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Let the normal reaction between the particle and the plane be \(R\). Resolving perpendicular to the plane: \(R = mg \cos\alpha\). Since \(\sin\alpha = 0.6\), we have \(\cos\alpha = \sqrt{1 - 0.6^2} = 0.8\), so \(R = 0.8mg\). The maximum frictional force is \(F = \mu R = 0.8\mu mg\). Case 1: The force of magnitude \(P\) acts up the plane and the particle is on the point of slipping down. Here, friction acts up the plane. Resolving parallel to the plane: \(P + F = mg \sin\alpha\), which gives \(P + 0.8\mu mg = 0.6mg\). Case 2: The force of magnitude \(2P\) acts up the plane and the particle is on the point of slipping up. Here, friction acts down the plane. Resolving parallel to the plane: \(2P = mg \sin\alpha + F\), which gives \(2P = 0.6mg + 0.8\mu mg\). Substituting the expression for \(P\) from the first equation, \(P = 0.6mg - 0.8\mu mg\), into the second equation: \(2(0.6mg - 0.8\mu mg) = 0.6mg + 0.8\mu mg\). Dividing both sides by \(mg\), we get \(2(0.6 - 0.8\mu) = 0.6 + 0.8\mu\). This simplifies to \(1.2 - 1.6\mu = 0.6 + 0.8\mu\), which gives \(0.6 = 2.4\mu\). Thus, \(\mu = 0.25\).

M1: For resolving forces perpendicular to the plane to obtain \( R = mg \cos\alpha \) and using \(\cos\alpha = 0.8\). M1: For setting up the equation of equilibrium parallel to the plane for Case 1 with friction acting up the plane: \(P + \mu R = mg \sin\alpha\). A1: Correct first equation: \( P + 0.8\mu mg = 0.6mg \) (or equivalent). M1: For setting up the equation of equilibrium parallel to the plane for Case 2 with friction acting down the plane, and solving the resulting equations to eliminate \(P\) and \(mg\). A1: For obtaining \(\mu = 0.25\) (or \(\frac{1}{4}\)).

(a) Let T be the tension in the string and a be the acceleration of the system. For A, the normal reaction is R = 3mg. The friction is F = \mu R = mg. The equation of motion is T - mg = 3ma. For B, the equation of motion is 2mg - T = 2ma. (b) Adding the equations: mg = 5ma, which gives a = 0.2g. (c) T = mg + 3m(0.2g) = 1.6mg. (d) The tension is the same in both vertical and horizontal parts of the string. (e) Speed v just before B hits the floor: v^2 = 2as = 2(0.2g)(1.5) = 0.6g. After B hits the floor, T = 0. Deceleration of A: -mg = 3ma' which gives a' = -g/3. Distance d to rest: 0 = 0.6g - 2(g/3)d which gives d = 0.9 m. Total distance = 1.5 + 0.9 = 2.4 m.

(a) M1: Resolving vertically for A and using F = \mu R. A1: Both equations correct: T - mg = 3ma and 2mg - T = 2ma. (b) M1: Adding equations. A1: Showing mg = 5ma. A1: Correctly obtaining a = 0.2g. (c) M1: Substituting a back. A1: T = 1.6mg. (d) B1: Tension is constant throughout the string. (e) M1: Finding v^2. A1: v^2 = 5.88. M1: New deceleration and finding d. A1: Total distance = 2.4 m.

(a) Setting v = 0 gives 3t^2 - 16t + 20 = 0, so (3t - 10)(t - 2) = 0, giving t = 2 and t = 10/3. (b) a = dv/dt = 6t - 16. At t = 4, a = 6(4) - 16 = 8 m/s^2. (c) Displacement s(t) = t^3 - 8t^2 + 20t. P changes direction at t = 2. s(0) = 0, s(2) = 16, s(3) = 15. Distance = |16 - 0| + |15 - 16| = 17 m.

(a) M1: setting v = 0 and solving. A1: one value. A1: both values. (b) M1: differentiating. A1: correct derivative. A1: finding 8. (c) M1: integrating. A1: correct integration. M1: identifying reversal at t = 2. A1: correct displacements. A1.4: total distance 17 m.

(a) At tilting about C, R_D = 0. Moments about C: 5g * 1.5 = 10g * (x - 1.5) => 7.5 = 10(x - 1.5) => x = 2.25. (b) At tilting about D, R_C = 0. AD = 5 m. Distance from center of mass to D is 5 - 2.25 = 2.75 m. Moments about D: 10g * 2.75 = Mg * 1 => M = 27.5. (c) Rigidity ensures the plank remains straight and doesn't bend, keeping distances constant.

(a) M1: setting R_D = 0. M1: taking moments. A1: correct equation. A1: solving. A1: x = 2.25. (b) M1: setting R_C = 0. M1: finding distance 2.75. M1: taking moments. A1: correct equation. A1: M = 27.5. (c) B1.4: explaining no bending.

(a) R = (2a+3)i + (b-5)j. Since R is parallel to i + 2j, (b-5)/(2a+3) = 2 => b-5 = 4a+6 => 4a-b = -11. (b) If a = 1, then b = 15. R = 5i + 10j. Using F = ma, 5i + 10j = 0.5a => a = 10i + 20j. (c) |a| = sqrt(10^2 + 20^2) = sqrt(500) = 22.4 m/s^2. Angle = arctan(20/10) = 63.4 degrees.

(a) M1: summing forces. A1: correct R. M1: setting up ratio. A1: correct proof. (b) M1: finding b. A1: finding R. M1: F=ma. A1: a = 10i + 20j. (c) M1: magnitude. A1: 22.4. A1.4: 63.4 degrees.

(a) Resolving horizontally: T cos theta = 18 => 0.8T = 18 => T = 22.5 N. (b) Resolving vertically: T sin theta = mg => 22.5 * 0.6 = 9.8m => 13.5 = 9.8m => m = 1.38. (c) (i) Resolving horizontally: 30 cos theta = S cos 30 => 24 = S sqrt(3)/2 => S = 16sqrt(3) = 27.7 N. (ii) Resolving vertically: 30 sin theta = Mg + S sin 30 => 18 = 9.8M + 8sqrt(3) => 9.8M = 18 - 13.856 => M = 4.144/9.8 = 0.423 kg.

(a) M1: resolving horizontally. A1: equation. A1: T = 22.5 N. (b) M1: resolving vertically. A1: equation. A1: m = 1.38. (c) M1: horizontal resolution. A1: equation. A1: S = 27.7 N. M1: vertical resolution. A1.4: M = 0.423 kg.

From \( \tan \alpha = \frac{3}{4} \), we obtain \( \sin \alpha = \frac{3}{5} = 0.6 \) and \( \cos \alpha = \frac{4}{5} = 0.8 \).

**(a)**
Let the tension in the string be \( T \) and the acceleration of the system be \( a \).
For particle \( B \) (moving downwards):
\( 4mg - T = 4ma \)

For particle \( A \) (moving up the plane):
\( T - 3mg \sin \alpha - F = 3ma \)

**(b)**
Resolving forces perpendicular to the plane for \( A \):
\( R = 3mg \cos \alpha = 3mg(0.8) = 2.4mg \)

Since \( A \) is moving, the friction force is at its maximum value:
\( F = \mu R = \frac{1}{4}(2.4mg) = 0.6mg \)

Substituting this and \( \sin \alpha = 0.6 \) into the equation of motion for \( A \):
\( T - 3mg(0.6) - 0.6mg = 3ma \implies T - 2.4mg = 3ma \)

Adding the equations of motion for \( A \) and \( B \):
\( (4mg - T) + (T - 2.4mg) = 4ma + 3ma \)
\( 1.6mg = 7ma \)
\( a = \frac{1.6}{7}g = \frac{8}{35}g \)

Using \( g = 9.8 \text{ m s}^{-2} \):
\( a = \frac{8}{35} \times 9.8 = 2.24 \text{ m s}^{-2} \) (as required).

**(c)**
Substituting \( a = 2.24 \text{ m s}^{-2} \) back into the equation of motion for \( B \):
\( T = 4m(g - a) = 4m(9.8 - 2.24) = 4m(7.56) = 30.24m \text{ N} \).

Rounding to 2 or 3 significant figures gives \( 30m \) or \( 30.2m \).

**(d)**
First, find the speed \( v \) of \( A \) at the instant the string breaks (after \( 1.5 \) seconds of motion from rest):
\( v = u + at = 0 + 2.24 \times 1.5 = 3.36 \text{ m s}^{-1} \)

When the string breaks, the tension \( T \) becomes \( 0 \).
Let \( a' \) be the new acceleration of \( A \). Since \( A \) is still moving up the plane, the friction force \( F \) still acts down the plane.
Resolving forces along the plane for \( A \):
\( -3mg \sin \alpha - F = 3ma' \)
\( -1.8mg - 0.6mg = 3ma' \implies -2.4mg = 3ma' \)
\( a' = -0.8g = -0.8 \times 9.8 = -7.84 \text{ m s}^{-2} \)

For the subsequent motion of \( A \) to its first point of rest, we use \( v_f^2 = u_f^2 + 2a's \) with \( v_f = 0 \), \( u_f = 3.36 \text{ m s}^{-1} \), and \( a' = -7.84 \text{ m s}^{-2} \):
\( 0^2 = 3.36^2 + 2(-7.84)s \)
\( 15.68s = 11.2896 \)
\( s = 0.72 \text{ m} \)

**(part a)**
- **M1**: Resolving forces parallel to the plane for \( A \) with tension, weight component, and friction present.
- **A1**: A correct equation of motion for \( A \): \( T - 3mg \sin \alpha - F = 3ma \) or equivalent.
- **B1**: A correct equation of motion for \( B \): \( 4mg - T = 4ma \).

