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The sum of the first 4 terms of the arithmetic sequence is given by the formula \(S_n = \frac{n}{2}(2u_1 + (n-1)d)\). Substituting the given values yields \(S_4 = \frac{4}{2}(2a + 3(2)) = 4a + 12\). The sum to infinity of the geometric sequence is given by the formula \(S_{\infty} = \frac{v_1}{1-r}\). Substituting the given values yields \(S_{\infty} = \frac{a}{1 - \frac{4}{5}} = 5a\). Setting these two sums equal to each other gives the equation \(4a + 12 = 5a\). Solving for \(a\) gives \(a = 12\).

M1 for attempting to use the arithmetic sum formula with \(n = 4\) and \(d = 2\). A1 for obtaining \(4a + 12\). M1 for attempting to use the geometric sum to infinity formula with \(r = \frac{4}{5}\). A1 for obtaining \(5a\). M1 for equating the two expressions and attempting to solve for \(a\). A1 for finding \(a = 12\).

For part (a), because \(f^{-1}(4) = 5\), it follows that \(f(5) = 4\). Substituting \(x = 5\) into the function gives \(f(5) = \frac{2(5) + b}{5 - 3} = \frac{10 + b}{2}\). Solving the equation \(10 + b = 8\) yields \(b = -2\). For part (b), we write the function as \(y = \frac{2x - 2}{x - 3}\) and swap variables to solve for \(y\), or rearrange directly for \(x\): \(y(x - 3) = 2x - 2\) which expands to \(xy - 3y = 2x - 2\). Rearranging terms gives \(xy - 2x = 3y - 2\), which factorizes to \(x(y - 2) = 3y - 2\). This simplifies to \(x = \frac{3y - 2}{y - 2}\). Thus, the inverse function is \(f^{-1}(x) = \frac{3x - 2}{x - 2}\).

M1 for applying the property \(f(5) = 4\). A1 for solving to find \(b = -2\). M1 for setting up the equation \(y = \frac{2x - 2}{x - 3}\) and multiplying by \(x - 3\). M1 for isolating the \(x\) terms on one side of the equation. A1 for factorizing \(x(y - 2)\). A1 for obtaining the correct final expression for \(f^{-1}(x)\).

We divide both sides of the equation by 2 to obtain \(\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x = \frac{1}{2}\). Using the compound angle identity \(\sin(x + \theta) = \sin x \cos \theta + \cos x \sin \theta\), we can write this expression as \(\sin(x + \frac{\pi}{6}) = \frac{1}{2}\). For the interval \(0 \le x \le 2\pi\), the domain of the argument is \(\frac{\pi}{6} \le x + \frac{\pi}{6} \le \frac{13\pi}{6\)}. Within this interval, \(\sin(x + \frac{\pi}{6}) = \frac{1}{2}\) has solutions at \(x + \frac{\pi}{6} = \frac{\pi}{6}\), \(x + \frac{\pi}{6} = \frac{5\pi}{6}\), and \(x + \frac{\pi}{6} = \frac{13\pi}{6}\). Solving for \(x\) in each case yields \(x = 0\), \(x = \frac{2\pi}{3}\), and \(x = 2\pi\).

M1 for attempting to divide by 2 or applying compound angle methods. A1 for expressing the equation in the form \(\sin(x + \frac{\pi}{6}) = \frac{1}{2}\) or equivalent. M1 for identifying the correct interval for the argument. A1 for finding the correct values of the argument: \(\frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}\). A1 for two correct solutions. A1 for finding all three correct solutions.

