2025 Cambridge IAL Mathematics (9709) Practice Paper with Answers

For the curve to lie entirely above the \( x \)-axis, two conditions must be met:

1. The coefficient of \( x^2 \) must be positive, which means \( k > 0 \).

2. The curve must not intersect or touch the \( x \)-axis, which means the discriminant of the quadratic equation \( kx^2 - 8x + (k+6) = 0 \) must be strictly negative (\( \Delta < 0 \)).

Calculate the discriminant:
\( \Delta = (-8)^2 - 4(k)(k+6) \)
\( \Delta = 64 - 4k^2 - 24k \)

Set \( \Delta < 0 \):
\( 64 - 4k^2 - 24k < 0 \)

Divide the entire inequality by \( -4 \) and reverse the inequality sign:
\( k^2 + 6k - 16 > 0 \)

Factorise the quadratic expression:
\( (k+8)(k-2) > 0 \)

The critical values are \( k = -8 \) and \( k = 2 \).
For the inequality to be greater than zero, we require:
\( k > 2 \) or \( k < -8 \)

Combining this with the condition \( k > 0 \), the only valid set of values is:
\( k > 2 \)

M1: Attempt to find the discriminant of the quadratic expression in terms of \( k \).
A1: Set the discriminant strictly less than zero to obtain \( 64 - 4k(k+6) < 0 \) (or equivalent).
M1: Correctly solve the quadratic inequality to find critical values \( k = -8, 2 \).
M1: State or imply that \( k > 0 \) is required for the curve to lie entirely above the \( x \)-axis.
A1: State the final correct range \( k > 2 \).

First, express \( \mathrm{f}(x) \) in completed square form:
\( \mathrm{f}(x) = 2(x^2 - 6x) + 13 \)
\( \mathrm{f}(x) = 2[(x-3)^2 - 9] + 13 \)
\( \mathrm{f}(x) = 2(x-3)^2 - 18 + 13 \)
\( \mathrm{f}(x) = 2(x-3)^2 - 5 \)

To find the inverse function, let \( y = 2(x-3)^2 - 5 \) and rearrange to make \( x \) the subject:
\( y + 5 = 2(x-3)^2 \)
\( \frac{y+5}{2} = (x-3)^2 \)
\( x - 3 = \pm\sqrt{\frac{y+5}{2}} \)

Since the domain of \( \mathrm{f} \) is \( x \le 2 \), we have \( x - 3 \le -1 \), which is negative. Therefore, we must choose the negative square root:
\( x - 3 = -\sqrt{\frac{y+5}{2}} \)
\( x = 3 - \sqrt{\frac{y+5}{2}} \)

So, the inverse function is:
\( \mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}} \)

To find the domain of \( \mathrm{f}^{-1} \), we find the range of \( \mathrm{f} \) for \( x \le 2 \):
At the boundary \( x = 2 \), \( \mathrm{f}(2) = 2(2-3)^2 - 5 = -3 \).
Since \( x \le 2 \) and the vertex is at \( x = 3 \), the function increases as \( x \) decreases from 2. Hence, the range of \( \mathrm{f} \) is \( \mathrm{f}(x) \ge -3 \).

Thus, the domain of \( \mathrm{f}^{-1} \) is \( x \ge -3 \).

M1: Attempt to express \( \mathrm{f}(x) \) in completed square form, obtaining \( 2(x-3)^2 - 5 \).
M1: Attempt to rearrange to make \( x \) the subject of \( y = 2(x-3)^2 - 5 \).
M1: Justify the choice of the negative square root using the domain constraint \( x \le 2 \).
A1: Obtain the correct inverse function expression \( \mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}} \).
A1: State the correct domain of \( \mathrm{f}^{-1} \) as \( x \ge -3 \) (or equivalent).

Let \( a \) be the first term and \( d \) be the common difference of the arithmetic progression (AP).
We are given \( a = 3 \).

The first three terms of the AP are:
\( T_1 = a = 3 \)
\( T_3 = a + 2d = 3 + 2d \)
\( T_{11} = a + 10d = 3 + 10d \)

These terms form a geometric progression, so the common ratio is constant:
\( \frac{3+2d}{3} = \frac{3+10d}{3+2d} \)

Cross-multiplying gives:
\( (3+2d)^2 = 3(3+10d) \)
\( 9 + 12d + 4d^2 = 9 + 30d \)
\( 4d^2 - 18d = 0 \)
\( 2d(2d - 9) = 0 \)

Since the common difference is non-zero (\( d \neq 0 \)):
\( 2d - 9 = 0 \implies d = 4.5 \) (or \( \frac{9}{2} \)).

Now, calculate the sum of the first \( 20 \) terms of the AP using the formula \( S_n = \frac{n}{2} [2a + (n-1)d] \):
\( S_{20} = \frac{20}{2} [2(3) + 19(4.5)] \)
\( S_{20} = 10 [6 + 85.5] \)
\( S_{20} = 10 [91.5] = 915 \)

M1: Write correct expressions for \( T_1 \), \( T_3 \), and \( T_{11} \) of the AP in terms of \( d \).
M1: Set up the geometric progression equation relating these three terms.
A1: Solve the quadratic equation in \( d \) to find the correct non-zero common difference \( d = 4.5 \).
M1: Apply the AP sum formula for \( n = 20 \) with \( a = 3 \) and their value of \( d \).
A1: Obtain the correct sum of \( 915 \).

To find the values of \( c \) for which the line is tangent to the circle, substitute \( y = 2x + c \) into the equation of the circle:
\( x^2 + (2x+c)^2 - 6x - 2(2x+c) + 5 = 0 \)

Expand the brackets:
\( x^2 + 4x^2 + 4cx + c^2 - 6x - 4x - 2c + 5 = 0 \)

Collect like terms to form a quadratic equation in \( x \):
\( 5x^2 + (4c - 10)x + (c^2 - 2c + 5) = 0 \)

Since the line is a tangent to the circle, this quadratic equation must have exactly one real root, meaning its discriminant must be zero (\( B^2 - 4AC = 0 \)):
\( (4c - 10)^2 - 4(5)(c^2 - 2c + 5) = 0 \)

Expand and simplify:
\( (16c^2 - 80c + 100) - 20(c^2 - 2c + 5) = 0 \)
\( 16c^2 - 80c + 100 - 20c^2 + 40c - 100 = 0 \)
\( -4c^2 - 40c = 0 \)

Factorise the equation:
\( -4c(c + 10) = 0 \)

This yields the solutions:
\( c = 0 \) or \( c = -10 \)

M1: Substitute \( y = 2x + c \) into the equation of the circle.
A1: Correctly expand and collect terms to obtain the quadratic equation \( 5x^2 + (4c-10)x + (c^2-2c+5) = 0 \).
M1: State that the discriminant is zero for tangency, and write down an equation in terms of \( c \).
M1: Simplify the discriminant equation to get a quadratic in \( c \), e.g. \( c^2 + 10c = 0 \).
A1: Obtain the two correct values: \( c = 0 \) and \( c = -10 \).

Rewrite the curve equation as:
\( y = 4x + 9(x-1)^{-1} \)

Differentiate with respect to \( x \):
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 4 - 9(x-1)^{-2} = 4 - \frac{9}{(x-1)^2} \)

To find the stationary point, set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\( 4 - \frac{9}{(x-1)^2} = 0 \)
\( 4 = \frac{9}{(x-1)^2} \)
\( (x-1)^2 = \frac{9}{4} \)
\( x - 1 = \pm 1.5 \)

Since \( x > 1 \), we must have \( x - 1 = 1.5 \), which gives:
\( x = 2.5 \) (or \( \frac{5}{2} \))

Substitute \( x = 2.5 \) back into the original curve equation to find the \( y \)-coordinate:
\( y = 4(2.5) + \frac{9}{2.5 - 1} = 10 + \frac{9}{1.5} = 10 + 6 = 16 \)

So the coordinates of the stationary point are \( (2.5, 16) \).

To determine the nature of the stationary point, find the second derivative:
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}[4 - 9(x-1)^{-2}] = 18(x-1)^{-3} = \frac{18}{(x-1)^3} \)

Substitute \( x = 2.5 \) into the second derivative:
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{18}{(2.5-1)^3} = \frac{18}{1.5^3} = \frac{18}{3.375} = \frac{16}{3} \)

Since \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \), the stationary point \( (2.5, 16) \) is a local minimum.

M1: Correctly differentiate to find \( \frac{\mathrm{d}y}{\mathrm{d}x} \).
M1: Set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) and solve for \( x \), selecting the value in the domain \( x > 1 \).
A1: Identify the coordinates of the stationary point as \( (2.5, 16) \) or equivalent fractional form.
M1: Find the second derivative expression \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{18}{(x-1)^3} \).
A1: Correctly evaluate the sign of the second derivative at \( x = 2.5 \) and conclude that the point is a minimum.

For (a), the terms of the arithmetic progression are \(x_1 = a\), \(x_2 = a + d\), and \(x_5 = a + 4d\). Since these correspond to the first three terms of a geometric progression, we have \(\frac{a+d}{a} = \frac{a+4d}{a+d}\). Cross-multiplying gives \((a+d)^2 = a(a+4d)\), which expands to \(a^2 + 2ad + d^2 = a^2 + 4ad\). Simplifying yields \(d^2 = 2ad\). Since \(d \ne 0\), we divide by \(d\) to get \(d = 2a\), or \(a = 0.5d\). The common ratio \(r\) is given by \(\frac{a+d}{a} = \frac{a+2a}{a} = \frac{3a}{a} = 3\). For (b), the sum of the first 8 terms of the arithmetic progression is \(S_8 = \frac{8}{2}(2a + 7d) = 4(2a + 7d) = 160\), so \(2a + 7d = 40\). Substituting \(a = 0.5d\) gives \(d + 7d = 40\), which simplifies to \(8d = 40\), so \(d = 5\). For (c), since \(d = 5\), we have \(a = 2.5\). The common ratio is \(r = 3\). The sum of the first 8 terms of the geometric progression is \(S_8 = \frac{a(r^8 - 1)}{r - 1} = \frac{2.5(3^8 - 1)}{3 - 1} = \frac{2.5(6561 - 1)}{2} = \frac{2.5 \times 6560}{2} = 8200\).

(a) M1: Setting up the geometric ratio equation \((a+d)/a = (a+4d)/(a+d)\). A1: Expanding and obtaining \(d = 2a\) or equivalent. M1: Showing that \(r = (a+d)/a\). A1: Correctly showing \(r = 3\) with clear algebraic steps. (b) M1: Using the sum of arithmetic progression formula for \(S_8 = 160\). M1: Substituting \(a = 0.5d\) into their equation. A1: Obtaining \(d = 5\). (c) M1: Using the geometric progression sum formula with \(r=3\) and their \(a\). A1: Correct substitution of \(a = 2.5\). A1: Obtaining 8200.

