2023 Cambridge IAL Mathematics (9709) 模擬試題連答案詳解

Let the terms of the arithmetic progression (AP) be \(u_n = a + (n-1)d\).

The 1st term of the AP is \(a\).
The 3rd term of the AP is \(a + 2d\).
The 7th term of the AP is \(a + 6d\).

These are the first three terms of a geometric progression (GP), so:
\(u_1 = a\)
\(u_2 = a + 2d\)
\(u_3 = a + 6d\)

Since these form a GP, the common ratio is constant:
\(\frac{a + 2d}{a} = \frac{a + 6d}{a + 2d}\)
\((a+2d)^2 = a(a+6d)\)
\(a^2 + 4ad + 4d^2 = a^2 + 6ad\)
\(4d^2 = 2ad\)

Since \(d \neq 0\), we can divide both sides by \(2d\):
\(2d = a \implies d = 0.5a\). (as required)

The common ratio \(r\) is given by:
\(r = \frac{a+2d}{a} = \frac{a + 2(0.5a)}{a} = \frac{2a}{a} = 2\).

(b) The sum of the first 8 terms of the AP is:
\(S_8 = \frac{8}{2}[2a + 7d] = 220\)
\(4[2a + 7(0.5a)] = 220\)
\(4[5.5a] = 220\)
\(22a = 220 \implies a = 10\).

For the GP, the first term is \(a = 10\) and the common ratio is \(r = 2\).
The sum of the first 5 terms is:
\(S_5 = \frac{a(r^5 - 1)}{r - 1} = \frac{10(2^5 - 1)}{2 - 1} = 10(32 - 1) = 310\).

(a)
M1: Attempt to write the terms of the GP in terms of \(a\) and \(d\).
M1: Apply the geometric progression property \((a+2d)^2 = a(a+6d)\).
A1: Show clearly that \(d = 0.5a\).
A1: State \(r = 2\).

(b)
M1: Use the sum of AP formula for \(S_8\) with \(d = 0.5a\).
A1: Solve to find \(a = 10\).
M1: Use the sum of GP formula for \(S_5\) with \(a = 10\) and \(r = 2\).
A1: Obtain 310.

(a) Write the equation as \(y = 4(2x+1)^{-1/4}\).
Differentiate using the chain rule:
\(\frac{\text{d}y}{\text{d}x} = 4 \left(-\frac{1}{4}\right)(2x+1)^{-5/4} \times 2 = -2(2x+1)^{-5/4}\).

At \(x = 0\):
\(y = 4(1)^{-1/4} = 4\).
\(\frac{\text{d}y}{\text{d}x} = -2(1)^{-5/4} = -2\).

The gradient of the tangent is \(-2\), so the gradient of the normal is \(m = -\frac{1}{-2} = \frac{1}{2}\).
The equation of the normal at \((0, 4)\) is:
\(y - 4 = \frac{1}{2}(x - 0) \implies y = \frac{1}{2}x + 4\).

(b) The volume \(V\) is given by:
\(V = \pi \int_{0}^{12} y^2 \text{d}x\)
\(y^2 = \left(\frac{4}{(2x+1)^{1/4}}\right)^2 = \frac{16}{(2x+1)^{1/2}} = 16(2x+1)^{-1/2}\).

Integrate:
\(V = \pi \int_{0}^{12} 16(2x+1)^{-1/2} \text{d}x\)
\(= 16\pi \left[ \frac{(2x+1)^{1/2}}{\frac{1}{2} \times 2} \right]_0^{12}\n= 16\pi \left[ \sqrt{2x+1} \right]_0^{12}\n= 16\pi [ \sqrt{25} - \sqrt{1} ]\n= 16\pi [ 5 - 1 ] = 64\pi\).

(a)
M1: Find \(\frac{\text{d}y}{\text{d}x}\) using chain rule.
A1: Correct derivative \(-2(2x+1)^{-5/4}\).
M1: Calculate normal gradient from negative reciprocal of tangent gradient at \(x=0\).
A1: State correct equation of normal \(y = \frac{1}{2}x + 4\).

(b)
M1: Set up volume integral \(V = \pi \int y^2 \text{d}x\).
A1: Correctly square \(y\) to obtain \(16(2x+1)^{-1/2}\).
M1: Integrate expression of the form \((2x+1)^{-1/2}\) to obtain \(k(2x+1)^{1/2}\).
A1: Correctly substitute limits and get \(64\pi\).

(a) Complete the square for \(\text{f}(x)\):
\(\text{f}(x) = 2(x^2 - 6x) + 13\)
\(= 2[(x-3)^2 - 9] + 13\)
\(= 2(x-3)^2 - 18 + 13\)
\(= 2(x-3)^2 - 5\).

(b) Since \(2(x-3)^2 \ge 0\) for all real \(x\), the minimum value of \(\text{f}(x)\) is \(-5\).
The range of \(\text{f}\) is \(\text{f}(x) \ge -5\).

(c) For a quadratic function to have an inverse, the domain must be restricted to one side of the vertex. The vertex is at \(x = 3\).
For the domain \(x \le k\), the largest possible value is \(k = 3\).

(d) Solve \(\text{gf}(x) = 23\):
\(\text{g}(\text{f}(x)) = 23 \implies 2\text{f}(x) - 3 = 23\)
\(2\text{f}(x) = 26 \implies \text{f}(x) = 13\).

Using the vertex form:
\(2(x-3)^2 - 5 = 13\)
\(2(x-3)^2 = 18\)
\((x-3)^2 = 9\)
\(x-3 = \pm 3\)
\(x = 0\) or \(x = 6\).

(a)
B1: Obtain \(2(x-3)^2\) part.
B1: Obtain the constant \(-5\).

(b)
B1: State \(\text{f}(x) \ge -5\) (accept \(y \ge -5\), reject \(x \ge -5\)).

(c)
B1: State \(k = 3\).

(d)
M1: Set up equation \(\text{g}(\text{f}(x)) = 23\) and simplify to \(\text{f}(x) = 13\).
M1: Solve \(2(x-3)^2 - 5 = 13\) (or the equivalent quadratic equation).
A1: Correct solutions \(x = 0\) and \(x = 6\).

(a) Combine the fractions on the left-hand side over a common denominator:
\(\text{LHS} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta(1 + \cos \theta)}\)

Since \(\sin^2 \theta + \cos^2 \theta \equiv 1\):
\(= \frac{1 + 1 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2(1 + \cos \theta)}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2}{\sin \theta}\) (as required).

(b) Using the identity from part (a) with \(\theta = 2x\), the equation becomes:
\(\frac{2}{\sin 2x} = \frac{4}{\sqrt{3}}\)
\(\sin 2x = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}\).

Since \(0^\circ < x < 180^\circ\), we have \(0^\circ < 2x < 360^\circ\).
\(2x = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^\circ \text{ or } 120^\circ\).
Therefore, \(x = 30^\circ \text{ or } 60^\circ\).

(a)
M1: Put LHS over a common denominator.
M1: Expand numerator and use \(\sin^2 \theta + \cos^2 \theta = 1\).
A1: Fully correct completion of identity with no steps omitted.

(b)
M1: Use identity to simplify equation to \(\frac{2}{\sin 2x} = \frac{4}{\sqrt{3}}\).
A1: Obtain \(\sin 2x = \frac{\sqrt{3}}{2}\).
M1: Find two values for \(2x\) in the interval \(0^\circ < 2x < 360^\circ\).
A1: State both correct final answers \(x = 30^\circ\) and \(x = 60^\circ\).

(a) When \(k = 1\), the curve is \(y = x^2 - 6x + 9\).
We set this equal to the line \(y = x + 3\):
\(x^2 - 6x + 9 = x + 3\)
\(x^2 - 7x + 6 = 0\)

Factorizing:
\((x - 1)(x - 6) = 0 \implies x = 1 \text{ or } x = 6\).

If \(x = 1\), \(y = 1 + 3 = 4\).
If \(x = 6\), \(y = 6 + 3 = 9\).

The points of intersection are \((1, 4)\) and \((6, 9)\).

(b) For the curve \(y = kx^2 - 6x + (k + 8)\) to lie entirely above the x-axis:
1) The coefficient of \(x^2\) must be positive, so \(k > 0\).
2) The curve must have no real roots, so the discriminant \(b^2 - 4ac < 0\):
\((-6)^2 - 4(k)(k+8) < 0\)
\(36 - 4k^2 - 32k < 0\)

Divide by \(-4\) (which reverses the inequality sign):
\(k^2 + 8k - 9 > 0\)
Factorizing:
\((k + 9)(k - 1) > 0\)
This gives \(k < -9\) or \(k > 1\).

Since we also require \(k > 0\) for the curve to open upwards, the only valid set of values is \(k > 1\).

(a)
M1: Equate curve and line equations with \(k=1\).
A1: Form the correct quadratic equation \(x^2 - 7x + 6 = 0\).
M1: Factorize or use formula to find \(x\)-values.
A1: State both correct coordinates \((1, 4)\) and \((6, 9)\).

(b)
M1: Identify that the discriminant \(b^2 - 4ac < 0\) is needed.
M1: Set up the inequality \(36 - 4k(k+8) < 0\) and simplify to \((k+9)(k-1) > 0\).
A1: Solve the quadratic inequality to get \(k < -9\) or \(k > 1\).
A1: Combine with the condition \(k > 0\) to obtain the final correct range \(k > 1\).

(a) Midpoint \(M\) of \(AB\):
\(M = \left(\frac{-2 + 6}{2}, \frac{3 + 9}{2}\right) = (2, 6)\).

