2023 Cambridge IAL Mathematics (9709) Practice Paper with Answers

To find the points of intersection, we set the equations equal to each other: \(kx - 5 = x^2 - 3x - 1\). Rearranging into a standard quadratic equation gives \(x^2 - (3+k)x + 4 = 0\). For the line and the curve not to intersect, this quadratic equation must have no real roots. Therefore, the discriminant must be negative: \(b^2 - 4ac < 0\). Substituting the coefficients: \((-(3+k))^2 - 4(1)(4) < 0\), which simplifies to \((k+3)^2 - 16 < 0\). This can be written as \((k+3)^2 < 16\), leading to \(-4 < k + 3 < 4\). Subtracting 3 from all parts yields \(-7 < k < 1\).

M1: For setting up the quadratic equation and attempting to use the discriminant \(b^2 - 4ac < 0\). A1: For obtaining the correct inequality \((k+3)^2 < 16\) or equivalent. A0.25: For the final correct range \(-7 < k < 1\).

To find when the function has an inverse, we must determine the range of \(x\) for which \(\mathrm{f}(x)\) is one-to-one. We can express the quadratic expression in completed square form: \(\mathrm{f}(x) = 2(x^2 - 6x) + 13 = 2((x-3)^2 - 9) + 13 = 2(x-3)^2 - 5\). This shows that the vertex of the parabola is at \(x = 3\). Since the domain is defined as \(x \le a\), the function will be one-to-one as long as \(a\) does not exceed the \(x\)-coordinate of the vertex. Thus, the largest value of \(a\) is \(3\).

M1: For attempting to complete the square or using differentiation to find the x-coordinate of the vertex. A1: For obtaining \(x = 3\) as the vertex coordinate. A0.25: For concluding that the maximum value of \(a\) is \(3\).

The formula for the gradient is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Substituting the given values: \(\frac{q - (-3)}{8 - 2} = \frac{4}{3}\), which simplifies to \(\frac{q + 3}{6} = \frac{4}{3}\). Multiplying both sides by 6 gives \(q + 3 = 8\), so \(q = 5\). Thus, the coordinates of \(B\) are \((8, 5)\). The midpoint of \(AB\) is given by \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{2 + 8}{2}, \frac{-3 + 5}{2}\right) = (5, 1)\).

M1: For using the gradient formula to set up an equation and solve for \(q\). A1: For finding \(q = 5\) and calculating the midpoint coordinates. A0.25: For expressing the final midpoint correctly as \((5, 1)\).

The perimeter of the sector is given by \(P = 2r + r\theta = 32\). The area of the sector is given by \(A = \frac{1}{2}r^2\theta = 48\), which implies \(\theta = \frac{96}{r^2}\). Substituting this expression for \(\theta\) into the perimeter equation gives \(2r + r\left(\frac{96}{r^2}\right) = 32\), which simplifies to \(2r + \frac{96}{r} = 32\). Multiplying the entire equation by \(r\) yields \(2r^2 - 32r + 96 = 0\). Dividing by 2, we get \(r^2 - 16r + 48 = 0\). Factoring the quadratic equation gives \((r - 4)(r - 12) = 0\), which results in \(r = 4\) or \(r = 12\).

M1: For setting up the system of equations for perimeter and area and substituting to form a quadratic in \(r\). A1: For obtaining a correct quadratic equation in \(r\), e.g., \(r^2 - 16r + 48 = 0\). A0.25: For obtaining both correct values of \(r\), namely \(4\) and \(12\).

Using the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\), we substitute into the equation to get \(3(1 - \cos^2 \theta) + 5 \cos \theta - 5 = 0\). This simplifies to \(3\cos^2 \theta - 5\cos \theta + 2 = 0\). Factoring this quadratic in \(\cos \theta\) gives \((3\cos \theta - 2)(\cos \theta - 1) = 0\). This yields \(\cos \theta = \frac{2}{3}\) or \(\cos \theta = 1\). Within the interval \(0^\circ \le \theta \le 180^\circ\): if \(\cos \theta = 1\), then \(\theta = 0^\circ\); if \(\cos \theta = \frac{2}{3}\), then \(\theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ\).

M1: For using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to obtain a quadratic in \(\cos \theta\) and attempting to factorise it. A1: For obtaining the correct values \(\cos \theta = 1\) and \(\cos \theta = \frac{2}{3}\). A0.25: For providing both correct angles, \(0^\circ\) and \(48.2^\circ\) (rounded to 1 decimal place).

Since the terms are in a geometric progression, the common ratio is constant: \(\frac{k}{k+2} = \frac{k-1.5}{k}\). Cross-multiplying gives \(k^2 = (k+2)(k-1.5)\), which expands to \(k^2 = k^2 + 0.5k - 3\). This simplifies to \(0.5k = 3\), so \(k = 6\). Substituting \(k = 6\) back into the terms gives: the first term \(a = 8\), and the common ratio \(r = \frac{6}{8} = 0.75\). Since \(|r| < 1\), the sum to infinity is given by \(S_\infty = \frac{a}{1-r} = \frac{8}{1-0.75} = 32\).

M1: For setting up the common ratio equation and solving for \(k\). A1: For finding \(k = 6\), and identifying \(a = 8\) and \(r = 0.75\). A0.25: For calculating the correct sum to infinity, \(32\).

First, we find the derivative of \(y\) with respect to \(x\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = 4 - \frac{9}{x^2}\). To find the stationary points, we set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\), which gives \(4 - \frac{9}{x^2} = 0\). Solving for \(x\) yields \(x^2 = \frac{9}{4}\), giving \(x = 1.5\) or \(x = -1.5\). Substituting these values back into the original curve equation: for \(x = 1.5\), \(y = 4(1.5) + \frac{9}{1.5} = 12\); for \(x = -1.5\), \(y = 4(-1.5) + \frac{9}{-1.5} = -12\). Thus, the stationary points are \((1.5, 12)\) and \((-1.5, -12)\).

M1: For differentiating \(y\) correctly to get \(4 - \frac{9}{x^2}\) and setting it equal to 0. A1: For finding both \(x\)-coordinates, \(x = \pm 1.5\). A0.25: For finding both correct \(y\)-coordinates and writing the coordinates of the two points as \((1.5, 12)\) and \((-1.5, -12)\).

We can rewrite the integrand as \(3x^{1/2} + 2x^{-2}\). Integrating term-by-term gives: \(\int \left(3x^{1/2} + 2x^{-2}\right) \mathrm{d}x = \left[ 2x^{3/2} - \frac{2}{x} \right]_{1}^{4}\). Now, substituting the upper limit \(x = 4\): \(2(4^{3/2}) - \frac{2}{4} = 16 - 0.5 = 15.5\). Substituting the lower limit \(x = 1\): \(2(1^{3/2}) - \frac{2}{1} = 2 - 2 = 0\). The value of the definite integral is therefore \(15.5 - 0 = 15.5\).

M1: For integrating at least one term correctly. A1: For obtaining the correct integrated expression \(2x^{3/2} - \frac{2}{x}\). A0.25: For substituting limits correctly and obtaining the final answer of \(15.5\).

For the arithmetic progression:
First term \(a = 12\).
Using the sum formula \(S_k = \frac{k}{2}[2a + (k-1)d]\):
\(S_{10} = \frac{10}{2}[2(12) + 9d] = 255\)
\(5[24 + 9d] = 255\)
\(24 + 9d = 51\)
\(9d = 27 \implies d = 3\).

Now, we find the terms of the arithmetic progression that form the geometric progression:
First term: \(u_1 = 12\)
Third term: \(u_3 = 12 + 2(3) = 18\)
\(n\)th term: \(u_n = 12 + (n-1)3 = 9 + 3n\).

Since \(u_1\), \(u_3\), and \(u_n\) are consecutive terms of a geometric progression, the common ratio \(r\) is:
\(r = \frac{u_3}{u_1} = \frac{18}{12} = 1.5\).

Therefore, the third term of the geometric progression is:
\(u_n = u_3 \times r = 18 \times 1.5 = 27\).

Equating this to the expression for \(u_n\):
\(9 + 3n = 27\)
\(3n = 18 \implies n = 6\).

M1: Attempts to use the sum formula for an AP to find \(d\).
A1: Obtains \(d = 3\).
M1: Sets up the geometric progression terms and calculates the common ratio \(r\) to find \(u_n\).
A1: Solves for \(n\) to obtain \(6\).

Given: \(3 \sin^2 \theta + 5 \sin \theta \cos \theta = 2 \cos^2 \theta\).

Since \(\cos \theta = 0\) is not a solution to the equation (as it would imply \(\sin \theta = 0\), which is impossible since \(\sin^2\theta + \cos^2\theta = 1\)), we can divide the entire equation by \(\cos^2 \theta\):
\(3 \tan^2 \theta + 5 \tan \theta = 2\)
\(3 \tan^2 \theta + 5 \tan \theta - 2 = 0\).

We can factorise this quadratic in terms of \(\tan \theta\):
\((3 \tan \theta - 1)(\tan \theta + 2) = 0\).

This gives two cases:
1) \(3 \tan \theta - 1 = 0 \implies \tan \theta = \frac{1}{3}\).
Since \(\tan \theta > 0\) and \(0^\circ \le \theta \le 180^\circ\):
\(\theta = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.4^\circ\) (to 1 d.p.).

2) \(\tan \theta + 2 = 0 \implies \tan \theta = -2\).
Since \(\tan \theta < 0\) and \(0^\circ \le \theta \le 180^\circ\):
\(\theta = 180^\circ - \tan^{-1}(2) \approx 180^\circ - 63.43^\circ = 116.6^\circ\) (to 1 d.p.).

Thus, the solutions are \(\theta = 18.4^\circ\) and \(\theta = 116.6^\circ\).

M1: Divides by \(\cos^2 \theta\) to form a quadratic equation in \(\tan \theta\).
A1: Obtains correct factorisation or values: \(\tan \theta = \frac{1}{3}\) and \(\tan \theta = -2\).
B1: Finds one correct angle (e.g., \(18.4^\circ\) or \(116.6^\circ\)).
A1: Finds both correct angles to 1 decimal place with no extra solutions in range.

To find the stationary points, we first find the first derivative \(\frac{dy}{dx}\) using the chain rule:
\(\frac{dy}{dx} = 3(2x - 3)^2 \times \frac{d}{dx}(2x - 3) - 24\)
\(\frac{dy}{dx} = 6(2x - 3)^2 - 24\).

At a stationary point, \(\frac{dy}{dx} = 0\):
\(6(2x - 3)^2 - 24 = 0\)
\((2x - 3)^2 = 4\)
\(2x - 3 = \pm 2\).

Case 1:
\(2x - 3 = 2 \implies 2x = 5 \implies x = 2.5\).
Substitute \(x = 2.5\) back into the equation of the curve to find the corresponding \(y\)-coordinate:
\(y = (2(2.5) - 3)^3 - 24(2.5) = 2^3 - 60 = 8 - 60 = -52\).
So the first stationary point is \((2.5, -52)\).

Case 2:
\(2x - 3 = -2 \implies 2x = 1 \implies x = 0.5\).
Substitute \(x = 0.5\) back into the equation of the curve:
\(y = (2(0.5) - 3)^3 - 24(0.5) = (-2)^3 - 12 = -8 - 12 = -20\).
So the second stationary point is \((0.5, -20)\).

M1: Evaluates the derivative using the chain rule (or by expansion first).
A1: Obtains the correct derivative \(6(2x-3)^2 - 24\).
M1: Sets derivative to 0 and solves for \(x\).
A1: Obtains both points: \((2.5, -52)\) and \((0.5, -20)\).

To find the equation of the curve, we integrate the derivative with respect to \(x\):
\(y = \int \frac{6}{\sqrt{2x+5}} \, dx = \int 6(2x+5)^{-\frac{1}{2}} \, dx\).

Using the reverse chain rule:
\(y = 6 \times \frac{(2x+5)^{\frac{1}{2}}}{\frac{1}{2} \times 2} + C\)
\(y = 6\sqrt{2x+5} + C\).

We are given that the curve passes through the point \((2, 15)\). Substitute these coordinates into the integrated equation to solve for the constant of integration \(C\):
\(15 = 6\sqrt{2(2)+5} + C\)
\(15 = 6\sqrt{9} + C\)
\(15 = 18 + C \implies C = -3\).

Therefore, the equation of the curve is:
\(y = 6\sqrt{2x+5} - 3\).

M1: Integrates \((2x+5)^{-\frac{1}{2}}\) to obtain the form \(k(2x+5)^{\frac{1}{2}}\).
A1: Obtains the correct integrated expression \(6\sqrt{2x+5} + C\).
M1: Substitutes \((2, 15)\) to solve for \(C\).
A1: Correctly identifies \(C = -3\) and states the full equation.

First, find the length of the arc \(AB\):
\(\text{Arc } AB = r\theta = 12 \times \frac{\pi}{3} = 4\pi\text{ cm}\).

Since \(BC\) is perpendicular to \(OA\), triangle \(OCB\) is right-angled at \(C\).
The hypotenuse of the triangle is the radius \(OB = 12\text{ cm}\).
The angle at the centre is \(\angle COB = \frac{\pi}{3}\).

Now, calculate \(OC\) and \(BC\) using trigonometry:
\(OC = OB \cos\left(\frac{\pi}{3}\right) = 12 \times \frac{1}{2} = 6\text{ cm}\),
\(BC = OB \sin\left(\frac{\pi}{3}\right) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}\text{ cm}\).

Next, calculate the length of the segment \(AC\):
\(AC = OA - OC = 12 - 6 = 6\text{ cm}\).

The perimeter of the shaded region is the sum of arc \(AB\), segment \(AC\), and segment \(BC\):
\(\text{Perimeter} = 4\pi + 6 + 6\sqrt{3}\text{ cm}\).

This is in the requested form with \(a = 4\), \(b = 6\), and \(c = 6\).

