two equations in terms of \(a\) and \(r\).
- M1: Eliminates \(a\) to obtain an equation in terms of \(r\) only.
- A1*: Completes the algebraic proof to show that \(27r^2 = 12\) with no errors.

Part (b):
- M1: Solves \(r^2 = \frac{4}{9}\) to find at least one value for \(r\).
- A1: Obtains both \(r = \frac{2}{3}\) and \(r = -\frac{2}{3}\) (or \(\pm\frac{2}{3}\)).
- M1: Substitutes at least one value of \(r\) back into an equation to find a value for \(a\).
- A1: Obtains one correct pair, either \(r = \frac{2}{3}, a = 9\) or \(r = -\frac{2}{3}, a = 45\).
- A1: Obtains both correct pairs.

(a) We use the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\).
Substitute this into the given equation:
\[3(1 - \sin^2 \theta) - 4 \sin \theta = 4\]
\[3 - 3 \sin^2 \theta - 4 \sin \theta = 4\]
Rearranging to make the leading coefficient positive:
\[3 \sin^2 \theta + 4 \sin \theta + 4 - 3 = 0\]
\[3 \sin^2 \theta + 4 \sin \theta + 1 = 0\]
This is the required form.

(b) Factorise the quadratic equation from part (a):
\[(3 \sin \theta + 1)(\sin \theta + 1) = 0\]
This gives two possible cases:
1) \(\sin \theta = -\frac{1}{3}\)
2) \(\sin \theta = -1\)

For \(\sin \theta = -\frac{1}{3}\):
The principal value is \(\theta = \arcsin\left(-\frac{1}{3}\right) \approx -19.47^\circ\).
Within the range \(0 \le \theta < 360^\circ\):
\(\theta = 180^\circ - (-19.47^\circ) = 199.47^\circ \approx 199.5^\circ\)
\(\theta = 360^\circ + (-19.47^\circ) = 340.53^\circ \approx 340.5^\circ\)

For \(\sin \theta = -1\):
Within the range \(0 \le \theta < 360^\circ\):
\(\theta = 270^\circ\)

Therefore, the solutions in the interval \(0 \le \theta < 360^\circ\) are:
\(\theta = 199.5^\circ, 270^\circ, 340.5^\circ\).

Part (a):
- M1: Attempts to use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to write the equation in terms of \(\sin \theta\) only.
- A1*: Correctly simplifies to show \(3 \sin^2 \theta + 4 \sin \theta + 1 = 0\) with no algebraic errors.

Part (b):
- M1: Attempts to factorise or solve the quadratic equation to find values for \(\sin \theta\).
- A1: Obtains \(\sin \theta = -\frac{1}{3}\) and \(\sin \theta = -1\).
- B1: Obtains \(\theta = 270^\circ\) as one solution.
- M1: Uses a correct method to find at least one angle in the correct range for \(\sin \theta = -\frac{1}{3}\).
- A1: Obtains either \(\theta \approx 199.5^\circ\) or \(\theta \approx 340.5^\circ\).
- A1: Obtains both \(\theta \approx 199.5^\circ\) and \(\theta \approx 340.5^\circ\), with no extra solutions in the range.

(a) The radius \(R\) is the distance between the centre \(P(3, -2)\) and the point on the circle \(Q(7, 1)\).
Using the distance formula:
\[R^2 = (7 - 3)^2 + (1 - (-2))^2\]
\[R^2 = 4^2 + 3^2 = 16 + 9 = 25\]
So the radius is \(R = 5\).

The equation of the circle is:
\[(x - 3)^2 + (y + 2)^2 = 25\]

(b) First, find the gradient of the radius \(PQ\):
\[m_{PQ} = \frac{1 - (-2)}{7 - 3} = \frac{3}{4}\]

Since the tangent is perpendicular to the radius at the point of contact \(Q\):
\[m_{\text{tangent}} = -\frac{1}{m_{PQ}} = -\frac{4}{3}\]

The equation of the tangent passing through \(Q(7, 1)\) is:
\[y - 1 = -\frac{4}{3}(x - 7)\]

Multiply through by 3 to clear the fraction:
\[3(y - 1) = -4(x - 7)\]
\[3y - 3 = -4x + 28\]

Rearrange into the form \(ax + by + c = 0\):
\[4x + 3y - 31 = 0\]
(where \(a = 4\), \(b = 3\), \(c = -31\)).

Part (a):
- M1: Attempts to use the distance formula to find \(R^2\) using coordinates of \(P\) and \(Q\).
- A1: Obtains \(R^2 = 25\) (or \(R = 5\)).
- A1: Writes down the correct equation of the circle: \((x - 3)^2 + (y + 2)^2 = 25\) (accept alternative correct expanded forms, but must be fully simplified).

Part (b):
- M1: Attempts to find the gradient of the radius \(PQ\).
- A1: Correct gradient of \(PQ = \frac{3}{4}\).
- M1: Uses the perpendicular gradient rule \(m_1 m_2 = -1\) to find the gradient of the tangent.
- M1: Uses their tangent gradient and coordinates of \(Q\) to write down an equation of the straight line.
- A1: Obtains \(4x + 3y - 31 = 0\) or equivalent integer form (e.g. \(-4x - 3y + 31 = 0\)).

(a) We use the quotient rule with \( u = \ln(2x - 1) \) and \( v = x^2 \). Differentiating both parts with respect to \( x \): \( \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{2x - 1} \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = 2x \). Applying the quotient rule formula: \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{x^2 \left(\frac{2}{2x - 1}\right) - 2x\ln(2x - 1)}{(x^2)^2} = \frac{\frac{2x^2}{2x - 1} - 2x\ln(2x - 1)}{x^4} \). Multiplying the numerator and denominator by \( (2x - 1) \): \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x^2 - 2x(2x - 1)\ln(2x - 1)}{x^4(2x - 1)} \). Factoring out \( x \) from the numerator and cancelling with one \( x \) in the denominator gives the required form: \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x - 2(2x - 1)\ln(2x - 1)}{x^3(2x - 1)} \). (b) At the stationary point of the curve \( C \), \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \). Since \( x > \frac{1}{2} \), the denominator is non-zero, so we set the numerator to zero: \( 2\alpha - 2(2\alpha - 1)\ln(2\alpha - 1) = 0 \). Dividing by 2: \( \alpha - (2\alpha - 1)\ln(2\alpha - 1) = 0 \implies (2\alpha - 1)\ln(2\alpha - 1) = \alpha \implies \ln(2\alpha - 1) = \frac{\alpha}{2\alpha - 1} \). Taking the exponential of both sides: \( 2\alpha - 1 = \exp\left(\frac{\alpha}{2\alpha - 1}\right) \implies 2\alpha = 1 + \exp\left(\frac{\alpha}{2\alpha - 1}\right) \implies \alpha = \frac{1}{2} + \frac{1}{2}\exp\left(\frac{\alpha}{2\alpha - 1}\right) \).

Part (a): M1: Attempts to differentiate \( u = \ln(2x-1) \) to get \( \frac{k}{2x-1} \) (where \( k \) is a constant). A1: Correctly finds \( \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{2x-1} \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = 2x \). M1: Applies the quotient rule correctly with their derivatives. M1: Simplifies the algebraic fraction by obtaining a common denominator. A1: Fully correct simplified expression matching the given answer. Part (b): M1: Sets the numerator of their derivative equal to 0. M1: Isolates the natural logarithm term and correctly applies the exponential function. A1: Fully correct algebraic steps leading to the given expression.

(a) We expand the target expression: \( R\sin(2\theta - \alpha) = R\sin(2\theta)\cos(\alpha) - R\cos(2\theta)\sin(\alpha) \). Equating coefficients with \( 4\sin(2\theta) - 3\cos(2\theta) \) gives: \( R\cos(\alpha) = 4 \) and \( R\sin(\alpha) = 3 \). Squaring and adding these: \( R^2 = 4^2 + 3^2 = 25 \implies R = 5 \). Dividing the equations: \( \tan(\alpha) = \frac{3}{4} = 0.75 \implies \alpha = \arctan(0.75) \approx 0.643501... \) radians. To 3 decimal places, \( \alpha = 0.644 \) radians. (b) Using the result from part (a): \( 5\sin(2\theta - 0.6435) = 2.5 \implies \sin(2\theta - 0.6435) = 0.5 \). Since \( 0 \le \theta < \pi \), the interval for \( 2\theta - 0.6435 \) is \( -0.6435 \le 2\theta - 0.6435 < 5.640 \). The solutions for \( \sin(X) = 0.5 \) in this interval are: \( 2\theta - 0.6435 = \frac{\pi}{6} \approx 0.5236 \) and \( 2\theta - 0.6435 = \frac{5\pi}{6} \approx 2.6180 \). Solving for \( \theta \): 1) \( 2\theta = 0.5236 + 0.6435 = 1.1671 \implies \theta \approx 0.58 \) radians. 2) \( 2\theta = 2.6180 + 0.6435 = 3.2615 \implies \theta \approx 1.63 \) radians.

Part (a): B1: Correctly identifies \( R = 5 \). M1: Uses \( \tan\alpha = 0.75 \) or equivalent. A1: Correctly identifies \( \alpha \approx 0.644 \) (accept 0.643). Part (b): M1: Formulates the equation \( \sin(2\theta - \alpha) = 0.5 \). M1: Finds at least one correct value for the bracketed angle (e.g., \( 0.524 \) or \( 2.62 \)). A1: Obtains both correct values for the bracketed angle. M1: Correctly processes their value of \( \alpha \) to solve for \( \theta \). A1.33: Obtains both \( \theta = 0.58 \) and \( \theta = 1.63 \).

(a) Let \( u = \sqrt{2x - 1} \). Rearranging for \( x \) gives: \( u^2 = 2x - 1 \implies x = \frac{u^2 + 1}{2} \). Differentiating with respect to \( u \) gives: \( 2u \mathrm{d}u = 2 \mathrm{d}x \implies \mathrm{d}x = u \mathrm{d}u \). Changing the limits of integration: When \( x = 1 \), \( u = \sqrt{2(1) - 1} = 1 \). When \( x = 5 \), \( u = \sqrt{2(5) - 1} = 3 \). Substituting these terms into the integral: \( \int_{1}^{5} \frac{x}{\sqrt{2x - 1}} \mathrm{d}x = \int_{1}^{3} \frac{\frac{u^2 + 1}{2}}{u} (u \mathrm{d}u) = \int_{1}^{3} \frac{1}{2}(u^2 + 1) \mathrm{d}u \), as required. (b) Integrating the expression: \( \int_{1}^{3} \frac{1}{2}(u^2 + 1) \mathrm{d}u = \left[ \frac{1}{2} \left( \frac{u^3}{3} + u \right) \right]_{1}^{3} \). Substituting the upper limit \( u = 3 \) gives: \( \frac{1}{2}\left(\frac{27}{3} + 3\right) = 6 \). Substituting the lower limit \( u = 1 \) gives: \( \frac{1}{2}\left(\frac{1}{3} + 1\right) = \frac{2}{3} \). Subtracting the values: \( 6 - \frac{2}{3} = \frac{16}{3} \).

Part (a): M1: Correctly rearranges the substitution to express \( x \) in terms of \( u \). M1: Differentiates to find the relationship between \( \mathrm{d}x \) and \( \mathrm{d}u \). B1: Correctly determines the new limits of integration to be 1 and 3. M1: Performs substitution of all variables into the integral. A1: Achieves the given simplified expression without algebraic errors. Part (b): M1: Integrates the polynomial expression successfully. M1: Correctly substitutes the limits of integration into their integrated function and subtracts. A1.33: Obtains the correct exact final answer of \( \frac{16}{3} \).

(a) We use the quotient rule with \(u = \ln(2x-1)\) and \(v = x^2\). Differentiating both parts: \(\frac{\text{d}u}{\text{d}x} = \frac{2}{2x-1}\) and \(\frac{\text{d}v}{\text{d}x} = 2x\). Applying the quotient rule: \(\frac{\text{d}y}{\text{d}x} = \frac{v \frac{\text{d}u}{\text{d}x} - u \frac{\text{d}v}{\text{d}x}}{v^2} = \frac{x^2 \left(\frac{2}{2x-1}\right) - 2x\ln(2x-1)}{x^4}\). Multiplying the numerator and denominator by \(2x-1\), we get: \(\frac{\text{d}y}{\text{d}x} = \frac{2x^2 - 2x(2x-1)\ln(2x-1)}{x^4(2x-1)} = \frac{2[x - (2x-1)\ln(2x-1)]}{x^3(2x-1)}\). (b) At the stationary point of C, \(\frac{\text{d}y}{\text{d}x} = 0\). Since \(x > \frac{1}{2}\), the denominator is non-zero, so: \(2[x - (2x-1)\ln(2x-1)] = 0 \implies x - (2x-1)\ln(2x-1) = 0 \implies x = (2x-1)\ln(2x-1)\). Dividing by \(2x-1\) gives: \(\ln(2x-1) = \frac{x}{2x-1}\). Taking the exponential of both sides: \(2x-1 = \exp\left(\frac{x}{2x-1}\right) \implies 2x = 1 + \exp\left(\frac{x}{2x-1}\right) \implies x = \frac{1}{2}\left(1 + \exp\left(\frac{x}{2x-1}\right)\right)\).

(a) M1: Attempts to differentiate \(\ln(2x-1)\) using the chain rule to obtain \(\frac{k}{2x-1}\). B1: Correct derivative of \(x^2\) is \(2x\). M1: Applies the quotient rule correctly, with correct signs, or the product rule on \(x^{-2}\ln(2x-1)\). A1: Correct simplified derivative in any equivalent form. (b) M1: Sets their \(\frac{\text{d}y}{\text{d}x} = 0\) to find the stationary point equation. M1: Rearranges the equation to isolate the log term, e.g., \(\ln(2x-1) = \frac{x}{2x-1}\). M1: Eliminates the logarithm by raising both sides to the power of e. A1: Correctly completes the algebraic steps to show the given result with no errors seen (cso).

(a) To integrate \(\int x \sec^2(3x) \, \text{d}x\), we use integration by parts: \(\int u \frac{\text{d}v}{\text{d}x} \, \text{d}x = uv - \int v \frac{\text{d}u}{\text{d}x} \, \text{d}x\). Let \(u = x \implies \frac{\text{d}u}{\text{d}x} = 1\) and let \(\frac{\text{d}v}{\text{d}x} = \sec^2(3x) \implies v = \frac{1}{3}\tan(3x)\). Substituting these in: \(\int x \sec^2(3x) \, \text{d}x = \frac{1}{3}x\tan(3x) - \int \frac{1}{3}\tan(3x) \, \text{d}x\). Since \(\int \tan(3x) \, \text{d}x = \frac{1}{3}\ln|\sec(3x)|\), we have: \(\int x \sec^2(3x) \, \text{d}x = \frac{1}{3}x\tan(3x) - \frac{1}{9}\ln|\sec(3x)| + C\). (b) Multiplying the integral by 9, we have: \(\int_{0}^{\frac{\pi}{12}} 9x \sec^2(3x) \, \text{d}x = [3x\tan(3x) - \ln|\sec(3x)|]_{0}^{\frac{\pi}{12}}\). At \(x = \frac{\pi}{12}\), the value is: \(3\left(\frac{\pi}{12}\right)\tan\left(\frac{\pi}{4}\right) - \ln\left|\sec\left(\frac{\pi}{4}\right)\right| = \frac{\pi}{4}(1) - \ln(\sqrt{2}) = \frac{\pi}{4} - \frac{1}{2}\ln 2\). At \(x = 0\), the value is: \(3(0)\tan(0) - \ln|\sec(0)| = 0 - \ln(1) = 0\). Therefore, the exact value is \(\frac{1}{4}\pi - \frac{1}{2}\ln 2\).

(a) B1: Correctly identifies \(u = x \implies \frac{\text{d}u}{\text{d}x} = 1\). M1: Integrates \(\sec^2(3x)\) to obtain \(k\tan(3x)\) where \(k \neq 0\). A1: Correctly obtains \(v = \frac{1}{3}\tan(3x)\). M1: Applies the integration by parts formula correctly to get \(\frac{1}{3}x\tan(3x) - \int \frac{1}{3}\tan(3x) \, \text{d}x\). A1: Fully correct derivation, including the integration of \(\tan(3x)\) to get \(-\frac{1}{9}\ln|\sec(3x)|\) plus constant \(+ C\). (b) M1: Multiplies by 9 and attempts to substitute the limits \(x = \frac{\pi}{12}\) and \(x = 0\). M1: Applies exact trigonometric values \(\tan(\frac{\pi}{4}) = 1\) and \(\sec(\frac{\pi}{4}) = \sqrt{2}\), and uses logarithm laws to simplify \(\ln(\sqrt{2})\) to \(\frac{1}{2}\ln 2\). A1: Correct final exact expression \(\frac{1}{4}\pi - \frac{1}{2}\ln 2\) (or clearly identified values \(a = \frac{1}{4}\) and \(b = -\frac{1}{2}\)).

(a) Expanding \(R\cos(2\theta + \alpha) = R\cos 2\theta \cos\alpha - R\sin 2\theta \sin\alpha\). Comparing coefficients with \(3\cos 2\theta - 4\sin 2\theta\), we get \(R\cos\alpha = 3\) and \(R\sin\alpha = 4\). Therefore, \(R = \sqrt{3^2 + 4^2} = 5\), and \(\tan\alpha = \frac{4}{3} \implies \alpha = \arctan\left(\frac{4}{3}\right) \approx 0.927295... \approx 0.9273\) radians. Thus, \(3\cos 2\theta - 4\sin 2\theta = 5\cos(2\theta + 0.9273)\). (b) Setting the expression equal to 2: \(5\cos(2\theta + 0.9273) = 2 \implies \cos(2\theta + 0.9273) = 0.4\). Since \(0 \le \theta < \pi\), the interval for \(2\theta + 0.9273\) is \(0.9273 \le 2\theta + 0.9273 < 2\pi + 0.9273 \approx 7.2105\). The solutions for \(2\theta + 0.9273\) are: 1) \(\arccos(0.4) \approx 1.1593\) radians. Solving for \(\theta\): \(2\theta = 1.1593 - 0.9273 = 0.2320 \implies \theta \approx 0.12\) radians (to 2 d.p.). 2) \(2\pi - 1.1593 \approx 5.1239\) radians. Solving for \(\theta\): \(2\theta = 5.1239 - 0.9273 = 4.1966 \implies \theta \approx 2.10\) radians (to 2 d.p.).