Gradient of \(AB\):
\(m = \frac{9 - 3}{6 - (-2)} = \frac{6}{8} = \frac{3}{4}\).

The gradient of the perpendicular bisector is the negative reciprocal:
\(m_{\perp} = -\frac{4}{3}\).

The equation of the perpendicular bisector is:
\(y - 6 = -\frac{4}{3}(x - 2)\)
\(3y - 18 = -4x + 8 \implies 4x + 3y = 26\).

(b) Since the circle passes through \(A\) and \(B\), its centre \(C\) must lie on the perpendicular bisector of \(AB\).
Since the centre also lies on the line \(y = x - 3\), we solve the simultaneous equations:
\(4x + 3y = 26\) and \(y = x - 3\).

Substitute the second equation into the first:
\(4x + 3(x - 3) = 26\)
\(4x + 3x - 9 = 26\)
\(7x = 35 \implies x = 5\).

Then \(y = 5 - 3 = 2\).
So the centre of the circle is \(C(5, 2)\).

The radius \(R\) is the distance from \(C(5, 2)\) to \(A(-2, 3)\):
\(R^2 = (5 - (-2))^2 + (2 - 3)^2 = 7^2 + (-1)^2 = 49 + 1 = 50\).

The equation of the circle is:
\((x - 5)^2 + (y - 2)^2 = 50\).

(a)
B1: Find the midpoint \((2, 6)\).
M1: Find gradient of \(AB\) and then negative reciprocal gradient.
A1: State perpendicular gradient \(-\frac{4}{3}\).
A1: Formulate the equation \(4x + 3y = 26\) (or equivalent linear form).

(b)
M1: State or imply that the centre is the intersection of the perpendicular bisector and the line \(y = x - 3\).
M1: Solve the simultaneous equations to find the coordinates of the centre.
A1: Obtain centre \((5, 2)\).
A1: Calculate \(R^2 = 50\) and write the correct equation \((x-5)^2 + (y-2)^2 = 50\).

(a) Area of the shaded region is the area of sector \(OAB\) minus the area of the right-angled triangle \(OAC\).

Area of sector \(OAB\) = \(\frac{1}{2} r^2 \theta = \frac{1}{2} (12^2) \theta = 72\theta\).

In the right-angled triangle \(OAC\):
\(OC = 12 \cos\theta\) and \(AC = 12 \sin\theta\).

Area of triangle \(OAC\) = \(\frac{1}{2} \times OC \times AC = \frac{1}{2} (12\cos\theta)(12\sin\theta) = 72 \sin\theta\cos\theta\).

Subtracting the triangle area from the sector area:
Shaded Area = \(72\theta - 72\sin\theta\cos\theta = 72(\theta - \sin\theta\cos\theta)\). (as required)

(b)(i) For \(\theta = 1.2\):
Area = \(72(1.2 - \sin(1.2)\cos(1.2))\)
Using a calculator in radian mode:
\(\sin(1.2) \approx 0.93204\)
\(\cos(1.2) \approx 0.36236\)
Area \(\approx 72(1.2 - 0.93204 \times 0.36236) \approx 72(1.2 - 0.33773) = 72(0.86227) = 62.1\text{ cm}^2\) (to 3 s.f.).

(b)(ii) The perimeter \(P\) is the sum of arc length \(AB\), line \(AC\), and line \(BC\).
Arc length \(AB = r\theta = 12(1.2) = 14.4\text{ cm}\).
Line \(AC = 12\sin(1.2) \approx 11.184\text{ cm}\).
Line \(BC = OB - OC = 12 - 12\cos(1.2) \approx 12 - 12(0.36236) = 12 - 4.348 = 7.652\text{ cm}\).

Total Perimeter = \(14.4 + 11.184 + 7.652 \approx 33.2\text{ cm}\) (to 3 s.f.).

(a)
M1: Express shaded area as sector area minus triangle area.
M1: Express \(AC = 12\sin\theta\) and \(OC = 12\cos\theta\) correctly.
A1: Show clearly the final given expression \(72(\theta - \sin\theta\cos\theta)\).

(b)(i)
M1: Substitute \(\theta = 1.2\) into the formula with calculator in radian mode.
A1: State \(62.1\text{ cm}^2\) (accept \(62.0\) to \(62.2\)).

(b)(ii)
M1: Write expression for perimeter \(r\theta + r\sin\theta + r(1-\cos\theta)\).
A1: Calculate correct numerical value of \(33.2\text{ cm}\) (accept \(33.1\) to \(33.3\)).

(a) Write \(y = kx^{-1} + 2x^2\).
\(\frac{\text{d}y}{\text{d}x} = -kx^{-2} + 4x = -\frac{k}{x^2} + 4x\).
\(\frac{\text{d}^2y}{\text{d}x^2} = 2kx^{-3} + 4 = \frac{2k}{x^3} + 4\).

(b) Since there is a stationary point at \(x = 2\):
\(\frac{\text{d}y}{\text{d}x}\Big|_{x=2} = 0 \implies -\frac{k}{2^2} + 4(2) = 0\)
\(-\frac{k}{4} + 8 = 0 \implies k = 32\).

The y-coordinate \(y_0\) of \(P\) is:
\(y_0 = \frac{32}{2} + 2(2^2) = 16 + 8 = 24\).

The coordinates of \(P\) are \((2, 24)\).

(c) Substitute \(k = 32\) and \(x = 2\) into \(\frac{\text{d}^2y}{\text{d}x^2}\):
\(\frac{\text{d}^2y}{\text{d}x^2} = \frac{2(32)}{2^3} + 4 = \frac{64}{8} + 4 = 8 + 4 = 12\).

Since \(\frac{\text{d}^2y}{\text{d}x^2} = 12 > 0\), the stationary point \(P\) is a minimum.

(a)
M1: Differentiate to find \(\frac{\text{d}y}{\text{d}x}\), containing at least one correct term.
A1: Correct first derivative \(-kx^{-2} + 4x\).
A1: Correct second derivative \(2kx^{-3} + 4\).

(b)
M1: Equate first derivative to 0 at \(x = 2\) and solve for \(k\).
A1: State \(k = 32\) and coordinates \(P(2, 24)\).

(c)
M1: Evaluate second derivative at \(x = 2, k = 32\).
A1: Conclude minimum since \(\frac{\text{d}^2y}{\text{d}x^2} = 12 > 0\).

(a) The first term of the AP is \(T_1 = a\), the fourth term is \(T_4 = a + 3d\), and the fifth term is \(T_5 = a + 4d\). Since these three terms form a geometric progression, we have: \((T_4)^2 = T_1 \times T_5\) which gives \((a + 3d)^2 = a(a + 4d)\). Expanding both sides, we get: \(a^2 + 6ad + 9d^2 = a^2 + 4ad\). Subtracting \(a^2\) and \(4ad\) from both sides yields: \(2ad + 9d^2 = 0\). Since \(d \neq 0\), we can divide through by \(d\) to obtain: \(9d + 2a = 0\). (b) The sum of the first 15 terms of the AP is given by: \(S_{15} = \frac{15}{2}[2a + 14d] = 150\), which simplifies to \(2a + 14d = 20\). From part (a), we have \(2a = -9d\). Substituting this in gives: \(-9d + 14d = 20 \implies 5d = 20\), so \(d = 4\). Then, \(2a = -36 \implies a = -18\). The first term of the GP is \(a_{GP} = a = -18\) and the second term is \(a_{GP}r = a + 3d = -18 + 12 = -6\). Thus, the common ratio is \(r = \frac{-6}{-18} = \frac{1}{3}\). Since \(|r| < 1\), the sum to infinity exists and is: \(S_\infty = \frac{-18}{1 - 1/3} = \frac{-18}{2/3} = -27\).

(a) M1: For attempting to write \((a + 3d)^2 = a(a + 4d)\). A1: For expanding and simplifying to a correct quadratic form \(2ad + 9d^2 = 0\). A1: For correctly dividing by \(d\) and showing the given equation \(9d + 2a = 0\) with clear steps. (b) M1: For using the AP sum formula to set up \(\frac{15}{2}(2a + 14d) = 150\). M1: For solving the simultaneous equations to find either \(a\) or \(d\). A1: For obtaining \(d = 4\) and \(a = -18\). A1: For calculating \(r = \frac{1}{3}\) and substituting into the sum to infinity formula to obtain \(-27\).

(a) The area \(A\) of the region \(R\) is given by: \(A = \int_{0}^{5} (3x + 1)^{-1/4} \, dx\). Integrating, we get: \(A = \left[ \frac{(3x + 1)^{3/4}}{3 \times \frac{3}{4}} \right]_{0}^{5} = \left[ \frac{4}{9} (3x + 1)^{3/4} \right]_{0}^{5}\). Substituting the upper and lower limits: \(A = \frac{4}{9} (3(5) + 1)^{3/4} - \frac{4}{9} (3(0) + 1)^{3/4} = \frac{4}{9} (16)^{3/4} - \frac{4}{9} (1)^{3/4}\). Since \(16^{3/4} = (16^{1/4})^3 = 2^3 = 8\), we get: \(A = \frac{4}{9}(8) - \frac{4}{9}(1) = \frac{28}{9}\). (b) The volume \(V\) of the solid of revolution is given by: \(V = \pi \int_{0}^{5} y^2 \, dx = \pi \int_{0}^{5} (3x + 1)^{-1/2} \, dx\). Integrating, we get: \(V = \pi \left[ \frac{(3x + 1)^{1/2}}{3 \times \frac{1}{2}} \right]_{0}^{5} = \pi \left[ \frac{2}{3} (3x + 1)^{1/2} \right]_{0}^{5}\). Substituting the limits: \(V = \pi \left( \frac{2}{3} (16)^{1/2} - \frac{2}{3} (1)^{1/2} \right) = \pi \left( \frac{2}{3}(4) - \frac{2}{3}(1) \right) = \pi \left( \frac{8}{3} - \frac{2}{3} \right) = 2\pi\).

