2024 Cambridge IAL Mathematics (9709) 模擬試題連答案詳解

First, we complete the square for \(\mathrm{f}(x)\):
\[\mathrm{f}(x) = 2(x^2 - 4x) + 3\]
\[\mathrm{f}(x) = 2\left((x - 2)^2 - 4\right) + 3\]
\[\mathrm{f}(x) = 2(x - 2)^2 - 5\]

Next, let \(y = 2(x - 2)^2 - 5\). We make \(x\) the subject:
\[y + 5 = 2(x - 2)^2\]
\[\frac{y+5}{2} = (x - 2)^2\]
Since the domain of \(\mathrm{f}\) is \(x \le 1\), we have \(x - 2 < 0\). Therefore, we must choose the negative square root when solving for \(x\):
\[x - 2 = -\sqrt{\frac{y+5}{2}}\]
\[x = 2 - \sqrt{\frac{y+5}{2}}\]

Replacing \(y\) with \(x\), the inverse function is:
\[\mathrm{f}^{-1}(x) = 2 - \sqrt{\frac{x+5}{2}}\]

To find the domain of \(\mathrm{f}^{-1}\), we find the range of \(\mathrm{f}\) for \(x \le 1\).
Since the vertex of the quadratic is at \(x = 2\) and the coefficient of \(x^2\) is positive, the function is decreasing for \(x \le 2\).
At the boundary \(x = 1\):
\[\mathrm{f}(1) = 2(1)^2 - 8(1) + 3 = -3\]
Since the function is decreasing as \(x\) increases up to 1, for \(x \le 1\) the range is \(\mathrm{f}(x) \ge -3\).
Thus, the domain of \(\mathrm{f}^{-1}\) is \(x \ge -3\).

M1: Attempt to complete the square to get \(2(x-a)^2 + b\) or make \(x\) the subject by another valid method.
A1: Correctly express \(x\) in terms of \(y\) with the negative root selected, e.g., \(x = 2 - \sqrt{\frac{y+5}{2}}\).
A1: Obtain the correct inverse function \(\mathrm{f}^{-1}(x) = 2 - \sqrt{\frac{x+5}{2}}\) (must be in terms of \(x\)).
B1: State the correct domain of \(\mathrm{f}^{-1}\) as \(x \ge -3\) (or equivalent).

From the line equation, we can express \(x\) in terms of \(y\):
\[x = 2y - k\]
Substitute this expression for \(x\) into the equation of the curve:
\[y^2 = 2(2y - k) + 5\]
\[y^2 = 4y - 2k + 5\]
Rearrange to form a standard quadratic equation in \(y\):
\[y^2 - 4y + (2k - 5) = 0\]
Since the line is a tangent to the curve, there must be exactly one point of intersection. Therefore, the discriminant of this quadratic equation must be equal to zero:
\[b^2 - 4ac = 0\]
\[(-4)^2 - 4(1)(2k - 5) = 0\]
\[16 - 8k + 20 = 0\]
\[36 - 8k = 0\]
\[8k = 36\]
\[k = 4.5 \quad \text{or} \quad k = \frac{9}{2}\]

M1: Substitute the equation of the line into the curve to obtain a quadratic equation in either \(x\) or \(y\).
A1: Form a correct quadratic equation, e.g., \(y^2 - 4y + (2k-5) = 0\) or \(x^2 + (2k-8)x + (k^2-20) = 0\).
M1: Apply the discriminant condition \(b^2 - 4ac = 0\) to their quadratic equation.
A1: Obtain the correct value of \(k = 4.5\) (or \(\frac{9}{2}\)).

To find the equation of the curve, we integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with respect to \(x\):
\[y = \int 6(3x + 4)^{-\frac{1}{2}} \, \mathrm{d}x\]
Using the reverse chain rule:
\[y = 6 \times \frac{(3x + 4)^{\frac{1}{2}}}{\frac{1}{2} \times 3} + C\]
\[y = 6 \times \frac{(3x + 4)^{\frac{1}{2}}}{\frac{3}{2}} + C\]
\[y = 4(3x + 4)^{\frac{1}{2}} + C\]
\[y = 4\sqrt{3x + 4} + C\]

We are given that the curve passes through the point \((4, 15)\). Substitute \(x = 4\) and \(y = 15\):
\[15 = 4\sqrt{3(4) + 4} + C\]
\[15 = 4\sqrt{16} + C\]
\[15 = 4(4) + C\]
\[15 = 16 + C\]
\[C = -1\]

Thus, the equation of the curve is:
\[y = 4\sqrt{3x + 4} - 1\]

M1: Attempt to integrate \(6(3x+4)^{-\frac{1}{2}}\) yielding a term of the form \(k(3x+4)^{\frac{1}{2}}\).
A1: Correct integration, obtaining \(4(3x+4)^{\frac{1}{2}}\) (condone omission of \(+C\) at this stage).
M1: Substitute the coordinates \((4, 15)\) into their integrated equation containing a constant of integration \(C\) to solve for \(C\).
A1: Obtain the correct final equation \(y = 4\sqrt{3x+4} - 1\) (or equivalent).

Using the trigonometric identity \
\cos^2\theta = 1 - \sin^2\theta\, we substitute into the equation:
\[3(1 - \sin^2\theta) + 5\sin\theta - 1 = 0\]
\[3 - 3\sin^2\theta + 5\sin\theta - 1 = 0\]
\[-3\sin^2\theta + 5\sin\theta + 2 = 0\]
Multiply the entire equation by \(-1\):
\[3\sin^2\theta - 5\sin\theta - 2 = 0\]

Factorise the quadratic equation in \(\sin\theta\):
\[(3\sin\theta + 1)(\sin\theta - 2) = 0\]

This gives two possible solutions:
1) \(\sin\theta = 2\) which has no solutions since the range of the sine function is \([-1, 1]\).
2) \(3\sin\theta + 1 = 0 \implies \sin\theta = -\frac{1}{3}\).

Since \(\sin\theta\) is negative, the solutions for \(\theta\) lie in the third and fourth quadrants.
First, calculate the basic angle:
\[\alpha = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ\]

Now find the values of \(\theta\) in the interval \([0^\circ, 360^\circ]\):
- In the third quadrant:
\[\theta = 180^\circ + 19.47^\circ = 199.5^\circ \quad \text{(to 1 d.p.)}\]
- In the fourth quadrant:
\[\theta = 360^\circ - 19.47^\circ = 340.5^\circ \quad \text{(to 1 d.p.)}\]

M1: Substitute \(\cos^2\theta = 1 - \sin^2\theta\) to obtain a quadratic equation in \(\sin\theta\).
A1: Solve the quadratic equation correctly to find \\sin\theta = -\frac{1}{3}\ and identify that \(\sin\theta = 2\) has no solutions.
M1: Calculate one correct angle in the range (e.g., \(199.5^\circ\) or \(340.5^\circ\), rounded to 1 decimal place).
A1: Obtain both correct solutions \(\theta = 199.5^\circ\) and \(\theta = 340.5^\circ\) and no other solutions in the interval.

(i) To find the vertex of the quadratic, we express \( f(x) \) in completed square form: \( f(x) = 2(x^2 - 6x) + 13 = 2((x-3)^2 - 9) + 13 = 2(x-3)^2 - 5 \). The vertex is at \( x = 3 \). For the function to have an inverse, it must be one-to-one. Since the domain is defined as \( x \le k \), the largest value of \( k \) is the x-coordinate of the vertex, so \( k = 3 \).

(ii) Let \( y = 2(x-3)^2 - 5 \). Rearranging to make \( x \) the subject: \( y + 5 = 2(x-3)^2 \) which gives \( \frac{y+5}{2} = (x-3)^2 \). Since \( x \le 3 \), we must choose the negative square root when solving: \( x - 3 = -\sqrt{\frac{y+5}{2}} \), so \( x = 3 - \sqrt{\frac{y+5}{2}} \). Replacing \( y \) with \( x \), we obtain \( f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}} \). The domain of \( f^{-1} \) is the range of \( f \). Since \( f(x) = 2(x-3)^2 - 5 \) for \( x \le 3 \), the minimum value of \( f(x) \) is \( -5 \), so the range of \( f \) is \( f(x) \ge -5 \). Thus, the domain of \( f^{-1} \) is \( x \ge -5 \).

(i) B1: State k = 3.

(ii) M1: Attempt to make x the subject of the formula using completing the square. A1: Correctly obtain \( (x-3)^2 = \frac{y+5}{2} \) or equivalent. A1: Correctly choose the negative root to write \( f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}} \) (must be in terms of x). B1: State domain as \( x \ge -5 \) (accept \( [-5, \infty) \)).

Using the sum formula for an arithmetic progression, \( S_n = \frac{n}{2}[2a + (n-1)d] \), we have: \( S_{10} = 5(2a + 9d) = 150 \). Dividing by 5 gives: \( 2a + 9d = 30 \) (Equation 1).

The first, fourth, and tenth terms of the arithmetic progression are \( a \, \), \( a + 3d \), and \( a + 9d \). Since these form a geometric progression, the ratio of successive terms is constant: \( \frac{a + 3d}{a} = \frac{a + 9d}{a + 3d} \). Cross-multiplying gives: \( (a + 3d)^2 = a(a + 9d) \). Expanding both sides: \( a^2 + 6ad + 9d^2 = a^2 + 9ad \). Subtracting \( a^2 \) and \( 6ad \) from both sides: \( 9d^2 = 3ad \). Since \( d \neq 0 \), we can divide by \( 3d \) to get: \( a = 3d \) (Equation 2).

Substitute Equation 2 into Equation 1: \( 2(3d) + 9d = 30 \implies 15d = 30 \implies d = 2 \). Substituting \( d = 2 \) back into Equation 2: \( a = 3(2) = 6 \). Therefore, \( a = 6 \) and \( d = 2 \).

M1: Use the AP sum formula to form the equation \( 5(2a+9d) = 150 \). A1: Correctly simplify to \( 2a + 9d = 30 \). M1: Use the GP relation to set up \( (a+3d)^2 = a(a+9d) \) and expand it. A1: Correctly simplify to obtain \( a = 3d \). A1: Correctly solve the simultaneous equations to find \( a = 6 \) and \( d = 2 \).

We divide the entire equation by \( \cos^2 \theta \) (noting that \( \cos \theta \neq 0 \) as it would lead to \( \sin \theta = 0 \), which cannot occur simultaneously):
\[ 3 + 5 \tan \theta - 2 \tan^2 \theta = 0 \]
Rearranging into standard quadratic form:
\[ 2 \tan^2 \theta - 5 \tan \theta - 3 = 0 \]
Factorising the quadratic:
\[ (2 \tan \theta + 1)(\tan \theta - 3) = 0 \]
This gives two cases:
1) \( \tan \theta = 3 \):
\( \theta = \tan^{-1}(3) \approx 71.57^\circ \approx 71.6^\circ \) (to 1 decimal place).
2) \( \tan \theta = -0.5 \):
The basic angle is \( \alpha = \tan^{-1}(0.5) \approx 26.57^\circ \). Since \( \tan \theta \) is negative and \( 0^\circ \le \theta \le 180^\circ \), \( \theta \) lies in the second quadrant:
\( \theta = 180^\circ - 26.57^\circ \approx 153.43^\circ \approx 153.4^\circ \) (to 1 decimal place).

Thus, the solutions are \( \theta = 71.6^\circ \) and \( \theta = 153.4^\circ \).

M1: Divide the equation by \( \cos^2 \theta \) to form a quadratic in \( \tan \theta \). A1: Correctly form the quadratic \( 2 \tan^2 \theta - 5 \tan \theta - 3 = 0 \) or equivalent. M1: Attempt to solve their quadratic to find values for \( \tan \theta \). A1: Obtain \( \theta = 71.6^\circ \) (accept 71.6). A1: Obtain \( \theta = 153.4^\circ \) (accept 153.4) and no other values in the range.

The volume \( V \) is given by \( V = \pi \int_{0}^{1} y^2 \, dx \).
First, find \( y^2 \):
\( y^2 = \left( \frac{2}{(3x+1)^{3/4}} \right)^2 = \frac{4}{(3x+1)^{3/2}} = 4(3x+1)^{-3/2} \).

Now, integrate with respect to \( x \):
\( \int 4(3x+1)^{-3/2} \, dx = 4 \left[ \frac{(3x+1)^{-1/2}}{(-1/2) \times 3} \right] = 4 \left[ -\frac{2}{3}(3x+1)^{-1/2} \right] = -\frac{8}{3\sqrt{3x+1}} \).

Apply the limits of integration from 0 to 1:
\( V = \pi \left[ -\frac{8}{3\sqrt{3x+1}} \right]_{0}^{1} \).
At \( x = 1 \): \( -\frac{8}{3\sqrt{4}} = -\frac{8}{6} = -\frac{4}{3} \).
At \( x = 0 \): \( -\frac{8}{3\sqrt{1}} = -\frac{8}{3} \).

Subtracting the lower limit value from the upper limit value:
\( V = \pi \left( -\frac{4}{3} - \left(-\frac{8}{3}\right) \right) = \pi \left( -\frac{4}{3} + \frac{8}{3} \right) = \frac{4}{3}\pi \).

