2026 Edexcel IAS-Level Mathematics (XMA01) 模擬試題連答案詳解

For the quadratic equation to have no real roots, its discriminant must be negative: \(b^2 - 4ac < 0\). Here, \(a = k - 2\), \(b = 2k\), and \(c = 3k - 4\). Substituting these into the discriminant gives \((2k)^2 - 4(k - 2)(3k - 4) < 0\). Expanding the terms yields \(4k^2 - 4(3k^2 - 10k + 8) < 0\). Dividing the entire inequality by 4, we get \(k^2 - (3k^2 - 10k + 8) < 0\), which simplifies to \(-2k^2 + 10k - 8 < 0\). Multiplying the entire inequality by \(-1\) and reversing the inequality sign gives \(2k^2 - 10k + 8 > 0\). Dividing by 2, we obtain \(k^2 - 5k + 4 > 0\). Factoring this quadratic gives \((k - 1)(k - 4) > 0\). The critical values are \(k = 1\) and \(k = 4\). For the inequality to hold, \(k\) must lie outside of these critical values. Therefore, the set of possible values for \(k\) is \(k < 1\) or \(k > 4\).

M1: Attempts to use the discriminant \(b^2 - 4ac\) with correct identification of \(a\), \(b\), and \(c\). A1: Correctly expands the discriminant to obtain a quadratic inequality, such as \(-2k^2 + 10k - 8 < 0\) or equivalent. M1: States or uses the condition \(b^2 - 4ac < 0\) for no real roots. M1: Attempts to find the critical values of their quadratic inequality in \(k\). A1: Correct critical values of \(k = 1\) and \(k = 4\). M1: Chooses the outside regions for a positive quadratic inequality. A1: Correct final answer: \(k < 1\) or \(k > 4\) (or equivalent set notation).

From the first equation, express \(y\) in terms of \(x\): \(y = 2x - 5\). Substitute this expression for \(y\) into the second equation: \(x^2 + 2x(2x - 5) = 15\). Expanding the brackets gives \(x^2 + 4x^2 - 10x = 15\), which simplifies to \(5x^2 - 10x - 15 = 0\). Dividing the entire equation by 5 yields \(x^2 - 2x - 3 = 0\). Factoring this quadratic equation gives \((x - 3)(x + 1) = 0\), which has solutions \(x = 3\) and \(x = -1\). To find the corresponding \(y\)-coordinates, substitute these values back into \(y = 2x - 5\). For \(x = 3\): \(y = 2(3) - 5 = 1\). For \(x = -1\): \(y = 2(-1) - 5 = -7\). Therefore, the solutions are \((3, 1)\) and \((-1, -7)\).

M1: Rearranges the linear equation to make \(y\) (or \(x\)) the subject, e.g., \(y = 2x - 5\). M1: Substitutes their linear expression into the quadratic equation to form an equation in a single variable. A1: Obtains a correct simplified quadratic equation, e.g., \(5x^2 - 10x - 15 = 0\) or \(x^2 - 2x - 3 = 0\). M1: Solves their 3-term quadratic equation by factoring, completing the square, or using the formula. A1: Correct \(x\)-values of \(x = 3\) and \(x = -1\). M1: Substitutes their \(x\)-values back into a suitable equation to find the corresponding \(y\)-values. A1: Correct paired solutions: \((3, 1)\) and \((-1, -7)\).

(a) The curve \(y = x(x-3)(x+4)\) is a cubic with a positive \(x^3\) coefficient, so it starts in quadrant 3 and ends in quadrant 1. The roots of \(f(x) = 0\) are \(x = 0\), \(x = 3\), and \(x = -4\). The y-intercept is at \((0,0)\). Thus, the sketch is a positive cubic curve passing through the points \((-4,0)\), \((0,0)\), and \((3,0)\). (b) The curve with equation \(y = f(x - 2)\) represents a horizontal translation of the curve \(C\) by 2 units to the right. The x-intercepts translate from \(-4, 0, 3\) to \(-4+2 = -2\), \(0+2 = 2\), and \(3+2 = 5\). Hence, the new x-intercepts are \((-2,0)\), \((2,0)\), and \((5,0)\). The new y-intercept is found by setting \(x = 0\): \(y = f(0-2) = f(-2) = -2(-2-3)(-2+4) = -2(-5)(2) = 20\). Therefore, the y-intercept is at \((0, 20)\).

Part (a): B1: Correct shape of a positive cubic (rising from left to right with one local maximum and one local minimum). B1: Identifies and marks the origin \((0,0)\) as an intersection point. B1: Identifies and marks the x-intercepts at \((-4,0)\) and \((3,0)\). Part (b): M1: Translates their cubic curve from part (a) to the right. A1: Identifies the correct new x-intercepts at \((-2,0)\), \((2,0)\), and \((5,0)\). B1: Calculates and marks the new y-intercept at \((0, 20)\). A1: Correctly sketched transformed curve showing all intersection points clearly labelled with coordinates.

First, rearrange the equation to isolate one of the square roots: \(\sqrt{4x + 9} = 3 + \sqrt{x}\). Square both sides of the equation: \(4x + 9 = (3 + \sqrt{x})^2\). Expanding the right-hand side gives: \(4x + 9 = 9 + 6\sqrt{x} + x\). Subtracting 9 and \(x\) from both sides to simplify: \(3x = 6\sqrt{x}\). Dividing both sides by 3 yields: \(x = 2\sqrt{x}\). Square both sides of the equation again to eliminate the remaining square root: \(x^2 = 4x\). Rearranging to form a quadratic equation gives: \(x^2 - 4x = 0\), which factors to \(x(x - 4) = 0\). This yields two potential solutions: \(x = 0\) and \(x = 4\). We must verify these solutions in the original equation. For \(x = 0\): \(\sqrt{4(0)+9} - \sqrt{0} = 3 - 0 = 3\) (which is valid). For \(x = 4\): \(\sqrt{4(4)+9} - \sqrt{4} = \sqrt{25} - 2 = 5 - 2 = 3\) (which is valid). Thus, the solutions are \(x = 0\) and \(x = 4\).

M1: Rearranges the equation to isolate a radical, e.g., \(\sqrt{4x + 9} = 3 + \sqrt{x}\). M1: Squares both sides, expanding the right-hand side correctly to obtain \(9 + 6\sqrt{x} + x\). A1: Simplifies the equation to a correct form with a single radical term, e.g., \(3x = 6\sqrt{x}\) or \(x = 2\sqrt{x}\). M1: Squares both sides again to eliminate the radical, leading to a polynomial equation, e.g., \(x^2 = 4x\). A1: Solves the equation to find the solution \(x = 4\). A1: Finds the other potential solution \(x = 0\). B1: Shows verification of both candidate solutions in the original equation, confirming both are valid.

(a) We divide \(2x^3 + 5x^2 - 4x - 12\) by \(x + 2\) using algebraic division or by equating coefficients. Let \(2x^3 + 5x^2 - 4x - 12 = (x + 2)(ax^2 + bx + c)\). Expanding the right-hand side gives \(ax^3 + (2a + b)x^2 + (2b + c)x + 2c\). Equating the coefficients of \(x^3\) gives \(a = 2\). Equating the constant terms gives \(2c = -12 \implies c = -6\). Equating the coefficients of \(x^2\) gives \(2a + b = 5 \implies 2(2) + b = 5 \implies b = 1\). Thus, the expression can be written as \(2x^2 + x - 6\). (b) To find the minimum value of \(2x^2 + x - 6\), we complete the square: \(2x^2 + x - 6 = 2(x^2 + \frac{1}{2}x) - 6 = 2[(x + \frac{1}{4})^2 - \frac{1}{16}] - 6 = 2(x + \frac{1}{4})^2 - \frac{1}{8} - 6 = 2(x + \frac{1}{4})^2 - \frac{49}{8}\). Since \(2(x + \frac{1}{4})^2 \ge 0\), the minimum value of the expression is \(-\frac{49}{8}\) (or \(-6.125\)), which occurs when \(x + \frac{1}{4} = 0\), giving \(x = -\frac{1}{4}\).

Part (a): M1: Attempts algebraic division or sets up an identity to equate coefficients, e.g., \(2x^3 + 5x^2 - 4x - 12 = (x+2)(ax^2 + bx + c)\). A1: Correctly identifies any two of the constants: \(a=2\), \(b=1\), or \(c=-6\). A1: Fully correct quadratic expression: \(2x^2 + x - 6\). Part (b): M1: Attempts to complete the square on their quadratic expression from part (a). A1: Obtains the correct completed square form: \(2(x + \frac{1}{4})^2 - \frac{49}{8}\) or equivalent. A1: Identifies the minimum value as \(-\frac{49}{8}\) (or \(-6.125\)). A1: Identifies that this minimum occurs at \(x = -\frac{1}{4}\) (or \(-0.25\)).