M1: Calculates the arc length \(AB = 4\pi\).
M1: Uses trigonometry on the right-angled triangle to find \(OC = 6\) and \(BC = 6\sqrt{3}\).
M1: Finds \(AC = 6\) and adds the three components to find the total perimeter.
A1: Obtains the correct exact perimeter \(4\pi + 6 + 6\sqrt{3}\).

For the curve to lie entirely above the \(x\)-axis, the quadratic expression \(2x^2 + kx + k + 6\) must always be positive.
Since the coefficient of \(x^2\) is \(2 > 0\) (the parabola opens upwards), this condition is met if and only if the equation \(2x^2 + kx + k + 6 = 0\) has no real roots.
Thus, the discriminant \(\Delta\) of the quadratic must be strictly negative:
\(\Delta = b^2 - 4ac < 0\),
where \(a = 2\), \(b = k\), and \(c = k + 6\).

Substitute these values into the discriminant:
\(k^2 - 4(2)(k + 6) < 0\)
\(k^2 - 8k - 48 < 0\).

To solve this inequality, find the roots of the corresponding quadratic equation:
\(k^2 - 8k - 48 = 0\)
\((k - 12)(k + 4) = 0\).

This yields critical values of \(k = 12\) and \(k = -4\).

For \(k^2 - 8k - 48 < 0\), we must select the region between the roots:
\(-4 < k < 12\).

M1: Understands and applies the condition that the discriminant must be less than 0.
A1: Formulates the correct discriminant inequality \(k^2 - 8k - 48 < 0\).
M1: Factorises or uses formula to find critical values \(-4\) and \(12\).
A1: States the correct final range \(-4 < k < 12\).

First, find an expression for the composite function \(\mathrm{fg}(x)\):
\(\mathrm{fg}(x) = \mathrm{f}\left(\frac{x+2}{x-1}\right)\)
\(\mathrm{fg}(x) = 2\left(\frac{x+2}{x-1}\right) - 3\)
\(\mathrm{fg}(x) = \frac{2(x+2) - 3(x-1)}{x-1}\)
\(\mathrm{fg}(x) = \frac{2x + 4 - 3x + 3}{x-1} = \frac{7-x}{x-1}\).

Next, find an expression for the inverse function \(\mathrm{g}^{-1}(x)\):
Let \(y = \mathrm{g}(x) = \frac{x+2}{x-1}\).
Rearrange to make \(x\) the subject:
\(y(x-1) = x+2\)
\(xy - y = x + 2\)
\(xy - x = y + 2\)
\(x(y-1) = y + 2\)
\(x = \frac{y+2}{y-1}\).
So, \(\mathrm{g}^{-1}(x) = \frac{x+2}{x-1}\).

Now, equate the two expressions:
\(\mathrm{fg}(x) = \mathrm{g}^{-1}(x)\)
\(\frac{7-x}{x-1} = \frac{x+2}{x-1}\).

Since \(x \ne 1\), we can multiply both sides by \(x-1\):
\(7 - x = x + 2\)
\(2x = 5\)
\(x = 2.5\).

M1: Obtains a simplified algebraic expression for \(\mathrm{fg}(x)\).
M1: Makes \(x\) the subject of \(y = \mathrm{g}(x)\) to find \(\mathrm{g}^{-1}(x)\).
M1: Equates their \(\mathrm{fg}(x)\) and \(\mathrm{g}^{-1}(x)\) and attempts to solve.
A1: Obtains the correct value \(x = 2.5\).

The equation of a circle with centre \((3, -2)\) and radius \(5\) is given by:
\((x - 3)^2 + (y + 2)^2 = 25\).

Substitute the equation of the line \(y = 2x + c\) into the equation of the circle:
\((x - 3)^2 + (2x + c + 2)^2 = 25\).

Expand the terms:
\((x^2 - 6x + 9) + \left[4x^2 + 4(c+2)x + (c+2)^2\right] = 25\)
\(5x^2 + (-6 + 4c + 8)x + 9 + (c^2 + 4c + 4) - 25 = 0\)
\(5x^2 + (4c + 2)x + (c^2 + 4c - 12) = 0\).

Since the line is a tangent, this quadratic equation has exactly one real root, meaning its discriminant \(\Delta\) must be zero:
\(\Delta = b^2 - 4ac = 0\),
where \(a = 5\), \(b = 4c + 2\), and \(c = c^2 + 4c - 12\).

Substitute these coefficients:
\((4c + 2)^2 - 4(5)(c^2 + 4c - 12) = 0\)
\(16c^2 + 16c + 4 - 20(c^2 + 4c - 12) = 0\)
\(16c^2 + 16c + 4 - 20c^2 - 80c + 240 = 0\)
\(-4c^2 - 64c + 244 = 0\).

Divide the entire equation by \(-4\):
\(c^2 + 16c - 61 = 0\).

Solve using the quadratic formula:
\(c = \frac{-16 \pm \sqrt{16^2 - 4(1)(-61)}}{2}\)
\(c = \frac{-16 \pm \sqrt{256 + 244}}{2}\)
\(c = \frac{-16 \pm \sqrt{500}}{2}\)
\(c = \frac{-16 \pm 10\sqrt{5}}{2}\)
\(c = -8 \pm 5\sqrt{5}\).

M1: Writes the equation of the circle and substitutes \(y = 2x+c\).
A1: Simplifies to obtain the correct quadratic equation in \(x\).
M1: Sets the discriminant \(b^2-4ac\) equal to zero to form a quadratic in \(c\).
A1: Solves the quadratic to find the exact values \(c = -8 \pm 5\sqrt{5}\).

First, we calculate the arc length of \(AB\): \(AB = r\theta = 6 \times 1.2 = 7.2\text{ cm}\). Next, we use right-angled trigonometry in triangle \(ACO\) to find the lengths of \(AC\) and \(OC\): \(AC = 6 \sin(1.2) \approx 5.592\text{ cm}\) and \(OC = 6 \cos(1.2) \approx 2.174\text{ cm}\). Since \(C\) lies on the radius \(OB\), we can find the length of \(BC\): \(BC = OB - OC = 6 - 2.174 = 3.826\text{ cm}\). Finally, the perimeter of the shaded region is the sum of these three lengths: \(\text{Perimeter} = AB + AC + BC = 7.2 + 5.592 + 3.826 = 16.618\text{ cm}\). Rounding to 3 significant figures gives \(16.6\text{ cm}\).

M1: State or use correct formula for arc length to get \(AB = 7.2\). M1: State or use correct trigonometry to find \(AC = 6\sin(1.2)\) or \(OC = 6\cos(1.2)\). M1: Subtract \(OC\) from \(6\) to find \(BC\). A1: Obtain \(16.6\) following correct working.

Let \(d\) be the common difference of the arithmetic progression (AP) and \(r\) be the common ratio of the geometric progression (GP). Given that the first term \(a = 5\), the terms of the AP are \(u_1 = 5\), \(u_2 = 5 + d\), and \(u_4 = 5 + 3d\). The terms of the GP are \(g_1 = 5\), \(g_2 = 5r\), and \(g_3 = 5r^2\). Since \(u_2 = g_2\) and \(u_4 = g_3\), we have \(5 + d = 5r \implies d = 5(r - 1)\) and \(5 + 3d = 5r^2 \implies 3d = 5(r^2 - 1)\). Substituting the first equation into the second gives \(3[5(r - 1)] = 5(r^2 - 1)\), which simplifies to \(15(r - 1) = 5(r - 1)(r + 1)\). Since the common ratio \(r \neq 1\), we can divide both sides by \(5(r - 1)\) to get \(3 = r + 1 \implies r = 2\). Substituting \(r = 2\) back to find \(d\) yields \(d = 5(2 - 1) = 5\). Finally, we find the sum of the first 20 terms of the AP: \(S_{20} = \frac{20}{2} [2(5) + (20 - 1)(5)] = 10 [10 + 95] = 1050\).

M1: Write down correct expressions for the terms of AP and GP and equate them, e.g., \(5+d = 5r\) and \(5+3d = 5r^2\). M1: Form and solve an equation in terms of \(r\) (or \(d\)) by eliminating the other variable. A1: Correctly find \(d = 5\) (or \(r = 2\)). A1: Use the correct AP sum formula to find \(S_{20} = 1050\).

(a) Let the first term of the arithmetic progression (AP) be \(a\) and the common difference be \(d\).

The 1st, 3rd and 4th terms of the AP are \(a\), \(a+2d\) and \(a+3d\) respectively.

Since these terms form a geometric progression (GP), the ratio between consecutive terms must be equal:
\((a+2d)^2 = a(a+3d)\)

Expanding both sides:
\(a^2 + 4ad + 4d^2 = a^2 + 3ad\)
\(ad + 4d^2 = 0\)
\(d(a + 4d) = 0\)

Since the common difference is non-zero (\(d \neq 0\)), we must have:
\(a + 4d = 0 \implies d = -0.25a\)

The common ratio, \(r\), of the GP is:
\(r = \frac{a+2d}{a} = 1 + \frac{2d}{a}\)

Substituting \(d = -0.25a\):
\(r = 1 + \frac{2(-0.25a)}{a} = 1 - 0.5 = 0.5\) (Showed)

(b) The sum of the first 10 terms of the AP is:
\(S_{10} = \frac{10}{2}[2a + 9d] = 85\)
\(5(2a + 9d) = 85 \implies 2a + 9d = 17\)

Substituting \(d = -0.25a\):
\(2a + 9(-0.25a) = 17\)
\(2a - 2.25a = 17\)
\(-0.25a = 17 \implies a = -68\)

Since the first term of the GP is equal to the first term of the AP, the first term of the GP is also \(a = -68\).

The sum to infinity of the GP, \(S_{\infty}\), is:
\(S_{\infty} = \frac{a}{1-r} = \frac{-68}{1 - 0.5} = -136\)

Part (a):
M1: Sets up the GP geometric mean relation \((a+2d)^2 = a(a+3d)\) or equivalent.
A1: Correctly simplifies to show \(d = -0.25a\) or \(a = -4d\) (rejecting \(d = 0\)).
A1: Correctly shows \(r = 0.5\).

Part (b):
M1: Sets up \(S_{10} = 85\) and substitutes \(d = -0.25a\) to find \(a\).
A1: Obtains \(a = -68\).
A1: Calculates the sum to infinity correctly to get \(-136\).

(a) Given:
\(\frac{3\sin\theta\tan\theta - 2}{1+\cos\theta} = 1\)

Multiply both sides by \(1+\cos\theta\):
\(3\sin\theta\tan\theta - 2 = 1 + \cos\theta\)

Substitute \(\tan\theta = \frac{\sin\theta}{\cos\theta}\):
\(3\sin\theta\left(\frac{\sin\theta}{\cos\theta}\right) - 2 = 1 + \cos\theta\)

Multiply through by \(\cos\theta\):
\(3\sin^2\theta - 2\cos\theta = \cos\theta + \cos^2\theta\)

Substitute the identity \(\sin^2\theta = 1 - \cos^2\theta\):
\(3(1 - \cos^2\theta) - 3\cos\theta - \cos^2\theta = 0\)
\(3 - 3\cos^2\theta - 3\cos\theta - \cos^2\theta = 0\)
\(4\cos^2\theta + 3\cos\theta - 3 = 0\) (Showed)

(b) To solve \(4\cos^2\theta + 3\cos\theta - 3 = 0\), we use the quadratic formula:
\(\cos\theta = \frac{-3 \pm \sqrt{3^2 - 4(4)(-3)}}{2(4)} = \frac{-3 \pm \sqrt{57}}{8}\)

This gives two values:
\(\cos\theta \approx 0.5687\) or \(\cos\theta \approx -1.319\)

Since \(\cos\theta\) must lie between \(-1\) and \(1\), we reject \(\cos\theta \approx -1.319\).

For \(\cos\theta = 0.5687\):
\(\theta = \cos^{-1}(0.5687) \approx 0.966\) radians (3 sf).

The other solution in the range \(0 \le \theta \le 2\pi\) is:
\(\theta = 2\pi - 0.966 \approx 5.32\) radians (3 sf).

Part (a):
M1: For eliminating the denominator and replacing \(\tan\theta\) with \(\frac{\sin\theta}{\cos\theta}\).
M1: For using the identity \(\sin^2\theta = 1 - \cos^2\theta\) and forming a quadratic in \(\cos\theta\).
A1: For obtaining \(4\cos^2\theta + 3\cos\theta - 3 = 0\) with clear intermediate steps.

Part (b):
M1: For solving the quadratic equation to get a valid value for \(\cos\theta\) (approx 0.569).
A1: For finding \(\theta = 0.966\) (or 0.965).
A1: For finding the second angle \(\theta = 5.32\) (or 5.31) and no other values in range.

(a) To find the points of intersection, we equate the two equations:
\(\frac{4}{\sqrt{2x+1}} = -\frac{4}{3}x + 4\)

Divide both sides by 4:
\(\frac{1}{\sqrt{2x+1}} = 1 - \frac{1}{3}x\)

Square both sides:
\(\frac{1}{2x+1} = \left(1 - \frac{1}{3}x\right)^2 = 1 - \frac{2}{3}x + \frac{1}{9}x^2\)

Multiply by \(9(2x+1)\):
\(9 = (2x+1)(9 - 6x + x^2)\)
\(9 = 18x - 12x^2 + 2x^3 + 9 - 6x + x^2\)
\(2x^3 - 11x^2 + 12x = 0\)

Factorizing the cubic equation:
\(x(2x^2 - 11x + 12) = 0\)
\(x(2x - 3)(x - 4) = 0\)

Since squaring can introduce extraneous solutions, we check \(x = 4\):
LHS of original equation = \(\frac{4}{3}\), RHS = \(-\frac{16}{3} + 4 = -\frac{4}{3}\). This is not a solution.