(a) B1: \(R = 5\). M1: Uses \(\tan\alpha = \pm \frac{4}{3}\) to find a value of \(\alpha\). A1: \(\alpha \approx 0.9273\) (must be in radians, correct to 4 decimal places). (b) M1: Sets \(R\cos(2\theta + \alpha) = 2\) and finds \(\arccos\left(\frac{2}{R}\right)\) to get a principal value of approximately 1.16 radians. M1: Finds the second value of the angle \((2\theta + \alpha)\) in the correct range, i.e., \(2\pi - 1.1593 \approx 5.12\). M1: Subtracts their \(\alpha\) and divides by 2 to solve for \(\theta\). A1: One correct solution \(\theta \approx 0.12\). A1: Second correct solution \(\theta \approx 2.10\). Deduct 1 mark overall if there are extra solutions within the range.

**(a)** We have \(R\cos(2\theta + \alpha) = R\cos(2\theta)\cos\alpha - R\sin(2\theta)\sin\alpha\).

Comparing this with \(7\cos(2\theta) - 24\sin(2\theta)\):
\(R\cos\alpha = 7\)
\(R\sin\alpha = 24\)

\(R = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\)

\(\tan\alpha = \frac{24}{7} \implies \alpha = \arctan\left(\frac{24}{7}\right) \approx 1.287002... \approx 1.287\) radians (to 3 decimal places).

Thus, \(7\cos(2\theta) - 24\sin(2\theta) = 25\cos(2\theta + 1.287)\).

**(b)** We want to solve:
\(25\cos(2\theta + 1.287) = 12.5 \implies \cos(2\theta + 1.287) = 0.5\)

Since \(0 \le \theta < \pi\), the range for \(2\theta + 1.287\) is:
\(1.287 \le 2\theta + 1.287 < 2\pi + 1.287 \approx 7.570\)

In this range, the solutions for \(\cos(X) = 0.5\) are:
\(X = \frac{5\pi}{3} \approx 5.236\)
\(X = \frac{7\pi}{3} \approx 7.330\)

Case 1:
\(2\theta + 1.287002 = 5.235988 \implies 2\theta = 3.948986 \implies \theta \approx 1.97\) (to 2 d.p.)

Case 2:
\(2\theta + 1.287002 = 7.330383 \implies 2\theta = 6.043381 \implies \theta \approx 3.02\) (to 2 d.p.)

**(a)**
* **M1**: Uses the identity \(\cos(A+B) = \cos A\cos B - \sin A\sin B\) to form equations \(R\cos\alpha = 7\) and \(R\sin\alpha = 24\), or finds \(R = 25\).
* **A1**: Correct value of \(R = 25\).
* **A1**: Correct value of \(\alpha \approx 1.287\) (accept 1.29).

**(b)**
* **M1**: Sets their \(R\cos(2\theta + \alpha) = 12.5\) and solves for \(\cos(2\theta + \alpha) = 0.5\).
* **M1**: Finds at least one correct principal value for \(2\theta + \alpha\) in the appropriate range (e.g. \(\frac{5\pi}{3}\) or \(5.24\)).
* **A1**: One correct value of \(\theta\) (either 1.97 or 3.02).
* **M1**: Employs a correct method to find the second value of \(\theta\) in the range by using \(\frac{7\pi}{3}\) (or equivalent).
* **A1**: Both \(\theta \approx 1.97\) and \(\theta \approx 3.02\) (to 2 d.p.). Deduct 1 mark if extra values in the range are present.

**(a)** Let \(y = \frac{2x+5}{x-3}\).
Rearranging to make \(x\) the subject:
\(y(x-3) = 2x + 5 \implies xy - 3y = 2x + 5\)
\(xy - 2x = 3y + 5 \implies x(y - 2) = 3y + 5\)
\(x = \frac{3y+5}{y-2}\)

Therefore, \(f^{-1}(x) = \frac{3x+5}{x-2}\).

The domain of \(f^{-1}\) is the range of \(f\). Since \(f(x) = \frac{2x+5}{x-3} = 2 + \frac{11}{x-3}\), we have \(f(x) \ne 2\).
So the domain of \(f^{-1}(x)\) is \(x \in \mathbb{R}, x \ne 2\).

**(b)** First find \(f(5)\):
\(f(5) = \frac{2(5) + 5}{5 - 3} = \frac{15}{2} = 7.5\)

Now find \(gf(5) = g(7.5)\):
\(g(7.5) = |7.5 - 1| - 4 = |6.5| - 4 = 6.5 - 4 = 2.5\) (or \(\frac{5}{2}\)).

**(c)** Solve \(g(x) = x - 3\):
\(|x - 1| - 4 = x - 3 \implies |x - 1| = x + 1\)

If \(x \ge 1\):
\(x - 1 = x + 1 \implies -1 = 1\) (no solution)

If \(x < 1\):
\(-(x - 1) = x + 1 \implies -x + 1 = x + 1 \implies 2x = 0 \implies x = 0\)

Since \(0 < 1\), \(x = 0\) is the only solution.

**(a)**
* **M1**: Attempts to change the subject of the formula \(y = \frac{2x+5}{x-3}\) by multiplying both sides by \((x-3)\) and collecting terms in \(x\).
* **A1**: Correct expression for \(f^{-1}(x) = \frac{3x+5}{x-2}\) (or equivalent).
* **B1**: Correct domain: \(x \in \mathbb{R}, x \ne 2\) (accept \(x \ne 2\)).

**(b)**
* **M1**: Attempts to evaluate \(f(5)\) and substitute the result into \(g(x)\).
* **A1**: Correct exact value of \(2.5\) (or \(\frac{5}{2}\)).

**(c)**
* **M1**: Sets up the equation \(|x-1| = x+1\) or equivalent, and attempts to solve at least one branch.
* **M1**: Solves the negative branch \(-(x-1) = x+1\) to find \(x=0\).
* **A1**: Shows that the other branch has no solution and concludes \(x = 0\) is the unique solution.

**(a)** Let \(y = (x-2)^2 e^{2x}\).
Using the product rule with \(u = (x-2)^2\) and \(v = e^{2x}\):
\(\frac{du}{dx} = 2(x-2)\)
\(\frac{dv}{dx} = 2e^{2x}\)

Applying the product rule:
\(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)
\(\frac{dy}{dx} = 2(x-2)^2 e^{2x} + 2(x-2) e^{2x}\)
\(\frac{dy}{dx} = 2(x-2)[(x-2) + 1] e^{2x}\)
\(\frac{dy}{dx} = 2(x-2)(x-1)e^{2x}\)
\(\frac{dy}{dx} = (2x^2 - 6x + 4)e^{2x}\)

So \(a = 2, b = -6, c = 4\).

**(b)** At the stationary points, \(\frac{dy}{dx} = 0\).
Since \(e^{2x} > 0\) for all \(x\), we have:
\(2x^2 - 6x + 4 = 0 \implies 2(x-1)(x-2) = 0\)

This gives \(x = 1\) and \(x = 2\).

When \(x = 1\):
\(y = (1-2)^2 e^{2(1)} = (-1)^2 e^2 = e^2\).
So one stationary point is \((1, e^2)\).

When \(x = 2\):
\(y = (2-2)^2 e^{2(2)} = 0\).
So the other stationary point is \((2, 0)\).

The exact coordinates of the stationary points are \((1, e^2)\) and \((2, 0)\).

**(c)** At \(x = 3\):
\(y = (3-2)^2 e^{2(3)} = e^6\)

The gradient of the tangent at \(x = 3\) is:
\(\frac{dy}{dx}\Big|_{x=3} = (2(3)^2 - 6(3) + 4)e^6 = (18 - 18 + 4)e^6 = 4e^6\)

The gradient of the normal is the negative reciprocal:
\(m = -\frac{1}{4e^6} = -\frac{1}{4}e^{-6}\)

Using the point-slope form \(y - y_1 = m(x - x_1)\):
\(y - e^6 = -\frac{1}{4}e^{-6}(x - 3)\)
\(y = -\frac{1}{4}e^{-6}x + \frac{3}{4}e^{-6} + e^6\)

**(a)**
* **M1**: Applies product rule correctly to differentiate \(y = (x-2)^2 e^{2x}\), with at least one part correctly differentiated.
* **A1**: Obtains a correct unsimplified expression: \(2(x-2)^2 e^{2x} + 2(x-2) e^{2x}\) (or equivalent).
* **A1**: Correctly expands and simplifies to \((2x^2 - 6x + 4)e^{2x}\).

**(b)**
* **M1**: Sets \(\frac{dy}{dx} = 0\) and solves the quadratic equation \(2x^2 - 6x + 4 = 0\) to find \(x = 1\) and \(x = 2\).
* **A1**: Finds one correct stationary point with exact coordinates (either \((1, e^2)\) or \((2, 0)\)).
* **A1**: Finds both correct stationary points: \((1, e^2)\) and \((2, 0)\).

**(c)**
* **M1**: Substitutes \(x = 3\) to find the gradient of the tangent (\(4e^6\)), and takes the negative reciprocal to find the gradient of the normal (\(-\frac{1}{4}e^{-6}\)).
* **A1**: Finds the correct equation of the normal in the form \(y = mx + d\), where \(m = -\frac{1}{4}e^{-6}\) and \(d = \frac{3}{4}e^{-6} + e^6\).

For (a): We find the derivatives with respect to t: \frac{dx}{dt} = 2 - 2\cos(2t) and \frac{dy}{dt} = -4\sin(t). Using the chain rule, \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-4\sin(t)}{2 - 2\cos(2t)}. Substitute t = \frac{\pi}{3}: \frac{dx}{dt} = 2 - 2\cos\left(\frac{2\pi}{3}\right) = 2 - 2\left(-\frac{1}{2}\right) = 3$ and \frac{dy}{dt} = -4\sin\left(\frac{\pi}{3}\right) = -2\sqrt{3}. Thus, \frac{dy}{dx} = -\frac{2\sqrt{3}}{3}.

For (b): At t = \frac{\pi}{3}, the coordinates of the point on the curve are x = 2\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} and y = 4\cos\left(\frac{\pi}{3}\right) = 2. The equation of the tangent is given by y - 2 = -\frac{2\sqrt{3}}{3}\left(x - \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)\right). Multiplying by 3 gives 3y - 6 = -2\sqrt{3}x + \frac{4\sqrt{3}\pi}{3} - 3. Rearranging and multiplying by 3 to clear the fraction yields 6\sqrt{3}x + 9y - 9 - 4\sqrt{3}\pi = 0.

Part (a): M1: Attempt to differentiate x and y with respect to t. A1: Both \frac{dx}{dt} = 2 - 2\cos(2t) and \frac{dy}{dt} = -4\sin(t) correct. M1: Substitute t = \frac{\pi}{3} into \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. A1: Correct exact gradient -\frac{2\sqrt{3}}{3} or equivalent.

Part (b): M1: Substitute t = \frac{\pi}{3} to find the exact coordinates of the point on the curve. M1: Form the equation of the tangent using their gradient and coordinates. A1: Any correct equivalent equation in the form ax + by + c = 0, such as 6\sqrt{3}x + 9y - 9 - 4\sqrt{3}\pi = 0 or 2\sqrt{3}x + 3y - 3 - \frac{4\sqrt{3}\pi}{3} = 0.

For (a): Let u = e^x, then \frac{du}{dx} = e^x \implies dx = \frac{1}{u} \, du. The limits of integration change as follows: when x = 0, u = e^0 = 1; when x = \ln(2), u = e^{\ln(2)} = 2. Substituting these into the integral gives \int_{1}^{2} \frac{u^3}{u^2 + 3u + 2} \cdot \frac{du}{u} = \int_{1}^{2} \frac{u^2}{(u+1)(u+2)} \, du, as required.

For (b): Use algebraic division on the integrand: \frac{u^2}{u^2 + 3u + 2} = 1 - \frac{3u+2}{(u+1)(u+2)}. Expressing the remainder in partial fractions: \frac{3u+2}{(u+1)(u+2)} = \frac{A}{u+1} + \frac{B}{u+2} \implies 3u+2 = A(u+2) + B(u+1). For u = -1, A = -1. For u = -2, B = 4. So the integrand is 1 - \left(\frac{-1}{u+1} + \frac{4}{u+2}\right) = 1 + \frac{1}{u+1} - \frac{4}{u+2}. Integrating gives \left[u + \ln(u+1) - 4\ln(u+2)\right]_{1}^{2} = (2 + \ln(3) - 4\ln(4)) - (1 + \ln(2) - 4\ln(3)) = 1 + 5\ln(3) - \ln(2) - 8\ln(2) = 1 + 5\ln(3) - 9\ln(2) = 1 + \ln(243) - \ln(512) = 1 + \ln\left(\frac{243}{512}\right).

Part (a): M1: Substitute u = e^x and find dx in terms of du. M1: Successfully change the integration limits to 1 and 2. A1: Complete proof showing all intermediate steps.

Part (b): M1: Perform algebraic division to obtain a constant term and a remainder. M1: Decompose the remainder into partial fractions with correct form. A1: Correctly obtain the simplified integrand 1 + \frac{1}{u+1} - \frac{4}{u+2}. M1: Integrate and substitute the limits 1 and 2, using log laws to combine log terms. A1: Correct final answer 1 + \ln\left(\frac{243}{512}\right) or equivalent with a = 1 and b = 243/512.

For (a): Since P lies on l_1, its position vector is \mathbf{p} = \begin{pmatrix} 2+\lambda \\ -1-2\lambda \\ 5+2\lambda \end{pmatrix}. The vector \mathbf{AP} is given by \mathbf{p} - \mathbf{a} = \begin{pmatrix} 2+\lambda \\ -1-2\lambda \\ 5+2\lambda \end{pmatrix} - \begin{pmatrix} 1 \\ -5 \\ 6 \end{pmatrix} = \begin{pmatrix} \lambda+1 \\ 4-2\lambda \\ 2\lambda-1 \end{pmatrix}. Since AP is perpendicular to l_1, the scalar product of \mathbf{AP} and the direction vector of l_1 is zero: \begin{pmatrix} \lambda+1 \\ 4-2\lambda \\ 2\lambda-1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = 0 \implies 1(\lambda+1) - 2(4-2\lambda) + 2(2\lambda-1) = 0 \implies 9\lambda - 9 = 0 \implies \lambda = 1. Substituting \lambda = 1 back gives \mathbf{p} = \begin{pmatrix} 3 \\ -3 \\ 7 \end{pmatrix}.

For (b): The shortest distance is the magnitude of \mathbf{AP} at \lambda = 1. Substituting \lambda = 1 into the expression for \mathbf{AP} yields \mathbf{AP} = \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}. The distance is |\mathbf{AP}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3.

Part (a): M1: Set up the vector \mathbf{AP} in terms of \lambda. M1: Use the perpendicularity condition by setting the dot product of \mathbf{AP} and the direction vector to 0. A1: Solve for \lambda to obtain \lambda = 1. A1: Substitute \lambda = 1 to find the correct position vector of P.

Part (b): M1: Find the vector \mathbf{AP} for their value of \lambda. M1: Use the 3D distance formula to find the magnitude of \mathbf{AP}. A1: Correct distance of 3.

Let \(u = e^x + 1\).
Then
\[\frac{du}{dx} = e^x \implies dx = \frac{du}{e^x} = \frac{du}{u-1}\]
Also, \(e^{2x} = (e^x)^2 = (u-1)^2\).

Substitute these into the integral:
\[\int \frac{e^{2x}}{e^x + 1} \, dx = \int \frac{(u-1)^2}{u} \cdot \frac{du}{u-1} = \int \frac{u-1}{u} \, du\]

Next, change the limits of integration:
- When \(x = 0\), \(u = e^0 + 1 = 2\).
- When \(x = \ln 3\), \(u = e^{\ln 3} + 1 = 3 + 1 = 4\).

Thus, the definite integral becomes:
\[\int_{2}^{4} \frac{u-1}{u} \, du = \int_{2}^{4} \left(1 - \frac{1}{u}\right) \, du\]

Integrate with respect to \(u\):
\[\left[ u - \ln|u| \right]_{2}^{4} = (4 - \ln 4) - (2 - \ln 2)\]

Simplify the expression:
\[= 4 - 2 - \ln 4 + \ln 2\]
\[= 2 - \ln\left(\frac{4}{2}\right)\]
\[= 2 - \ln 2\]

Thus, \(a = 2\) and \(b = 2\), and the exact value is \(2 - \ln 2\).

**M1**: For a complete substitution method, finding \(du = e^x dx\) or equivalent and attempting to replace \(dx\) and \(e^{2x}\).
**A1**: Correct simplified integrand in terms of \(u\): \(\int \frac{u-1}{u} \, du\).
**B1**: Correctly changing the limits of integration to \(u = 2\) and \(u = 4\).
**M1**: Integrates \(\left(1 - \frac{1}{u}\right)\) to obtain \(u - \ln u\) (or \(u - \ln|u|\)).
**A1**: Correct integrated expression \(u - \ln u\).
**M1**: Substitutes their limits \(2\) and \(4\) into their integrated expression.
**A1**: Obtain the final exact answer of \(2 - \ln 2\) (with \(a = 2, b = 2\)).

(a) To find the point of intersection, we equate the \(y\) and \(z\) components of \(l_1\) and \(l_2\):

For the \(y\)-component:
\[-5 + \lambda = -4 - 2\mu \implies \lambda + 2\mu = 1 \quad \text{(Equation 1)}\]

For the \(z\)-component:
\[2 - 2\lambda = -1 + 3\mu \implies 2\lambda + 3\mu = 3 \quad \text{(Equation 2)}\]

From Equation 1, we have \(\lambda = 1 - 2\mu\).

Substitute this into Equation 2:
\[2(1 - 2\mu) + 3\mu = 3\]
\[2 - 4\mu + 3\mu = 3\]
\[2 - \mu = 3 \implies \mu = -1\]

Substitute \(\mu = -1\) back into the expression for \(\lambda\):
\[\lambda = 1 - 2(-1) = 3\]

Thus, \(\lambda = 3\) and \(\mu = -1\).