(a) Differentiating \(y = \frac{1}{3}x^3 - 3x^2 + 5x + 7\) with respect to \(x\): \(\frac{\text{d}y}{\text{d}x} = 3 \times \frac{1}{3}x^2 - 2 \times 3x + 5 = x^2 - 6x + 5\). (b) For the gradient to be negative, we need \(\frac{\text{d}y}{\text{d}x} < 0\). So, \(x^2 - 6x + 5 < 0\). Factoring the quadratic: \((x-1)(x-5) < 0\). The critical values are \(x=1\) and \(x=5\). Since the coefficient of \(x^2\) is positive, the inequality is satisfied between the roots: \(1 < x < 5\). (c) Stationary points occur where \(\frac{\text{d}y}{\text{d}x} = 0\). This gives \(x^2 - 6x + 5 = 0\), so \(x=1\) or \(x=5\). Substitute these values back into the original curve equation: When \(x=1\): \(y = \frac{1}{3}(1)^3 - 3(1)^2 + 5(1) + 7 = \frac{1}{3} - 3 + 5 + 7 = \frac{28}{3}\). When \(x=5\): \(y = \frac{1}{3}(5)^3 - 3(5)^2 + 5(5) + 7 = \frac{125}{3} - 75 + 25 + 7 = -\frac{4}{3}\). Thus, the coordinates of the stationary points are \(\left(1, \frac{28}{3}\right)\) and \(\left(5, -\frac{4}{3}\right)\).

(a) M1: Attempts to differentiate with the power of at least two terms decreased by 1. A1: Correct derivative \(x^2 - 6x + 5\). (b) M1: Sets their derivative to 0 and attempts to find critical values. M1: Selects the inside region for their critical values. A1: Correct inequality \(1 < x < 5\) (or equivalent interval notation). (c) M1: Uses \(\frac{\text{d}y}{\text{d}x} = 0\) to find \(x = 1\) and \(x = 5\) (can be implied). A1: Correctly identifies both \(x\) values. M1: Substitutes at least one of their \(x\) values into the original equation for \(y\). A0.5: Both coordinates correct: \(\left(1, \frac{28}{3}\right)\) and \(\left(5, -\frac{4}{3}\right)\) (or exact equivalents).

(a) At the points of intersection with the \(x\)-axis, \(y = 0\). So, \(8\sqrt{x} - 2x - 6 = 0\). Dividing by 2 gives \(4\sqrt{x} - x - 3 = 0\), which rearranges to \(x - 4\sqrt{x} + 3 = 0\). Let \(u = \sqrt{x}\). The equation becomes \(u^2 - 4u + 3 = 0\). Factoring gives \((u-1)(u-3) = 0\), so \(u = 1\) or \(u = 3\). Since \(\sqrt{x} = u\), we have \(\sqrt{x} = 1 \implies x = 1\), and \(\sqrt{x} = 3 \implies x = 9\). Thus, the curve intersects the \(x\)-axis at \(x=1\) and \(x=9\). (b) The area of \(R\) is given by \(\\int_{1}^{9} (8x^{1/2} - 2x - 6) \\,\text{d}x = \left[ \frac{16}{3}x^{3/2} - x^2 - 6x \right]_{1}^{9}\). Evaluating this at the upper limit \(x=9\): \(\frac{16}{3}(9)^{3/2} - (9)^2 - 6(9) = \frac{16}{3}(27) - 81 - 54 = 144 - 135 = 9\). Evaluating this at the lower limit \(x=1\): \(\frac{16}{3}(1)^{3/2} - (1)^2 - 6(1) = \frac{16}{3} - 1 - 6 = -\frac{5}{3}\). Subtracting the lower limit value from the upper limit value: \(9 - \left(-\frac{5}{3}\right) = \frac{32}{3}\).

(a) M1: Sets \(y = 0\) to form the equation \(8\sqrt{x} - 2x - 6 = 0\). M1: Recognises this as a quadratic equation in \(\sqrt{x}\) (or squares terms to form a quadratic in \(x\)) and attempts to solve it. A0.5: Correctly completes the proof to show that \(x = 1\) and \(x = 9\). (b) M1: Integrates at least one term of \(8x^{1/2} - 2x - 6\) correctly (power increased by 1). A1: Correct integration to give \(\frac{16}{3}x^{3/2} - x^2 - 6x\). M1: Substitutes the limit 9 and the limit 1 into their integrated expression. A1: Obtains the values 9 and \(-\frac{5}{3}\) (or equivalent fractions). M1: Subtracts their value at 1 from their value at 9. A1: Correct exact area of \(\frac{32}{3}\) (or exact equivalent).

(a) The gradient of the line \(l_1\) is:
\[m_1 = \frac{2 - 5}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}\]
Using the point-slope form with point \(B(4, 2)\):
\[y - 2 = -\frac{1}{2}(x - 4)\]
\[2y - 4 = -x + 4\]
\[x + 2y - 8 = 0\]

(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 point-slope form with point \(C(1, 1)\):
\[y - 1 = 2(x - 1)\]
\[y - 1 = 2x - 2\]
\[y = 2x - 1\]

(c) To find the coordinates of \(P\), solve the simultaneous equations:
\[x + 2y - 8 = 0\]
\[y = 2x - 1\]
Substitute the expression for \(y\) from the second equation into the first equation:
\[x + 2(2x - 1) - 8 = 0\]
\[x + 4x - 2 - 8 = 0\]
\[5x - 10 = 0 \implies x = 2\]
Substitute \(x = 2\) back into \(y = 2x - 1\):
\[y = 2(2) - 1 = 3\]
So the coordinates of \(P\) are \((2, 3)\).

(a)
- M1: Attempts to find the gradient of \(l_1\) using \(\frac{y_2 - y_1}{x_2 - x_1}\).
- M1: Formulates the equation of the line using their gradient and one of the coordinates \(A\) or \(B\).
- A1: Correct equation in the form \(ax + by + c = 0\), e.g., \(x + 2y - 8 = 0\) (or any non-zero integer multiple thereof).

(b)
- M1: Identifies the gradient of the perpendicular line as \(2\) and attempts to write the equation of \(l_2\) passing through \(C(1, 1)\).
- A1: Correct equation \(y = 2x - 1\) (must be in the form \(y = mx + c\)).

(c)
- M1: Eliminates one variable to find either \(x\) or \(y\).
- A1: Finds either \(x = 2\) or \(y = 3\).
- A1: Correct coordinates of \(P\) as \((2, 3)\) or written clearly as \(x = 2, y = 3\).

(a) Using the trigonometric identity \(\cos^2 \theta + \sin^2 \theta = 1\), substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) into the given equation:
\[4(1 - \sin^2 \theta) + 7 \sin \theta - 7 = 0\]
\[4 - 4 \sin^2 \theta + 7 \sin \theta - 7 = 0\]
\[-4 \sin^2 \theta + 7 \sin \theta - 3 = 0\]
Multiplying the entire equation by \(-1\) yields:
\[4 \sin^2 \theta - 7 \sin \theta + 3 = 0 \quad \text{(as required)}\]

(b) Factorising the quadratic equation in \(\sin \theta\):
\[(4 \sin \theta - 3)(\sin \theta - 1) = 0\]
This gives two possible values for \(\sin \theta\):
\[\sin \theta = \frac{3}{4} \quad \text{or} \quad \sin \theta = 1\]

Case 1: \(\sin \theta = 1\)
For the range \(0 \le \theta < 360^\circ\):
\[\theta = 90^\circ\]

Case 2: \(\sin \theta = 0.75\)
\[\theta = \arcsin(0.75) \approx 48.59^\circ \implies \theta \approx 48.6^\circ\]
The second solution in the interval is:
\[\theta = 180^\circ - 48.59^\circ = 131.41^\circ \implies \theta \approx 131.4^\circ\]

Thus, the solutions in the interval are \(\theta = 48.6^\circ, 90^\circ, 131.4^\circ\).

(a)
- M1: Uses the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to eliminate \(\cos^2 \theta\) from the equation.
- A1*: Correctly simplifies to show the given quadratic equation \(4 \sin^2 \theta - 7 \sin \theta + 3 = 0\) with no algebraic errors.

(b)
- M1: Attempts to solve the quadratic equation to find values for \(\sin \theta\).
- A1: Correctly identifies \(\sin \theta = \frac{3}{4}\) and \(\sin \theta = 1\).
- B1: Obtains \(\theta = 90^\circ\).
- M1: Obtains one correct angle from \(\sin \theta = 0.75\).
- A1: Obtains \(\theta \approx 48.6^\circ\) (or \(48.59^\circ\)).
- A1: Obtains the other angle \(\theta \approx 131.4^\circ\) (or \(131.41^\circ\)) and no other solutions in the interval.

(a) The formula for the area of a triangle is:
\[\text{Area} = \frac{1}{2} ac \sin B\]
Substituting the given values:
\[\text{Area} = \frac{1}{2} (x)(x + 3) \sin 60^\circ\]
Since the area is \(5\sqrt{3}\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):
\[5\sqrt{3} = \frac{1}{2} x(x+3) \left(\frac{\sqrt{3}}{2}\right)\]
\[5\sqrt{3} = \frac{\sqrt{3}}{4} (x^2 + 3x)\]
Dividing both sides by \(\sqrt{3}\):
\[5 = \frac{1}{4} (x^2 + 3x)\]
Multiplying by 4:
\[20 = x^2 + 3x \implies x^2 + 3x - 20 = 0 \quad \text{(as required)}\]

(b) Applying the quadratic formula to \(x^2 + 3x - 20 = 0\):
\[x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-20)}}{2} = \frac{-3 \pm \sqrt{9 + 80}}{2} = \frac{-3 \pm \sqrt{89}}{2}\]
Since \(x\) represents a physical length, \(x > 0\). Therefore:
\[x = \frac{-3 + \sqrt{89}}{2}\]

(c) Using the Cosine Rule to find the length of \(AC\):
\[AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos B\]
\[AC^2 = x^2 + (x+3)^2 - 2x(x+3) \cos 60^\circ\]
Since \(\cos 60^\circ = \frac{1}{2}\):
\[AC^2 = x^2 + (x^2 + 6x + 9) - 2x(x+3)\left(\frac{1}{2}\right)\]
\[AC^2 = 2x^2 + 6x + 9 - x(x+3)\]
\[AC^2 = 2x^2 + 6x + 9 - x^2 - 3x\]
\[AC^2 = x^2 + 3x + 9\]
From part (a), we know that \(x^2 + 3x = 20\). Substituting this into our expression for \(AC^2\):
\[AC^2 = 20 + 9 = 29\]
Since length must be positive:
\[AC = \sqrt{29}\text{ cm}\]

(a)
- M1: Applies the area formula \(\frac{1}{2} ac \sin B = 5\sqrt{3}\) with given values.
- M1: Uses \(\sin 60^\circ = \frac{\sqrt{3}}{2}\) and attempts to simplify.
- A1*: Correctly simplifies to the given quadratic equation \(x^2 + 3x - 20 = 0\) with no algebraic errors.