The valid solutions are \(x = 0\) and \(x = 1.5\).
- At \(x = 0\), \(y = 4\). So the first point is \((0, 4)\).
- At \(x = 1.5\), \(y = 2\). So the second point is \((1.5, 2)\).

(b) Since the line lies above the curve on the interval \(0 \le x \le 1.5\), the area of the region is:
\(\text{Area} = \int_{0}^{1.5} \left( \left(-\frac{4}{3}x + 4\right) - 4(2x+1)^{-\frac{1}{2}} \right) dx\)

Integrating each term:
\(\int \left(-\frac{4}{3}x + 4\right) dx = \left[ -\frac{2}{3}x^2 + 4x \right]\)
\(\int 4(2x+1)^{-\frac{1}{2}} dx = \left[ \frac{4(2x+1)^{\frac{1}{2}}}{\frac{1}{2} \cdot 2} \right] = \left[ 4\sqrt{2x+1} \right]\)

Now evaluating from 0 to 1.5:
- For the linear part:
\(\left[ -\frac{2}{3}x^2 + 4x \right]_{0}^{1.5} = -\frac{2}{3}(2.25) + 4(1.5) = -1.5 + 6 = 4.5\)

- For the curve part:
\(\left[ 4\sqrt{2x+1} \right]_{0}^{1.5} = 4\sqrt{4} - 4\sqrt{1} = 8 - 4 = 4\)

Thus, the exact area is:
\(4.5 - 4 = 0.5\)

Part (a):
M1: Equates the curve and line, squares to remove the square root, and attempts to solve the resulting polynomial.
A1: Correctly identifies the quadratic factors to obtain \(x = 0\) and \(x = 1.5\).
A1: Finds both points of intersection: \((0, 4)\) and \((1.5, 2)\).

Part (b):
M1: Formulates the correct definite integral \(\int_{0}^{1.5} (\text{Line} - \text{Curve}) dx\) and attempts integration of both terms.
A1: Correctly integrates both terms, yielding \(-\frac{2}{3}x^2 + 4x\) and \(4\sqrt{2x+1}\).
A1: Substitutes the limits correctly and obtains the final exact area of \(0.5\) (or \(\frac{1}{2}\)).

Squaring both sides of the inequality gives \((2x - 3)^2 < (x + 5)^2\). Expanding this yields \(4x^2 - 12x + 9 < x^2 + 10x + 25\). Rearranging into standard quadratic form gives \(3x^2 - 22x - 16 < 0\). Factoring the quadratic, we get \((3x + 2)(x - 8) < 0\). The critical values are \(x = -\frac{2}{3}\) and \(x = 8\). Since the inequality is less than zero, the solution set is the interval between these critical values: \(-\frac{2}{3} < x < 8\).

M1: For squaring both sides and obtaining a 3-term quadratic inequality, or for finding both critical values.
A1: For the correct critical values \(-\frac{2}{3}\) and \(8\).
A1: For the correct final range \(-\frac{2}{3} < x < 8\).

We can rewrite the equation as \(3 \cdot (3^x)^2 - 10(3^x) + 3 = 0\). Let \(u = 3^x\). The equation becomes the quadratic \(3u^2 - 10u + 3 = 0\). Factoring this gives \((3u - 1)(u - 3) = 0\), which yields solutions \(u = \frac{1}{3}\) and \(u = 3\). Substituting back \(3^x = u\), we have \(3^x = 3^{-1}\) which gives \(x = -1\), and \(3^x = 3^1\) which gives \(x = 1\). Thus, the solutions are \(x = -1\) and \(x = 1\).

M1: For substituting \(u = 3^x\) to obtain a quadratic equation and attempting to solve it.
A1: For finding the correct values of \(u\) as \(\frac{1}{3}\) and \(3\).
A1: For obtaining both correct values of \(x\), which are \(x = -1\) and \(x = 1\).

Using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\), the equation becomes \(1 + \tan^2 \theta + \tan \theta = 3\). Rearranging gives the quadratic equation \(\tan^2 \theta + \tan \theta - 2 = 0\). Factoring this quadratic, we get \((\tan \theta + 2)(\tan \theta - 1) = 0\). This yields two possible values: \(\tan \theta = 1\) or \(\tan \theta = -2\). Within the interval \(0 \le \theta \le \pi\), the equation \(\tan \theta = 1\) gives the exact solution \(\theta = \frac{\pi}{4}\). The equation \(\tan \theta = -2\) has a negative tangent, which lies in the second quadrant for this interval, giving the exact solution \(\theta = \pi - \arctan(2)\).

M1: For using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to form a quadratic in \(\tan \theta\).
A1: For solving the quadratic to get \(\tan \theta = 1\) and \(\tan \theta = -2\).
A1: For both correct exact solutions in the given interval.

Using the quotient rule, where \(u = \ln x\) and \(v = x^2\), we have \(u' = \frac{1}{x}\) and \(v' = 2x\). The derivative is given by \(\frac{dy}{dx} = \frac{v u' - u v'}{v^2} = \frac{x^2 \cdot \frac{1}{x} - (\ln x)(2x)}{(x^2)^2} = \frac{x - 2x \ln x}{x^4} = \frac{1 - 2 \ln x}{x^3}\). Substituting \(x = e\) into the derivative gives \(\frac{dy}{dx} = \frac{1 - 2 \ln e}{e^3} = \frac{1 - 2(1)}{e^3} = -\frac{1}{e^3} = -e^{-3}\).

M1: For applying the quotient rule (or product rule) correctly to find the derivative.
A1: For obtaining the correct derivative \(\frac{1 - 2 \ln x}{x^3}\).
A1: For substituting \(x = e\) to obtain the exact value \(-e^{-3}\) or \(-\frac{1}{e^3}\).

To integrate \(\int \frac{2}{4x - 1} \, dx\), we use the standard integration result for linear denominators: \(\int \frac{2}{4x - 1} \, dx = 2 \cdot \frac{1}{4} \ln|4x - 1| = \frac{1}{2} \ln|4x - 1|\). Applying the limits from \(1\) to \(3\), we get: \(\left[ \frac{1}{2} \ln|4x - 1| \right]_{1}^{3} = \frac{1}{2} \ln|4(3) - 1| - \frac{1}{2} \ln|4(1) - 1| = \frac{1}{2} \ln 11 - \frac{1}{2} \ln 3 = \frac{1}{2} \ln \frac{11}{3}\).

M1: For integrating to obtain a form \(k \ln(4x - 1)\).
A1: For the correct integration \(\frac{1}{2} \ln(4x - 1)\).
A1: For substituting the limits and simplifying to the correct exact value \(\frac{1}{2} \ln \frac{11}{3}\).

We are given the iterative formula \(x_{n+1} = \sqrt[3]{5x_n + 3}\) with initial value \(x_1 = 2.5\).
First iteration: \(x_2 = \sqrt[3]{5(2.5) + 3} = \sqrt[3]{15.5} \approx 2.493297\).
Second iteration: \(x_3 = \sqrt[3]{5(2.493297) + 3} = \sqrt[3]{15.466485} \approx 2.491497\).
Rounding to 3 decimal places gives \(x_3 = 2.491\).

M1: For calculating the first iteration value \(x_2 \approx 2.4933\).
A1: For finding the second iteration value \(x_3 \approx 2.4915\).
A1: For rounding the final value of \(x_3\) correctly to \(2.491\).

Let \(p(x) = 2x^3 - x^2 + ax + b\).
Using the remainder theorem with the division by \(x - 1\), we have \(p(1) = 3\):
\(2(1)^3 - (1)^2 + a(1) + b = 3 \implies 2 - 1 + a + b = 3 \implies a + b = 2\) (Equation 1).
Using the remainder theorem with the division by \(x + 2\), we have \(p(-2) = -15\):
\(2(-2)^3 - (-2)^2 + a(-2) + b = -15 \implies -16 - 4 - 2a + b = -15 \implies -2a + b = 5\) (Equation 2).
Subtracting Equation 2 from Equation 1 gives:
\((a + b) - (-2a + b) = 2 - 5 \implies 3a = -3 \implies a = -1\).
Substituting \(a = -1\) back into Equation 1 gives:
\(-1 + b = 2 \implies b = 3\).
Therefore, the values are \(a = -1\) and \(b = 3\).

M1: For applying the remainder theorem to set up two simultaneous equations in \(a\) and \(b\).
A1: For obtaining both correct equations, e.g., \(a + b = 2\) and \(-2a + b = 5\).
A1: For solving the system of equations to find the correct values \(a = -1\) and \(b = 3\).

To solve the inequality \(|3x - 5| > 2|x + 1|\), we can square both sides because both sides are non-negative. This gives \((3x - 5)^2 > 4(x + 1)^2\). Expanding both sides, we get \(9x^2 - 30x + 25 > 4(x^2 + 2x + 1)\), which simplifies to \(9x^2 - 30x + 25 > 4x^2 + 8x + 4\). Rearranging all terms to one side gives the quadratic inequality \(5x^2 - 38x + 21 > 0\). Factorizing the quadratic expression yields \((5x - 3)(x - 7) > 0\). The critical values are \(x = 3/5\) and \(x = 7\). Since we require the expression to be strictly greater than 0, the solution set is \(x < 3/5\) or \(x > 7\).

M1: For squaring both sides and attempting expansion of both sides. A1: For obtaining the correct critical values of 3/5 and 7. M1: For choosing the outside regions of the critical values for a greater-than quadratic inequality. A1.8: For the correct final inequality x < 3/5 or x > 7 (or equivalent interval notation).

By taking the exponential of both sides of the equation, we get \(2e^{2x} + 5 = e^{x + \ln 7}\). Using exponent laws, we can rewrite the right-hand side as \(e^x \cdot e^{\ln 7} = 7e^x\). This yields the equation \(2e^{2x} + 5 = 7e^x\). Let \(u = e^x\), so the equation becomes the quadratic \(2u^2 - 7u + 5 = 0\). Factorizing this quadratic gives \((2u - 5)(u - 1) = 0\), which has solutions \(u = 5/2\) and \(u = 1\). Substituting back \(u = e^x\), we have \(e^x = 5/2\) which gives \(x = \ln(5/2)\), and \(e^x = 1\) which gives \(x = \ln 1 = 0\). Both solutions are valid because the exponential function is always positive.

M1: For using logarithm and exponent laws to remove the logarithms, leading to 2e^(2x) + 5 = 7e^x. M1: For setting up and solving a quadratic equation in terms of e^x. A1: For finding the correct values e^x = 5/2 and e^x = 1. A1.8: For obtaining the correct exact solutions x = 0 and x = ln(5/2) (or equivalent form such as ln 5 - ln 2).

Using the double angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\), we substitute into the equation to get \(3(1 - 2\sin^2\theta) - 7\sin\theta = 0\). Expanding and rearranging gives \(3 - 6\sin^2\theta - 7\sin\theta = 0\), which simplifies to the quadratic equation \(6\sin^2\theta + 7\sin\theta - 3 = 0\). Factorizing this quadratic gives \((2\sin\theta + 3)(3\sin\theta - 1) = 0\). This yields two possible values: \(\sin\theta = -3/2\) or \(\sin\theta = 1/3\). Since the range of the sine function is between -1 and 1, there are no real solutions for \(\sin\theta = -3/2\). For \(\sin\theta = 1/3\) in the interval \(0^\circ \le \theta \le 360^\circ\), the solutions are \(\theta = \sin^{-1}(1/3) \approx 19.5^\circ\) and \(\theta = 180^\circ - 19.5^\circ = 160.5^\circ\).

M1: For applying the correct double angle identity to express the equation in terms of sine only. M1: For factorizing or solving the resulting quadratic equation in terms of sin theta. A1: For identifying sin theta = 1/3 and noting that sin theta = -1.5 has no solutions. A1.8: For both correct angles 19.5 degrees and 160.5 degrees, rounded to 1 decimal place, with no extra solutions in the range.

We differentiate the given function using the quotient rule, where \(u = \ln(2x+3)\) and \(v = x^2\). The derivatives are \(du/dx = 2/(2x+3)\) and \(dv/dx = 2x\). Applying the quotient rule formula \(dy/dx = (v \cdot du/dx - u \cdot dv/dx) / v^2\), we get \(dy/dx = (x^2 \cdot \frac{2}{2x+3} - 2x \ln(2x+3)) / x^4\). Substituting \(x = -1\) into this derivative expression gives: \(2x+3 = 1\), so \(\ln(2x+3) = \ln 1 = 0\); \(du/dx = 2/1 = 2\); \(v = 1\); and \(dv/dx = -2\). Therefore, \(dy/dx = (1 \cdot 2 - (-2) \cdot 0) / 1^2 = 2/1 = 2\). The exact gradient of the curve at \(x = -1\) is 2.

M1: For applying the quotient rule (or product rule) to differentiate the function. A1: For finding the correct derivative of ln(2x+3) which is 2/(2x+3). A1: For the correct unsimplified expression for dy/dx. A1.8: For substituting x = -1 and obtaining the correct exact value of 2.

We integrate each term of the integrand individually. The integral of \(\cos 2x\) is \(\frac{1}{2} \sin 2x\), and the integral of \(\sec^2 x\) is \(\tan x\). Thus, the definite integral is \([\frac{1}{2} \sin 2x + \tan x]_{0}^{\frac{\pi}{3}}\). Evaluating at the upper limit \(x = \pi/3\) gives \(\frac{1}{2} \sin(2\pi/3) + \tan(\pi/3) = \frac{1}{2}(\frac{\sqrt{3}}{2}) + \sqrt{3} = \frac{\sqrt{3}}{4} + \sqrt{3} = \frac{5\sqrt{3}}{4}\). Evaluating at the lower limit \(x = 0\) gives \(\frac{1}{2} \sin 0 + \tan 0 = 0\). Subtracting the lower limit value from the upper limit value, we obtain the final exact value of \(\frac{5\sqrt{3}}{4}\).