For (a), \(\text{f}(x) = 2(x^2 - 6x) + 13 = 2[(x-3)^2 - 9] + 13 = 2(x-3)^2 - 18 + 13 = 2(x-3)^2 - 5\). Thus, \(p = 3\) and \(q = -5\). For (b), the vertex of the quadratic is at \(x = 3\). For a quadratic function on a domain \(x \le k\) to be one-to-one (and thus have an inverse), the domain must not include the vertex and the other branch. Therefore, the largest value of \(k\) is 3. For (c), let \(y = 2(x-3)^2 - 5\). Re-arranging gives \(y + 5 = 2(x-3)^2\), so \((x-3)^2 = \frac{y+5}{2}\). Taking the negative square root because \(x \le 3\) gives \(x - 3 = -\sqrt{\frac{y+5}{2}}\), hence \(x = 3 - \sqrt{\frac{y+5}{2}}\). Thus, \(\text{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\). The domain of \(\text{f}^{-1}\) is the range of \(\text{f}\), which is \(y \ge -5\). Thus, the domain is \(x \ge -5\). For (d), we want \(\text{f}(\text{g}(x)) = 13\). Let \(\text{g}(x) = u\), so \(\text{f}(u) = 13\). This gives \(2(u-3)^2 - 5 = 13 \implies 2(u-3)^2 = 18 \implies (u-3)^2 = 9 \implies u - 3 = \pm 3\). So \(u = 6\) or \(u = 0\). Since the domain of \(\text{f}\) is \(x \le 3\), we must have \(\text{g}(x) \le 3\). Thus, \(u = 6\) is rejected because it is outside the domain of \(\text{f}\). We use \(u = 0\), which means \(\text{g}(x) = 2 - 3x = 0 \implies x = \frac{2}{3}\).

(a) B1: Correctly identifying \(p = 3\). B1: Correctly identifying \(q = -5\). (b) B1: Stating \(k = 3\). (c) M1: Attempting to make \(x\) the subject of the formula. M1: Choosing the negative square root due to domain restriction. A1: Correct expression for \(\text{f}^{-1}(x)\). B1: Correct domain \(x \ge -5\). (d) M1: Setting up equation \(\text{f}(\text{g}(x)) = 13\) and finding values for \(\text{g}(x)\). M1: Recognizing that \(\text{g}(x) \le 3\) and rejecting \(\text{g}(x) = 6\). A1: Finding \(x = \frac{2}{3}\) (or \(0.667\)).

For (a), equating the curve and the line gives \(x^2 - 3x + 5 = mx + 1\), which simplifies to \(x^2 - (3+m)x + 4 = 0\). For two distinct points of intersection, the discriminant must be greater than zero: \(\Delta = [-(3+m)]^2 - 4(1)(4) > 0 \implies (3+m)^2 - 16 > 0 \implies (3+m)^2 > 16\). Taking square roots gives \(3+m > 4\) or \(3+m < -4\), which yields \(m > 1\) or \(m < -7\). For (b)(i), substituting \(m = 2\) gives the equation \(x^2 - 5x + 4 = 0 \implies (x-1)(x-4) = 0\), so \(x = 1\) or \(x = 4\). The corresponding \(y\)-coordinates are found by substituting into the line equation \(y = 2x + 1\): for \(x = 1, y = 3\), giving \(A(1, 3)\); for \(x = 4, y = 9\), giving \(B(4, 9)\). For (b)(ii), the midpoint of \(AB\) is \(M = (\frac{1+4}{2}, \frac{3+9}{2}) = (2.5, 6)\). The gradient of the line \(AB\) is \(2\), so the gradient of the perpendicular bisector is \(-\frac{1}{2}\). The equation of the perpendicular bisector is \(y - 6 = -\frac{1}{2}(x - 2.5) \implies y - 6 = -0.5x + 1.25 \implies 2y - 12 = -x + 2.5 \implies 2x + 4y = 29\).

(a) M1: Equating the line and curve equations. M1: Attempting to find the discriminant of the resulting quadratic. A1: Obtaining the inequality \((3+m)^2 - 16 > 0\) or equivalent. A1: Correctly solving to get \(m < -7\) or \(m > 1\) (do not accept weak inequalities). (b)(i) M1: Solving the quadratic for \(x\) with \(m=2\). A1: Correct \(x\)-values. A1: Correct coordinates for \(A\) and \(B\). (b)(ii) M1: Finding the midpoint of \(AB\). M1: Finding the perpendicular gradient. A1: Correct equation in the specified form \(2x + 4y = 29\).

For (a)(i), in the right-angled triangle \(ORQ\), \(\cos\theta = \frac{OR}{OQ} = \frac{6}{r}\). Since \(\theta = \frac{\pi}{3}\), we have \(\cos(\frac{\pi}{3}) = \frac{6}{r} \implies 0.5 = \frac{6}{r} \implies r = 12\). For (a)(ii), the perimeter of the shaded region is the sum of arc length \(PQ\), line segment \(PR\), and line segment \(RQ\). Arc length \(PQ = r\theta = 12 \times \frac{\pi}{3} = 4\pi\). Line segment \(PR = OP - OR = r - 6 = 12 - 6 = 6\). In triangle \(ORQ\), \(RQ = OR \tan\theta = 6 \tan(\frac{\pi}{3}) = 6\sqrt{3}\). Thus, the exact perimeter is \(4\pi + 6 + 6\sqrt{3}\). For (b), the area of sector \(OPQ\) is \(\frac{1}{2}r^2\theta = 27\). Since \(r = \frac{6}{\cos\theta}\), we can substitute this to get \(\frac{1}{2}\left(\frac{6}{\cos\theta}\right)^2\theta = 27 \implies \frac{18\theta}{\cos^2\theta} = 27 \implies 18\theta = 27\cos^2\theta \implies \theta = \frac{27}{18}\cos^2\theta = 1.5\cos^2\theta\). For (c), the area of the shaded region is the area of the sector \(OPQ\) minus the area of the right-angled triangle \(ORQ\). Area of sector \(OPQ = \frac{1}{2}r^2\theta = \frac{18\theta}{\cos^2\theta} = 18\theta\sec^2\theta\). Area of triangle \(ORQ = \frac{1}{2} \times OR \times RQ = \frac{1}{2} \times 6 \times 6\tan\theta = 18\tan\theta\). Therefore, the area of the shaded region is \(18\theta\sec^2\theta - 18\tan\theta = 18(\theta\sec^2\theta - \tan\theta)\).

(a)(i) B1: Correctly stating \(r = 12\). (a)(ii) B1: Correct arc length \(4\pi\). B1: Correct length \(PR = 6\). M1: Attempting to find \(RQ\) using trigonometry. A1: Correct exact perimeter \(4\pi + 6 + 6\sqrt{3}\). (b) M1: Using the formula for sector area. M1: Substituting \(r = \frac{6}{\cos\theta}\) into the area formula. A1: Correctly simplifying to obtain \(\theta = 1.5\cos^2\theta\). (c) M1: Writing down an expression for the area of triangle \(ORQ\). A1: Correctly subtracting the area of the triangle from the area of the sector to obtain the given expression.

For (a), we find \(y\) by integrating \(\frac{\text{d}y}{\text{d}x}\): \(y = \int \left(6(3x+4)^{-1/2} - 2\right) \text{d}x = \frac{6(3x+4)^{1/2}}{(\frac{1}{2}) \times 3} - 2x + C = 4\sqrt{3x+4} - 2x + C\). Since the curve passes through \((4, 5)\), we substitute these coordinates: \(5 = 4\sqrt{3(4)+4} - 2(4) + C \implies 5 = 4(4) - 8 + C \implies 5 = 8 + C \implies C = -3\). So the equation of the curve is \(y = 4\sqrt{3x+4} - 2x - 3\). For (b), at a stationary point, \(\frac{\text{d}y}{\text{d}x} = 0 \implies \frac{6}{\sqrt{3x+4}} - 2 = 0 \implies \sqrt{3x+4} = 3 \implies 3x+4 = 9 \implies 3x = 5 \implies x = \frac{5}{3}\). Substituting \(x = \frac{5}{3}\) into the curve equation gives \(y = 4\sqrt{3(\frac{5}{3})+4} - 2(\frac{5}{3}) - 3 = 12 - \frac{10}{3} - 3 = \frac{17}{3}\). To determine the nature, we find the second derivative: \(\frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}}{\text{d}x}\left(6(3x+4)^{-1/2} - 2\right) = 6(-\frac{1}{2})(3)(3x+4)^{-3/2} = -\frac{9}{(3x+4)^{3/2}}\). Since \(3x+4 = 9 > 0\) at the stationary point, \(\frac{\text{d}^2y}{\text{d}x^2} = -\frac{9}{27} = -\frac{1}{3} < 0\), which indicates a maximum point. For (c), the curve crosses the \(y\)-axis where \(x=0\). At \(x=0\), \(y = 4\sqrt{4} - 0 - 3 = 5\). The gradient of the tangent at \(x=0\) is \(\frac{\text{d}y}{\text{d}x} = \frac{6}{\sqrt{4}} - 2 = 1\). Thus, the gradient of the normal is \(-1\). The equation of the normal is \(y - 5 = -1(x - 0) \implies y = -x + 5\).

(a) M1: Attempting to integrate \(\frac{\text{d}y}{\text{d}x}\). A1: Correct integration including division by 3. M1: Substituting \((4, 5)\) to find the constant of integration \(C\). A1: Correct final equation. (b) M1: Setting \(\frac{\text{d}y}{\text{d}x} = 0\) and solving for \(x\). A1: Finding \(x = \frac{5}{3}\) and \(y = \frac{17}{3}\). M1: Finding the second derivative. A1: Confirming the second derivative is negative and concluding it is a maximum. (c) M1: Finding the \(y\)-coordinate and tangent gradient at \(x=0\). A1: Correct equation of the normal, \(y = -x + 5\) or equivalent.

First, solve the inequality \( |2x - 3| < |x + 4| \) by squaring both sides: \[(2x - 3)^2 < (x + 4)^2\] \[4x^2 - 12x + 9 < x^2 + 8x + 16\] \[3x^2 - 20x - 7 < 0\] Factorise the quadratic expression: \[(3x + 1)(x - 7) < 0\] Thus, the solution is \( -\frac{1}{3} < x < 7 \). For the second part, let \( x = e^{-y} \). Since \( e^{-y} > 0 \) for all real values of \( y \), the inequality \( e^{-y} > -\frac{1}{3} \) is always satisfied. Thus, we only need to solve: \[e^{-y} < 7\] Taking the natural logarithm of both sides: \[-y < \ln 7\] \[y > -\ln 7\]

M1: For squaring both sides and obtaining a 3-term quadratic inequality.
A1: For obtaining \( 3x^2 - 20x - 7 < 0 \).
A1: For correct critical values \( x = -1/3 \) and \( x = 7 \).
A1: For the correct range \( -1/3 < x < 7 \).
M1: For substituting \( x = e^{-y} \) and identifying that only \( e^{-y} < 7 \) needs to be solved.
A1: For using logarithm laws correctly to solve for \( y \).
A1: For obtaining the final range \( y > -\ln 7 \) (or equivalent).

Rewrite the equation \( 3^{2x-1} + 3^x = 10 \) as: \[\frac{1}{3}(3^x)^2 + 3^x - 10 = 0\] Let \( u = 3^x \): \[\frac{1}{3}u^2 + u - 10 = 0 \implies u^2 + 3u - 30 = 0\] Solve the quadratic equation using the quadratic formula: \[u = \frac{-3 \pm \sqrt{3^2 - 4(1)(-30)}}{2} = \frac{-3 \pm \sqrt{129}}{2}\] Since \( u = 3^x \) must be positive, we reject the negative root: \[u = \frac{-3 + \sqrt{129}}{2} \approx 4.1789\] Now solve for \( x \): \[3^x = 4.1789\] \[x \ln 3 = \ln 4.1789\] \[x = \frac{\ln 4.1789}{\ln 3} \approx 1.30 \text{ (correct to 3 s.f.)}\]

M1: For expressing \( 3^{2x-1} \) as \( \frac{1}{3}(3^x)^2 \) or equivalent.
M1: For setting up a quadratic equation in terms of \( u = 3^x \) (e.g. \( u^2 + 3u - 30 = 0 \)).
A1: For obtaining the correct quadratic equation.
M1: For solving the quadratic equation to find a positive value for \( u \).
A1: For \( u \approx 4.179 \).
M1: For applying logarithms to solve \( 3^x = u \) for \( x \).
A1: For obtaining the final answer \( x = 1.30 \) correct to 3 significant figures.