(b) Now equate the \(x\)-components to find the value of \(a\):
\[4 + 2\lambda = a + \mu\]

Using \(\lambda = 3\) and \(\mu = -1\):
\[4 + 2(3) = a + (-1)\]
\[10 = a - 1 \implies a = 11 \quad \text{(as required)}\]

(c) Substitute \(\lambda = 3\) into the equation for \(l_1\) to find the position vector of \(P\):

\[\mathbf{r}_P = \begin{pmatrix} 4 \\ -5 \\ 2 \end{pmatrix} + 3 \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 + 6 \\ -5 + 3 \\ 2 - 6 \end{pmatrix} = \begin{pmatrix} 10 \\ -2 \\ -4 \end{pmatrix}\]

Alternatively, substituting \(\mu = -1\) into \(l_2\) gives:
\[\mathbf{r}_P = \begin{pmatrix} 11 \\ -4 \\ -1 \end{pmatrix} - 1 \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 10 \\ -2 \\ -4 \end{pmatrix}\]

Therefore, the coordinates of \(P\) are \((10, -2, -4)\).

**(a)**
**M1**: Equating the \(y\) and \(z\) components of \(l_1\) and \(l_2\) to form two equations in \(\lambda\) and \(\mu\).
**A1**: Correct equations: \(\lambda + 2\mu = 1\) and \(2\lambda + 3\mu = 3\) (or equivalent).
**M1**: Solving the simultaneous equations to find a value for \(\lambda\) or \(\mu\).
**A1**: Both \(\lambda = 3\) and \(\mu = -1\).

**(b)**
**M1**: Equates \(x\)-components using their values of \(\lambda\) and \(\mu\) to form an equation for \(a\).
**A1***: Correctly shows \(a = 11\) with clear algebraic steps and no errors.

**(c)**
**M1**: Substitutes their value of \(\lambda\) (or \(\mu\)) into \(l_1\) (or \(l_2\)) to find the position vector of \(P\).
**A1**: Correct coordinates \((10, -2, -4)\) or equivalent vector form.

(a) Differentiate \(x\) and \(y\) with respect to \(\theta\):
\[\frac{dx}{d\theta} = 2 - 2\cos(2\theta)\]
\[\frac{dy}{d\theta} = -4\sin\theta\]

Using the chain rule:
\[\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-4\sin\theta}{2 - 2\cos(2\theta)} = \frac{-4\sin\theta}{2(1 - \cos(2\theta))}\]

Recall the double angle identity for cosine:
\[\cos(2\theta) = 1 - 2\sin^2\theta \implies 1 - \cos(2\theta) = 2\sin^2\theta\]

Substitute this into the expression for \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{-4\sin\theta}{2(2\sin^2\theta)} = \frac{-4\sin\theta}{4\sin^2\theta} = -\frac{1}{\sin\theta} = -\csc\theta \quad \text{(as required)}\]

(b) At the point where \(\theta = \frac{\pi}{3}\):

Calculate the \(x\) and \(y\) coordinates:
\[x = 2\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}\]
\[y = 4\cos\left(\frac{\pi}{3}\right) = 4\left(\frac{1}{2}\right) = 2\]

Calculate the gradient of the tangent, \(m\):
\[m = \left. \frac{dy}{dx} \right|_{\theta=\frac{\pi}{3}} = -\csc\left(\frac{\pi}{3}\right) = -\frac{1}{\sin(\pi/3)} = -\frac{1}{\sqrt{3}/2} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}\]

Using the equation of a straight line \(y - y_1 = m(x - x_1)\):
\[y - 2 = -\frac{2\sqrt{3}}{3} \left( x - \left( \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right) \right)\]
\[y - 2 = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} + \frac{2\sqrt{3}\sqrt{3}}{6}\]

Since
\[\frac{2\sqrt{3}\sqrt{3}}{6} = \frac{2 \times 3}{6} = 1\]

We have:
\[y - 2 = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} + 1\]
\[y + \frac{2\sqrt{3}}{3}x = 3 + \frac{4\sqrt{3}\pi}{9}\]

So \(m = \frac{2\sqrt{3}}{3}\) and \(c = 3 + \frac{4\sqrt{3}\pi}{9}\).

**(a)**
**M1**: Attempting to differentiate \(x\) to find \(\frac{dx}{d\theta}\). Look for \(2 \pm 2\cos(2\theta)\).
**A1**: Correct derivatives \(\frac{dx}{d\theta} = 2 - 2\cos(2\theta)\) and \(\frac{dy}{d\theta} = -4\sin\theta\).
**M1**: Using \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) and substituting a double-angle identity for \(\cos(2\theta)\) into the denominator.
**A1***: Showing \(\frac{dy}{dx} = -\csc\theta\) with complete, correct intermediate steps.

**(b)**
**B1**: Finding the exact values of \(x\) and \(y\) at \(\theta = \frac{\pi}{3}\): \(x = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}\) and \(y = 2\).
**M1**: Substituting \(\theta = \frac{\pi}{3}\) into \(\frac{dy}{dx}\) to obtain the gradient \(m_t = -\frac{2}{\sqrt{3}}\) or equivalent.
**M1**: Applying the straight line equation \(y - y_1 = m_t(x - x_1)\) using their coordinates and gradient.
**A1**: Expressing the tangent equation in the form \(y + mx = c\) (e.g., \(y + \frac{2\sqrt{3}}{3}x = 3 + \frac{4\sqrt{3}\pi}{9}\) or equivalent exact forms).

(a) Separating the variables:

\[\int \frac{1}{y^2} \mathrm{d}y = \int \sin(2x) \mathrm{d}x\]

Integrating both sides:

\[-\frac{1}{y} = -\frac{1}{2}\cos(2x) + k\]

where \(k\) is an arbitrary constant.

Multiplying by \(-1\) and letting \(C = -k\):

\[\frac{1}{y} = \frac{1}{2}\cos(2x) + C\]

(b) Using the boundary conditions \(y = 2\) when \(x = \frac{\pi}{6}\):

\[\frac{1}{2} = \frac{1}{2}\cos\left(2 \cdot \frac{\pi}{6}\right) + C\]

\[\frac{1}{2} = \frac{1}{2}\cos\left(\frac{\pi}{3}\right) + C\]

Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\):

\[\frac{1}{2} = \frac{1}{2}\left(\frac{1}{2}\right) + C \implies \frac{1}{2} = \frac{1}{4} + C \implies C = \frac{1}{4}\]

Substituting \(C = \frac{1}{4}\) back into the general solution:

\[\frac{1}{y} = \frac{1}{2}\cos(2x) + \frac{1}{4}\]

\[\frac{1}{y} = \frac{2\cos(2x) + 1}{4}\]

Taking the reciprocal to express \(y\) in terms of \(\cos(2x)\):

\[y = \frac{4}{2\cos(2x) + 1}\]

(a)
- M1: Separating the variables to obtain \(\int \frac{1}{y^2} \mathrm{d}y = \int \sin(2x) \mathrm{d}x\). (Condone missing integral signs).
- A1: Correct integration of the LHS to obtain \(-\frac{1}{y}\).
- A1: Correct integration of the RHS to obtain \(-\frac{1}{2}\cos(2x)\).
- A1: Inclusion of a constant of integration and rearranging to the form \(\frac{1}{y} = \frac{1}{2}\cos(2x) + C\) or equivalent (e.g. \(\frac{1}{y} = \frac{1}{2}\cos(2x) - k\)).

(b)
- M1: Substitutes \(x = \frac{\pi}{6}\) and \(y = 2\) into their general solution to find the constant \(C\).
- A1: Obtains \(C = \frac{1}{4}\) or equivalent constant value consistent with their general solution.
- M1: Rearranges their equation to make \(y\) the subject.
- A1: Fully correct particular solution: \(y = \frac{4}{2\cos(2x) + 1}\).

(a) Express \(\mathrm{f}(x)\) as \((2 + x)(4 - 3x)^{-\frac{1}{2}}\).

First, expand \((4 - 3x)^{-\frac{1}{2}}\):

\[(4 - 3x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}}\left(1 - \frac{3}{4}x\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1 - \frac{3}{4}x\right)^{-\frac{1}{2}}\]

Using the binomial expansion formula with \(n = -\frac{1}{2}\) and \(u = -\frac{3}{4}x\):

\[\left(1 - \frac{3}{4}x\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{3}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{3}{4}x\right)^2 + \dots\]

\[= 1 + \frac{3}{8}x + \frac{3}{8}\left(\frac{9}{16}x^2\right) + \dots\]

\[= 1 + \frac{3}{8}x + \frac{27}{128}x^2 + \dots\]

Now, multiply by \(\frac{1}{2}\):

\[(4 - 3x)^{-\frac{1}{2}} = \frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2 + \dots\]

Now multiply this expansion by \((2 + x)\):

\[\mathrm{f}(x) = (2 + x)\left(\frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2 + \dots\right)\]

\[= 2\left(\frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2\right) + x\left(\frac{1}{2} + \frac{3}{16}x\right)\]

\[= 1 + \frac{3}{8}x + \frac{27}{128}x^2 + \frac{1}{2}x + \frac{3}{16}x^2\]

Combine the terms:

\[\text{Constant term: } 1\]

\[\text{Coefficient of } x: \frac{3}{8} + \frac{1}{2} = \frac{7}{8}\]

\[\text{Coefficient of } x^2: \frac{27}{128} + \frac{3}{16} = \frac{27}{128} + \frac{24}{128} = \frac{51}{128}\]

Therefore,

\[\mathrm{f}(x) \approx 1 + \frac{7}{8}x + \frac{51}{128}x^2\]

(b) For the expansion of \((1 + u)^n\) to be valid, we require \(|u| < 1\).

Here, \(u = -\frac{3}{4}x\), so:

\[\left|-\frac{3}{4}x\right| < 1 \implies |x| < \frac{4}{3}\]

(a)
- B1: Correctly takes out the factor of \(4^{-\frac{1}{2}}\) or \(\frac{1}{2}\).
- M1: Applies the binomial expansion to \((1 + ku)^{-\frac{1}{2}}\) with at least the \(x\) and \(x^2\) term correct in form (must have correct power of \(k\)).
- A1: Correct simplified expansion of \(\left(1 - \frac{3}{4}x\right)^{-\frac{1}{2}}\) up to \(x^2\), which is \(1 + \frac{3}{8}x + \frac{27}{128}x^2\).
- A1: Correct expansion of \((4 - 3x)^{-\frac{1}{2}}\), which is \(\frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2\).
- M1: Multiplies their expansion by \((2+x)\) to find terms up to \(x^2\).
- A1: Fully correct simplified expression \(1 + \frac{7}{8}x + \frac{51}{128}x^2\).

(b)
- B1: Correct range, accept \(|x| < \frac{4}{3}\) or \(-\frac{4}{3} < x < \frac{4}{3}\).

(a) To find \(\frac{\mathrm{d}y}{\mathrm{d}x}\), first find \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\):

\[\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{t + 2}\]

Using the quotient rule for \(y = \frac{t}{t + 1}\):

\[\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{(t+1)(1) - t(1)}{(t+1)^2} = \frac{1}{(t + 1)^2}\]

Using the chain rule:

\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{\frac{1}{(t + 1)^2}}{\frac{1}{t + 2}} = \frac{t + 2}{(t + 1)^2}\]

(b) At \(t = 0\):

\[x = \ln(0 + 2) = \ln 2\]

\[y = \frac{0}{0 + 1} = 0\]

The gradient of the tangent at \(t = 0\) is:

\[m = \left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{t=0} = \frac{0 + 2}{(0 + 1)^2} = 2\]

Using the equation of a straight line:

\[y - y_1 = m(x - x_1) \implies y - 0 = 2(x - \ln 2) \implies y = 2x - 2\ln 2\]

(c) To find the Cartesian equation, rearrange the equation for \(x\) to make \(t\) the subject:

\[x = \ln(t + 2) \implies \mathrm{e}^x = t + 2 \implies t = \mathrm{e}^x - 2\]

Substitute this expression for \(t\) into the equation for \(y\):

\[y = \frac{\mathrm{e}^x - 2}{(\mathrm{e}^x - 2) + 1} = \frac{\mathrm{e}^x - 2}{\mathrm{e}^x - 1}\]

Comparing this to the required form gives \(a = 2\) and \(b = 1\).

Since \(t > -1\), we have:

\[t + 2 > 1 \implies \ln(t + 2) > \ln(1) \implies x > 0\]

So the domain of the function is \(x > 0\).

(a)
- M1: Differentiates \(x\) with respect to \(t\) to get \(\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{t+2}\).
- M1: Differentiates \(y\) with respect to \(t\) using the quotient rule (or product rule) to get \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{k}{(t+1)^2}\).
- A1: Correctly applies \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\) to obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{t+2}{(t+1)^2}\).

(b)
- M1: Identifies the coordinates of the point at \(t = 0\) as \((\ln 2, 0)\), and substitutes \(t = 0\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the gradient \(m\).
- A1: Obtains \(y = 2x - 2\ln 2\) (accept \(y = 2x - \ln 4\)).

(c)
- M1: Rearranges \(x = \ln(t + 2)\) to make \(t\) the subject, obtaining \(t = \mathrm{e}^x - 2\) (or equivalent).
- A1: Substitutes into \(y\) to find \(y = \frac{\mathrm{e}^x - 2}{\mathrm{e}^x - 1}\).
- B1: Correctly states the domain as \(x > 0\) (or \(x \in \mathbb{R}, x > 0\)).

(a) To find the limits of integration in terms of \(t\), we set \(x = 0 \implies \ln(t+1) = 0 \implies t = 0\), and \(x = \ln 3 \implies \ln(t+1) = \ln 3 \implies t = 2\). Differentiating \(x\) with respect to \(t\), we get \(\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{t+1}\). The area of \(R\) is given by \(\int y \, \mathrm{d}x = \int_{0}^{2} y \frac{\mathrm{d}x}{\mathrm{d}t} \, \mathrm{d}t = \int_{0}^{2} \left(\frac{t}{t+2}\right) \left(\frac{1}{t+1}\right) \, \mathrm{d}t = \int_{0}^{2} \frac{t}{(t+1)(t+2)} \, \mathrm{d}t\). (b) We express the integrand in partial fractions: \(\frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2} \implies t = A(t+2) + B(t+1)\). Setting \(t = -1\) gives \(A = -1\). Setting \(t = -2\) gives \(B = 2\). Thus, the integral becomes \(\int_{0}^{2} \left( \frac{2}{t+2} - \frac{1}{t+1} \right) \mathrm{d}t = \left[ 2\ln(t+2) - \ln(t+1) \right]_{0}^{2}\). Evaluating at the limits: upper limit \(t = 2\) gives \(2\ln 4 - \ln 3\); lower limit \(t = 0\) gives \(2\ln 2 - \ln 1 = 2\ln 2\). Subtracting the lower limit from the upper limit: \((2\ln 4 - \ln 3) - 2\ln 2 = 4\ln 2 - \ln 3 - 2\ln 2 = 2\ln 2 - \ln 3 = \ln 4 - \ln 3 = \ln\left(\frac{4}{3}\right)\).

Part (a): M1: Attempts to find both limits for \(t\) by setting \(\ln(t+1) = 0\) and \(\ln(t+1) = \ln 3\) to obtain \(t = 0\) and \(t = 2\). B1: Correctly differentiates \(x\) to find \(\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{t+1}\). A1*: Completes the proof by substituting \(y\), \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and limits into the parametric integration formula to obtain the given integral. Part (b): M1: Sets up the correct form for partial fractions \(\frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}\) and attempts to find at least one constant. A1: Correctly finds \(A = -1\) and \(B = 2\). M1: Integrates their partial fractions to obtain an expression of the form \(P\ln(t+2) + Q\ln(t+1)\). M1: Substitutes the limits \(2\) and \(0\) and applies laws of logarithms to simplify the expression to a single logarithm. A1: Obtains the correct final answer of \(\ln\left(\frac{4}{3}\right)\) or equivalent single logarithm form.

(a) Since \(\tan \alpha = \frac{3}{4}\), we have \(\sin \alpha = \frac{3}{5} = 0.6\) and \(\cos \alpha = \frac{4}{5} = 0.8\).
Resolving perpendicular to the inclined plane for particle \(A\):
\(R = 3mg \cos \alpha = 3mg(0.8) = 2.4mg\).
Since \(A\) is moving, the friction is limiting:
\(F = \mu R = \frac{1}{6} \times 2.4mg = 0.4mg\) (as required).

(b) Particle \(A\) moves up the plane with acceleration \(a = 0.2g\).
Writing the equation of motion for \(A\) parallel to the plane:
\(T - 3mg \sin \alpha - F = 3ma\)
\(T - 3mg(0.6) - 0.4mg = 3m(0.2g)\)
\(T - 1.8mg - 0.4mg = 0.6mg\)
\(T - 2.2mg = 0.6mg \implies T = 2.8mg\).

(c) Particle \(B\) moves downwards with acceleration \(a = 0.2g\).
Writing the equation of motion for \(B\):
\(kmg - T = kma\)
\(kmg - 2.8mg = km(0.2g)\)
Dividing through by \(mg\):
\(k - 2.8 = 0.2k\)
\(0.8k = 2.8 \implies k = 3.5\).

(a)
M1: Resolving perpendicular to the plane to find \(R\), using \(\cos\alpha = 0.8\).
A1: Correct reaction force \(R = 2.4mg\) (or equivalent).
A1*: Correctly applies \(F = \mu R\) to show \(F = 0.4mg\).

(b)
M1: Writing an equation of motion for \(A\) parallel to the plane with three forces and acceleration term.
A1: Correct unsimplified equation, e.g., \(T - 3mg\sin\alpha - F = 3ma\).
M1: Substituting \(\sin\alpha = 0.6\), \(F = 0.4mg\), and \(a = 0.2g\) into their equation.
A1: \(T = 2.8mg\).

(c)
M1: Writing an equation of motion for \(B\), with difference of gravity and tension equal to mass times acceleration.
A1: Correct equation \(kmg - T = kma\).
M1: Substituting \(T = 2.8mg\) and \(a = 0.2g\) to form an equation in \(k\).
A1: \(k = 3.5\).

(a) When the person of mass \(60\text{ kg}\) stands at \(A\), the beam is on the point of tilting about \(C\).
This implies the reaction force at support \(D\) is zero, so \(R_D = 0\).
Taking moments about \(C\) for the beam:
\(60g \times AC = 30g \times (x - AC)\)
\(60g \times 1.2 = 30g \times (x - 1.2)\)
Divide both sides by \(30g\):
\(2 \times 1.2 = x - 1.2\)
\(2.4 = x - 1.2 \implies x = 3.6\text{ m}\) (as required).