M1: For integrating cos 2x to obtain a sin 2x (where a is a constant and a is not 1). M1: For integrating sec^2 x to obtain tan x. A1: For obtaining the correct integrated expression of 1/2 * sin 2x + tan x. A1.8: For substituting the limits correctly and simplifying to the correct exact final answer of 5*sqrt(3)/4.

To find the coordinates of the stationary point, we first differentiate \(y\) with respect to \(x\).

Using the quotient rule with \(u = \ln x\) and \(v = x^3\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\)
\(\frac{\mathrm{d}v}{\mathrm{d}x} = 3x^2\)

\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{x^3\left(\frac{1}{x}\right) - (\ln x)(3x^2)}{(x^3)^2}\]

Simplifying the expression:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 - 3x^2\ln x}{x^6} = \frac{x^2(1 - 3\ln x)}{x^6} = \frac{1 - 3\ln x}{x^4}\]

At a stationary point, \[\frac{\mathrm{d}y}{\mathrm{d}x} = 0\]. Therefore:
\[1 - 3\ln x = 0\]
\[3\ln x = 1\]
\[\ln x = \frac{1}{3}\]
\[x = e^{1/3}\]

Substitute \(x = e^{1/3}\) back into the original equation to find the corresponding \(y\)-coordinate:
\[y = \frac{\ln\left(e^{1/3}\right)}{\left(e^{1/3}\right)^3} = \frac{\frac{1}{3}}{e} = \frac{1}{3e}\]

Thus, the exact coordinates of the stationary point are \(\left(e^{1/3}, \frac{1}{3e}\right)\).

**M1**: Attempt to use the quotient rule or product rule to differentiate \(y = \frac{\ln x}{x^3}\). For the quotient rule, accept \(\frac{x^3 \cdot \frac{1}{x} - \ln x \cdot 3x^2}{(x^3)^2}\).

**A1**: Obtain a correct simplified derivative of \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1-3\ln x}{x^4}\) (or equivalent).

**M1**: Set their derivative equal to 0 and solve for \(x\) using exponential functions.

**A1**: Obtain \(x = e^{1/3}\) (or equivalent exact form, e.g., \(\sqrt[3]{e}\)).

**A1**: Obtain \(y = \frac{1}{3e}\) (or equivalent exact form, e.g., \(\frac{1}{3}e^{-1}\)) and state the coordinates.

Squaring both sides of the inequality: \((2x - 3)^2 > (x + 4)^2\). Expanding both sides: \(4x^2 - 12x + 9 > x^2 + 8x + 16\). Rearranging gives the quadratic inequality \(3x^2 - 20x - 7 > 0\). Factoring the quadratic: \((3x + 1)(x - 7) > 0\). This gives the critical values \(x = -\frac{1}{3}\) and \(x = 7\). Since we require the quadratic to be greater than zero, the solution is \(x < -\frac{1}{3}\) or \(x > 7\).

M1: For squaring both sides and attempting to collect terms into a three-term quadratic. A1: For finding the correct critical values -1/3 and 7. A1: For obtaining the correct final range.

Using the laws of logarithms: \(2\ln x = \ln(x^2)\). The equation becomes \(\ln(x + 4) = \ln(x^2) + \ln 2\), which simplifies to \(\ln(x + 4) = \ln(2x^2)\). Taking exponentials of both sides: \(x + 4 = 2x^2\). Rearranging to a quadratic equation: \(2x^2 - x - 4 = 0\). Using the quadratic formula: \(x = \frac{1 \pm \sqrt{33}}{4}\). Since the term \(\ln x\) requires \(x > 0\), we reject the negative root, leaving \(x = \frac{1 + \sqrt{33}}{4} \approx 1.69\).

M1: For applying laws of logarithms correctly to eliminate logs. M1: For attempting to solve the resulting quadratic equation. A1: For obtaining the correct positive solution x = 1.69 and rejecting the negative root.

Using the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\), the equation becomes \(1 + \tan^2 \theta + 2\tan \theta = 4\). Rearranging gives the quadratic equation \(\tan^2 \theta + 2\tan \theta - 3 = 0\). Factoring the quadratic: \((\tan \theta - 1)(\tan \theta + 3) = 0\), which yields \(\tan \theta = 1\) or \(\tan \theta = -3\). In the interval \(0^\circ \le \theta \le 180^\circ\), the solution for \(\tan \theta = 1\) is \(\theta = 45^\circ\), and for \(\tan \theta = -3\) is \(\theta = 180^\circ - 71.57^\circ = 108.4^\circ\).

M1: For using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to obtain a quadratic in \(\tan \theta\). M1: For solving the quadratic equation to find values of \(\tan \theta\). A1: For obtaining both correct angles 45 and 108.4 degrees.

We find the derivative of \(y\) with respect to \(x\) using the product rule: \(\frac{dy}{dx} = \frac{d}{dx}(e^{2x})\cos x + e^{2x}\frac{d}{dx}(\cos x) = 2e^{2x}\cos x - e^{2x}\sin x\). Substituting \(x = 0\) into the derivative: \(\frac{dy}{dx} = 2e^0 \cos 0 - e^0 \sin 0 = 2(1)(1) - 1(0) = 2\).

M1: For attempting to differentiate using the product rule. A1: For obtaining the correct gradient of 2 at x = 0.

Integrating the expression: \(\int_0^2 \frac{1}{2x+3} \, dx = \left[ \frac{1}{2} \ln(2x+3) \right]_0^2\). Substituting the upper and lower limits: \(\left( \frac{1}{2} \ln(2(2)+3) \right) - \left( \frac{1}{2} \ln(2(0)+3) \right) = \frac{1}{2} \ln 7 - \frac{1}{2} \ln 3 = \frac{1}{2} \ln \left(\frac{7}{3}\right)\).

M1: For integrating to obtain a term of the form \(k \ln(2x+3)\). A1: For obtaining the correct exact value \(\frac{1}{2} \ln (7/3)\) or equivalent.

Using the iterative formula: \(x_1 = 2\), \(x_2 = \sqrt[3]{5(2) + 2} = \sqrt[3]{12} \approx 2.2894\), \(x_3 = \sqrt[3]{5(2.2894) + 2} \approx 2.3780\), \(x_4 = \sqrt[3]{5(2.3780) + 2} \approx 2.4039\). Rounding the final value to 3 decimal places gives \(x_4 = 2.404\).

M1: For calculating the values of \(x_2\) and \(x_3\) to at least 4 decimal places. A1: For obtaining the correct final value of 2.404 for \(x_4\).

The angle \(\theta\) is found using the formula \(\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}\). First, the scalar product is \(\mathbf{u} \cdot \mathbf{v} = 2(3) + (-1)(0) + 2(4) = 14\). The magnitude of \(\mathbf{u}\) is \(|\mathbf{u}| = \sqrt{2^2 + (-1)^2 + 2^2} = 3\). The magnitude of \(\mathbf{v}\) is \(|\mathbf{v}| = \sqrt{3^2 + 0^2 + 4^2} = 5\). Thus, \(\cos \theta = \frac{14}{(3)(5)} = \frac{14}{15}\), which gives \(\theta = \cos^{-1}\left(\frac{14}{15}\right) \approx 21.039^\circ \approx 21.0^\circ\).

M1: For finding the correct scalar product of 14. M1: For finding the correct magnitudes 3 and 5 and substituting into the cosine formula. A1: For obtaining the correct angle 21.0 degrees.

First, express \(z\) in the form \(x + \text{i}y\) by multiplying the numerator and denominator by the conjugate of the denominator, \(1 - 2\text{i}\): \(z = \frac{5(1-2\text{i})}{(1+2\text{i})(1-2\text{i})} = \frac{5-10\text{i}}{5} = 1 - 2\text{i}\). The modulus is \(|z| = \sqrt{1^2 + (-2)^2} = \sqrt{5}\). The argument of \(z\) (which lies in the fourth quadrant) is \(\arg(z) = \tan^{-1}\left(\frac{-2}{1}\right) \approx -1.11\) radians.

M1: For attempting to multiply the numerator and denominator by the conjugate to simplify \(z\) to the form \(x + \text{i}y\). A1: For obtaining the correct modulus \(\sqrt{5}\) (or 2.24). A1: For obtaining the correct argument -1.11 radians.

We can solve the inequality by squaring both sides: \((2x - 5)^2 < 9(x + 1)^2\). Expanding both sides gives: \(4x^2 - 20x + 25 < 9(x^2 + 2x + 1)\), which simplifies to: \(4x^2 - 20x + 25 < 9x^2 + 18x + 9\). Rearranging all terms to one side gives: \(5x^2 + 38x - 16 > 0\). We can factorise this quadratic: \((5x - 2)(x + 8) > 0\). The critical values are \(x = 0.4\) and \(x = -8\). Since the quadratic must be strictly greater than zero, the solution is \(x < -8\) or \(x > 0.4\).

M1: For squaring both sides and attempting expansion of terms. A1: For obtaining a correct three-term quadratic in any form, e.g., \(5x^2 + 38x - 16 > 0\). M1: For finding the critical values of their quadratic equation. A1: For the correct solution intervals: \(x < -8\) or \(x > 0.4\).

Let \(y = 5^x\). Since \(5^{2x} = (5^x)^2 = y^2\) and \(5^{x+1} = 5 \times 5^x = 5y\), the equation can be rewritten as: \(y^2 - 5y + 4 = 0\). Factorising this quadratic equation gives: \((y - 4)(y - 1) = 0\), which yields solutions \(y = 4\) or \(y = 1\). We now substitute back \(5^x\): For \(5^x = 1\), we find \(x = 0\). For \(5^x = 4\), taking logarithms on both sides gives \(x \ln 5 = \ln 4 \implies x = \frac{\ln 4}{\ln 5} \approx 0.861\).

M1: For substituting \(y = 5^x\) to obtain a quadratic equation in \(y\). A1: For finding correct roots \(y = 4\) and \(y = 1\). M1: For solving \(5^x = y\) using logarithms for the positive root. A1: For correct answers \(x = 0\) and \(x = 0.861\).

Using the compound angle formula, we have: \(R \cos(\theta + \alpha) = R \cos \theta \cos \alpha - R \sin \theta \sin \alpha\). Comparing coefficients with \(3 \cos \theta - 2 \sin \theta\), we obtain: \(R \cos \alpha = 3\) and \(R \sin \alpha = 2\). Squaring and adding gives \(R^2 = 3^2 + 2^2 = 13 \implies R = \sqrt{13}\). Dividing the equations gives \(\tan \alpha = \frac{2}{3} \implies \alpha = \tan^{-1}\left(\frac{2}{3}\right) \approx 0.5880\) radians. Now we solve the equation: \(\sqrt{13} \cos(\theta + 0.5880) = 1.5 \implies \cos(\theta + 0.5880) = \frac{1.5}{\sqrt{13}}\). Thus, \(\theta + 0.5880 = \cos^{-1}\left(\frac{1.5}{\sqrt{13}}\right) \approx 1.1414\) radians. Finding the smallest positive value of \(\theta\) gives: \(\theta = 1.1414 - 0.5880 = 0.553\) (to 3 s.f.).

M1: For calculating \(R = \sqrt{13}\) and setting up \(\tan \alpha = \frac{2}{3}\). A1: For obtaining \(\alpha \approx 0.588\) radians. M1: For setting up and solving the equation \(\cos(\theta + \alpha) = \frac{1.5}{R}\) to obtain a positive value for \(\theta\). A1: For obtaining the correct smallest positive value \(\theta = 0.553\).

Differentiating both sides of the equation \(x^3 + y^3 - 3xy = 3\) implicitly with respect to \(x\): \(3x^2 + 3y^2 \frac{\text{d}y}{\text{d}x} - 3\left(y + x \frac{\text{d}y}{\text{d}x}\right) = 0\). Simplifying by dividing all terms by 3: \(x^2 + y^2 \frac{\text{d}y}{\text{d}x} - y - x \frac{\text{d}y}{\text{d}x} = 0\). Rearranging terms to make \(\frac{\text{d}y}{\text{d}x}\) the subject: \(\frac{\text{d}y}{\text{d}x}(y^2 - x) = y - x^2 \implies \frac{\text{d}y}{\text{d}x} = \frac{y - x^2}{y^2 - x}\). Substituting \(x = 2\) and \(y = 1\) into this expression: \(\frac{\text{d}y}{\text{d}x} = \frac{1 - 2^2}{1^2 - 2} = \frac{1 - 4}{1 - 2} = \frac{-3}{-1} = 3\). Thus, the gradient of the curve at \((2, 1)\) is 3.

M1: For attempting implicit differentiation with respect to \(x\), with at least one term involving \(\frac{\text{d}y}{\text{d}x}\) processed correctly. A1: For correct differentiation of all terms: \(3x^2 + 3y^2 \frac{\text{d}y}{\text{d}x} - 3y - 3x \frac{\text{d}y}{\text{d}x} = 0\). M1: For substituting \(x = 2\) and \(y = 1\) into their expression. A1: For obtaining the final answer of 3.

We use integration by parts, which states \(\int u \frac{\text{d}v}{\text{d}x} \text{d}x = uv - \int v \frac{\text{d}u}{\text{d}x} \text{d}x\). Let \(u = x \implies \frac{\text{d}u}{\text{d}x} = 1\). Let \(\frac{\text{d}v}{\text{d}x} = e^{2x} \implies v = \frac{1}{2} e^{2x}\). Applying the formula: \(\int_{0}^{1} x e^{2x} \text{d}x = \left[ \frac{1}{2} x e^{2x} \right]_0^1 - \int_{0}^{1} \frac{1}{2} e^{2x} \text{d}x\). Evaluating the first term and integrating the second term: \(= \left( \frac{1}{2}(1)e^{2} - 0 \right) - \left[ \frac{1}{4} e^{2x} \right]_0^1\). Substituting limits: \(= \frac{1}{2} e^2 - \left( \frac{1}{4} e^2 - \frac{1}{4} e^0 \right) = \frac{1}{2} e^2 - \frac{1}{4} e^2 + \frac{1}{4} = \frac{1}{4} e^2 + \frac{1}{4} = \frac{1}{4}(e^2 + 1)\).