M1: Attempt to square \( y \) to find \( y^2 \). A1: Correct expression \( y^2 = 4(3x+1)^{-3/2} \). M1: Integrate a term of the form \( k(3x+1)^{-3/2} \) to obtain \( c(3x+1)^{-1/2} \). A1: Correct integral \( -\frac{8}{3}(3x+1)^{-1/2} \). A1: Obtain the correct final volume of \( \frac{4}{3}\pi \) (or equivalent exact form).

To find the intersection of the line and the curve, we equate their expressions:
\[ (m+2)x^2 + 4x + m = 2mx + 1 \]
Rearranging into standard quadratic form \( Ax^2 + Bx + C = 0 \):
\[ (m+2)x^2 + (4-2m)x + (m-1) = 0 \]
For the line to be a tangent to the curve, the quadratic equation must have exactly one real root, meaning the discriminant \( B^2 - 4AC \) must equal zero:
\[ (4-2m)^2 - 4(m+2)(m-1) = 0 \]
Expanding the terms:
\[ (16 - 16m + 4m^2) - 4(m^2 + m - 2) = 0 \]
\[ 16 - 16m + 4m^2 - 4m^2 - 4m + 8 = 0 \]
Simplifying by combining like terms:
\[ 24 - 20m = 0 \]
\[ 20m = 24 \implies m = \frac{24}{20} = 1.2 \]
Since \( m = 1.2 \neq -2 \), the curve remains a valid quadratic and the tangent exists. Thus, the value of \( m \) is \( 1.2 \) (or \( \frac{6}{5} \)).

M1: Equate the line and curve equations and collect terms to form a quadratic equation in x. A1: Correctly identify the coefficients as \( A = m+2 \), \( B = 4-2m \), and \( C = m-1 \). M1: Set up the discriminant equation \( B^2 - 4AC = 0 \). A1: Correctly expand and simplify to get the linear equation \( 24 - 20m = 0 \) or equivalent. A1: Obtain \( m = 1.2 \) (or \( \frac{6}{5} \)).

(a) Completing the square for \( f(x) \) gives \( f(x) = (x-2)^2 - 1 \). For \( f \) to have an inverse, it must be a one-to-one function. The line of symmetry of the quadratic curve is \( x = 2 \). Since the domain is defined for \( x \le k \), the largest value of \( k \) is 2. (b)(i) Let \( y = (x-2)^2 - 1 \). Since the domain is restricted to \( x \le 2 \), we have \( x - 2 \le 0 \). Rearranging for \( x \): \( y + 1 = (x-2)^2 \implies x - 2 = -\sqrt{y+1} \implies x = 2 - \sqrt{y+1} \). Hence, \( f^{-1}(x) = 2 - \sqrt{x+1} \). The domain of \( f^{-1} \) is the range of \( f \). For \( x \le 2 \), \( (x-2)^2 \ge 0 \), so \( f(x) \ge -1 \). Thus, the domain of \( f^{-1} \) is \( x \ge -1 \). (b)(ii) Solve \( f^{-1}(x) = 0 \implies 2 - \sqrt{x+1} = 0 \implies \sqrt{x+1} = 2 \implies x+1 = 4 \implies x = 3 \). (Alternatively, \( f^{-1}(0) = 0 \implies f(0) = x \implies 0^2 - 4(0) + 3 = 3 \)). (c)(i) \( gf(x) = g(f(x)) = \frac{2}{2(x^2-4x+3)+3} = \frac{2}{2x^2 - 8x + 6 + 3} = \frac{2}{2x^2 - 8x + 9} \). For the domain of \( gf \), we require \( x \) to be in the domain of \( f \) (\( x \le 1 \)) and \( f(x) \) to be in the domain of \( g \) (\( f(x) > -1.5 \)). For \( x \le 1 \), \( f(x) = (x-2)^2 - 1 \ge 0 \), which is always greater than \( -1.5 \). Thus, the domain of \( gf \) is \( x \le 1 \). (c)(ii) To find the range of \( gf \) for \( x \le 1 \), we complete the square on the denominator: \( 2x^2 - 8x + 9 = 2(x-2)^2 + 1 \). For \( x \le 1 \), we have \( x-2 \le -1 \implies (x-2)^2 \ge 1 \). Therefore, the denominator is at least \( 2(1) + 1 = 3 \). Since the denominator increases without bound as \( x \to -\infty \) and has a minimum value of 3, the range of \( gf(x) = \frac{2}{\text{denominator}} \) is \( 0 < gf(x) \le \frac{2}{3} \).

(a) M1: For attempting to find the vertex of the quadratic. A1: For \( k = 2 \). (b)(i) M1: For completing the square. M1: For rearranging to make \( x \) the subject. M1: For choosing the negative root correctly based on the domain. A1: For \( f^{-1}(x) = 2 - \sqrt{x+1} \). A1: For domain \( x \ge -1 \). (b)(ii) M1: For setting \( f^{-1}(x) = 0 \) and solving or using \( f(0) = x \). A1: For \( x = 3 \). (c)(i) M1: For substituting \( f(x) \) into \( g(x) \). A1: For showing the given expression with no errors. A1: For stating the domain is \( x \le 1 \) with valid reason. (c)(ii) M1: For completing the square of the denominator or using the range of \( f \). A1: For finding the minimum value of the denominator is 3. M1: For realizing the maximum value of the fraction occurs when the denominator is minimized. A1: For upper bound of \( \frac{2}{3} \). A1: For lower bound of 0 and expressing the final range as \( 0 < gf(x) \le \frac{2}{3} \).

(a) Write \( y = 3(2x-3)^{1/2} \). Differentiating with respect to \( x \) using the chain rule: \( \frac{dy}{dx} = 3 \cdot \frac{1}{2}(2x-3)^{-1/2} \cdot 2 = 3(2x-3)^{-1/2} \). At \( x = 6 \), the gradient of the tangent is \( \frac{dy}{dx} = 3(12-3)^{-1/2} = 3(9)^{-1/2} = 1 \). The gradient of the normal is the negative reciprocal of 1, which is \( -1 \). The equation of the normal is \( y - 9 = -1(x - 6) \implies y - 9 = -x + 6 \implies x + y - 15 = 0 \). (b) First find the coordinates of the points of intersection: \( 3\sqrt{2x-3} = 2x-3 \). Let \( u = \sqrt{2x-3} \), then \( 3u = u^2 \implies u(u-3) = 0 \). Thus \( u = 0 \implies x = 1.5 \), and \( u = 3 \implies 2x - 3 = 9 \implies x = 6 \). The area is given by \( \int_{1.5}^{6} (3(2x-3)^{1/2} - (2x-3)) \, dx \). Integrating: \( \int 3(2x-3)^{1/2} \, dx = 3 \cdot \frac{(2x-3)^{3/2}}{\frac{3}{2} \cdot 2} = (2x-3)^{3/2} \). \( \int (2x-3) \, dx = x^2 - 3x \). Evaluating the limits: At \( x = 6 \): \( (12-3)^{3/2} - (36-18) = 27 - 18 = 9 \). At \( x = 1.5 \): \( (3-3)^{3/2} - (2.25-4.5) = 2.25 \). The area is \( 9 - 2.25 = 6.75 \). (c) The volume of the solid is \( V = \pi \int_{2}^{6} y^2 \, dx = \pi \int_{2}^{6} 9(2x-3) \, dx = \pi \int_{2}^{6} (18x - 27) \, dx \). Integrating: \( \int (18x - 27) \, dx = 9x^2 - 27x \). Substituting limits: At \( x = 6 \): \( 9(36) - 27(6) = 162 \). At \( x = 2 \): \( 9(4) - 27(2) = -18 \). Thus, \( V = \pi (162 - (-18)) = 180\pi \).

(a) M1: For differentiating with the chain rule. A1: For correct derivative \( 3(2x-3)^{-1/2} \). M1: For finding the normal gradient \( -1 \). M1: For using line formula with point \( (6, 9) \). A1: For \( x + y - 15 = 0 \). (b) M1: For attempting to solve the intersection equation. A1: For finding \( x = 1.5 \) and \( x = 6 \). M1: For correct integration of \( 3(2x-3)^{1/2} \) to get \( (2x-3)^{3/2} \). M1: For correct integration of \( 2x-3 \) to get \( x^2-3x \) (or equivalent). M1: For substituting limits \( 6 \) and \( 1.5 \) into their integrated expression. A1: For \( 6.75 \). (c) M1: For using \( V = \pi \int y^2 \, dx \) with limits 2 and 6. B1: For obtaining \( y^2 = 18x - 27 \). M1: For integrating to obtain \( 9x^2 - 27x \) (or equivalent). A1: For correct integration. M1: For substituting limits 6 and 2. A1: For \( 180\pi \).

We begin by taking the natural logarithm of both sides of the relation: \(y^2 = a b^x\) leads to \(\ln(y^2) = \ln(a b^x)\). Applying the laws of logarithms, we get \(2 \ln y = \ln a + x \ln b\), which simplifies to the linear form \(\ln y = \left(\frac{1}{2} \ln b\right) x + \frac{1}{2} \ln a\). Comparing this with the standard equation of a straight line, \(Y = mX + c\), we see that the gradient is \(m = \frac{1}{2} \ln b\) and the vertical intercept is \(c = \frac{1}{2} \ln a\). The line passes through the points \((1, 1.45)\) and \((4, 2.95)\). The gradient of the line is calculated as: \(m = \frac{2.95 - 1.45}{4 - 1} = \frac{1.5}{3} = 0.5\). Equating the gradient to the coefficient of \(x\): \(\frac{1}{2} \ln b = 0.5\), which gives \(\ln b = 1\), hence \(b = e^1 \approx 2.72\) (to 2 decimal places). To find \(a\), we can substitute the point \((1, 1.45)\) into the line equation: \(1.45 = 0.5(1) + c\), giving \(c = 0.95\). Equating the intercept to the constant term: \(\frac{1}{2} \ln a = 0.95\), which gives \(\ln a = 1.9\), hence \(a = e^{1.9} \approx 6.69\) (to 2 decimal places).

M1: For taking logarithms of both sides and expressing in a linear form, e.g., \(2 \ln y = \ln a + x \ln b\). M1: For calculating the gradient of the line from the coordinates, \(m = 0.5\), or establishing simultaneous equations. A1: For obtaining \(b = e \approx 2.72\). M1: For finding the vertical intercept \(c = 0.95\) and equating it to \(\frac{1}{2} \ln a\) (or solving equivalent equations for \(a\)). A1: For obtaining \(a = e^{1.9} \approx 6.69\).

To solve the inequality \(|2x - 5| < 3|x + 1|\), we can square both sides because both sides are non-negative: \[(2x - 5)^2 < 9(x + 1)^2\] Expanding both sides gives: \[4x^2 - 20x + 25 < 9(x^2 + 2x + 1)\] \[4x^2 - 20x + 25 < 9x^2 + 18x + 9\] Rearranging the terms to one side: \[5x^2 + 38x - 16 > 0\] Factorising the quadratic expression: \[(5x - 2)(x + 8) > 0\] The critical values are \(x = -8\) and \(x = \frac{2}{5} = 0.4\). Since the quadratic must be strictly greater than zero, we choose the regions outside of these critical values. Thus, the solution is \(x < -8\) or \(x > 0.4\) (or \(x > \frac{2}{5}\)).

M1: For squaring both sides and expanding to obtain a 3-term quadratic, or for solving the two linear equations \(2x - 5 = 3(x + 1)\) and \(2x - 5 = -3(x + 1)\). A1: For obtaining both critical values \(x = -8\) and \(x = 0.4\) (or \(x = \frac{2}{5}\)). M1: For identifying that the solution set lies outside the critical values (i.e., \(x < \text{smaller value}\) and \(x > \text{larger value}\)). A1: For the correct final solution: \(x < -8\) or \(x > 0.4\) (or equivalent).

We make the substitution \(u = \cos x\).
Then \(du = -\sin x \, dx\), which means \(\sin x \, dx = -du\).

We rewrite the numerator:
\(\sin^3 x \, dx = \sin^2 x (\sin x \, dx) = (1 - \cos^2 x)(-\sin x \, dx) = -(1 - u^2) \, du = (u^2 - 1) \, du\).

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

The integral becomes:
\[ \int_{1}^{\frac{1}{2}} \frac{u^2 - 1}{2 + u} \, du = \int_{\frac{1}{2}}^{1} \frac{1 - u^2}{u + 2} \, du \]

Using algebraic division to split the integrand:
\[ \frac{1 - u^2}{u + 2} = -u + 2 - \frac{3}{u + 2} \]

Now we integrate term by term:
\[ \int_{\frac{1}{2}}^{1} \left( -u + 2 - \frac{3}{u + 2} \right) \, du = \left[ 2u - \frac{1}{2}u^2 - 3\ln|u + 2| \right]_{\frac{1}{2}}^{1} \]

Substitute the upper limit \(u = 1\):
\[ \left( 2(1) - \frac{1}{2}(1)^2 - 3\ln(3) \right) = \frac{3}{2} - 3\ln(3) \]

Substitute the lower limit \(u = \frac{1}{2}\):
\[ \left( 2\left(\frac{1}{2}\right) - \frac{1}{2}\left(\frac{1}{2}\right)^2 - 3\ln\left(\frac{5}{2}\right) \right) = 1 - \frac{1}{8} - 3\ln\left(\frac{5}{2}\right) = \frac{7}{8} - 3\ln\left(\frac{5}{2}\right) \]

Subtract the values:
\[ \left(\frac{3}{2} - 3\ln(3)\right) - \left(\frac{7}{8} - 3\ln\left(\frac{5}{2}\right)\right) = \frac{5}{8} - 3\ln(3) + 3\ln\left(\frac{5}{2}\right) \]

Combine the logarithmic terms:
\[ \frac{5}{8} + 3\ln\left(\frac{5/2}{3}\right) = \frac{5}{8} + 3\ln\left(\frac{5}{6}\right) \]

Comparing with \(a + b\ln\left(\frac{5}{6}\right)\), we find \(a = \frac{5}{8}\) and \(b = 3\).