(a) M1: For attempting to integrate \((3x + 1)^{-1/4}\) to obtain \(k(3x + 1)^{3/4}\). A1: For the correct integrated expression \(\frac{4}{9} (3x + 1)^{3/4}\). M1: For substituting limits 5 and 0 and subtracting. A1: For the correct exact area \(\frac{28}{9}\) (or equivalent). (b) M1: For stating and applying the volume formula \(V = \pi \int y^2 \, dx\). A1: For obtaining the correct integrated expression \(\frac{2}{3} (3x + 1)^{1/2}\) (ignoring \(\pi\) for now). M1: For substituting limits 5 and 0 and subtracting with \(\pi\) present. A1: For the correct exact volume \(2\pi\).

Let \(\frac{3x^2 - x + 2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx+C}{1+x^2}\).
Multiplying by the common denominator, we get:
\(3x^2 - x + 2 = A(1+x^2) + (Bx+C)(1-x)\).
Let \(x=1\):
\(3(1)^2 - 1 + 2 = A(1+1) \implies 4 = 2A \implies A = 2\).
Let \(x=0\):
\(2 = A + C \implies 2 = 2 + C \implies C = 0\).
Equating coefficients of \(x^2\):
\(3 = A - B \implies 3 = 2 - B \implies B = -1\).
Thus, the partial fractions are:
\(\frac{2}{1-x} - \frac{x}{1+x^2}\).

Now, expand each term binomialy:
\(2(1-x)^{-1} = 2(1 + x + x^2 + x^3 + \dots) = 2 + 2x + 2x^2 + 2x^3 + \dots\)
\(x(1+x^2)^{-1} = x(1 - x^2 + x^4 - \dots) = x - x^3 + \dots\)

Subtracting the two series gives:
\(\mathrm{f}(x) = (2 + 2x + 2x^2 + 2x^3) - (x - x^3) = 2 + x + 2x^2 + 3x^3\).

M1: State or imply the form \(\frac{A}{1-x} + \frac{Bx+C}{1+x^2}\) and find at least one constant.
A1: Obtain \(A = 2, B = -1, C = 0\) (allow expression as \(\frac{2}{1-x} - \frac{x}{1+x^2}\)).
M1: Attempt binomial expansion of \((1-x)^{-1}\) up to the term in \(x^3\).
M1: Attempt binomial expansion of \((1+x^2)^{-1}\) up to the term in \(x^2\) (yielding term in \(x^3\) when multiplied by \(x\)).
A1: Obtain correct final expansion: \(2 + x + 2x^2 + 3x^3\).

We rewrite the equation using logarithmic properties:
\(\ln(e^{2x} - 4) - \ln(e^x) = \ln(3)\)
\(\ln\left(\frac{e^{2x} - 4}{e^x}\right) = \ln(3)\)
Taking the exponential of both sides:
\(\frac{e^{2x} - 4}{e^x} = 3\)
\(e^{2x} - 4 = 3e^x\)
\(e^{2x} - 3e^x - 4 = 0\)
Let \(u = e^x\). This yields the quadratic equation:
\(u^2 - 3u - 4 = 0\)
\((u-4)(u+1) = 0\)
Thus, \(u = 4\) or \(u = -1\).
Since \(e^x > 0\) for all real \(x\), we discard \(u = -1\).
Thus, \(e^x = 4 \implies x = \ln(4)\).

M1: Express \(x\) as \(\ln(e^x)\) and use the subtraction law of logarithms correctly.
M1: Remove logarithms to obtain a polynomial equation in terms of \(e^x\).
A1: Obtain a correct quadratic equation: \(e^{2x} - 3e^x - 4 = 0\).
M1: Solve the quadratic equation to find two roots for \(e^x\).
A1: Discard the negative root and solve for \(x\) to obtain \(x = \ln(4)\).

Using the double-angle identity \(\cos(2\theta) = 1 - 2\sin^2\theta\):
\(3(1 - 2\sin^2\theta) + 17\sin\theta = 10\)
\(3 - 6\sin^2\theta + 17\sin\theta = 10\)
\(6\sin^2\theta - 17\sin\theta + 7 = 0\)
Factoring the quadratic in terms of \(\sin\theta\):
\((2\sin\theta - 1)(3\sin\theta - 7) = 0\)
This gives:
\(\sin\theta = \frac{1}{2}\) or \(\sin\theta = \frac{7}{3}\).
Since \(\frac{7}{3} > 1\), \(\sin\theta = \frac{7}{3}\) has no real solutions.
For \(\sin\theta = \frac{1}{2}\) in the interval \(0 \le \theta \le 2\pi\):
\(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

M1: Use the identity \(\cos(2\theta) = 1 - 2\sin^2\theta\) to form an equation in \(\sin\theta\).
A1: Obtain the correct quadratic equation \(6\sin^2\theta - 17\sin\theta + 7 = 0\).
M1: Attempt to solve the quadratic equation to find values of \(\sin\theta\).
A1: Identify \(\sin\theta = 1/2\) and state that \(\sin\theta = 7/3\) yields no solution.
A1: Obtain \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\) (and no others in the given range).

We differentiate \(y\) with respect to \(x\) using the product rule:
\(\frac{dy}{dx} = -\sqrt{3}e^{-\sqrt{3}x} \sin x + e^{-\sqrt{3}x} \cos x\)
\(\frac{dy}{dx} = e^{-\sqrt{3}x} (\cos x - \sqrt{3}\sin x)\)
To find the stationary points, we set \(\frac{dy}{dx} = 0\):
\(e^{-\sqrt{3}x} (\cos x - \sqrt{3}\sin x) = 0\)
Since \(e^{-\sqrt{3}x} \neq 0\) for all real \(x\):
\(\cos x - \sqrt{3}\sin x = 0\)
\(\tan x = \frac{1}{\sqrt{3}}\)
In the interval \(0 < x < \pi\), the solution is:
\(x = \frac{\pi}{6}\)
We find the corresponding \(y\)-coordinate:
\(y = e^{-\sqrt{3}(\frac{\pi}{6})} \sin\left(\frac{\pi}{6}\right) = e^{-\frac{\sqrt{3}\pi}{6}} \cdot \frac{1}{2} = \frac{1}{2}e^{-\frac{\sqrt{3}\pi}{6}}\)
So the stationary point has coordinates \(\left(\frac{\pi}{6}, \frac{1}{2}e^{-\frac{\sqrt{3}\pi}{6}}\right)\).

M1: Apply the product rule to differentiate \(y = e^{-\sqrt{3}x} \sin x\).
A1: Obtain a correct expression for the first derivative.
M1: Equate their derivative to zero and solve for \(\tan x\).
A1: Obtain \(x = \frac{\pi}{6}\).
M1: Substitute their \(x\)-value back into the original curve equation to find the exact \(y\)-coordinate.
A1: Provide the exact coordinates \(\left(\frac{\pi}{6}, \frac{1}{2}e^{-\frac{\sqrt{3}\pi}{6}}\right)\).

Let \(u = \cos x \implies du = -\sin x \, dx \implies \sin x \, dx = -du\).
Now we transform the limits:
When \(x = 0\), \(u = \cos 0 = 1\).
When \(x = \frac{\pi}{3}\), \(u = \cos\frac{\pi}{3} = \frac{1}{2}\).
Rewrite the integrand:
\(\int_{0}^{\frac{\pi}{3}} \frac{\sin^2 x}{2 - \cos x} (\sin x \, dx) = \int_{1}^{1/2} \frac{1 - u^2}{2 - u} (-du) = \int_{1/2}^{1} \frac{1 - u^2}{2 - u} \, du\).
Perform algebraic division:
\(\frac{1 - u^2}{2 - u} = \frac{u^2 - 1}{u - 2} = u + 2 + \frac{3}{u - 2}\).
Now, integrate each term:
\(\int_{1/2}^{1} \left( u + 2 + \frac{3}{u-2} \right) du = \left[ \frac{1}{2}u^2 + 2u + 3\ln|u - 2| \right]_{1/2}^{1}\)
At \(u = 1\):
\(\frac{1}{2}(1)^2 + 2(1) + 3\ln|1-2| = \frac{5}{2} + 0 = \frac{5}{2}\)
At \(u = 1/2\):
\(\frac{1}{2}\left(\frac{1}{4}\right) + 2\left(\frac{1}{2}\right) + 3\ln|\frac{1}{2} - 2| = \frac{1}{8} + 1 + 3\ln\left(\frac{3}{2}\right) = \frac{9}{8} + 3\ln\left(\frac{3}{2}\right)\)
Subtracting the lower limit value from the upper limit value:
\(\frac{5}{2} - \left( \frac{9}{8} + 3\ln\left(\frac{3}{2}\right) \right) = \frac{11}{8} - 3\ln\left(\frac{3}{2}\right)\).

M1: Perform change of variables with \(u = \cos x\) and express the differential correctly.
A1: Change limits correctly to \(u = 1\) and \(u = 1/2\), obtaining a correct integral in terms of \(u\).
M1: Carry out algebraic division or use partial fractions on \(\frac{1-u^2}{2-u}\).
A1: Obtain \(u + 2 + \frac{3}{u-2}\) (or equivalent).
M1: Integrate to obtain terms of the form \(a u^2 + b u + c \ln|u-2|\) and substitute limits.
A1: Obtain the exact answer \(\frac{11}{8} - 3\ln\left(\frac{3}{2}\right)\) (or equivalent exact expression).