M1: For identifying and attempting integration by parts in the correct form, obtaining \(a x e^{2x} - \int b e^{2x} \text{d}x\). A1: For obtaining the correct integrated expression \(\frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x}\). M1: For substituting the limits \(0\) and \(1\) correctly into their integrated expression. A1: For obtaining the exact final answer \rac{1}{4}(e^2 + 1)\) or equivalent exact form.

Let \(f(x) = x^3 - 5x + 1\). Since \(f(x)\) is a continuous function, we test the sign at \(x = 0\) and \(x = 1\): \(f(0) = 0^3 - 5(0) + 1 = 1 > 0\) and \(f(1) = 1^3 - 5(1) + 1 = -3 < 0\). Because there is a change of sign between \(x = 0\) and \(x = 1\), a root must lie in this interval. Now, using the iteration formula with \(x_1 = 0.2\): \(x_2 = \frac{0.2^3 + 1}{5} = \frac{0.008 + 1}{5} = 0.20160\), \(x_3 = \frac{0.20160^3 + 1}{5} \approx 0.20164\), \(x_4 = \frac{0.20164^3 + 1}{5} \approx 0.20164\). Since successive values round to \(0.202\), the root is \(0.202\) correct to 3 decimal places.

M1: For evaluating \(f(0)\) and \(f(1)\) to show a change of sign and stating that a root exists. M1: For calculating \(x_2\) and \(x_3\) using the given formula. A1: For showing iterations clearly to 5 decimal places, specifically reaching \(x_3 \approx 0.20164\). A1: For stating the root as \(0.202\) based on consistent rounded iterations.

We standardise the denominator of \(u\) by multiplying numerator and denominator by the complex conjugate of the denominator, \(2 + 3\text{i}\): \(u = \frac{(5 + \text{i})(2 + 3\text{i})}{(2 - 3\text{i})(2 + 3\text{i})} = \frac{10 + 15\text{i} + 2\text{i} - 3}{4 + 9} = \frac{7 + 17\text{i}}{13} = \frac{7}{13} + \frac{17}{13}\text{i}\). To find the modulus: \(|u| = \sqrt{\left(\frac{7}{13}\right)^2 + \left(\frac{17}{13}\right)^2} = \sqrt{\frac{49 + 289}{169}} = \sqrt{\frac{338}{169}} = \sqrt{2} \approx 1.41\). To find the argument (since \(u\) is in the first quadrant): \(\arg(u) = \tan^{-1}\left(\frac{17/13}{7/13}\right) = \tan^{-1}\left(\frac{17}{7}\right) \approx 1.18\) radians.

M1: For attempting to multiply numerator and denominator by the conjugate \(2 + 3\text{i}\). A1: For obtaining \(u = \frac{7}{13} + \frac{17}{13}\text{i}\) (or equivalent). A1: For calculating the modulus \(|u| = \sqrt{2}\) (or 1.41). A1: For calculating the argument \(\arg(u) = 1.18\) radians.

The parametric equations for coordinates on the line are: \(x = 1 + 2\lambda\), \(y = 2 - \lambda\), \(z = -1 + \lambda\). We substitute these coordinates into the plane equation \(2x + y - z = 7\): \(2(1 + 2\lambda) + (2 - \lambda) - (-1 + \lambda) = 7\). Expanding and simplifying: \(2 + 4\lambda + 2 - \lambda + 1 - \lambda = 7 \implies 5 + 2\lambda = 7 \implies 2\lambda = 2 \implies \lambda = 1\). Substituting \\lambda = 1\) back into the equations for the coordinates: \(x = 1 + 2(1) = 3\), \(y = 2 - 1 = 1\), \(z = -1 + 1 = 0\). Therefore, the coordinates of the intersection point are \((3, 1, 0)\).

M1: For writing general coordinate components in terms of \(\lambda\). M1: For substituting components into the plane equation and setting up an equation in \(\lambda\). A1: For solving to find \(\lambda = 1\). A1: For obtaining the correct coordinates \((3, 1, 0)\).

We first rewrite the expression as \((2-x)(4+3x)^{-\frac{1}{2}}\).

Step 1: Expand \((4+3x)^{-\frac{1}{2}}\):
\[(4+3x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \left(1 + \frac{3}{4}x\right)^{-\frac{1}{2}} = \frac{1}{2} \left(1 + \frac{3}{4}x\right)^{-\frac{1}{2}}\]
Using the binomial theorem:
\[\left(1 + \frac{3}{4}x\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(\frac{3}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(\frac{3}{4}x\right)^2 + \dots\]
\[= 1 - \frac{3}{8}x + \frac{3}{8}\left(\frac{9}{16}x^2\right) + \dots = 1 - \frac{3}{8}x + \frac{27}{128}x^2 + \dots\]

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

Step 3: Multiply by \(2-x\) and find the term in \(x^2\):
\[(2-x)\left(\frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2 + \dots\right)\]
The term in \(x^2\) is:
\[2 \left(\frac{27}{256}x^2\right) - x \left(-\frac{3}{16}x\right) = \frac{27}{128}x^2 + \frac{3}{16}x^2\]
\[= \left(\frac{27}{128} + \frac{24}{128}\right)x^2 = \frac{51}{128}x^2\]

Thus, the coefficient of \(x^2\) is \(\frac{51}{128}\).

M1: For taking out a factor of \(4^{-\frac{1}{2}}\) (or \(\frac{1}{2}\)) and attempting to expand \((1+kx)^{-\frac{1}{2}}\).
A1: For a correct simplified expansion of \((4+3x)^{-\frac{1}{2}}\) up to the \(x^2\) term, i.e., \(\frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2\).
M1: For identifying and combining the two terms that contribute to the coefficient of \(x^2\) when multiplied by \(2-x\).
A1: For the final correct simplified fraction \(\frac{51}{128}\).

We express \(u\) in the form \(x + \mathrm{i}y\) by multiplying the numerator and denominator by the complex conjugate of the denominator, \(2 + \mathrm{i}\):
\[u = \frac{a + 3\mathrm{i}}{2 - \mathrm{i}} \cdot \frac{2 + \mathrm{i}}{2 + \mathrm{i}}\]
\[= \frac{2a + a\mathrm{i} + 6\mathrm{i} - 3}{4 - \mathrm{i}^2}\]
\[= \frac{(2a - 3) + (a + 6)\mathrm{i}}{5}\]

So the real and imaginary parts of \(u\) are:
\[\operatorname{Re}(u) = \frac{2a - 3}{5}, \quad \operatorname{Im}(u) = \frac{a + 6}{5}\]

Since \(\arg(u) = \frac{3}{4}\pi\), the complex number lies in the second quadrant. This means \(\operatorname{Re}(u) < 0\) and \(\operatorname{Im}(u) > 0\).

Additionally, we have:
\[\tan\left(\frac{3}{4}\pi\right) = \frac{\operatorname{Im}(u)}{\operatorname{Re}(u)}\]
\[-1 = \frac{\frac{a + 6}{5}}{\frac{2a - 3}{5}}\]
\[-1 = \frac{a + 6}{2a - 3}\]
\[-(2a - 3) = a + 6\]
\[-2a + 3 = a + 6\]
\[3a = -3 \implies a = -1\]

We check the quadrant conditions with \(a = -1\):
\[\operatorname{Re}(u) = \frac{2(-1) - 3}{5} = -1 < 0\]
\[\operatorname{Im}(u) = \frac{-1 + 6}{5} = 1 > 0\]
This is consistent with the argument being in the second quadrant. Thus, \(a = -1\).

M1: For multiplying numerator and denominator by \(2+\mathrm{i}\) to express \(u\) in Cartesian form.
A1: For obtaining correct real part \(\frac{2a-3}{5}\) and imaginary part \(\frac{a+6}{5}\).
M1: For setting up the equation \(\frac{\operatorname{Im}(u)}{\operatorname{Re}(u)} = \tan\left(\frac{3}{4}\pi\right)\) (or equivalent) and solving for \(a\).
A1: For obtaining \(a = -1\) and confirming it satisfies the quadrant conditions (or ruling out invalid solutions if any).

Let \(u = \ln x\) and \(\frac{\text{d}v}{\text{d}x} = x^{-1/2}\). Then \(\frac{\text{d}u}{\text{d}x} = \frac{1}{x}\) and \(v = 2x^{1/2}\). Using the integration by parts formula: \(\int u \frac{\text{d}v}{\text{d}x} \text{d}x = uv - \int v \frac{\text{d}u}{\text{d}x} \text{d}x\), we get \(\int \frac{\ln x}{\sqrt{x}} \text{d}x = 2\sqrt{x}\ln x - \int 2\sqrt{x} \cdot \frac{1}{x} \text{d}x = 2\sqrt{x}\ln x - \int 2x^{-1/2} \text{d}x = 2\sqrt{x}\ln x - 4\sqrt{x} + C\). Substituting the limits 1 and 9: \(\left[ 2\sqrt{x}\ln x - 4\sqrt{x} \right]_{1}^{9} = (2\sqrt{9}\ln 9 - 4\sqrt{9}) - (2\sqrt{1}\ln 1 - 4\sqrt{1}) = (6\ln(3^2) - 12) - (-4) = (12\ln 3 - 12) + 4 = 12\ln 3 - 8\). Thus, \(a = 12\) and \(b = -8\).

M1: Attempt integration by parts with \(u = \ln x\) and \(\text{d}v = x^{-1/2}\text{d}x\). A1: Obtain correct derivatives and integrals: \(\text{d}u = \frac{1}{x}\text{d}x\) and \(v = 2\sqrt{x}\). A1: Obtain correct term \(2\sqrt{x}\ln x\). M1: Integrate the remaining term to obtain \(4\sqrt{x}\). A1: Correct fully integrated expression \(2\sqrt{x}\ln x - 4\sqrt{x}\). M1: Substitute limits 9 and 1 into their integrated expression. A1: Obtain final answer \(12\ln 3 - 8\).

\( \text{(a)} \) Let \(z = x + \text{i}y\), so \(z^* = x - \text{i}y\). Substituting these into the equation: \((2+\text{i})(x+\text{i}y) + (1+\text{i})(x-\text{i}y) = 6 + 7\text{i}\). Expanding: \((2x - y) + \text{i}(x + 2y) + (x + y) + \text{i}(x - y) = 6 + 7\text{i}\). Grouping real and imaginary parts: Real part: \(3x\), Imaginary part: \(2x + y\). Equating components: \(3x = 6 \implies x = 2\) and \(2x + y = 7 \implies 4 + y = 7 \implies y = 3\). Thus, \(z = 2 + 3\text{i}\). \( \text{(b)} \) The inequality represents the interior and boundary of a circle of radius \(R = \sqrt{2}\) centered at \(z = 2 + 3\text{i}\). The argument of the center is \(\theta = \arg(z) = \arctan(3/2) \approx 0.9828\) radians. The distance from the origin to the center is \(|z| = \sqrt{2^2 + 3^2} = \sqrt{13}\). The maximum argument is achieved when the line from the origin is tangent to the circle. The angle \(\alpha\) between this tangent and the line to the center satisfies \(\sin\alpha = \frac{R}{|z|} = \frac{\sqrt{2}}{\sqrt{13}}\), giving \(\alpha \approx 0.4029\) radians. The maximum value of \(\arg(w)\) is \(\theta + \alpha \approx 0.9828 + 0.4029 = 1.3857\) radians, which is \(1.39\) to 3 significant figures.

Part (a): M1: Substitute \(z = x + \text{i}y\) and expand. A1: Obtain correct real and imaginary parts. M1: Equate components and attempt to solve for \(x\) and \(y\). A1: Obtain \(z = 2 + 3\text{i}\). Part (b): M1: Calculate \(|z| = \sqrt{13}\) and \(\arg(z) \approx 0.983\). M1: Set up the trigonometric equation for the tangent angle \(\alpha\), e.g., \(\sin\alpha = \sqrt{2}/\sqrt{13}\). A1: Obtain final answer \(1.39\).

Let the initial direction of motion of \(A\) be positive. The initial momentum of the system is given by: \(P_{\text{initial}} = m_A u_A + m_B u_B = 0.4 \times 5 + 0.6 \times (-2) = 2.0 - 1.2 = 0.8\text{ kg m s}^{-1}\). After the collision, the velocity of \(A\) is \(-1\text{ m s}^{-1}\). Let \(v_B\) be the velocity of \(B\) after the collision. The final momentum of the system is given by: \(P_{\text{final}} = m_A v_A + m_B v_B = 0.4 \times (-1) + 0.6 v_B = -0.4 + 0.6 v_B\). Using conservation of linear momentum: \(P_{\text{initial}} = P_{\text{final}} \implies 0.8 = -0.4 + 0.6 v_B \implies 1.2 = 0.6 v_B \implies v_B = 2\text{ m s}^{-1}\). Thus, the speed of \(B\) after the collision is \(2\text{ m s}^{-1}\).

M1: Attempt to write a conservation of linear momentum equation with correct sign changes for opposing directions. A1: Correctly solve for the velocity of \(B\) to obtain the speed of \(2\text{ m s}^{-1}\).

The forces acting on the ring are: Weight \(W = mg = 1.2 \times 10 = 12\text{ N}\) downwards, Normal reaction \(R\) upwards, Frictional force \(F\) horizontally opposing the motion, and the pulling force \(T\) at an angle \(\alpha\) above the horizontal. Since \(\sin \alpha = 0.6\), we have \(\cos \alpha = 0.8\). Resolving forces vertically: \(R + T \sin \alpha = W \implies R = 12 - 0.6 T\). Since the ring is on the point of slipping, the friction is limiting: \(F = \mu R = \frac{1}{3}(12 - 0.6 T) = 4 - 0.2 T\). Resolving forces horizontally: \(T \cos \alpha = F \implies 0.8 T = 4 - 0.2 T \implies 1.0 T = 4 \implies T = 4\).