M1: Substitute \(u = \cos x\) and find correct differential relation \(du = -\sin x \, dx\).
A1: Obtain the correct limits of integration: \(1\) and \(0.5\).
M1: Perform correct polynomial division of the integrand to obtain \(-u + 2 - \frac{3}{u+2}\).
A1: Integrate correctly to get \(2u - \frac{1}{2}u^2 - 3\ln|u+2|\).
M1: Substitute the limits correctly into their integrated expression.
A1: Apply logarithmic laws correctly to combine \(\ln\) terms into a single term of the form \(\ln\left(\frac{5}{6}\right)\).
A1.2: Obtain correct final values \(a = \frac{5}{8}\) and \(b = 3\).

To find the stationary points, we first calculate the derivative \(\frac{dy}{dx}\) using the product rule:
\[ \frac{dy}{dx} = -2e^{-2x} \sin(2x) + 2e^{-2x} \cos(2x) \]
\[ \frac{dy}{dx} = 2e^{-2x} (\cos(2x) - \sin(2x)) \]

At a stationary point, \(\frac{dy}{dx} = 0\):
Since \(2e^{-2x} \neq 0\) for all real \(x\), we must have:
\[ \cos(2x) - \sin(2x) = 0 \implies \tan(2x) = 1 \]

Given that \(0 < x < \pi\), the interval for \(2x\) is \(0 < 2x < 2\pi\).
In this interval, \(\tan(2x) = 1\) has solutions:
\[ 2x = \frac{\pi}{4} \implies x = \frac{\pi}{8} \]
\[ 2x = \frac{5\pi}{4} \implies x = \frac{5\pi}{8} \]

To determine the nature of the stationary points, we find the second derivative \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} [2e^{-2x} (\cos(2x) - \sin(2x))] \]
\[ = -4e^{-2x} (\cos(2x) - \sin(2x)) + 2e^{-2x} (-2\sin(2x) - 2\cos(2x)) \]
\[ = e^{-2x} (-4\cos(2x) + 4\sin(2x) - 4\sin(2x) - 4\cos(2x)) \]
\[ = -8e^{-2x} \cos(2x) \]

Now evaluate the second derivative at each critical point:
- At \(x = \frac{\pi}{8}\):
\(\frac{d^2y}{dx^2} = -8e^{-\pi/4} \cos\left(\frac{\pi}{4}\right) = -8e^{-\pi/4} \left(\frac{1}{\sqrt{2}}\right) < 0\).
Thus, the point at \(x = \frac{\pi}{8}\) is a local maximum.

- At \(x = \frac{5\pi}{8}\):
\(\frac{d^2y}{dx^2} = -8e^{-5\pi/4} \cos\left(\frac{5\pi}{4}\right) = -8e^{-5\pi/4} \left(-\frac{1}{\sqrt{2}}\right) > 0\).
Thus, the point at \(x = \frac{5\pi}{8}\) is a local minimum.

M1: Apply product rule correctly to find \(\frac{dy}{dx}\).
A1: Obtain correct derivative \(\frac{dy}{dx} = 2e^{-2x}(\cos(2x) - \sin(2x))\).
M1: Set derivative to zero and solve \(\tan(2x) = 1\).
A1: Obtain first correct value \(x = \frac{\pi}{8}\).
A1: Obtain second correct value \(x = \frac{5\pi}{8}\).
M1: Differentiate again to find \(\frac{d^2y}{dx^2}\) (or use a valid first derivative sign test).
A1: Obtain correct second derivative \(-8e^{-2x}\cos(2x)\).
A1.2: Correctly identify \(x = \frac{\pi}{8}\) as a maximum and \(x = \frac{5\pi}{8}\) as a minimum with supporting calculations.

(i) We expand \(R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha\).

Comparing coefficients with \(3\sin\theta - 2\cos\theta\):
\(R\cos\alpha = 3\)
\(R\sin\alpha = 2\)

Squaring and adding:
\(R^2 = 3^2 + 2^2 = 13 \implies R = \sqrt{13}\).

Dividing the equations:
\(\tan\alpha = \frac{2}{3} \implies \alpha = \arctan\left(\frac{2}{3}\right) \approx 0.5880\) radians.

Thus, \(3\sin\theta - 2\cos\theta = \sqrt{13}\sin(\theta - 0.5880)\).

(ii) To solve \(3\sin(2x) - 2\cos(2x) = 1.5\), we use the result from part (i) with \(\theta = 2x\):
\[ \sqrt{13}\sin(2x - 0.5880) = 1.5 \]
\[ \sin(2x - 0.5880) = \frac{1.5}{\sqrt{13}} \approx 0.41596 \]

Let \(\phi = 2x - 0.5880\). Since \(0 < x < \pi\), the range for \(2x\) is \(0 < 2x < 2\pi\), which means \(-0.5880 < \phi < 2\pi - 0.5880 \approx 5.695\).

Finding the principal value:
\(\phi = \arcsin(0.41596) \approx 0.4289\) radians.

The second value in the interval is:
\(\phi = \pi - 0.4289 \approx 2.7127\) radians.

Now we solve for \(x\):
1) \(2x - 0.5880 = 0.4289 \implies 2x = 1.0169 \implies x \approx 0.508\)
2) \(2x - 0.5880 = 2.7127 \implies 2x = 3.3007 \implies x \approx 1.65\)

Part (i):
M1: Set up equations \(R\cos\alpha = 3\) and \(R\sin\alpha = 2\) or calculate \(R = \sqrt{3^2 + 2^2}\).
A1: Obtain \(R = \sqrt{13}\).
A1: Obtain \(\alpha = 0.5880\) (accept 0.588).

Part (ii):
M1: Substitute \(2x\) into their expression and equate to \(1.5\).
M1: Calculate the principal value of the inverse sine function (approx. 0.429).
A1: Identify the second valid angle in the range (approx. 2.713).
M1: Solve for both values of \(x\) by adding \(0.5880\) and dividing by 2.
A1.2: Correctly state \(x = 0.508\) and \(x = 1.65\).

(i) First, we find the gradient function by differentiating \(y = x \cos(2x)\) using the product rule:
\[ \frac{dy}{dx} = 1 \cdot \cos(2x) + x \cdot (-2\sin(2x)) = \cos(2x) - 2x\sin(2x) \]

At the origin \((0,0)\), we substitute \(x = 0\):
\[ m = \cos(0) - 2(0)\sin(0) = 1 \]

The equation of the tangent line at \((0,0)\) with gradient \(m = 1\) is:
\[ y - 0 = 1(x - 0) \implies y = x \]

(ii) The area of the region is given by the definite integral:
\[ A = \int_{0}^{\frac{\pi}{4}} x \cos(2x) \, dx \]

We integrate by parts using \(\int u \, dv = uv - \int v \, du\):
Let \(u = x \implies du = dx\)
Let \(dv = \cos(2x) \, dx \implies v = \frac{1}{2}\sin(2x)\)

Applying the integration by parts formula:
\[ A = \left[ \frac{1}{2}x\sin(2x) \right]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} \frac{1}{2}\sin(2x) \, dx \]
\[ A = \left[ \frac{1}{2}x\sin(2x) \right]_{0}^{\frac{\pi}{4}} - \left[ -\frac{1}{4}\cos(2x) \right]_{0}^{\frac{\pi}{4}} \]
\[ A = \left[ \frac{1}{2}x\sin(2x) + \frac{1}{4}\cos(2x) \right]_{0}^{\frac{\pi}{4}} \]

Now substitute the limits:
- Upper limit \(x = \frac{\pi}{4}\):
\[ \frac{1}{2}\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{2}\right) + \frac{1}{4}\cos\left(\frac{\pi}{2}\right) = \frac{\pi}{8}(1) + \frac{1}{4}(0) = \frac{\pi}{8} \]
- Lower limit \(x = 0\):
\[ \frac{1}{2}(0)\sin(0) + \frac{1}{4}\cos(0) = 0 + \frac{1}{4}(1) = \frac{1}{4} \]

Subtracting the lower limit from the upper limit:
\[ A = \frac{\pi}{8} - \frac{1}{4} \]

Part (i):
M1: Use product rule to differentiate \(x \cos(2x)\).
A1: Obtain \(dy/dx = \cos(2x) - 2x\sin(2x)\) and show gradient at \(x=0\) is \(1\), hence confirming \(y = x\).

Part (ii):
M1: State the correct integral for area \(\int_0^{\pi/4} x \cos(2x) \, dx\).
M1: Apply integration by parts correctly with suitable choices for \(u\) and \(dv\).
A1: Obtain the first part of the integration: \(\frac{1}{2}x\sin(2x)\).
A1: Obtain the second part of the integration: \(\frac{1}{4}\cos(2x)\).
M1: Substitute the limits of \(0\) and \(\frac{\pi}{4}\) correctly.
A1.2: Obtain the correct exact area \(\frac{\pi}{8} - \frac{1}{4}\).

(i) We expand \(\sin(3\theta)\) using the compound angle identity:
\[ \sin(3\theta) = \sin(2\theta + \theta) = \sin(2\theta)\cos\theta + \cos(2\theta)\sin\theta \]

Using the double angle identities \(\sin(2\theta) = 2\sin\theta\cos\theta\) and \(\cos(2\theta) = 1 - 2\sin^2\theta\):
\[ \sin(3\theta) = (2\sin\theta\cos\theta)\cos\theta + (1 - 2\sin^2\theta)\sin\theta \]
\[ \sin(3\theta) = 2\sin\theta\cos^2\theta + \sin\theta - 2\sin^3\theta \]

Substitute \(\cos^2\theta = 1 - \sin^2\theta\):
\[ \sin(3\theta) = 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta \]
\[ \sin(3\theta) = 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta \]
\[ \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \]

This completes the proof.

(ii) Rearranging the identity from part (i) to express \(\sin^3\theta\) in terms of single-power trigonometric functions:
\[ 4\sin^3\theta = 3\sin\theta - \sin(3\theta) \implies \sin^3\theta = \frac{3}{4}\sin\theta - \frac{1}{4}\sin(3\theta) \]

Now we integrate this expression between the limits \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\):
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^3 \theta \, d\theta = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{3}{4}\sin\theta - \frac{1}{4}\sin(3\theta) \right) d\theta \]
\[ = \left[ -\frac{3}{4}\cos\theta + \frac{1}{12}\cos(3\theta) \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \]

Substitute upper limit \(\theta = \frac{\pi}{3}\):
\[ -\frac{3}{4}\cos\left(\frac{\pi}{3}\right) + \frac{1}{12}\cos(\pi) = -\frac{3}{4}\left(\frac{1}{2}\right) + \frac{1}{12}(-1) = -\frac{3}{8} - \frac{1}{12} = -\frac{11}{24} \]

Substitute lower limit \(\theta = \frac{\pi}{6}\):
\[ -\frac{3}{4}\cos\left(\frac{\pi}{6}\right) + \frac{1}{12}\cos\left(\frac{\pi}{2}\right) = -\frac{3}{4}\left(\frac{\sqrt{3}}{2}\right) + 0 = -\frac{3\sqrt{3}}{8} = -\frac{9\sqrt{3}}{24} \]

Subtracting lower limit from upper limit:
\[ -\frac{11}{24} - \left( -\frac{9\sqrt{3}}{24} \right) = \frac{9\sqrt{3} - 11}{24} \]

Part (i):
M1: Apply compound angle identity to rewrite \(\sin(2\theta + \theta)\).
M1: Substitute double angle formulas for \(\sin(2\theta)\) and \(\cos(2\theta)\).
M1: Substitute \(\cos^2\theta = 1 - \sin^2\theta\) to express everything in terms of \(\sin\theta\).
A1: Obtain the final form \(3\sin\theta - 4\sin^3\theta\) showing all intermediate steps.

Part (ii):
M1: Rearrange identity to make \(\sin^3\theta\) the subject.
A1: Integrate correctly to get \(-\frac{3}{4}\cos\theta + \frac{1}{12}\cos(3\theta)\).
M1: Correctly substitute limits \(\frac{\pi}{3}\) and \(\frac{\pi}{6}\).
A1.2: Obtain the exact answer \(\frac{9\sqrt{3} - 11}{24}\) in its simplest form.