**(part b)**
- **M1**: Resolving perpendicular to the plane to find \( R \) and attempting to calculate the limiting friction \( F = \mu R \).
- **A1**: Finding \( R = 2.4mg \) and \( F = 0.6mg \).
- **M1**: Eliminating \( T \) from the simultaneous equations to find an expression for \( a \).
- **A1* (cso)**: Showing clearly that \( a = \frac{8}{35}g \) and substituting \( g = 9.8 \) to obtain the given value \( 2.24 \text{ m s}^{-2} \).

**(part c)**
- **M1**: Substituting their value of \( a \) into one of the equations of motion to find \( T \) in terms of \( m \).
- **A1**: Finding \( T = 30.24m \) (accept \( 30m \) or \( 30.2m \)).

**(part d)**
- **M1**: Using \( v = u + at \) with their \( a \) from (b) and \( t = 1.5 \) to find the speed when the string breaks.
- **A1**: Finding \( v = 3.36 \text{ m s}^{-1} \).
- **M1**: Setting up the new equation of motion for \( A \) with \( T = 0 \) and finding the deceleration \( a' = -7.84 \text{ m s}^{-2} \) (or \( -0.8g \)).
- **A1**: Using \( v^2 = u^2 + 2as \) with \( v = 0 \) to find \( s = 0.72 \text{ m} \).

For the quadratic equation \(kx^2 + (k+3)x + k = 0\) to have real roots, its discriminant must be greater than or equal to zero:
\(\Delta = b^2 - 4ac \ge 0\)

Identify the coefficients:
\(a = k\), \(b = k+3\), \(c = k\)

Substitute these into the discriminant formula:
\((k+3)^2 - 4(k)(k) \ge 0\)
\(k^2 + 6k + 9 - 4k^2 \ge 0\)
\(-3k^2 + 6k + 9 \ge 0\)

Divide the entire inequality by \(-3\) (remembering to reverse the inequality sign):
\(k^2 - 2k - 3 \le 0\)

Factorise the quadratic expression:
\((k-3)(k+1) \le 0\)

The critical values are \(k = -1\) and \(k = 3\).
Since we require the expression to be less than or equal to zero, the solution range is:
\(-1 \le k \le 3\)

However, because the original equation is specified to be a quadratic equation, the coefficient of \(x^2\) cannot be zero. Therefore, \(k \neq 0\).

Combining these conditions, the set of possible values for \(k\) is:
\(-1 \le k \le 3\) and \(k \neq 0\) (or written as \([-1, 0) \cup (0, 3]\)).

M1: Attempts to find the discriminant \(b^2 - 4ac\) with \(a = k\), \(b = k+3\), and \(c = k\).
A1: Obtains a correct simplified expression for the discriminant or a correct quadratic inequality, e.g., \(-3k^2 + 6k + 9 \ge 0\) or \(k^2 - 2k - 3 \le 0\).
M1: Solves their 3-term quadratic equation or inequality to find critical values (must see factors or use of quadratic formula showing \(k = -1\) and \(k = 3\)).
A1: Correct outside boundaries, identifying the closed interval \(-1 \le k \le 3\).
A1: Correctly states or implies that \(k \neq 0\) because the equation is quadratic, giving the final complete set of values \(-1 \le k \le 3, k \neq 0\).

(a) Rearrange the equation of \(l_1\) to find its gradient:
\(2y = 3x + 8 \implies y = \frac{3}{2}x + 4\)

The gradient of \(l_1\) is \(m_1 = \frac{3}{2}\).
Since \(l_2\) is perpendicular to \(l_1\), its gradient \(m_2\) satisfies \(m_1 \times m_2 = -1\):
\(m_2 = -\frac{2}{3}\)

Using the point-gradient formula with point \(P(6, -1)\):
\(y - y_1 = m(x - x_1)\)
\(y - (-1) = -\frac{2}{3}(x - 6)\)
\(y + 1 = -\frac{2}{3}x + 4\)

Multiply through by 3 to eliminate the fraction:
\(3(y + 1) = -2(x - 6)\)
\(3y + 3 = -2x + 12\)
\(3y + 2x - 9 = 0\)
This is in the required form \(ay + bx + c = 0\) where \(a = 3\), \(b = 2\), and \(c = -9\).

(b) To find the coordinates of \(Q\), set \(x = 0\) in the equation of \(l_2\):
\(3y + 2(0) - 9 = 0 \implies 3y = 9 \implies y = 3\)
So, \(Q\) has coordinates \((0, 3)\).

Triangle \(OPQ\) has vertices at \(O(0,0)\), \(P(6, -1)\), and \(Q(0,3)\).
Taking \(OQ\) as the base along the \(y\)-axis:
\(\text{Base} = 3\)
The height is the perpendicular distance from \(P(6, -1)\) to the \(y\)-axis, which is the absolute value of the \(x\)-coordinate of \(P\):
\(\text{Height} = 6\)

Therefore, the area of triangle \(OPQ\) is:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 6 = 9\)

Part (a):
M1: Identifies the gradient of \(l_1\) is \(\frac{3}{2}\) and uses the perpendicular rule \(m_2 = -\frac{1}{m_1}\) to find the gradient of \(l_2\) is \(-\frac{2}{3}\).
M1: Uses their perpendicular gradient and point \((6, -1)\) to construct a linear equation.
A1: Obtains a correct equation in the specified form, e.g., \(3y + 2x - 9 = 0\) (or any non-zero integer multiple thereof).

Part (b):
B1: Identifies the coordinates of \(Q\) as \((0, 3)\).
M1: Employs a valid method to find the area of triangle \(OPQ\), such as \(\frac{1}{2} \times y_Q \times |x_P|\).
A1: Obtains the correct area of \(9\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to rewrite the equation in terms of \(\cos \theta\) only:
\(6(1 - \cos^2 \theta) + 5 \cos \theta - 5 = 0\)
\(6 - 6 \cos^2 \theta + 5 \cos \theta - 5 = 0\)
\(-6 \cos^2 \theta + 5 \cos \theta + 1 = 0\)

Multiply through by \(-1\) to write as a standard quadratic equation:
\(6 \cos^2 \theta - 5 \cos \theta - 1 = 0\)

Factorise the quadratic expression:
\((6 \cos \theta + 1)(\cos \theta - 1) = 0\)

This gives two possible equations:
1) \(\cos \theta = 1\)
2) \(\cos \theta = -\frac{1}{6}\)

Solve \(\cos \theta = 1\) in the range \(0 \le \theta < 360^\circ\):
\(\theta = 0^\circ\) (note that \(360^\circ\) is excluded by the inequality \(\theta < 360^\circ\)).

Solve \(\cos \theta = -\frac{1}{6}\) in the range \(0 \le \theta < 360^\circ\):
First, find the principal value:
\(\cos^{-1}\left(\frac{1}{6}\right) \approx 80.41^\circ\)
Since the cosine value is negative, solutions are in the second and third quadrants:
Second quadrant: \(\theta = 180^\circ - 80.41^\circ = 99.59^\circ \approx 99.6^\circ\)
Third quadrant: \(\theta = 180^\circ + 80.41^\circ = 260.41^\circ \approx 260.4^\circ\)

Thus, the solutions in the interval are \(\theta = 0^\circ, 99.6^\circ, 260.4^\circ\).

M1: Substitutes \(\sin^2 \theta = 1 - \cos^2 \theta\) to form an equation in \(\cos \theta\).
A1: Obtains a correct three-term quadratic equation, e.g., \(6 \cos^2 \theta - 5 \cos \theta - 1 = 0\) or equivalent.
M1: Solves their quadratic to obtain values for \(\cos \theta\) (must see factors \((6 \cos \theta + 1)(\cos \theta - 1) = 0\) or formula).
B1: Identifies \(\theta = 0^\circ\) as a solution (and does not include \(360^\circ\)).
M1: Uses a correct quadrant method to find at least one solution to 1 d.p. from \(\cos \theta = -\frac{1}{6}\).
A1: Both \(\theta = 99.6^\circ\) and \(\theta = 260.4^\circ\) are given, with no extra solutions in the range.

First, find the \(y\)-coordinate of the point \(P\) by substituting \(x = 4\) into the equation of the curve:
\(y = 2(4)^{\frac{3}{2}} - \frac{16}{4} + 5\)
\(y = 2(8) - 4 + 5\)
\(y = 16 - 4 + 5 = 17\)
So, the coordinates of \(P\) are \((4, 17)\).

Next, express the equation of the curve in a form ready for differentiation:
\(y = 2x^{\frac{3}{2}} - 16x^{-1} + 5\)

Differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = 2 \left(\frac{3}{2}\right)x^{\frac{1}{2}} - 16(-1)x^{-2} + 0\)
\(\frac{dy}{dx} = 3x^{\frac{1}{2}} + 16x^{-2}\)
\(\frac{dy}{dx} = 3\sqrt{x} + \frac{16}{x^2}\)

Substitute \(x = 4\) into the derivative to find the gradient of the tangent at \(P\):
\(m = 3\sqrt{4} + \frac{16}{4^2} = 3(2) + \frac{16}{16} = 6 + 1 = 7\)

Now, use the equation of a straight line with gradient \(m = 7\) and passing through \(P(4, 17)\):
\(y - y_1 = m(x - x_1)\)
\(y - 17 = 7(x - 4)\)
\(y - 17 = 7x - 28\)
\(y = 7x - 11\)

Thus, the equation of the tangent is \(y = 7x - 11\).

B1: Correctly evaluates the \(y\)-coordinate of \(P\) as \(17\).
M1: Attempts to differentiate the given expression with at least one power of \(x\) decreased by 1 (i.e. \(x^{3/2} \to k_1 x^{1/2}\) or \(x^{-1} \to k_2 x^{-2}\)).
A1: Correct derivative \(\frac{dy}{dx} = 3x^{\frac{1}{2}} + 16x^{-2}\) (unsimplified is acceptable).
M1: Substitutes \(x = 4\) into their \(\frac{dy}{dx}\) to find the gradient of the tangent.
M1: Employs their point \(P\) and gradient to construct the straight line equation.
A1: Obtains the correct equation in the required form: \(y = 7x - 11\).