M1: Resolve forces vertically to express the normal reaction in terms of \(T\), and apply the friction formula \(F = \mu R\). A1: Set up horizontal force balance and solve for \(T\) to obtain \(T = 4\).

The forces acting parallel and perpendicular to the inclined plane are as follows: Weight component parallel to the plane: \(W_{\parallel} = mg \sin \theta = 4 \times 10 \times 0.6 = 24\text{ N}\) downwards. Weight component perpendicular to the plane: \(W_{\perp} = mg \cos \theta\). Since \(\sin \theta = 0.6\), \(\cos \theta = 0.8\), so \(W_{\perp} = 4 \times 10 \times 0.8 = 32\text{ N}\). The normal reaction is \(R = 32\text{ N}\). The frictional force opposing the motion is \(F = \mu R = 0.25 \times 32 = 8\text{ N}\). Applying Newton's second law parallel to the slope: \(P - W_{\parallel} - F = m a \implies 40 - 24 - 8 = 4 a \implies 8 = 4 a \implies a = 2\text{ m s}^{-2}\).

M1: Find the normal reaction force and the limiting friction force. M1: Set up the equation of motion along the plane using Newton's second law. A1: Correctly calculate the acceleration as \(2\text{ m s}^{-2}\).

Since the velocity of the cyclist is constant, the acceleration is zero and the forces along the slope are in equilibrium. The driving force \(D\) must balance the resistance force and the component of weight down the slope: \(D = F_{\text{resistance}} + mg \sin \alpha = 15 + 80 \times 10 \times 0.05 = 15 + 40 = 55\text{ N}\). The power \(P\) is given by \(P = D v\), where \(v = 4\text{ m s}^{-1}\): \(P = 55 \times 4 = 220\text{ W}\).

M1: Formulate the driving force equation by summing the constant resistance and the component of weight down the slope. A1: Calculate the power using \(P = Dv\) to obtain \(220\text{ W}\).

The particle is at instantaneous rest when its velocity \(v = 0\). Setting the velocity expression to zero: \(1.5t^2 - 12t + 18 = 0 \implies t^2 - 8t + 12 = 0\). Factoring the quadratic: \((t-2)(t-6) = 0\). The positive times at which the particle is at rest are \(t=2\) and \(t=6\). Therefore, it first comes to rest when \(t = 2\text{ s}\). The acceleration \(a\) is the derivative of velocity with respect to time: \(a = \frac{\mathrm{d}v}{\mathrm{d}t} = 3t - 12\). Substituting \(t = 2\) into the acceleration formula: \(a = 3(2) - 12 = -6\text{ m s}^{-2}\).

M1: Solve \(v = 0\) to find the first positive value of \(t\) when the particle is at rest. M1: Differentiate velocity with respect to time to find the acceleration function. A1: Correctly calculate the acceleration at this instant as \(-6\text{ m s}^{-2}\).

Let \( R \) be the normal reaction force and \( F \) be the frictional force.

Resolving forces perpendicular to the plane:
\( R = mg \cos 30^\circ = 5 \times 10 \times \cos 30^\circ = 25\sqrt{3} \approx 43.301 \text{ N} \)

Since the block is on the point of sliding up the plane, the frictional force \( F \) acts down the plane and is at its maximum value:
\( F = \mu R = 25\sqrt{3}\mu \)

Resolving forces parallel to the plane:
\( P = mg \sin 30^\circ + F \)
\( 40 = 5 \times 10 \times \sin 30^\circ + 25\sqrt{3}\mu \)
\( 40 = 25 + 25\sqrt{3}\mu \)
\( 15 = 25\sqrt{3}\mu \)
\( \mu = \frac{15}{25\sqrt{3}} = \frac{3}{5\sqrt{3}} = \frac{\sqrt{3}}{5} \approx 0.346 \)

M1: Resolving forces parallel to the plane with \( P \), \( mg \sin 30^\circ \) and friction \( F \).
A1: Correct parallel equation: \( 40 = 25 + F \).
M1: Resolving forces perpendicular to the plane to find \( R = 25\sqrt{3} \) and using \( F = \mu R \).
A1.5: Obtains \( \mu = \frac{\sqrt{3}}{5} \approx 0.346 \) (3 s.f.).

First, find when the velocity is zero to determine when the particle changes direction:
\( 3t^2 - 12t + 9 = 0 \Rightarrow 3(t^2 - 4t + 3) = 0 \Rightarrow 3(t-1)(t-3) = 0 \)
So, the velocity changes sign at \( t = 1 \) and \( t = 3 \).

Integrate the velocity to find the displacement function \( s(t) \), assuming \( s(0) = 0 \):
\( s(t) = \int (3t^2 - 12t + 9) \mathrm{d}t = t^3 - 6t^2 + 9t \)

Evaluate the displacement at the key times:
At \( t = 0 \): \( s(0) = 0 \)
At \( t = 1 \): \( s(1) = 1^3 - 6(1)^2 + 9(1) = 4 \)
At \( t = 3 \): \( s(3) = 3^3 - 6(3)^2 + 9(3) = 0 \)
At \( t = 4 \): \( s(4) = 4^3 - 6(4)^2 + 9(4) = 4 \)

Calculate the distance in each interval:
From \( t = 0 \) to \( t = 1 \): \( |4 - 0| = 4 \text{ m} \)
From \( t = 1 \) to \( t = 3 \): \( |0 - 4| = 4 \text{ m} \)
From \( t = 3 \) to \( t = 4 \): \( |4 - 0| = 4 \text{ m} \)

Total distance travelled = \( 4 + 4 + 4 = 12 \text{ m} \).

M1: Solves \( v = 0 \) to identify the times at which the direction of motion changes (\( t = 1, 3 \)).
M1: Integrates \( v \) to find the displacement function.
A1: Correct displacement expression \( s(t) = t^3 - 6t^2 + 9t \) (with or without constant of integration).
M1: Calculates displacement at \( t = 0, 1, 3, 4 \) and sums the individual distances.
A0.5: Correct total distance of \( 12 \text{ m} \).

Let the initial direction of motion of \( A \) be the positive direction.
Initial velocities:
\( u_A = 6 \text{ m/s} \)
\( u_B = -2 \text{ m/s} \)

Final velocity of \( A \):
\( v_A = -1 \text{ m/s} \)

Let \( v_B \) be the final velocity of \( B \). Applying the principle of conservation of momentum:
\( m_A u_A + m_B u_B = m_A v_A + m_B v_B \)
\( 0.4(6) + 0.6(-2) = 0.4(-1) + 0.6 v_B \)
\( 2.4 - 1.2 = -0.4 + 0.6 v_B \)
\( 1.2 = -0.4 + 0.6 v_B \)
\( 1.6 = 0.6 v_B \Rightarrow v_B = \frac{8}{3} \approx 2.67 \text{ m/s} \)
So the final speed of \( B \) is \( 2.67 \text{ m/s} \).

The impulse \( I \) exerted by \( A \) on \( B \) is equal to the change in momentum of \( B \):
\( I = m_B (v_B - u_B) = 0.6 \left(\frac{8}{3} - (-2)\right) = 0.6 \left(\frac{14}{3}\right) = 2.8 \text{ N s} \)
(Alternatively, using sphere \( A \): \( |I| = |0.4(-1 - 6)| = 2.8 \text{ N s} \)).

M1: Applies the principle of conservation of linear momentum with consistent signs.
A1: Correct momentum equation: \( 0.4(6) + 0.6(-2) = 0.4(-1) + 0.6 v_B \).
A1: Correct final speed of \( B \) as \( 2.67 \text{ m/s} \) (or \( \frac{8}{3} \)).
M1: Uses \( I = m \Delta v \) to find the change in momentum of either particle.
A0.5: Correct magnitude of impulse of \( 2.8 \text{ N s} \).

Let \( T \) be the tension in the string and \( a \) be the acceleration of the system. Since \( m_Q > m_P \), \( Q \) moves downwards and \( P \) moves upwards.

Applying Newton's second law to each particle:
For \( Q \): \( 5g - T = 5a \Rightarrow 50 - T = 5a \)
For \( P \): \( T - 3g = 3a \Rightarrow T - 30 = 3a \)

Adding these two equations:
\( 20 = 8a \Rightarrow a = 2.5 \text{ m/s}^2 \)

To find the time taken for \( Q \) to reach the floor, we use the equation of motion:
\( s = ut + \frac{1}{2}at^2 \)
With \( s = 2 \text{ m} \), \( u = 0 \text{ m/s} \), and \( a = 2.5 \text{ m/s}^2 \):
\( 2 = 0 + \frac{1}{2}(2.5)t^2 \)
\( 1.25t^2 = 2 \)
\( t^2 = 1.6 \Rightarrow t = \sqrt{1.6} \approx 1.26 \text{ s} \)

M1: Sets up Newton's second law for both particles \( P \) and \( Q \).
A1: Correct equations: \( 50 - T = 5a \) and \( T - 30 = 3a \).
A1: Correct acceleration \( a = 2.5 \text{ m/s}^2 \).
M1: Uses a constant acceleration formula to find \( t \) with \( s = 2 \).
A0.5: Correct time \( t \approx 1.26 \text{ s} \) (3 s.f.).

Let the constant power of the engine in watts be \( P_{watts} = 1000P \).
The driving force is given by \( F_D = \frac{1000P}{v} \).

Newton's second law along the direction of motion gives:
\( F_D - R = ma \Rightarrow \frac{1000P}{v} - R = 1200 a \)

Substitute the two given conditions into the equation:
1) When \( v = 15 \) and \( a = 0.8 \):
\( \frac{1000P}{15} - R = 1200(0.8) \Rightarrow \frac{200P}{3} - R = 960 \) --- (Equation 1)

2) When \( v = 25 \) and \( a = 0.2 \):
\( \frac{1000P}{25} - R = 1200(0.2) \Rightarrow 40P - R = 240 \) --- (Equation 2)

Subtract Equation 2 from Equation 1 to eliminate \( R \):
\( \left(\frac{200}{3} - 40\right)P = 960 - 240 \)
\( \frac{80}{3}P = 720 \)
\( P = \frac{720 \times 3}{80} = 27 \)

Substitute \( P = 27 \) into Equation 2:
\( 40(27) - R = 240 \)
\( 1080 - R = 240 \Rightarrow R = 840 \)

Thus, \( P = 27 \) and \( R = 840 \).

M1: Applies \( \text{Driving Force} - R = ma \) with \( \text{Driving Force} = \frac{1000P}{v} \).
A1: Obtains two correct simultaneous equations in \( P \) and \( R \).
M1: Solves the system of equations by eliminating one variable.
A1: Correctly finds \( P = 27 \).
A0.5: Correctly finds \( R = 840 \).

Since the system is in equilibrium, the sum of the components in both the horizontal and vertical directions must equal zero.

Resolving components horizontally (along the \( x \)-axis):
\( 12 - F \cos \theta = 0 \Rightarrow F \cos \theta = 12 \) --- (Equation 1)

Resolving components vertically (along the \( y \)-axis):
\( F \sin \theta - 15 = 0 \Rightarrow F \sin \theta = 15 \) --- (Equation 2)

Divide Equation 2 by Equation 1:
\( \tan \theta = \frac{15}{12} = 1.25 \)
\( \theta = \arctan(1.25) \approx 51.34^\circ \approx 51.3^\circ \)

Using the Pythagorean identity on Equations 1 and 2 to find \( F \):
\( F^2 = 12^2 + 15^2 = 144 + 225 = 369 \)
\( F = \sqrt{369} \approx 19.2 \text{ N} \)

M1: Resolves forces horizontally to find \( F \cos \theta = 12 \).
M1: Resolves forces vertically to find \( F \sin \theta = 15 \).
A1: Finds the correct angle \( \theta \approx 51.3^\circ \) (or \( 0.896 \text{ radians} \)).
M1: Uses squaring and adding or trigonometric substitution to solve for \( F \).
A0.5: Finds the correct magnitude \( F \approx 19.2 \) (or \( 3\sqrt{41} \)).

First, find the time taken for the third stage (deceleration phase):
\( t_3 = \text{Total Time} - t_1 - t_2 = 20 - 6 - 10 = 4 \text{ s} \).

Now set up the distance expressions for each of the three stages:
1) Acceleration stage: \( s_1 = \frac{u + v}{2} t_1 = \frac{3 + v}{2} \times 6 = 3(3 + v) = 9 + 3v \).
2) Constant speed stage: \( s_2 = v t_2 = 10v \).
3) Deceleration stage: \( s_3 = \frac{v + 0}{2} t_3 = \frac{v}{2} \times 4 = 2v \).

The total distance is \( 135 \text{ m} \):
\( s_1 + s_2 + s_3 = 135 \)
\( (9 + 3v) + 10v + 2v = 135 \)
\( 9 + 15v = 135 \)
\( 15v = 126 \Rightarrow v = 8.4 \text{ m/s} \).

Now, find the acceleration \( a \) in the first stage:
\( v = u + a t_1 \)
\( 8.4 = 3 + 6a \)
\( 5.4 = 6a \Rightarrow a = 0.9 \text{ m/s}^2 \).

M1: Deduces the time for the deceleration phase is \( 4 \text{ s} \).
M1: Formulates expressions for the distances of all three stages in terms of \( v \).
A1: Equates the total distance expression to \( 135 \) and solves to get \( v = 8.4 \text{ m/s} \).
M1: Uses constant acceleration formula \( v = u + at \) (or equivalent) to find \( a \).
A0.5: Correct acceleration \( a = 0.9 \text{ m/s}^2 \).