Use the identity \( \sec^2 \theta = 1 + \tan^2 \theta \): \[3(1 + \tan^2 \theta) + 5\tan \theta = 7\] \[3 + 3\tan^2 \theta + 5\tan \theta - 7 = 0\] \[3\tan^2 \theta + 5\tan \theta - 4 = 0\] Let \( t = \tan \theta \): \[3t^2 + 5t - 4 = 0\] Solve for \( t \) using the quadratic formula: \[t = \frac{-5 \pm \sqrt{5^2 - 4(3)(-4)}}{2(3)} = \frac{-5 \pm \sqrt{73}}{6}\] This gives: \[t_1 = \frac{-5 + \sqrt{73}}{6} \approx 0.5907\] \[t_2 = \frac{-5 - \sqrt{73}}{6} \approx -2.2573\] Now find the angles \( \theta \) in the range \( 0^\circ \le \theta \le 180^\circ \): For \( \tan \theta = 0.5907 \): \[\theta = \tan^{-1}(0.5907) \approx 30.6^\circ\] For \( \tan \theta = -2.2573 \): \[\theta = 180^\circ - \tan^{-1}(2.2573) \approx 180^\circ - 66.1^\circ = 113.9^\circ\]

M1: For using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \) to obtain an equation in terms of \( \tan \theta \) only.
A1: For obtaining the correct quadratic equation \( 3\tan^2 \theta + 5\tan \theta - 4 = 0 \).
M1: For solving the quadratic equation to find two values of \( \tan \theta \).
A1: For obtaining \( \tan \theta \approx 0.591 \) and \( \tan \theta \approx -2.257 \).
M1: For finding one correct angle in the given range.
A1: For \( \theta = 30.6^\circ \) (accept \( 30.6 \)).
A1: For \( \theta = 113.9^\circ \) (accept \( 113.9 \)) and no other solutions in range.

First, differentiate both \( x \) and \( y \) with respect to \( t \): \[\frac{\mathrm{d}x}{\mathrm{d}t} = 2e^{2t} - 3\] \[\frac{\mathrm{d}y}{\mathrm{d}t} = 4e^t + 1\] Using the chain rule, find \( \frac{\mathrm{d}y}{\mathrm{d}x} \): \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{4e^t + 1}{2e^{2t} - 3}\] Substitute \( t = 0 \) to find the gradient of the tangent, \( m \): \[m = \frac{4(1) + 1}{2(1) - 3} = \frac{5}{-1} = -5\] Find the coordinates of the point on the curve when \( t = 0 \): \[x = e^{0} - 3(0) = 1\] \[y = 4e^0 + 0 = 4\] Use the point-gradient form to find the equation of the tangent: \[y - 4 = -5(x - 1) \implies y = -5x + 9\]

M1: For differentiating \( x \) to obtain \( \frac{\mathrm{d}x}{\mathrm{d}t} = ae^{2t} - b \).
A1: For correct derivative \( \frac{\mathrm{d}x}{\mathrm{d}t} = 2e^{2t} - 3 \).
A1: For correct derivative \( \frac{\mathrm{d}y}{\mathrm{d}t} = 4e^t + 1 \).
M1: For using \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} \) and substituting \( t = 0 \).
A1: For finding the correct gradient \( m = -5 \).
M1: For finding the point \( (1, 4) \) by substituting \( t = 0 \) into parametric equations.
A1: For the correct tangent equation \( y = -5x + 9 \) (or equivalent format).

Integrate each term of the integrand: \[\int (2\sin 2x + \cos x) \, \mathrm{d}x = \left[ -\cos 2x + \sin x \right]\] Now substitute the limits \( 0 \) and \( \frac{1}{3}\pi \): At the upper limit \( x = \frac{1}{3}\pi \): \[-\cos\left(2 \cdot \frac{1}{3}\pi\right) + \sin\left(\frac{1}{3}\pi\right) = -\cos\left(\frac{2}{3}\pi\right) + \sin\left(\frac{1}{3}\pi\right)\] Since \( \cos\left(\frac{2}{3}\pi\right) = -\frac{1}{2} \) and \( \sin\left(\frac{1}{3}\pi\right) = \frac{\sqrt{3}}{2} \): \[-\left(-\frac{1}{2}\right) + \frac{\sqrt{3}}{2} = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}\] At the lower limit \( x = 0 \): \[-\cos(0) + \sin(0) = -1 + 0 = -1\] Subtract the lower limit evaluation from the upper limit evaluation: \[\left(\frac{1 + \sqrt{3}}{2}\right) - (-1) = \frac{1 + \sqrt{3}}{2} + 1 = \frac{3 + \sqrt{3}}{2}\]

M1: For attempting to integrate \( 2\sin 2x \) to get \( -a\cos 2x \).
A1: For correct term \( -\cos 2x \).
A1: For correct term \( \sin x \).
M1: For substituting the limits \( \frac{1}{3}\pi \) and \( 0 \) correctly into an integrated expression.
M1: For using exact trigonometric values for both \( \cos(2\pi/3) \) and \( \sin(\pi/3) \).
A1: For evaluating at \( x = 0 \) to get \( -1 \).
A1: For obtaining the correct exact simplified answer \( \frac{3 + \sqrt{3}}{2} \) (or equivalent).

(a) Let \( f(x) = x^3 + 4x^2 - 10 \). Evaluate \( f(x) \) at the boundaries: \[f(1.3) = (1.3)^3 + 4(1.3)^2 - 10 = 2.197 + 6.76 - 10 = -1.043\] \[f(1.4) = (1.4)^3 + 4(1.4)^2 - 10 = 2.744 + 7.84 - 10 = +0.584\] Since \( f(1.3) < 0 \) and \( f(1.4) > 0 \), there is a sign change. Since \( f(x) \) is continuous, a root must lie between 1.3 and 1.4. (b) Using the iterative formula \( x_{n+1} = \sqrt{\frac{10}{x_n + 4}} \) with \( x_1 = 1.35 \): \[x_2 = \sqrt{\frac{10}{1.35 + 4}} = \sqrt{\frac{10}{5.35}} \approx 1.36717\] \[x_3 = \sqrt{\frac{10}{1.36717 + 4}} \approx 1.36498\] \[x_4 = \sqrt{\frac{10}{1.36498 + 4}} \approx 1.36526\] \[x_5 = \sqrt{\frac{10}{1.36526 + 4}} \approx 1.36523\] The values are converging to \( 1.365 \) to 3 decimal places. Thus, the root is \( 1.365 \).

M1: For evaluating \( f(1.3) \) and \( f(1.4) \) with a continuous function \( f(x) = x^3+4x^2-10 \).
A1: For obtaining correct values \(-1.043\) and \(0.584\), and concluding with reference to sign change.
M1: For substituting \( x_1 = 1.35 \) into the iterative formula to find \( x_2 \).
A1: For obtaining \( x_2 = 1.36717 \) and \( x_3 = 1.36498 \).
A1: For obtaining \( x_4 = 1.36526 \) and \( x_5 = 1.36523 \).
M1: For showing sufficient iterations to justify accuracy to 3 decimal places.
A1: For the final value \( 1.365 \).

(a) By the Remainder Theorem: Since \( P(x) \) is divisible by \( 2x - 1 \), we have \( P(1/2) = 0 \): \[2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) - 6 = 0\] \[\frac{2}{8} + \frac{a}{4} + \frac{b}{2} - 6 = 0 \implies 1 + a + 2b - 24 = 0 \implies a + 2b = 23 \quad \text{(Equation 1)}\] Since the remainder is \(-30\) when divided by \( x + 2 \), we have \( P(-2) = -30 \): \[2(-2)^3 + a(-2)^2 + b(-2) - 6 = -30\] \[-16 + 4a - 2b - 6 = -30 \implies 4a - 2b - 22 = -30 \implies 4a - 2b = -8 \implies 2a - b = -4 \quad \text{(Equation 2)}\] From Equation 2, \( b = 2a + 4 \). Substitute this into Equation 1: \[a + 2(2a + 4) = 23 \implies 5a + 8 = 23 \implies 5a = 15 \implies a = 3\] Then find \( b \): \[b = 2(3) + 4 = 10\] (b) The polynomial is \( P(x) = 2x^3 + 3x^2 + 10x - 6 \). Since \( 2x - 1 \) is a factor: \[2x^3 + 3x^2 + 10x - 6 = (2x - 1)(x^2 + cx + 6)\] Equating coefficients of \( x^2 \): \[3 = 2c - 1 \implies 2c = 4 \implies c = 2\] So the quadratic factor is \( x^2 + 2x + 6 \). Checking the discriminant of \( x^2 + 2x + 6 \): \[\Delta = 2^2 - 4(1)(6) = -20 < 0\] Since the discriminant is negative, the quadratic cannot be factorised further over real numbers. Thus, the complete factorization is \( (2x - 1)(x^2 + 2x + 6) \).

M1: For applying the Remainder Theorem with \( P(1/2) = 0 \) to form an equation in \( a \) and \( b \).
A1: For obtaining a correct equation equivalent to \( a + 2b = 23 \).
M1: For applying the Remainder Theorem with \( P(-2) = -30 \) to form a second equation in \( a \) and \( b \).
A1: For obtaining a correct equation equivalent to \( 2a - b = -4 \).
M1: For solving the simultaneous equations to find \( a \) and \( b \).
A1: For \( a = 3 \) and \( b = 10 \).
A1: For completing the factorization to find \( (2x - 1)(x^2 + 2x + 6) \).

(i) To express \(w\) in the form \(x + \mathrm{i}y\), multiply the numerator and denominator by the conjugate of the denominator:
\[ w = \frac{1 + 7\mathrm{i}}{3 + \mathrm{i}} = \frac{(1 + 7\mathrm{i})(3 - \mathrm{i})}{(3 + \mathrm{i})(3 - \mathrm{i})} = \frac{3 - \mathrm{i} + 21\mathrm{i} - 7\mathrm{i}^2}{9 - \mathrm{i}^2} \]
Since \(\mathrm{i}^2 = -1\):
\[ w = \frac{3 + 20\mathrm{i} + 7}{10} = \frac{10 + 20\mathrm{i}}{10} = 1 + 2\mathrm{i} \]

(ii) Since \(\arg(z) = \frac{\pi}{4}\), the real and imaginary parts of \(z\) must be equal and positive.
Let \(z = a + a\mathrm{i}\) where \(a > 0\).
Using the condition \(|z - w| = \sqrt{5}\) with \(w = 1 + 2\mathrm{i}\):
\[ |(a - 1) + (a - 2)\mathrm{i}| = \sqrt{5} \]
\[ (a - 1)^2 + (a - 2)^2 = 5 \]
\[ a^2 - 2a + 1 + a^2 - 4a + 4 = 5 \]
\[ 2a^2 - 6a + 5 = 5 \]
\[ 2a(a - 3) = 0 \]
Since \(a > 0\), we reject \(a = 0\) and obtain \(a = 3\).
Thus, \(z = 3 + 3\mathrm{i}\).