First, simplify the expression for \(f'(x)\) by dividing each term in the numerator by the denominator:
\(f'(x) = \frac{4x^2}{x} - \frac{5x^{\frac{1}{2}}}{x}\)
\(f'(x) = 4x - 5x^{-\frac{1}{2}}\)

To find \(f(x)\), integrate \(f'(x)\) with respect to \(x\):
\(f(x) = \int (4x - 5x^{-\frac{1}{2}}) \, dx\)
\(f(x) = \frac{4x^2}{2} - \frac{5x^{\frac{1}{2}}}{\frac{1}{2}} + C\)
\(f(x) = 2x^2 - 10x^{\frac{1}{2}} + C\)
\(f(x) = 2x^2 - 10\sqrt{x} + C\)

Using the given boundary condition that the curve passes through the point \((9, 10)\), substitute \(x = 9\) and \(y = f(x) = 10\) into the integrated equation to determine \(C\):
\(10 = 2(9)^2 - 10\sqrt{9} + C\)
\(10 = 2(81) - 10(3) + C\)
\(10 = 162 - 30 + C\)
\(10 = 132 + C\)
\(C = 10 - 132 = -122\)

Substitute the value of \(C\) back into the equation of the curve:
\(f(x) = 2x^2 - 10\sqrt{x} - 122\)

M1: Attempts to rewrite the expression for \(f'(x)\) as individual terms of \(x\) to some power, with at least one power correct.
A1: Correctly simplified derivative: \(f'(x) = 4x - 5x^{-\frac{1}{2}}\).
M1: Integrates their simplified expression, raising the power by 1 in at least one term.
A1: Correct integrated expression: \(2x^2 - 10x^{\frac{1}{2}}\) (constant of integration not required for this mark).
M1: Substitutes \(x = 9\) and \(y = 10\) into their integrated function containing a constant \(C\) to find the value of \(C\).
A1: Correct final equation: \(f(x) = 2x^2 - 10\sqrt{x} - 122\) or equivalent.

(a) To find the points of intersection, set the equation of the curve equal to the equation of the line:
\(x^3 - 2x^2 - 5x + 6 = 3x + 6\)

Subtract \(3x + 6\) from both sides to form a polynomial equation:
\(x^3 - 2x^2 - 8x = 0\)

Factor out \(x\):
\(x(x^2 - 2x - 8) = 0\)

Factorise the quadratic part:
\(x(x - 4)(x + 2) = 0\)

This gives three solutions for \(x\):
\(x = -2\), \(x = 0\), \(x = 4\)

Now, substitute each \(x\)-value back into the simpler equation (the line \(y = 3x + 6\)) to get the corresponding \(y\)-coordinates:
For \(x = -2\): \(y = 3(-2) + 6 = 0\) \(\implies (-2, 0)\)
For \(x = 0\): \(y = 3(0) + 6 = 6\) \(\implies (0, 6)\)
For \(x = 4\): \(y = 3(4) + 6 = 18\) \(\implies (4, 18)\)

So, the coordinates of the points of intersection are \((-2, 0)\), \((0, 6)\), and \((4, 18)\).

(b) The inequality \(x^3 - 2x^2 - 5x + 6 > 3x + 6\) is equivalent to:
\(x^3 - 2x^2 - 8x > 0\)
\(x(x - 4)(x + 2) > 0\)

Consider the signs of the three factors in different intervals defined by the critical values \(-2\), \(0\), and \(4\):
- For \(x < -2\): All three factors are negative \((- \times - \times - = -)\), so the expression is negative.
- For \(-2 < x < 0\): \(x+2\) is positive, other two are negative \((- \times - \times + = +)\), so the expression is positive.
- For \(0 < x < 4\): \(x\) and \(x+2\) are positive, \(x-4\) is negative \((+ \times - \times + = -)\), so the expression is negative.
- For \(x > 4\): All factors are positive \((+ \times + \times + = +)\), so the expression is positive.

Thus, the solution to the inequality is \(-2 < x < 0\) or \(x > 4\).

Part (a):
M1: Equates the curve and line to set up an equation: \(x^3 - 2x^2 - 5x + 6 = 3x + 6\).
A1: Formulates \(x^3 - 2x^2 - 8x = 0\) and solves it to obtain the roots \(x = -2, 0, 4\).
M1: Substitutes at least one of their \(x\) values into either equation to calculate the \(y\)-coordinate.
A1: Provides all three correct coordinate pairs: \((-2, 0)\), \((0, 6)\), and \((4, 18)\).

Part (b):
M1: Uses their critical values to identify intervals or sketches the cubic/line intersection regions to attempt solving the inequality.
A1: Correct final answer: \(-2 < x < 0\) or \(x > 4\) (accept equivalent set notation).

(a) Since the line \(L: y = 3x - 4\) is tangent to \(C: y = 2x^2 + px + q\) at \(x = 1\), the gradient of \(C\) at \(x = 1\) is equal to the gradient of \(L\), which is 3. Differentiating the equation of the curve: \(\frac{dy}{dx} = 4x + p\). At \(x = 1\), we have \(4(1) + p = 3\), which gives \(p = -1\). Since the point \(A(1, -1)\) lies on the curve \(C\), we can substitute its coordinates into the equation of \(C\): \(-1 = 2(1)^2 + p(1) + q\). Substituting \(p = -1\): \(-1 = 2 - 1 + q\), which simplifies to \(q = -2\). (b) The equation of \(C\) is \(y = 2x^2 - x - 2\). Completing the square: \(y = 2(x^2 - 0.5x) - 2 = 2(x - 0.25)^2 - 0.125 - 2 = 2(x - 1/4)^2 - 17/8\). Thus, the vertex of \(C\) is at \((1/4, -17/8)\). The \(y\)-intercept of the curve is at \((0, -2)\). The sketch is a U-shaped parabola with vertex at \((1/4, -17/8)\) and crossing the \(y\)-axis at \((0, -2)\).

(a) M1: Differentiates the curve equation to find the gradient function. A1: Correct derivative \(\frac{dy}{dx} = 4x + p\). M1: Equates their gradient function at \(x = 1\) to 3. A1: Obtains \(p = -1\). A1: Uses the point \((1, -1)\) on the curve to find \(q = -2\). (b) M1: Completes the square or uses \(x = -b/(2a)\) to find the vertex coordinates. A1: Correct vertex coordinates \((1/4, -17/8)\) or equivalent. A1: Correct U-shaped sketch showing the vertex and the \(y\)-intercept.

(a) Find the midpoint \(M\) of \(AB\): \(M = ((-2+4)/2, (5-3)/2) = (1, 1)\). Find the gradient of \(AB\): \(m_{AB} = (-3-5)/(4-(-2)) = -8/6 = -4/3\). The gradient of the perpendicular bisector \(L_1\) is the negative reciprocal: \(m_1 = -1/(-4/3) = 3/4\). Using the point-slope formula with point \((1, 1)\) and gradient \(3/4\): \(y - 1 = (3/4)(x - 1)\). Multiplying by 4: \(4y - 4 = 3x - 3\). Rearranging gives \(3x - 4y + 1 = 0\). (b) Substitute \(y = 2x - 1\) into the equation of \(L_1\): \(3x - 4(2x - 1) + 1 = 0\). Expanding: \(3x - 8x + 4 + 1 = 0\), which simplifies to \(-5x + 5 = 0\), so \(x = 1\). Substituting \(x = 1\) into \(y = 2x - 1\) gives \(y = 2(1) - 1 = 1\). Thus, the coordinates of \(P\) are \((1, 1)\).

(a) M1: Correct method to find midpoint of \(AB\). A1: Midpoint is \((1, 1)\). M1: Finds gradient of \(AB\) and uses negative reciprocal rule. M1: Forms line equation using their midpoint and perpendicular gradient. A1: Correct equation in integer form: \(3x - 4y + 1 = 0\) (or equivalent). (b) M1: Substitutes \(y = 2x - 1\) into their line equation. M1: Solves for \(x\) (or \(y\)). A1: Correct coordinates \((1, 1)\).

Using the identity \(\sin^2 x = 1 - \cos^2 x\), substitute into the equation: \(6(1 - \cos^2 x) - \cos x - 4 = 0\). This simplifies to: \(6 - 6\cos^2 x - \cos x - 4 = 0\), which rearranges to: \(6\cos^2 x + \cos x - 2 = 0\). Factoring the quadratic: \((3\cos x + 2)(2\cos x - 1) = 0\). This yields two cases. Case 1: \(\cos x = 1/2\), which gives \(x = 60^\circ\) and \(x = 360^\circ - 60^\circ = 300^\circ\). Case 2: \(\cos x = -2/3\). The principal value is \(x = \arccos(-2/3) \approx 131.81^\circ\). The other angle in the range is \(x = 360^\circ - 131.81^\circ \approx 228.19^\circ\). Therefore, the solutions to 1 decimal place are \(x = 60.0^\circ, 131.8^\circ, 228.2^\circ, 300.0^\circ\).

M1: Substitution of \(\sin^2 x = 1 - \cos^2 x\) to form a quadratic in \(\cos x\). A1: Correct quadratic equation: \(6\cos^2 x + \cos x - 2 = 0\). M1: Factorising or using formula to solve the quadratic equation. A1: Finding both values: \(\cos x = 1/2\) and \(\cos x = -2/3\). M1: Solves \(\cos x = 1/2\) to find at least one solution. A1: Obtains \(60^\circ\) and \(300^\circ\). M1: Solves \(\cos x = -2/3\) to find at least one solution. A1: Obtains \(131.8^\circ\) and \(228.2^\circ\). Note: Deduct 1 mark overall for extra solutions within the range.