For part (a), using the product rule with \(u = x\) and \(v = e^{-2x}\), we find the derivatives \(u' = 1\) and \(v' = -2e^{-2x}\). Substituting into the product rule formula gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = (1)e^{-2x} + x(-2e^{-2x}) = (1 - 2x)e^{-2x}\). For part (b), a horizontal tangent occurs when \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\). Setting \((1 - 2x)e^{-2x} = 0\) and noting that \(e^{-2x} > 0\) for all real \(x\), we find \(1 - 2x = 0\), which gives \(x = \frac{1}{2}\). For part (c), when \(x = 0\), the corresponding \(y\)-value is \(y = 0\). The gradient of the tangent line is given by substituting \(x = 0\) into the derivative, which gives \(m = (1 - 0)e^0 = 1\). Using the point-slope formula \(y - y_1 = m(x - x_1)\), we obtain \(y - 0 = 1(x - 0)\), simplifying to \(y = x\).

M1 for applying the product rule. A1 for obtaining the correct derivative \((1-2x)e^{-2x}\). M1 for setting their derivative to 0. A1 for solving to get \(x = \frac{1}{2}\). M1 for finding the tangent slope \(m = 1\) and point \((0,0)\). A1 for the correct equation of the tangent line \(y = x\).

For part (a), the sum of all probabilities in a probability distribution must equal 1, which gives \(0.1 + p + q + 0.3 = 1 \implies p + q = 0.6\). The expected value is given by \(\mathrm{E}(X) = 1(0.1) + 2p + 3q + 4(0.3) = 2.7\), which simplifies to \(0.1 + 2p + 3q + 1.2 = 2.7 \implies 2p + 3q = 1.4\). Solving the system of equations, we multiply the first equation by 2 to get \(2p + 2q = 1.2\), and subtracting this from the second equation yields \(q = 0.2\). Substituting this back into \(p + q = 0.6\) gives \(p = 0.4\). For part (b), the variance is given by \(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2\). First, calculate \(\mathrm{E}(X^2) = 1^2(0.1) + 2^2(0.4) + 3^2(0.2) + 4^2(0.3) = 0.1 + 1.6 + 1.8 + 4.8 = 8.3\). Finally, compute the variance: \(\mathrm{Var}(X) = 8.3 - (2.7)^2 = 8.3 - 7.29 = 1.01\).

M1 for setting up the sum of probabilities equation \(p + q = 0.6\). M1 for setting up the expectation equation \(2p + 3q = 1.4\). A1 for finding the correct values \(p = 0.4\) and \(q = 0.2\). M1 for attempting to calculate \(\mathrm{E}(X^2)\). A1 for obtaining \(\mathrm{E}(X^2) = 8.3\). A1 for the correct final variance value of \(1.01\).

Letting \(u = x^2\), we find the derivative \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\), which implies \(x \, \mathrm{d}x = \frac{1}{2} \, \mathrm{d}u\). Next, we transform the limits of integration. When \(x = 0\), \(u = 0^2 = 0\). When \(x = \sqrt{\frac{\pi}{3}}\), \(u = (\sqrt{\frac{\pi}{3}})^2 = \frac{\pi}{3}\). Substituting these into the original integral yields \(\int_{0}^{\frac{\pi}{3}} \sin(u) \cdot \frac{1}{2} \, \mathrm{d}u = \frac{1}{2} \left[ -\cos(u) \right]_0^{\frac{\pi}{3}}\). Evaluating this expression gives \(\frac{1}{2} \left( -\cos(\frac{\pi}{3}) - (-\cos(0)) \right) = \frac{1}{2} \left( -\frac{1}{2} + 1 \right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\).

M1 for writing down the relationship \(\mathrm{d}u = 2x \, \mathrm{d}x\) or equivalent. A1 for correctly determining the new integration limits as \(0\) and \(\frac{\pi}{3}\). A1 for writing the correct transformed integral \(\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \sin(u) \, \mathrm{d}u\). M1 for integrating \(\sin(u)\) to obtain \(-\cos(u)\). M1 for substituting the limits of integration into their integrated expression. A1 for the correct final exact value of \(\frac{1}{4}\).