First, simplify \(u\) by multiplying the numerator and denominator by the complex conjugate of the denominator:
\(u = \frac{(1+3\mathrm{i})(2-\mathrm{i})}{(2+\mathrm{i})(2-\mathrm{i})} = \frac{2 - \mathrm{i} + 6\mathrm{i} - 3\mathrm{i}^2}{4 - \mathrm{i}^2} = \frac{5 + 5\mathrm{i}}{5} = 1 + \mathrm{i}\)
Now find the modulus \(r\) and argument \(\theta\):
\(r = |u| = \sqrt{1^2 + 1^2} = \sqrt{2}\)
\(\theta = \arg(u) = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}\)
So, \(u = \sqrt{2}\left(\cos\frac{\pi}{4} + \mathrm{i}\sin\frac{\pi}{4}\right)\).
Using de Moivre's theorem:
\(u^n = (\sqrt{2})^n \left(\cos\frac{n\pi}{4} + \mathrm{i}\sin\frac{n\pi}{4}\right)\)
For \(u^n\) to be real, the imaginary part must be zero:
\(\sin\frac{n\pi}{4} = 0 \implies \frac{n\pi}{4} = k\pi\) for some integer \(k\).
Since we require the smallest positive integer \(n\), we set \(k = 1\):
\(\frac{n\pi}{4} = \pi \implies n = 4\).

M1: Multiply numerator and denominator by \(2 - \mathrm{i}\) to simplify \(u\).
A1: Obtain \(u = 1 + \mathrm{i}\).
M1: Calculate the modulus and argument of \(1+\mathrm{i}\).
A1: Express \(u\) in the correct polar form \(\sqrt{2}\left(\cos\frac{\pi}{4} + \mathrm{i}\sin\frac{\pi}{4}\right)\).
M1: Apply de Moivre's theorem and set the imaginary part of \(u^n\) equal to zero.
A1: Correctly deduce that the smallest positive integer is \(n = 4\).

Separate variables to obtain:
\(\int \frac{1}{y^2} \, dy = \int \frac{\ln x}{x} \, dx\)
Integrate the LHS:
\(\int y^{-2} \, dy = -\frac{1}{y}\)
Integrate the RHS using substitution \(u = \ln x\) or by inspection:
\(\int \frac{\ln x}{x} \, dx = \frac{1}{2}(\ln x)^2\)
Adding a constant of integration, we get:
\(-\frac{1}{y} = \frac{1}{2}(\ln x)^2 + C\)
Using the initial condition \(y = -1\) when \(x = 1\):
\(-\frac{1}{-1} = \frac{1}{2}(\ln 1)^2 + C \implies 1 = 0 + C \implies C = 1\)
Thus:
\(-\frac{1}{y} = \frac{1}{2}(\ln x)^2 + 1\)
\(-\frac{1}{y} = \frac{(\ln x)^2 + 2}{2}\)
Taking the reciprocal and multiplying by \(-1\):
\(y = -\frac{2}{(\ln x)^2 + 2}\).

M1: Separate variables correctly.
A1: Integrate the LHS to obtain \(-\frac{1}{y}\).
M1: Integrate the RHS to obtain \(\frac{1}{2}(\ln x)^2\) (or equivalent).
A1: Introduce a constant of integration and form a correct equation.
M1: Substitute \(x = 1\) and \(y = -1\) to find \(C\).
A1: Express the final relation explicitly as \(y = -\frac{2}{(\ln x)^2 + 2}\).

The vector equation of the line \(l\) is:
\(\mathbf{r} = (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) + \lambda(\mathbf{i} + 2\mathbf{j} - \mathbf{k})\)
Since \(P\) lies on \(l\), the position vector of \(P\) is:
\(\vec{OP} = (2+\lambda)\mathbf{i} + (-1+2\lambda)\mathbf{j} + (3-\lambda)\mathbf{k}\)
The vector \(\vec{BP}\) is given by:
\(\vec{BP} = \vec{OP} - \vec{OB} = (2+\lambda-1)\mathbf{i} + (-1+2\lambda-2)\mathbf{j} + (3-\lambda-2)\mathbf{k}\)
\(\vec{BP} = (1+\lambda)\mathbf{i} + (-3+2\lambda)\mathbf{j} + (1-\lambda)\mathbf{k}\)
The direction vector of the line \(l\) is \(\mathbf{u} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}\).
Since \(BP\) is perpendicular to \(l\), we have \(\vec{BP} \cdot \mathbf{u} = 0\):
\((1+\lambda)(1) + (-3+2\lambda)(2) + (1-\lambda)(-1) = 0\)
\(1 + \lambda - 6 + 4\lambda - 1 + \lambda = 0\)
\(6\lambda - 6 = 0 \implies \lambda = 1\)
Substituting \(\lambda = 1\) back into \(\vec{OP}\):
\(\vec{OP} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\).

M1: State the general position vector of \(P\) on the line \(l\) using a parameter \(\lambda\).
M1: Form the vector \(\vec{BP}\) in terms of \(\lambda\).
M1: State that the dot product of \(\vec{BP}\) and the direction vector \(\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) is 0.
A1: Set up a correct linear equation in \(\lambda\).
A1: Solve to find \(\lambda = 1\).
A1: Obtain the correct position vector \(3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\).

Let \(u = \cos x\). Then \(du = -\sin x \, dx\), which means \(\sin x \, dx = -du\). We also need to change the limits of integration: When \(x = 0\), \(u = \cos 0 = 1\). When \(x = \frac{\pi}{3}\), \(u = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\). Substituting these into the integral gives: \(\int_{1}^{1/2} \frac{-du}{1 + u^2} = \int_{1/2}^{1} \frac{1}{1 + u^2} \, du\). The antiderivative of \(\frac{1}{1+u^2}\) is \(\arctan u\). Evaluating this from \(1/2\) to \(1\): \([\arctan u]_{1/2}^{1} = \arctan(1) - \arctan(1/2)\). We know that \(\arctan(1) = \frac{\pi}{4}\). Let \(\theta = \arctan(1) - \arctan(1/2)\). Taking the tangent of both sides: \(\tan \theta = \tan\left(\arctan(1) - \arctan(1/2)\right)\). Using the subtraction formula for tangent, \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\), we have: \(\tan \theta = \frac{1 - 1/2}{1 + 1 \cdot (1/2)} = \frac{1/2}{3/2} = \frac{1}{3}\). Since \(\theta\) is in the first quadrant, \(\theta = \arctan\left(\frac{1}{3}\right)\). Thus, \(k = \frac{1}{3}\).

M1: Attempt to use the substitution \(u = \cos x\) to find the differential relationship \(du = -\sin x \, dx\). A1: Correctly transform the limits of integration to \(1\) and \(1/2\). A1: Obtain the simplified integral in terms of \(u\): \(\int_{1/2}^{1} \frac{1}{1 + u^2} \, du\). M1: Integrate to obtain \(\arctan u\) and substitute limits. A1: Obtain \(\arctan 1 - \arctan(1/2)\). M1: Use the tangent subtraction identity to combine the terms. A1: Correctly deduce that \(k = 1/3\).

(i) The complex number \(w\) can be written in polar form as: \(w = \sqrt{2} \left( \cos\left(\frac{3\pi}{4}\right) + \mathrm{i}\sin\left(\frac{3\pi}{4}\right) \right)\). Since \(\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}\) and \(\sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}\), we have: \(w = \sqrt{2}\left(-\frac{1}{\sqrt{2}} + \mathrm{i}\frac{1}{\sqrt{2}}\right) = -1 + \mathrm{i}\). (ii) We substitute \(w = -1 + \mathrm{i}\) into the given equation: \(\frac{z + (-1 + \mathrm{i})}{z - \mathrm{i}} = 2\mathrm{i}\). Multiply both sides by \(z - \mathrm{i}\): \(z - 1 + \mathrm{i} = 2\mathrm{i}(z - \mathrm{i})\). Expand the right-hand side: \(z - 1 + \mathrm{i} = 2\mathrm{i}z - 2\mathrm{i}^2 = 2\mathrm{i}z + 2\) (since \(\mathrm{i}^2 = -1\)). Group the terms with \(z\) on one side and the constant terms on the other: \(z - 2\mathrm{i}z = 2 - (-1 + \mathrm{i})\), which simplifies to \(z(1 - 2\mathrm{i}) = 3 - \mathrm{i}\). Solve for \(z\): \(z = \frac{3 - \mathrm{i}}{1 - 2\mathrm{i}}\). Multiply the numerator and the denominator by the complex conjugate of the denominator, \(1 + 2\mathrm{i}\): \(z = \frac{(3 - \mathrm{i})(1 + 2\mathrm{i})}{(1 - 2\mathrm{i})(1 + 2\mathrm{i})} = \frac{3 + 6\mathrm{i} - \mathrm{i} - 2\mathrm{i}^2}{1^2 + 2^2} = \frac{3 + 5\mathrm{i} + 2}{5} = \frac{5 + 5\mathrm{i}}{5} = 1 + \mathrm{i}\).