(a) Write \(y = 16x^{-1} + x^2 + 4\). Differentiating with respect to \(x\) gives: \(\frac{dy}{dx} = -16x^{-2} + 2x = -\frac{16}{x^2} + 2x\). Differentiating a second time gives: \(\frac{d^2y}{dx^2} = 32x^{-3} + 2 = \frac{32}{x^3} + 2\). (b) At the stationary point, \(\frac{dy}{dx} = 0\), so: \(-\frac{16}{x^2} + 2x = 0 \implies 2x^3 = 16 \implies x^3 = 8 \implies x = 2\). Substitute \(x = 2\) back into the equation of the curve to find \(y\): \(y = \frac{16}{2} + 2^2 + 4 = 8 + 4 + 4 = 16\). The coordinates of the stationary point are \((2, 16)\). (c) Substitute \(x = 2\) into the second derivative: \(\frac{d^2y}{dx^2} = \frac{32}{2^3} + 2 = 4 + 2 = 6\). Since \(\frac{d^2y}{dx^2} = 6 > 0\), the stationary point is a local minimum.

(a) M1: Differentiates \(16/x\) to obtain a term of the form \(k x^{-2}\) where \(k \neq 16\). A1: Correct first derivative \(\frac{dy}{dx} = -16x^{-2} + 2x\). A1: Correct second derivative \(\frac{d^2y}{dx^2} = 32x^{-3} + 2\). (b) M1: Sets their first derivative to 0 and attempts to solve for \(x\). A1: Correct value \(x = 2\). A1: Correct coordinates \((2, 16)\). (c) M1: Evaluates their second derivative at \(x = 2\). A1: Obtains \(\frac{d^2y}{dx^2} = 6\) and concludes it is a local minimum since \(6 > 0\) (with correct justification).

(a) Express the gradient function using index notation: \(f'(x) = 5x^{3/2} - 4x^{-1/2}\). Integrating with respect to \(x\): \(f(x) = \int (5x^{3/2} - 4x^{-1/2}) \, dx = \frac{5x^{5/2}}{5/2} - \frac{4x^{1/2}}{1/2} + C = 2x^{5/2} - 8x^{1/2} + C\). Using the point \(P(1, 5)\): \(5 = 2(1)^{5/2} - 8(1)^{1/2} + C \implies 5 = 2 - 8 + C \implies C = 11\). Thus, the equation of the curve is \(y = 2x^{5/2} - 8x^{1/2} + 11\). (b) The definite integral is: \(\int_{1}^{4} f'(x) \, dx = [2x^{5/2} - 8x^{1/2}]_{1}^{4}\). At \(x = 4\): \(2(4)^{5/2} - 8(4)^{1/2} = 2(32) - 8(2) = 64 - 16 = 48\). At \(x = 1\): \(2(1)^{5/2} - 8(1)^{1/2} = 2 - 8 = -6\). Subtracting the values: \(48 - (-6) = 54\). Alternative method: \(\int_{1}^{4} f'(x) \, dx = f(4) - f(1) = [2(4)^{5/2} - 8(4)^{1/2} + 11] - 5 = [48 + 11] - 5 = 59 - 5 = 54\).

(a) M1: Converts radical terms to fractional exponents. M1: Integrates to find \(f(x)\), increasing powers by 1. A1: Correct integrated form \(2x^{5/2} - 8x^{1/2}\) (constant not required). M1: Substitutes the point \((1, 5)\) to find \(C\). A1: Correct curve equation \(y = 2x^{5/2} - 8x^{1/2} + 11\). (b) M1: Integrates or uses their result from part (a) to substitute limits 4 and 1. M1: Evaluates both limits correctly. A1: Correct value of 54.

We use the power law of logarithms on the first term:
\[2\log_3(x - 2) = \log_3(x - 2)^2\]

Next, use the subtraction (division) law of logarithms to combine the terms:
\[\log_3\left(\frac{(x - 2)^2}{x + 4}\right) = 1\]

Now, rewrite the logarithmic equation in exponential form:
\[\frac{(x - 2)^2}{x + 4} = 3^1\]
\[\frac{x^2 - 4x + 4}{x + 4} = 3\]

Multiply both sides by \((x + 4)\):
\[x^2 - 4x + 4 = 3(x + 4)\]
\[x^2 - 4x + 4 = 3x + 12\]

Rearrange into a quadratic equation:
\[x^2 - 7x - 8 = 0\]

Factorise the quadratic equation:
\[(x - 8)(x + 1) = 0\]

This gives possible solutions of:
\[x = 8 \quad \text{or} \quad x = -1\]

We must check the domain of the original logarithmic functions. For \(\log_3(x - 2)\) to be defined, we require:
\[x - 2 > 0 \implies x > 2\]

Therefore, \(x = -1\) is an invalid solution because we cannot take the logarithm of a negative number.

Thus, the only valid solution is:
\[x = 8\]

M1: Uses the power law to express \(2\log_3(x-2)\) as \[\log_3(x-2)^2\\] and then applies the division law to combine the terms into a single logarithm.
M1: Removes logarithms correctly to form a linear or quadratic equation in \(x\), i.e., \((x-2)^2 = 3(x+4)\).
A1: Obtains the correct simplified quadratic equation \(x^2 - 7x - 8 = 0\) and solves to find the values \(x = 8\) and \(x = -1\).
A1: Explicitly or implicitly rejects the solution \(x = -1\) to give \(x = 8\) as the only final answer.

The formula for the sum to infinity of a geometric series is:
\[S_{\infty} = \frac{a}{1 - r}\]

The sum of the first two terms is:
\[S_2 = a + ar = a(1 + r)\]

We are given that:
\[S_{\infty} = 3 S_2\]

Substituting the formulas into this equation:
\[\frac{a}{1 - r} = 3a(1 + r)\]

Since \(a \neq 0\), we can divide both sides by \(a\):
\[\frac{1}{1 - r} = 3(1 + r)\]

Multiply both sides by \((1 - r)\):
\[1 = 3(1 + r)(1 - r)\]

Expand the brackets using the difference of two squares:
\[1 = 3(1 - r^2)\]
\[1 = 3 - 3r^2\]

Rearrange to solve for \(r^2\):
\[3r^2 = 2\]
\[r^2 = \frac{2}{3}\]

Since \(r > 0\), we take the positive square root:
\[r = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}\]

Rationalising the denominator:
\[r = \frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{\sqrt{6}}{3}\]

This is in the form \(\frac{\sqrt{p}}{q}\) where \(p = 6\) and \(q = 3\).

M1: Sets up a correct algebraic equation representing the given condition using the correct formulas for \(S_{\infty}\) and \(S_2\), e.g., \(\frac{a}{1-r} = 3(a + ar)\).
M1: Divides by \(a\) and multiplies by \((1-r)\) to obtain an equation in terms of \(r\) only, e.g., \(1 = 3(1-r^2)\).
A1: Solves the equation to find \(r^2 = \frac{2}{3}\).
A1: Correctly rationalises and simplifies to find the exact value \(r = \frac{\sqrt{6}}{3}\). (Accept \(\frac{1}{3}\sqrt{6}\); reject negative root due to \(r > 0\)).

(a) Using the formula for the sum to infinity: \(S_{\infty} = \frac{a}{1-r} = 18 \implies a = 18(1-r)\).
Using the formula for the sum of the first \(3\) terms: \(S_3 = \frac{a(1-r^3)}{1-r} = 15.75\).
Substituting \(a = 18(1-r)\) into the expression for \(S_3\) gives:
\(\frac{18(1-r)(1-r^3)}{1-r} = 15.75 \implies 18(1-r^3) = 15.75\).
Dividing both sides by \(18\):
\(1-r^3 = \frac{15.75}{18} = \frac{7}{8} \implies r^3 = 1 - \frac{7}{8} = \frac{1}{8}\) (as required).

(b) From \(r^3 = \frac{1}{8}\), we have \(r = \frac{1}{2}\).
Substituting \(r = \frac{1}{2}\) back into \(a = 18(1-r)\):
\(a = 18\left(1 - \frac{1}{2}\right) = 9\).

(c) The sum of the first \(5\) terms is given by:
\(S_5 = \frac{9\left(1 - (0.5)^5\right)}{1 - 0.5} = 18\left(1 - \frac{1}{32}\right) = 18 \times \frac{31}{32} = \frac{279}{16} = 17.4375\).
To \(3\) decimal places, this is \(17.438\).

(a)
- M1: Attempts to use the sum to infinity formula to express \(a\) in terms of \(r\).
- M1: Uses the sum of \(3\) terms formula with \(S_3 = 15.75\).
- dM1: Substitutes \(a\) into the second equation to eliminate \(a\).
- A1*: Completes the proof to show \(r^3 = \frac{1}{8}\) with no errors.

(b)
- M1: Solves \(r^3 = \frac{1}{8}\) to find \(r = 0.5\) and attempts to find \(a\).
- A1: Obtains \(a = 9\).

(c)
- M1: Applies the sum formula \(S_n\) for \(n=5\) using their \(a\) and \(r\).
- A1: Obtains \(17.438\) (or \(17.4375\)).

(a) To find the intersections, we set the curve and line equations equal:
\(3\sqrt{x} - x = x \implies 3\sqrt{x} = 2x\).
Squaring both sides:
\(9x = 4x^2 \implies 4x^2 - 9x = 0 \implies x(4x - 9) = 0\).
This yields \(x = 0\) or \(x = \frac{9}{4} = 2.25\).
When \(x = 0\), \(y = 0\), which is the origin \((0,0)\).
When \(x = 2.25\), \(y = 2.25\). Thus, the intersection points are \((0,0)\) and \((2.25, 2.25)\).