First, we find the driving force \(F\) of the engine at the instant when the speed \(v = 16\text{ m s}^{-1}\). Using the power formula \(P = Fv\), we get \(F = \frac{P}{v} = \frac{32000\text{ W}}{16\text{ m s}^{-1}} = 2000\text{ N}\). Next, we calculate the component of the weight acting down the slope: \(W_{\text{slope}} = mg \sin \theta = 1200 \times 10 \times 0.05 = 600\text{ N}\). We apply Newton's second law along the direction of motion up the slope: \(F - R - mg \sin \theta = ma\), where \(R = 400\text{ N}\) is the resistance force and \(m = 1200\text{ kg}\). Substituting the values, we obtain \(2000 - 400 - 600 = 1200a\), which simplifies to \(1000 = 1200a\). Solving for \(a\) gives \(a = \frac{5}{6} \approx 0.833\text{ m s}^{-2}\).

M1: Use \(P = Fv\) to find the driving force \(F\). A1: Obtain \(F = 2000\text{ N}\). M1: Apply Newton's second law up the plane: \(F - R - mg \sin \theta = ma\). A1: Correctly substitute all values: \(2000 - 400 - 600 = 1200a\). A1: Obtain \(a = 0.833\text{ m s}^{-2}\) (or \(\frac{5}{6}\)).

We know that for any two events \(A\) and \(B\):
\(\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B)\)

Substituting the given values:
\(0.76 = 0.6 + 0.4 - \text{P}(A \cap B)\)
\(\text{P}(A \cap B) = 1.0 - 0.76 = 0.24\)

Now, we test for independence:
\(\text{P}(A) \times \text{P}(B) = 0.6 \times 0.4 = 0.24\)

Since \(\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B)\), the events \(A\) and \(B\) are independent.

M1: Use the addition rule to find \(\text{P}(A \cap B)\).
A1: Correctly calculate \(\text{P}(A \cap B) = 0.24\).
A1: Compare with \(\text{P}(A) \times \text{P}(B)\) and conclude they are independent.

The heights are already arranged in ascending order:
8, 9, 12, 14, 15, 17, 20, 21, 23

There are \(n = 9\) observations. The position of the median is:
\(\frac{n + 1}{2} = \frac{10}{2} = 5\)th value.

The 5th value in the ordered list is 15.

M1: Identify the middle value or 5th term in the ordered list.
A1: State the correct median of 15.

Treat the group of vowels (E, E, A) as a single block.
This leaves the 5 consonants: L, P, H, N, T.

Now we have 6 entities to arrange: the block (EEA) and the 5 consonants.
These 6 entities can be arranged in \(6! = 720\) ways.

Within the block, the letters E, E, A can be arranged in:
\(\frac{3!}{2!} = 3\) ways (since there are two identical Es).

Total number of arrangements = \(720 \times 3 = 2160\).

M1: Treat the vowels as a block and calculate \(6!\).
M1: Calculate the arrangements of the vowels within the block as \(\frac{3!}{2!}\) or 3.
A1: Correct final answer of 2160.

First, we find \(k\) using the property that the sum of probabilities is 1:
\(0.3 + k + 0.5 = 1 \implies k = 0.2\)

Next, we calculate the expected value \(\text{E}(X)\):
\(\text{E}(X) = \sum x \cdot \text{P}(X=x)\)
\(\text{E}(X) = (1 \times 0.3) + (2 \times 0.2) + (3 \times 0.5)\)
\(\text{E}(X) = 0.3 + 0.4 + 1.5 = 2.2\)

M1: Find the value of \(k = 0.2\).
M1: Use the formula \(\sum x \cdot \text{P}(X=x)\) with their value of \(k\).
A1: Obtain 2.2.

We standardise the variable \(W\):
\(\text{P}\left(Z > \frac{135 - 120}{\sigma}\right) = 0.1587\)

Since this probability is less than 0.5, the z-score must be positive.
\(\Phi\left(\frac{15}{\sigma}\right) = 1 - 0.1587 = 0.8413\)

Using normal distribution tables:
\(\Phi(1.0) = 0.8413\)

Therefore, we set:
\(\frac{15}{\sigma} = 1\)
\(\sigma = 15\)

M1: Standardise and set equal to a z-value from the normal table.
A1: Identify the correct z-value as 1.0 (or 1.00).
A1: Correctly calculate \(\sigma = 15\).

We identify the possible combinations of men and women that meet the condition "at least 3 women":

Case 1: 3 women and 1 man
Number of ways = \(\binom{4}{3} \times \binom{5}{1} = 4 \times 5 = 20\)

Case 2: 4 women and 0 men
Number of ways = \(\binom{4}{4} \times \binom{5}{0} = 1 \times 1 = 1\)

Total number of committees = \(20 + 1 = 21\).

M1: Identify the two cases (3W, 1M and 4W) and show addition.
M1: Calculate combinations for at least one case correctly.
A1: Obtain the correct total of 21.

Let \(X\) be the number of seeds that germinate. \(X \sim \text{B}(6, 0.8)\).

We want to find \(\text{P}(X = 5)\):
\(\text{P}(X = 5) = \binom{6}{5} (0.8)^5 (0.2)^1\)
\(\text{P}(X = 5) = 6 \times 0.32768 \times 0.2 = 0.393216\)

To 3 significant figures, the probability is 0.393.

M1: Apply the binomial probability formula with \(n=6\), \(p=0.8\), and \(x=5\).
A1: Calculate \(0.393216\) or 0.393.

First, find the total number of arrangements of the 8 letters of the word PARABOLA. The word contains 8 letters, with the letter 'A' repeating 3 times. Total arrangements = \( \frac{8!}{3!} = \frac{40320}{6} = 6720 \). Next, find the number of arrangements where all three 'A's are next to each other. We can treat the three 'A's as a single entity, (AAA). The entities to arrange are (AAA), B, L, O, P, R. There are 6 entities, which can be arranged in \( 6! = 720 \) ways. To find the number of arrangements where the three 'A's are not all next to each other, subtract this from the total: Number of arrangements = \( 6720 - 720 = 6000 \).

M1: For calculating total arrangements \( \frac{8!}{3!} \). A1: For obtaining 6720. M1: For treating the three 'A's as a single unit and calculating \( 6! \). A1: For subtracting to get the final correct answer of 6000.

Let \( A \) be the event that at least one of the dice shows a 2. Let \( B \) be the event that the sum \( X \ge 5 \). We want to find \( \mathrm{P}(B | A) = \frac{\mathrm{P}(B \cap A)}{\mathrm{P}(A)} \. First, list all outcomes where at least one of the dice shows a 2: \) \{(2, 1), (2, 2), (2, 3), (2, 4), (1, 2), (3, 2), (4, 2)\} \). There are 7 such outcomes, so \( \mathrm{P}(A) = \frac{7}{16} \). Now, from these 7 outcomes, identify those where the sum \( X \ge 5 \): \( (2, 3) \), \( (2, 4) \), \( (3, 2) \), and \( (4, 2) \). There are 4 such outcomes, so \( \mathrm{P}(B \cap A) = \frac{4}{16} \). Thus, the conditional probability is \( \mathrm{P}(B | A) = \frac{4/16}{7/16} = \frac{4}{7} \).

M1: For listing or identifying the set of outcomes where at least one die shows a 2 (7 outcomes). A1: For stating \( \mathrm{P}(A) = \frac{7}{16} \) (or equivalent). M1: For identifying the subset of these outcomes that sum to 5 or more (4 outcomes). A1: For the correct probability of \( \frac{4}{7} \) (or exact equivalent).

First, find the value of \( k \) using the fact that the sum of probabilities is 1: \( \mathrm{P}(X = 0) = k \), \( \mathrm{P}(X = 1) = 2k \), \( \mathrm{P}(X = 2) = 5k \), \( \mathrm{P}(X = 3) = 10k \). Sum of probabilities: \( k + 2k + 5k + 10k = 18k = 1 \implies k = \frac{1}{18} \). Next, calculate the expectation \( \mathrm{E}(X) \): \( \mathrm{E}(X) = 0(k) + 1(2k) + 2(5k) + 3(10k) = 42k = \frac{42}{18} = \frac{7}{3} \). Next, calculate \( \mathrm{E}(X^2) \): \( \mathrm{E}(X^2) = 0^2(k) + 1^2(2k) + 2^2(5k) + 3^2(10k) = 2k + 20k + 90k = 112k = \frac{112}{18} = \frac{56}{9} \). Finally, calculate the variance \( \mathrm{Var}(X) \): \( \mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = \frac{56}{9} - (\frac{7}{3})^2 = \frac{56}{9} - \frac{49}{9} = \frac{7}{9} \).

M1: For summing the probabilities in terms of \( k \), setting to 1, and finding \( k = \frac{1}{18} \). M1: For calculating \( \mathrm{E}(X) \) using their \( k \). M1: For calculating \( \mathrm{E}(X^2) \) using their \( k \). A1: For obtaining the correct variance of \( \frac{7}{9} \).

Let \( L \) represent the length of a metal rod. We are given \( L \sim \mathrm{N}(12.0, \sigma^2) \) and \( \mathrm{P}(L < 11.2) = 0.12 \). Standardizing this equation gives \( \mathrm{P}(Z < \frac{11.2 - 12.0}{\sigma}) = 0.12 \), which simplifies to \( \Phi(\frac{-0.8}{\sigma}) = 0.12 \). Since \( 0.12 < 0.5 \), the z-score must be negative. Thus: \( \frac{-0.8}{\sigma} = -\Phi^{-1}(1 - 0.12) = -\Phi^{-1}(0.88) \). From standard normal tables, \( \Phi^{-1}(0.88) \approx 1.175 \). So: \( \frac{-0.8}{\sigma} = -1.175 \implies \sigma = \frac{0.8}{1.175} \approx 0.68085 \). To 3 significant figures, \( \sigma = 0.681 \).

M1: For standardizing and setting up an equation with a z-value (e.g., \( \frac{11.2 - 12}{\sigma} \)). M1: For finding the correct critical value from standard normal tables corresponding to 0.88 (accept 1.17 to 1.18). M1: For equating standardisation expression to negative of their z-value. A1: For obtaining 0.681.

First, find the corrected sum of times: \( \sum t_{\text{new}} = 640 - 45 + 25 = 620 \). Next, find the corrected sum of squares of times: \( \sum t^2_{\text{new}} = 21120 - 45^2 + 25^2 = 21120 - 2025 + 625 = 19720 \). Now calculate the corrected mean: \( \bar{t}_{\text{new}} = \frac{620}{20} = 31\text{ seconds} \). Next, calculate the corrected variance: \( \text{Variance} = \frac{\sum t^2_{\text{new}}}{20} - (\bar{t}_{\text{new}})^2 = \frac{19720}{20} - 31^2 = 986 - 961 = 25 \). The corrected standard deviation is the square root of the corrected variance: \( \text{Standard Deviation} = \sqrt{25} = 5\text{ seconds} \).

M1: For calculating the corrected sum of times \( \sum t = 620 \). M1: For calculating the corrected sum of squares \( \sum t^2 = 19720 \). A1: For the corrected mean of 31. M1: For using the formula \( \frac{\sum t^2}{n} - (\bar{t})^2 \) with their corrected values. A1: For the corrected standard deviation of 5.

Let \( X \) be the number of germinating seeds. Since the seeds are independent and each has a germination probability of 0.8, we have \( X \sim \mathrm{B}(8, 0.8) \). We want to find \( \mathrm{P}(X \ge 7) = \mathrm{P}(X = 7) + \mathrm{P}(X = 8) \). Using the binomial probability formula: \( \mathrm{P}(X = 7) = \binom{8}{7} (0.8)^7 (0.2)^1 = 8 \times 0.2097152 \times 0.2 = 0.33554432 \). \( \mathrm{P}(X = 8) = (0.8)^8 = 0.16777216 \). Summing these probabilities: \( \mathrm{P}(X \ge 7) = 0.33554432 + 0.16777216 = 0.50331648 \). To 3 significant figures, the probability is 0.503.

M1: For recognizing and using the binomial distribution model \( \mathrm{B}(8, 0.8) \). M1: For expressing the required probability as \( \mathrm{P}(X = 7) + \mathrm{P}(X = 8) \). A1: For calculating both individual probabilities correctly. A1: For obtaining the correct final probability of 0.503.

To ensure that no two brothers stand next to each other, we place the 5 friends first and then place the brothers in the gaps between them.

1. Arrange the 5 friends in a line: there are \(5! = 120\) ways to do this.

2. This arrangement of friends creates 6 potential gaps (including the two outer ends): \(_ F_1 _ F_2 _ F_3 _ F_4 _ F_5 _\).

3. Select and arrange the 3 brothers into 3 of these 6 gaps: there are \(^{6}P_{3} = 6 \times 5 \times 4 = 120\) ways (which is equivalent to \(\binom{6}{3} \times 3! = 20 \times 6 = 120\)).

4. Multiply the two components to get the total number of arrangements: \(120 \times 120 = 14400\).

M1 for calculating the number of ways to arrange the 5 friends (\(5!\) or 120)
M1 for identifying 6 gaps and calculating the number of ways to place the 3 brothers in these gaps (\(^{6}P_{3}\) or \(\binom{6}{3} \times 3!\) or 120)
M1 for multiplying their two components together
A1 for the correct final answer of 14400

First, we find the mean rate for a 30-minute period: \(\lambda = 3.2 \div 2 = 1.6\) inquiries. Let \(X\) be the number of inquiries received in 30 minutes, so \(X \sim \text{Po}(1.6)\). The probability of receiving exactly 2 inquiries is given by: \(\text{P}(X = 2) = \frac{e^{-1.6} \times 1.6^2}{2!} = e^{-1.6} \times 1.28 \approx 0.2584\). To 3 significant figures, this is 0.258.

M1 for finding the correct mean \(\lambda = 1.6\) and attempting to calculate \(\text{P}(X=2)\) using the Poisson formula. A1 for obtaining 0.258.