(i)
M1: For multiplying numerator and denominator by \(3 - \mathrm{i}\) and expanding.
A1: Obtain \(1 + 2\mathrm{i}\) or equivalent.

(ii)
M1: State that \(z = a + a\mathrm{i}\) (or \(x + x\mathrm{i}\)) with \(a > 0\) from the argument condition.
M1: Set up the modulus equation \((a - 1)^2 + (a - 2)^2 = 5\) and solve for \(a\).
A1: Obtain the quadratic equation \(2a^2 - 6a = 0\) or equivalent.
A1: Identify \(a = 3\) (rejecting \(a = 0\)) and state the final complex number \(z = 3 + 3\mathrm{i}\).

Separate the variables to integrate both sides:
\[ \int \frac{1}{y^2} \;\mathrm{d}y = \int \frac{\ln x}{x} \;\mathrm{d}x \]

Integrating the left-hand side:
\[ \int y^{-2} \;\mathrm{d}y = -\frac{1}{y} \]

Integrating the right-hand side using the substitution \(u = \ln x\), \(\mathrm{d}u = \frac{1}{x} \;\mathrm{d}x\):
\[ \int \frac{\ln x}{x} \;\mathrm{d}x = \int u \;\mathrm{d}u = \frac{1}{2}u^2 + C = \frac{1}{2}(\ln x)^2 + C \]

Thus, the general solution is:
\[ -\frac{1}{y} = \frac{1}{2}(\ln x)^2 + C \]

Using the initial condition \(y = -2\) when \(x = 1\):
\[ -\frac{1}{-2} = \frac{1}{2}(\ln 1)^2 + C \]
\[ \frac{1}{2} = 0 + C \implies C = \frac{1}{2} \]

Substitute \(C = \frac{1}{2}\) back into the equation:
\[ -\frac{1}{y} = \frac{1}{2}(\ln x)^2 + \frac{1}{2} \]
Multiply by \(-1\) and take the reciprocal:
\[ \frac{1}{y} = -\frac{(\ln x)^2 + 1}{2} \implies y = -\frac{2}{(\ln x)^2 + 1} \]

M1: Separate variables correctly and attempt integration of both sides.
A1: Obtain \(-\frac{1}{y}\) on the left.
M1: Integrate \(\frac{\ln x}{x}\) to obtain \(\frac{1}{2}(\ln x)^2\).
A1: Obtain correct general equation including constant of integration \(C\).
M1: Substitute \(x = 1\) and \(y = -2\) to find the value of \(C\).
A1: Express \(y\) correctly in terms of \(x\) as \(y = -\frac{2}{(\ln x)^2 + 1}\).

Let \(u = \cos x\), then \(\mathrm{d}u = -\sin x \;\mathrm{d}x\), which means \(\sin x \;\mathrm{d}x = -\mathrm{d}u\).

Change the limits of integration:
When \(x = 0\), \(u = \cos 0 = 1\).
When \(x = \frac{\pi}{3}\), \(u = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\).

Rewrite \(\sin^3 x\) as \(\sin^2 x \sin x = (1 - \cos^2 x) \sin x = (1 - u^2) \sin x\).
Now substitute everything into the integral:
\[ \int_{0}^{\frac{\pi}{3}} \sin^3 x \cos^2 x \;\mathrm{d}x = \int_{1}^{\frac{1}{2}} (1 - u^2) u^2 (-\mathrm{d}u) \]
Reverse the limits to absorb the negative sign:
\[ = \int_{\frac{1}{2}}^{1} (u^2 - u^4) \;\mathrm{d}u \]
Integrate with respect to \(u\):
\[ = \left[ \frac{1}{3}u^3 - \frac{1}{5}u^5 \right]_{\frac{1}{2}}^{1} \]
Evaluate at the limits:
At \(u = 1\):
\[ \frac{1}{3}(1)^3 - \frac{1}{5}(1)^5 = \frac{1}{3} - \frac{1}{5} = \frac{2}{15} \]
At \(u = \frac{1}{2}\):
\[ \frac{1}{3}\left(\frac{1}{2}\right)^3 - \frac{1}{5}\left(\frac{1}{2}\right)^5 = \frac{1}{24} - \frac{1}{160} = \frac{20 - 3}{480} = \frac{17}{480} \]

Subtract the lower limit evaluation from the upper limit evaluation:
\[ \frac{2}{15} - \frac{17}{480} = \frac{64}{480} - \frac{17}{480} = \frac{47}{480} \]

M1: Substitute \(u = \cos x\) and obtain the differential relation \(\mathrm{d}u = -\sin x \;\mathrm{d}x\) or equivalent.
M1: Express the integrand entirely in terms of \(u\) as \((u^2 - u^4)\) (ignoring limits and signs at this stage).
A1: Obtain correct limits \(u = 1\) and \(u = 0.5\), and correctly handle the negative sign.
M1: Integrate \(u^2 - u^4\) to get \(\frac{1}{3}u^3 - \frac{1}{5}u^5\).
M1: Substitute the limits correctly and perform the arithmetic.
A1: Obtain the exact fraction \\frac{47}{480}\\.

(i) Define \(f(x) = \ln(x + 2) + x^2 - 3\).
Evaluate \(f(1.3)\) and \(f(1.4)\):
\[ f(1.3) = \ln(1.3 + 2) + 1.3^2 - 3 = \ln(3.3) + 1.69 - 3 \approx 1.1939 - 1.31 = -0.1161 \]
\[ f(1.4) = \ln(1.4 + 2) + 1.4^2 - 3 = \ln(3.4) + 1.96 - 3 \approx 1.2238 - 1.04 = 0.1838 \]
Since there is a change of sign between \(f(1.3) < 0\) and \(f(1.4) > 0\) and the function is continuous, a root must lie in the interval \((1.3, 1.4)\).

(ii) Let \(\lim_{n\to\infty} x_n = L\). If the sequence converges, then:
\[ L = \sqrt{3 - \ln(L + 2)} \]
Squaring both sides:
\[ L^2 = 3 - \ln(L + 2) \]
Rearranging terms:
\[ \ln(L + 2) = 3 - L^2 \]
which is the original equation \(\ln(x+2) = 3 - x^2\).

(iii) Let \(x_1 = 1.35\):
\[ x_2 = \sqrt{3 - \ln(1.35 + 2)} = \sqrt{3 - 1.2090} \approx 1.3383 \]
\[ x_3 = \sqrt{3 - \ln(1.3383 + 2)} = \sqrt{3 - 1.2055} \approx 1.3396 \]
\[ x_4 = \sqrt{3 - \ln(1.3396 + 2)} = \sqrt{3 - 1.2058} \approx 1.3395 \]
\[ x_5 = \sqrt{3 - \ln(1.3395 + 2)} = \sqrt{3 - 1.2058} \approx 1.3395 \]
Thus, the root is \(1.34\) correct to 2 decimal places.

(i)
M1: Evaluate \(f(x)\) at \(x = 1.3\) and \(x = 1.4\).
A1: Obtain correct values showing a sign change, and conclude.

(ii)
B1: Set \(x_{n+1} = x_n = L\) (or \(x\)) and show the steps to obtain \(\ln(x+2) = 3-x^2\).

(iii)
M1: Perform iterations starting with an initial value in the interval \((1.3, 1.4)\).
A1: Obtain at least three correct iterations to 4 decimal places (e.g., \(1.3383\), \(1.3396\), \(1.3395\)).
A1: State final answer \(1.34\) and show that iterations have converged to this value.

(i) To find the intersection, equate the components of \(l_1\) and \(l_2\):
\[ 1 + 2\lambda = -1 + \mu \implies \mu - 2\lambda = 2 \quad \text{(1)} \]
\[ 3 - \lambda = 7 + \mu \implies \mu + \lambda = -4 \quad \text{(2)} \]
\[ 10 + 3\lambda = 2 - \mu \implies \mu + 3\lambda = -8 \quad \text{(3)} \]

Subtract equation (2) from equation (1):
\[ -3\lambda = 6 \implies \lambda = -2 \]
Substitute \(\lambda = -2\) into equation (2):
\[ \mu - 2 = -4 \implies \mu = -2 \]

Now substitute \(\lambda = -2\) and \(\mu = -2\) into equation (3) to check for consistency:
\[ \text{LHS} = -2 + 3(-2) = -8 \]
Since LHS = RHS, the system is consistent and the lines intersect.

The position vector of the point of intersection is:
\[ \mathbf{r} = \mathbf{i} + 3\mathbf{j} + 10\mathbf{k} - 2(2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = -3\mathbf{i} + 5\mathbf{j} + 4\mathbf{k} \]

(ii) The direction vectors of the lines are \(\mathbf{d}_1 = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and \(\mathbf{d}_2 = \mathbf{i} + \mathbf{j} - \mathbf{k}\).
Let \(\theta\) be the acute angle between the lines:
\[ \cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|} \]
Evaluate the dot product:
\[ \mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-1)(1) + (3)(-1) = 2 - 1 - 3 = -2 \]
Evaluate the magnitudes:
\[ |\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14} \]
\[ |\mathbf{d}_2| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \]
Thus:
\[ \cos\theta = \frac{|-2|}{\sqrt{14}\sqrt{3}} = \frac{2}{\sqrt{42}} \]
\[ \theta = \arccos\left(\frac{2}{\sqrt{42}}\right) \approx 72.025^\circ \]
To 1 decimal place, \(\theta = 72.0^\circ\).

(i)
M1: State three equations in \(\lambda\) and \(\mu\) from the component equations.
M1: Solve a pair of equations to find values for \(\lambda\) and \(\mu\).
A1: Show that the values satisfy the third equation (consistency check).
A1: Obtain the correct position vector \(-3\mathbf{i} + 5\mathbf{j} + 4\mathbf{k}\) (or equivalent coordinate form).

(ii)
M1: Use the scalar product of the two direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\).
M1: Divide by the product of their magnitudes.
A1: Obtain \(72.0^\circ\) (accept \(1.26\) radians if specified in radians, but penalize missing 1 d.p. accuracy).

(i) Differentiate both sides of the equation implicitly with respect to \(x\):
\[ \frac{\mathrm{d}}{\mathrm{d}x}(x^2 + xy + y^2) = \frac{\mathrm{d}}{\mathrm{d}x}(12) \]
\[ 2x + \left(y + x\frac{\mathrm{d}y}{\mathrm{d}x}\right) + 2y\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \]
Group the terms containing \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\[ (x + 2y)\frac{\mathrm{d}y}{\mathrm{d}x} = -(2x + y) \]
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2x + y}{x + 2y} \]

(ii) The tangent is parallel to the \(y\)-axis (vertical) where \(\frac{\mathrm{d}x}{\mathrm{d}y} = 0\), which corresponds to the denominator of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) being zero (provided the numerator is non-zero).
Set the denominator equal to zero:
\[ x + 2y = 0 \implies x = -2y \]

Substitute \(x = -2y\) back into the equation of the curve:
\[ (-2y)^2 + (-2y)y + y^2 = 12 \]
\[ 4y^2 - 2y^2 + y^2 = 12 \]
\[ 3y^2 = 12 \implies y^2 = 4 \implies y = \pm 2 \]

Using \(x = -2y\):
If \(y = 2\), then \(x = -2(2) = -4\).
If \(y = -2\), then \(x = -2(-2) = 4\).