(b) The area \(A\) is given by the integral of \(y_C - y_L\) from \(x = 0\) to \(x = 2.25\):
\(A = \int_{0}^{2.25} (3x^{1/2} - x - x) \, dx = \int_{0}^{2.25} (3x^{1/2} - 2x) \, dx\).
Integrating term by term:
\(\int (3x^{1/2} - 2x) \, dx = \left[ 3 \times \frac{2}{3}x^{3/2} - x^2 \right] = \left[ 2x^{3/2} - x^2 \right]\).
Applying the limits:
\(A = \left( 2(2.25)^{3/2} - (2.25)^2 \right) - (0)\).
Since \(2.25 = \frac{9}{4}\):
\(2\left(\frac{9}{4}\right)^{3/2} - \left(\frac{9}{4}\right)^2 = 2\left(\frac{27}{8}\right) - \frac{81}{16} = \frac{27}{4} - \frac{81}{16} = \frac{108 - 81}{16} = \frac{27}{16} = 1.6875\).

(a)
- M1: Equates the curve and the line to set up an equation in terms of \(x\).
- A1: Solves the equation to obtain the non-zero root \(x = 2.25\) (or \(x = \frac{9}{4}\)).
- A1: Confirms both coordinates of the intersection points are \((0,0)\) and \((2.25, 2.25)\).

(b)
- M1: Formulates the integral for the area with correct upper limit \(2.25\) (or their non-zero limit) and lower limit \(0\).
- M1: Integrates at least one term correctly (increasing power of \(x\) by 1).
- A1: Correct integration resulting in \(2x^{3/2} - x^2\).
- dM1: Substitutes the limits \(2.25\) and \(0\) into their integrated function.
- A1: Obtains the exact area of \(1.6875\) (or \(\frac{27}{16}\)).

(a) (i) & (ii) We complete the square on the given equation of the circle:
\(x^2 - 8x + y^2 + 6y = 0\)
\((x - 4)^2 - 16 + (y + 3)^2 - 9 = 0\)
\((x - 4)^2 + (y + 3)^2 = 25\).
Thus, the centre of the circle is \((4, -3)\) and the radius is \(\sqrt{25} = 5\).

(b) The gradient of the radius from the centre \((4, -3)\) to the point \(P(8, 0)\) is:
\(m_{\text{radius}} = \frac{0 - (-3)}{8 - 4} = \frac{3}{4}\).
Since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent \(l\) is:
\(m_{\text{tangent}} = -\frac{4}{3}\).
Using the point-slope formula for line \(l\) at point \(P(8, 0)\):
\(y - 0 = -\frac{4}{3}(x - 8)\)
\(3y = -4x + 32\)
\(4x + 3y - 32 = 0\).

(a)
- M1: Attempts to complete the square for \(x^2 - 8x\) and \(y^2 + 6y\).
- A1: Correct centre at \((4, -3)\).
- A1: Correct radius of \(5\).

(b)
- M1: Calculates the gradient of the radius connecting the centre and \(P(8, 0)\).
- A1: Correct radius gradient of \(\frac{3}{4}\).
- M1: Uses the perpendicular gradient rule \(m_1 m_2 = -1\) to find the gradient of the tangent.
- dM1: Formulates the equation of the line using their perpendicular gradient and point \((8, 0)\).
- A1: Expresses the final line equation in the form \(4x + 3y - 32 = 0\) (or any integer multiple).

(a) Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):
\(5(1 - \sin^2 \theta) + 2\sin \theta = 4\)
\(5 - 5\sin^2 \theta + 2\sin \theta = 4\)
\(5\sin^2 \theta - 2\sin \theta - 1 = 0\) (as required).

(b) Let \(\theta = 2x - 30^\circ\). The equation becomes \(5\sin^2 \theta - 2\sin \theta - 1 = 0\).
Using the quadratic formula for \(\sin \theta\):
\(\sin \theta = \frac{2 \pm \sqrt{(-2)^2 - 4(5)(-1)}}{10} = \frac{2 \pm \sqrt{24}}{10} = \frac{1 \pm \sqrt{6}}{5}\).
This gives two cases:
Case 1: \(\sin \theta \approx 0.6899\)
\(\theta \approx 43.62^\circ, 136.38^\circ, 403.62^\circ, 496.38^\circ\).
Case 2: \(\sin \theta \approx -0.2899\)
\(\theta \approx -16.85^\circ, 196.85^\circ, 343.15^\circ, 556.85^\circ\).

Using \(x = \frac{\theta + 30^\circ}{2}\), we find the values of \(x\):
- From \(\theta = -16.85^\circ \implies x = 6.575^\circ \approx 6.6^\circ\)
- From \(\theta = 43.62^\circ \implies x = 36.81^\circ \approx 36.8^\circ\)
- From \(\theta = 136.38^\circ \implies x = 83.19^\circ \approx 83.2^\circ\)
- From \(\theta = 196.85^\circ \implies x = 113.43^\circ \approx 113.4^\circ\)
- From \(\theta = 343.15^\circ \implies x = 186.58^\circ \approx 186.6^\circ\)
- From \(\theta = 403.62^\circ \implies x = 216.81^\circ \approx 216.8^\circ\)
- From \(\theta = 496.38^\circ \implies x = 263.19^\circ \approx 263.2^\circ\)
- From \(\theta = 556.85^\circ \implies x = 293.43^\circ \approx 293.4^\circ\)

(a)
- M1: Substitutes \(\cos^2 \theta = 1 - \sin^2 \theta\) into the equation.
- A1*: Simplifies correctly to the given quadratic equation.

(b)
- M1: Solves the quadratic equation to find the two values of \(\sin(2x-30^\circ)\).
- A1: Correct values for \(\sin(2x-30^\circ) = \frac{1 \pm \sqrt{6}}{5}\) (or approx. \(0.690\) and \(-0.290\)).
- M1: Obtains at least two correct values for \(\theta = 2x-30^\circ\).
- M1: Finds at least four values of \(2x-30^\circ\) within the interval \([-30^\circ, 690^\circ)\).
- dM1: Solves for \(x\) by adding \(30^\circ\) and dividing by \(2\).
- A1: Any four correct solutions for \(x\) to 1 d.p.
- A1: All eight correct solutions: \(6.6^\circ, 36.8^\circ, 83.2^\circ, 113.4^\circ, 186.6^\circ, 216.8^\circ, 263.2^\circ, 293.4^\circ\).

(a) We start by applying log laws to the equation:
\(\log_3 (x+2) - \log_3 \sqrt{x-4} = \log_3 3 + \log_3 2\).
Combining terms using the division and multiplication laws:
\(\log_3 \left( \frac{x+2}{\sqrt{x-4}} \right) = \log_3 (3 \times 2) = \log_3 6\).
Removing logs from both sides:
\(\frac{x+2}{\sqrt{x-4}} = 6 \implies x+2 = 6\sqrt{x-4}\).
Squaring both sides:
\((x+2)^2 = 36(x-4)\)
\(x^2 + 4x + 4 = 36x - 144\)
\(x^2 - 32x + 148 = 0\) (as required).

(b) Solving the quadratic equation \(x^2 - 32x + 148 = 0\) using the quadratic formula:
\(x = \frac{32 \pm \sqrt{(-32)^2 - 4(1)(148)}}{2} = \frac{32 \pm \sqrt{1024 - 592}}{2} = \frac{32 \pm \sqrt{432}}{2}\).
Since \(\sqrt{432} = \sqrt{144 \times 3} = 12\sqrt{3}\):
\(x = \frac{32 \pm 12\sqrt{3}}{2} = 16 \pm 6\sqrt{3}\).

Check for validity:
For the logs to be defined, we require \(x+2 > 0\) and \(x-4 > 0\), so \(x > 4\).
- \(16 + 6\sqrt{3} \approx 26.4 > 4\) (valid)
- \(16 - 6\sqrt{3} \approx 5.6 > 4\) (valid)
Thus, both solutions are valid.

(a)
- M1: Writes \(\frac{1}{2} \log_3 (x-4)\) as \(\log_3 \sqrt{x-4}\).
- M1: Replaces \(1\) with \(\log_3 3\).
- M1: Combines the logarithms correctly on both sides of the equation.
- dM1: Equates arguments to get a non-logarithmic equation and squares both sides.
- A1*: Simplifies with no errors to reach \(x^2 - 32x + 148 = 0\).

(b)
- M1: Solves the quadratic equation to find the roots in surd form.
- A1: Obtains the roots \(16 \pm 6\sqrt{3}\).
- A1: States or demonstrates that both solutions are valid since both are greater than \(4\).

(a) The volume of the box is given by \(V = x^2 h = 108 \implies h = \frac{108}{x^2}\).
The surface area of the open-topped box consists of the square base and four vertical sides:
\(A = x^2 + 4xh\).
Substituting the expression for \(h\):
\(A = x^2 + 4x\left(\frac{108}{x^2}\right) = x^2 + \frac{432}{x}\) (as required).

(b) To find the minimum surface area, we differentiate \(A\) with respect to \(x\):
\(\frac{dA}{dx} = 2x - 432x^{-2} = 2x - \frac{432}{x^2}\).
Setting \(\frac{dA}{dx} = 0\) for a stationary point:
\(2x - \frac{432}{x^2} = 0 \implies 2x^3 = 432 \implies x^3 = 216 \implies x = 6\).
Substituting \(x = 6\) back into the surface area equation:
\(A = 6^2 + \frac{432}{6} = 36 + 72 = 108\text{ cm}^2\).

To justify that this is a minimum, we find the second derivative:
\(\frac{d^2A}{dx^2} = 2 + 864x^{-3} = 2 + \frac{864}{x^3}\).
When \(x = 6\):
\(\frac{d^2A}{dx^2} = 2 + \frac{864}{216} = 2 + 4 = 6 > 0\).
Since \(\frac{d^2A}{dx^2} > 0\), the value \(108\text{ cm}^2\) is indeed a minimum.