(b) As the person walks past \(D\) towards \(B\), the beam is on the point of tilting about \(D\).
This implies the reaction force at support \(C\) is zero, so \(R_C = 0\).
Let \(d\) be the distance from \(A\) to the person's position.
The distance of support \(D\) from \(A\) is \(6 - 1.5 = 4.5\text{ m}\).
Taking moments about \(D\):
\(30g \times (4.5 - x) = 60g \times (d - 4.5)\)
Using \(x = 3.6\):
\(30g \times (4.5 - 3.6) = 60g \times (d - 4.5)\)
\(30g \times 0.9 = 60g \times (d - 4.5)\)
Divide both sides by \(30g\):
\(0.9 = 2(d - 4.5)\)
\(0.45 = d - 4.5 \implies d = 4.95\text{ m}\).
So the furthest distance from \(A\) is \(4.95\text{ m}\).

(c) The assumption that the person is a particle means that their weight acts at a single point, allowing us to specify their position as a single precise distance from \(A\).

(a)
B1: State or imply that the reaction force at \(D\) is zero (\(R_D = 0\)).
M1: Formulation of a moments equation about \(C\) (or another valid point), with correct terms.
A1: Correct moments equation, e.g., \(60g \times 1.2 = 30g \times (x - 1.2)\).
M1: Solving the linear equation for \(x\).
A1*: Correctly obtains \(x = 3.6\).

(b)
B1: State or imply that the reaction force at \(C\) is zero (\(R_C = 0\)).
M1: Formulation of a moments equation about \(D\) (or another valid point) involving the person's distance \(d\).
A1: Correct moments equation, e.g., \(30g \times 0.9 = 60g \times (d - 4.5)\).
A1: \(d = 4.95\text{ m}\) (or equivalent).

(c)
B1: Explain that the weight acts at a single concentrated point.
B1: Explain that the position/distance of the person can be measured precisely.

(a) Since the force is constant, the acceleration \(\mathbf{a}\) is constant.
Using \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\):
\(\mathbf{a} = \frac{\mathbf{v}(4) - \mathbf{v}(0)}{4}\)
\(\mathbf{a} = \frac{(6\mathbf{i} + 5\mathbf{j}) - (2\mathbf{i} - 3\mathbf{j})}{4} = \frac{4\mathbf{i} + 8\mathbf{j}}{4} = (\mathbf{i} + 2\mathbf{j})\text{ m/s}^2\).

(b) Using Newton's second law \(\mathbf{F} = m\mathbf{a}\):
\(\mathbf{F} = 0.5(\mathbf{i} + 2\mathbf{j}) = (0.5\mathbf{i} + \mathbf{j})\text{ N}\).

(c) Using \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with \(t = 6\):
\(\mathbf{r}(6) = (-3\mathbf{i} + 5\mathbf{j}) + (2\mathbf{i} - 3\mathbf{j})(6) + \frac{1}{2}(\mathbf{i} + 2\mathbf{j})(6^2)\)
\(\mathbf{r}(6) = (-3\mathbf{i} + 5\mathbf{j}) + (12\mathbf{i} - 18\mathbf{j}) + 18(\mathbf{i} + 2\mathbf{j})\)
\(\mathbf{r}(6) = (-3 + 12 + 18)\mathbf{i} + (5 - 18 + 36)\mathbf{j} = (27\mathbf{i} + 23\mathbf{j})\text{ m}\).

(d) The velocity at time \(t\) is:
\(\mathbf{v}(t) = \mathbf{u} + \mathbf{a}t = (2 + t)\mathbf{i} + (-3 + 2t)\mathbf{j}\).
If \(P\) is moving in the direction of \(3\mathbf{i} + 4\mathbf{j}\), then:
\(\mathbf{v}(t) = \lambda (3\mathbf{i} + 4\mathbf{j})\) for some scalar \(\lambda > 0\).
This gives:
\(2 + t = 3\lambda\)
\(-3 + 2t = 4\lambda\)
From the first equation, \(t = 3\lambda - 2\).
Substituting into the second equation:
\(-3 + 2(3\lambda - 2) = 4\lambda\)
\(-3 + 6\lambda - 4 = 4\lambda\)
\(2\lambda = 7 \implies \lambda = 3.5\).
Since \(\lambda = 3.5 > 0\), the particle is moving in the same (not opposite) direction.
At this instant, the velocity is:
\(\mathbf{v} = 3.5(3\mathbf{i} + 4\mathbf{j}) = 10.5\mathbf{i} + 14\mathbf{j}\).
The speed is the magnitude of the velocity:
\(\text{Speed} = \sqrt{10.5^2 + 14^2} = \sqrt{110.25 + 196} = \sqrt{306.25} = 17.5\text{ m/s}\).

(a)
M1: Applies \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) or equivalent to find \(\mathbf{a}\).
A1: Correct numerator \(4\mathbf{i} + 8\mathbf{j}\).
A1: Correct acceleration \(\mathbf{i} + 2\mathbf{j}\) (or \(\begin{pmatrix}1\\2\end{pmatrix}\)).

(b)
M1: Applies \(\mathbf{F} = m\mathbf{a}\).
A1: Correct force \(0.5\mathbf{i} + \mathbf{j}\) (or \(\frac{1}{2}\mathbf{i} + \mathbf{j}\)).

(c)
M1: Applies \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with their \(\mathbf{a}\) and \(t=6\).
A1: Correct substitution of values.
A1: Correct position vector \(27\mathbf{i} + 23\mathbf{j}\).

(d)
M1: Formulates velocity expression \(\mathbf{v} = (2 + t)\mathbf{i} + (-3 + 2t)\mathbf{j}\) and sets up a ratio/equation to find when it is parallel to \(3\mathbf{i} + 4\mathbf{j}\).
A1: Finds \(\lambda = 3.5\) or \(t = 8.5\).
A1: Calculates speed as \(17.5\text{ m/s}\).

Let \( g = 9.8 \text{ m/s}^2 \).
From \( \tan\alpha = \frac{3}{4} \), we find \( \sin\alpha = 0.6 \) and \( \cos\alpha = 0.8 \).

(a) Consider the motion of block \( A \) of mass \( m_A = 3 \text{ kg} \).
As \( B \) accelerates downwards, \( A \) accelerates up the inclined plane with acceleration \( a = 1.4 \text{ m/s}^2 \).

The normal reaction force \( R \) on \( A \) is given by:
\[ R = m_A g \cos\alpha = 3 \times 9.8 \times 0.8 = 23.52 \text{ N} \]

Since \( A \) is in motion, the friction force \( F \) is at its limiting value:
\[ F = \mu R = 0.25 \times 23.52 = 5.88 \text{ N} \]

Applying Newton's second law along the plane for block \( A \) (moving up the plane):
\[ T - m_A g \sin\alpha - F = m_A a \]
\[ T - 3 \times 9.8 \times 0.6 - 5.88 = 3 \times 1.4 \]
\[ T - 17.64 - 5.88 = 4.2 \]
\[ T - 23.52 = 4.2 \]
\[ T = 27.72 \text{ N} \approx 27.7 \text{ N} \text{ (or } 28 \text{ N)} \]

(b) Consider the motion of particle \( B \) of mass \( M \text{ kg} \).
Applying Newton's second law vertically downwards for \( B \):
\[ M g - T = M a \]
\[ M(9.8) - 27.72 = M(1.4) \]
\[ 8.4 M = 27.72 \]
\[ M = 3.3 \]

(c) Particle \( B \) starts from rest and falls \( 2 \text{ m} \) with constant acceleration \( a = 1.4 \text{ m/s}^2 \).
Using \( v^2 = u^2 + 2as \):
\[ v^2 = 0^2 + 2 \times 1.4 \times 2 = 5.6 \]
\[ v = \sqrt{5.6} \approx 2.37 \text{ m/s} \text{ (or } 2.4 \text{ m/s to 2 s.f.)} \]
Since the string remains taut until \( B \) hits the ground, the speed of \( A \) at this instant is also \( 2.37 \text{ m/s} \).

(d) The assumption that the string is inextensible ensures that both block \( A \) and particle \( B \) move with the same magnitude of acceleration (and same speed at any instant).

(a)
- M1: For resolving forces parallel to the inclined plane for block \( A \) with friction \( F = \mu R \) and writing an equation of motion of the form \( T - m_A g \sin\alpha - F = m_A a \).
- A1: Correctly calculating normal reaction \( R = 23.52 \text{ N} \) and friction \( F = 5.88 \text{ N} \).
- A1: Correctly obtaining \( T = 27.72 \text{ N} \) or rounded to \( 27.7 \text{ N} \) (3 s.f.) or \( 28 \text{ N} \) (2 s.f.).

(b)
- M1: Writing the equation of motion for particle \( B \): \( M g - T = M a \).
- A1: Correct substitution of values into the equation of motion: \( M(9.8) - 27.72 = M(1.4) \).
- M1: Solving the linear equation for \( M \).
- A1: Finding \( M = 3.3 \).

(c)
- M1: Using \( v^2 = u^2 + 2as \) with \( u = 0 \), \( a = 1.4 \), and \( s = 2 \).
- A1: Finding \( v = \sqrt{5.6} \approx 2.37 \text{ m/s} \) (accept \( 2.4 \text{ m/s} \)).

(d)
- B1: Explaining that the accelerations of the two bodies are equal in magnitude.
- B1: Stating that this allows the same value of \( a = 1.4 \text{ m/s}^2 \) to be used for both particles.

(a) When the load of \( 3 \text{ kg} \) is at \( A \) and the beam is on the point of tilting about \( C \), the reaction force at the support \( D \) is zero (\( R_D = 0 \)).
Taking moments about \( C \):
\[ 3g \times AC = 9g \times (x - AC) \]
\[ 3g \times 1 = 9g \times (x - 1) \]
Dividing both sides by \( 3g \):
\[ 1 = 3(x - 1) \]
\[ 3x - 3 = 1 \implies x = \frac{4}{3} \approx 1.33 \]

(b) When the load is removed and a load of mass \( W \text{ kg} \) is placed at \( B \), the beam is on the point of tilting about \( D \), so the reaction force at \( C \) is zero (\( R_C = 0 \)).

The distance of the support \( D \) from \( A \) is:
\[ AD = AB - BD = 6 - 1.5 = 4.5 \text{ m} \]

The distance of the centre of mass \( G \) from \( D \) is:
\[ GD = AD - x = 4.5 - \frac{4}{3} = \frac{9}{2} - \frac{4}{3} = \frac{19}{6} \text{ m} \]

Taking moments about \( D \):
\[ 9g \times GD = Wg \times BD \]
\[ 9g \times \frac{19}{6} = Wg \times 1.5 \]
\[ 1.5 W = \frac{3 \times 19}{2} = 28.5 \]
\[ W = \frac{28.5}{1.5} = 19 \]

(c) The load of mass \( W \) is replaced by a load of mass \( 5 \text{ kg} \) at \( B \). Since \( 5 < 19 \), the beam remains in equilibrium resting on both supports \( C \) and \( D \).
Let \( R_C \) and \( R_D \) be the normal reactions at \( C \) and \( D \) respectively.

Taking moments about \( D \) for the entire system:
\[ R_C \times CD + 5g \times BD = 9g \times GD \]
Here, \( CD = AD - AC = 4.5 - 1 = 3.5 \text{ m} \).
\[ R_C \times 3.5 + 5g \times 1.5 = 9g \times \frac{19}{6} \]
\[ 3.5 R_C + 7.5g = 28.5g \]
\[ 3.5 R_C = 21g \]
\[ R_C = 6g = 6 \times 9.8 = 58.8 \text{ N} \]

(a)
- M1: Identifying that \( R_D = 0 \) and writing a moment equation about \( C \).
- A1: Correctly substituted moment equation: \( 3g \times 1 = 9g \times (x - 1) \) (or equivalent).
- M1: Solving for \( x \).
- A1: Finding \( x = \frac{4}{3} \) (accept \( 1.3 \) or \( 1.33 \)).

(b)
- M1: Identifying that \( R_C = 0 \) and calculating the distance \( GD = \frac{19}{6} \text{ m} \) or equivalent.
- A1: Writing a moment equation about \( D \): \( 9g \times GD = Wg \times BD \).
- M1: Correct substitution: \( 9 \times \frac{19}{6} = 1.5W \).
- A1: Finding \( W = 19 \).

(c)
- M1: Setting up a moment equation about \( D \) (or using vertical forces and moments about \( C \)) with both \( R_C \) and \( R_D \) acting.
- A1: Correctly substituting distances and forces: \( 3.5 R_C + 7.5g = 28.5g \).
- A1: Finding \( R_C = 58.8 \text{ N} \) (or \( 59 \text{ N} \) or \( 6g \text{ N} \)).

(a) The position vector of a particle moving with constant velocity is given by \( \mathbf{r} = \mathbf{r}_0 + \mathbf{v}t \).
For ship \( P \):
\[ \mathbf{r}_P = (-3\mathbf{i} + 5\mathbf{j}) + (6\mathbf{i} + 8\mathbf{j})t = (-3 + 6t)\mathbf{i} + (5 + 8t)\mathbf{j} \]

For ship \( Q \):
\[ \mathbf{r}_Q = (12\mathbf{i} - 5\mathbf{j}) + (-4\mathbf{i} + 13\mathbf{j})t = (12 - 4t)\mathbf{i} + (-5 + 13t)\mathbf{j} \]

(b) If \( Q \) is due west of \( P \), then they have the same \( \mathbf{j} \)-component (latitude) and the \( \mathbf{i} \)-component of \( Q \) is less than that of \( P \).
Equating the \( \mathbf{j} \)-components:
\[ 5 + 8t = -5 + 13t \]
\[ 5t = 10 \implies t = 2 \text{ hours} \]

Checking \( \mathbf{i} \)-components at \( t = 2 \):
\[ x_P = -3 + 6(2) = 9 \]
\[ x_Q = 12 - 4(2) = 4 \]
Since \( 4 < 9 \), \( Q \) is indeed due west of \( P \).

(c) The vector \( \vec{PQ} \) representing the displacement from \( P \) to \( Q \) is:
\[ \vec{PQ} = \mathbf{r}_Q - \mathbf{r}_P \]
\[ \vec{PQ} = \left[(12 - 4t) - (-3 + 6t)\right]\mathbf{i} + \left[(-5 + 13t) - (5 + 8t)\right]\mathbf{j} \]
\[ \vec{PQ} = (15 - 10t)\mathbf{i} + (-10 + 5t)\mathbf{j} \]

The square of the distance between them, \( d^2 \), is given by:
\[ d^2 = (15 - 10t)^2 + (5t - 10)^2 \]
To minimize \( d^2 \), we differentiate with respect to \( t \):
\[ \frac{\text{d}(d^2)}{\text{d}t} = 2(15 - 10t)(-10) + 2(5t - 10)(5) \]
\[ = -300 + 200t + 50t - 100 \]
\[ = 250t - 400 \]

Setting the derivative to zero for the minimum:
\[ 250t - 400 = 0 \implies t = 1.6 \text{ hours} \]

Since the second derivative \( \frac{\text{d}^2(d^2)}{\text{d}t^2} = 250 > 0 \), this confirms a minimum.

Substituting \( t = 1.6 \) back into the displacement vector \( \vec{PQ} \):
\[ \vec{PQ} = (15 - 16)\mathbf{i} + (8 - 10)\mathbf{j} = -\mathbf{i} - 2\mathbf{j} \]

The minimum distance is:
\[ d = \sqrt{(-1)^2 + (-2)^2} = \sqrt{5} \approx 2.24 \text{ km} \]

(a)
- B1: Correct position vector for \( P \): \( (-3 + 6t)\mathbf{i} + (5 + 8t)\mathbf{j} \).
- B1: Correct position vector for \( Q \): \( (12 - 4t)\mathbf{i} + (-5 + 13t)\mathbf{j} \).

(b)
- M1: Equating the \( \mathbf{j} \)-components of \( \mathbf{r}_P \) and \( \mathbf{r}_Q \).
- A1: Finding \( t = 2 \).
- A1: Verifying that at \( t = 2 \), the \( \mathbf{i} \)-component of \( Q \) is less than that of \( P \) (or demonstrating understanding of 'due west').

(c)
- M1: Finding the relative displacement vector \( \vec{PQ} = \mathbf{r}_Q - \mathbf{r}_P \).
- A1: Obtaining the correct expression \( (15 - 10t)\mathbf{i} + (5t - 10)\mathbf{j} \).
- M1: Expressing the distance squared \( d^2 \) as a quadratic in \( t \).
- A1: Differentiating \( d^2 \) (or completing the square) and setting to 0 to find the minimum point \( t = 1.6 \).
- M1: Substituting \( t = 1.6 \) back into the distance expression.
- A1: Correctly finding the minimum distance as \( \sqrt{5} \text{ km} \) (or \( 2.24 \text{ km} \) to 3 s.f.).

(a) Let \(a\) be the acceleration of the system and \(T\) be the tension in the string.

For particle \(A\), the vertical forces are in equilibrium, so the normal reaction is:
\[ R = 3mg \]

The frictional force \(F\) acting on \(A\) is at its maximum value:
\[ F = \mu R = \frac{1}{3} (3mg) = mg \]

Applying Newton's second law of motion (\(F = ma\)) for each particle:

For \(A\) (moving horizontally):
\[ T - F = 3ma \implies T - mg = 3ma \quad (1) \]

For \(B\) (moving vertically downwards):
\[ 2mg - T = 2ma \quad (2) \]

Adding equations (1) and (2) gives:
\[ 2mg - mg = 5ma \implies mg = 5ma \]
\[ a = 0.2g \quad \text{(or } 1.96\text{ m s}^{-2}\text{)} \]

(b) Substitute \(a = 0.2g\) into equation (2):
\[ T = 2mg - 2ma = 2mg - 2m(0.2g) = 1.6mg \]

(c) During the first \(1.5\text{ s}\), particle \(A\) accelerates at \(a = 1.96\text{ m s}^{-2}\) from rest (\(u = 0\)):
\[ s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(1.96)(1.5)^2 = 2.205\text{ m} \]

The velocity of both particles at \(t = 1.5\text{ s}\) is:
\[ v = u + at = 0 + (1.96)(1.5) = 2.94\text{ m s}^{-1} \]

After \(B\) hits the floor, the string becomes slack (\(T = 0\)). The only horizontal force acting on \(A\) is the frictional force \(F = mg\).