Check if the numerator \(2x + y\) is non-zero at these points:
For \((-4, 2)\): \(2(-4) + 2 = -6 \neq 0\).
For \((4, -2)\): \(2(4) - 2 = 6 \neq 0\).

Thus, the points are \((-4, 2)\) and \((4, -2)\).

(i)
M1: Attempt implicit differentiation of \(xy\) using the product rule.
A1: Correctly differentiate \(x^2 + xy + y^2\) to obtain \(2x + y + x\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\frac{\mathrm{d}y}{\mathrm{d}x}\).
M1: Rearrange to make \\frac{\mathrm{d}y}{\mathrm{d}x}\\ the subject.
A1: Obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2x + y}{x + 2y}\) or equivalent.

(ii)
M1: State that the denominator is zero, \(x + 2y = 0\).
M1: Substitute \(x = -2y\) into the original equation of the curve.
A1: Solve for \(y\) to get \(y = \pm 2\).
A1: Obtain both points \((-4, 2)\) and \((4, -2)\).

(i) We want to express \(3 \cos \theta - 4 \sin \theta = R \cos(\theta + \alpha) = R \cos \theta \cos \alpha - R \sin \theta \sin \alpha\).
Equating coefficients of \\cos \theta\\ and \\sin \theta\\:
\[ R \cos \alpha = 3 \]
\[ R \sin \alpha = 4 \]
Squaring and adding:
\[ R^2 = 3^2 + 4^2 = 25 \implies R = 5 \]
Dividing the equations:
\[ \tan \alpha = \frac{4}{3} \implies \alpha = \arctan\left(\frac{4}{3}\right) \approx 0.927295 \text{ radians} \]
Thus, \(R = 5\) and \(\alpha = 0.9273\) (correct to 4 s.f.).

(ii) Rewrite the equation using part (i) with \(\theta = 2x\):
\[ 5 \cos(2x + 0.9273) = 2 \]
\[ \cos(2x + 0.9273) = 0.4 \]

Since \(0 < x < \pi\), the interval for \(2x + 0.9273\) is:
\[ 0.9273 < 2x + 0.9273 < 2\pi + 0.9273 \approx 7.2105 \]

The basic angle is \(\arccos(0.4) \approx 1.1593\) radians.

Case 1:
\[ 2x + 0.9273 = 1.1593 \implies 2x = 0.2320 \implies x \approx 0.116 \text{ radians} \]

Case 2:
\[ 2x + 0.9273 = 2\pi - 1.1593 \approx 5.1239 \implies 2x = 4.1966 \implies x \approx 2.10 \text{ radians} \]

Thus, the solutions are \(x = 0.116\) and \(x = 2.10\).

(i)
B1: State \(R = 5\) exactly.
M1: Use \(\tan \alpha = \frac{4}{3}\) or equivalent to find \(\alpha\).
A1: Obtain \(\alpha = 0.9273\) to 4 s.f.

(ii)
M1: Use the result from (i) to set up \(5 \cos(2x + \alpha) = 2\).
M1: Find one correct value of \(2x + \alpha\) inside the allowed interval.
A1: Obtain \(x = 0.116\) to 3 s.f.
A1: Obtain \(x = 2.10\) to 3 s.f. (and no other values in the interval).

(i) Write the partial fraction expansion:
\[ \frac{7x^2 + x - 2}{(1 - x)(1 + 2x^2)} = \frac{A}{1 - x} + \frac{Bx + C}{1 + 2x^2} \]
Multiply both sides by \((1 - x)(1 + 2x^2)\):
\[ 7x^2 + x - 2 = A(1 + 2x^2) + (Bx + C)(1 - x) \]

Let \(x = 1\):
\[ 7(1)^2 + 1 - 2 = A(1 + 2(1)^2) + 0 \implies 6 = 3A \implies A = 2 \]

Let \(x = 0\):
\[ -2 = A(1) + C(1) \implies -2 = 2 + C \implies C = -4 \]

Equate the coefficients of \(x^2\):
\[ 7 = 2A - B \implies 7 = 2(2) - B \implies B = -3 \]

Therefore, the partial fractions are:
\[ f(x) = \frac{2}{1 - x} + \frac{-3x - 4}{1 + 2x^2} = \frac{2}{1 - x} - \frac{3x + 4}{1 + 2x^2} \]

(ii) Express \(f(x)\) using binomial series expansions:
\[ f(x) = 2(1 - x)^{-1} - (3x + 4)(1 + 2x^2)^{-1} \]

Expand the first term:
\[ 2(1 - x)^{-1} = 2(1 + x + x^2 + \dots) = 2 + 2x + 2x^2 + \dots \]

Expand the second term:
\[ (1 + 2x^2)^{-1} = 1 - 2x^2 + \dots \]
Multiply this by \((4 + 3x)\):
\[ (4 + 3x)(1 - 2x^2) = 4 - 8x^2 + 3x - 6x^3 \approx 4 + 3x - 8x^2 \]

Now, subtract the two expansions:
\[ f(x) \approx (2 + 2x + 2x^2) - (4 + 3x - 8x^2) \]
\[ f(x) \approx (2 - 4) + (2x - 3x) + (2x^2 - (-8x^2)) \]
\[ f(x) \approx -2 - x + 10x^2 \]

(i)
M1: Set up the correct form of partial fractions with a linear numerator over \(1 + 2x^2\).
M1: Substitute a value or equate coefficients to find at least one constant.
A1: Obtain \(A = 2\).
A1: Obtain \(B = -3\) and \(C = -4\) (or equivalent expression).

(ii)
M1: Use binomial theorem to expand \((1 - x)^{-1}\) up to the \(x^2\) term.
M1: Use binomial theorem to expand \((1 + 2x^2)^{-1}\) up to the \(x^2\) term.
M1: Multiply by \((3x + 4)\) and correctly collect terms up to \(x^2\).
A1: Obtain the final correct expansion \(-2 - x + 10x^2\).

We use the quadratic formula to solve \( z^2 - (2+4\text{i})z - 6 + 8\text{i} = 0 \):

\( z = \frac{(2+4\text{i}) \pm \sqrt{(2+4\text{i})^2 - 4(1)(-6+8\text{i})}}{2} \)

First, evaluate the discriminant \( D \):
\( D = (2+4\text{i})^2 - 4(-6+8\text{i}) \)
\( D = (4 - 16 + 16\text{i}) + (24 - 32\text{i}) \)
\( D = 12 - 16\text{i} \)

Next, find the square roots of \( 12 - 16\text{i} \).
Let \( (u+\text{i}v)^2 = 12 - 16\text{i} \), where \( u \) and \( v \) are real.
Equating real and imaginary parts:
1) \( u^2 - v^2 = 12 \)
2) \( 2uv = -16 \implies uv = -8 \)

Using the modulus relation:
\( u^2 + v^2 = \sqrt{12^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \)

Adding 1) and this relation:
\( 2u^2 = 32 \implies u^2 = 16 \implies u = \pm 4 \)

Since \( uv = -8 \) is negative, \( u \) and \( v \) have opposite signs:
If \( u = 4 \), then \( v = -2 \).
If \( u = -4 \), then \( v = 2 \).
So, the square roots of the discriminant are \( \pm(4 - 2\text{i}) \).

Substitute these back into the quadratic formula:
\( z = \frac{2+4\text{i} \pm (4-2\text{i})}{2} \)

For the positive sign:
\( z = \frac{2+4\text{i} + 4-2\text{i}}{2} = \frac{6+2\text{i}}{2} = 3+\text{i} \)

For the negative sign:
\( z = \frac{2+4\text{i} - (4-2\text{i})}{2} = \frac{-2+6\text{i}}{2} = -1+3\text{i} \)

M1: Apply the quadratic formula or complete the square on the given equation.
A1: Obtain the correct discriminant \( 12 - 16\text{i} \).
M1: Set up and solve the system \( u^2 - v^2 = 12 \) and \( 2uv = -16 \) (or equivalent method) to find the square roots of the discriminant.
A1: Obtain the square roots \( \pm(4 - 2\text{i}) \).
M1: Use these square roots to find the two solutions for \( z \).
A1: Obtain both correct final answers \( z = 3 + \text{i} \) and \( z = -1 + 3\text{i} \).

Let \( u = \mathrm{e}^x \). Then \( \mathrm{d}u = \mathrm{e}^x \, \mathrm{d}x \).

We rewrite the integral as:
\( \int_{0}^{\ln 2} \frac{\mathrm{e}^x}{\mathrm{e}^x + 3} \cdot \mathrm{e}^x \, \mathrm{d}x \)

Now find the limits of integration in terms of \( u \):
When \( x = 0 \), \( u = \mathrm{e}^0 = 1 \).
When \( x = \ln 2 \), \( u = \mathrm{e}^{\ln 2} = 2 \).

Substituting \( u \) and \( \mathrm{d}u \) into the integral:
\( \int_{1}^{2} \frac{u}{u+3} \, \mathrm{d}u \)

Perform algebraic division or express the integrand as:
\( \frac{u}{u+3} = \frac{u+3-3}{u+3} = 1 - \frac{3}{u+3} \)

Integrate with respect to \( u \):
\( \int_{1}^{2} \left(1 - \frac{3}{u+3}\right) \, \mathrm{d}u = \left[ u - 3\ln(u+3) \right]_1^2 \)

Substitute the limits:
\( = \left(2 - 3\ln(5)\right) - \left(1 - 3\ln(4)\right) \)
\( = 2 - 3\ln 5 - 1 + 3\ln 4 \)
\( = 1 - 3\ln 5 + 3\ln 4 \)
\( = 1 - 3\ln\left(\frac{5}{4}\right) \)
\( = 1 - \ln\left(\left(\frac{5}{4}\right)^3\right) \)
\( = 1 - \ln\left(\frac{125}{64}\right) \)
\( = 1 + \ln\left(\frac{64}{125}\right) \)

Here, \( a = 1 \) and \( b = \frac{64}{125} \) are rational numbers.

M1: Attempt to use the substitution \( u = \mathrm{e}^x \) (or equivalent) to find \( \mathrm{d}u \).
A1: Obtain the correct transformed integral in terms of \( u \) with correct limits.
M1: Write the integrand in an integrable form, e.g., by using division to get \( 1 - \frac{3}{u+3} \).
A1: Correctly integrate to get \( u - 3\ln(u+3) \).
M1: Substitute limits of integration and apply laws of logarithms to combine terms.
A1: Obtain the correct final exact value \( 1 + \ln\left(\frac{64}{125}\right) \) (or identify \( a = 1, b = \frac{64}{125} \)).