(a)
- M1: Writes down a correct volume relation, e.g., \(x^2 h = 108\).
- M1: Formulates a correct expression for the open-box surface area \(A = x^2 + 4xh\).
- A1*: Eliminates \(h\) to obtain the given expression for \(A\) with no errors shown.

(b)
- M1: Differentiates \(A\) to obtain \(\frac{dA}{dx} = ax + bx^{-2}\).
- A1: Correct first derivative \(2x - \frac{432}{x^2}\).
- M1: Sets \(\frac{dA}{dx} = 0\) and solves for \(x\).
- A1: Identifies \(x = 6\) and finds \(A = 108\).
- M1: Finds \(\frac{d^2A}{dx^2}\) and evaluates it at \(x = 6\) to confirm \(\frac{d^2A}{dx^2} > 0\) (or uses an equivalent valid method).

(a) Complete the square for both \(x\) and \(y\):
\[x^2 - 16x = (x-8)^2 - 64\]
\[y^2 - 12y = (y-6)^2 - 36\]
Substitute these into the equation of \(C\):
\[(x-8)^2 - 64 + (y-6)^2 - 36 + 75 = 0\]
\[(x-8)^2 + (y-6)^2 - 25 = 0\]
\[(x-8)^2 + (y-6)^2 = 25\]
Hence:
(i) The center is \(P(8, 6)\).
(ii) The radius is \(\sqrt{25} = 5\).

(b) The line \(l\) passes through \(O(0,0)\) and the center \(P(8,6)\).
The gradient of \(l\) is:
\[m = \frac{6 - 0}{8 - 0} = \frac{3}{4}\]
Since the line passes through the origin, the equation of \(l\) is:
\[y = \frac{3}{4}x \quad (\text{or } 3x - 4y = 0)\]

(c) Substitute \(y = \frac{3}{4}x\) into the equation of \(C\):
\[x^2 + \left(\frac{3}{4}x\right)^2 - 16x - 12\left(\frac{3}{4}x\right) + 75 = 0\]
\[x^2 + \frac{9}{16}x^2 - 16x - 9x + 75 = 0\]
\[\frac{25}{16}x^2 - 25x + 75 = 0\]
Multiply the entire equation by \(\frac{16}{25}\):
\[x^2 - 16x + 48 = 0\]
Factor the quadratic:
\[(x - 4)(x - 12) = 0\]
This gives \(x = 4\) or \(x = 12\).

Find the corresponding \(y\)-coordinates:
If \(x = 4\), \(y = \frac{3}{4}(4) = 3 \implies (4,3)\).
If \(x = 12\), \(y = \frac{3}{4}(12) = 9 \implies (12,9)\).

Since \(A\) lies between \(O(0,0)\) and \(P(8,6)\), its \(x\)-coordinate must be less than \(8\).
Thus, \(A\) is \((4, 3)\) and \(B\) is \((12, 9)\).

(d) The tangent to \(C\) at \(A(4,3)\) is perpendicular to the radius \(PA\).
The gradient of the radius \(PA\) (which lies on \(l\)) is \(\frac{3}{4}\).
Therefore, the gradient of the tangent at \(A\) is:
\[m_t = -\frac{4}{3}\]
The equation of the tangent is:
\[y - 3 = -\frac{4}{3}(x - 4)\]
\[y = -\frac{4}{3}x + \frac{16}{3} + 3 = -\frac{4}{3}x + \frac{25}{3}\]
To find the intersection with the \(y\)-axis (point \(T\)), set \(x = 0\):
\[y = \frac{25}{3} \implies T\left(0, \frac{25}{3}\right)\]
The triangle \(OAT\) has vertices \(O(0,0)\), \(T\left(0, \frac{25}{3}\right)\), and \(A(4,3)\).
Taking \(OT\) as the base along the \(y\)-axis:
\[\text{Base} = \frac{25}{3}\]
The height is the perpendicular distance from \(A\) to the \(y\)-axis, which is the \(x\)-coordinate of \(A\):
\[\text{Height} = 4\]
The area of the triangle is:
\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{25}{3} \times 4 = \frac{50}{3}\]

(a)
- M1: Attempts to complete the square for both x and y. Allow one sign or arithmetic error.
- A1: Correct coordinates of the center: (8, 6) or x = 8, y = 6.
- A1: Correct radius: 5 (do not accept sqrt(25)).
(b)
- M1: Attempts to find the gradient of the line connecting (0,0) and their center, and uses it to write a linear equation.
- A1: Correct equation of the line, e.g., y = 3/4x or 3x - 4y = 0.
(c)
- M1: Substitutes their equation of l into the equation of C.
- A1: Obtains a correct simplified quadratic in x, e.g., x^2 - 16x + 48 = 0 (or in y, e.g., y^2 - 12y + 27 = 0).
- M1: Solves their quadratic to find two values of x (or y) and finds the corresponding coordinates.
- A1: Correct coordinates for A(4,3) and B(12,9) clearly identified.
(d)
- M1: Uses the negative reciprocal of the gradient of l to find the gradient of the tangent, and finds the equation of the tangent at A.
- A1: Finds the correct coordinates of T(0, 25/3).
- A1: Correct exact area of 50/3 (or 16 2/3 or 16.7).

(a) For the series to be geometric, the ratio of successive terms must be constant:
\[r_1 = \frac{u_2}{u_1} = \frac{2\sin 2x}{4\cos^2 x}\]
Using the double angle identity \(\sin 2x = 2\sin x \cos x\):
\[r_1 = \frac{2(2\sin x \cos x)}{4\cos^2 x} = \frac{4\sin x \cos x}{4\cos^2 x} = \frac{\sin x}{\cos x} = \tan x\]
Similarly:
\[r_2 = \frac{u_3}{u_2} = \frac{4\sin^2 x}{2\sin 2x} = \frac{4\sin^2 x}{4\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x\]
Since \(\frac{u_2}{u_1} = \frac{u_3}{u_2} = \tan x\), the series is geometric with common ratio \(\tan x\).

(b) For \(x = \frac{\pi}{6}\):
The first term is:
\[a = u_1 = 4\cos^2\left(\frac{\pi}{6}\right) = 4 \left(\frac{\sqrt{3}}{2}\right)^2 = 4 \times \frac{3}{4} = 3\]
The common ratio is:
\[r = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\]
Since \(|r| = \frac{1}{\sqrt{3}} < 1\), the sum to infinity exists:
\[S_{\infty} = \frac{a}{1 - r} = \frac{3}{1 - \frac{1}{\sqrt{3}}} = \frac{3\sqrt{3}}{\sqrt{3} - 1}\]
Rationalize the denominator:
\[S_{\infty} = \frac{3\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{9 + 3\sqrt{3}}{3 - 1} = \frac{9 + 3\sqrt{3}}{2}\]

(c)(i) The sum of the first two terms is:
\[u_1 + u_2 = \frac{8}{5}\]
\[4\cos^2 x + 2\sin 2x = \frac{8}{5}\]
Using \(\sin 2x = 2\sin x\cos x\):
\[4\cos^2 x + 4\sin x\cos x = \frac{8}{5}\]
We can use the identity \(\sin^2 x + \cos^2 x = 1\) to write:
\[4\cos^2 x + 4\sin x\cos x = \frac{8}{5}(\sin^2 x + \cos^2 x)\]
Multiply both sides by 5:
\[20\cos^2 x + 20\sin x\cos x = 8\sin^2 x + 8\cos^2 x\]
Rearrange all terms to one side:
\[8\sin^2 x - 20\sin x\cos x - 12\cos^2 x = 0\]
Since \(0 < x < \frac{\pi}{2}\), \(\cos x \neq 0\). Divide the entire equation by \(\cos^2 x\):
\[8\left(\frac{\sin^2 x}{\cos^2 x}\right) - 20\left(\frac{\sin x \cos x}{\cos^2 x}\right) - 12 = 0\]
\[8\tan^2 x - 20\tan x - 12 = 0\]
Divide by 4:
\[2\tan^2 x - 5\tan x - 3 = 0\]

(c)(ii) Factor the quadratic:
\[(2\tan x + 1)(\tan x - 3) = 0\]
This gives:
\[\tan x = -\frac{1}{2} \quad \text{or} \quad \tan x = 3\]
Since \(x\) lies in the first quadrant (\(0 < x < \frac{\pi}{2}\)), \(\tan x\) must be positive.
Therefore, the negative solution \(\tan x = -\frac{1}{2}\) is rejected.
The exact value is:
\[\tan x = 3\]

(a)
- M1: Uses the double angle formula sin 2x = 2sin x cos x on either u2/u1 or u3/u2.
- A1: Correctly shows that u2/u1 = tan x.
- A1: Correctly shows that u3/u2 = tan x and concludes that the series is geometric since ratios are equal.
(b)
- M1: Evaluates a and r for x = pi/6 to find a = 3 and r = 1/sqrt(3) (or equivalent).
- M1: Applies the formula S_inf = a/(1-r) with their a and r (where |r| < 1).
- A1: Correctly rationalizes to obtain (9 + 3sqrt(3))/2 (or equivalent form with a=9, b=3, c=2).
(c)(i)
- M1: Sets up the equation 4cos^2 x + 4sin x cos x = 8/5.
- M1: Multiplies the constant term by (sin^2 x + cos^2 x) (or divides by cos^2 x and replaces 1/cos^2 x with 1 + tan^2 x).
- M1: Obtains a homogenous equation in sin x and cos x (or an equation in tan x) and divides by cos^2 x (or simplifies) to get a quadratic in tan x.
- A1*: Correctly arrives at the given equation 2tan^2 x - 5tan x - 3 = 0 with no errors in working.
(c)(ii)
- M1: Solves the quadratic to find tan x = -1/2 and tan x = 3.
- A1: States tan x = 3 and explains that tan x = -1/2 is rejected because 0 < x < pi/2 implies tan x > 0.