Applying \(F = ma\) for \(A\) when the string is slack:
\[ -mg = 3ma' \implies a' = -\frac{1}{3}g = -\frac{9.8}{3}\text{ m s}^{-2} \]

Let \(s_2\) be the additional distance traveled by \(A\) before coming to rest:
\[ 0^2 = v^2 + 2a's_2 \implies 0 = 2.94^2 + 2\left(-\frac{9.8}{3}\right)s_2 \]
\[ s_2 = \frac{2.94^2 \times 3}{19.6} = 1.323\text{ m} \]

Total distance traveled by \(A\) is:
\[ s = s_1 + s_2 = 2.205 + 1.323 = 3.528\text{ m} \]

To 3 significant figures, this is \(3.53\text{ m}\) (or \(3.5\text{ m}\) to 2 significant figures).

(a)
- M1: For resolving vertically for \(A\) to find \(R = 3mg\) and using \(F = \mu R\) to get \(F = mg\).
- M1: For setting up the equations of motion for both particles (at least one equation correct).
- A1: Both equations of motion correct: \(T - mg = 3ma\) and \(2mg - T = 2ma\).
- A1: Solving equations to show \(a = 0.2g\) (or equivalent).

(b)
- M1: Attempting to substitute their value of \(a\) back into either equation of motion to find \(T\).
- A1: Correct expression for \(T\) in terms of \(m\) and \(g\) (e.g. \(1.6mg\) or \(\frac{8}{5}mg\)).
- A1: Correctly simplified answer with no rounding errors.

(c)
- M1: Finding the distance \(s_1\) traveled in the first \(1.5\text{ s}\) or finding the velocity \(v\) when \(B\) hits the ground.
- A1: Finding both \(s_1 = 2.205\text{ m}\) and \(v = 2.94\text{ m s}^{-1}\) (allow in terms of \(g\): \(s_1 = 0.225g\), \(v = 0.3g\)).
- M1: Finding the deceleration of \(A\) once the string goes slack (\(a' = -\frac{1}{3}g\)) and using \(v^2 = u^2 + 2as\) to find \(s_2\).
- A1: Adding the two distances to obtain total distance \(s = 3.53\text{ m}\) (accept \(3.5\text{ m}\) or \(3.528\text{ m}\)).

See above

See above

Detailed solution:

(a) When moving up the hill at a constant speed of \(15\text{ m s}^{-1}\):
- The driving force is \(F = \frac{P}{v} = \frac{30000}{15} = 2000\text{ N}\).
- The component of weight down the slope is \(mg \sin\theta = 1200 \times 9.8 \times \frac{1}{14} = 840\text{ N}\).
- Since the speed is constant, the forces along the slope are in equilibrium:
\(F = R + mg \sin\theta \implies 2000 = R + 840 \implies R = 1160\text{ N}\).

(b) When moving down the hill at a speed of \(20\text{ m s}^{-1}\) with a rate of \(24\text{ kW}\):
- The driving force is \(F_d = \frac{24000}{20} = 1200\text{ N}\).
- The component of weight down the slope is \(mg \sin\theta = 840\text{ N}\) (acting down the slope).
- The resistance is \(R = 1160\text{ N}\) (acting up the slope).
- Applying Newton's second law down the slope:
\(F_{\text{net}} = F_d + mg \sin\theta - R = 1200 + 840 - 1160 = 880\text{ N}\).
- Therefore, the acceleration \(a\) is given by:
\(F_{\text{net}} = ma \implies 880 = 1200a \implies a = \frac{880}{1200} = \frac{11}{15} \approx 0.733\text{ m s}^{-2}\) (or \(0.73\text{ m s}^{-2}\) to 2 significant figures).

(a)
- M1: For calculating the driving force \(F = \frac{30000}{15} = 2000\).
- M1: For calculating the weight component \(mg \sin\theta = 1200 \times 9.8 \times \frac{1}{14} = 840\).
- M1: For setting up the equation of motion/equilibrium along the slope: \(F = R + mg \sin\theta\).
- A1: For obtaining the given answer \(R = 1160\) with no errors shown.

(b)
- M1: For calculating the new driving force \(F_d = \frac{24000}{20} = 1200\).
- M1: For setting up the equation of motion down the slope: \(F_d + mg \sin\theta - R = ma\).
- A1: Correctly substituting the values into the equation of motion: \(1200 + 840 - 1160 = 1200a\).
- M1: Solving for \(a\).
- A1: Correct value of \(a = \frac{11}{15} \approx 0.733\text{ m s}^{-2}\) (accept \(0.73\) or \(0.733\)).

Detailed solution:

(a) Let the initial direction of motion of \(A\) be the positive direction.
- Before collision:
\(u_A = 2u\), \(u_B = -u\).
- Let \(v_A\) and \(v_B\) be the velocities of \(A\) and \(B\) respectively after the collision.

Applying Conservation of Linear Momentum:
\(3m(2u) + 2m(-u) = 3m v_A + 2m v_B\)
\(6mu - 2mu = 3m v_A + 2m v_B \implies 4u = 3v_A + 2v_B\) (Equation 1)

Applying Newton's Law of Restitution:
\(v_B - v_A = e(u_A - u_B)\)
\(v_B - v_A = e(2u - (-u)) = 3eu\) (Equation 2)

From Equation 2, \(v_B = v_A + 3eu\).

Substitute \(v_B\) into Equation 1:
\(4u = 3v_A + 2(v_A + 3eu)\)
\(4u = 5v_A + 6eu \implies 5v_A = 4u - 6eu\)
\(v_A = \frac{u}{5}(4 - 6e)\)

Now, substitute \(v_A\) back to find \(v_B\):
\(v_B = \frac{u}{5}(4 - 6e) + 3eu = \frac{u}{5}(4 - 6e + 15e) = \frac{u}{5}(4 + 9e)\).

(b) Since the direction of motion of \(A\) is reversed:
- \(v_A < 0\)
- \(\frac{u}{5}(4 - 6e) < 0 \implies 4 - 6e < 0 \implies 6e > 4 \implies e > \frac{2}{3}\).

Since the coefficient of restitution \(e\) must satisfy \(e \le 1\), the range of possible values of \(e\) is:
\(\frac{2}{3} < e \le 1\).

(a)
- M1: Attempts conservation of linear momentum equation with correct mass and velocity terms.
- A1: Correct momentum equation: \(4u = 3v_A + 2v_B\) (or equivalent).
- M1: Attempts Newton's Law of Restitution equation with correct signs and terms.
- A1: Correct restitution equation: \(v_B - v_A = 3eu\) (or equivalent).
- M1: Solves the simultaneous equations to express \(v_A\) or \(v_B\) in terms of \(u\) and \(e\).
- A1: Correct expression for \(v_A = \frac{u}{5}(4 - 6e)\).
- A1: Correct expression for \(v_B = \frac{u}{5}(4 + 9e)\).

(b)
- M1: Sets up the condition for reversal of direction of motion of \(A\): \(v_A < 0\).
- M1: Solves the inequality \(4 - 6e < 0\) to find \(e > \frac{2}{3}\).
- A1: Recognizes the upper bound of \(e \le 1\).
- A1: Correct final range: \(\frac{2}{3} < e \le 1\).

Detailed solution:

(a) Let us split the lamina into two uniform parts: the rectangle \(OABC\) and the triangle \(CDE\).

1. For the rectangular sheet \(OABC\):
- Area \(A_1 = 12 \times 6 = 72\).
- The coordinates of its centre of mass \(G_1\) are \((6, 3)\).

2. For the triangular sheet \(CDE\) with vertices \(C(0,6)\), \(D(6,6)\), and \(E(3,10)\):
- The base along \(CD\) is of length \(6\).
- The perpendicular height is \(10 - 6 = 4\).
- Area \(A_2 = \frac{1}{2} \times 6 \times 4 = 12\).
- The coordinates of its centroid \(G_2\) are given by:
\(x_2 = \frac{0 + 6 + 3}{3} = 3\)
\(y_2 = \frac{6 + 6 + 10}{3} = \frac{22}{3}\).

3. Combining the two parts to find the centre of mass \(G(\bar{x}, \bar{y})\) of the lamina \(L\):
- Total area \(A = A_1 + A_2 = 72 + 12 = 84\).
- Taking moments about the y-axis:
\(A \bar{x} = A_1 x_1 + A_2 x_2 \implies 84\bar{x} = 72(6) + 12(3) = 432 + 36 = 468\)
\(\bar{x} = \frac{468}{84} = \frac{39}{7}\).
- Taking moments about the x-axis:
\(A \bar{y} = A_1 y_1 + A_2 y_2 \implies 84\bar{y} = 72(3) + 12\left(\frac{22}{3}\right) = 216 + 88 = 304\)
\(\bar{y} = \frac{304}{84} = \frac{76}{21}\).

Thus, the coordinates of the centre of mass of \(L\) are \(\left(\frac{39}{7}, \frac{76}{21}\right)\).

(b) When the lamina is suspended from \(O\), the line \(OG\) is vertical.
- Let \(\theta\) be the angle that the side \(OA\) (which lies along the positive x-axis) makes with the downward vertical \(OG\).
- In the right-angled triangle formed by the coordinates of \(G\):
\(\tan \theta = \frac{\bar{y}}{\bar{x}} = \frac{76/21}{39/7} = \frac{76}{21} \times \frac{7}{39} = \frac{76}{117}\).
- \(\theta = \arctan\left(\frac{76}{117}\right) \approx 33.0076^{\circ}\).
- To the nearest degree, the angle is \(33^{\circ}\).

(a)
- B1: Correct area of rectangle (72) and triangle (12).
- B1: Correct coordinates of the centre of mass of the rectangle: \((6, 3)\).
- M1: For finding the centroid of the triangle \(CDE\).
- A1: Correct coordinates of the centre of mass of the triangle: \((3, \frac{22}{3})\).
- M1: Setting up moment equation about the y-axis to find \(\bar{x}\).
- A1: Showing \(\bar{x} = \frac{39}{7}\) with no errors.
- M1: Setting up moment equation about the x-axis to find \(\bar{y}\).
- A1: Showing \(\bar{y} = \frac{76}{21}\) with no errors.

(b)
- M1: Identifying that \(\tan \theta = \frac{\bar{y}}{\bar{x}}\).
- A1: Correct substitution \(\tan \theta = \frac{76}{117}\).
- A1: Correctly calculating \(\theta \approx 33^{\circ}\) (accept \(33\) or \(33.0\)).

(a)
Using the work-energy principle:
Loss in Gravitational Potential Energy (GPE) = \( m g h = 80 \times 9.8 \times (150 \sin\theta) = 80 \times 9.8 \times 150 \times \frac{1}{14} = 8400 \text{ J} \).
Gain in Kinetic Energy (KE) = \( \frac{1}{2} m (v^2 - u^2) = \frac{1}{2} \times 80 \times (8^2 - 4^2) = 40 \times (64 - 16) = 1920 \text{ J} \).
By the work-energy principle:
Loss in GPE = Gain in KE + Work done against resistance
\( 8400 = 1920 + 150R \)
\( 150R = 6480 \implies R = 43.2 \).

(b)
At the maximum speed of \( 12 \text{ m s}^{-1} \) on the horizontal road, the acceleration is zero, so the driving force \( F \) equals the resistance \( R \):
\( F = R = 43.2 \text{ N} \).
Power \( P = F \times v_{\max} = 43.2 \times 12 = 518.4 \text{ W} \).

(c)
At the instant speed \( v = 10 \text{ m s}^{-1} \), the driving force is:
\( F = \frac{P}{v} = \frac{518.4}{10} = 51.84 \text{ N} \).
Using Newton's second law on the horizontal road:
\( F - R = m a \)
\( 51.84 - 43.2 = 80 a \)
\( 8.64 = 80 a \implies a = 0.108 \text{ m s}^{-2} \).

(a)
M1: Formulating a work-energy equation of the form \( mgh = \frac{1}{2}m(v^2-u^2) + R \times AB \).
A1: Correct loss in GPE (\( 8400 \text{ J} \)) or gain in KE (\( 1920 \text{ J} \)).
A1: Fully correct equation with numerical values substituted.
A1: \( R = 43.2 \) (or 43 with 2 sf).

(b)
M1: Stating or using that at maximum speed, driving force equals resistance, and using \( P = Fv \).
A1: Correct substitution of values.
A1: \( P = 518.4 \) (or 520 with 2 sf).

(c)
M1: Finding the driving force at speed \( 10 \text{ m s}^{-1} \) using \( F = \frac{P}{10} \).
M1: Setting up the equation of motion \( F - R = ma \).
A1: Correct substitution of values.
A1: \( a = 0.108 \text{ m s}^{-2} \) (or 0.11 with 2 sf).

(a)
Acceleration \( \mathbf{a} = \frac{d\mathbf{v}}{dt} \).
\( \mathbf{a} = (6t - 6)\mathbf{i} + 4\mathbf{j} \).
When \( t = 4 \):
\( \mathbf{a} = (6(4) - 6)\mathbf{i} + 4\mathbf{j} = 18\mathbf{i} + 4\mathbf{j} \text{ m s}^{-2} \).

(b)
If \( \mathbf{v} \) is perpendicular to \( (2\mathbf{i} - \mathbf{j}) \), their scalar product is zero:
\( \mathbf{v} \cdot (2\mathbf{i} - \mathbf{j}) = 0 \)
\( 2(3t^2 - 6t) - (4t + 6) = 0 \)
\( 6t^2 - 12t - 4t - 6 = 0 \)
\( 6t^2 - 16t - 6 = 0 \)
\( 3t^2 - 8t - 3 = 0 \)
\( (3t + 1)(t - 3) = 0 \).
Since \( t \ge 0 \), we have \( t = 3 \).

(c)
Position vector \( \mathbf{r} = \int \mathbf{v} \, dt \).
\( \mathbf{r} = (t^3 - 3t^2 + c_1)\mathbf{i} + (2t^2 + 6t + c_2)\mathbf{j} \).
At \( t = 0 \), \( \mathbf{r} = 5\mathbf{i} - 2\mathbf{j} \), so \( c_1 = 5 \) and \( c_2 = -2 \).
Thus, \( \mathbf{r} = (t^3 - 3t^2 + 5)\mathbf{i} + (2t^2 + 6t - 2)\mathbf{j} \).
At \( t = 2 \):
\( \mathbf{r} = (2^3 - 3(2^2) + 5)\mathbf{i} + (2(2^2) + 6(2) - 2)\mathbf{j} = (8 - 12 + 5)\mathbf{i} + (8 + 12 - 2)\mathbf{j} = \mathbf{i} + 18\mathbf{j} \).
Distance from origin \( = \sqrt{1^2 + 18^2} = \sqrt{325} = 5\sqrt{13} \approx 18.0 \text{ m} \).

(a)
M1: Differentiating the velocity vector to find expression for acceleration.
A1: Correct vector expression \( (6t - 6)\mathbf{i} + 4\mathbf{j} \).
A1: Correct acceleration vector \( 18\mathbf{i} + 4\mathbf{j} \text{ m s}^{-2} \) (or magnitude \( 2\sqrt{85} \approx 18.4 \text{ m s}^{-2} \)).

(b)
M1: Setting up the dot product equation \( \mathbf{v} \cdot (2\mathbf{i} - \mathbf{j}) = 0 \).
A1: Obtaining the quadratic equation \( 3t^2 - 8t - 3 = 0 \) (or equivalent).
M1: Solving the quadratic equation for \( t \).
A1: \( t = 3 \) (rejecting \( t = -1/3 \)).

(c)
M1: Integrating \( \mathbf{v} \) to find \( \mathbf{r} \) (at least one term integrated correctly).
A1: Correct integration including constants of integration.
M1: Substituting boundary conditions to find constants, and finding position vector at \( t = 2 \).
A1: \( 5\sqrt{13} \text{ m} \) (or \( 18.0 \text{ m} \)).

(a)
Let the initial velocity of \( A \) be \( 3u \) (positive direction) and the initial velocity of \( B \) be \( -u \) (negative direction).
Let \( v_A \) and \( v_B \) be the velocities of \( A \) and \( B \) after the collision in the positive direction.

By conservation of linear momentum:
\[ (2m)(3u) + (3m)(-u) = 2m v_A + 3m v_B \]
\[ 6mu - 3mu = 2m v_A + 3m v_B \]
\[ 3u = 2v_A + 3v_B \quad \text{--- (Equation 1)} \]

By Newton's law of restitution:
\[ v_B - v_A = e(3u - (-u)) \]
\[ v_B - v_A = 4eu \quad \text{--- (Equation 2)} \]

From Equation 2, \( v_A = v_B - 4eu \).
Substitute this into Equation 1:
\[ 3u = 2(v_B - 4eu) + 3v_B \]
\[ 3u = 5v_B - 8eu \]
\[ 5v_B = 3u + 8eu \]
\[ v_B = \frac{u}{5}(3 + 8e) \]
Since \( u > 0 \) and \( e > 0 \), this is positive, so the speed of \( B \) is indeed \( \frac{u}{5}(3 + 8e) \).

(b)
To find when the direction of motion of \( A \) is reversed, we need its velocity after the collision to be negative:
\( v_A < 0 \).
Using \( v_A = v_B - 4eu \):
\[ v_A = \frac{u}{5}(3 + 8e) - 4eu = \frac{u}{5}(3 + 8e - 20e) = \frac{u}{5}(3 - 12e) \]
We require:
\[ \frac{u}{5}(3 - 12e) < 0 \]
Since \( u > 0 \):
\[ 3 - 12e < 0 \implies 12e > 3 \implies e > \frac{1}{4} \]
Since the coefficient of restitution \( e \) cannot exceed 1, the range of values is:
\[ \frac{1}{4} < e \le 1 \]

(a)
M1: Setting up conservation of linear momentum equation with correct signs.
A1: Correct CLM equation: \( 3u = 2v_A + 3v_B \).
M1: Setting up Newton's law of restitution with correct signs.
A1: Correct NLR equation: \( v_B - v_A = 4eu \).
M1: Solving the simultaneous equations to eliminate \( v_A \).
A1*: Showing the given expression for speed of \( B \) clearly.

(b)
M1: Expressing the velocity of \( A \) after the collision in terms of \( u \) and \( e \).
A1: Correct expression \( v_A = \frac{u}{5}(3 - 12e) \).
M1: Setting up the inequality \( v_A < 0 \).
A1: Obtaining \( e > \frac{1}{4} \).
A1: Final range \( \frac{1}{4} < e \le 1 \) (must include the upper limit \( \le 1 \)).