First, separate the variables:
\( \frac{1}{y^2} \, \mathrm{d}y = 2 \cos^2 x \sin x \, \mathrm{d}x \)

Integrate both sides:
\( \int y^{-2} \, \mathrm{d}y = \int 2 \cos^2 x \sin x \, \mathrm{d}x \)

For the LHS:
\( \int y^{-2} \, \mathrm{d}y = -\frac{1}{y} \)

For the RHS, use integration by substitution or inspection. Let \( u = \cos x \), so \( \mathrm{d}u = -\sin x \, \mathrm{d}x \):
\( \int 2 \cos^2 x \sin x \, \mathrm{d}x = \int 2 u^2 (-\mathrm{d}u) = -\frac{2}{3}u^3 = -\frac{2}{3}\cos^3 x \)

Combining these results with a constant of integration \( C \):
\( -\frac{1}{y} = -\frac{2}{3}\cos^3 x + C \)

Use the boundary condition \( y = 1 \) when \( x = 0 \):
\( -\frac{1}{1} = -\frac{2}{3}\cos^3(0) + C \)
\( -1 = -\frac{2}{3}(1) + C \)
\( C = -1 + \frac{2}{3} = -\frac{1}{3} \)

Substitute the value of \( C \) back into the equation:
\( -\frac{1}{y} = -\frac{2}{3}\cos^3 x - \frac{1}{3} \)

Multiply the entire equation by \( -1 \):
\( \frac{1}{y} = \frac{2}{3}\cos^3 x + \frac{1}{3} \)
\( \frac{1}{y} = \frac{2\cos^3 x + 1}{3} \)

Invert to solve for \( y \):
\( y = \frac{3}{2\cos^3 x + 1} \)

M1: Separate variables and attempt integration of both sides.
A1: Correct integration of LHS to get \( -y^{-1} \) (or equivalent).
M1: Integrate RHS using substitution or inspection, obtaining a form of \( k\cos^3 x \).
A1: Obtain the correct RHS integration as \( -\frac{2}{3}\cos^3 x \).
M1: Apply the initial condition \( y = 1 \) at \( x = 0 \) to find the constant of integration.
A1: Express \( y \) correctly in terms of \( x \) as \( y = \frac{3}{2\cos^3 x + 1} \) (or equivalent form).

Let the direction of the initial motion of particle \(A\) be the positive direction.

Initial velocities:
\(u_A = 4\text{ m s}^{-1}\)
\(u_B = -v\text{ m s}^{-1}\)

Final velocities after the collision:
\(v_A = -2\text{ m s}^{-1}\)
\(v_B = 1\text{ m s}^{-1}\) (since the initial direction of \(B\) was negative, its reversed direction is positive)

Using the principle of conservation of linear momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)

Substitute the given values:
\(0.3(4) + 0.5(-v) = 0.3(-2) + 0.5(1)\)
\(1.2 - 0.5v = -0.6 + 0.5\)
\(1.2 - 0.5v = -0.1\)
\(0.5v = 1.3\)
\(v = 2.6\)

M1: For using conservation of linear momentum with appropriate signs.
A1: For a correct unsimplified equation, e.g., \(0.3(4) - 0.5v = 0.3(-2) + 0.5(1)\).
M1: For attempting to solve the linear equation for \(v\).
A1: For obtaining the correct value \(v = 2.6\).

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

Resolving forces vertically for equilibrium:
\(R + 30 \sin 25^\circ = mg\)
\(R + 30 \sin 25^\circ = 6 \times 10\)
\(R = 60 - 30 \sin 25^\circ \approx 47.322\text{ N}\)

Resolving forces horizontally for equilibrium (on the point of sliding):
\(F = 30 \cos 25^\circ \approx 27.189\text{ N}\)

Since the block is on the point of sliding, we use the friction relation:
\(F = \mu R\)

\(\mu = \frac{F}{R} = \frac{30 \cos 25^\circ}{60 - 30 \sin 25^\circ}\)

\(\mu \approx \frac{27.1892}{47.3215} \approx 0.575\) (to 3 s.f.)

M1: For resolving forces vertically to find the normal reaction \(R\).
A1: For obtaining a correct expression for \(R\), e.g., \(R = 60 - 30 \sin 25^\circ\).
M1: For resolving forces horizontally to find the frictional force \(F = 30 \cos 25^\circ\).
M1: For using \(F = \mu R\) to solve for \(\mu\).
A1: For obtaining \(\mu \approx 0.575\) (accept 0.574 to 0.575).

For particle \(P\) on the horizontal table, the normal reaction is:
\(R = m_P g = 0.8 \times 10 = 8\text{ N}\)

The frictional force is:
\(F = \mu R = 0.3 \times 8 = 2.4\text{ N}\)

Using Newton's second law for \(P\):
\(T - F = m_P a \implies T - 2.4 = 0.8a\) (Equation 1)

For particle \(Q\) hanging vertically:
\(m_Q g - T = m_Q a \implies 0.4(10) - T = 0.4a \implies 4 - T = 0.4a\) (Equation 2)

Adding Equation 1 and Equation 2:
\((T - 2.4) + (4 - T) = 0.8a + 0.4a\)

\(1.6 = 1.2a\)

\(a = \frac{1.6}{1.2} = \frac{4}{3} \approx 1.33\text{ m s}^{-2}\) (to 3 s.f.)

M1: For calculating the frictional force \(F = 0.3 \times 8 = 2.4\text{ N}\).
M1: For writing Newton's second law for \(P\), e.g., \(T - F = 0.8a\).
M1: For writing Newton's second law for \(Q\), e.g., \(4 - T = 0.4a\).
M1: For solving the simultaneous equations to eliminate \(T\).
A1: For obtaining \(a = \frac{4}{3}\) or \(1.33\) (accept 1.33).

Let \(DF\) be the driving force of the engine.

The forces acting down the incline are the component of the weight of the car and the resistance force.

Component of weight down the slope:
\(W_{\parallel} = mg \sin \alpha = 1200 \times 10 \times 0.05 = 600\text{ N}\)

Resistance force:
\(R = 400\text{ N}\)

Using Newton's second law along the incline:
\(DF - W_{\parallel} - R = ma\)
\(DF - 600 - 400 = 1200 \times 0.4\)
\(DF - 1000 = 480\)
\(DF = 1480\text{ N}\)

The power \(P\) is related to the driving force \(DF\) and speed \(v\) by:
\(P = DF \times v\)

Given \(P = 30\text{ kW} = 30000\text{ W}\):
\(30000 = 1480 \times v\)

\(v = \frac{30000}{1480} = \frac{750}{37} \approx 20.3\text{ m s}^{-1}\) (to 3 s.f.)

M1: For finding the component of weight down the slope, \(1200 \times 10 \times 0.05 = 600\text{ N}\).
M1: For applying Newton's second law along the incline, \(DF - 600 - 400 = 1200 \times 0.4\).
A1: For obtaining \(DF = 1480\text{ N}\).
M1: For using \(P = DF \times v\) with \(P = 30000\).
A1: For obtaining \(v \approx 20.3\).

To find the total distance travelled, we need to check if the velocity changes sign in the interval \(0 \le t \le 3\).

Set \(v(t) = 0\):
\(3t^2 - 12t + 9 = 0\)
\(t^2 - 4t + 3 = 0\)
\((t-1)(t-3) = 0\)

So the particle is instantaneously at rest at \(t = 1\) and \(t = 3\).

In the interval \(0 \le t < 1\), \(v(t) > 0\).
In the interval \(1 < t < 3\), \(v(t) < 0\).

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

For the first interval \(0 \le t \le 1\):
\(\text{Distance}_1 = s(1) - s(0) = (1^3 - 6(1)^2 + 9(1)) - 0 = 4\text{ m}\)

For the second interval \(1 \le t \le 3\):
\(\text{Distance}_2 = |s(3) - s(1)| = |(3^3 - 6(3)^2 + 9(3)) - 4| = |0 - 4| = 4\text{ m}\)

Total distance travelled \(= 4 + 4 = 8\text{ m}\).

M1: For finding when \(v(t) = 0\) inside the interval \(0 \le t \le 3\).
A1: For identifying the critical time \(t = 1\).
M1: For integrating the velocity function to obtain \(s(t) = t^3 - 6t^2 + 9t\).
M1: For calculating the distance in the interval \([0, 1]\) or \([1, 3]\).
A1: For obtaining both distance components as \(4\) (or showing a sum of magnitudes).
A1: For obtaining the total distance of \(8\text{ m}\).

Let \(R\) be the normal reaction force perpendicular to the plane. Resolving forces perpendicular to the incline:
\(R = mg \cos \theta = 2.5 \times 10 \times 0.8 = 20\text{ N}\)

The maximum frictional force \(F\) opposing the motion up the plane is:
\(F = \mu R = 0.2 \times 20 = 4\text{ N}\)

The component of the particle's weight acting down the plane is:
\(W_{\parallel} = mg \sin \theta = 2.5 \times 10 \times 0.6 = 15\text{ N}\)

Using Newton's second law for the motion up the plane:
\(P - W_{\parallel} - F = ma\)

where \(P = 30\text{ N}\) is the pulling force.

\(30 - 15 - 4 = 2.5a\)

\(11 = 2.5a\)

\(a = \frac{11}{2.5} = 4.4\text{ m s}^{-2}\)

M1: For calculating the normal reaction \(R = 2.5 \times 10 \times 0.8 = 20\text{ N}\).
M1: For calculating the frictional force \(F = 0.2 \times 20 = 4\text{ N}\).
M1: For calculating the weight component down the slope, \(25 \times 0.6 = 15\text{ N}\).
M1: For applying Newton's second law up the plane: \(30 - 15 - 4 = 2.5a\).
A1: For obtaining \(a = 4.4\).

Let us use the principle of conservation of energy.

The vertical height \(h\) through which the child descends is:
\(h = 8 \sin 30^\circ = 4\text{ m}\)

The loss in gravitational potential energy (\(\Delta PE\)) during the descent is:
\(\Delta PE = mgh = 35 \times 10 \times 4 = 1400\text{ J}\)

The gain in kinetic energy (\(\Delta KE\)) is:
\(\Delta KE = \frac{1}{2} m v^2 - 0 = \frac{1}{2} \times 35 \times (5.2)^2 = 17.5 \times 27.04 = 473.2\text{ J}\)

According to the work-energy principle, the loss in mechanical energy is equal to the work done against friction (\(W_f\)):

\(W_f = \Delta PE - \Delta KE\)

\(W_f = 1400 - 473.2 = 926.8\text{ J}\)

M1: For calculating the vertical height \(h = 8 \sin 30^\circ = 4\text{ m}\).
M1: For calculating the loss in gravitational potential energy, \(mgh = 1400\text{ J}\).
M1: For calculating the gain in kinetic energy, \(\frac{1}{2} m v^2 = 473.2\text{ J}\).
M1: For applying the work-energy relation, \(W_f = \Delta PE - \Delta KE\).
A1: For obtaining the correct answer of \(926.8\) (or 927).

(i) Let \(u = x - 50\).
We are given \(\sum u = 240\) and \(\sum u^2 = 4120\) for \(n = 80\).
Mean of \(u\) is \(\bar{u} = \frac{\sum u}{n} = \frac{240}{80} = 3\).
Therefore, the mean of \(x\) is \(\bar{x} = \bar{u} + 50 = 3 + 50 = 53\) grams.

Variance of \(u\) is \(\text{Var}(u) = \frac{\sum u^2}{n} - \bar{u}^2 = \frac{4120}{80} - 3^2 = 51.5 - 9 = 42.5\).
Since standard deviation is unaffected by subtracting a constant, the standard deviation of \(x\) is equal to the standard deviation of \(u\):
\(\text{SD}(x) = \sqrt{42.5} \approx 6.52\) grams.

(ii) For the first 80 packages, \(\sum x = 80 \times 53 = 4240\).
For the next 20 packages, the mean is 54, so \(\sum y = 20 \times 54 = 1080\).

The total mass of all 100 packages is \(\sum x + \sum y = 4240 + 1080 = 5320\).
The combined mean is \(\frac{5320}{100} = 53.2\) grams.