(a) Since the sum of all probabilities must equal 1:
\(\sum P(X=x) = k(3 - (-1)) + k(3 - 0) + k(3 - 1) + k(3 - 2) + c = 1\)
\(4k + 3k + 2k + k + c = 1 \implies 10k + c = 1 \implies c = 1 - 10k\) [Equation 1]

Now, we use the expectation formula:
\(E(X) = \sum x P(X=x)\)
\(E(X) = (-1)(4k) + (0)(3k) + (1)(2k) + (2)(k) + (3)(c) = 0.6\)
\(-4k + 0 + 2k + 2k + 3c = 0.6\)
\(3c = 0.6 \implies c = 0.2\)

Substitute \(c = 0.2\) back into Equation 1:
\(10k + 0.2 = 1 \implies 10k = 0.8 \implies k = 0.08\) (as required).

(b) We first calculate \(E(X^2)\):
\(E(X^2) = \sum x^2 P(X=x)\)
\(P(X = -1) = 4(0.08) = 0.32\)
\(P(X = 0) = 3(0.08) = 0.24\)
\(P(X = 1) = 2(0.08) = 0.16\)
\(P(X = 2) = 0.08\)
\(P(X = 3) = 0.20\)

\(E(X^2) = (-1)^2(0.32) + (0)^2(0.24) + (1)^2(0.16) + (2)^2(0.08) + (3)^2(0.20)\)
\(E(X^2) = 1(0.32) + 0 + 1(0.16) + 4(0.08) + 9(0.20)\)
\(E(X^2) = 0.32 + 0.16 + 0.32 + 1.80 = 2.60\)

Now find \(Var(X)\):
\(Var(X) = E(X^2) - [E(X)]^2 = 2.60 - 0.6^2 = 2.60 - 0.36 = 2.24\)

(c) Using the properties of variance:
\(Var(3 - 2X) = (-2)^2 Var(X) = 4 \times 2.24 = 8.96\)

(d) We want \(P(X \ge 0.6)\). The values of \(X\) that satisfy this are \(1, 2, 3\):
\(P(X \ge 0.6) = P(X = 1) + P(X = 2) + P(X = 3) = 0.16 + 0.08 + 0.20 = 0.44\)

(a)
M1: Writes an expression for sum of probabilities equal to 1 to find a relation between k and c.
A1: Correctly derives \(10k + c = 1\) (or equivalent).
M1: Writes an expression for \(E(X) = 0.6\) in terms of k and c.
A1: Obtains \(c = 0.2\) (or equivalent).
A1*: Shows clearly that \(k = 0.08\) using the values of c.

(b)
M1: Attempts to find \(E(X^2)\) with at least 3 correct terms of \(x^2 P(X=x)\).
A1: Correctly calculates \(E(X^2) = 2.60\).
M1: Applies the correct formula \(Var(X) = E(X^2) - [E(X)]^2\).
A1: Obtains \(2.24\).

(c)
B1: Obtains \(8.96\) (accept \(4 \times \text{their } Var(X)\)).

(d)
M1: Identifies the correct outcomes for \(X \ge 0.6\), i.e., \(X=1, X=2, X=3\).
A1: Obtains \(0.44\).

(a) We standardize the given percentages:
\(P(M < 110) = 0.15 \implies P\left(Z < \frac{110 - \mu}{\sigma}\right) = 0.15\)
From standard normal tables, \(z = -1.0364\) for the 15th percentile.
So, \(\frac{110 - \mu}{\sigma} = -1.0364 \implies \mu - 1.0364\sigma = 110\) [Equation 1]

\(P(M > 165) = 0.08 \implies P\left(Z > \frac{165 - \mu}{\sigma}\right) = 0.08\)
From standard normal tables, \(z = 1.4051\) for the 92nd percentile.
So, \(\frac{165 - \mu}{\sigma} = 1.4051 \implies \mu + 1.4051\sigma = 165\) [Equation 2]

(b) Subtracting Equation 1 from Equation 2:
\((\mu + 1.4051\sigma) - (\mu - 1.0364\sigma) = 165 - 110\)
\(2.4415\sigma = 55 \implies \sigma = \frac{55}{2.4415} \approx 22.527\text{ g}\)
So \(\sigma = 22.5\text{ g}\) (to 1 d.p.)

Substitute \(\sigma = 22.527\) into Equation 2:
\(\mu = 165 - 1.4051(22.527) = 165 - 31.653 \approx 133.347\text{ g}\)
So \(\mu = 133.3\text{ g}\) (to 1 d.p.)

(c) We want to find \(P(120 < M < 150)\) using \(\mu = 133.35\) and \(\sigma = 22.53\):
Standardizing the values:
\(z_1 = \frac{120 - 133.35}{22.53} \approx -0.59\)
\(z_2 = \frac{150 - 133.35}{22.53} \approx 0.74\)

Now find the probability:
\(P(-0.59 < Z < 0.74) = \Phi(0.74) - \Phi(-0.59)\)
\(\Phi(0.74) = 0.7704\)
\(\Phi(-0.59) = 1 - \Phi(0.59) = 1 - 0.7224 = 0.2776\)
\(P(-0.59 < Z < 0.74) = 0.7704 - 0.2776 = 0.4928\)
(Allow answers in range 0.491 to 0.495 depending on rounding of z-values)

(a)
M1: Standardizes \(110\) or \(165\) setting equal to a z-value.
B1: Finds correct z-values \(-1.0364\) (accept \(-1.04\)) and \(1.4051\) (accept \(1.41\)) from tables.
A1: Correct equation 1: \(\mu - 1.0364\sigma = 110\) (or equivalent).
A1: Correct equation 2: \(\mu + 1.4051\sigma = 165\) (or equivalent).

(b)
M1: Eliminates one variable from their two linear simultaneous equations.
A1: Obtains a correct value for \(\sigma\) (approx. 22.5).
A1: Obtains a correct value for \(\mu\) (approx. 133.3).
A1: Correctly rounds \(\sigma\) to \(22.5\) (accept 22.5 or 22.6 depending on z-values used).
A1: Correctly rounds \(\mu\) to \(133.3\) (accept 133.3 or 133.4).

(c)
M1: Standardizes both \(120\) and \(150\) with their mean and standard deviation.
M1: Attempts \(\Phi(z_2) - (1 -
\Phi(z_1))\) for their z-values.
A1: Correct final probability in the range \(0.491 - 0.495\).

(a) Since \(A\) and \(B\) are independent:
\(P(A \cap B) = P(A) \times P(B) = 0.5 \times 0.35 = 0.175\)
Now use the addition rule:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.35 - 0.175 = 0.675\)

(b) From part (a):
\(P(A \cap B) = 0.175\)

(c) To construct the Venn diagram, we need the individual disjoint regions:
- Since \(B\) and \(C\) are mutually exclusive, there is no intersection between \(B\) and \(C\). Thus \(P(B \cap C) = 0\) and \(P(A \cap B \cap C) = 0\).
- \(P(A \cap B) = 0.175\) (lies completely inside \(A\) and \(B\) but outside \(C\))
- \(P(A \cap C) = 0.15\) (lies completely inside \(A\) and \(C\) but outside \(B\))
- \(P(A \text{ only}) = P(A) - P(A \cap B) - P(A \cap C) = 0.5 - 0.175 - 0.15 = 0.175\)
- \(P(B \text{ only}) = P(B) - P(A \cap B) = 0.35 - 0.175 = 0.175\)
- \(P(C \text{ only}) = P(C) - P(A \cap C) = 0.4 - 0.15 = 0.25\)
- Sum of all probabilities inside circles: \(0.175 + 0.175 + 0.25 + 0.175 + 0.15 = 0.925\)
- Outside region: \(1 - 0.925 = 0.075\)
Draw three circles with \(B\) and \(C\) separated, and \(A\) intersecting both. Label regions with these calculated probabilities.

(d) \(P(A' | C) = \frac{P(A' \cap C)}{P(C)}\)
From the Venn diagram, \(P(A' \cap C) = P(C \text{ only}) = 0.25\).
So,
\(P(A' | C) = \frac{0.25}{0.4} = \frac{5}{8} = 0.625\)

(a)
M1: Uses independence property \(P(A \cap B) = P(A)P(B)\).
M1: Applies the addition rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
A1: Correctly finds \(0.675\) (or equivalent fraction).

(b)
B1: Correct value \(0.175\).

(c)
M1: Draws three circles with \(B\) and \(C\) mutually exclusive (non-intersecting circles, or intersection with 0 label).
A1: Correctly places \(0.175\) for \(P(A \cap B)\) and \(0.15\) for \(P(A \cap C)\).
A1: Correctly calculates and places \(0.175\) for \(A\) only, \(0.175\) for \(B\) only, and \(0.25\) for \(C\) only.
A1: Correctly calculates and places \(0.075\) for the outside region.
B1: Venn diagram has labels \(A, B, C\) and is enclosed in a box representing the sample space.

(d)
M1: Uses conditional probability formula \(P(A' | C) = \frac{P(A' \cap C)}{P(C)}\).
A1: Correctly identifies \(P(A' \cap C) = 0.25\) (or \(0.4 - 0.15\)).
A1: Correct final answer of \(0.625\) (or \(\frac{5}{8}\)).

(a) Total frequency \(N = 80\). The median corresponds to the 40th value.
Cumulative frequencies are:
- \(< 18\): 8
- \(< 20\): 23
- \(< 22\): 47
- \(< 25\): 65
- \(< 30\): 80
The 40th value lies in the class \(20 \le t < 22\).
Using linear interpolation:
\(\text{Median} = 20 + \frac{40 - 23}{24} \times (22 - 20) = 20 + \frac{17}{24} \times 2 = 20 + 1.417 \approx 21.4\) °C (to 3 s.f.).