(a) Using the product rule:
\( f'(x) = \frac{d}{dx}(x^2)\ln(x) + x^2\frac{d}{dx}(\ln(x)) \)
\( f'(x) = 2x\ln(x) + x^2\left(\frac{1}{x}\right) = 2x\ln(x) + x = x(2\ln(x) + 1) \).

(b) For a stationary point, set \( f'(x) = 0 \):
\( x(2\ln(x) + 1) = 0 \).
Since \( x > 0 \), we have \( 2\ln(x) + 1 = 0 \implies \ln(x) = -1/2 \implies x = e^{-1/2} \).
Substituting back into \( f(x) \):
\( f(e^{-1/2}) = (e^{-1/2})^2 \ln(e^{-1/2}) = e^{-1} (-1/2) = -\frac{1}{2e} \).
So the stationary point is \( \left(e^{-1/2}, -\frac{1}{2e}\right) \).
Using the second derivative test:
\( f''(x) = 2\ln(x) + 3 \).
\( f''(e^{-1/2}) = 2(-1/2) + 3 = 2 > 0 \), which confirms it is a local minimum.

(c) Using the product rule or derivative of \( 2x\ln(x) + x \):
\( f''(x) = 2\ln(x) + 2x\left(\frac{1}{x}\right) + 1 = 2\ln(x) + 3 \).
For the point of inflection, set \( f''(x) = 0 \):
\( 2\ln(x) + 3 = 0 \implies \ln(x) = -3/2 \implies x = e^{-3/2} \).
Substituting back into \( f(x) \):
\( f(e^{-3/2}) = (e^{-3/2})^2 \ln(e^{-3/2}) = e^{-3} (-3/2) = -\frac{3}{2e^3} \).
So the point of inflection is \( \left(e^{-3/2}, -\frac{3}{2e^3}\right) \).

(d) The area \( A \) is given by \( \int_{1}^{e} x^2 \ln(x) \, dx \).
Using integration by parts, let \( u = \ln(x) \) and \( dv = x^2 \, dx \).
Then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^3}{3} \).
\( \int x^2 \ln(x) \, dx = \frac{x^3}{3}\ln(x) - \int \frac{x^2}{3} \, dx = \frac{x^3}{3}\ln(x) - \frac{x^3}{9} \).
Evaluating this from 1 to \( e \):
\( \left[ \frac{x^3}{3}\ln(x) - \frac{x^3}{9} \right]_{1}^{e} = \left(\frac{e^3}{3}\ln(e) - \frac{e^3}{9}\right) - \left(\frac{1^3}{3}\ln(1) - \frac{1^3}{9}\right) \)
\( = \left(\frac{e^3}{3} - \frac{e^3}{9}\right) - \left(0 - \frac{1}{9}\right) \)
\( = \frac{2e^3}{9} + \frac{1}{9} = \frac{2e^3 + 1}{9} \).

(a)
M1 for attempting to use the product rule.
A1 for one correct term (either \( 2x\ln(x) \) or \( x \)).
A1 for the fully correct derivative \( x(2\ln(x) + 1) \).

(b)
M1 for setting \( f'(x) = 0 \).
A1 for finding \( x = e^{-1/2} \).
A1 for finding the y-coordinate \( -\frac{1}{2e} \).
R1 for verifying that it is a local minimum (using first or second derivative test).

(c)
M1 for finding a correct derivative of \( f'(x) \).
A1 for \( f''(x) = 2\ln(x) + 3 \).
A1 for coordinates \( \left(e^{-3/2}, -\frac{3}{2e^3}\right) \).

(d)
M1 for attempting integration by parts.
A1 for correct integration terms \( \frac{x^3}{3}\ln(x) - \frac{x^3}{9} \).
M1 for correct substitution of limits \( 1 \) and \( e \).
A1 for the final exact answer \( \frac{2e^3 + 1}{9} \).