B1: Correctly find \(w = -1 + \mathrm{i}\). M1: Substitute \(w\) and multiply both sides by \(z - \mathrm{i}\). A1: Obtain correct expanded equation, e.g., \(z - 1 + \mathrm{i} = 2\mathrm{i}z + 2\). M1: Collect terms in \(z\) and factorise to get \(z(1 - 2\mathrm{i}) = 3 - \mathrm{i}\). M1: Attempt to divide complex numbers by multiplying numerator and denominator by the conjugate \(1 + 2\mathrm{i}\). A1: Correctly calculate the denominator as \(5\) and numerator as \(5 + 5\mathrm{i}\). A1: Obtain final answer \(z = 1 + \mathrm{i}\).

To solve the differential equation, we separate the variables: \(\frac{1}{y \ln y} \, dy = \frac{1}{x} \, dx\). Integrating both sides: \(\int \frac{1}{y \ln y} \, dy = \int \frac{1}{x} \, dx\). For the left-hand side, use the substitution \(u = \ln y\), then \(du = \frac{1}{y} \, dy\). The integral becomes: \(\int \frac{1}{u} \, du = \ln|u| = \ln(\ln y)\) (since \(y > 1\), \(\ln y > 0\)). The right-hand side integrates to: \(\ln x + C\) (since \(x > 0\)). Therefore, we have: \(\ln(\ln y) = \ln x + C\). Use the initial condition \(y = \mathrm{e}^2\) when \(x = 1\): \(\ln(\ln(\mathrm{e}^2)) = \ln(1) + C \implies \ln(2) = 0 + C \implies C = \ln 2\). Substitute \(C = \ln 2\) back into the equation: \(\ln(\ln y) = \ln x + \ln 2\). Using the addition rule for logarithms: \(\ln(\ln y) = \ln(2x)\). Taking the exponential of both sides: \(\ln y = 2x\). Taking the exponential once more gives the final solution: \(y = \mathrm{e}^{2x}\).

M1: Separate variables correctly to obtain \(\int \frac{1}{y \ln y} \, dy = \int \frac{1}{x} \, dx\). M1: Integrate the LHS to obtain \(\ln(\ln y)\). A1: Integrate the RHS to obtain \(\ln x\). M1: Include a constant of integration \(C\) and substitute boundary conditions to find \(C\). A1: Find \(C = \ln 2\). M1: Rearrange the equation with logarithms to make \(y\) the subject. A1: Obtain the final exact solution \(y = \mathrm{e}^{2x}\).

To find the velocity of the particle, we integrate the acceleration function with respect to time:

\[ v(t) = \int a \, dt = \int (6t - 12) \, dt = 3t^2 - 12t + c \]

Given that the velocity at \(t = 0\) is \(9 \text{ m s}^{-1}\):

\[ 9 = 3(0)^2 - 12(0) + c \implies c = 9 \]

So, the velocity function is:

\[ v(t) = 3t^2 - 12t + 9 \]

To find the distance traveled, we must check if the velocity changes sign in the interval \(0 \le t \le 3\). We set \(v(t) = 0\):

\[ 3t^2 - 12t + 9 = 0 \implies 3(t^2 - 4t + 3) = 0 \implies 3(t - 1)(t - 3) = 0 \]

This gives \(t = 1\) and \(t = 3\). Thus, the velocity is positive for \(0 \le t < 1\) and negative for \(1 < t < 3\).

Now, we find the displacement function \(s(t)\) by integrating the velocity function, assuming \(s(0) = 0\):

\[ s(t) = \int (3t^2 - 12t + 9) \, dt = t^3 - 6t^2 + 9t \]

We evaluate the displacement at critical times:
- At \(t = 0\): \(s(0) = 0\)
- At \(t = 1\): \(s(1) = 1^3 - 6(1)^2 + 9(1) = 4\)
- At \(t = 3\): \(s(3) = 3^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0\)

The distance traveled in \(0 \le t \le 1\) is:

\[ |s(1) - s(0)| = |4 - 0| = 4 \text{ m} \]

The distance traveled in \(1 \le t \le 3\) is:

\[ |s(3) - s(1)| = |0 - 4| = 4 \text{ m} \]

Thus, the total distance traveled is:

\[ 4 + 4 = 8 \text{ m} \]

M1: For integrating the acceleration function to find the velocity, including a constant of integration.
A1: For correctly determining \(c = 9\) and writing the expression \(v(t) = 3t^2 - 12t + 9\).
M1: For setting \(v(t) = 0\) and solving to find critical values of \(t\) at \(t=1\) and \(t=3\).
M1: For integrating the velocity function to obtain the displacement expression \(s(t) = t^3 - 6t^2 + 9t\).
A1: For calculating the correct displacement values at \(t = 1\) and \(t = 3\).
M1: For summing the absolute values of the separate parts of the path to obtain total distance.
A1: For the final correct distance of \(8\) m.

Let \(a\) be the acceleration of the system and \(T\) be the tension in the string.
For particle \(B\) (moving downwards):
\[ m_B g - T = m_B a \implies 0.6(10) - T = 0.6a \implies 6 - T = 0.6a \quad \text{(Equation 1)} \]

For particle \(A\) (moving upwards):
\[ T - m_A g = m_A a \implies T - 0.4(10) = 0.4a \implies T - 4 = 0.4a \quad \text{(Equation 2)} \]

Adding Equation 1 and Equation 2:
\[ 6 - 4 = 1.0a \implies a = 2 \text{ m s}^{-2} \]

Substitute \(a = 2\) into Equation 2 to find tension \(T\):
\[ T - 4 = 0.4(2) \implies T = 4.8 \text{ N} \]

Initially, both particles are at the same horizontal level. When they are \(1.2\text{ m}\) vertically apart, each particle has moved a distance \(s\) of:
\[ s = \frac{1.2}{2} = 0.6 \text{ m} \]

Using the equations of constant acceleration with \(u = 0\), \(a = 2\), and \(s = 0.6\):
\[ v^2 = u^2 + 2as \implies v^2 = 0 + 2(2)(0.6) = 2.4 \]
\[ v = \sqrt{2.4} \approx 1.55 \text{ m s}^{-1} \text{ (to 3 s.f.)} \]

M1: For applying Newton's second law to both particles to obtain two equations of motion.
A1: For both equations of motion being correct (e.g., \(6 - T = 0.6a\) and \(T - 4 = 0.4a\)).
M1: For solving the system of equations to find acceleration \(a\).
A1: For finding the correct acceleration \(a = 2 \text{ m s}^{-2}\) and the correct tension \(T = 4.8\text{ N}\).
B1: For recognizing that each particle travels a distance of \(s = 0.6\text{ m}\).
M1: For using \(v^2 = u^2 + 2as\) with their calculated acceleration and distance.
A1: For finding the correct speed \(v = 1.55\text{ m s}^{-1}\) (accept \(\sqrt{2.4}\) or \(1.55\)).

Let the initial direction of motion of \(A\) be positive.
Initial velocities:
\[ u_A = 4 \text{ m s}^{-1}, \quad u_B = -2 \text{ m s}^{-1} \]

Final velocity of \(A\):
\[ v_A = -1 \text{ m s}^{-1} \quad \text{(since its direction is reversed)} \]

By conservation of momentum:
\[ m_A u_A + m_B u_B = m_A v_A + m_B v_B \]
\[ 0.3(4) + 0.5(-2) = 0.3(-1) + 0.5 v_B \]
\[ 1.2 - 1.0 = -0.3 + 0.5 v_B \]
\[ 0.2 = -0.3 + 0.5 v_B \]
\[ 0.5 v_B = 0.5 \implies v_B = 1 \text{ m s}^{-1} \]

Since \(v_B = 1 \text{ m s}^{-1}\) is positive, \(B\) moves in the positive direction (which is the direction in which \(A\) was originally traveling). This means the direction of motion of \(B\) is also reversed.

Now, let us calculate the magnitude of the impulse \(I\) exerted by \(B\) on \(A\):
\[ I = |m_A v_A - m_A u_A| \]
\[ I = |0.3(-1) - 0.3(4)| = |-0.3 - 1.2| = |-1.5| = 1.5 \text{ N s} \]

M1: For using the principle of conservation of momentum with appropriate signs.
A1: For the correct formulation: \(0.3(4) + 0.5(-2) = 0.3(-1) + 0.5 v_B\).
A1: For calculating the correct final velocity of \(B\), \(v_B = 1 \text{ m s}^{-1}\).
B1: For stating that the direction of \(B\)'s motion is reversed (or moves in the direction of \(A\)'s initial motion).
M1: For using the formula for impulse \(I = |\Delta p|\) on particle \(A\).
A1: For calculating the correct magnitude of impulse as \(1.5 \text{ N s}\).
B0.1: For correct units overall.

Given \(\sin\alpha = 0.6\), we find \(\cos\alpha = \sqrt{1 - 0.6^2} = 0.8\).

The normal contact force \(R\) is perpendicular to the plane:
\[ R = mg\cos\alpha = 5(10)(0.8) = 40 \text{ N} \]

The maximum possible frictional force \(F_{\text{max}}\) is:
\[ F_{\text{max}} = \mu R = 0.4(40) = 16 \text{ N} \]

The component of the weight acting down the incline is:
\[ W_{\parallel} = mg\sin\alpha = 5(10)(0.6) = 30 \text{ N} \]

We analyze the two limiting cases of equilibrium:

Case 1: The block is on the point of slipping down the plane.
Here, the frictional force acts up the plane to oppose the motion.
\[ P + F_{\text{max}} = W_{\parallel} \implies P + 16 = 30 \implies P = 14 \text{ N} \]

Case 2: The block is on the point of slipping up the plane.
Here, the frictional force acts down the plane to oppose the motion.
\[ P = W_{\parallel} + F_{\text{max}} \implies P = 30 + 16 = 46 \text{ N} \]

Thus, the block remains in equilibrium for:
\[ 14 \le P \le 46 \]

B1: For identifying \(\cos\alpha = 0.8\).
M1: For resolving forces perpendicular to the plane to find the normal contact force \(R = 40 \text{ N}\).
A1: For calculating the limiting friction \(F_{\text{max}} = 16 \text{ N}\).
M1: For setting up the equation of equilibrium when the block is on the point of sliding down the plane.
A1: For finding the lower limit \(P = 14 \text{ N}\).
M1: For setting up the equation of equilibrium when the block is on the point of sliding up the plane.
A1: For finding the upper limit \(P = 46 \text{ N}\) and stating the range clearly.