(i) Multiplying the numerator and denominator by the complex conjugate of the denominator:
\[ z = \frac{(a + 3\mathrm{i})(2 + \mathrm{i})}{(2 - \mathrm{i})(2 + \mathrm{i})} = \frac{2a + a\mathrm{i} + 6\mathrm{i} - 3}{5} = \frac{(2a - 3) + (a + 6)\mathrm{i}}{5} \]
Since \( \arg(z) = \frac{3\pi}{4} \), we must have \( \operatorname{Re}(z) < 0 \), \( \operatorname{Im}(z) > 0 \) and \( \operatorname{Im}(z) = -\operatorname{Re}(z) \).
This gives:
\[ a + 6 = -(2a - 3) \implies a + 6 = -2a + 3 \implies 3a = -3 \implies a = -1 \]
Checking the signs:
For \( a = -1 \), \( \operatorname{Re}(z) = \frac{-5}{5} = -1 < 0 \) and \( \operatorname{Im}(z) = \frac{5}{5} = 1 > 0 \). Thus \( a = -1 \) is verified.

(ii) With \( a = -1 \), \( z = -1 + \mathrm{i} \).
Let the square roots of \( z \) be \( x + \mathrm{i}y \), where \( x, y \in \mathbb{R} \).
\[ (x + \mathrm{i}y)^2 = -1 + \mathrm{i} \implies x^2 - y^2 + 2xy\mathrm{i} = -1 + \mathrm{i} \]
Equating real and imaginary parts:
\[ x^2 - y^2 = -1 \quad (1) \]
\[ 2xy = 1 \implies y = \frac{1}{2x} \quad (2) \]
Substitute (2) into (1):
\[ x^2 - \frac{1}{4x^2} = -1 \implies 4x^4 + 4x^2 - 1 = 0 \]
Using the quadratic formula for \( x^2 \):
\[ x^2 = \frac{-4 \pm \sqrt{16 - 4(4)(-1)}}{8} = \frac{-4 \pm \sqrt{32}}{8} = \frac{-4 \pm 4\sqrt{2}}{8} = -\frac{1}{2} \pm \frac{\sqrt{2}}{2} \]
Since \( x \) is real, \( x^2 \ge 0 \). Therefore:
\[ x^2 = \frac{\sqrt{2} - 1}{2} \implies x = \pm \sqrt{\frac{\sqrt{2}-1}{2}} \]
From (2), \( y \) has the same sign as \( x \):
\[ y = \frac{1}{2x} = \pm \frac{1}{\sqrt{2(\sqrt{2}-1)}} = \pm \sqrt{\frac{\sqrt{2}+1}{2}} \]
Thus, the square roots are:
\[ \pm \left( \sqrt{\frac{\sqrt{2}-1}{2}} + \mathrm{i}\sqrt{\frac{\sqrt{2}+1}{2}} \right) \]

(i)
M1: Expresses \( z \) in the form \( X + \mathrm{i}Y \) by multiplying numerator and denominator by \( 2 + \mathrm{i} \).
A1: Obtains \( z = \frac{(2a-3) + (a+6)\mathrm{i}}{5} \).
M1: Uses \( \arg(z) = \frac{3\pi}{4} \) to set up the equation \( Y = -X \) (or equivalent tangent relationship with attention to quadrant).
A1: Solves correctly to obtain \( a = -1 \).

(ii)
M1: Sets up equations \( x^2 - y^2 = -1 \) and \( 2xy = 1 \).
M1: Eliminates one variable to form a quadratic in \( x^2 \) (or \( y^2 \)).
A1: Solves the quadratic to find exact positive value of \( x^2 = \frac{\sqrt{2}-1}{2} \).
A1: States both roots correctly in exact form.

(i) Let \( u = 1 + \sqrt{x} \).
Then \( \sqrt{x} = u - 1 \implies x = (u - 1)^2 \).
Differentiating with respect to \( u \):
\[ \frac{\mathrm{d}x}{\mathrm{d}u} = 2(u - 1) \implies \mathrm{d}x = 2(u - 1) \mathrm{d}u \]
Now find the limits of integration:
When \( x = 0 \), \( u = 1 + \sqrt{0} = 1 \).
When \( x = 4 \), \( u = 1 + \sqrt{4} = 3 \).
Substitute into the integral:
\[ I = \int_{1}^{3} \frac{1}{u} \cdot 2(u-1) \mathrm{d}u = \int_{1}^{3} \frac{2u - 2}{u} \mathrm{d}u = \int_{1}^{3} \left( 2 - \frac{2}{u} \right) \mathrm{d}u \]
as required.

(ii) Integrating the expression term-by-term:
\[ I = \left[ 2u - 2\ln u \right]_{1}^{3} \]
Evaluate at the upper and lower limits:
\[ I = (2(3) - 2\ln 3) - (2(1) - 2\ln 1) \]
\[ I = 6 - 2\ln 3 - 2 + 0 = 4 - 2\ln 3 \]
Here, \( a = 4 \) and \( b = 2 \).

(i)
M1: Expresses \( x \) in terms of \( u \) as \( x = (u-1)^2 \).
M1: Differentiates to find expression for \( \mathrm{d}x \) in terms of \( u \) and \( \mathrm{d}u \).
A1: Correctly obtains \( \mathrm{d}x = 2(u-1)\mathrm{d}u \).
B1: Correctly converts limits to \( 1 \) and \( 3 \).
A1: Substitutes all components correctly to obtain the given integrand.

(ii)
M1: Integrates \( 2 - \frac{2}{u} \) to obtain \( 2u - 2\ln u \).
M1: Applies the limits \( 1 \) and \( 3 \) correctly.
A1: Obtains \( 4 - 2\ln 3 \).

(i) We require:
\[ 3\sin 2x - \cos 2x = R\sin(2x - \alpha) = R\sin 2x\cos\alpha - R\cos 2x\sin\alpha \]
Comparing coefficients:
\[ R\cos\alpha = 3 \quad \text{and} \quad R\sin\alpha = 1 \]
Squaring and adding:
\[ R^2 = 3^2 + 1^2 = 10 \implies R = \sqrt{10} \]
Dividing the equations:
\[ \tan\alpha = \frac{1}{3} \implies \alpha = \arctan\left(\frac{1}{3}\right) \approx 0.3217505... \]
Rounding to 4 decimal places gives \( \alpha = 0.3218 \) radians.

(ii) Using the result from part (i):
\[ \sqrt{10}\sin(2x - 0.32175) = 1 \implies \sin(2x - 0.32175) = \frac{1}{\sqrt{10}} \]
Let \( \theta = 2x - 0.32175 \). Since \( 0 \le x \le \pi \), we have \( -0.32175 \le 2x - 0.32175 \le 2\pi - 0.32175 \).
Find the principal angle:
\[ \theta_{\text{principal}} = \arcsin\left(\frac{1}{\sqrt{10}}\right) \approx 0.32175 \]
Within the range, the solutions for \( \theta \) are:
\[ \theta_1 = 0.32175 \]
\[ \theta_2 = \pi - 0.32175 \approx 2.81984 \]
Solve for \( x \):
\[ 2x - 0.32175 = 0.32175 \implies 2x = 0.64350 \implies x \approx 0.32175 \approx 0.322 \]
\[ 2x - 0.32175 = 2.81984 \implies 2x = 3.14159 \implies x \approx 1.5708 \approx 1.57 \] (or \( \frac{\pi}{2} \)).
So the solutions are \( x = 0.322 \) and \( x = 1.57 \).

(i)
M1: Identifies \( R^2 = 3^2 + 1^2 \) or uses trig expansion to form equations for \( R \) and \( \alpha \).
A1: Obtains \( R = \sqrt{10} \).
M1: Uses \( \tan\alpha = \frac{1}{3} \) to find \( \alpha \).
A1: Obtains \( \alpha = 0.3218 \) (or better, e.g. 0.32175).

(ii)
M1: Sets up the equation \( \sin(2x - \alpha) = \frac{1}{\sqrt{10}} \) and finds at least one value for \( 2x - \alpha \).
A1: Obtains \( 2x - 0.3218 = 0.3218 \) or \( 2x - 0.3218 = 2.820 \).
M1: Correctly solves for both values of \( x \) in the interval.
A1: States both solutions \( x = 0.322 \) and \( x = 1.57 \) (accept \( \pi/2 \)).

(i) For \( l_1 \) and \( l_2 \) to intersect, there must exist scalars \( \lambda \) and \( \mu \) such that:
\[ 1 + 2\lambda = a + \mu \quad (1) \]
\[ 2 - \lambda = 1 + \mu \quad (2) \]
\[ -1 + 3\lambda = 4 - 2\mu \quad (3) \]
From equation (2):
\[ \mu = 1 - \lambda \]
Substitute this expression for \( \mu \) into equation (3):
\[ -1 + 3\lambda = 4 - 2(1 - \lambda) \implies -1 + 3\lambda = 2 + 2\lambda \implies \lambda = 3 \]
Then substitute \( \lambda = 3 \) back into the expression for \( \mu \):
\[ \mu = 1 - 3 = -2 \]
Using equation (1) to find \( a \):
\[ 1 + 2(3) = a + (-2) \implies 7 = a - 2 \implies a = 9 \]
Using \( \lambda = 3 \) in \( l_1 \) (or \( \mu = -2 \) in \( l_2 \)), the position vector of the intersection point is:
\[ \mathbf{r} = \begin{pmatrix} 1 + 2(3) \\ 2 - 3 \\ -1 + 3(3) \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \\ 8 \end{pmatrix} \]

(ii) The direction vectors of the lines are \( \mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \) and \( \mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} \).
The angle \( \theta \) between them is given by:
\[ \cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|} \]
Evaluate the dot product and magnitudes:
\[ \mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-1)(1) + (3)(-2) = 2 - 1 - 6 = -5 \]
\[ |\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14} \]
\[ |\mathbf{d}_2| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6} \]
\[ \cos\theta = \frac{-5}{\sqrt{14}\sqrt{6}} = \frac{-5}{\sqrt{84}} \]
For the acute angle, we take the absolute value:
\[ \cos\theta = \frac{5}{\sqrt{84}} \implies \theta = \arccos\left(\frac{5}{\sqrt{84}}\right) \approx 56.94^\circ \approx 56.9^\circ \]

(i)
M1: Sets up three simultaneous equations in \( \lambda \), \( \mu \) and \( a \).
M1: Solves two equations (without \( a \)) to find \( \lambda \) and \( \mu \).
A1: Obtains \( \lambda = 3 \) and \( \mu = -2 \) (or equivalent).
A1: Shows that \( a = 9 \).
A1: Obtains the correct position vector \( \begin{pmatrix} 7 \\ -1 \\ 8 \end{pmatrix} \).

(ii)
M1: Employs the scalar product formula \( \cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|} \) using the direction vectors.
A1: Obtains \( \cos\theta = \pm\frac{5}{\sqrt{84}} \).
A1: Finds acute angle \( 56.9^\circ \).

(i) Separate variables:
\[ \int \frac{1}{y^2} \mathrm{d}y = \int \frac{\ln x}{x} \mathrm{d}x \]
The left hand side integrates to:
\[ \int y^{-2} \mathrm{d}y = -\frac{1}{y} \]
For the right hand side, let \( u = \ln x \implies \mathrm{d}u = \frac{1}{x} \mathrm{d}x \):
\[ \int u \, \mathrm{d}u = \frac{1}{2} u^2 + C = \frac{1}{2}(\ln x)^2 + C \]
Combining these gives:
\[ -\frac{1}{y} = \frac{1}{2}(\ln x)^2 + C \]
Using the boundary condition \( y = -1 \) when \( x = 1 \):
\[ -\frac{1}{-1} = \frac{1}{2}(\ln 1)^2 + C \implies 1 = 0 + C \implies C = 1 \]
Thus, the equation relating \( y \) and \( x \) is:
\[ -\frac{1}{y} = \frac{1}{2}(\ln x)^2 + 1 \]

(ii) To express \( y \) explicitly in terms of \( x \), rearrange the expression:
\[ -\frac{1}{y} = \frac{(\ln x)^2 + 2}{2} \]
Taking the reciprocal and multiplying by \( -1 \):
\[ y = -\frac{2}{(\ln x)^2 + 2} \]

(i)
M1: Separates variables correctly to obtain \( \int \frac{1}{y^2} \mathrm{d}y = \int \frac{\ln x}{x} \mathrm{d}x \).
A1: Integrates LHS to get \( -\frac{1}{y} \).
M1: Integrates RHS using substitution or inspection.
A1: Integrates RHS to get \( \frac{1}{2}(\ln x)^2 \).
M1: Substitutes \( y = -1 \) and \( x = 1 \) to evaluate the constant of integration \( C \).
A1: Obtains correct relation, e.g. \( -\frac{1}{y} = \frac{1}{2}(\ln x)^2 + 1 \).

(ii)
M1: Algebrically manipulates the equation to make \( y \) the subject.
A1: Obtains the correct explicit form \( y = -\frac{2}{(\ln x)^2 + 2} \).