(a) To find \( f^{-1}(x) \), set \( y = \frac{x+3}{x-1} \).
Swap variables to get \( x = \frac{y+3}{y-1} \).
Rearrange for \( y \):
\( x(y-1) = y+3 \implies xy - x = y + 3 \implies xy - y = x + 3 \implies y(x-1) = x+3 \).
Thus, \( f^{-1}(x) = \frac{x+3}{x-1} \).
Domain of \( f^{-1}(x) \) is \( x \in \mathbb{R}, x \neq 1 \).
Range of \( f^{-1}(x) \) is \( y \in \mathbb{R}, y \neq 1 \).

(b) Set \( f(x) = x \):
\( \frac{x+3}{x-1} = x \implies x+3 = x^2 - x \implies x^2 - 2x - 3 = 0 \).
Factorizing the quadratic equation:
\( (x-3)(x+1) = 0 \implies x = 3 \text{ or } x = -1 \).
Since they lie on the line \( y = x \), the coordinates are \( (3, 3) \) and \( (-1, -1) \).

(c) Substituting \( f(x) \) into itself:
\( f(f(x)) = \frac{f(x)+3}{f(x)-1} = \frac{\frac{x+3}{x-1} + 3}{\frac{x+3}{x-1} - 1} \).
Multiply numerator and denominator by \( (x-1) \):
\( f(f(x)) = \frac{(x+3) + 3(x-1)}{(x+3) - (x-1)} = \frac{x+3+3x-3}{x+3-x+1} = \frac{4x}{4} = x \).

(d) Set \( f(g(x)) = 2 \):
\( \frac{g(x)+3}{g(x)-1} = 2 \implies g(x)+3 = 2(g(x)-1) \implies g(x)+3 = 2g(x)-2 \implies g(x) = 5 \).
Since \( g(x) = e^{2x} \), we have:
\( e^{2x} = 5 \implies 2x = \ln(5) \implies x = \frac{1}{2}\ln(5) \).

(a)
M1 for exchanging \( x \) and \( y \) (or setting up equation to solve for \( x \)).
A1 for correctly isolating the variable.
A1 for \( f^{-1}(x) = \frac{x+3}{x-1} \).
A1 for stating the correct domain \( x \neq 1 \) and range \( y \neq 1 \).

(b)
M1 for setting \( f(x) = x \).
A1 for setting up the correct quadratic equation \( x^2 - 2x - 3 = 0 \).
A1 for finding \( x = 3 \) and \( x = -1 \).
A1 for the coordinates \( (3, 3) \) and \( (-1, -1) \).

(c)
M1 for substituting \( f(x) \) into the expression.
A1 for simplifying the numerator and denominator correctly.
AG for showing that it simplifies to \( x \).

(d)
M1 for setting up the equation \( \frac{g(x)+3}{g(x)-1} = 2 \) and solving for \( g(x) \).
A1 for finding \( g(x) = 5 \).
A1 for the final exact solution \( x = \frac{1}{2}\ln(5) \).

(a) To find the x-intercepts, we set \( f(x) = 0 \):
\( 2\sin(2x) - \sqrt{3} = 0 \implies \sin(2x) = \frac{\sqrt{3}}{2} \).
Since \( 0 \le x \le \pi \), we have \( 0 \le 2x \le 2\pi \).
The solutions for \( 2x \) are:
\( 2x = \frac{\pi}{3} \) and \( 2x = \frac{2\pi}{3} \).
Thus, \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{3} \).
The x-intercepts are \( \left(\frac{\pi}{6}, 0\right) \) and \( \left(\frac{\pi}{3}, 0\right) \).

(b) For a local maximum, \( \sin(2x) = 1 \):
\( 2x = \frac{\pi}{2} \implies x = \frac{\pi}{4} \).
\( y = 2(1) - \sqrt{3} = 2 - \sqrt{3} \).
So, the local maximum is at \( \left(\frac{\pi}{4}, 2-\sqrt{3}\right) \).
For a local minimum, \( \sin(2x) = -1 \):
\( 2x = \frac{3\pi}{2} \implies x = \frac{3\pi}{4} \).
\( y = 2(-1) - \sqrt{3} = -2 - \sqrt{3} \).
So, the local minimum is at \( \left(\frac{3\pi}{4}, -2-\sqrt{3}\right) \).