(i) To find the acceleration of the sledge, we resolve forces horizontally:
\[ T\cos(20^\circ) - R_f = ma \]
where \(T = 80\text{ N}\), \(R_f = 50\text{ N}\), and \(m = 25\text{ kg}\).
\[ 80\cos(20^\circ) - 50 = 25a \]
\[ 80(0.93969) - 50 = 25a \]
\[ 75.175 - 50 = 25a \implies 25.175 = 25a \implies a \approx 1.01 \text{ m s}^{-2} \text{ (to 3 s.f.)} \]

(ii) To find the normal contact force \(N\), we resolve forces vertically:
\[ N + T\sin(20^\circ) - mg = 0 \]
\[ N + 80\sin(20^\circ) - 25(10) = 0 \]
\[ N + 80(0.34202) - 250 = 0 \]
\[ N + 27.36 - 250 = 0 \implies N = 222.64 \approx 223 \text{ N} \text{ (to 3 s.f.)} \]

M1: For resolving forces horizontally to form an equation of motion.
A1: For the correct equation: \(80\cos(20^\circ) - 50 = 25a\).
A1: For finding the correct acceleration \(a = 1.01 \text{ m s}^{-2}\).
M1: For resolving forces vertically to form an equation for the normal contact force.
A1: For the correct vertical equation: \(N + 80\sin(20^\circ) - 250 = 0\).
A1: For finding the correct normal contact force \(N = 223 \text{ N}\) (or \(222.6\)).
B0.1: For consistent accuracy across both parts.

Let us break down the motion into three stages:

**Stage 1: Constant acceleration**
- Initial velocity \(u_1 = 0\), final velocity \(v_1 = V\), time \(t_1 = 10\text{ s}\).
- Acceleration \(a = \frac{V}{10}\).
- Distance traveled \(s_1 = \frac{1}{2}(u_1 + v_1)t_1 = \frac{1}{2}(0 + V)(10) = 5V\).

**Stage 2: Constant speed**
- Constant speed is \(V\), duration is \(T\).
- Distance traveled \(s_2 = VT\).

**Stage 3: Constant deceleration**
- Deceleration is \(1\text{ m s}^{-2}\) (so acceleration is \(-1\text{ m s}^{-2}\)), starting speed is \(V\), ending speed is \(0\).
- Time taken \(t_3 = \frac{V - 0}{1} = V\).
- Distance traveled \(s_3 = \frac{1}{2}(V + 0)t_3 = 0.5V^2\).

**Equations for the whole journey:**
1. **Total time:**
\[ t_1 + T + t_3 = 38 \implies 10 + T + V = 38 \implies T = 28 - V \]

2. **Total distance:**
\[ s_1 + s_2 + s_3 = 232 \implies 5V + VT + 0.5V^2 = 232 \]

Substitute \(T = 28 - V\) into the distance equation:
\[ 5V + V(28 - V) + 0.5V^2 = 232 \]
\[ 5V + 28V - V^2 + 0.5V^2 = 232 \]
\[ 33V - 0.5V^2 = 232 \]

Multiply the entire equation by 2 to clear the fraction:
\[ 66V - V^2 = 464 \implies V^2 - 66V + 464 = 0 \]

Using the quadratic formula to solve for \(V\):
\[ V = \frac{-(-66) \pm \sqrt{(-66)^2 - 4(1)(464)}}{2(1)} \]
\[ V = \frac{66 \pm \sqrt{4356 - 1856}}{2} \]
\[ V = \frac{66 \pm \sqrt{2500}}{2} = \frac{66 \pm 50}{2} \]

This gives two possible values for \(V\):
- \(V = 58\)
- \(V = 8\)

If \(V = 58\), then \(T = 28 - 58 = -30\), which is impossible because time \(T\) must be positive.
Thus, we select:
\[ V = 8 \text{ m s}^{-1} \]

Now, we can find \(T\) and \(a\):
\[ T = 28 - 8 = 20 \text{ s} \]
\[ a = \frac{V}{10} = \frac{8}{10} = 0.8 \text{ m s}^{-2} \]

M1: For expressing distances \(s_1\) and \(s_3\) in terms of \(V\).
A1: For correct expressions: \(s_1 = 5V\), \(t_3 = V\), and \(s_3 = 0.5V^2\).
M1: For setting up the total time equation to get \(T\) in terms of \(V\) (\(T = 28 - V\)).
M1: For substituting \(T\) into the total distance equation to form a quadratic equation in terms of \(V\).
A1: For the correct quadratic equation \(V^2 - 66V + 464 = 0\).
A1: For solving the quadratic equation to find \(V = 8\) (rejecting \(V = 58\)).
A1: For determining both \(a = 0.8\text{ m s}^{-2}\) and \(T = 20\text{ s}\) correctly.
B0.1: For consistent accuracy across all parts.

(i) We use the constant acceleration equation with \(u = 8\text{ m s}^{-1}\), \(v = 0\text{ m s}^{-1}\), and \(s = 4\text{ m}\):
\[ v^2 = u^2 + 2as \]
\[ 0^2 = 8^2 + 2a(4) \]
\[ 0 = 64 + 8a \implies 8a = -64 \implies a = -8 \text{ m s}^{-2} \]
Thus, the deceleration of the particle is \(8\text{ m s}^{-2}\).

(ii) Let us set up the equation of motion for the particle moving up the inclined plane. The forces acting parallel to the plane in the downward direction are the component of gravity down the slope and the friction force \(F\):
\[ -mg\sin(30^\circ) - F = ma \]
Using \(m = 4\text{ kg}\), \(g = 10\text{ m s}^{-2}\), and \(a = -8\text{ m s}^{-2}\):
\[ -4(10)\sin(30^\circ) - F = 4(-8) \]
\[ -40(0.5) - F = -32 \]
\[ -20 - F = -32 \implies F = 12 \text{ N} \]

The normal reaction force \(R\) perpendicular to the slope is:
\[ R = mg\cos(30^\circ) = 4(10)\cos(30^\circ) = 40 \left(\frac{\sqrt{3}}{2}\right) = 20\sqrt{3} \text{ N} \]

Since \(F = \mu R\):
\[ 12 = \mu (20\sqrt{3}) \]
\[ \mu = \frac{12}{20\sqrt{3}} = \frac{3}{5\sqrt{3}} = \frac{\sqrt{3}}{5} \approx 0.346 \text{ (to 3 s.f.)} \]

M1: For using \(v^2 = u^2 + 2as\) with appropriate values.
A1: For finding the correct deceleration of \(8\text{ m s}^{-2}\).
M1: For setting up the equation of motion parallel to the plane, including weight and friction components.
A1: For the correct force equation: \(40\sin(30^\circ) + F = 32\).
A1: For determining the correct frictional force \(F = 12\text{ N}\).
M1: For calculating the normal reaction force \(R = 20\sqrt{3}\) and applying \(F = \mu R\).
A1: For finding the correct coefficient of friction \(\mu = 0.346\) (accept \(\frac{\sqrt{3}}{5}\)).
B0.1: For consistent accuracy across both parts.

(i) Let \(y = x - 5\). The mean of \(y\) is \(\bar{y} = 32 / 40 = 0.8\). The mean of \(x\) is \(\bar{x} = \bar{y} + 5 = 5.8\) hours. The variance of \(y\) is \(\frac{\sum y^2}{n} - (\bar{y})^2 = \frac{184}{40} - 0.8^2 = 4.6 - 0.64 = 3.96\). The standard deviation of \(x\) is the same as the standard deviation of \(y\), which is \(\sqrt{3.96} \approx 1.99\) hours. (ii) For the first group of 40 students, the sum of times is \(40 \times 5.8 = 232\) hours. For the second group of 10 students, the sum of times is \(10 \times 6.2 = 62\) hours. The combined mean is \(\frac{232 + 62}{50} = \frac{294}{50} = 5.88\) hours.

(i) M1 for finding the mean of \(y\) and adding 5. A1 for 5.8. M1 for using the variance formula. A1 for 1.99 (or \(\sqrt{3.96}\)). (ii) M1 for finding the sum of times for both groups. M1 for dividing the total sum by 50. A1 for 5.88.

(i) Treat the 3 men as a single block. There are now 6 items to arrange (5 women and 1 block), which can be done in \(6! = 720\) ways. Within the block, the 3 men can be arranged in \(3! = 6\) ways. Thus, the total number of arrangements is \(6! \times 3! = 720 \times 6 = 4320\). (ii) First, arrange the 5 women in a row, which can be done in \(5! = 120\) ways. This creates 6 available spaces (including the ends) for the 3 men: _ W _ W _ W _ W _ W _. The number of ways to arrange the 3 men in these 6 spaces is \(^{6}P_{3} = 6 \times 5 \times 4 = 120\) ways. The total number of arrangements is \(120 \times 120 = 14400\).