(i) Since the denominator has a linear factor and an irreducible quadratic factor, we write:
\[ \frac{4x^2 - x + 3}{(x-1)(2x^2+1)} = \frac{A}{x-1} + \frac{Bx + C}{2x^2+1} \]
Multiply both sides by the denominator:
\[ 4x^2 - x + 3 = A(2x^2+1) + (Bx + C)(x-1) \]
Let \( x = 1 \):
\[ 4(1)^2 - (1) + 3 = A(2(1)^2+1) \implies 6 = 3A \implies A = 2 \]
Comparing coefficients of \( x^2 \):
\[ 4 = 2A + B \implies 4 = 4 + B \implies B = 0 \]
Comparing constant terms:
\[ 3 = A - C \implies 3 = 2 - C \implies C = -1 \]
Thus, the partial fractions representation is:
\[ f(x) = \frac{2}{x-1} - \frac{1}{2x^2+1} \]

(ii) Expand each term using binomial expansion:
First term:
\[ \frac{2}{x-1} = -2(1-x)^{-1} = -2(1 + x + x^2 + \dots) = -2 - 2x - 2x^2 - \dots \]
Second term:
\[ -\frac{1}{2x^2+1} = -(1 + 2x^2)^{-1} = -(1 - 2x^2 + \dots) = -1 + 2x^2 - \dots \]
Adding the expansions together:
\[ f(x) = (-2 - 2x - 2x^2) + (-1 + 2x^2) + \dots = -3 - 2x + 0x^2 \]
Thus, the expansion of \( f(x) \) up to and including the term in \( x^2 \) is \( -3 - 2x \).

(i)
M1: Expresses in correct form \( \frac{A}{x-1} + \frac{Bx+C}{2x^2+1} \).
M1: Uses a valid method to find at least one constant.
A1: Obtains \( A = 2 \).
A1: Obtains \( B = 0 \) and \( C = -1 \).

(ii)
M1: Expands \( \frac{2}{x-1} \) as \( -2(1-x)^{-1} \) to get \( -2 - 2x - 2x^2 \).
M1: Expands \( -\frac{1}{2x^2+1} \) as \( -(1+2x^2)^{-1} \) to get \( -1 + 2x^2 \).
A1: Combines terms correctly to yield \( -3 - 2x \).

(i) Let \( f(x) = x^3 - 3x - 1 \).
Evaluate \( f(x) \) at \( x = 1.5 \) and \( x = 2.0 \):
\[ f(1.5) = (1.5)^3 - 3(1.5) - 1 = 3.375 - 4.5 - 1 = -2.125 \]
\[ f(2.0) = (2.0)^3 - 3(2.0) - 1 = 8 - 6 - 1 = 1 \]
Since \( f(1.5) < 0 \) and \( f(2.0) > 0 \), and \( f \) is continuous, there must be a root of \( f(x) = 0 \) in the interval \( (1.5, 2.0) \).

(ii) Let the sequence converge to a limit \( L \). Then:
\[ L = \sqrt[3]{3L + 1} \]
Cube both sides:
\[ L^3 = 3L + 1 \implies L^3 - 3L - 1 = 0 \]
This is the same equation as \( x^3 - 3x - 1 = 0 \), so the limit is indeed the root \( \alpha \).

(iii) Choose \( x_1 = 1.8 \):
\[ x_2 = \sqrt[3]{3(1.8) + 1} = \sqrt[3]{6.4} \approx 1.85664 \]
\[ x_3 = \sqrt[3]{3(1.85664) + 1} \approx 1.87291 \]
\[ x_4 = \sqrt[3]{3(1.87291) + 1} \approx 1.87754 \]
\[ x_5 = \sqrt[3]{3(1.87754) + 1} \approx 1.87885 \]
\[ x_6 = \sqrt[3]{3(1.87885) + 1} \approx 1.87922 \]
\[ x_7 = \sqrt[3]{3(1.87922) + 1} \approx 1.87933 \]
\[ x_8 = \sqrt[3]{3(1.87933) + 1} \approx 1.87936 \]
Since the values are converging to \( 1.879 \), \( \alpha \approx 1.879 \) to 3 decimal places.

(i)
M1: Evaluates \( f(1.5) \) and \( f(2.0) \) with at least one correct calculation.
A1: Obtains correct signs and concludes that a root exists in the interval.

(ii)
M1: Substitutes \( L \) into \( L = \sqrt[3]{3L+1} \) (or equivalent with \( x \)).
A1: Cubes and rearranges to show equivalence to original equation.

(iii)
M1: Initiates iteration with a suitable value in the interval, e.g., \( 1.8 \).
A1: Obtains at least 3 correct iterations to 5 decimal places.
A1: Shows iterations continuing to convergence.
A1: States final answer \( 1.879 \) correct to 3 d.p.

(i) Using the quotient rule with \( u = \ln x \) and \( v = x^2 + 1 \):
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x} \quad \text{and} \quad \frac{\mathrm{d}v}{\mathrm{d}x} = 2x \]
Applying the formula:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{u'v - uv'}{v^2} = \frac{\frac{1}{x}(x^2 + 1) - 2x\ln x}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2\ln x}{x(x^2 + 1)^2} \]

(ii) At the stationary point, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\[ \frac{x^2 + 1 - 2x^2\ln x}{x(x^2 + 1)^2} = 0 \]
Since the denominator is non-zero for \( x > 0 \), we equate the numerator to zero:
\[ x^2 + 1 - 2x^2\ln x = 0 \]
Rearranging to make \( \ln x \) the subject:
\[ 2x^2\ln x = x^2 + 1 \implies \ln x = \frac{x^2 + 1}{2x^2} \]
Taking exponentials on both sides:
\[ x = \mathrm{e}^{\frac{x^2 + 1}{2x^2}} \]
as required.

(i)
M1: Applies quotient rule (or product rule) correctly.
A1: Correct derivative of \( \ln x \) and \( x^2+1 \).
A1: Obtains correct unsimplified or simplified derivative.

(ii)
M1: Sets derivative equal to 0.
A1: Obtains numerator equal to 0 correctly: \( x^2 + 1 - 2x^2\ln x = 0 \).
M1: Rearranges the equation to express \( \ln x \) in terms of \( x \).
A1: Completes the proof to show \( x = \mathrm{e}^{\frac{x^2+1}{2x^2}} \) convincingly.

(i) To prove the identity, we use the double angle formulae:
\[\cos 2\theta = 1 - 2\sin^2 \theta = 2\cos^2 \theta - 1\]
and
\[\sin 2\theta = 2\sin \theta \cos \theta\]

Substitute these into the numerator of the left-hand side:
\[1 - \cos 2\theta + \sin 2\theta = 1 - (1 - 2\sin^2 \theta) + 2\sin \theta \cos \theta\]
\[= 2\sin^2 \theta + 2\sin \theta \cos \theta\]
\[= 2\sin \theta (\sin \theta + \cos \theta)\]

Substitute into the denominator of the left-hand side:
\[1 + \cos 2\theta + \sin 2\theta = 1 + (2\cos^2 \theta - 1) + 2\sin \theta \cos \theta\]
\[= 2\cos^2 \theta + 2\sin \theta \cos \theta\]
\[= 2\cos \theta (\cos \theta + \sin \theta)\]

Combining these, we obtain:
\[\frac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta} = \frac{2\sin \theta (\sin \theta + \cos \theta)}{2\cos \theta (\cos \theta + \sin \theta)} = \frac{\sin \theta}{\cos \theta} = \tan \theta\]

(ii) Using the identity proven in part (i), the equation simplifies to:
\[\tan \theta = 3 \sec^2 \theta - 5\]

We use the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\):
\[\tan \theta = 3(1 + \tan^2 \theta) - 5\]
\[\tan \theta = 3\tan^2 \theta - 2\]
\[3\tan^2 \theta - \tan \theta - 2 = 0\]

Letting \(u = \tan \theta\), we can factorise the quadratic equation:
\[(3u + 2)(u - 1) = 0\]

This gives two cases:
1. \(\tan \theta = 1\)
Since \(0 < \theta < \pi\), the solution is:
\[\theta = \frac{\pi}{4} \approx 0.785\]

2. \(\tan \theta = -\frac{2}{3}\)
Since \(0 < \theta < \pi\), the angle must be in the second quadrant:
\[\theta = \pi - \arctan\left(\frac{2}{3}\right) \approx 3.1416 - 0.5880 = 2.55\]

Part (i):
M1: Use double angle formulae \(\cos 2\theta = 1 - 2\sin^2 \theta\) or \(\cos 2\theta = 2\cos^2 \theta - 1\) and \(\sin 2\theta = 2\sin\theta\cos\theta\) correctly in LHS.
A1: Obtain \(2\sin\theta(\sin\theta + \cos\theta)\) for the numerator and \(2\cos\theta(\cos\theta + \sin\theta)\) for the denominator.
M1: Factorise and cancel the common factor \(2(\sin\theta + \cos\theta)\).
A1: Arrive at \ an \theta\ with no errors seen.

Part (ii):
M1: Substitute the result from (i) and use \(\sec^2\theta = 1 + \tan^2\theta\) to form a quadratic equation in \(\tan\theta\).
A1: Obtain \(3\tan^2 \theta - \tan \theta - 2 = 0\) and find \(\tan\theta = 1, -2/3\).
B1: For obtaining \(\theta = \frac{\pi}{4}\) (or 0.785).
B1: For obtaining \(\theta = 2.55\) (allow 2.54 to 2.56; reject 2.5 since 3 s.f. is required).

(i) The equation \(|z - (3+4\mathrm{i})| = 2\) represents a circle in the Argand diagram.
- The centre of the circle is at \(C(3, 4)\).
- The radius of the circle is \(r = 2\).
The sketch must show a circle with centre in the first quadrant. Since the distance from the centre to the real axis is 4 (which is greater than 2) and to the imaginary axis is 3 (which is greater than 2), the circle must not intersect either axis.

(ii) The distance from the origin \(O(0,0)\) to the centre of the circle \(C(3,4)\) is:
\[OC = \sqrt{3^2 + 4^2} = 5\]
The maximum distance from the origin to a point on the circle is:
\[|z|_{\text{max}} = OC + r = 5 + 2 = 7\]
The minimum distance from the origin to a point on the circle is:
\[|z|_{\text{min}} = OC - r = 5 - 2 = 3\]

(iii) Let \(\theta\) be the argument of the centre \(C\):
\[\theta = \arctan\left(\frac{4}{3}\right) \approx 0.9273\text{ radians}\]
Let \(\phi\) be the angle between the line \(OC\) and the tangent from the origin to the circle. In the right-angled triangle formed by the origin, the centre, and the point of tangency:
\[\sin \phi = \frac{\text{radius}}{\text{hypotenuse}} = \frac{2}{5} = 0.4\]
\[\phi = \arcsin(0.4) \approx 0.4115\text{ radians}\]
The maximum value of \(\arg z\) is given by the upper tangent:
\[\arg z_{\text{max}} = \theta + \phi = \arctan\left(\frac{4}{3}\right) + \arcsin(0.4)\]
\[\approx 0.9273 + 0.4115 = 1.3388 \approx 1.34\text{ radians}\]

Part (i):
M1: Sketch a circle with centre in the first quadrant.
A1: Identify the centre as \((3, 4)\) and draw the circle such that it does not cross or touch either axis.

Part (ii):
M1: Calculate the distance from the origin to the centre \(OC = 5\).
A1: Find the exact maximum value \(7\) and exact minimum value \(3\).

Part (iii):
M1: Identify that the maximum argument is given by \(\theta + \phi\) where \(\theta = \arg(3+4\mathrm{i})\) and \(\sin\phi = 2/5\).
A1: Calculate either \(\theta \approx 0.927\) or \(\phi \approx 0.412\).
A1: Obtain the final answer of \(1.34\) (allow 1.33 to 1.35).

First, calculate the components of the weight of the block:
- Component down the plane: \( W_{\parallel} = mg \sin\alpha = 8 \times 10 \times 0.6 = 48 \text{ N} \)
- Component perpendicular to the plane: \( W_{\perp} = mg \cos\alpha = 8 \times 10 \times 0.8 = 64 \text{ N} \)

Since the block is in equilibrium, the normal reaction \( R \) is equal to \( W_{\perp} \), so \( R = 64 \text{ N} \).
The maximum frictional force that can act is \( F_{\max} = \mu R = 0.25 \times 64 = 16 \text{ N} \).

We consider two limiting cases:
1. The block is on the point of sliding down the plane. In this case, friction acts upwards:
\( P + F_{\max} = mg \sin\alpha \implies P + 16 = 48 \implies P = 32 \text{ N} \).

2. The block is on the point of sliding up the plane. In this case, friction acts downwards:
\( P - F_{\max} = mg \sin\alpha \implies P - 16 = 48 \implies P = 64 \text{ N} \).

Therefore, the range of values for which the block remains in equilibrium is \( 32 \le P \le 64 \).

M1: Resolving forces perpendicular to the plane to find the normal reaction \( R = 64 \text{ N} \).
A1: Calculating maximum friction force \( F_{\max} = 16 \text{ N} \).
M1: Resolving forces parallel to the plane for the case of sliding downwards to find the lower bound of \( P \).
A1: Finding \( P = 32 \text{ N} \).
M1: Resolving forces parallel to the plane for the case of sliding upwards to find the upper bound of \( P \).
A1.25: Correctly stating the range \( 32 \le P \le 64 \).

(i) To find the acceleration, we resolve forces horizontally:
\( T \cos 20^\circ - R_f = m a \)
\( 80 \cos 20^\circ - 15 = 25 a \)
\( 75.175 - 15 = 25 a \implies 60.175 = 25 a \implies a \approx 2.41 \text{ m s}^{-2} \) (to 3 s.f.).