(c) The sketch should show:
- A sine wave starting at \( (0, -\sqrt{3}) \) and ending at \( (\pi, -\sqrt{3}) \).
- A maximum point at \( \left(\frac{\pi}{4}, 2-\sqrt{3}\right) \).
- X-intercepts at \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{3} \).
- A minimum point at \( \left(\frac{3\pi}{4}, -2-\sqrt{3}\right) \).

(d) The region lies entirely above the x-axis because the peak of the curve occurs at \( x = \frac{\pi}{4} \) with positive height \( 2-\sqrt{3} > 0 \).
The area is given by:
\( A = \int_{\pi/6}^{\pi/3} (2\sin(2x) - \sqrt{3}) \, dx \)
\( A = \left[ -\cos(2x) - \sqrt{3}x \right]_{\pi/6}^{\pi/3} \)
Evaluating at the upper limit \( \frac{\pi}{3} \):
\( -\cos\left(\frac{2\pi}{3}\right) - \sqrt{3}\left(\frac{\pi}{3}\right) = -\left(-\frac{1}{2}\right) - \frac{\pi\sqrt{3}}{3} = \frac{1}{2} - \frac{\pi\sqrt{3}}{3} \).
Evaluating at the lower limit \( \frac{\pi}{6} \):
\( -\cos\left(\frac{\pi}{3}\right) - \sqrt{3}\left(\frac{\pi}{6}\right) = -\frac{1}{2} - \frac{\pi\sqrt{3}}{6} \).
Subtracting:
\( A = \left(\frac{1}{2} - \frac{\pi\sqrt{3}}{3}\right) - \left(-\frac{1}{2} - \frac{\pi\sqrt{3}}{6}\right) \)
\( A = 1 - \frac{2\pi\sqrt{3}}{6} + \frac{\pi\sqrt{3}}{6} = 1 - \frac{\pi\sqrt{3}}{6} \).

(a)
M1 for setting \( f(x) = 0 \).
A1 for realizing \( \sin(2x) = \frac{\sqrt{3}}{2} \).
A1 for finding the correct values for \( 2x \).
A1 for the final x-intercepts \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{3} \).

(b)
M1 for recognizing that maximum occurs when \( \sin(2x) = 1 \) and minimum when \( \sin(2x) = -1 \).
A1 for correct x-values \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \).
A1 for the maximum coordinates \( \left(\frac{\pi}{4}, 2-\sqrt{3}\right) \).
A1 for the minimum coordinates \( \left(\frac{3\pi}{4}, -2-\sqrt{3}\right) \).

(c)
N1 for a correctly shaped sine wave with correct domain.
N1 for showing correct intercepts and y-intercept.
N1 for clearly marked local maximum and minimum positions.

(d)
M1 for setting up the correct definite integral with correct limits.
A1 for correct integration \( -\cos(2x) - \sqrt{3}x \).
A1 for finding the exact area \( 1 - \frac{\pi\sqrt{3}}{6} \).

(a) Let \(W\) be the weight of an apple. \(W \sim N(150, 12^2)\).
Using a GDC to find \(P(W > 165)\):
\(P(W > 165) \approx 0.105649 \approx 0.106\).

(b) Let \(X\) be the number of "Premium" apples in a sample of 10.
\(X \sim B(10, p)\) where \(p = 0.105649\).
Using a GDC to find \(P(X = 3)\):
\(P(X = 3) = \binom{10}{3} (0.105649)^3 (1 - 0.105649)^7 \approx 0.0646\).