(a) The mass of the particle is \( m = 0.5 \text{ kg} \). Given \( \tan \theta = \frac{3}{4} \, \) we have \( \sin \theta = 0.6 \) and \( \cos \theta = 0.8 \). The initial kinetic energy at \( A \) is \( \text{KE}_A = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.5 \times 8^2 = 16 \text{ J} \). At \( B \, \) the particle is at rest, so \( \text{KE}_B = 0 \). The vertical height gained is \( h = 4 \sin \theta = 2.4 \text{ m} \). The potential energy gained is \( \text{PE}_B = mgh = 0.5 \times 9.8 \times 2.4 = 11.76 \text{ J} \). The normal reaction is \( R = mg \cos \theta = 0.5 \times 9.8 \times 0.8 = 3.92 \text{ N} \). The frictional force is \( F = \mu R = 3.92 \mu \). The work done against friction is \( W_f = F \times 4 = 15.68 \mu \text{ J} \). By the work-energy principle, \( \text{KE}_A = \text{PE}_B + W_f \, \) which gives \( 16 = 11.76 + 15.68 \mu \, \) so \( 15.68 \mu = 4.24 \, \) yielding \( \mu \approx 0.2704 \). To 2 significant figures, \( \mu = 0.27 \). (b) For the descent from \( B \) to \( A \, \) let the speed of \( P \) at \( A \) be \( v \text{ m s}^{-1} \). By the work-energy principle, \( \text{Loss in PE} = \text{Gain in KE} + \text{Work done against friction} \, \) so \( 11.76 = \frac{1}{2} m v^2 + 15.68 \mu \). Using \( 15.68 \mu = 4.24 \, \) we have \( 11.76 = 0.25 v^2 + 4.24 \, \) which simplifies to \( 0.25 v^2 = 7.52 \, \) so \( v^2 = 30.08 \, \) giving \( v \approx 5.485 \text{ m s}^{-1} \). To 2 significant figures, \( v = 5.5 \text{ m s}^{-1} \). (c) The work done by friction on each leg of the journey is resistive, so the total work done by friction is \( 2 \times W_f = 2 \times 4.24 = 8.48 \text{ J} \). To 2 significant figures, this is \( 8.5 \text{ J} \).

(a) B1: Correct initial Kinetic Energy of \( 16 \text{ J} \). M1: Formulates the normal reaction \( R \) and uses \( F = \mu R \). A1: Correct potential energy gained \( 11.76 \text{ J} \) or correct work done expression \( 15.68 \mu \). M1: Sets up a complete work-energy equation for the ascent. A1: Correct value of \( \mu = 0.27 \) (accept 0.270). (b) M1: Sets up a complete work-energy equation for the descent. A1: Correct intermediate equation, e.g., \( 0.25 v^2 = 7.52 \) or \( v^2 = 30.08 \). A1: Correct speed \( v = 5.5 \text{ m s}^{-1} \) (accept 5.48). (c) M1: Recognizes work done by friction is resistive in both directions and must be summed. M1: Attempts to calculate total work as \( 2 \times 4.24 \) or equivalent. A1: Correct total work done of \( 8.5 \text{ J} \) (accept 8.48).

(a) Since the sum of the probabilities must equal 1:
\(p + q + 0.3 + r = 1 \implies p + q + r = 0.7\) (Equation 1)

Using the formula for the expectation \(\text{E}(X) = \sum x \text{P}(X=x)\):
\(\text{E}(X) = 1(p) + 2(q) + 4(0.3) + 5(r) = 3.1 \implies p + 2q + 5r = 1.9\) (Equation 2)

Using the formula for variance \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\):
\(1.89 = \text{E}(X^2) - 3.1^2 \implies \text{E}(X^2) = 1.89 + 9.61 = 11.5\)

Since \(\text{E}(X^2) = \sum x^2 \text{P}(X=x)\):
\(\text{E}(X^2) = 1^2(p) + 2^2(q) + 4^2(0.3) + 5^2(r) = 11.5 \implies p + 4q + 25r = 6.7\) (Equation 3)

We now solve the system of three equations:
Subtracting Equation 1 from Equation 2:
\(q + 4r = 1.2 \implies q = 1.2 - 4r\)

Subtracting Equation 1 from Equation 3:
\(3q + 24r = 6.0 \implies q + 8r = 2.0\)

Subtracting these two derived equations:
\((q + 8r) - (q + 4r) = 2.0 - 1.2 \implies 4r = 0.8 \implies r = 0.2\)

Substituting \(r = 0.2\) back into \(q = 1.2 - 4r\):
\(q = 1.2 - 0.8 = 0.4\)

Substituting \(q = 0.4\) and \(r = 0.2\) into Equation 1:
\(p = 0.7 - 0.4 - 0.2 = 0.1\)

Thus, \(p = 0.1\), \(q = 0.4\), and \(r = 0.2\).

(b)(i) Using the linear properties of expectation:
\(\text{E}(Y) = \text{E}(3X - 5) = 3\text{E}(X) - 5 = 3(3.1) - 5 = 4.3\)

(b)(ii) Using the properties of variance:
\(\text{Var}(Y) = \text{Var}(3X - 5) = 3^2 \text{Var}(X) = 9(1.89) = 17.01\)

(b)(iii) We want to find \(\text{P}(Y > X)\):
\(\text{P}(3X - 5 > X) = \text{P}(2X > 5) = \text{P}(X > 2.5)\)

Since the possible values of \(X\) are \(1, 2, 4, 5\), the values greater than \(2.5\) are \(4\) and \(5\):
\(\text{P}(X > 2.5) = \text{P}(X=4) + \text{P}(X=5) = 0.3 + r = 0.3 + 0.2 = 0.5\)

**Part (a)** (5 Marks)
- **M1**: Sets up a correct equation for the sum of probabilities, i.e., \(p + q + r = 0.7\).
- **M1**: Sets up a correct equation for the expectation, i.e., \(p + 2q + 5r = 1.9\).
- **M1**: Uses \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\) to correctly find \(\text{E}(X^2) = 11.5\) and forms the equation \(p + 4q + 25r = 6.7\).
- **A1**: Solves the simultaneous equations to find any one of \(p\), \(q\), or \(r\) correctly.
- **A1**: Finds all three correct values: \(p = 0.1\), \(q = 0.4\), and \(r = 0.2\).

**Part (b)** (4.375 Marks)
- **(i) B1**: Correctly calculates \(\text{E}(Y) = 4.3\).
- **(ii) B1**: Correctly calculates \(\text{Var}(Y) = 17.01\).
- **(iii) M1**: Rewrites \(\text{P}(Y > X)\) as \(\text{P}(X > 2.5)\) or equivalent inequality.
- **A1.375**: Obtains the final probability of \(0.5\).

(a) Let \(Z \sim \text{N}(0, 1)\).

From the first given fact, we have:
\(\text{P}(M < 95) = 0.1587 \implies \text{P}\left(Z < \frac{95 - \mu}{\sigma}\right) = 0.1587\)

Since this probability is less than 0.5, the z-value must be negative:
\(\Phi\left(-\frac{95 - \mu}{\sigma}\right) = 1 - 0.1587 = 0.8413\)

Using the standard normal distribution table, \(\Phi(1.00) = 0.8413\), so:
\(-\left(\frac{95 - \mu}{\sigma}\right) = 1.00 \implies \mu - \sigma = 95\) (Equation 1)

From the second given fact, we have:
\(\text{P}(M > 108) = 0.0548 \implies \text{P}\left(Z > \frac{108 - \mu}{\sigma}\right) = 0.0548\)

By symmetry, this means:
\(\text{P}\left(Z < \frac{108 - \mu}{\sigma}\right) = 1 - 0.0548 = 0.9452\)

Using the standard normal distribution table, \(\Phi(1.60) = 0.9452\), so:
\(\frac{108 - \mu}{\sigma} = 1.60 \implies \mu + 1.6\sigma = 108\) (Equation 2)

Subtracting Equation 1 from Equation 2:
\((\mu + 1.6\sigma) - (\mu - \sigma) = 108 - 95\)
\(2.6\sigma = 13 \implies \sigma = 5\) (as required)

Substituting \(\sigma = 5\) back into Equation 1:
\(\mu - 5 = 95 \implies \mu = 100\) (as required)

(b) We want to find \(\text{P}(92 < M < 104)\):
\(\text{P}(92 < M < 104) = \text{P}\left(\frac{92 - 100}{5} < Z < \frac{104 - 100}{5}\right)\)
\(= \text{P}(-1.6 < Z < 0.8)\)
\(= \Phi(0.8) - \Phi(-1.6)\)
\(= \Phi(0.8) - [1 - \Phi(1.6)]\)

From the tables:
\(\Phi(0.8) = 0.7881\)
\(\Phi(1.6) = 0.9452 \implies \Phi(-1.6) = 1 - 0.9452 = 0.0548\)

Thus:
\(\text{P}(-1.6 < Z < 0.8) = 0.7881 - 0.0548 = 0.7333\)

(c) First, find the probability that a single chocolate bar has a mass less than \(92\text{ g}\):
\(p = \text{P}(M < 92) = \text{P}(Z < -1.6) = 0.0548\)

Let \(Y\) be the number of bars out of 3 that have a mass less than \(92\text{ g}\). Then \(Y \sim \text{B}(3, 0.0548)\).
We want \(\text{P}(Y = 2)\):
\(\text{P}(Y = 2) = \binom{3}{2} p^2 (1 - p)^1\)
\(= 3 \times (0.0548)^2 \times (1 - 0.0548)\)
\(= 3 \times 0.00300304 \times 0.9452\)
\(= 0.008515... \approx 0.00852\) (to 3 s.f.)

**Part (a)** (5 Marks)
- **M1**: Standardises 95 with mean and standard deviation, and equates to a z-value (sign can be incorrect for this mark).
- **A1**: Obtains the correct equation \(\mu - \sigma = 95\) (using the correct negative z-value \(-1.00\)).
- **M1**: Standardises 108 with mean and standard deviation, and equates to a positive z-value.
- **A1**: Obtains the correct equation \(\mu + 1.6\sigma = 108\) (using \(1.60\)).
- **A1**: Solves the simultaneous equations to show \(\sigma = 5\) and \(\mu = 100\) (must show clear algebraic steps).

**Part (b)** (2 Marks)
- **M1**: Standardises both 92 and 104 using \(\mu=100\) and \(\sigma=5\), and attempts to find the area between them.
- **A1**: Obtains \(0.7333\) (accept answers in range \([0.733, 0.734]\)).

**Part (c)** (2.375 Marks)
- **M1**: Identifies the probability of one bar being less than \(92\text{ g}\) is \(0.0548\) and applies the binomial probability formula for exactly 2 out of 3.
- **A1.375**: Obtains \(0.00852\) (accept \(0.00851\) to \(0.00853\) depending on rounding).

(a) Let the events be \(A\) (Art), \(D\) (Drama), and \(M\) (Music).
From the information given:
- No student who attends Art also attends Drama, so \(A \cap D = \emptyset\). Thus, the circles for \(A\) and \(D\) do not overlap.
- \(\text{P}(A) = 0.40\)
- \(\text{P}(D) = 0.35\)
- \(\text{P}(M) = 0.30\)
- Of those who attend Music, \(30\%\) also attend Art, which means:
\(\text{P}(A | M) = 0.30 \implies \text{P}(A \cap M) = \text{P}(M) \times \text{P}(A | M) = 0.30 \times 0.30 = 0.09\)
- Of those who attend Music, \(40\%\) also attend Drama, which means:
\(\text{P}(D | M) = 0.40 \implies \text{P}(D \cap M) = \text{P}(M) \times \text{P}(D | M) = 0.30 \times 0.40 = 0.12\)

Now we determine the remaining regions for the Venn diagram:
- Since \(A \cap D = \emptyset\), there is no intersection of all three, so \(\text{P}(A \cap D \cap M) = 0\).
- The region representing 'Art only' is:
\(\text{P}(A \cap M') = \text{P}(A) - \text{P}(A \cap M) = 0.40 - 0.09 = 0.31\)
- The region representing 'Drama only' is:
\(\text{P}(D \cap M') = \text{P}(D) - \text{P}(D \cap M) = 0.35 - 0.12 = 0.23\)
- The region representing 'Music only' is:
\(\text{P}(M \cap A' \cap D') = \text{P}(M) - \text{P}(A \cap M) - \text{P}(D \cap M) = 0.30 - 0.09 - 0.12 = 0.09\)
- The probability of a student attending none of these clubs is:
\(1 - [\text{P}(A \text{ only}) + \text{P}(D \text{ only}) + \text{P}(M \text{ only}) + \text{P}(A \cap M) + \text{P}(D \cap M)]\)
\(= 1 - [0.31 + 0.23 + 0.09 + 0.09 + 0.12] = 1 - 0.84 = 0.16\)

The Venn diagram should show three circles, with \(A\) and \(D\) disjoint, both intersecting \(M\), with the following values inside the regions:
- Inside \(A\) only: \(0.31\)
- Intersection of \(A\) and \(M\): \(0.09\)
- Inside \(M\) only: \(0.09\)
- Intersection of \(D\) and \(M\): \(0.12\)
- Inside \(D\) only: \(0.23\)
- Outside all three circles: \(0.16\)

(b)(i) We want to find the probability that a student attends Drama or Music but not both.
This is represented by the regions:
- Drama only: \(0.23\)
- Music only: \(0.09\)
- Art and Music: \(0.09\) (since they attend Music but not Drama)

Sum of these probabilities:
\(0.23 + 0.09 + 0.09 = 0.41\)
(Alternatively: \(\text{P}(D \cup M) - \text{P}(D \cap M) = [\text{P}(D) + \text{P}(M) - \text{P}(D \cap M)] - \text{P}(D \cap M) = 0.35 + 0.30 - 2(0.12) = 0.41\))

(b)(ii) We want to find the conditional probability \(\text{P}(M | A')\):
\(\text{P}(M | A') = \frac{\text{P}(M \cap A')}{\text{P}(A')}\)

We know:
\(\text{P}(A') = 1 - \text{P}(A) = 1 - 0.40 = 0.60\)
\(\text{P}(M \cap A') = \text{P}(M) - \text{P}(M \cap A) = 0.30 - 0.09 = 0.21\)

Thus:
\(\text{P}(M | A') = \frac{0.21}{0.60} = 0.35\)

**Part (a)** (4 Marks)
- **M1**: Recognises that \(A\) and \(D\) are disjoint (i.e. no overlap on Venn diagram between \(A\) and \(D\)).
- **M1**: Calculates the intersection probabilities correctly: \(\text{P}(A \cap M) = 0.09\) and \(\text{P}(D \cap M) = 0.12\).
- **A1**: Calculates the single-club regions correctly: Art only = \(0.31\), Drama only = \(0.23\), Music only = \(0.09\).
- **A1**: Draws a fully correct Venn diagram including the outside region labeled with \(0.16\) and all probabilities correct.

**Part (b)** (5.375 Marks)
- **(i) M1**: Identifies the correct regions representing 'Drama or Music but not both' (e.g. \(0.23 + 0.09 + 0.09\) or \(\text{P}(D \cup M) - \text{P}(D \cap M)\)).
- **A1**: Obtains \(0.41\).
- **(ii) M1**: Uses the conditional probability formula \(\text{P}(M | A') = \frac{\text{P}(M \cap A')}{\text{P}(A')}\).
- **M1**: Correctly identifies \(\text{P}(A') = 0.60\) or \(\text{P}(M \cap A') = 0.21\).
- **A1.375**: Obtains \(0.35\) (or equivalent fraction \(\frac{7}{20}\)).

(a) Using the property that the sum of probabilities is 1:
\[ a + b + c + 0.1 = 1 \implies a + b + c = 0.9 \quad \text{--- (Equation 1)} \]

Using the definition of the expectation \(\text{E}(X)\):
\[ \text{E}(X) = \sum x \text{P}(X=x) = 2.4 \]
\[ 1(a) + 2(b) + 3(c) + 4(0.1) = 2.4 \implies a + 2b + 3c = 2.0 \quad \text{--- (Equation 2)} \]

Using the definition of the expectation \(\text{E}(X^2)\):
\[ \text{E}(X^2) = \sum x^2 \text{P}(X=x) = 6.4 \]
\[ 1^2(a) + 2^2(b) + 3^2(c) + 4^2(0.1) = 6.4 \implies a + 4b + 9c = 4.8 \quad \text{--- (Equation 3)} \]

Subtracting Equation 1 from Equation 2:
\[ b + 2c = 1.1 \quad \text{--- (Equation 4)} \]

Subtracting Equation 2 from Equation 3:
\[ 2b + 6c = 2.8 \implies b + 3c = 1.4 \quad \text{--- (Equation 5)} \]

Subtracting Equation 4 from Equation 5:
\[ (b + 3c) - (b + 2c) = 1.4 - 1.1 \implies c = 0.3 \]

Substituting \(c = 0.3\) into Equation 4:
\[ b + 2(0.3) = 1.1 \implies b = 0.5 \]

Substituting \(b = 0.5\) and \(c = 0.3\) into Equation 1:
\[ a + 0.5 + 0.3 = 0.9 \implies a = 0.1 \]

So, \(a = 0.1\), \(b = 0.5\), and \(c = 0.3\).

(b) Using the formulas for variance:
\[ \text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 \]
\[ \text{Var}(X) = 6.4 - (2.4)^2 = 6.4 - 5.76 = 0.64 \]

Now, applying \(\text{Var}(g + hX) = h^2 \text{Var}(X)\):
\[ \text{Var}(3 - 2X) = (-2)^2 \text{Var}(X) = 4 \times 0.64 = 2.56 \]

(c) Rearranging the inequality inside the probability:
\[ \text{P}(2X - 3 > 1) = \text{P}(2X > 4) = \text{P}(X > 2) \]
\[ \text{P}(X > 2) = \text{P}(X = 3) + \text{P}(X = 4) = c + 0.1 = 0.3 + 0.1 = 0.4 \]

(a)
- M1: Setting up Equation 1 using \(\sum \text{P}(X=x) = 1\).
- M1: Setting up Equation 2 using the formula for \(\text{E}(X)\).
- M1: Setting up Equation 3 using the formula for \(\text{E}(X^2)\).
- M1: Solving the simultaneous equations systematically to find at least one of the variables \(a\), \(b\), or \(c\).
- A1: Correctly obtaining all three values: \(a = 0.1\), \(b = 0.5\), and \(c = 0.3\).