(ii) To find the normal reaction, we resolve forces vertically:
\( R + T \sin 20^\circ - mg = 0 \)
\( R + 80 \sin 20^\circ - 25(10) = 0 \)
\( R + 27.361 - 250 = 0 \implies R \approx 223 \text{ N} \) (to 3 s.f.).

M1: Resolving forces horizontally to set up the equation of motion.
A1: Finding the correct acceleration \( a = 2.41 \text{ m s}^{-2} \).
M1: Resolving forces vertically to include weight, reaction, and vertical component of tension.
A1: Setting up correct vertical equation \( R + T \sin 20^\circ = mg \).
A2.25: Correct calculation of normal reaction \( R = 223 \text{ N} \) (or 222.6 N).

(i) Let the initial direction of motion of particle \( A \) be the positive direction.
- Initial velocity of \( A \): \( v_{A1} = 4 \text{ m s}^{-1} \)
- Initial velocity of \( B \): \( v_{B1} = -u \text{ m s}^{-1} \)
- Final velocity of \( A \): \( v_{A2} = -2 \text{ m s}^{-1} \)
- Final velocity of \( B \): \( v_{B2} = 1 \text{ m s}^{-1} \)

By conservation of momentum:
\( m_A v_{A1} + m_B v_{B1} = m_A v_{A2} + m_B v_{B2} \)
\( 0.3(4) + 0.5(-u) = 0.3(-2) + 0.5(1) \)
\( 1.2 - 0.5u = -0.6 + 0.5 \)
\( 1.2 - 0.5u = -0.1 \implies 0.5u = 1.3 \implies u = 2.6 \text{ m s}^{-1} \).

(ii) The impulse exerted on \( B \) is equal to the change in momentum of \( B \):
\( I = m_B (v_{B2} - v_{B1}) = 0.5(1 - (-2.6)) = 0.5(3.6) = 1.8 \text{ N s} \).
Alternatively, from \( A \)'s change of momentum: \( |0.3(-2 - 4)| = 1.8 \text{ N s} \).

M1: Applying the conservation of momentum with consistent signs.
A1: Finding correct conservation equation \( 1.2 - 0.5u = -0.1 \).
A1: Solving to find \( u = 2.6 \).
M1: Stating formula for impulse (change in momentum of either particle).
A2.25: Correctly calculating the impulse magnitude as \( 1.8 \text{ N s} \).

(i) Power \( P = 36000 \text{ W} \). At speed \( v = 15 \text{ m s}^{-1} \), the driving force \( F_D \) is:
\( F_D = \frac{P}{v} = \frac{36000}{15} = 2400 \text{ N} \).

The resistance force is \( R_f = 450 \text{ N} \), and the component of gravity acting down the incline is:
\( F_g = mg \sin\theta = 1200 \times 10 \times 0.05 = 600 \text{ N} \).

Using Newton's second law along the incline:
\( F_D - R_f - F_g = m a \)
\( 2400 - 450 - 600 = 1200 a \implies 1350 = 1200 a \implies a = 1.125 \text{ m s}^{-2} \).

(ii) At steady speed, the acceleration is zero, so the driving force equals the sum of resistance and gravitational components:
\( F_D = 450 + 600 = 1050 \text{ N} \).

Using \( P = F_D v_{\max} \):
\( v_{\max} = \frac{36000}{1050} = \frac{240}{7} \approx 34.3 \text{ m s}^{-1} \).

M1: Calculating the driving force at \( v = 15 \) using \( P/v \).
A1: Calculating gravity component down the slope as \( 600 \text{ N} \).
M1: Applying Newton's second law along the incline to solve for acceleration.
A1: Finding acceleration \( a = 1.125 \text{ m s}^{-2} \).
M1: Setting driving force equal to total resistance to represent steady speed.
A1.25: Correctly calculating steady speed as \( 34.3 \text{ m s}^{-1} \).

(i) The particle is instantaneously at rest when \( v(t) = 0 \):
\( 3t^2 - 14t + 8 = 0 \implies (3t - 2)(t - 4) = 0 \implies t = \frac{2}{3} \text{ s} \) and \( t = 4 \text{ s} \).

(ii) Acceleration is the derivative of velocity:
\( a(t) = \frac{dv}{dt} = 6t - 14 \).
At \( t = 3 \):
\( a(3) = 6(3) - 14 = 4 \text{ m s}^{-2} \).

(iii) The particle changes direction at \( t = \frac{2}{3} \) and \( t = 4 \). To find the total distance travelled from \( t = 0 \) to \( t = 4 \), we integrate to find the displacement function \( s(t) \):
\( s(t) = \int (3t^2 - 14t + 8) dt = t^3 - 7t^2 + 8t + c \).
Since \( s(0) = 0 \), \( c = 0 \), so \( s(t) = t^3 - 7t^2 + 8t \).

Now calculate the displacement at the key times:
- \( s(0) = 0 \)
- \( s(2/3) = (2/3)^3 - 7(2/3)^2 + 8(2/3) = \frac{8}{27} - \frac{28}{9} + \frac{16}{3} = \frac{68}{27} \approx 2.519 \text{ m} \)
- \( s(4) = 4^3 - 7(4)^2 + 8(4) = 64 - 112 + 32 = -16 \text{ m} \)

Total distance travelled:
\( D = |s(2/3) - s(0)| + |s(4) - s(2/3)| = \frac{68}{27} + |-16 - \frac{68}{27}| = \frac{68}{27} + \frac{500}{27} = \frac{568}{27} \approx 21.0 \text{ m} \) (to 3 s.f.).

M1: Equating velocity to zero and solving the quadratic equation.
A1: Finding \( t = 2/3 \) and \( t = 4 \).
M1: Differentiating velocity to find acceleration, and evaluating at \( t = 3 \) to get \( 4 \text{ m s}^{-2} \).
M1: Integrating velocity to find the displacement function \( s(t) \).
A1: Correctly evaluating displacement at the turning point \( t = 2/3 \).
A1.25: Summing the absolute values of displacement intervals to obtain the correct total distance \( 21.0 \text{ m} \) (or \( 568/27 \)).

(i) For block \( B \) (moving downwards):
\( m_B g - T = m_B a \implies 45 - T = 4.5 a \)

For block \( A \) (moving upwards):
\( T - m_A g = m_A a \implies T - 35 = 3.5 a \)

Adding the two equations:
\( 10 = 8 a \implies a = 1.25 \text{ m s}^{-2} \).

Substituting \( a \) back to find \( T \):
\( T = 35 + 3.5(1.25) = 39.375 \approx 39.4 \text{ N} \).

(ii) Since \( B \) descends through \( 1.6 \text{ m} \) before hitting the floor, we use equations of motion with constant acceleration:
\( v^2 = u^2 + 2 a s \)
\( v^2 = 0 + 2(1.25)(1.6) = 4 \implies v = 2 \text{ m s}^{-1} \).

(iii) After \( B \) hits the floor, the string becomes slack. Block \( A \) continues to move upwards under gravity alone with initial velocity \( u = 2 \text{ m s}^{-1} \):
\( v^2 = u^2 - 2 g y \)
\( 0 = 2^2 - 2(10)y \implies 20y = 4 \implies y = 0.2 \text{ m} \).
So \( A \) rises a further height of \( 0.2 \text{ m} \).

M1: Writing correct equations of motion for both blocks \( A \) and \( B \).
A1: Solving simultaneously to find \( a = 1.25 \text{ m s}^{-2} \) and tension \( T = 39.4 \text{ N} \).
M1: Applying constant acceleration formula to find velocity when \( B \) hits the floor.
A1: Correctly calculating velocity \( v = 2 \text{ m s}^{-1} \).
M1: Setting up equation under gravity alone for block \( A \) after string goes slack.
A1.25: Correctly calculating further height as \( 0.2 \text{ m} \).

(i) The vertical drop height \( h \) between \( A \) and \( B \) is:
\( h = AB \sin\theta = 150 \times 0.04 = 6 \text{ m} \).

The loss in potential energy is:
\( \Delta PE = mgh = 70 \times 10 \times 6 = 4200 \text{ J} \).

(ii) The kinetic energy at \( B \) is:
\( KE_B = \frac{1}{2} m v^2 = \frac{1}{2} \times 70 \times 8^2 = 2240 \text{ J} \).

By conservation of energy, the work done against resistive forces \( W_R \) is the difference between the initial mechanical energy and the final mechanical energy:
\( W_R = \Delta PE - KE_B = 4200 - 2240 = 1960 \text{ J} \).

(iii) If the resistance force \( R_f \) is constant:
\( W_R = R_f \times d \implies 1960 = R_f \times 150 \implies R_f = \frac{1960}{150} \approx 13.1 \text{ N} \) (to 3 s.f.).

M1: Calculating vertical height drop \( h = 6 \text{ m} \) and potential energy loss.
A1: Finding correct potential energy loss of \( 4200 \text{ J} \).
M1: Calculating kinetic energy at \( B \) and formulating energy balance equation.
A1: Correct work done against resistance of \( 1960 \text{ J} \).
M1: Using \( W_R = R_f \times d \) to find resistance force.
A1.25: Correct calculation of resistance force as \( 13.1 \text{ N} \).

The pulling force is \( T = 1.5 \text{ N} \) at \( 30^\circ \).
Horizontal component of force: \( F_H = T \cos 30^\circ = 1.5 \cos 30^\circ \approx 1.299 \text{ N} \).
Vertical component of force: \( F_V = T \sin 30^\circ = 1.5 \sin 30^\circ = 0.75 \text{ N} \).

Since the ring is on the point of sliding, the horizontal driving force equals maximum friction: \( F_H = F_{\max} = \mu R \).

(a) When the vertical component acts downwards:
\( R = mg + F_V = 0.5 \times 10 + 0.75 = 5.75 \text{ N} \).
\( \mu = \frac{F_H}{R} = \frac{1.299}{5.75} \approx 0.226 \) (to 3 s.f.).

(b) When the vertical component acts upwards:
\( R = mg - F_V = 5 - 0.75 = 4.25 \text{ N} \).
\( \mu = \frac{F_H}{R} = \frac{1.299}{4.25} \approx 0.306 \) (to 3 s.f.).

M1: Resolving forces horizontally to find the driving force component \( 1.5 \cos 30^\circ \).
M1: Formulating normal reaction equation for case (a) with downwards vertical component.
A1: Correctly calculating \( \mu = 0.226 \) for case (a).
M1: Formulating normal reaction equation for case (b) with upwards vertical component.
A1: Finding correct normal reaction \( R = 4.25 \text{ N} \) for case (b).
A1.25: Correctly calculating \( \mu = 0.306 \) for case (b).

(i) To construct a back-to-back stem-and-leaf diagram, we use stems 2, 3, and 4. The leaves for the men are listed on the left (ordered outwards from the stem), and the leaves for the women are listed on the right (ordered outwards from the stem).

$$\begin{array}{r|c|l}
\text{Men} & \text{Stem} & \text{Women} \\
\hline
8\; 7\; 5\; 5\; 4\; 3\; 1 & 2 & 4\; 6\; 8\; 8\; 9 \\
6\; 5\; 3\; 2\; 2\; 1 & 3 & 0\; 1\; 3\; 4\; 5\; 6\; 8 \\
4\; 0 & 4 & 1\; 2\; 7
\end{array}$$

Key: $1 \mid 2 \mid 4$ means 21 minutes for a man and 24 minutes for a woman.

(ii) For the 15 women's times:
The data sorted in ascending order is: 24, 26, 28, 28, 29, 30, 31, 33, 34, 35, 36, 38, 41, 42, 47.
- The median is the $\frac{15 + 1}{2} = 8\text{th}$ value: $33$ minutes.
- The lower quartile ($Q_1$) is the median of the lower 7 values (the $4\text{th}$ value): $28$ minutes.
- The upper quartile ($Q_3$) is the median of the upper 7 values (the $12\text{th}$ value): $38$ minutes.
- The interquartile range (IQR) is $Q_3 - Q_1 = 38 - 28 = 10$ minutes.

(i)
- M1: For correct stems (2, 3, 4) and some attempt at back-to-back ordering.
- A1: For correct leaves, fully ordered on both sides.
- A1: For a correct key, including both sides with labels or units.

(ii)
- B1: For correct median of 33.
- M1: For finding the correct quartiles $Q_1 = 28$ and $Q_3 = 38$ (or showing evidence of identifying them).
- A1: For correct IQR of 10.

The letters of the word ELEVATION are: E, L, E, V, A, T, I, O, N.
There are 9 letters in total, with 2 E's, and all other 7 letters distinct.

(i) The number of arrangements of the 9 letters where 2 are identical E's is:
$$\frac{9!}{2!} = \frac{362\,880}{2} = 181\,440$$

(ii) To find the number of arrangements where the two E's are not next to each other, we subtract the arrangements where they are together from the total arrangements.
If the two E's are together, we treat them as a single unit (EE).
Then there are 8 units to arrange: (EE), L, V, A, T, I, O, N.
The number of ways to arrange these 8 distinct units is:
$$8! = 40\,320$$
So, the number of arrangements where the E's are not next to each other is:
$$181\,440 - 40\,320 = 141\,120$$

(iii) To choose 4 letters from {E, E, L, V, A, T, I, O, N}, we consider three cases based on the number of E's selected:
- Case 1: No E's are selected. We choose 4 letters from the remaining 7 distinct letters {L, V, A, T, I, O, N}:
$$\binom{7}{4} = 35$$
- Case 2: Exactly 1 E is selected. We choose the remaining 3 letters from the 7 distinct letters:
$$\binom{7}{3} = 35$$
- Case 3: Both E's are selected. We choose the remaining 2 letters from the 7 distinct letters:
$$\binom{7}{2} = 21$$

Total number of selections is $35 + 35 + 21 = 91$.