The formula for the unbiased estimate of the population variance is: \(s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)\). Substituting the given values: \(s^2 = \frac{1}{80-1} \left( 3620 - \frac{520^2}{80} \right) = \frac{1}{79} \left( 3620 - 3380 \right) = \frac{240}{79} \approx 3.03797\). To 3 significant figures, the unbiased estimate of the population variance is 3.04.

M1 for a correct substitution into the formula for the unbiased variance estimate. A1 for obtaining 3.04.

Since \(f(x)\) is a probability density function, the total area under the curve must equal 1: \(\int_{0}^{3} kx^2 \text{d}x = 1\). Integrating the function: \(\left[ \frac{kx^3}{3} \right]_{0}^{3} = 9k = 1\). Solving for \(k\) gives \(k = \frac{1}{9}\).

M1 for integrating \(kx^2\) from 0 to 3 and setting the result equal to 1. A1 for obtaining 1/9 (or 0.111).

Using the properties of variance for independent random variables: \(\text{Var}(W) = \text{Var}(3X - 2Y) = 3^2 \text{Var}(X) + (-2)^2 \text{Var}(Y) = 9 \text{Var}(X) + 4 \text{Var}(Y)\). Substituting the given variances: \(\text{Var}(W) = 9(5) + 4(8) = 45 + 32 = 77\).

M1 for applying the variance formula \(\text{Var}(aX - bY) = a^2 \text{Var}(X) + b^2 \text{Var}(Y)\). A1 for obtaining 77.

A Type I error occurs when the null hypothesis is rejected when it is actually true. Under \(H_0\), the number of heads \(X\) has a binomial distribution: \(X \sim \text{B}(10, 0.5)\). The probability of a Type I error is \(\text{P}(X \ge 9) = \text{P}(X = 9) + \text{P}(X = 10)\). Calculating these probabilities: \(\text{P}(X = 9) = \binom{10}{9} (0.5)^{10} = \frac{10}{1024}\) and \(\text{P}(X = 10) = \binom{10}{10} (0.5)^{10} = \frac{1}{1024}\). Thus, \(\text{P}(X \ge 9) = \frac{11}{1024} \approx 0.010742\). To 3 significant figures, the probability is 0.0107.

M1 for identifying and calculating the binomial terms for \(X = 9\) and \(X = 10\) with \(p=0.5\). A1 for obtaining 0.0107.

For a large sample of size 150, the test statistic \(z\) is assumed to follow a standard normal distribution. A one-tailed test with \(H_1: \mu < \mu_0\) places the entire rejection region in the lower tail. At a 5% significance level, the critical value \(z\) is the value such that \(\Phi(z) = 0.05\). Using standard normal distribution tables, this gives \(z = -1.645\) (or -1.64).

B1 for a negative value. B1 for the correct magnitude of 1.645 or 1.64.

Let \(Y\) be the number of flaws in a 5-metre length. The mean rate is \(\lambda = 0.4 \times 5 = 2\). Thus, \(Y \sim \text{Po}(2)\). We want to find \(\text{P}(Y < 2) = \text{P}(Y = 0) + \text{P}(Y = 1) = e^{-2} + 2e^{-2} = 3e^{-2} \approx 0.406006\). To 3 significant figures, this is 0.406.

M1 for finding the correct mean \(\lambda = 2\) and expressing \(\text{P}(Y < 2)\) as \(\text{P}(Y=0) + \text{P}(Y=1)\). A1 for obtaining 0.406.

A \(95\%\) confidence interval for the population mean is \(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\), where \(z = 1.96\) is the critical value. The width of this interval is \(2 \times z \frac{\sigma}{\sqrt{n}}\). Substituting the given values: \(\text{Width} = 2 \times 1.96 \times \frac{4.5}{\sqrt{120}} = 3.92 \times \frac{4.5}{10.95445} \approx 1.6103\). To 3 significant figures, the width is 1.61.

M1 for identifying the correct formula for the width, \(2 \times 1.96 \times \frac{4.5}{\sqrt{120}}\). A1 for obtaining 1.61.

We are given that \(X \sim \text{Po}(\lambda)\), so the probability mass function is given by:
\[\text{P}(X = r) = \frac{e^{-\lambda} \lambda^r}{r!}\] for \(r = 0, 1, 2, \dots\]

Using this formula for \)r = 2\) and \(r = 3\), we have:
\[\text{P}(X = 2) = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^2}{2}\]
\[\text{P}(X = 3) = \frac{e^{-\lambda} \lambda^3}{3!} = \frac{e^{-\lambda} \lambda^3}{6}\]

Substitute these into the given equation \(3\text{P}(X = 2) = 2\text{P}(X = 3)\):
\[3 \left(\frac{e^{-\lambda} \lambda^2}{2}\right) = 2 \left(\frac{e^{-\lambda} \lambda^3}{6}\right)\]

Simplify both sides:
\[\frac{3}{2} e^{-\lambda} \lambda^2 = \frac{1}{3} e^{-\lambda} \lambda^3\]

Since \(\lambda > 0\), we know that \(\lambda^2 \neq 0\) and \(e^{-\lambda} \neq 0\). We can therefore divide both sides by \(e^{-\lambda} \lambda^2\):
\[\frac{3}{2} = \frac{1}{3} \lambda\]

Multiply both sides by 3 to solve for \(\lambda\):
\[\lambda = \frac{9}{2} = 4.5\]

M1: For substituting the correct Poisson probability formula for \(\text{P}(X = 2)\) and \(\text{P}(X = 3)\) into the given equation.
A1: For obtaining \(\lambda = 4.5\) (or \(\frac{9}{2}\)).

Let \(X\) be the number of calls in a 30-minute period. Since calls occur randomly and at a constant average rate, \(X\) can be modeled by a Poisson distribution. The mean rate for a 10-minute period is 1.8, so for a 30-minute period, the mean rate is \(\lambda = 1.8 \times 3 = 5.4\). We need to find \(P(X \le 3)\). Using the Poisson probability formula, \(P(X \le 3) = e^{-5.4} (1 + 5.4 + \frac{5.4^2}{2!} + \frac{5.4^3}{3!})\). Evaluating the terms inside the parentheses gives \(1 + 5.4 + 14.58 + 26.244 = 47.224\). Thus, \(P(X \le 3) = e^{-5.4} \times 47.224 \approx 0.00451658 \times 47.224 = 0.213\) (to 3 significant figures).

M1 for calculating the correct mean \(\lambda = 5.4\) for the 30-minute period. M1 for expressing \(P(X \le 3)\) as the sum of 4 Poisson terms with their \(\lambda\). A1 for correct evaluation of the sum inside the brackets to 47.224 (or equivalent). A1.5 for the correct final probability of 0.213 (accept answers in the range 0.213 to 0.214).

First, we determine the constant \(k\) by integrating the probability density function over its non-zero domain and setting it equal to 1: \(\int_0^2 k(4 - x^2) dx = 1\). Integrating gives \(k [4x - \frac{x^3}{3}]_0^2 = 1\), which simplifies to \(k (8 - \frac{8}{3}) = 1\), so \(k \frac{16}{3} = 1\), yielding \(k = \frac{3}{16}\). Next, we find the cumulative distribution function \(F(x)\) for \(0 \le x \le 2\) by integrating \(f(t)\) from 0 to \(x\): \(F(x) = \int_0^x \frac{3}{16}(4 - t^2) dt = \frac{3}{16} [4t - \frac{t^3}{3}]_0^x = \frac{3}{16} (4x - \frac{x^3}{3}) = \frac{3}{4}x - \frac{1}{16}x^3\).

M1 for integrating the pdf from 0 to 2 and setting equal to 1 to find \(k\). A1 for obtaining \(k = \frac{3}{16}\). M1 for integrating their \(f(t)\) from 0 to \(x\) (or using indefinite integration with \(F(0)=0\) or \(F(2)=1\)). A1.5 for the correct final expression \(F(x) = \frac{3}{4}x - \frac{1}{16}x^3\) (or equivalent factored form).

We are given \(n = 80\), \(\bar{x} = 112.4\), and the unbiased variance estimate \(s^2 = 240.5\). For a 98% confidence interval, the critical value \(z\) is obtained from the standard normal distribution such that \(\Phi(z) = 0.99\), which gives \(z \approx 2.326\). The confidence interval is calculated using the formula \(\bar{x} \pm z \sqrt{\frac{s^2}{n}}\). Substituting the values, we get \(112.4 \pm 2.326 \sqrt{\frac{240.5}{80}} = 112.4 \pm 2.326 \sqrt{3.00625} \approx 112.4 \pm 2.326 \times 1.73385 \approx 112.4 \pm 4.033\). This gives a lower limit of \(112.4 - 4.033 = 108.367\) and an upper limit of \(112.4 + 4.033 = 116.433\). Therefore, the 98% confidence interval is \([108.4, 116.4]\) (to 1 decimal place) or \([108, 116]\) (to 3 significant figures).

B1 for stating or using the correct critical value \(z = 2.326\) (or 2.327). M1 for substituting their values into the confidence interval formula \(\bar{x} \pm z \sqrt{\frac{s^2}{n}}\). A1 for correct standard error of \(\sqrt{3.00625} \approx 1.734\). A1.5 for the correct final confidence interval of \([108.4, 116.4]\) or \([108, 116]\) (accept limits in the range [108.3, 108.4] and [116.4, 116.5]).

Let \(L_1, L_2\) be two independent large boxes, and \(S_1, S_2, S_3, S_4\) be four independent small boxes. We define a new random variable \(D = L_1 + L_2 - (S_1 + S_2 + S_3 + S_4)\). Since \(L\) and \(S\) are normally distributed, \(D\) is also normally distributed. The mean of \(D\) is \(E(D) = 2 \times 750 - 4 \times 380 = 1500 - 1520 = -20\). The variance of \(D\) is \(\text{Var}(D) = 2 \times \text{Var}(L) + 4 \times \text{Var}(S) = 2 \times 12^2 + 4 \times 8^2 = 288 + 256 = 544\). Thus, \(D \sim \text{N}(-20, 544)\). We want to find \(P(D > 0)\). Standardizing gives \(P(D > 0) = P(Z > \frac{0 - (-20)}{\sqrt{544}}) = P(Z > \frac{20}{23.3238}) \approx P(Z > 0.8575)\). Using the standard normal table, \(P(Z > 0.8575) = 1 - \Phi(0.8575) \approx 1 - 0.8044 = 0.1956\). To 3 significant figures, the probability is 0.196.

M1 for finding the mean of the difference, \(E(D) = -20\). M1 for finding the variance of the difference, \(\text{Var}(D) = 544\). M1 for standardizing correctly with their mean and variance to get a \(z\)-value of 0.858. A1.5 for the correct final probability of 0.196 (accept 0.195 to 0.197).

(i) Since the total probability under the probability density function is 1, we must have:\
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\\int_{0}^{2} c(4x - x^3) dx = 1\
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c \\left[ 2x^2 - \\frac{x^4}{4} \\right]_{0}^{2} = 1\
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c \\left( \\left( 2(2)^2 - \\frac{2^4}{4} \\right) - 0 \\right) = 1\
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c (8 - 4) = 1 \\implies 4c = 1 \\implies c = \\frac{1}{4}.\
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(ii) Let \(m\) be the median of \(X\). Then:\
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\\int_{0}^{m} \\frac{1}{4}(4x - x^3) dx = 0.5\
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\\frac{1}{4} \\left[ 2x^2 - \\frac{x^4}{4} \\right]_{0}^{m} = 0.5\
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2m^2 - \\frac{m^4}{4} = 2\
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m^4 - 8m^2 + 8 = 0\
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Using the quadratic formula to solve for \(m^2\):\
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m^2 = \\frac{8 \\pm \\sqrt{64 - 32}}{2} = 4 \\pm 2\\sqrt{2}\
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Since the domain of \(X\) is \([0, 2]\), we must have \(0 \le m^2 \le 4\). Therefore, \(m^2 = 4 - 2\sqrt{2} \approx 1.1716\).\
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m = \\sqrt{4 - 2\\sqrt{2}} \\approx 1.08\
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(iii) Using conditional probability:\
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P(X > 1.5 \\mid X > 1) = \\frac{P(X > 1.5)}{P(X > 1)}\
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P(X > 1.5) = \\int_{1.5}^{2} \\frac{1}{4}(4x - x^3) dx = \\frac{1}{4} \\left[ 2x^2 - \\frac{x^4}{4} \\right]_{1.5}^{2} = \\frac{1}{4} \\left( 4 - \\left( 2(1.5)^2 - \\frac{(1.5)^4}{4} \\right) \\right) = \\frac{49}{256} \\approx 0.1914\
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P(X > 1) = \\int_{1}^{2} \\frac{1}{4}(4x - x^3) dx = \\frac{1}{4} \\left[ 2x^2 - \\frac{x^4}{4} \\right]_{1}^{2} = \\frac{1}{4} \\left( 4 - \\left( 2(1)^2 - \\frac{1}{4} \\right) \\right) = \\frac{9}{16} = 0.5625\
\
P(X > 1.5 \\mid X > 1) = \\frac{49/256}{9/16} = \\frac{49}{144} \\approx 0.340

(i) \
- M1: Attempts to integrate the probability density function and equate to 1.\
- A1: Shows sufficient integration and algebra to obtain \(c = \frac{1}{4}\) correctly.\
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(ii) \
- M1: Equates the integral from 0 to \(m\) to 0.5 (or equivalent) and integrates.\
- M1: Sets up and solves a quadratic in \(m^2\).\
- A1: Obtains \(m = \sqrt{4-2\sqrt{2}} \approx 1.08\).\
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(iii) \
- M1: Applies the conditional probability formula \\frac{P(X > 1.5)}{P(X > 1)} with an attempt to calculate both probabilities.\
- A1: Correctly calculates \\frac{49}{144} or \\approx 0.340\\.