(b) Midpoints (\(x\)) and frequencies (\(f\)):
- \(15-18\): midpoint \(x = 16.5\), \(f = 8\), \(fx = 132\), \(fx^2 = 2178\)
- \(18-20\): midpoint \(x = 19.0\), \(f = 15\), \(fx = 285\), \(fx^2 = 5415\)
- \(20-22\): midpoint \(x = 21.0\), \(f = 24\), \(fx = 504\), \(fx^2 = 10584\)
- \(22-25\): midpoint \(x = 23.5\), \(f = 18\), \(fx = 423\), \(fx^2 = 9940.5\)
- \(25-30\): midpoint \(x = 27.5\), \(f = 15\), \(fx = 412.5\), \(fx^2 = 11343.75\)

Sum of frequencies \(\sum f = 80\)
\(\sum fx = 132 + 285 + 504 + 423 + 412.5 = 1756.5\)
\(\sum fx^2 = 2178 + 5415 + 10584 + 9940.5 + 11343.75 = 39461.25\)

Estimate for the mean:
\(\bar{t} = \frac{\sum fx}{N} = \frac{1756.5}{80} = 21.95625 \approx 22.0\) °C.

Estimate for the standard deviation:
\(\sigma = \sqrt{\frac{\sum fx^2}{N} - \bar{t}^2} = \sqrt{\frac{39461.25}{80} - (21.95625)^2}\)
\(\sigma = \sqrt{493.265625 - 482.076914} = \sqrt{11.188711} \approx 3.34495 \approx 3.34\) °C.

(c) \(Q_1 = 19.3\), \(Q_3 = 24.2\).
\(\text{IQR} = Q_3 - Q_1 = 24.2 - 19.3 = 4.9\) °C.
Upper outlier boundary \(= Q_3 + 1.5 \times \text{IQR} = 24.2 + 1.5 \times 4.9 = 24.2 + 7.35 = 31.55\) °C.
Since the maximum recorded temperature in our distribution is less than \(30\) °C (the upper bound of the highest class), there are no values greater than \(31.55\) °C. Thus, there are no outliers at the upper end of the distribution.

(a)
M1: Identifies the 40th value and uses a correct interpolation fraction, e.g., \(20 + \frac{17}{24} \times 2\).
A1: Correct numerator and denominator in the fraction.
A1: \(21.4\) (accept \(21.42\)).

(b)
M1: Attempts to find midpoints for all classes (at least 4 correct).
M1: Calculates \(\sum fx\) and \(\sum fx^2\) using their midpoints.
A1: Mean \(= 22.0\) (accept \(21.96\)).
M1: Uses correct formula for standard deviation with their values.
A1: Standard deviation \(= 3.34\) (accept \(3.34 - 3.35\)).

(c)
M1: Calculates \(\text{IQR} = 4.9\).
M1: Calculates upper boundary \(24.2 + 1.5 \times 4.9 = 31.55\).
A1: Explicitly compares with the maximum limit of the last class (30) and concludes there are no outliers at the upper end.

(a) We calculate the sum of squares:
\(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 204 - \frac{36^2}{8} = 204 - 162 = 42\)
\(S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 1754 - \frac{112^2}{8} = 1754 - 1568 = 186\)
\(S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 424 - \frac{36 \times 112}{8} = 424 - 504 = -80\)

(b) Calculate \(r\):
\(r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{-80}{\sqrt{42 \times 186}} = \frac{-80}{\sqrt{7812}} \approx -0.9051 \approx -0.905\) (to 3 s.f.).
This strong negative correlation indicates that as the age of the car increases, the selling price decreases.

(c) Regression line \(y = a + bx\):
\(b = \frac{S_{xy}}{S_{xx}} = \frac{-80}{42} \approx -1.9048 \approx -1.90\) (to 3 s.f.)

Find \(\bar{x}\) and \(\bar{y}\):
\(\bar{x} = \frac{36}{8} = 4.5\)
\(\bar{y} = \frac{112}{8} = 14\)

Now find \(a\):
\(a = \bar{y} - b\bar{x} = 14 - (-1.9048 \times 4.5) = 14 + 8.5716 = 22.5716 \approx 22.6\) (to 3 s.f.)
So the regression equation is:
\(y = 22.6 - 1.90x\)

(d) For \(x = 5.5\):
\(y = 22.5716 - 1.9048(5.5) = 12.0952\) thousands of dollars, which is approximately \(\$12,100\).
This estimate is reliable because \(5.5\) is within the range of the given data (interpolation) and the correlation between age and price is very strong (\(r \approx -0.905\)).

(a)
M1: Attempt to use correct formulae for at least two of \(S_{xx}\), \(S_{yy}\), \(S_{xy}\).
A1: Two of \(S_{xx} = 42\), \(S_{yy} = 186\), \(S_{xy} = -80\) correct.
A1: All three correct.

(b)
M1: Correct formula for \(r\) using their values from (a).
A1: \(r = -0.905\) (or \(-0.905\) to \(-0.91\)).
B1: Describes negative correlation and explains it in context (selling price decreases as age increases).

(c)
M1: Calculates gradient \(b = \frac{S_{xy}}{S_{xx}}\).
A1: \(b \approx -1.90\).
M1: Method for calculating intercept \(a = \bar{y} - b\bar{x}\).
A1: \(y = 22.6 - 1.90x\) (accept \(y = 22.6 - 1.9x\); must be in terms of \(y\) and \(x\)).

(d)
B1: Correct price estimate: \(12.1\) thousand dollars (or \(\$12,100\)).
B1: Correctly identifies that it is reliable due to interpolation and strong correlation.

**(a)**
From the standard normal tables:
\(P(Z < -1.75) = 0.0401 \implies z_1 = -1.75\)
\(P(Z > 1.25) = 0.1056 \implies z_2 = 1.25\)

Using the standardisation formula \(Z = \frac{L - \mu}{\sigma}\):
\[\frac{147 - \mu}{\sigma} = -1.75 \implies 147 - \mu = -1.75\sigma \quad [1]\]
\[\frac{156 - \mu}{\sigma} = 1.25 \implies 156 - \mu = 1.25\sigma \quad [2]\]

Subtracting equation [1] from equation [2]:
\[(156 - \mu) - (147 - \mu) = 1.25\sigma - (-1.75\sigma)\]
\[9 = 3\sigma \implies \sigma = 3\]

Substituting \(\sigma = 3\) back into equation [2]:
\[156 - \mu = 1.25(3)\]
\[156 - \mu = 3.75 \implies \mu = 152.25 \quad \text{(as required)}\]

**(b)**
A rod is accepted if \(147.75 \le L \le 158.25\).
\[P(147.75 \le L \le 158.25) = P\left(\frac{147.75 - 152.25}{3} \le Z \le \frac{158.25 - 152.25}{3}\right)\]
\[= P(-1.5 \le Z \le 2)\]
\[= \Phi(2) - \Phi(-1.5) = \Phi(2) - (1 - \Phi(1.5))\]
\[= 0.9772 - (1 - 0.9332) = 0.9772 - 0.0668 = 0.9104\]

**(c)**
Let \(A\) be the event that a rod is accepted.
We want to find \(P(L > 152.25 | A)\).
\[P(L > 152.25 | A) = \frac{P((L > 152.25) \cap A)}{P(A)}\]
Since \(A\) is the event \(147.75 \le L \le 158.25\), the intersection \((L > 152.25) \cap A\) is the interval \(152.25 < L \le 158.25\).

\[P(152.25 < L \le 158.25) = P\left(\frac{152.25 - 152.25}{3} < Z \le \frac{158.25 - 152.25}{3}\right)\]
\[= P(0 < Z \le 2) = \Phi(2) - \Phi(0) = 0.9772 - 0.5 = 0.4772\]

Thus, the conditional probability for one rod is:
\[P(L > 152.25 | A) = \frac{0.4772}{0.9104} \approx 0.524165\]

For three independent rods, the probability that all three satisfy this is:
\[(0.524165)^3 \approx 0.14402 \approx 0.144 \quad \text{(3 s.f.)}\]

**(a)**
* **B1**: For \(z = -1.75\) (or \(-1.75\)) seen or implied.
* **B1**: For \(z = 1.25\) (or \(1.25\)) seen or implied.
* **M1**: For standardising both 147 and 156 and setting equal to their \(z\)-values (signs must be consistent with the probabilities given).
* **M1**: Solving the simultaneous equations to eliminate either \(\mu\) or \(\sigma\).
* **A1**: For \(\sigma = 3\).
* **A1**: For showing \(\mu = 152.25\) with clear algebraic steps.

**(b)**
* **M1**: Standardising at least one of the boundaries \(147.75\) or \(158.25\) with their \(\mu\) and \(\sigma\).
* **M1**: Standardising both boundaries correctly: \(\frac{147.75 - 152.25}{3}\) and \(\frac{158.25 - 152.25}{3}\).
* **M1**: Finding \(P(-1.5 \le Z \le 2)\) using standard normal tables: \(\Phi(2) - (1 - \Phi(1.5))\) or equivalent.
* **A1**: For \(0.9104\) (accept \(0.910\) from correct working).

**(c)**
* **M1**: Writing down or using the conditional probability formula \(\frac{P(152.25 < L \le 158.25)}{P(\text{accepted})}\).
* **A1**: For finding \(P(152.25 < L \le 158.25) = 0.4772\).
* **M1**: For calculating the conditional probability \(\frac{0.4772}{\text{their (b)}}\) (must be \(< 1\)).
* **M1**: For cubing their conditional probability.
* **A1**: For an answer in the range \([0.144, 0.145]\) (e.g., \(0.144\)).