(i)
- M1: For calculating $\frac{9!}{2!}$ or equivalent.
- A1: For $181\,440$ (accept standard form).

(ii)
- M1: For calculating $8!$ as the number of arrangements with E's together.
- A1: For subtracting to get $141\,120$.

(iii)
- M1: For identifying and summing cases (e.g. 0 E's, 1 E, 2 E's).
- A1: For writing down correct combinations: $\binom{7}{4}$, $\binom{7}{3}$, and $\binom{7}{2}$ (or equivalent list of terms).
- A1: For obtaining final answer of $91$.

Let $D_1$ be the event that the score on the die is 1 or 2, so $\text{P}(D_1) = \frac{2}{6} = \frac{1}{3}$.
Let $D_2$ be the event that the score on the die is 3, 4, 5, or 6, so $\text{P}(D_2) = \frac{4}{6} = \frac{2}{3}$.
Let $R_2$ be the event that the second ball drawn is red.

(i) Under event $D_1$ (draw from A, put in B, draw from B):
- Probability of drawing Red from Bag A is $\frac{4}{10} = 0.4$. Bag B now has 6 Red and 3 Blue (total 9). The probability of drawing Red from Bag B is $\frac{6}{9}$.
- Probability of drawing Blue from Bag A is $\frac{6}{10} = 0.6$. Bag B now has 5 Red and 4 Blue (total 9). The probability of drawing Red from Bag B is $\frac{5}{9}$.
So, the conditional probability is:
$$\text{P}(R_2 \mid D_1) = 0.4 \times \frac{6}{9} + 0.6 \times \frac{5}{9} = \frac{2.4 + 3}{9} = \frac{5.4}{9} = 0.6$$

Under event $D_2$ (draw from B, put in A, draw from A):
- Probability of drawing Red from Bag B is $\frac{5}{8}$. Bag A now has 5 Red and 6 Blue (total 11). The probability of drawing Red from Bag A is $\frac{5}{11}$.
- Probability of drawing Blue from Bag B is $\frac{3}{8}$. Bag A now has 4 Red and 7 Blue (total 11). The probability of drawing Red from Bag A is $\frac{4}{11}$.
So, the conditional probability is:
$$\text{P}(R_2 \mid D_2) = \frac{5}{8} \times \frac{5}{11} + \frac{3}{8} \times \frac{4}{11} = \frac{25 + 12}{88} = \frac{37}{88}$$

Now, the total probability is:
$$\text{P}(R_2) = \text{P}(D_1)\text{P}(R_2 \mid D_1) + \text{P}(D_2)\text{P}(R_2 \mid D_2)$$
$$\text{P}(R_2) = \frac{1}{3} \times 0.6 + \frac{2}{3} \times \frac{37}{88} = \frac{1}{5} + \frac{37}{132} = \frac{132 + 185}{660} = \frac{317}{660} \approx 0.480$$

(ii) Using Bayes' Theorem:
$$\text{P}(D_1 \mid R_2) = \frac{\text{P}(D_1 \cap R_2)}{\text{P}(R_2)} = \frac{\frac{1}{3} \times 0.6}{\frac{317}{660}} = \frac{0.2}{\frac{317}{660}} = \frac{132}{317} \approx 0.416$$

(i)
- M1: For calculating the probability of a red ball under $D_1$: $0.4 \times \frac{6}{9} + 0.6 \times \frac{5}{9}$.
- M1: For calculating the probability of a red ball under $D_2$: $\frac{5}{8} \times \frac{5}{11} + \frac{3}{8} \times \frac{4}{11}$.
- M1: For combining the probabilities correctly with die outcomes: $\frac{1}{3}\text{P}(R_2 \mid D_1) + \frac{2}{3}\text{P}(R_2 \mid D_2)$.
- A1: For correct final probability $\frac{317}{660}$ (accept $0.480$ to 3 significant figures).

(ii)
- M1: For utilizing conditional probability formula $\frac{\text{P}(D_1 \cap R_2)}{\text{P}(R_2)}$.
- M1: For substituting correct values into numerator ($\frac{1}{5}$) and denominator ($\text{P}(R_2)$ from part (i)).
- A1: For correct final answer $\frac{132}{317}$ (accept $0.416$ to 3 significant figures).

(i) Since the sum of all probabilities in a probability distribution is 1:
$$a + b + 0.3 + 2a + 0.1 = 1 \implies 3a + b + 0.4 = 1 \implies 3a + b = 0.6 \quad (\text{Eq } 1)$$

Using the given expectation $\text{E}(X) = 0.1$:
$$\text{E}(X) = \sum x \cdot \text{P}(X=x) = (-2)(a) + (-1)(b) + (0)(0.3) + (1)(2a) + (2)(0.1) = 0.1$$
$$-2a - b + 2a + 0.2 = 0.1 \implies -b + 0.2 = 0.1 \implies b = 0.1$$

Substitute $b = 0.1$ into (Eq 1):
$$3a + 0.1 = 0.6 \implies 3a = 0.5 \implies a = \frac{1}{6}$$

(ii) To find the variance, we first calculate $\text{E}(X^2)$:
$$\text{E}(X^2) = \sum x^2 \cdot \text{P}(X=x) = (-2)^2(a) + (-1)^2(b) + (0)^2(0.3) + (1)^2(2a) + (2)^2(0.1)$$
$$\text{E}(X^2) = 4\left(\frac{1}{6}\right) + 1(0.1) + 0 + 1\left(\frac{2}{6}\right) + 4(0.1)$$
$$\text{E}(X^2) = \frac{4}{6} + 0.1 + \frac{2}{6} + 0.4 = 1 + 0.5 = 1.5$$

Now, we calculate the variance:
$$\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = 1.5 - (0.1)^2 = 1.5 - 0.01 = 1.49$$

(i)
- M1: For setting the sum of probabilities equal to 1, obtaining $3a + b = 0.6$.
- M1: For setting up the expectation equation $\text{E}(X) = 0.1$, leading to $-b + 0.2 = 0.1$.
- A1: For finding $b = 0.1$ (or equivalent).
- A1: For finding $a = \frac{1}{6}$ (or equivalent).

(ii)
- M1: For calculating $\text{E}(X^2)$ with their values of $a$ and $b$.
- M1: For using the formula $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$ with their values.
- A1: For correct final variance of $1.49$.

Let $X$ represent the mass of a randomly chosen packet of tea. $X \sim \text{N}(\mu, \sigma^2)$.

(i) We are given:
$$\text{P}(X < 242) = 0.15 \implies \text{P}\left(Z < \frac{242 - \mu}{\sigma}\right) = 0.15$$
Since 0.15 < 0.5, the z-value is negative. Using the standard normal table:
$$\frac{242 - \mu}{\sigma} = -\Phi^{-1}(0.85) = -1.036 \implies 242 - \mu = -1.036\sigma \implies \mu - 1.036\sigma = 242 \quad (\text{Eq } 1)$$

We are also given:
$$\text{P}(X > 265) = 0.08 \implies \text{P}\left(Z > \frac{265 - \mu}{\sigma}\right) = 0.08 \implies \text{P}\left(Z < \frac{265 - \mu}{\sigma}\right) = 0.92$$
Using the standard normal table:
$$\frac{265 - \mu}{\sigma} = \Phi^{-1}(0.92) = 1.405 \implies 265 - \mu = 1.405\sigma \implies \mu + 1.405\sigma = 265 \quad (\text{Eq } 2)$$

Subtracting (Eq 1) from (Eq 2):
$$(1.405 + 1.036)\sigma = 265 - 242 \implies 2.441\sigma = 23$$
$$\sigma = \frac{23}{2.441} \approx 9.4223 \approx 9.42 \text{ g}$$

Substitute $\sigma$ back into (Eq 1):
$$\mu = 242 + 1.036(9.4223) \approx 251.76 \approx 252 \text{ g}$$

(ii) We want to find $\text{P}(245 < X < 260)$ using $\mu = 251.76$ and $\sigma = 9.422$:
$$\text{P}(245 < X < 260) = \text{P}\left(\frac{245 - 251.76}{9.422} < Z < \frac{260 - 251.76}{9.422}\right)$$
$$= \text{P}(-0.7175 < Z < 0.8745)$$
$$= \Phi(0.875) - \Phi(-0.718) = \Phi(0.875) - (1 - \Phi(0.718)) = \Phi(0.875) + \Phi(0.718) - 1$$

Using the normal tables:
- $\Phi(0.875) = 0.8092$
- $\Phi(0.718) = 0.7636$

$$\text{P}(245 < X < 260) = 0.8092 + 0.7636 - 1 = 0.5728 \approx 0.573$$

(i)
- M1: For standardizing and setting equal to $-1.036$ (or $-1.037$).
- M1: For standardizing and setting equal to $1.405$ (or $1.406$).
- M1: For solving the simultaneous equations to find $\sigma$ or $\mu$.
- A1: For correct standard deviation $\sigma = 9.42$ (or $9.41$).
- A1: For correct mean $\mu = 252$.

(ii)
- M1: For standardizing both 245 and 260 using their mean and standard deviation and subtracting the probabilities.
- A1: For correct final answer $0.573$.

Let $X$ represent the number of heads obtained in 80 spins. $X \sim \text{B}(80, 0.2)$.

(i) The probability of obtaining exactly 15 heads is:
$$\text{P}(X = 15) = \binom{80}{15} (0.2)^{15} (0.8)^{65} \approx 0.0988$$

(ii) To approximate this binomial distribution with a normal distribution, we verify the conditions:
- $n = 80$ is large.
- $np = 80 \times 0.2 = 16 > 5$.
- $n(1-p) = 80 \times 0.8 = 64 > 5$.
Both conditions are satisfied, so we can approximate $X \sim \text{B}(80, 0.2)$ with a normal distribution $Y \sim \text{N}(\mu, \sigma^2)$ where:
- $\mu = np = 16$
- $\sigma^2 = np(1-p) = 80 \times 0.2 \times 0.8 = 12.8$
- $\sigma = \sqrt{12.8} \approx 3.578$

We want to find $\text{P}(X > 20)$. Applying the continuity correction:
$$\text{P}(X > 20) \approx \text{P}(Y > 20.5)$$
$$\text{P}(Y > 20.5) = \text{P}\left(Z > \frac{20.5 - 16}{3.578}\right) = \text{P}\left(Z > \frac{4.5}{3.578}\right) = \text{P}(Z > 1.258)$$
$$= 1 - \Phi(1.258)$$

Using standard normal tables:
- $\Phi(1.258) \approx 0.8958$

$$\text{P}(X > 20) \approx 1 - 0.8958 = 0.1042 \approx 0.104$$

(i)
- M1: For substituting correctly into the binomial formula $\binom{80}{15}(0.2)^{15}(0.8)^{65}$.
- A1: For correct probability $0.0988$.

(ii)
- B1: For justifying the normal approximation by showing $np = 16 > 5$ and $nq = 64 > 5$.
- B1: For calculating correct mean $\mu = 16$ and variance $\sigma^2 = 12.8$.
- M1: For standardizing with a continuity correction (using $20.5$).
- M1: For calculating $1 - \Phi(z)$ for their $z$-value.
- A1: For correct final probability $0.104$.

(i) The debate team must have 6 students, and there must be more girls than boys. Since the team size is 6, the possible combinations of (girls, boys) are:
- Case 1: 4 girls, 2 boys
$$\text{Ways} = \binom{8}{4} \times \binom{5}{2} = 70 \times 10 = 700$$
- Case 2: 5 girls, 1 boy
$$\text{Ways} = \binom{8}{5} \times \binom{5}{1} = 56 \times 5 = 280$$
- Case 3: 6 girls, 0 boys
$$\text{Ways} = \binom{8}{6} \times \binom{5}{0} = 28 \times 1 = 28$$

Total number of ways to select the team is:
$$700 + 280 + 28 = 1008$$

(ii) The team consists of 4 distinct girls and 2 distinct boys. We wish to arrange these 6 students in a line such that the 2 boys are not adjacent.

Method 1 (Subtraction):
- Total possible arrangements of 6 distinct students: $6! = 720$.
- Arrangements where the 2 boys are together: treat the 2 boys as a single unit. There are 5 units to arrange: the (boy-unit) and 4 girls. These 5 units can be arranged in $5! = 120$ ways. Within the unit, the 2 boys can be arranged in $2! = 2$ ways.
So, arrangements with boys together: $120 \times 2 = 240$.
- Arrangements where boys are not together: $720 - 240 = 480$.

Method 2 (Gap Method):
- First, arrange the 4 girls in a line. This can be done in $4! = 24$ ways.
- This leaves 5 possible spaces (gaps) to place the 2 boys: $\_\, G_1 \,\_\, G_2 \,\_\, G_3 \,\_\, G_4 \,\_$.
- The number of ways to place 2 boys into these 5 spaces is $P^5_2 = 5 \times 4 = 20$ ways.
- Total arrangements: $24 \times 20 = 480$.