(i)
M1: For calculating the mean of \(u\) and adding 50 (or equivalent overall sum approach).
A1: For mean of 53.
M1: For correct substitution into the variance formula \(\frac{\sum u^2}{n} - \bar{u}^2\).
A1: For standard deviation of 6.52 (accept 6.519).

(ii)
M1: For finding the total sum of the first 80 packages (4240).
M1: For finding the total sum of the next 20 packages (1080).
M1: For dividing the combined sum by 100.
A1: For 53.2.

(i) The word PARAMETRIC has 10 letters, with 2 As, 2 Rs, and 6 other distinct letters (P, M, E, T, I, C).
Number of arrangements = \(\frac{10!}{2! \times 2!} = 907200\).

(ii) We can use the complement method: Total arrangements minus arrangements where the two Rs are adjacent.
If the two Rs are adjacent, treat 'RR' as a single entity.
There are now 9 entities to arrange: (RR), A, A, P, M, E, T, I, C.
Number of arrangements = \(\frac{9!}{2!} = 181440\).
Arrangements where Rs are not adjacent = \(907200 - 181440 = 725760\).

(iii) The pool of letters consists of: 2 Rs, 2 As, and 6 other distinct letters.
We want to find the number of selections of 4 letters with at least one R.
We can count the complement: total selections of 4 letters minus selections with 0 Rs.
Let's count selections with 0 Rs (i.e., choosing 4 letters from 2 As and 6 distinct letters):
- Case 1: 2 As and 2 distinct: \(\binom{6}{2} = 15\) ways.
- Case 2: 1 A and 3 distinct: \(\binom{6}{3} = 20\) ways.
- Case 3: 0 As and 4 distinct: \(\binom{6}{4} = 15\) ways.
Total selections with 0 Rs = \(15 + 20 + 15 = 50\) ways.

Now let's count total selections of 4 letters from the entire pool:
- Case A: 2 Rs, 2 As -> 1 way.
- Case B: 2 Rs, 1 A, 1 distinct -> \(\binom{6}{1} = 6\) ways.
- Case C: 2 Rs, 0 As, 2 distinct -> \(\binom{6}{2} = 15\) ways.
- Case D: 1 R, 2 As, 1 distinct -> \(\binom{6}{1} = 6\) ways.
- Case E: 1 R, 1 A, 2 distinct -> \(\binom{6}{2} = 15\) ways.
- Case F: 1 R, 0 As, 3 distinct -> \(\binom{6}{3} = 20\) ways.
- Case G: 0 Rs (already shown to be 50 ways).
Total selections overall = \(1 + 6 + 15 + 6 + 15 + 20 + 50 = 113\) ways.

Selections with at least one R = \(113 - 50 = 63\) ways.

(i)
M1: For \(\frac{10!}{2! \times 2!}\) or equivalent expression.
A1: For 907200.

(ii)
M1: For identifying and attempting to subtract 'Rs adjacent' from total arrangements (or using slotting method).
M1: For calculating \(\frac{9!}{2!} = 181440\) (or slotting expression \(\frac{8!}{2!} \times \binom{9}{2}\)).
A1: For 725760.

(iii)
M1: For identifying a systematic way to calculate selections (either direct cases or complement).
M1: For correct calculation of either the complement (50 ways) or the sum of cases containing R (41 + 22).
A1: For 63.

Let \(I\) be the event that a student plays an instrument, and \(S\) be the event that a student plays a sport.
We are given:
\(P(I) = 0.60\), so \(P(I') = 0.40\).
\(P(S|I) = 0.40\).
\(P(S|I') = 0.75\).

(i) \(P(S) = P(S \cap I) + P(S \cap I') = P(I)P(S|I) + P(I')P(S|I')\)
\(P(S) = (0.60 \times 0.40) + (0.40 \times 0.75) = 0.24 + 0.30 = 0.54\).

(ii) We want to find \(P(I|S)\):
\(P(I|S) = \frac{P(I \cap S)}{P(S)} = \frac{0.24}{0.54} = \frac{4}{9} \approx 0.444\).

(iii) Let \(X\) be the number of students who play a sport among 3 randomly chosen students.
\(X \sim B(3, 0.54)\).
\(P(X = 2) = \binom{3}{2} (0.54)^2 (0.46)^1 = 3 \times 0.2916 \times 0.46 = 0.402\) (3 s.f.).

(i)
M1: For calculating and summing two two-stage probabilities: \(0.60 \times 0.40\) and \(0.40 \times 0.75\).
A1: For one correct term (either 0.24 or 0.30 seen).
A1: For 0.54.

(ii)
M1: For using \(P(I|S) = \frac{P(I \cap S)}{P(S)}\) with their values.
A1: For 4/9 or 0.444 (3 s.f.).

(iii)
M1: For binomial structure of the form \(\binom{3}{2} p^2 (1-p)\).
M1: For substituting \(p = 0.54\).
A1: For 0.402 (accept 0.4024).

There are 16 equally likely outcomes when spinning the spinner twice:
\((1,1), (1,2), (1,3), (1,4)\)
\((2,1), (2,2), (2,3), (2,4)\)
\((3,1), (3,2), (3,3), (3,4)\)
\((4,1), (4,2), (4,3), (4,4)\)

(i) The outcomes with an absolute difference of 1 are:
\((1,2), (2,1), (2,3), (3,2), (3,4), (4,3)\).
There are 6 such outcomes.
Thus, \(P(X = 1) = \frac{6}{16} = 0.375\).

(ii) Let's find the probability for each possible value of \(X\):
- For \(X = 0\): \((1,1), (2,2), (3,3), (4,4)\) (4 outcomes) \(\implies P(X=0) = \frac{4}{16} = 0.25\).
- For \(X = 1\): 6 outcomes \(\implies P(X=1) = 0.375\).
- For \(X = 2\): \((1,3), (3,1), (2,4), (4,2)\) (4 outcomes) \(\implies P(X=2) = \frac{4}{16} = 0.25\).
- For \(X = 3\): \((1,4), (4,1)\) (2 outcomes) \(\implies P(X=3) = \frac{2}{16} = 0.125\).

Probability distribution table:
\(\begin{array}{|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 \\ \hline P(X=x) & 0.25 & 0.375 & 0.25 & 0.125 \\ \hline \end{array}\)

(iii) Expectation:
\(E(X) = \sum x \cdot P(X=x) = 0(0.25) + 1(0.375) + 2(0.25) + 3(0.125) = 0 + 0.375 + 0.50 + 0.375 = 1.25\).

Variance:
\(E(X^2) = \sum x^2 \cdot P(X=x) = 0^2(0.25) + 1^2(0.375) + 2^2(0.25) + 3^2(0.125) = 0 + 0.375 + 1.0 + 1.125 = 2.5\).
\(\text{Var}(X) = E(X^2) - [E(X)]^2 = 2.5 - 1.25^2 = 2.5 - 1.5625 = 0.9375\).

(i)
M1: For listing at least 4 of the 6 successful outcomes (or drawing sample space diagram).
A1: For showing \(P(X=1) = 6/16 = 0.375\).

(ii)
M1: For drawing a table with 4 columns for \(X = 0, 1, 2, 3\) and obtaining correct probabilities for at least two of them.
A1: For all correct probabilities in the table.

(iii)
M1: For correct formula for \(E(X)\) applied to their distribution.
A1: For \(E(X) = 1.25\).
M1: For calculating \(E(X^2)\) and using \(\text{Var}(X) = E(X^2) - [E(X)]^2\).
A1: For \(\text{Var}(X) = 0.9375\) (or 15/16).

Let \(W\) be the weight of a randomly chosen dog. \(W \sim N(\mu, \sigma^2)\).

(i) We are given:
\(P(W > 35) = 0.15 \implies P(W < 35) = 0.85\).
Standardising:
\(\frac{35 - \mu}{\sigma} = z_{0.85}\).
Using standard normal tables, \(z_{0.85} \approx 1.036\).
So, \(35 - \mu = 1.036\sigma\) (Equation 1)

We are also given:
\(P(W > 25) = 0.60 \implies P(W < 25) = 0.40\).
Standardising:
\(\frac{25 - \mu}{\sigma} = z_{0.40}\).
Using standard normal tables, \(z_{0.40} = -z_{0.60} \approx -0.253\).
So, \(25 - \mu = -0.253\sigma\) (Equation 2)

Subtract Equation 2 from Equation 1:
\(10 = 1.289\sigma \implies \sigma \approx 7.758\) kg.

Substitute \(\sigma\) back into Equation 2:
\(\mu = 25 + 0.253(7.758) \approx 26.96\) kg.

So, \(\mu = 27.0\) and \(\sigma = 7.76\) (to 3 s.f.).

(ii) We want to find \(P(W < 22)\):
\(P(W < 22) = P\left(Z < \frac{22 - 26.96}{7.758}\right) = P(Z < -0.6393)\).
\(P(Z < -0.6393) = 1 - \Phi(0.6393) \approx 1 - 0.7387 = 0.261\) (3 s.f.).

(i)
M1: For standardising 35 and equating to a positive z-value (from normal tables).
A1: For \(35 - \mu = 1.036\sigma\) (allow 1.037).
M1: For standardising 25 and equating to a negative z-value.
A1: For \(25 - \mu = -0.253\sigma\) (allow -0.253 to -0.254).
M1: For solving simultaneous equations to find at least one variable.
A1: For \(\mu = 27.0\) and \(\sigma = 7.76\) (allow 26.9 or 27.0, and 7.75 to 7.77).

(ii)
M1: For standardising 22 using their \(\mu\) and \(\sigma\).
M1: For calculating the correct area region (1 - \(\Phi\)).
A1: For 0.261 (accept 0.260 to 0.262).

(i) Let \(X\) be the number of defective bulbs in a sample of 15. Then \(X \sim B(15, 0.08)\).
\(P(X = 2) = \binom{15}{2} (0.08)^2 (0.92)^{13}\)
\(P(X = 2) = 105 \times 0.0064 \times 0.3382 = 0.227\) (3 s.f.).

(ii) Let \(Y\) be the number of defective bulbs in a batch of 250. Then \(Y \sim B(250, 0.08)\).
Since \(n = 250\) is large and \(np = 20 > 5\), \(nq = 230 > 5\), we can approximate \(Y\) with a normal distribution \(N(\mu, \sigma^2)\):
\(\mu = np = 250 \times 0.08 = 20\)
\(\sigma^2 = npq = 20 \times 0.92 = 18.4\)

We want to find \(P(Y < 18) = P(Y \le 17)\).
Using a continuity correction, this becomes:
\(P(Y < 17.5)\).
Standardising:
\(P\left(Z < \frac{17.5 - 20}{\sqrt{18.4}}\right) = P\left(Z < \frac{-2.5}{4.2895}\right) = P(Z < -0.5828)\).
\(P(Z < -0.5828) = 1 - \Phi(0.5828) \approx 1 - 0.7200 = 0.280\) (3 s.f.).

(i)
M1: For binomial term of the form \(\binom{15}{2} p^2 (1-p)^{13}\).
A1: For substituting \(p = 0.08\).
A1: For 0.227 (accept 0.227 to 0.228).

(ii)
M1: For calculating both mean (20) and variance (18.4) of the normal approximation.
M1: For applying a continuity correction to get 17.5.
M1: For standardising 17.5 using their mean and variance.
M1: For finding the correct area region for negative z.
A1: For 0.280 (accept 0.279 to 0.281).