(b)
- M1: Applies the quadratic formula (or completes the square) to solve the equation.
- A1: Identifies the correct positive root \(x = \frac{-3 + \sqrt{89}}{2}\) and rejects the negative root.

(c)
- M1: Applies the Cosine Rule to find \(AC^2\).
- M1: Uses \(\cos 60^\circ = \frac{1}{2}\), expands correctly, and simplifies the expression to \(AC^2 = x^2 + 3x + 9\).
- A1: Obtains \(AC = \sqrt{29}\text{ cm}\) (must be exact; do not accept decimal approximations).

For (a): The second term is given by \( u_2 = ar = 12 \implies a = \frac{12}{r} \). The sum to infinity is given by \( S_{\infty} = \frac{a}{1 - r} = 64 \). Substituting \( a = \frac{12}{r} \) into the sum to infinity formula gives: \( \frac{12}{r(1-r)} = 64 \implies 12 = 64r(1-r) \implies 12 = 64r - 64r^2 \). Dividing by 4 and rearranging gives \( 16r^2 - 16r + 3 = 0 \) as required. For (b): Factorising the quadratic equation: \( (4r - 1)(4r - 3) = 0 \), which gives \( r = \frac{1}{4} \) or \( r = \frac{3}{4} \). Since both values are in the range \( 0 < r < 1 \), we find the corresponding values of \( a \): For \( r = \frac{1}{4} \), \( a = \frac{12}{1/4} = 48 \). For \( r = \frac{3}{4} \), \( a = \frac{12}{3/4} = 16 \).

(a) M1: Uses \( u_2 = ar = 12 \) to write \( a \) in terms of \( r \). M1: Substitutes their expression for \( a \) into \( S_{\infty} = \frac{a}{1-r} = 64 \) and attempts to eliminate fractions. A1*: Correctly shows the given quadratic equation. (b) M1: Attempts to solve the quadratic equation to find two values of \( r \). A1: Correct values of \( r = \frac{1}{4} \) and \( r = \frac{3}{4} \). M1: Uses their values of \( r \) to find the corresponding values of \( a \). A0.25: Correct pairs \( (r = 1/4, a = 48) \) and \( (r = 3/4, a = 16) \).

Using the laws of logarithms: \( 2 \log_2 (x - 2) = \log_2 (x - 2)^2 \) and \( \log_2 (x - 2)^2 - \log_2 (x + 4) = \log_2 \left( \frac{(x - 2)^2}{x + 4} \right) \). Therefore, \( \log_2 \left( \frac{(x - 2)^2}{x + 4} \right) = -1 \). Rewriting in exponential form: \( \frac{(x - 2)^2}{x + 4} = 2^{-1} = \frac{1}{2} \). Multiplying both sides by \( 2(x + 4) \) gives: \( 2(x^2 - 4x + 4) = x + 4 \implies 2x^2 - 8x + 8 = x + 4 \implies 2x^2 - 9x + 4 = 0 \). Factorising the quadratic equation: \( (2x - 1)(x - 4) = 0 \), which gives \( x = \frac{1}{2} \) or \( x = 4 \). Checking the domain of the original logarithmic terms: we require \( x - 2 > 0 \implies x > 2 \). Since \( x = \frac{1}{2} \) is not greater than 2, this value is rejected. Thus, the only valid solution is \( x = 4 \).

M1: Applies the power law to obtain \( \log_2 (x-2)^2 \). M1: Applies the subtraction law to form a single logarithm: \( \log_2 \left( \frac{(x-2)^2}{x+4} \right) \). M1: Removes logarithms correctly to obtain \( \frac{(x-2)^2}{x+4} = 2^{-1} \) or equivalent. A1: Obtains the correct simplified quadratic equation \( 2x^2 - 9x + 4 = 0 \). M1: Solves the quadratic equation to find two values of \( x \). A1.25: Identifies and rejects \( x = \frac{1}{2} \) with a valid reason, and gives the final answer \( x = 4 \) only.

For (a): The general term in the expansion of \( (2 + kx)^6 \) is given by \( \binom{6}{r} 2^{6-r} (kx)^r \). The term containing \( x^2 \) is \( \binom{6}{2} 2^4 (kx)^2 = 15 \times 16 \times k^2 x^2 = 240 k^2 x^2 \). The term containing \( x^3 \) is \( \binom{6}{3} 2^3 (kx)^3 = 20 \times 8 \times k^3 x^3 = 160 k^3 x^3 \). We are given that the coefficient of \( x^3 \) is twice the coefficient of \( x^2 \), so \( 160 k^3 = 2 \times 240 k^2 \implies 160 k^3 = 480 k^2 \). Since \( k \neq 0 \), we can divide by \( 160 k^2 \) to obtain \( k = 3 \). For (b): Using \( k = 3 \), the term containing \( x^4 \) is \( \binom{6}{4} 2^2 (3x)^4 = 15 \times 4 \times 81 x^4 = 4860 x^4 \). Thus, the coefficient of \( x^4 \) is \( 4860 \).

(a) M1: Attempts to find the terms in \( x^2 \) or \( x^3 \) with correct binomial coefficients. A1: Correctly identifies both coefficients: \( 240 k^2 \) and \( 160 k^3 \). M1: Sets up the equation \( 160 k^3 = 2 \times 240 k^2 \) or equivalent. A1: Obtains \( k = 3 \). (b) M1: Uses their value of \( k \) to find the term containing \( x^4 \): \( \binom{6}{4} 2^2 (kx)^4 \). A1.25: Obtains the correct coefficient \( 4860 \).

For (a): The sum of the first 8 terms is given by \( S_8 = \frac{8}{2} [2a + 7d] = 104 \implies 4[2a + 7d] = 104 \implies 2a + 7d = 26 \) (Equation 1). The 12th term is given by \( u_{12} = a + 11d = 28 \) (Equation 2). Multiplying Equation 2 by 2 gives \( 2a + 22d = 56 \). Subtracting Equation 1 from this yields: \( (2a + 22d) - (2a + 7d) = 56 - 26 \implies 15d = 30 \implies d = 2 \). Substituting \( d = 2 \) into Equation 2: \( a + 11(2) = 28 \implies a + 22 = 28 \implies a = 6 \). For (b): The sum of the first 20 terms is given by \( S_{20} = \frac{20}{2} [2a + 19d] = 10 [2(6) + 19(2)] = 10 [12 + 38] = 10 [50] = 500 \).

(a) M1: Uses \( S_8 = 104 \) to form an equation in \( a \) and \( d \). M1: Uses \( u_{12} = 28 \) to form an equation in \( a \) and \( d \). M1: Solves the two equations simultaneously to find at least one variable. A1: Obtains both \( a = 6 \) and \( d = 2 \). (b) M1: Substitutes their \( a \) and \( d \) into the arithmetic sum formula for \( S_{20} \). A1.25: Obtains \( 500 \).

(a) The radius of the circle is the distance \(AB\):
\(r = \sqrt{(7-3)^2 + (7-4)^2} = \sqrt{16+9} = \sqrt{25} = 5\).
Using the circle equation formula \((x-h)^2 + (y-k)^2 = r^2\) with centre \((3, 4)\):
\((x-3)^2 + (y-4)^2 = 25\).

(b) The gradient of the radius \(AB\) is:
\(m_{AB} = \frac{7-4}{7-3} = \frac{3}{4}\).
Since the tangent is perpendicular to the radius, the gradient of the tangent is:
\(m_t = -\frac{1}{m_{AB}} = -\frac{4}{3}\).
The equation of the tangent at \(B(7, 7)\) is:
\(y - 7 = -\frac{4}{3}(x - 7)\)
\(3(y - 7) = -4(x - 7)\)
\(3y - 21 = -4x + 28\)
\(4x + 3y - 49 = 0\).

(c) The tangent to a circle is perpendicular to the radius at the point of contact. Therefore, angle \(ABP\) is \(90^\circ\), and triangle \(ABP\) is a right-angled triangle at \(B\).
Using Pythagoras' theorem on triangle \(ABP\):
\(AP^2 = AB^2 + BP^2\)
Since \(AP = 10\) and \(AB = r = 5\):
\(10^2 = 5^2 + BP^2\)
\(100 = 25 + BP^2 \implies BP^2 = 75 \implies BP = \sqrt{75} = 5\sqrt{3}\).
The area of the right-angled triangle \(ABP\) is:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BP = \frac{1}{2} \times 5 \times 5\sqrt{3} = \frac{25\sqrt{3}}{2}\).