(i) M1 for treating the 3 men as a single block (showing \(6!\)). M1 for multiplying by \(3!\). A1 for 4320. (ii) M1 for arranging 5 women in \(5!\) ways. M1 for identifying 6 slots and choosing/arranging 3 men (i.e. \(^{6}P_{3}\) or \(^{6}C_{3} \times 3!\)). M1 for multiplying both parts. A1 for 14400.

(i) The probability of selecting Bag A is \(P(A) = 2/6 = 1/3\) and Bag B is \(P(B) = 4/6 = 2/3\). The probability of drawing a blue ball from Bag A is \(P(\text{Blue}|A) = 3/7\) and from Bag B is \(P(\text{Blue}|B) = 5/7\). Using the law of total probability, \(P(\text{Blue}) = P(A)P(\text{Blue}|A) + P(B)P(\text{Blue}|B) = \frac{1}{3} \times \frac{3}{7} + \frac{2}{3} \times \frac{5}{7} = \frac{3}{21} + \frac{10}{21} = \frac{13}{21} \approx 0.619\). (ii) We require the conditional probability \(P(A|\text{Blue})\). Using Bayes' Theorem: \(P(A|\text{Blue}) = \frac{P(A \cap \text{Blue})}{P(\text{Blue})} = \frac{\frac{1}{3} \times \frac{3}{7}}{\frac{13}{21}} = \frac{\frac{3}{21}}{\frac{13}{21}} = \frac{3}{13} \approx 0.231\).

(i) M1 for finding \(P(A) = 1/3\) and \(P(B) = 2/3\). M1 for adding \(P(A \cap \text{Blue})\) and \(P(B \cap \text{Blue})\). A1 for \(13/21\) or 0.619. (ii) M1 for setting up the conditional probability formula. M1 for numerator \(3/21\). A1 for denominator \(13/21\). A1 for \(3/13\) or 0.231.

(i) Since the sum of probabilities is 1, we have \(a + 0.3 + b + 0.2 = 1 \implies a + b = 0.5 \implies b = 0.5 - a\). The expectation is given by \(E(X) = -2a + 0(0.3) + 1b + 3(0.2) = 0.5 \implies -2a + b + 0.6 = 0.5\). Substituting \(b = 0.5 - a\) yields \(-2a + 0.5 - a + 0.6 = 0.5 \implies -3a + 1.1 = 0.5 \implies -3a = -0.6 \implies a = 0.2\). Then \(b = 0.5 - 0.2 = 0.3\). (ii) First, find \(E(X^2) = (-2)^2(0.2) + 0^2(0.3) + 1^2(0.3) + 3^2(0.2) = 4(0.2) + 0 + 1(0.3) + 9(0.2) = 0.8 + 0.3 + 1.8 = 2.9\). The variance is \(Var(X) = E(X^2) - [E(X)]^2 = 2.9 - 0.5^2 = 2.9 - 0.25 = 2.65\).

(i) M1 for setting sum of probabilities to 1. M1 for setting up expectation equation. M1 for solving simultaneous equations. A1 for \(a = 0.2\) and \(b = 0.3\). (ii) M1 for finding \(E(X^2)\). M1 for using the formula \(Var(X) = E(X^2) - [E(X)]^2\). A1 for 2.65.

(i) From the given probabilities: \(P(X < 495) = 0.10 \implies P(Z < \frac{495 - \mu}{\sigma}) = 0.10 \implies \frac{495 - \mu}{\sigma} = -1.282 \implies \mu - 1.282\sigma = 495\). Also, \(P(X > 515) = 0.05 \implies P(Z > \frac{515 - \mu}{\sigma}) = 0.05 \implies \frac{515 - \mu}{\sigma} = 1.645 \implies \mu + 1.645\sigma = 515\). Subtracting the first equation from the second equation: \(2.927\sigma = 20 \implies \sigma = 6.833 \approx 6.8\) (1 d.p.). Substituting back: \(\mu = 495 + 1.282(6.833) = 503.76 \approx 503.8\) (1 d.p.). (ii) We require \(P(500 < X < 510)\). Standardising gives: \(P(\frac{500 - 503.76}{6.833} < Z < \frac{510 - 503.76}{6.833}) = P(-0.550 < Z < 0.913) = \Phi(0.913) - \Phi(-0.550) = \Phi(0.913) - (1 - \Phi(0.550)) = 0.8194 - (1 - 0.7088) = 0.5282 \approx 0.528\).

(i) M1 for standardising with 495 and setting to \(-1.282\). M1 for standardising with 515 and setting to \(1.645\). M1 for solving the simultaneous equations. A1 for showing \(\sigma = 6.8\) with clear steps. A1 for showing \(\mu = 503.8\) with clear steps. (ii) M1 for standardising both 500 and 510. A1 for 0.528 (accept 0.527 to 0.529).

(i) The class width of \(20 < t \le 40\) is 20 units, represented by a bar width of 2 cm. Hence, 10 units of class width is represented by 1 cm. For the class \(40 < t \le 60\), the class width is 20, so its bar width is also 2 cm. The frequency density for \(20 < t \le 40\) is \(\frac{40}{20} = 2.0\), represented by a height of 4 cm. Thus, 1 unit of frequency density is represented by 2 cm of height. For the class \(40 < t \le 60\), the frequency density is \(\frac{30}{20} = 1.5\), so the height of the bar is \(1.5 \times 2 = 3\) cm. (ii) Midpoints are 5, 15, 30, and 50 respectively. The sum of \(f \times \text{midpoint} = 15(5) + 35(15) + 40(30) + 30(50) = 75 + 525 + 1200 + 1500 = 3300\). The estimated mean is \(\frac{3300}{120} = 27.5\) minutes.

(i) M1 for finding class width scaling factor (10 units = 1 cm) or width = 2 cm. M1 for finding frequency densities (2.0 and 1.5). M1 for finding height scaling factor (1 unit of FD = 2 cm). A1 for correctly stating width is 2 cm and height is 3 cm. (ii) M1 for identifying correct midpoints. M1 for calculating \(\sum f \times \text{midpoint}\) and dividing by 120. A1 for 27.5.

Let \(X\) be the number of faulty components. Then \(X \sim B(15, 0.08)\). (i) The probability of exactly 2 faulty components is \(P(X = 2) = ^{15}C_{2} (0.08)^2 (0.92)^{13} = 105 \times 0.0064 \times 0.338239 \approx 0.227\). (ii) The probability of more than 1 but fewer than 4 faulty components is \(P(1 < X < 4) = P(X = 2) + P(X = 3)\). We calculate \(P(X = 3) = ^{15}C_{3} (0.08)^3 (0.92)^{12} = 455 \times 0.000512 \times 0.367651 \approx 0.0856\). Thus, \(P(1 < X < 4) = 0.2273 + 0.0856 = 0.3129 \approx 0.313\).

(i) M1 for identifying binomial distribution with \(n = 15, p = 0.08\). M1 for using binomial formula for \(P(X=2)\). A1 for 0.227. (ii) M1 for recognizing \(P(1 < X < 4) = P(X = 2) + P(X = 3)\). M1 for calculating \(P(X=3)\). M1 for adding both probabilities. A1 for 0.313.

First, we state the hypotheses:
\(H_0: \mu = 1500\)
\(H_1: \mu < 1500\)

Since the sample size \(n = 80\) is large, we can use the Central Limit Theorem. Under \(H_0\), the sample mean \(\bar{X}\) is approximately normally distributed:
\(\bar{X} \sim N\left(1500, \frac{110^2}{80}\right)\)

We calculate the standardized test statistic \(z\):
\(z = \frac{1475 - 1500}{110 / \sqrt{80}} = \frac{-25}{12.30} \approx -2.033\)

For a one-tailed test at the 5% significance level, the critical value is \(-1.645\).

Since the test statistic \(-2.033 < -1.645\), it falls in the critical region. Therefore, we reject \(H_0\). There is significant evidence at the 5% level to suggest that the mean lifetime of the components is less than 1500 hours.

M1: For stating correct null and alternative hypotheses.
M1: For calculating the standard error of the mean \(110 / \sqrt{80} \approx 12.30\).
M1: For standardizing to find the \(z\)-value of \(-2.033\).
A1: For comparing with the critical value \(-1.645\) (or comparing area \(0.0210\) with \(0.05\)).
A1.25: For correct conclusion in context, rejecting \(H_0\).

(a) Let \(X\) be the number of typos per page, so \(X \sim \text{Po}(0.6)\).
\(P(X \ge 2) = 1 - P(X = 0) - P(X = 1) = 1 - e^{-0.6} - 0.6 e^{-0.6} = 1 - 1.6 e^{-0.6} \approx 0.1219\) (or \(0.122\) to 3 s.f.).

(b) Let \(Y\) be the total number of typos in 10 pages. Since the pages are independent, \(Y \sim \text{Po}(10 \times 0.6) = \text{Po}(6)\).
\(P(Y = 4) = \frac{e^{-6} \times 6^4}{4!} = \frac{1296 e^{-6}}{24} = 54 e^{-6} \approx 0.1339\) (or \(0.134\) to 3 s.f.).

M1: For expressing the probability in (a) as \(1 - P(X=0) - P(X=1)\).
A1: For obtaining the correct answer \(0.122\) in (a).
M1: For stating the new parameter \(\lambda = 6\) for the 10-page chapter in (b).
M1: For using the Poisson probability formula for \(P(Y=4)\).
A2.25: For obtaining the correct answer \(0.134\) in (b).