(i) A Poisson distribution is suitable because the number of trials, \( n = 600 \), is large (\( n > 50 \)) and the probability of success, \( p = 0.004 \), is small (\( p < 0.1 \)). The mean \( \lambda = np = 2.4 \), which is constant and less than 5.
(ii)(a) Let \( X \) be the number of cracked tiles, where \( X \sim \text{Po}(2.4) \).
\( P(X = 3) = \frac{e^{-2.4} \times 2.4^3}{3!} = \frac{e^{-2.4} \times 13.824}{6} \approx 0.209 \).
(b) \( P(X \ge 2) = 1 - P(X = 0) - P(X = 1) = 1 - e^{-2.4} - 2.4 e^{-2.4} = 1 - 3.4 e^{-2.4} \approx 0.692 \).

M1: For explaining why Poisson is appropriate (stating large \( n \) and small \( p \)).
A1: For finding the mean \( \lambda = 2.4 \).
M1: For using Poisson formula with \( \lambda = 2.4 \) and \( x = 3 \) for (ii)(a).
A1.14: For correct value 0.209 (accept 0.209 to 0.210).
M1: For identifying \( P(X \ge 2) = 1 - P(X = 0) - P(X = 1) \).
A1: For correct calculation of \( 1 - 3.4e^{-2.4} \).
A1: For correct value 0.692 (accept 0.691 to 0.693).

We set up the hypotheses:
\( H_0: \mu = 80 \)
\( H_1: \mu < 80 \) (one-tailed test)
The sample size \( n = 40 \) is large, so by the Central Limit Theorem, the sample mean \( \bar{X} \) can be assumed to follow a normal distribution.
Using the unbiased variance estimate \( s^2 = 25.0 \), the standard error is \( \frac{s}{\sqrt{n}} = \sqrt{\frac{25}{40}} \approx 0.7906 \).
The test statistic is \( z = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{77.8 - 80}{0.7906} \approx -2.78 \).
At the 5% level of significance, the critical value for a one-tailed test is \( -1.645 \).
Since the calculated \( z \)-value of \( -2.78 \) is less than \( -1.645 \), it lies in the critical region.
Therefore, we reject \( H_0 \). There is significant evidence at the 5% level of significance to support the consumer group's suspicion that the mean lifetime is less than 80 hours.

B1: For stating both hypotheses correctly.
M1.14: For finding standard error \( \sqrt{25/40} \approx 0.7906 \).
M2: For calculating the test statistic \( z \approx -2.78 \).
B1: For identifying the critical value as \( -1.645 \) (or comparing area \( 0.0027 \) with \( 0.05 \)).
A1: For comparing the test statistic with the critical value correctly.
A1: For a correct conclusion in context (rejecting \( H_0 \) and confirming suspicion).

Let \( A \sim N(120, 12^2) \) and \( B \sim N(140, 15^2) \).
(i) Let \( W = A_1 + A_2 + A_3 \).
\( E(W) = 3 \times 120 = 360 \).
\( \operatorname{Var}(W) = 3 \times 12^2 = 432 \).
\( P(W < 340) = P\left(Z < \frac{340 - 360}{\sqrt{432}}\right) = P(Z < -0.962) = 1 - \Phi(0.962) \approx 1 - 0.8320 = 0.168 \).
(ii) We require \( P(B - A > 15) \). Let \( D = B - A \).
\( E(D) = E(B) - E(A) = 140 - 120 = 20 \).
\( \operatorname{Var}(D) = \operatorname{Var}(B) + \operatorname{Var}(A) = 15^2 + 12^2 = 369 \).
\( P(D > 15) = P\left(Z > \frac{15 - 20}{\sqrt{369}}\right) = P(Z > -0.260) = \Phi(0.260) \approx 0.603 \).

M1: For finding correct mean of \( 3A \) as 360 and variance as 432.
A1.14: For standardizing and finding correct probability of 0.168.
M1: For finding mean of \( B - A \) as 20.
M1: For finding variance of \( B - A \) as \( 15^2 + 12^2 = 369 \).
M1: For standardizing with their mean and variance for \( B - A > 15 \).
A1: For obtaining correct probability of 0.603.

(i) We have \( n = 80 \), \( \bar{x} = 4.35 \), and \( s^2 = 0.64 \), so \( s = 0.8 \).
For a 98% confidence interval, the critical value \( z \) is obtained from normal distribution tables where \( \Phi(z) = 0.99 \), giving \( z = 2.326 \).
The interval is given by \( \bar{x} \pm z \frac{s}{\sqrt{n}} = 4.35 \pm 2.326 \frac{0.8}{\sqrt{80}} = 4.35 \pm 0.208 \).
So the interval is \( [4.14, 4.56] \) (to 3 s.f.).
(ii) Yes, it was necessary because the distribution of the individual concentrations in the river is not stated to be normal. Since the sample size \( n = 80 \) is large (\( n \ge 30 \)), the Central Limit Theorem allows us to assume that the sample mean is approximately normally distributed.

M1: For identifying critical value \( z = 2.326 \) (or 2.33).
M1: For calculating the standard error \( \frac{0.8}{\sqrt{80}} \approx 0.0894 \).
M2.14: For correct application of the confidence interval formula \( 4.35 \pm z \times \text{SE} \).
A1: For correct interval \( [4.14, 4.56] \) (accept \( [4.14, 4.56] \)).
B1: For stating 'Yes' and referencing that the population distribution is unknown.
B1: For explaining that because \( n = 80 \) is large, the CLT ensures the sample mean is approximately normal.

(i) For \( f(x) \) to be a valid pdf, we must have \( \int_{0}^{2} f(x) dx = 1 \).
\( k \int_{0}^{2} (4x - x^3) dx = 1 \implies k \left[ 2x^2 - \frac{x^4}{4} \right]_0^2 = 1 \).
Evaluating the limit: \( k \left( 2(4) - \frac{16}{4} \right) = 1 \implies k(8 - 4) = 1 \implies 4k = 1 \implies k = \frac{1}{4} \). (Shown)
(ii) Let the median be \( m \).
\( \int_{0}^{m} \frac{1}{4}(4x - x^3) dx = 0.5 \implies \left[ 2x^2 - \frac{x^4}{4} \right]_0^m = 2 \).
\( 2m^2 - \frac{m^4}{4} = 2 \implies m^4 - 8m^2 + 8 = 0 \).
Let \( u = m^2 \), then \( u^2 - 8u + 8 = 0 \).
Using the quadratic formula: \( u = \frac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2} \).
Since \( m \in [0, 2] \), \( m^2 \in [0, 4] \). Thus we select \( u = 4 - 2\sqrt{2} \approx 1.1716 \).
Hence, \( m = \sqrt{4 - 2\sqrt{2}} \approx 1.08 \) (to 3 s.f.).

M1: For integrating \( f(x) \) and setting the integral from 0 to 2 equal to 1.
A1: For finding the value of the integral as 4.
A1: For concluding \( k = 1/4 \).
M1: For setting up the integral for the median \( \int_0^m f(x) dx = 0.5 \) (or from \( m \) to 2).
A1.14: For deriving the quartic equation \( m^4 - 8m^2 + 8 = 0 \) or equivalent quadratic in \( m^2 \).
M1: For solving the quadratic in \( m^2 \) to find a valid root.
A1: For correct median of 1.08 (accept 1.08 to 1.09).

(i) Let \( X_1 \) be the number of complaints received by the online team in 2 hours.
Since the hourly rate is 1.2, the mean for a 2-hour period is \( \lambda = 1.2 \times 2 = 2.4 \).
Thus, \( X_1 \sim \text{Po}(2.4) \).
\( P(X_1 = 2) = \frac{e^{-2.4} \times 2.4^2}{2!} = 2.88 e^{-2.4} \approx 0.261 \).
(ii) Let \( Y \) be the total complaints in 1 hour.
\( Y = X_{\text{online}} + X_{\text{phone}} \).
Since the teams receive complaints independently, the sum of independent Poisson variables is also Poisson distributed:
\( Y \sim \text{Po}(1.2 + 1.8) \implies Y \sim \text{Po}(3.0) \).
We want \( P(Y \le 2) = P(Y=0) + P(Y=1) + P(Y=2) \).
\( P(Y \le 2) = e^{-3.0} + e^{-3.0}(3.0) + \frac{e^{-3.0} \times 3.0^2}{2!} = e^{-3.0} (1 + 3 + 4.5) = 8.5 e^{-3.0} \approx 0.423 \).

M1: For adjusting the mean to \( 2.4 \) for the 2-hour period.
A1.14: For correct calculation of \( P(X_1 = 2) \approx 0.261 \).
M1: For summing the two independent Poisson parameters to get \( \lambda = 3.0 \).
M2: For finding the sum of terms for \( P(Y \le 2) \): \( P(0) + P(1) + P(2) \).
A1: For correct expansion \( e^{-3}(1 + 3 + 4.5) \).
A1: For correct probability of 0.423.

(i) Let \( p \) be the population proportion of defective components.
\( H_0: p = 0.05 \)
\( H_1: p > 0.05 \)
(ii) A Type I error is rejecting \( H_0 \) when \( H_0 \) is true.
Under \( H_0 \), \( X \sim B(30, 0.05) \).
The critical region is \( X \ge 4 \).
\( P(\text{Type I error}) = P(X \ge 4 \mid p=0.05) = 1 - P(X \le 3 \mid p=0.05) \).
\( P(X \le 3) = \binom{30}{0}(0.05)^0(0.95)^{30} + \binom{30}{1}(0.05)^1(0.95)^{29} + \binom{30}{2}(0.05)^2(0.95)^{28} + \binom{30}{3}(0.05)^3(0.95)^{27} \).
\( P(X=0) \approx 0.2146 \)
\( P(X=1) \approx 0.3389 \)
\( P(X=2) \approx 0.2586 \)
\( P(X=3) \approx 0.1271 \)
Sum \( = 0.9392 \).
\( P(\text{Type I error}) = 1 - 0.9392 = 0.0608 \) (to 3 s.f.).
(iii) A Type II error is failing to reject \( H_0 \) when \( H_1 \) is true.
Under \( H_1 \) where \( p = 0.15 \), the distribution is \( X \sim B(30, 0.15) \).
We fail to reject \( H_0 \) if \( X \le 3 \).
\( P(\text{Type II error}) = P(X \le 3 \mid p=0.15) \).
\( P(X \le 3) = \binom{30}{0}(0.15)^0(0.85)^{30} + \binom{30}{1}(0.15)^1(0.85)^{29} + \binom{30}{2}(0.15)^2(0.85)^{28} + \binom{30}{3}(0.15)^3(0.85)^{27} \).
\( P(X=0) \approx 0.0076 \)
\( P(X=1) \approx 0.0404 \)
\( P(X=2) \approx 0.1034 \)
\( P(X=3) \approx 0.1703 \)
Sum \( = 0.3217 \approx 0.322 \) (to 3 s.f.).

B1: For stating hypotheses correctly.
M1: For setting up the expression for \( 1 - P(X \le 3) \) under \( B(30, 0.05) \).
A1.14: For calculating the individual binomial probabilities and finding the sum 0.9392.
A1: For correct Type I error probability 0.0608 (or 0.061).
M1: For identifying that Type II error is \( P(X \le 3) \) under \( B(30, 0.15) \).
M1: For calculating terms for \( P(X \le 3) \) with \( p = 0.15 \).
A1: For correct Type II error probability of 0.322.