(a) M1: Attempts to find the radius using the distance formula with points \((3,4)\) and \((7,7)\). A1: Correct equation for \(C\): \((x-3)^2 + (y-4)^2 = 25\) (or equivalent).
(b) M1: Finds the gradient of the radius \(AB\). M1: Uses the perpendicular gradient rule to find the gradient of the tangent. M1: Employs \(y - y_1 = m(x - x_1)\) with their gradient and coordinates of \(B\). A1: Correct equation in the form \(4x + 3y - 49 = 0\) (or any integer multiple thereof).
(c) M1: Recognises that angle \(ABP = 90^\circ\) and applies Pythagoras' theorem to find the length \(BP\). A1: Correct length \(BP = 5\sqrt{3}\) or \(\sqrt{75}\). M1: Uses \(\frac{1}{2} \times \text{base} \times \text{height}\) to find the area of triangle \(ABP\). A1: Correct exact area of \(\frac{25\sqrt{3}}{2}\).

(a) To find the stationary point, we first find \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{d}{dx}(3x^2 - 12x^{1/2} + 12) = 6x - 6x^{-1/2}\).
Setting \(\frac{dy}{dx} = 0\) for a stationary point:
\(6x - \frac{6}{\sqrt{x}} = 0 \implies 6x = \frac{6}{\sqrt{x}} \implies x\sqrt{x} = 1\).
Squaring both sides or writing as powers: \(x^{3/2} = 1 \implies x = 1\).
When \(x = 1\):
\(y = 3(1)^2 - 12(1)^{1/2} + 12 = 3 - 12 + 12 = 3\).
So the stationary point is \((1, 3)\).
To determine its nature, find the second derivative:
\(\frac{d^2y}{dx^2} = \frac{d}{dx}(6x - 6x^{-1/2}) = 6 + 3x^{-3/2}\).
At \(x = 1\):
\(\frac{d^2y}{dx^2} = 6 + 3(1)^{-3/2} = 9\).
Since \(\frac{d^2y}{dx^2} > 0\), the stationary point \((1, 3)\) is a local minimum.

(b) The area \(A\) of the region is given by the definite integral:
\(A = \int_{1}^{4} (3x^2 - 12x^{1/2} + 12) \, dx\)
\(A = \left[ x^3 - \frac{12x^{3/2}}{3/2} + 12x \right]_{1}^{4} = \left[ x^3 - 8x^{3/2} + 12x \right]_{1}^{4}\).
Evaluating at the upper limit \(x = 4\):
\(4^3 - 8(4)^{3/2} + 12(4) = 64 - 8(8) + 48 = 64 - 64 + 48 = 48\).
Evaluating at the lower limit \(x = 1\):
\(1^3 - 8(1)^{3/2} + 12(1) = 1 - 8 + 12 = 5\).
Subtracting the limits:
\(A = 48 - 5 = 43\).

(a) M1: Attempts to differentiate to find \(\frac{dy}{dx}\) (at least one power decreased by 1). A1: Correct derivative \(6x - 6x^{-1/2}\). M1: Sets their \(\frac{dy}{dx} = 0\) and solves for \(x\). A1: Correct coordinates \((1, 3)\). M1: Differentiates again to find \(\frac{d^2y}{dx^2}\) and evaluates at \(x=1\). A1: Obtains \(9\) and concludes it is a local minimum based on \(\frac{d^2y}{dx^2} > 0\).
(b) M1: Attempts to integrate the function (at least one power increased by 1). A1: Correct integration: \(x^3 - 8x^{3/2} + 12x\) (constant of integration not needed). M1: Substitutes the limits \(4\) and \(1\) into their integrated expression and subtracts. A1: Correct exact area of \(43\).

(a) The perimeter \(P\) of a sector with radius \(r\) and angle \(\theta\) (in radians) is given by:
\(P = 2r + r\theta\).
Given \(P = 40\):
\(2r + r\theta = 40 \implies r\theta = 40 - 2r \implies \theta = \frac{40 - 2r}{r}\).
The area \(A\) of the sector is:
\(A = \frac{1}{2}r^2\theta\).
Substituting the expression for \(\theta\):
\(A = \frac{1}{2}r^2 \left(\frac{40 - 2r}{r}\right) = \frac{1}{2}r(40 - 2r) = 20r - r^2\) (as required).

(b) To find the maximum value of \(A\), we differentiate with respect to \(r\):
\(\frac{dA}{dr} = 20 - 2r\).
Setting \(\frac{dA}{dr} = 0\) for a stationary point:
\(20 - 2r = 0 \implies r = 10\).
To justify that it is a maximum, find the second derivative:
\(\frac{d^2A}{dr^2} = -2\).
Since \(\frac{d^2A}{dr^2} < 0\), the value is indeed a maximum.
The maximum area is:
\(A = 20(10) - 10^2 = 200 - 100 = 100\, \text{cm}^2\).

(c) Using the equation \(r\theta = 40 - 2r\) with \(r = 10\):
\(10\theta = 40 - 2(10)\)
\(10\theta = 20 \implies \theta = 2\) radians.

(a) B1: Writes down correct formula for perimeter: \(2r + r\theta = 40\) (or equivalent). M1: Expresses \(\theta\) in terms of \(r\) and substitutes into \(A = \frac{1}{2}r^2\theta\). A1*: Fully correct proof with no errors leading to \(A = 20r - r^2\).
(b) M1: Differentiates \(A\) with respect to \(r\). A1: Correct derivative \(20 - 2r\). M1: Sets their derivative to \(0\) to find \(r = 10\), and evaluates \(A\) at this value. M1: Finds \(\frac{d^2A}{dr^2}\) and uses its sign to justify a maximum. A1: Correct maximum area of \(100\).
(c) M1: Uses their value of \(r\) to calculate \(\theta\). A1: Correct angle \(\theta = 2\) (radians).

(a) To find the points of intersection, set the equation of the curve equal to the equation of the line:
\(8x^{1/2} - 2x = 6\)
\(2x - 8x^{1/2} + 6 = 0\)
Dividing by 2:
\(x - 4x^{1/2} + 3 = 0\)
Letting \(u = x^{1/2}\), we obtain a quadratic equation:
\(u^2 - 4u + 3 = 0\)
\((u-1)(u-3) = 0\)
So \(u = 1\) or \(u = 3\).
Therefore, \(x^{1/2} = 1 \implies x = 1\) and \(x^{1/2} = 3 \implies x = 9\).
Since both points lie on the line \(y = 6\), the coordinates of the points of intersection are \((1, 6)\) and \((9, 6)\).

(b) The area of the region is given by the integral of the upper curve minus the lower line between the limits of intersection:
\(\text{Area} = \int_{1}^{9} ((8x^{1/2} - 2x) - 6) \, dx\)
\(\text{Area} = \left[ \frac{8x^{3/2}}{3/2} - x^2 - 6x \right]_{1}^{9} = \left[ \frac{16}{3}x^{3/2} - x^2 - 6x \right]_{1}^{9}\).
Evaluating at the upper limit \(x = 9\):
\(\frac{16}{3}(9)^{3/2} - 9^2 - 6(9) = \frac{16}{3}(27) - 81 - 54 = 16(9) - 135 = 144 - 135 = 9\).
Evaluating at the lower limit \(x = 1\):
\(\frac{16}{3}(1)^{3/2} - 1^2 - 6(1) = \frac{16}{3} - 1 - 6 = \frac{16}{3} - 7 = -\frac{5}{3}\).
Subtracting the lower limit value from the upper limit value:
\(\text{Area} = 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{32}{3}\).

(a) M1: Equates curve and line equations to form an equation in \(x\) or \(x^{1/2}\). M1: Solves the resulting quadratic equation in terms of \(x^{1/2}\) or otherwise. A1: Correct coordinates \((1, 6)\) and \((9, 6)\).
(b) M1: Sets up the integral \(\int (8x^{1/2} - 2x - 6) \, dx\) or equivalent. M1: Attempts to integrate (at least one power increased by 1). A1: Correct integration: \(\frac{16}{3}x^{3/2} - x^2 - 6x\). M1: Substitutes their limits \(9\) and \(1\) into their integrated expression and subtracts. A1: Correct exact area of \(\frac{32}{3}\) (or \(10\frac{2}{3}\)).

(a) Since \( (2x - 3) \) is a factor of \( f(x) \), by the Factor Theorem, \( f\left(\frac{3}{2}\right) = 0 \).

Substituting \( x = \frac{3}{2} \) into \( f(x) \):

\( 2\left(\frac{3}{2}\right)^3 + a\left(\frac{3}{2}\right)^2 + b\left(\frac{3}{2}\right) - 12 = 0 \)

\( 2\left(\frac{27}{8}\right) + \frac{9}{4}a + \frac{3}{2}b - 12 = 0 \)

Multiply the entire equation by 4 to clear the denominators:

\( 27 + 9a + 6b - 48 = 0 \implies 9a + 6b = 21 \implies 3a + 2b = 7 \) (Equation 1)

Since the remainder when \( f(x) \) is divided by \( (x + 2) \) is 14, by the Remainder Theorem, \( f(-2) = 14 \).

Substituting \( x = -2 \) into \( f(x) \):

\( 2(-2)^3 + a(-2)^2 + b(-2) - 12 = 14 \)

\( -16 + 4a - 2b - 12 = 14 \)

\( 4a - 2b - 28 = 14 \implies 4a - 2b = 42 \implies 2a - b = 21 \) (Equation 2)

From Equation 2, we have \( b = 2a - 21 \). Substituting this into Equation 1:

\( 3a + 2(2a - 21) = 7 \)

\( 3a + 4a - 42 = 7 \implies 7a = 49 \implies a = 7 \)

Substituting \( a = 7 \) back into the expression for \( b \):

\( b = 2(7) - 21 = -7 \).