(a) Since \(f(x)\) is a probability density function, the total area under the curve must equal 1:
\(\int_0^2 k(4x - x^3) \, dx = 1\)
\(k \left[ 2x^2 - \frac{x^4}{4} \right]_0^2 = 1\)
\(k \left( 2(2)^2 - \frac{2^4}{4} \right) = 1\)
\(k (8 - 4) = 1 \implies 4k = 1 \implies k = \frac{1}{4}\). (Shown)

(b) The expectation \(E(X)\) is given by:
\(E(X) = \int_0^2 x f(x) \, dx = \frac{1}{4} \int_0^2 (4x^2 - x^4) \, dx\)
\(E(X) = \frac{1}{4} \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_0^2 = \frac{1}{4} \left( \frac{32}{3} - \frac{32}{5} \right)\)
\(E(X) = \frac{1}{4} \left( \frac{160 - 96}{15} \right) = \frac{1}{4} \left( \frac{64}{15} \right) = \frac{16}{15} \approx 1.07\).

M1: For integrating \(4x - x^3\) from 0 to 2 and setting equal to 1.
A1: For showing \(k = 1/4\) correctly with clear steps.
M1: For writing down the integral expression for \(E(X)\), multiplying by \(x\).
M1: For integrating to get \(\frac{4x^3}{3} - \frac{x^5}{5}\).
A2.25: For obtaining \(\frac{16}{15}\) or \(1.07\).

Let \(X\) be the number of users who experience drowsiness. Under the null hypothesis, \(X \sim B(20, 0.1)\).

(a) We are testing:
\(H_0: p = 0.1\)
\(H_1: p > 0.1\)

We look for a critical region of the form \(X \ge c\) such that \(P(X \ge c) \le 0.05\).
Let's calculate cumulative probabilities using the Binomial formula \(P(X = r) = \binom{20}{r} 0.1^r 0.9^{20-r}\):
\(P(X = 0) = 0.9^{20} \approx 0.1216\)
\(P(X = 1) = 20 \times 0.1 \times 0.9^{19} \approx 0.2702\)
\(P(X = 2) = 190 \times 0.01 \times 0.9^{18} \approx 0.2852\)
\(P(X = 3) = 1140 \times 0.001 \times 0.9^{17} \approx 0.1901\)
\(P(X = 4) = 4845 \times 0.0001 \times 0.9^{16} \approx 0.0898\)

Now we find:
\(P(X \le 3) = 0.1216 + 0.2702 + 0.2852 + 0.1901 = 0.8671 \implies P(X \ge 4) = 1 - 0.8671 = 0.1329 > 0.05\)
\(P(X \le 4) = 0.8671 + 0.0898 = 0.9569 \implies P(X \ge 5) = 1 - 0.9569 = 0.0431 \le 0.05\)

Thus, the critical region is \(X \ge 5\).

(b) A Type I error is rejecting \(H_0\) when it is true. This occurs if \(X \ge 5\) when \(p = 0.1\).
Therefore, the probability of a Type I error is \(P(X \ge 5 \mid p = 0.1) = 0.0431\).

M1: For identifying the binomial distribution \(B(20, 0.1)\) under \(H_0\).
M1: For calculating probabilities near the tail (such as \(P(X \ge 4)\) and \(P(X \ge 5)\)).
A1: For identifying the correct critical value \(c = 5\) and stating the critical region \(X \ge 5\).
M1: For defining Type I error probability as the size of the critical region.
A2.25: For obtaining \(0.0431\).

(a) Let \(T = A + B\) be the total weekly sales. Since \(A \sim \text{Po}(4.2)\) and \(B \sim \text{Po}(3.5)\) are independent, we can add their means:
\(T \sim \text{Po}(4.2 + 3.5) = \text{Po}(7.7)\).
We want to find \(P(T > 10) = 1 - P(T \le 10)\).
\(P(T \le 10) = e^{-7.7} \sum_{k=0}^{10} \frac{7.7^k}{k!} \approx 0.8423\).
Therefore, \(P(T > 10) = 1 - 0.8423 = 0.1577 \approx 0.158\) (to 3 s.f.).

(b) Let \(A_2\) be the sales of brand A over a 2-week period. \(A_2 \sim \text{Po}(2 \times 4.2) = \text{Po}(8.4)\).
We want to find \(P(A_2 = 8)\):
\(P(A_2 = 8) = \frac{e^{-8.4} \times 8.4^8}{8!} \approx 0.1382 \approx 0.138\) (to 3 s.f.).

M1: For adding means to get \(\lambda_T = 7.7\) for total weekly sales.
M1: For writing the correct expression for \(P(T > 10)\) as \(1 - P(T \le 10)\).
A1: For obtaining the correct probability in (a) \(0.158\).
M1: For multiplying the weekly mean by 2 to get \(\lambda_{A_2} = 8.4\) for the 2-week period.
M1: For using the Poisson probability formula for \(P(A_2 = 8)\).
A1.25: For obtaining the correct probability in (b) \(0.138\).

(a) To find the probability density function \(f(x)\), we differentiate the cumulative distribution function \(F(x)\) with respect to \(x\):
For \(1 \le x \le 4\):
\(f(x) = \frac{d}{dx} \left[ \frac{1}{15}(x^2 - 1) \right] = \frac{2x}{15}\).
For other values of \(x\), the derivative of the constant functions 0 and 1 is 0.
Thus,
\[f(x) = \begin{cases} \frac{2x}{15} & 1 \le x \le 4 \\ 0 & \text{otherwise} \end{cases}\]

(b) The median \(m\) is the value of \(x\) for which \(F(m) = 0.5\):
\(\frac{1}{15}(m^2 - 1) = 0.5\)
\(m^2 - 1 = 7.5\)
\(m^2 = 8.5\)
\(m = \sqrt{8.5} \approx 2.915 \approx 2.92\) (to 3 s.f.).

M1: For differentiating \(F(x)\) to find \(f(x)\).
A1: For stating the correct expression \(2x/15\).
A1: For including the range \(1 \le x \le 4\) and stating 0 elsewhere.
M1: For setting \(F(m) = 0.5\) (or integrating \(f(x)\) from 1 to \(m\) and setting equal to 0.5).
A2.25: For finding the correct median \(\sqrt{8.5}\) or \(2.92\).

Let \(p\) be the population proportion of non-fiction orders.
State the hypotheses:
\(H_0: p = 0.35\)
\(H_1: p > 0.35\) (one-tailed test)

Under \(H_0\), the number of non-fiction orders in a sample of 150 is \(X \sim B(150, 0.35)\).
Since \(np = 150 \times 0.35 = 52.5 > 5\) and \(n(1-p) = 150 \times 0.65 = 97.5 > 5\), we can approximate the binomial distribution with a normal distribution:
\(X \sim N(\mu, \sigma^2)\) where:
\(\mu = np = 52.5\)
\(\sigma^2 = np(1-p) = 150 \times 0.35 \times 0.65 = 34.125\)

We observe \(X = 61\). Applying a continuity correction to find \(P(X \ge 61)\), we calculate \(P(X > 60.5)\):
\(z = \frac{60.5 - 52.5}{\sqrt{34.125}} = \frac{8}{5.8417} \approx 1.369\)

At the 5% significance level, the critical value for a one-tailed test is \(1.645\).

Since the test statistic \(1.369 < 1.645\), we do not reject \(H_0\). There is insufficient evidence at the 5% level to suggest that the proportion of non-fiction book orders has increased.

M1: For stating correct hypotheses \(H_0: p = 0.35\) and \(H_1: p > 0.35\).
M1: For calculating the correct mean \(52.5\) and variance \(34.125\) of the normal approximation.
M1: For standardizing with continuity correction (using \(60.5\)).
A1: For obtaining \(z = 1.37\) (accept \(1.369\) or \(1.38\) if no continuity correction used, but must lose accuracy mark if no continuity correction).
A2.25: For comparing with \(1.645\) and making a correct, contextual conclusion not rejecting \(H_0\).

Let \(X\) be the number of breakdowns in a fortnight. Under the null hypothesis, since the weekly rate is 2.8, the rate for a fortnight (2 weeks) is \(\lambda = 2 \times 2.8 = 5.6\).

State the hypotheses:
\(H_0: \lambda = 5.6\)
\(H_1: \lambda < 5.6\) (one-tailed test for a decrease)

We observe \(X = 6\). We calculate the probability of obtaining 6 or fewer breakdowns under \(H_0\):
\(P(X \le 6) = e^{-5.6} \sum_{k=0}^6 \frac{5.6^k}{k!}\)
\(P(X \le 6) = e^{-5.6} \left( 1 + 5.6 + \frac{5.6^2}{2!} + \frac{5.6^3}{3!} + \frac{5.6^4}{4!} + \frac{5.6^5}{5!} + \frac{5.6^6}{6!} \right)\)
\(P(X \le 6) = e^{-5.6} (1 + 5.6 + 15.68 + 29.2693 + 40.9770 + 45.8943 + 42.8347)\)
\(P(X \le 6) = e^{-5.6} (181.2553) \approx 0.003698 \times 181.2553 \approx 0.6703\)

Since the p-value \(0.6703 > 0.10\), the result is not significant.

Therefore, we do not reject \(H_0\). There is insufficient evidence at the 10% significance level to suggest that the rate of computer breakdowns has decreased.

M1: For stating the hypotheses with correct parameter \(\lambda = 5.6\) for a 2-week period.
M1: For stating that \(P(X \le 6)\) is needed to find the p-value.
M1: For setting up the sum of Poisson probabilities from 0 to 6.
A1: For calculating the sum correctly to obtain \(0.670\) (or \(0.6703\)).
A2.25: For comparing with \(0.10\) and drawing a correct conclusion in context.