(b)
- M1: Correct application of \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\) and the scaling rule \(\text{Var}(a + bX) = b^2 \text{Var}(X)\).
- A1: Correct answer of \(2.56\) (or equivalent fraction, e.g., \(\frac{64}{25}\)).

(c)
- M1: Simplifying the inequality inside the probability to \(\text{P}(X > 2)\) and attempting to sum the correct probabilities from their distribution.
- A1: Correct answer of \(0.4\) (or equivalent fraction, e.g., \(\frac{2}{5}\)).

(a) We standardise the given percentages. Let \(Z \sim \text{N}(0, 1)\).

For the lower tail:
\[ \text{P}(L < 48.5) = 0.15 \implies \text{P}\left(Z < \frac{48.5 - \mu}{\sigma}\right) = 0.15 \]
From standard normal tables, \(z = -1.0364\) (or \(-1.04\) from standard tables).
\[ \frac{48.5 - \mu}{\sigma} = -1.0364 \implies \mu - 1.0364\sigma = 48.5 \quad \text{--- (Equation 1)} \]

For the upper tail:
\[ \text{P}(L > 51.5) = 0.10 \implies \text{P}\left(Z > \frac{51.5 - \mu}{\sigma}\right) = 0.10 \implies \text{P}\left(Z < \frac{51.5 - \mu}{\sigma}\right) = 0.90 \]
From standard normal tables, \(z = 1.2816\) (or \(1.28\) from standard tables).
\[ \frac{51.5 - \mu}{\sigma} = 1.2816 \implies \mu + 1.2816\sigma = 51.5 \quad \text{--- (Equation 2)} \]

Subtracting Equation 1 from Equation 2:
\[ (1.2816 + 1.0364)\sigma = 3.0 \implies 2.318\sigma = 3.0 \implies \sigma \approx 1.2942 \]
Thus, to 3 significant figures, \(\sigma = 1.29\).

Substituting \(\sigma = 1.2942\) back into Equation 1:
\[ \mu = 48.5 + 1.0364(1.2942) \approx 49.841 \]
Thus, to 3 significant figures, \(\mu = 49.8\).

(b) We want to find \(\text{P}(49.0 < L < 51.0)\):
\[ \text{P}(49.0 < L < 51.0) = \text{P}\left(\frac{49.0 - 49.841}{1.2942} < Z < \frac{51.0 - 49.841}{1.2942}\right) \]
\[ = \text{P}(-0.6498 < Z < 0.8955) \]
\[ \approx \Phi(0.90) - \Phi(-0.65) \]
\[ = \Phi(0.90) - (1 - \Phi(0.65)) \]
Using tables:
\[ \Phi(0.90) = 0.8159 \quad \text{and} \quad \Phi(0.65) = 0.7422 \]
\[ \text{Probability} = 0.8159 - (1 - 0.7422) = 0.8159 - 0.2578 = 0.5581 \approx 0.558 \]

(a)
- M1: Standardises using \(48.5\), \(\mu\), and \(\sigma\), setting equal to a negative z-value (with range \([-1.05, -1.03]\)).
- B1: Awarded for a correct z-value of \(-1.0364\) (or \(-1.04\)) used or seen.
- M1: Standardises using \(51.5\), \(\mu\), and \(\sigma\), setting equal to a positive z-value (with range \([1.27, 1.29]\)).
- B1: Awarded for a correct z-value of \(1.2816\) (or \(1.28\)) used or seen.
- M1: Eliminates one variable from their two equations to obtain a value for \(\mu\) or \(\sigma\).
- A1: Both \(\mu = 49.8\) and \(\sigma = 1.29\) correct to 3 s.f. (Accept answers resulting from standard z-values of 1.04 and 1.28: \(\sigma \approx 1.29\), \(\mu \approx 49.8\)).

(b)
- M1: Attempts to standardise both \(49.0\) and \(51.0\) using their \(\mu\) and \(\sigma\).
- M1: Correct method to calculate \(\text{P}(z_1 < Z < z_2)\) using standard normal tables, i.e., \(\Phi(z_{\text{upper}}) - [1 - \Phi(|z_{\text{lower}}|)]\).
- A1: Correct probability in the range \([0.557, 0.560]\).

(a) Since \(A\) and \(B\) are independent:
\[ \text{P}(A \cap B) = \text{P}(A) \times \text{P}(B) = 0.4 \times 0.5 = 0.2 \]

Since \(B\) and \(C\) are mutually exclusive:
\[ \text{P}(B \cap C) = 0 \quad \text{and} \quad \text{P}(A \cap B \cap C) = 0 \]

We are given:
\[ \text{P}(A \cap C) = 0.15 \]

Let's determine the remaining probability regions in the Venn diagram:
- Region \(A\) only: \(\text{P}(A) - \text{P}(A \cap B) - \text{P}(A \cap C) = 0.4 - 0.2 - 0.15 = 0.05\).
- Region \(B\) only: \(\text{P}(B) - \text{P}(A \cap B) = 0.5 - 0.2 = 0.3\).
- Region \(C\) only: \(\text{P}(C) - \text{P}(A \cap C) = 0.35 - 0.15 = 0.2\).
- Region outside all three events: \(1 - (0.05 + 0.3 + 0.2 + 0.2 + 0.15) = 1 - 0.9 = 0.1\).

The diagram consists of three circles \(A\), \(B\), and \(C\). Circles \(B\) and \(C\) must not overlap, while circle \(A\) overlaps both \(B\) and \(C\).

(b)(i) Using the Venn diagram or the addition rule:
\[ \text{P}(A' \cap B') = 1 - \text{P}(A \cup B) \]
\[ \text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B) = 0.4 + 0.5 - 0.2 = 0.7 \]
\[ \text{P}(A' \cap B') = 1 - 0.7 = 0.3 \]
Alternatively, summing regions outside both \(A\) and \(B\) from the Venn diagram:
\[ \text{P}(C \text{ only}) + \text{P}(\text{Outside}) = 0.2 + 0.1 = 0.3 \]

(b)(ii) Using the conditional probability formula:
\[ \text{P}(A \cup C' | B') = \frac{\text{P}((A \cup C') \cap B')}{\text{P}(B')} \]

We have:
\[ \text{P}(B') = 1 - \text{P}(B) = 1 - 0.5 = 0.5 \]

The intersection \((A \cup C') \cap B'\) represents outcomes that are not in \(B\) AND are either in \(A\) or not in \(C\).
Looking at the Venn diagram, the regions not in \(B\) are:
- \(A\) only: \(0.05\) (This is in \(A\), so it satisfies the condition)
- \(A \cap C\): \(0.15\) (This is in \(A\), so it satisfies the condition)
- \(C\) only: \(0.2\) (This is in \(C\), so it does not satisfy \(C'\); it is also not in \(A\). Thus, it is excluded)
- Outside all: \(0.1\) (This is in \(C'\) and not in \(B\), so it satisfies the condition)

Thus, the sum of these valid regions is:
\[ \text{P}((A \cup C') \cap B') = 0.05 + 0.15 + 0.1 = 0.3 \]

Therefore:
\[ \text{P}(A \cup C' | B') = \frac{0.3}{0.5} = 0.6 \]

(a)
- M1: Shows awareness that \(\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B) = 0.2\) due to independence.
- M1: Draws a Venn diagram with three circles where \(B\) and \(C\) do not intersect, and \(A\) intersects both \(B\) and \(C\).
- A1: Correctly places intersection probabilities \(0.2\) for \(A \cap B\) and \(0.15\) for \(A \cap C\).
- A1: Correctly calculates individual-event regions: \(A\) only = \(0.05\), \(B\) only = \(0.3\), and \(C\) only = \(0.2\).
- A1: Correctly determines the outside probability as \(0.1\).

(b)(i)
- M1: Expresses \(\text{P}(A' \cap B')\) as \(1 - \text{P}(A \cup B)\) or sums the correct regions from their diagram.
- A1: Correct answer of \(0.3\) (or equivalent fraction).

(b)(ii)
- M1: Correct application of the conditional probability formula, with a denominator of \(\text{P}(B') = 0.5\) (or their \(1 - \text{P}(B)\)).
- A1: Correct answer of \(0.6\) (or equivalent fraction, e.g., \(\frac{3}{5}\)).

(a) To fill in the Venn diagram, we start from the center and work outwards:
- \(P(A \cap B \cap C) = 0.08\)
- Only \(A\) and \(B\): \(P(A \cap B) - 0.08 = 0.25 - 0.08 = 0.17\)
- Only \(A\) and \(C\): \(P(A \cap C) - 0.08 = 0.20 - 0.08 = 0.12\)
- Only \(B\) and \(C\): \(P(B \cap C) - 0.08 = 0.15 - 0.08 = 0.07\)
- Only \(A\): \(P(A) - 0.17 - 0.12 - 0.08 = 0.55 - 0.37 = 0.18\)
- Only \(B\): \(P(B) - 0.17 - 0.07 - 0.08 = 0.40 - 0.32 = 0.08\)
- Only \(C\): \(P(C) - 0.12 - 0.07 - 0.08 = 0.45 - 0.27 = 0.18\)
- Outside all three circles: \(1 - (0.18 + 0.08 + 0.18 + 0.17 + 0.12 + 0.07 + 0.08) = 1 - 0.88 = 0.12\)

(b) The probability of taking at least two modules is:
\(P(\text{at least two}) = 0.17 + 0.12 + 0.07 + 0.08 = 0.44\)

(c) We want \(P(B' | A) = \frac{P(A \cap B')}{P(A)}\).
From the diagram, \(P(A \cap B') = P(\text{only } A) + P(\text{only } A \text{ and } C) = 0.18 + 0.12 = 0.30\).
Alternatively, \(P(A \cap B') = P(A) - P(A \cap B) = 0.55 - 0.25 = 0.30\).
Thus, \(P(B' | A) = \frac{0.30}{0.55} = \frac{6}{11} \approx 0.545\).

(d) Two events are independent if \(P(A \cap C) = P(A) \times P(C)\).
\(P(A) \times P(C) = 0.55 \times 0.45 = 0.2475\)
Since \(P(A \cap C) = 0.20 \neq 0.2475\), the events \(A\) and \(C\) are not independent.

(a)
M1: Attempt to use the intersection value of 0.08 to find at least two of the two-event-only regions (e.g., 0.17, 0.12, 0.07).
A1: Correct values for all interior regions of the circles (0.18, 0.08, 0.18, 0.17, 0.12, 0.07, 0.08).
A1: Correct label and value for the region outside the circles (0.12), with all eight probabilities summing to 1.

(b)
M1: For summing the four relevant intersection regions: \(0.17 + 0.12 + 0.07 + 0.08\) (or equivalent formula).
A1: Correct answer of 0.44.

(c)
M1: A correct ratio formula for conditional probability with their values, i.e., \(\frac{P(A \cap B')}{P(A)}\).
A1: Correct answer of \(\frac{6}{11}\) (or 0.545 or 0.55 with 3 sig fig).

(d)
M1: Correctly calculating \(P(A) \times P(C)\) and comparing it to \(P(A \cap C)\).
A1: Clearly stating that they are not independent with a correct numerical comparison.

(a) The expectation of \(X\) is given by:
\(E(X) = \sum x P(X = x)\)
\(E(X) = -2(a) - 1(b) + 0(0.3) + 1(b) + 2(c)\)
\(E(X) = -2a - b + b + 2c = 2c - 2a\)
Given that \(E(X) = 0.2\):
\(2c - 2a = 0.2 \implies c - a = 0.1\) (as required).

(b) The sum of the probabilities must equal 1:
\(a + b + 0.3 + b + c = 1 \implies a + 2b + c = 0.7\) [Equation 1]

Using the variance formula \(Var(X) = E(X^2) - [E(X)]^2\):
\(1.56 = E(X^2) - (0.2)^2\)
\(E(X^2) = 1.56 + 0.04 = 1.60\)

We can express \(E(X^2)\) as:
\(E(X^2) = (-2)^2(a) + (-1)^2(b) + 0^2(0.3) + 1^2(b) + 2^2(c)\)
\(E(X^2) = 4a + b + b + 4c = 4a + 2b + 4c = 1.60\)
Dividing by 2 gives:
\(2a + b + 2c = 0.8\) [Equation 2]

From part (a), we have \(c = a + 0.1\). Substituting this into Equation 1 and Equation 2:
1) \(a + 2b + a + 0.1 = 0.7 \implies 2a + 2b = 0.6 \implies a + b = 0.3 \implies b = 0.3 - a\)
2) \(2a + b + 2(a + 0.1) = 0.8 \implies 4a + b = 0.6\)

Substituting \(b = 0.3 - a\) into the simplified Equation 2:
\(4a + 0.3 - a = 0.6 \implies 3a = 0.3 \implies a = 0.1\)

Then, \(b = 0.3 - 0.1 = 0.2\) and \(c = 0.1 + 0.1 = 0.2\).
Thus, \(a = 0.1\), \(b = 0.2\), and \(c = 0.2\).

(c) Using the properties of variance:
\(Var(3 - 2X) = (-2)^2 Var(X) = 4 Var(X)\)
\(Var(3 - 2X) = 4 \times 1.56 = 6.24\).

(d) We want to find:
\(P(X \le 0 | X \ge -1) = \frac{P(-1 \le X \le 0)}{P(X \ge -1)}\)
\(P(-1 \le X \le 0) = P(X = -1) + P(X = 0) = b + 0.3 = 0.2 + 0.3 = 0.5\)
\(P(X \ge -1) = P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) = 0.2 + 0.3 + 0.2 + 0.2 = 0.9\)
So,
\(P(X \le 0 | X \ge -1) = \frac{0.5}{0.9} = \frac{5}{9} \approx 0.556\).

(a)
M1: Writing a correct algebraic expression for \(E(X)\) in terms of \(a, b, c\).
A1: Show clearly that this leads to \(c - a = 0.1\).

(b)
M1: Using the sum of probabilities equal to 1 to write a correct equation in \(a, b, c\).
M1: Using \(Var(X) = E(X^2) - [E(X)]^2\) to find \(E(X^2) = 1.60\) and writing an equation for \(E(X^2)\).
M1: Setting up and solving a system of simultaneous equations to find at least one constant.
A1: Finding all three values correctly: \(a = 0.1, b = 0.2, c = 0.2\).

(c)
M1: Stating or using \(Var(3 - 2X) = 4 Var(X)\).
A1: Correct answer of 6.24.

(d)
M1: Writing a correct conditional probability statement, identifying the correct numerator and denominator.
A1: Correct answer of \(\frac{5}{9}\) or any equivalent fraction or decimal (to at least 3 sig fig).

(a) Let \(X\) be the number of emergency calls received in a 2-hour period. Under \(H_0\), \(X \sim \text{Po}(9)\). Hypotheses: \(H_0: \lambda = 9\) (or \(\mu = 4.5\) per hour) and \(H_1: \lambda < 9\) (or \(\mu < 4.5\) per hour). We calculate \(P(X \le 4)\) using cumulative Poisson tables for \(\lambda = 9\): \(P(X \le 4) = 0.0550\). Since \(0.0550 > 0.05\), the result is not in the critical region. We fail to reject \(H_0\). There is insufficient evidence at the 5% level of significance to support the manager's belief that the mean rate of emergency calls has decreased. (b) Let \(Y\) be the number of emergency calls in a 4-hour period. Under \(H_0\), \(Y \sim \text{Po}(18)\) since the rate is \(4 \times 4.5 = 18\). We require the largest critical value \(c\) such that \(P(Y \le c) \le 0.05\). From cumulative Poisson tables for \(\lambda = 18\): \(P(Y \le 10) = 0.0301\) and \(P(Y \le 11) = 0.0652\). Since \(P(Y \le 10) \le 0.05\) and \(P(Y \le 11) > 0.05\), the critical region is \(Y \le 10\). (c) The actual significance level of the test is the probability of rejecting \(H_0\) when it is true, which is \(P(Y \le 10) = 0.0301\) or 3.01%.

(a) B1: Both hypotheses stated correctly in terms of the rate parameter (\(\lambda\) or \(\mu\)). M1: Recognising the test statistic follows \(X \sim \text{Po}(9)\) and attempting to evaluate \(P(X \le 4)\). A1: Correct probability \(P(X \le 4) = 0.0550\). M1: Comparing their probability to 0.05 (or comparing the test statistic 4 to the critical value 3). A1: Correct non-contextual decision followed by a correct contextual conclusion. (b) M1: Realising the model for a 4-hour period is \(Y \sim \text{Po}(18)\). M1: Attempting to find cumulative probabilities near 0.05 from Poisson tables for \(\lambda = 18\). A1: Finding \(P(Y \le 10) = 0.0301\). A1: Finding \(P(Y \le 11) = 0.0652\). A1: Deducing the correct critical region \(Y \le 10\) (or \(0 \le Y \le 10\)). (c) B1: Correct actual significance level of 0.0301 or 3.01% (allow awrt 0.030).