So, \( a = 7 \) and \( b = -7 \).

(b) Substituting \( a = 7 \) and \( b = -7 \) into \( f(x) \) gives:

\( f(x) = 2x^3 + 7x^2 - 7x - 12 \).

To show the required identity, we can expand \( (2x - 3)(x^2 + 5x + 4) \):

\( (2x - 3)(x^2 + 5x + 4) = 2x(x^2 + 5x + 4) - 3(x^2 + 5x + 4) \)

\( = 2x^3 + 10x^2 + 8x - 3x^2 - 15x - 12 \)

\( = 2x^3 + 7x^2 - 7x - 12 = f(x) \) (as required).

(c) To factorise \( f(x) \) completely, we factorise the quadratic part \( x^2 + 5x + 4 \):

\( x^2 + 5x + 4 = (x + 1)(x + 4) \).

Therefore, \( f(x) = (2x - 3)(x + 1)(x + 4) \).

(a)
- M1: Attempts to use the factor theorem with \( x = 1.5 \) by setting \( f(1.5) = 0 \).
- A1: Obtains a correct simplified linear equation in terms of \( a \) and \( b \), e.g., \( 3a + 2b = 7 \) or equivalent.
- M1: Attempts to use the remainder theorem with \( x = -2 \) by setting \( f(-2) = 14 \).
- A1: Obtains a correct simplified linear equation in terms of \( a \) and \( b \), e.g., \( 2a - b = 21 \) or equivalent.
- A1: Solves the simultaneous equations to find both \( a = 7 \) and \( b = -7 \).

(b)
- M1: Attempts to expand \( (2x - 3)(x^2 + 5x + 4) \) showing at least four terms, or attempts algebraic division of \( 2x^3 + 7x^2 - 7x - 12 \) by \( 2x-3 \).
- A0.5: Correctly completes the proof with no errors seen.

(c)
- M1: Attempts to factorise the quadratic factor \( x^2 + 5x + 4 \) into two linear brackets.
- A1: Correct fully factorised expression \( (2x - 3)(x + 1)(x + 4) \).

(a) Using the binomial expansion formula, the general term is given by \( \binom{6}{r} a^{6-r} (3x)^r \).

To find the term in \( x^2 \), set \( r = 2 \):

\( \text{Term in } x^2 = \binom{6}{2} a^{4} (3x)^2 = 15 \cdot a^4 \cdot 9x^2 = 135 a^4 x^2 \)

So, the coefficient of \( x^2 \) is \( 135 a^4 \).

To find the term in \( x^3 \), set \( r = 3 \):

\( \text{Term in } x^3 = \binom{6}{3} a^{3} (3x)^3 = 20 \cdot a^3 \cdot 27x^3 = 540 a^3 x^3 \)

So, the coefficient of \( x^3 \) is \( 540 a^3 \).

We are given that:

\( \text{Coefficient of } x^3 = 8 \times \text{Coefficient of } x^2 \)

\( 540 a^3 = 8 \cdot 135 a^4 \)

\( 540 a^3 = 1080 a^4 \)

Since \( a > 0 \), we can divide both sides by \( 540 a^3 \):

\( 1 = 2a \implies a = 0.5 \) (or \( \frac{1}{2} \)).

(b) Using the value of \( a = 0.5 \), we find the term in \( x^4 \) by setting \( r = 4 \):

\( \text{Term in } x^4 = \binom{6}{4} a^2 (3x)^4 = 15 \cdot a^2 \cdot 81x^4 = 1215 a^2 x^4 \)

So, the coefficient of \( x^4 \) is \( 1215 a^2 \).

Substituting \( a = 0.5 \) into the coefficient:

\( 1215 \cdot (0.5)^2 = 1215 \cdot 0.25 = \frac{1215}{4} = 303.75 \).

(a)
- M1: Attempts to find the term in \( x^2 \) using the binomial expansion, identifying the correct binomial coefficient \( \binom{6}{2} = 15 \) and powers of \( a \) and \( 3x \).
- A1: Correctly simplified coefficient of \( x^2 \) as \( 135 a^4 \).
- M1: Attempts to find the term in \( x^3 \) using the binomial expansion, identifying the correct binomial coefficient \( \binom{6}{3} = 20 \) and powers of \( a \) and \( 3x \).
- A1: Correctly simplified coefficient of \( x^3 \) as \( 540 a^3 \).
- A0.5: Equates and solves the equation \( 540 a^3 = 8 \times 135 a^4 \) to find \( a = 0.5 \) (or equivalent fraction).

(b)
- M1: Writes down the formula for the term in \( x^4 \) with the correct binomial coefficient \( \binom{6}{4} = 15 \).
- A1: Correct expression for the coefficient of \( x^4 \) in terms of \( a \), i.e., \( 1215 a^2 \).
- M1: Substitutes their value of \( a \) from part (a) into their expression for the coefficient of \( x^4 \).
- A1: Obtains the correct final answer \( 303.75 \) or \( \frac{1215}{4} \).

(a)
Given \(\tan \alpha = \frac{3}{4}\), we have \(\sin \alpha = 0.6\) and \(\cos \alpha = 0.8\).

Resolving perpendicular to the plane:
\[R = mg \cos \alpha = 3 \times 9.8 \times 0.8 = 23.52\text{ N}\]

The maximum frictional force is:
\[F_{\max} = \mu R = 0.5 \times 23.52 = 11.76\text{ N}\]

The component of weight down the plane is:
\[W_p = mg \sin \alpha = 3 \times 9.8 \times 0.6 = 17.64\text{ N}\]

For equilibrium, the net force must be balanced by friction:
Case 1: Particle is on the point of sliding down the plane. The frictional force acts up the plane.
\[F + F_{\max} = W_p \implies F + 11.76 = 17.64 \implies F = 5.88\text{ N}\]

Case 2: Particle is on the point of slipping up the plane. The frictional force acts down the plane.
\[F = W_p + F_{\max} \implies F = 17.64 + 11.76 = 29.4\text{ N}\]

Thus, the range of values of \(F\) for equilibrium is:
\[5.88 \le F \le 29.4\]
(or \(5.9 \le F \le 29\) using 2 s.f.)

(b)
When \(F = 40\text{ N}\), \(F > 29.4\text{ N}\), so the particle moves up the plane and friction opposes motion (acting down the plane).
Using \(F_{\text{net}} = ma\) up the plane:
\[F - W_p - F_{\max} = ma\]
\[40 - 17.64 - 11.76 = 3a\]
\[10.6 = 3a \implies a \approx 3.533...\text{ m s}^{-2}\]

To 3 s.f., the acceleration is \(3.53\text{ m s}^{-2}\) (or \(3.5\text{ m s}^{-2}\) to 2 s.f.).

(a)
- M1: For calculating the normal reaction force \(R = mg \cos \alpha\) and finding \(R = 23.52\text{ N}\).
- B1: For finding \(F_{\max} = 11.76\text{ N}\).
- M1: For setting up the lower limit of equilibrium where friction acts up the slope, i.e., \(F + F_{\max} = mg \sin \alpha\).
- A1: For finding the lower limit \(5.88\text{ N}\) (or \(5.9\text{ N}\)).
- M1: For setting up the upper limit where friction acts down the slope, i.e., \(F - F_{\max} = mg \sin \alpha\).
- A1: For finding the upper limit \(29.4\text{ N}\) (or \(29\text{ N}\)) and stating the final range.

(b)
- M1: For setting up the equation of motion up the plane: \(F - mg \sin \alpha - F_{\max} = ma\).
- A1: For substituting the correct values: \(40 - 17.64 - 11.76 = 3a\).
- M1: For solving for \(a\).
- A1: For finding \(3.53\text{ m s}^{-2}\) or \(3.5\text{ m s}^{-2}\).

(a)
The speed-time graph should consist of:
- A straight line starting at \((0, 0)\) with a positive gradient up to \((t_1, V)\).
- A horizontal line of length \(40\text{ s}\) from \((t_1, V)\) to \((t_1 + 40, V)\).
- A straight line with a negative gradient from \((t_1 + 40, V)\) down to \((t_1 + 40 + t_3, 0)\).

(b)
Let \(t_1\) be the acceleration time. Since \(a = 2\text{ m s}^{-2}\):
\[V = 0 + 2t_1 \implies t_1 = \frac{V}{2}\]

Let \(t_3\) be the deceleration time. Since the deceleration is \(3\text{ m s}^{-2}\):
\[0 = V - 3t_3 \implies t_3 = \frac{V}{3}\]

The constant speed phase lasts for \(t_2 = 40\text{ seconds}\).

Therefore, the total time \(T\) is:
\[T = t_1 + t_2 + t_3 = \frac{V}{2} + 40 + \frac{V}{3} = \frac{5V}{6} + 40\text{ seconds}\]

(c)
The total distance is represented by the area under the speed-time graph (area of a trapezium):
\[D = \frac{1}{2}(t_2 + T)V\]
\[1575 = \frac{1}{2}\left(40 + \frac{5V}{6} + 40\right)V\]
\[1575 = \frac{1}{2}\left(80 + \frac{5V}{6}\right)V\]
\[3150 = 80V + \frac{5V^2}{6}\]

Multiply through by \(\frac{6}{5}\):
\[V^2 + 96V - 3780 = 0\]

Factorising this quadratic equation:
\[(V - 30)(V + 126) = 0\]

Since \(V > 0\), we take the positive root:
\[V = 30\]

(a)
- M1: For a graph showing three distinct phases (acceleration, constant speed, deceleration).
- A1: Correctly labelled axes, with key points showing \(V\) on the vertical axis and showing the shape starting from the origin.