(i)
- M1: For calculating the combinations for at least one correct case.
- M1: For identifying and summing the three correct cases (4g/2b, 5g/1b, 6g/0b).
- A1: For correct calculations: $700$, $280$, $28$.
- A1: For obtaining final answer of $1008$.

(ii)
- M1: For calculating total unrestricted arrangements ($6! = 720$) or arranging girls ($4! = 24$).
- M1: For arranging with boys together ($5! \times 2 = 240$) or placing boys in gaps ($5 \times 4 = 20$).
- A1: For correct final answer of $480$.

Let \( W_i \) be the total mass of a single packed box, so \( W_i = T_i + B_i \).
Since \( T_i \sim N(250, 4^2) \) and \( B_i \sim N(20, 1.5^2) \), we have:
\( E(W_i) = 250 + 20 = 270 \) g
\( \text{Var}(W_i) = 4^2 + 1.5^2 = 16 + 2.25 = 18.25 \) \( \text{g}^2 \)

Let \( S \) be the total mass of 4 independent packed boxes: \( S = W_1 + W_2 + W_3 + W_4 \).
\( E(S) = 4 \times E(W_i) = 4 \times 270 = 1080 \) g
\( \text{Var}(S) = 4 \times \text{Var}(W_i) = 4 \times 18.25 = 73 \) \( \text{g}^2 \)

We need to find \( P(S < 1090) \):
\( P(S < 1090) = P\left( Z < \frac{1090 - 1080}{\sqrt{73}} \right) \)
\( Z < \frac{10}{8.544} \approx 1.170 \)
Using the normal cumulative distribution tables:
\( P(Z < 1.170) = 0.8790 \)

Thus, the probability is 0.879.

M1: For calculating the mean of 4 boxes (1080 g).
M1: For calculating the variance of 1 box (18.25) or 4 boxes (73).
A1: For correctly identifying the distribution of the sum \( S \sim N(1080, 73) \).
M1: For standardizing to find the Z-score with their mean and variance.
A1: For obtaining a Z-value of approximately 1.17.
M1: For correctly finding the probability from normal tables corresponding to their positive Z-score.
A1: For the final correct probability 0.879 (accept 0.8790).

Let \( X \) be the number of calls received in a 2-minute interval.
Since the average rate is 4.2 calls per minute, the mean rate for 2 minutes is:
\( \lambda = 4.2 \times 2 = 8.4 \) calls.
Therefore, \( X \sim \text{Po}(8.4) \).

We need to find \( P(6 \le X \le 10) \):
\( P(6 \le X \le 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \)
Using the Poisson probability mass function \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \):
\( P(X = 6) = \frac{e^{-8.4} 8.4^6}{6!} \approx 0.10972 \)
\( P(X = 7) = \frac{e^{-8.4} 8.4^7}{7!} \approx 0.13166 \)
\( P(X = 8) = \frac{e^{-8.4} 8.4^8}{8!} \approx 0.13824 \)
\( P(X = 9) = \frac{e^{-8.4} 8.4^9}{9!} \approx 0.12903 \)
\( P(X = 10) = \frac{e^{-8.4} 8.4^{10}}{10!} \approx 0.10838 \)

Summing these probabilities:
\( P(6 \le X \le 10) \approx 0.10972 + 0.13166 + 0.13824 + 0.12903 + 0.10838 = 0.61703 \)

To 3 significant figures, the probability is 0.617.

B1: For identifying the correct parameter \( \lambda = 8.4 \).
M1: For expressing the required probability as the sum of 5 terms from \( X = 6 \) to \( X = 10 \).
M1: For applying the Poisson formula correctly for at least two terms.
A1: For calculating individual probabilities correctly (at least 3 decimal places).
M1: For summing the correct 5 individual probabilities.
A1: For the final correct probability of 0.617 (accept 0.6170 to 0.6171).

(i) To show that \( k = \frac{3}{4} \), we use the property that the total area under the probability density function must equal 1:
\( \int_{0}^{2} f(x) \, dx = 1 \)
\( \int_{0}^{2} k(2x - x^2) \, dx = k \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} = 1 \)
Evaluating the limits:
\( k \left( \left( 2^2 - \frac{2^3}{3} \right) - 0 \right) = 1 \)
\( k \left( 4 - \frac{8}{3} \right) = 1 \)
\( k \left( \frac{4}{3} \right) = 1 \implies k = \frac{3}{4} \).

(ii) Now we calculate \( P(X > 1.5) \):
\( P(X > 1.5) = \int_{1.5}^{2} \frac{3}{4}(2x - x^2) \, dx \)
\( P(X > 1.5) = \frac{3}{4} \left[ x^2 - \frac{x^3}{3} \right]_{1.5}^{2} \)
Evaluating at the upper limit \( x = 2 \):
\( 4 - \frac{8}{3} = \frac{4}{3} \)
Evaluating at the lower limit \( x = 1.5 = \frac{3}{2} \):
\( \left( \frac{3}{2} \right)^2 - \frac{(3/2)^3}{3} = \frac{9}{4} - \frac{27}{24} = \frac{9}{4} - \frac{9}{8} = \frac{9}{8} \)

Subtracting the values:
\( P(X > 1.5) = \frac{3}{4} \left( \frac{4}{3} - \frac{9}{8} \right) = \frac{3}{4} \left( \frac{32 - 27}{24} \right) = \frac{3}{4} \left( \frac{5}{24} \right) = \frac{5}{32} = 0.15625 \).

To 3 significant figures, the probability is 0.156.

Part (i):
M1: For integrating the PDF and setting the result equal to 1.
A1: For correct integration with limits substituted.
A1: For obtaining \( k = 3/4 \) clearly showing the steps.

Part (ii):
M1: For setting up the correct integral from 1.5 to 2 with their value of \( k \).
A1: For correct integration and substitution of limits.
A1: For obtaining \( 5/32 \) or 0.156 (accept 0.15625).

(i) Let \( p \) be the population proportion of defective items.
Null hypothesis: \( H_0: p = 0.12 \)
Alternative hypothesis: \( H_1: p < 0.12 \)

(ii) Under \( H_0 \), let \( X \) be the number of defective items in the sample of 25, so \( X \sim B(25, 0.12) \).
We want to find the critical value \( c \) such that \( P(X \le c) \le 0.05 \).
Let's calculate the cumulative probabilities:
\( P(X = 0) = (0.88)^{25} \approx 0.0409 \)
\( P(X \le 1) = P(X=0) + \binom{25}{1} (0.12)^1 (0.88)^{24} \approx 0.0409 + 25 \times 0.12 \times 0.0465 \approx 0.1804 \)
Since \( P(X \le 0) = 0.0409 \le 0.05 \) and \( P(X \le 1) = 0.1804 > 0.05 \), the critical region is \( X = 0 \).

(iii) A Type II error occurs when we fail to reject \( H_0 \) when \( H_0 \) is false. This happens when the sample observation does not fall in the critical region, i.e., \( X \ge 1 \), given that the actual probability is \( p = 0.05 \).
Under the new probability, \( X \sim B(25, 0.05) \).
\( P(\text{Type II Error}) = P(X \ge 1 \mid p = 0.05) = 1 - P(X = 0 \mid p = 0.05) \)
\( P(X = 0) = (0.95)^{25} \approx 0.2774 \)
\( P(\text{Type II Error}) = 1 - 0.2774 = 0.7226 \approx 0.723 \).

Part (i):
B1: For stating both hypotheses correctly using \( p \).

Part (ii):
M1: For calculating cumulative Binomial probabilities under \( H_0 \) starting from \( X = 0 \).
A1: For identifying the critical region as \( X = 0 \) with a probability of 0.0409.

Part (iii):
M1: For realizing that Type II error corresponds to \( X \ge 1 \) (not in critical region).
M1: For using \( B(25, 0.05) \) to find the probability of being outside the critical region.
A1: For the final answer of 0.723 (accept 0.7226).

(i) Unbiased estimate of the population mean, \( \bar{x} \):
\( \bar{x} = \frac{\sum x}{n} = \frac{14240}{80} = 178 \) g

Unbiased estimate of the population variance, \( s^2 \):
\( s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right) \)
\( s^2 = \frac{1}{79} \left( 2561800 - \frac{14240^2}{80} \right) \)
\( s^2 = \frac{1}{79} \left( 2561800 - 2534720 \right) = \frac{27080}{79} \approx 342.785 \) \( \text{g}^2 \)

(ii) For a 95% confidence interval, the critical value \( z = 1.96 \).
The formula for the confidence interval for the mean is:
\( \bar{x} \pm z \frac{s}{\sqrt{n}} \)
Standard error: \( \frac{s}{\sqrt{n}} = \sqrt{\frac{342.785}{80}} \approx 2.070 \)

Confidence interval limits:
\( 178 \pm 1.96 \times 2.070 = 178 \pm 4.057 \)
Lower limit: \( 178 - 4.057 = 173.943 \approx 173.9 \)
Upper limit: \( 178 + 4.057 = 182.057 \approx 182.1 \)

The 95% confidence interval is [173.9, 182.1].

Part (i):
B1: For finding the unbiased mean estimate of 178.
M1: For using the correct formula for the unbiased variance estimate.
A1: For the correct unbiased variance estimate of 342.8 (or 343).

Part (ii):
B1: For identifying the critical value \( z = 1.96 \).
M1: For calculating the standard error and applying the CI formula \( \bar{x} \pm z \frac{s}{\sqrt{n}} \).
A1: For the correct lower limit of 173.9.
A1: For the correct upper limit of 182.1.

(i) For a 10-minute period, the average rate of arrivals is:
\( \lambda = 15.6 \times \frac{10}{60} = 2.6 \) arrivals.
Let \( X \) be the number of arrivals in a 10-minute period, so \( X \sim \text{Po}(2.6) \).
We want \( P(X \le 2) \):
\( P(X \le 2) = e^{-2.6} \left( 1 + 2.6 + \frac{2.6^2}{2} \right) \)
\( P(X \le 2) = e^{-2.6} \left( 1 + 2.6 + 3.38 \right) = e^{-2.6} \times 6.98 \approx 0.07427 \times 6.98 \approx 0.518 \).

(ii) For a 10-hour period, the average rate of arrivals is:
\( \lambda = 15.6 \times 10 = 156 \) arrivals.
Since \( \lambda > 15 \), we can approximate the Poisson distribution \( Y \sim \text{Po}(156) \) using a normal distribution:
\( Y \approx N(156, 156) \)
We need to find \( P(Y > 170) \).
Applying continuity correction:
\( P(Y > 170) = P(Y_{normal} > 170.5) \)

Standardizing:
\( Z = \frac{170.5 - 156}{\sqrt{156}} = \frac{14.5}{12.49} \approx 1.161 \)

Using the normal distribution table:
\( P(Z > 1.161) = 1 - \Phi(1.161) \approx 1 - 0.8772 = 0.1228 \approx 0.123 \).

To 3 significant figures, the probability is 0.123.

Part (i):
B1: For finding the correct lambda value of 2.6.
M1: For correctly calculating \( P(X \le 2) \) using the Poisson formula.
A1: For obtaining 0.518.

Part (ii):
B1: For stating the normal distribution approximation \( N(156, 156) \).
M1: For applying a continuity correction (using 170.5).
M1: For standardizing using their mean and standard deviation.
A1: For the correct probability of 0.123.

(i) Let \( \mu \) be the population mean lifetime of the light bulbs.
Null hypothesis: \( H_0: \mu = 1200 \)
Alternative hypothesis: \( H_1: \mu < 1200 \)

(ii) Under \( H_0 \), the sample mean \( \bar{X} \) has distribution:
\( \bar{X} \sim N\left(1200, \frac{80^2}{60}\right) \)
Standard error of the mean: \( \frac{\sigma}{\sqrt{n}} = \frac{80}{\sqrt{60}} \approx 10.328 \) hours.

Calculate the test statistic, \( z \):
\( z = \frac{1182 - 1200}{10.328} = \frac{-18}{10.328} \approx -1.743 \)

The test is one-tailed, so the p-value is:
\( p\text{-value} = P(Z < -1.743) = 1 - \Phi(1.743) \)
From the normal cumulative table, \( \Phi(1.743) \approx 0.9593 \).
\( p\text{-value} = 1 - 0.9593 = 0.0407 \).

(iii) Compare the p-value with the significance level \( \alpha = 0.025 \):
Since \( 0.0407 > 0.025 \), we do not reject \( H_0 \).
There is insufficient evidence at the 2.5% significance level to support the consumer association's suspicion that the mean lifetime of the bulbs is less than 1200 hours.

Part (i):
B1: For stating both hypotheses correctly with \( \mu \).

Part (ii):
M1: For calculating the correct standard error \( 80 / \sqrt{60} \approx 10.33 \).
M1: For calculating the test statistic \( z \).
A1: For getting \( z \approx -1.743 \).
A1: For finding the correct p-value of 0.0407 (or 0.0406).

Part (iii):
M1: For comparing their p-value with 0.025 (or their z-value with -1.96).
A1: For a correct decision in context (do not reject \( H_0 \), insufficient evidence to support the claim).