(a) Since \(f(x)\) is a probability density function, the total area under the curve is 1: \(\int_{0}^{3} kx(4-x) \, dx + \int_{3}^{4} k(12-3x) \, dx = 1\). Evaluating the first integral: \(\int_{0}^{3} (4x-x^2) \, dx = \left[ 2x^2 - \frac{x^3}{3} \right]_0^3 = (18-9) - 0 = 9\). Evaluating the second integral: \(\int_{3}^{4} (12-3x) \, dx = \left[ 12x - \frac{3x^2}{2} \right]_3^4 = (48-24) - (36-13.5) = 24 - 22.5 = 1.5\). Thus, \(k(9 + 1.5) = 1 \implies 10.5k = 1 \implies k = \frac{2}{21}\) (as required). (b) For \(x < 0\), \(F(x) = 0\). For \(0 \le x \le 3\): \(F(x) = \int_{0}^{x} \frac{2}{21}(4t-t^2) \, dt = \frac{2}{21}\left[ 2t^2 - \frac{t^3}{3} \right]_0^x = \frac{2}{21}\left(2x^2 - \frac{x^3}{3}\right)\). Note that \(F(3) = \frac{2}{21}(18-9) = \frac{18}{21} = \frac{6}{7}\). For \(3 < x \le 4\): \(F(x) = F(3) + \int_{3}^{x} \frac{2}{21}(12-3t) \, dt = \frac{6}{7} + \frac{2}{21}\left[ 12t - \frac{3t^2}{2} \right]_3^x = \frac{18}{21} + \frac{2}{21}\left( (12x - 1.5x^2) - (36 - 13.5) \right) = \frac{18}{21} + \frac{24x - 3x^2 - 45}{21} = \frac{-3x^2 + 24x - 27}{21} = \frac{-x^2 + 8x - 9}{7}\). For \(x > 4\), \(F(x) = 1\). Thus, \(F(x) = \begin{cases} 0 & x < 0 \\ \frac{2}{21}\left(2x^2 - \frac{x^3}{3}\right) & 0 \le x \le 3 \\ \frac{1}{7}(-x^2 + 8x - 9) & 3 < x \le 4 \\ 1 & x > 4 \end{cases}\). (c) \(P(1 < X < 3.5) = F(3.5) - F(1)\). Using the CDF: \(F(3.5) = \frac{1}{7}(-(3.5)^2 + 8(3.5) - 9) = \frac{1}{7}(-12.25 + 28 - 9) = \frac{6.75}{7} = \frac{27}{28}\). \(F(1) = \frac{2}{21}(2(1)^2 - \frac{1^3}{3}) = \frac{2}{21}(\frac{5}{3}) = \frac{10}{63}\). \(P(1 < X < 3.5) = \frac{27}{28} - \frac{10}{63} = \frac{243}{252} - \frac{40}{252} = \frac{203}{252} = \frac{29}{36} \approx 0.806\).

(a) M1: Correct setup of total area integral equal to 1 with splitting at \(x=3\). A1: Correctly integrating the first part to obtain \(9k\). A1: Correctly integrating the second part to obtain \(1.5k\). A1* cso: Showing \(k = \frac{2}{21}\) with no errors shown. (b) M1: Integrating the first interval with a variable upper limit. A1: Correct expression for \(0 \le x \le 3\). M1: Integrating the second interval using a variable upper limit and adding \(F(3)\), or using a constant of integration and matching. A1: Fully correct piecewise CDF including domains and boundary values. (c) M1: Expressing the probability as \(F(3.5) - F(1)\) or as a sum of two integrals. A1: Finding \(F(3.5) = \frac{27}{28}\) (or awrt 0.964) and \(F(1) = \frac{10}{63}\) (or awrt 0.159). A1: Obtaining the exact fraction \rac{29}{36}\ or awrt 0.806.

(a) The total length of the continuous uniform distribution interval is \(3.5 - (-2.5) = 6\). A rod is acceptable if its error lies in the interval \([-1.5, 1.5]\), which has a length of \(1.5 - (-1.5) = 3\). The probability that a rod is acceptable is \(P(-1.5 \le X \le 1.5) = \frac{3}{6} = 0.5\). Therefore, the probability that a rod is defective is \(1 - 0.5 = 0.5\). (b) For a continuous uniform distribution \(U(a, b)\): Mean \(\mu = \frac{a+b}{2} = \frac{-2.5 + 3.5}{2} = 0.5\). Variance \(\sigma^2 = \frac{(b-a)^2}{12} = \frac{(3.5 - (-2.5))^2}{12} = \frac{6^2}{12} = 3\). Hence, standard deviation \(\sigma = \sqrt{3} \approx 1.73\) (to 3 s.f.). (c) Let \(Y\) be the number of defective rods in a sample of 10. Since the rods are randomly and independently selected, \(Y \sim \text{B}(10, 0.5)\). (i) The probability that exactly 3 rods are defective is: \(P(Y = 3) = \binom{10}{3} (0.5)^3 (0.5)^7 = 120 \times (0.5)^{10} = \frac{120}{1024} = \frac{15}{128} \approx 0.117\) (to 3 s.f.). (ii) The probability that more than 3 rods are defective is: \(P(Y > 3) = 1 - P(Y \le 3)\). \(P(Y \le 3) = P(Y=0) + P(Y=1) + P(Y=2) + P(Y=3) = \frac{1 + 10 + 45 + 120}{1024} = \frac{176}{1024} = \frac{11}{64} = 0.171875\). Therefore, \(P(Y > 3) = 1 - 0.171875 = 0.828125 \approx 0.828\) (to 3 s.f.).

(a) M1: Attempting to find the length of the interval for acceptable error or using integration. A1: Correct acceptable probability of 0.5. A1: Finding defective probability is 0.5. (b) B1: Correct mean of 0.5 (or \(\frac{1}{2}\)). M1: Attempting to use the variance formula \(\frac{(b-a)^2}{12}\). A1: Correct standard deviation of \(\sqrt{3}\) or awrt 1.73. (c) M1: Recognizing the binomial distribution model \(Y \sim \text{B}(10, 0.5)\). M1: Using the binomial formula for \(P(Y = 3)\). A1: Finding exact value \(\frac{15}{128}\) or awrt 0.117. M1: Expressing the probability for more than 3 as \(1 - P(Y \le 3)\) and attempting to calculate. A1: Correct probability of \(\frac{53}{64}\) or awrt 0.828.

Let \(W\) be the number of outages in a given week. \(W \sim \text{Po}(1.5)\).

**(a)** Let \(X\) be the number of outages in a 2-week period.
\(X \sim \text{Po}(2 \times 1.5) \implies X \sim \text{Po}(3)\).
\[\text{P}(X = 3) = \frac{e^{-3} 3^3}{3!} = 4.5 e^{-3} \approx 0.22404 \approx 0.224\ (3\text{ s.f.})\]

**(b)** Let \(Y\) be the number of outages in a 4-week period.
\(Y \sim \text{Po}(4 \times 1.5) \implies Y \sim \text{Po}(6)\).
We require \(\text{P}(Y > 4) = 1 - \text{P}(Y \le 4)\).
Using cumulative Poisson tables or direct calculation:
\[\text{P}(Y \le 4) = 0.2851\]
\[\text{P}(Y > 4) = 1 - 0.2851 = 0.7149 \approx 0.715\ (3\text{ s.f.})\]

**(c)** We want to find the conditional probability \(\text{P}(W \le 4 \mid W \ge 2)\).
\[\text{P}(W \le 4 \mid W \ge 2) = \frac{\text{P}(2 \le W \le 4)}{\text{P}(W \ge 2)}\]
First, find \(\text{P}(W \ge 2)\):
\[\text{P}(W \ge 2) = 1 - \text{P}(W \le 1) = 1 - e^{-1.5}(1 + 1.5) = 1 - 2.5 e^{-1.5} \approx 1 - 0.557825 = 0.442175\]
Next, find \(\text{P}(2 \le W \le 4)\):
\[\text{P}(2 \le W \le 4) = \text{P}(W \le 4) - \text{P}(W \le 1)\]
\[\text{P}(W \le 4) = e^{-1.5} \left(1 + 1.5 + \frac{1.5^2}{2} + \frac{1.5^3}{6} + \frac{1.5^4}{24}\right) = e^{-1.5} (1 + 1.5 + 1.125 + 0.5625 + 0.2109375) = 4.3984375 e^{-1.5}\]
\[\text{P}(2 \le W \le 4) = 4.3984375 e^{-1.5} - 2.5 e^{-1.5} = 1.8984375 e^{-1.5} \approx 0.423599\]
Now find the conditional probability:
\[\text{P}(W \le 4 \mid W \ge 2) = \frac{0.423599}{0.442175} \approx 0.95799 \approx 0.958\ (3\text{ s.f.})\]

**(a)**
* **M1:** Identifies the correct distribution \(X \sim \text{Po}(3)\) and attempts \(\text{P}(X=3)\).
* **A1:** Correct expression or substitution: \(\frac{e^{-3} 3^3}{6}\).
* **A1:** awrt 0.224

**(b)**
* **M1:** Identifies the correct distribution \(Y \sim \text{Po}(6)\) and writes the expression \(1 - \text{P}(Y \le 4)\).
* **A1:** Correct use of tables or calculation of \(\text{P}(Y \le 4) \approx 0.2851\).
* **A1:** awrt 0.715

**(c)**
* **M1:** Identifies the required probability as a conditional probability: \(\frac{\text{P}(2 \le W \le 4)}{\text{P}(W \ge 2)}\).
* **M1:** Attempts to find \(\text{P}(W \ge 2) = 1 - \text{P}(W \le 1)\) with correct values.
* **A1:** \(\text{P}(W \ge 2) \approx 0.442\).
* **M1:** Attempts to find \(\text{P}(2 \le W \le 4)\) using Poisson probabilities.
* **A1:** awrt 0.958

Let \(X\) be the number of customers in the sample of 40 who purchased a snack. Under the null hypothesis, \(X \sim \text{B}(40, 0.35)\).

**(a)**
Hypotheses:
\(H_0: p = 0.35\)
\(H_1: p > 0.35\)

We observe \(X = 19\).
We calculate the probability of obtaining a value at least as extreme as 19 under \(H_0\):
\[\text{P}(X \ge 19) = 1 - \text{P}(X \le 18)\]
From the binomial cumulative distribution tables for \(n = 40, p = 0.35\):
\[\text{P}(X \le 18) = 0.9301\]
Therefore,
\[\text{P}(X \ge 19) = 1 - 0.9301 = 0.0699\]
Since \(0.0699 > 0.05\) (the \(5\%\) significance level), the result is not statistically significant.
Therefore, we fail to reject \(H_0\). There is insufficient evidence at the \(5\%\) level of significance to suggest that the marketing campaign was successful in increasing the proportion of customers who purchase a snack.

**(b)**
For a 5% level of significance, we require the critical region to be in the upper tail:
\[\text{P}(X \ge c) \le 0.05 \implies 1 - \text{P}(X \le c-1) \le 0.05 \implies \text{P}(X \le c-1) \ge 0.95\]
From the tables for \(n = 40, p = 0.35\):
\[\text{P}(X \le 18) = 0.9301 < 0.95\]
\[\text{P}(X \le 19) = 0.9646 \ge 0.95\]
So \(c - 1 = 19 \implies c = 20\).
Thus, the critical region is \(X \ge 20\).

**(c)**
The actual significance level is the probability of rejecting \(H_0\) when \(H_0\) is true, which is:
\[\text{P}(X \ge 20) = 1 - \text{P}(X \le 19) = 1 - 0.9646 = 0.0354 = 3.54\%\]

**(a)**
* **B1:** Both hypotheses correctly stated using \(p\) or \(\pi\).
* **M1:** Attempting to calculate or write down the correct probability statement: \(\text{P}(X \ge 19)\) or \(1 - \text{P}(X \le 18)\).
* **A1:** Finding the correct probability \(0.0699\) (accept awrt 0.070).
* **M1:** Comparing their probability with 0.05 (or comparing 19 with critical value 20).
* **A1:** Correct non-contextual conclusion: Do not reject \(H_0\) (or fail to reject \(H_0\)).
* **A1:** Fully correct contextual conclusion: Insufficient evidence to suggest the marketing campaign was successful.

**(b)**
* **M1:** Attempting to find \(c\) such that \(\text{P}(X \le c-1) \ge 0.95\) using table values.
* **A1:** Identifying \(\text{P}(X \le 18) = 0.9301\) and \(\text{P}(X \le 19) = 0.9646\).
* **A1:** Correct critical region: \(X \ge 20\) (accept \(X \in [20, 40]\)).

**(c)**
* **M1:** Attempting \(1 - \text{P}(X \le 19)\) using their table values.
* **A1:** Correct answer of 0.0354 or \(3.54\%\).

**(a)** Since \(f(x)\) is a probability density function, the total area under the curve is 1:
\[\int_1^4 k(x^2 - 2x + 1) \, dx = 1\]
Recognising that \(x^2 - 2x + 1 = (x-1)^2\):
\[k \int_1^4 (x-1)^2 \, dx = k \left[ \frac{(x-1)^3}{3} \right]_1^4 = 1\]
Evaluating the limits:
\[k \left( \frac{(4-1)^3}{3} - \frac{(1-1)^3}{3} \right) = k \left( \frac{27}{3} - 0 \right) = 9k\]
Since \(9k = 1\), we have \(k = \frac{1}{9}\) (as required).

**(b)** The cumulative distribution function \(\text{F}(x) = \int_{-\infty}^x f(t) \, dt\).
For \(x < 1\), \(\text{F}(x) = 0\).
For \(1 \le x \le 4\):
\[\text{F}(x) = \int_1^x \frac{1}{9}(t-1)^2 \, dt = \left[ \frac{(t-1)^3}{27} \right]_1^x = \frac{(x-1)^3}{27}\]
For \(x > 4\), \(\text{F}(x) = 1\).

Thus, the cumulative distribution function is:
\[\text{F}(x) = \begin{cases} 0 & x < 1 \\ \frac{(x-1)^3}{27} & 1 \le x \le 4 \\ 1 & x > 4 \end{cases}\]

**(c)** Let \(m\) be the median of \(X\). Then:
\[\text{F}(m) = 0.5\]
\[\frac{(m-1)^3}{27} = 0.5\]
\[(m-1)^3 = 13.5\]
Taking the cube root of both sides:
\[m - 1 = \sqrt[3]{13.5}\]
\[m - 1 \approx 2.3811\]
\[m \approx 3.3811 \approx 3.38\ (3\text{ s.f.})\]

**(a)**
* **M1:** Sets up the integral \(\int_1^4 k(x^2 - 2x + 1) \, dx = 1\) (condone missing \(k\) or limits for M1).
* **A1:** Correct integration of \((x-1)^2\) or \(x^2-2x+1\) to obtain \(\frac{(x-1)^3}{3}\) or \(\frac{x^3}{3} - x^2 + x\).
* **A1*:** Shows clearly that \(9k = 1 \implies k = \frac{1}{9}\) with no errors.

**(b)**
* **M1:** Attempts to find \(\text{F}(x) = \int_1^x \frac{1}{9}(t-1)^2 \, dt\).
* **A1:** Obtains \(\frac{(x-1)^3}{27}\) (or equivalent polynomial expression).
* **A1:** Writes a fully correct piecewise definition including the domains \(x < 1\) and \(x > 4\) with correct functional values.

**(c)**
* **M1:** Sets their \(\text{F}(m) = 0.5\).
* **M1:** Solves to get \((m-1)^3 = C\) where \(C\) is a positive constant (e.g., 13.5).
* **A1:** Obtains \(m - 1 = \sqrt[3]{13.5}\) or awrt 2.38.
* **A1:** awrt 3.38 (must be 3 s.f.).

\( \text{(a)} \) For a Poisson distribution to be a suitable model, the following conditions must hold in context:
- Phone calls must occur independently of one another.
- Phone calls must occur at a constant average rate.
- Phone calls must occur singly (i.e., no two calls are received at the exact same instant).
- The probability of a call arriving in a very short interval is proportional to the length of that interval.
*(Any two of these are acceptable)*

\( \text{(b)} \) Let \( X \) be the number of phone calls received in a 10-minute interval on a weekday.
Thus, \( X \sim \text{Po}(3.6) \).

\( \text{(i)} \) We require \( P(X = 4) \):
\[ P(X = 4) = \frac{e^{-3.6} \times 3.6^4}{4!} \]
\[ P(X = 4) = \frac{e^{-3.6} \times 167.9616}{24} \approx 0.1912 \approx 0.191 \text{ (to 3 s.f.)} \]
Alternatively, using statistical tables:
\[ P(X = 4) = P(X \le 4) - P(X \le 3) = 0.7064 - 0.5152 = 0.1912 \]

\( \text{(ii)} \) We require \( P(X \ge 3) \):
\[ P(X \ge 3) = 1 - P(X \le 2) \]
\[ P(X \le 2) = e^{-3.6} \left(1 + 3.6 + \frac{3.6^2}{2!}\right) \]
\[ P(X \le 2) = e^{-3.6} (1 + 3.6 + 6.48) = 11.08 \times e^{-3.6} \approx 0.30275 \]
\[ P(X \ge 3) = 1 - 0.30275 = 0.69725 \approx 0.697 \text{ (to 3 s.f.)} \]

\( \text{(c)} \) Let \( Y \) be the number of phone calls received in a 5-minute interval on a weekday.
Since the rate is 3.6 per 10 minutes, the mean rate for 5 minutes is:
\[ \mu = 3.6 \times \frac{5}{10} = 1.8 \]
So \( Y \sim \text{Po}(1.8) \).
We want \( P(Y = 1) \):
\[ P(Y = 1) = e^{-1.8} \times 1.8^1 \approx 0.165299 \times 1.8 = 0.2975 \approx 0.298 \text{ (to 3 s.f.)} \]

\( \text{(d)} \) Let \( W \) be the number of phone calls received in a 20-minute interval on a weekend.
Since the rate is \( \lambda \) per 10 minutes, the mean rate for 20 minutes is \( 2\lambda \).
So \( W \sim \text{Po}(2\lambda) \).
We are given \( P(W = 0) = 0.0821 \).
\[ P(W = 0) = e^{-2\lambda} = 0.0821 \]
Taking the natural logarithm of both sides:
\[ -2\lambda = \ln(0.0821) \]
\[ -2\lambda \approx -2.4998 \]
\[ \lambda \approx 1.2499 \approx 1.25 \text{ (to 2 d.p.)} \]

\( \text{(a)} \)
- **B1**: One correct condition stated in context (e.g., 'calls arrive independently').
- **B1**: A second correct condition stated in context (e.g., 'calls arrive at a constant average rate').
*Note: Allow 1 mark max if conditions are correct but not in context (e.g., 'events are independent').*

\( \text{(b)(i)} \)
- **M1**: For using or writing \( \frac{e^{-3.6} \times 3.6^4}{4!} \) or equivalent, e.g., table subtraction \( P(X \le 4) - P(X \le 3) \).
- **A1**: awrt 0.191

\( \text{(b)(ii)} \)
- **M1**: For translating \( P(X \ge 3) \) as \( 1 - P(X \le 2) \) and attempting to evaluate \( P(X \le 2) \).
- **A1**: awrt 0.697

\( \text{(c)} \)
- **M1**: For identifying the new mean as \( 1.8 \) and calculating \( e^{-1.8} \times 1.8 \).
- **A1**: awrt 0.298

\( \text{(d)} \)
- **M1**: For identifying the mean as \( 2\lambda \) and setting up the equation \( e^{-2\lambda} = 0.0821 \).
- **M1**: For correctly taking natural logarithms of both sides to solve for \( \lambda \).
- **A1**: awrt 1.25