(b)
- M1: For expressing \(t_1\) or \(t_3\) in terms of \(V\) using \(v = u + at\).
- A1: For finding \(t_1 = \frac{V}{2}\) and \(t_3 = \frac{V}{3}\).
- A1: For adding \(t_1\), \(t_2 = 40\), and \(t_3\) to show the given total time expression.

(c)
- M1: For using the formula for the area of a trapezium or splitting into three parts and setting the sum equal to 1575.
- A1: For obtaining a correct equation in \(V\), e.g., \(\frac{1}{2}(80 + \frac{5V}{6})V = 1575\).
- M1: For forming a 3-term quadratic equation in \(V\), e.g., \(V^2 + 96V - 3780 = 0\).
- M1: For solving the quadratic equation (by factorisation, formula, or calculator).
- A1: For \(V = 30\) (rejecting the negative root \(-126\)).

(a)
For particle \(Q\) moving downwards:
\(s = 3.92\text{ m}\), \(u = 0\), \(t = 2\text{ s}\).

Using \(s = ut + \frac{1}{2}at^2\):
\[3.92 = 0 + \frac{1}{2}a(2^2)\]
\[3.92 = 2a \implies a = 1.96\text{ m s}^{-2}\]

(b)
Using equations of motion for both particles:
For \(Q\) (moving downwards):
\[mg - T = ma\]

For \(P\) (moving upwards):
\[T - 4g = 4a\]

Adding these two equations:
\[(m - 4)g = (m + 4)a\]

Given \(a = 1.96\text{ m s}^{-2}\) and \(g = 9.8\text{ m s}^{-2}\):
\[(m - 4) \times 9.8 = (m + 4) \times 1.96\]
\[9.8m - 39.2 = 1.96m + 7.84\]
\[7.84m = 47.04\]
\[m = 6\]

(c)
When \(Q\) hits the floor, \(P\) has travelled \(3.92\text{ m}\) upwards, so its height above the floor is:
\[H_0 = 3.92 + 3.92 = 7.84\text{ m}\]

At this instant, the string becomes slack. The speed of \(P\) at this point is:
\[v_0 = u + at = 0 + 1.96 \times 2 = 3.92\text{ m s}^{-1}\]

Now, \(P\) moves freely under gravity:
\(u = 3.92\text{ m s}^{-1}\), \(a = -9.8\text{ m s}^{-2}\), final velocity \(v = 0\).

Using \(v^2 = u^2 + 2as_1\):
\[0 = 3.92^2 - 2 \times 9.8 \times s_1\]
\[19.6s_1 = 15.3664 \implies s_1 = 0.784\text{ m}\]

Thus, the greatest height above the floor reached by \(P\) is:
\[H_{\max} = 7.84 + 0.784 = 8.624\text{ m}\]

To either 2 or 3 s.f., this is \(8.6\text{ m}\) or \(8.62\text{ m}\).

(a)
- M1: For using \(s = ut + \frac{1}{2}at^2\) with the given parameters \(s = 3.92\) and \(t = 2\).
- A1: For obtaining a correct equation in \(a\).
- A1: For finding \(a = 1.96\text{ m s}^{-2}\).

(b)
- M1: For setting up equations of motion for both particles, \(mg - T = ma\) and \(T - 4g = 4a\).
- M1: For eliminating \(T\) to obtain an equation in terms of \(m\), \(a\), and \(g\).
- A1: For substituting the correct values of \(a\) and \(g\).
- A1: For obtaining \(m = 6\).

(c)
- B1: For finding the speed of \(P\) when \(Q\) hits the floor, \(v_0 = 3.92\text{ m s}^{-1}\).
- M1: For using \(v^2 = u^2 + 2as\) to find the extra displacement \(s_1\) under gravity.
- A1: For adding \(2 \times 3.92 + s_1\) to get \(8.62\text{ m}\) or \(8.6\text{ m}\).

(a)
Using the vector constant acceleration formula \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\):
\[\mathbf{v} - \mathbf{u} = \mathbf{a}t\]
\[(6\mathbf{i} - 5\mathbf{j}) - (-2\mathbf{i} + 3\mathbf{j}) = 4\mathbf{a}\]
\[8\mathbf{i} - 8\mathbf{j} = 4\mathbf{a}\]
\[\mathbf{a} = (2\mathbf{i} - 2\mathbf{j})\text{ m s}^{-2}\]

(b)
Using Newton's Second Law \(\mathbf{F} = m\mathbf{a}\):
\[\mathbf{F} = 0.5(2\mathbf{i} - 2\mathbf{j}) = (\mathbf{i} - \mathbf{j})\text{ N}\]

Now, find the magnitude of \(\mathbf{F}\):
\[|\mathbf{F}| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \approx 1.41\text{ N}\]
(or \(1.4\text{ N}\) to 2 s.f.)

(c)
Using the displacement vector formula \(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) where \(\mathbf{s}\) is the displacement from the position at \(t=0\):
At \(t = 6\text{ seconds}\):
\[\mathbf{s} = (-2\mathbf{i} + 3\mathbf{j})(6) + \frac{1}{2}(2\mathbf{i} - 2\mathbf{j})(6^2)\]
\[\mathbf{s} = (-12\mathbf{i} + 18\mathbf{j}) + \frac{1}{2}(2\mathbf{i} - 2\mathbf{j})(36)\]
\[\mathbf{s} = (-12\mathbf{i} + 18\mathbf{j}) + (36\mathbf{i} - 36\mathbf{j})\]
\[\mathbf{s} = (24\mathbf{i} - 18\mathbf{j})\text{ m}\]

The distance is the magnitude of the displacement vector \(\mathbf{s}\):
\[|\mathbf{s}| = \sqrt{24^2 + (-18)^2} = \sqrt{576 + 324} = \sqrt{900} = 30\text{ m}\]

(a)
- M1: For using \(\mathbf{a} = \frac{\mathbf{v} - \mathbf{u}}{t}\) with vector inputs.
- A1: For obtaining \(\mathbf{a} = 2\mathbf{i} - 2\mathbf{j}\).

(b)
- M1: For using \(\mathbf{F} = m\mathbf{a}\) to find the force vector.
- M1: For calculating the magnitude of their force vector using Pythagoras' theorem.
- A1: For finding \(\sqrt{2}\text{ N}\), \(1.41\text{ N}\) or \(1.4\text{ N}\).

(c)
- M1: For attempting to use the vector formula \(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with \(t=6\).
- A1: For substituting their \(\mathbf{a}\) and \(\mathbf{u}\) correctly.
- A1: For finding \(\mathbf{s} = 24\mathbf{i} - 18\mathbf{j}\).
- M1: For finding the magnitude of their displacement vector.
- A1: For \(30\text{ m}\) exactly.

Let the normal reaction of the plane on the particle be and the frictional force be . Since , we have and . Resolving perpendicular to the inclined plane: . Since the particle is on the point of slipping up the plane, the frictional force acts down the plane and is at its limiting value: . Resolving parallel to the inclined plane: . Substituting the expression for : . Solving for : . Thus, the coefficient of friction is 0.366 (to 3 s.f.).

M1: Resolves perpendicular to the plane to obtain an expression for . A1: Correct equation or . M1: Resolves parallel to the plane with acting down the plane. A1: Correct equation or . B1: Uses the limiting friction condition . M1: Substitutes and to solve for . A1: Correct value of (accept 0.37).

At time , the position vector of is given by . At , the position vector of is . At time , the position vector of is given by . At , the position vector of is . Since they collide at , we equate their position vectors: . Equating the components: . Equating the components: . (b) The velocity of is . The speed of is the magnitude of its velocity: .

Part (a): M1: Expression for the position vector of or at any time or specifically at . A1: Correct position of at , which is . M1: Equates the position vectors at to set up simultaneous equations in and . A1: Correct value of . A1: Correct value of . Part (b): M1: Applies the magnitude formula to their velocity vector of . A1: Correct speed of .

Let the tension in be and the tension in be . (a) Resolving the forces horizontally: . Thus, (to 3 s.f.) or 36 (to 2 s.f.). (b) Resolving the forces vertically: . Substituting the values: . Using : . Thus, (to 3 s.f.) or 4.8 (to 2 s.f.).

Part (a): M1: Resolves horizontally with both tension components. A1: Correct equation . M1: Solves for . A1: or 36. Part (b): M1: Resolves vertically with both tension components and weight . A1: Correct equation . A1: Correct mass or 4.8.

\((a)\) When the rod is on the point of tilting about \( C \), the reaction at \( D \) is \( R_D = 0 \).

Taking moments about \( C \):
\( 12g \times 1 = 15g \times (x - 1) \)

Divide both sides by \( g \):
\( 12 = 15(x - 1) \)
\( 12 = 15x - 15 \)
\( 15x = 27 \)
\( x = 1.8 \)

\((b)\) When the rod is on the point of tilting about \( D \), the reaction at \( C \) is \( R_C = 0 \).
The centre of mass is \( 1.8\text{ m} \) from \( A \). Since \( D \) is \( 4.5\text{ m} \) from \( A \) (as \( BD = 1.5\text{ m} \)), the distance of the centre of mass from \( D \) is:
\( 4.5 - 1.8 = 2.7\text{ m} \)

The particle of mass \( W\text{ kg} \) is at \( B \), which is \( 1.5\text{ m} \) to the right of \( D \).

Taking moments about \( D \):
\( 15g \times 2.7 = Wg \times 1.5 \)

Divide both sides by \( g \):
\( 40.5 = 1.5W \)
\( W = 27 \)

\((c)\) For the rod to remain in equilibrium, both reactions at \( C \) and \( D \) must be non-negative (\( R_C \ge 0 \) and \( R_D \ge 0 \)).

**To prevent tilting about \( C \) (i.e., \( R_D \ge 0 \)):**
Taking moments about \( C \):
\( R_D \times 3.5 + 15g \times 0.8 + Kg \times 5 = 20g \times 1 \)

Since \( R_D \ge 0 \):
\( 15g \times 0.8 + Kg \times 5 \ge 20g \times 1 \)
\( 12 + 5K \ge 20 \)
\( 5K \ge 8 \implies K \ge 1.6 \)

**To prevent tilting about \( D \) (i.e., \( R_C \ge 0 \)):**
Taking moments about \( D \):
\( R_C \times 3.5 + Kg \times 1.5 = 20g \times 4.5 + 15g \times 2.7 \)

Since \( R_C \ge 0 \):
\( Kg \times 1.5 \le 20g \times 4.5 + 15g \times 2.7 \)
\( 1.5K \le 90 + 40.5 \)
\( 1.5K \le 130.5 \implies K \le 87 \)

Thus, the range of possible values for \( K \) is \( 1.6 \le K \le 87 \).

\((a)\)
- **M1**: Realising that \( R_D = 0 \) (implied by setting up a moments equation with no reaction force at \( D \)).
- **M1**: Taking moments about \( C \) with the correct term for the mass of the rod, with consistent units and \( g \) included/omitted consistently.
- **A1**: Correct equation, e.g. \( 12 = 15(x - 1) \) or equivalent.
- **A1**: \( x = 1.8 \)

\((b)\)
- **M1**: Realising that \( R_C = 0 \) (implied by setting up a moments equation with no reaction force at \( C \)).
- **M1**: Taking moments about \( D \) with the correct distance from the centre of mass to \( D \) (i.e., \( 4.5 - \text{their } x \)).
- **A1**: \( W = 27 \)

\((c)\)
- **M1**: Formulating the condition for tilting about \( C \) (e.g. taking moments about \( C \) with \( R_D \ge 0 \) or using \( R_D = 0 \) to find the limit).
- **A1**: Finding the lower limit \( K = 1.6 \) (or \( K \ge 1.6 \)).
- **M1**: Formulating the condition for tilting about \( D \) (e.g. taking moments about \( D \) with \( R_C \ge 0 \) or using \( R_C = 0 \) to find the limit).
- **A1**: Finding the upper limit \( K = 87 \) (or \( K \le 87 \)).
- **A1**: Writing the final correct range \( 1.6 \le K \le 87 \) (accept interval notation \( [1.6, 87] \)).

(a) A Venn diagram is constructed with three intersecting circles labeled \(A\), \(B\), and \(C\) inside a bounding box representing the sample space.

- The central intersection of all three events is \(\mathrm{P}(A \cap B \cap C) = 0.04\).
- The region representing \(A \cap B\) only is \(0.15 - 0.04 = 0.11\).
- The region representing \(A \cap C\) only is \(0.12 - 0.04 = 0.08\).
- The region representing \(B \cap C\) only is \(0.08 - 0.04 = 0.04\).
- The region representing \(A\) only is \(0.45 - 0.11 - 0.08 - 0.04 = 0.22\).
- The region representing \(B\) only is \(0.35 - 0.11 - 0.04 - 0.04 = 0.16\).
- The region representing \(C\) only is \(0.30 - 0.08 - 0.04 - 0.04 = 0.14\).
- The region outside all three circles is \(1 - (0.22 + 0.16 + 0.14 + 0.11 + 0.08 + 0.04 + 0.04) = 1 - 0.79 = 0.21\).

(b) \(\mathrm{P}(A \cup B') = \mathrm{P}(A) + \mathrm{P}(B') - \mathrm{P}(A \cap B')\).
From the diagram, \(\mathrm{P}(B') = 1 - 0.35 = 0.65\).
And \(\mathrm{P}(A \cap B') = \mathrm{P}(\text{only } A) + \mathrm{P}(\text{only } A \cap C) = 0.22 + 0.08 = 0.30\).
So \(\mathrm{P}(A \cup B') = 0.45 + 0.65 - 0.30 = 0.80\).

(c) \(\mathrm{P}(A \cap B' | C) = \frac{\mathrm{P}(A \cap B' \cap C)}{\mathrm{P}(C)}\).
From the diagram, \(\mathrm{P}(A \cap B' \cap C)\) corresponds to the region \(A \cap C\) only, which is \(0.08\).
Therefore, \(\mathrm{P}(A \cap B' | C) = \frac{0.08}{0.30} = \frac{8}{30} = \frac{4}{15} \approx 0.267\) (to 3 s.f.).

(d) For independence, we must have \(\mathrm{P}(A) \times \mathrm{P}(B) = \mathrm{P}(A \cap B)\).
\(\mathrm{P}(A) \times \mathrm{P}(B) = 0.45 \times 0.35 = 0.1575\).
Since \(\mathrm{P}(A \cap B) = 0.15 \neq 0.1575\), the events \(A\) and \(B\) are not independent.

Part (a):
- M1: Draw three intersecting circles with a box.
- A1: Central region correctly labeled 0.04 and at least three other correct intersection regions.
- A1: All regions inside the circles correctly calculated.
- A1: Value 0.21 outside the circles correctly calculated.

Part (b):
- M1: Attempting to use a valid method to find \(\mathrm{P}(A \cup B')\), e.g., summing correct regions or using formula.
- A1: 0.80 or equivalent fraction.

Part (c):
- M1: Using conditional probability formula with a correct numerator and denominator of 0.30.
- A1: \(\frac{4}{15}\) or 0.267.

Part (d):
- M1: Evaluating \(\mathrm{P}(A) \times \mathrm{P}(B)\) and making a direct comparison with \(\mathrm{P}(A \cap B)\).
- A1: Correct conclusion based on a sound numerical comparison.

(a) Since the sum of all probabilities must equal 1:
\[0.1 + a + b + 0.25 + 0.15 = 1 \implies a + b = 0.5\]
The expectation is given as:
\[\mathrm{E}(X) = -2(0.1) + 0(a) + 1(b) + k(0.25) + 5(0.15) = 1.35\]
\[-0.2 + b + 0.25k + 0.75 = 1.35 \implies b + 0.25k = 0.8\]

(b) Since \(k\) is an integer such that \(1 < k < 5\), the possible values of \(k\) are 2, 3, or 4.
- If \(k = 2\):
\[b + 0.25(2) = 0.8 \implies b = 0.3\]
\[a = 0.5 - 0.3 = 0.2\]
We calculate \(\mathrm{E}(X^2)\):
\[\mathrm{E}(X^2) = (-2)^2(0.1) + 0^2(0.2) + 1^2(0.3) + 2^2(0.25) + 5^2(0.15) = 0.4 + 0.3 + 1.0 + 3.75 = 5.45\]
\[\mathrm{Var}(X) = 5.45 - 1.35^2 = 3.6275 \neq 4.6275\]

- If \(k = 3\):
\[b + 0.25(3) = 0.8 \implies b = 0.05\]
\[a = 0.5 - 0.05 = 0.45\]
We calculate \(\mathrm{E}(X^2)\):
\[\mathrm{E}(X^2) = (-2)^2(0.1) + 0^2(0.45) + 1^2(0.05) + 3^2(0.25) + 5^2(0.15) = 0.4 + 0.05 + 2.25 + 3.75 = 6.45\]
\[\mathrm{Var}(X) = 6.45 - 1.35^2 = 4.6275\]
This matches the given value of the variance.

- If \(k = 4\):
\[b + 0.25(4) = 0.8 \implies b = -0.2\], which is impossible since probabilities cannot be negative.

Thus, \(k = 3\), \(a = 0.45\), and \(b = 0.05\).

(c) Using the linear properties of expectation and variance:
\[\mathrm{E}(3 - 2X) = 3 - 2\mathrm{E}(X) = 3 - 2(1.35) = 0.3\]
\[\mathrm{Var}(3 - 2X) = (-2)^2 \mathrm{Var}(X) = 4 \times 4.6275 = 18.51\]

Part (a):
- M1: Equating the sum of probabilities to 1 to find a relation between \(a\) and \(b\).
- M1: Using \(\mathrm{E}(X) = \sum x \mathrm{P}(X=x)\) to form an equation in \(b\) and \(k\).
- A1: Correct algebraic completion to show \(b + 0.25k = 0.8\).

Part (b):
- M1: Setting up trial values for the integer \(k\) (2, 3, or 4).
- M1: Finding candidate values for \(a\) and \(b\) for at least one choice of \(k\).
- A1: Correctly calculating \(\mathrm{E}(X^2)\) for at least one of the choices.
- M1: Checking the variance constraint \(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2\).
- A1: Correct values: \(k = 3\), \(a = 0.45\), \(b = 0.05\).

Part (c):
- M1: Applying the formula \(\mathrm{E}(aX+b) = a\mathrm{E}(X)+b\).
- A1: 0.3.
- M1: Applying the formula \(\mathrm{Var}(aX+b) = a^2 \mathrm{Var}(X)\).
- A1: 18.51.

(a) We are given \(\mathrm{P}(L < 47.5) = 0.15\) and \(\mathrm{P}(L > 53.2) = 0.08\).

Standardizing the first probability:
\[\mathrm{P}\left(Z < \frac{47.5 - \mu}{\sigma}\right) = 0.15\]
Since this probability is less than 0.5, the z-value must be negative. From standard normal tables, \(\Phi(-1.0364) = 0.15\).
Thus:
\[47.5 - \mu = -1.0364\sigma \implies \mu - 1.0364\sigma = 47.5\quad (1)\]

Standardizing the second probability:
\[\mathrm{P}\left(Z > \frac{53.2 - \mu}{\sigma}\right) = 0.08 \implies \mathrm{P}\left(Z < \frac{53.2 - \mu}{\sigma}\right) = 0.92\]
From standard normal tables, \(\Phi(1.4051) = 0.92\).
Thus:
\[53.2 - \mu = 1.4051\sigma \implies \mu + 1.4051\sigma = 53.2\quad (2)\]

(b) Subtracting equation (1) from equation (2):
\[(1.4051 + 1.0364)\sigma = 53.2 - 47.5\]
\[2.4415\sigma = 5.7 \implies \sigma = 2.3346...\quad \implies \sigma \approx 2.33\text{ (to 3 s.f.)}\]

Substituting this back into equation (1):
\[\mu = 47.5 + 1.0364(2.3346) = 49.920...\quad \implies \mu \approx 49.9\text{ (to 3 s.f.)}\]

(c) We want to find \(\mathrm{P}(48.0 < L < 52.5)\) using \(\mu = 49.920\) and \(\sigma = 2.335\):
Standardizing both limits:
\[z_1 = \frac{48.0 - 49.920}{2.335} \approx -0.822\]
\[z_2 = \frac{52.5 - 49.920}{2.335} \approx 1.105\]

Using standard normal tables:
- \(\Phi(1.10) = 0.8643\) (or \(\Phi(1.105) \approx 0.8654\) with calculator/interpolation)
- \(\Phi(-0.82) = 1 - 0.7939 = 0.2061\) (or \(\Phi(-0.822) \approx 0.2056\))

Calculating the probability:
\[\mathrm{P}(48.0 < L < 52.5) = \Phi(z_2) - \Phi(z_1)\]
Using 2 d.p. tables: \(0.8643 - 0.2061 = 0.6582\).
Using exact calculator value: \(0.8654 - 0.2056 = 0.6598\).
Both methods yield a probability of approximately \(0.658\) to \(0.660\).

Part (a):
- M1: Attempting to standardize either 47.5 or 53.2, setting equal to a z-value.
- B1: Finding z-values of \(-1.0364\) (accept \(-1.04\)) and \(1.4051\) (accept \(1.41\)) from tables.
- A1: Equation 1 correct: \(\mu - 1.0364\sigma = 47.5\) (or equivalent).
- A1: Equation 2 correct: \(\mu + 1.4051\sigma = 53.2\) (or equivalent).

Part (b):
- M1: Subtracting or substituting to eliminate one variable.
- A1: \(\sigma = 2.33\) (accept answers in range 2.32 to 2.35).
- M1: Substituting their \(\sigma\) back into one equation to find \(\mu\).
- A1: \(\mu = 49.9\) (accept answers in range 49.8 to 50.0).

Part (c):
- M1: Attempting to standardize both 48.0 and 52.5 with their mean and standard deviation.
- M1: Correct method to find the area between these two limits, e.g., \(\Phi(z_2) - (1 - \Phi(|z_1|))\).
- A1: Accept any value in the range \([0.658, 0.660]\).

(a) \(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 148 - \frac{32^2}{8} = 148 - 128 = 20\). \(S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 422 - \frac{56^2}{8} = 422 - 392 = 30\). \(S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 246 - \frac{32 \times 56}{8} = 246 - 224 = 22\). (b) \(r = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}} = \frac{22}{\sqrt{20 \times 30}} = \frac{22}{\sqrt{600}} \approx 0.898\) (to 3 s.f.). (c) Since \(r = 0.898\) is close to 1, there is a strong positive linear correlation, so a linear regression model is suitable. (d) \(b = \frac{S_{xy}}{S_{xx}} = \frac{22}{20} = 1.1\). \(\bar{x} = 4\), \(\bar{y} = 7\). \(a = \bar{y} - b\bar{x} = 7 - 1.1(4) = 2.6\). Equation: \(y = 2.6 + 1.1x\). (e) For each additional year of the machinery part's age, the annual maintenance cost increases by 1.1 hundred pounds (or £110).

(a) M1: For a correct method to find any of \(S_{xx}\), \(S_{yy}\) or \(S_{xy}\). A1: Two of \(S_{xx} = 20\), \(S_{yy} = 30\), \(S_{xy} = 22\) correct. A1: All three correct. (b) M1: Correct expression for \(r\) with their values. A1: \(r \approx 0.898\). (c) B1: Identifies strong positive correlation. B0.75: Concludes suitability of linear model based on this. (d) M1: Correct method for \(b\). M1: Correct method for \(a\). A1: Correct equation \(y = 2.6 + 1.1x\). (e) B1: Correct interpretation referring to increase in maintenance cost of £110 per year of age.

(a) \(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 285 - \frac{45^2}{10} = 82.5\). \(S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 185 - \frac{35^2}{10} = 62.5\). \(S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 210 - \frac{45 \times 35}{10} = 52.5\). (b) \(r = \frac{52.5}{\sqrt{82.5 \times 62.5}} \approx 0.731\). (c) The PMCC between \(h\) and \(w\) is also \(0.731\). Coding of the form \(x = px + q\) (with positive scale factor) does not affect the product moment correlation coefficient. (d) For \(y\) on \(x\), \(b = \frac{52.5}{82.5} = \frac{7}{11}\). \(\bar{x} = 4.5\), \(\bar{y} = 3.5\). \(a = 3.5 - \frac{7}{11}(4.5) = \frac{7}{11}\). So, \(y = \frac{7}{11} + \frac{7}{11}x\). Substituting the coded variables: \(2w - 3 = \frac{7}{11} + \frac{7}{11}(h - 5)\) which simplifies to \(2w = \frac{7}{11}h + \frac{5}{11}\), hence \(w = \frac{7}{22}h + \frac{5}{22}\), which is \(w \approx 0.227 + 0.318h\).

(a) M1: Correct method for any of \(S_{xx}\), \(S_{yy}\), \(S_{xy}\). A1: Two of \(S_{xx} = 82.5\), \(S_{yy} = 62.5\), \(S_{xy} = 52.5\) correct. A1: All three correct. (b) M1: Correct formula for \(r\). A1: \(r \approx 0.731\). (c) B1: Correct value of \(0.731\). B0.75: States coding does not affect PMCC. (d) M1: Correct method for gradient of \(y\) on \(x\). M1: Correct method for intercept of \(y\) on \(x\). M1: Correct substitution of coded variables. A1: Correct final equation.

(a) \(\bar{t} = 20\), \(\bar{s} = 13\). \(S_{tt} = 5020 - \frac{240^2}{12} = 220\). \(S_{ts} = 3380 - \frac{240 \times 156}{12} = 260\). \(b = \frac{260}{220} = \frac{13}{11} \approx 1.18\). \(a = 13 - \frac{13}{11}(20) = -\frac{117}{11} \approx -10.6\). Regression line: \(s = -10.6 + 1.18t\). (b) (i) For \(t = 21\), \(s = -\frac{117}{11} + \frac{13}{11}(21) \approx 14.2\) (or £1420). (ii) For \(t = 32\), \(s = -\frac{117}{11} + \frac{13}{11}(32) \approx 27.2\) (or £2720). (c) For \(t = 21\) °C, the estimate is reliable because 21 is within the range of observed data (interpolation). For \(t = 32\) °C, the estimate is unreliable because 32 is outside the range of observed data (extrapolation).

(a) M1: Correct method for \(S_{tt}\). M1: Correct method for \(S_{ts}\). M1: Correct method for \(b\). M1: Correct method for \(a\). A1: Correct equation. (b) B1: Correct estimate for 21 °C. B1: Correct estimate for 32 °C. (c) B1.75: States reliable and mentions interpolation/within range. B2: States unreliable and mentions extrapolation/outside range.

(a) \(r = \frac{S_{pq}}{\sqrt{S_{pp} S_{qq}}} = \frac{-180}{\sqrt{144 \times 324}} = \frac{-180}{12 \times 18} = \frac{-180}{216} = -\frac{5}{6}\). (b) \(r^2 = (-\frac{5}{6})^2 = \frac{25}{36} \approx 0.694\) (or 69.4%). Approximately 69.4% of the variation in the weekly demand is explained by the linear relationship with price. (c) \(d = \frac{-180}{144} = -1.25\). \(c = 40 - (-1.25)(12) = 55\). So, \(q = 55 - 1.25p\). (d) Price is the independent variable and demand is the dependent variable. To estimate demand from price, we should minimize residuals in the demand direction, which is done by the regression of \(q\) on \(p\), not \(p\) on \(q\).

(a) M1: Formula for \(r\). A1: Correct substitution. A0.75: Correct simplification to \(-\frac{5}{6}\). (b) B1: \(r^2 \approx 0.694\). M1: References variation in demand explained by linear relationship with price. A1: Correctly interprets percentage. (c) M1: Correct method for \(d\). M1: Correct method for \(c\). A1: Correct equation. (d) B1: Identifies independent vs dependent variables. B1: Explains that regression of \(q\) on \(p\) minimizes residuals in the direction of the estimated variable.