(c) Let \(n\) be the sample size. We require \(P(X \ge 1) > 0.99\).
\(1 - P(X = 0) > 0.99 \implies (1 - p)^n < 0.01\).
Using \(1 - p \approx 0.894351\):
\(0.894351^n < 0.01\)
\(n \ln(0.894351) < \ln(0.01)\)
\(n > \frac{\ln(0.01)}{\ln(0.894351)} \approx 41.25\).
Since \(n\) must be an integer, the minimum sample size is \(42\).

(a)
M1 for setting up the normal distribution probability \(P(W > 165)\).
A1 for \(0.106\) (accept \(0.105649...\)).

(b)
M1 for recognizing the binomial distribution \(B(10, 0.1056)\) and setting up \(P(X = 3)\).
A1 for \(0.0646\) (accept \(0.0645\) from rounded \(p\)).

(c)
M1 for setting up the inequality \((1 - p)^n < 0.01\) or equivalent equation.
A1 for \(n = 42\).

(a) The acceleration \(a(t)\) is the derivative of the velocity function: \(a(t) = v'(t)\).
Using the GDC numerical derivative function at \(t = 2\):
\(a(2) = v'(2) \approx 13.2\text{ m s}^{-2}\) (or exactly \(12\ln(3)\)).

(b) The particle is at rest when \(v(t) = 0\):
\(3t^2 \ln(t+1) - 4t = 0\).
Using GDC to solve for \(t > 0\) in the interval \(0 \le t \le 5\):
\(t \approx 1.47\text{ s}\) (or \(1.4727...\)).

(c) The total distance travelled is the integral of the speed: \(\text{Distance} = \int_{0}^{3} |v(t)| \text{ d}t\).
Using GDC to compute this definite integral:
\(\int_{0}^{3} |3t^2 \ln(t+1) - 4t| \text{ d}t \approx 17.3\text{ m}\) (or \(17.304...\)).

(a)
M1 for recognizing that \(a(t) = v'(t)\) and attempting to evaluate at \(t=2\).
A1 for \(13.2\) (or \(12\ln(3)\)).

(b)
M1 for setting \(v(t) = 0\).
A1 for \(t \approx 1.47\).

(c)
M1 for setting up the integral \(\int_{0}^{3} |v(t)| \text{ d}t\).
A1 for \(17.3\).

(a) Let North at \(B\) be represented by a vertical line. The bearing of \(B\) from \(A\) is \(075^\circ\), so the line \(AB\) makes an angle of \(75^\circ\) with the North line at \(B\) pointing towards the South-West (i.e., \(180^\circ + 75^\circ = 255^\circ\)).
The bearing of \(C\) from \(B\) is \(160^\circ\).
The interior angle \(\angle ABC\) is the difference between these directions:
\(\angle ABC = 180^\circ - (160^\circ - 75^\circ) = 95^\circ\).
Using the Cosine Rule on triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 12^2 + 18^2 - 2(12)(18)\cos(95^\circ)\)
\(AC^2 = 144 + 324 - 432\cos(95^\circ) \approx 505.651\)
\(AC \approx 22.5\text{ km}\).

(b) Using the Sine Rule to find the interior angle \(\angle ACB\) (or \(C\)):
\(\frac{\sin C}{12} = \frac{\sin(95^\circ)}{22.487}\)
\(\sin C \approx 0.5317 \implies C \approx 32.12^\circ\).
The bearing of \(B\) from \(C\) is \(160^\circ + 180^\circ = 340^\circ\).
Since point \(A\) lies to the West of the line \(CB\), we subtract \(C\) from this bearing:
\(\text{Bearing} = 340^\circ - 32.12^\circ \approx 308^\circ\).

(a)
M1 for finding the interior angle \(\angle ABC = 95^\circ\) (or equivalent geometric construction).
M1 for substituting correctly into the cosine rule.
A1 for \(22.5\text{ km}\) (accept \(22.487...\)).

(b)
M1 for attempting to use the sine rule to find angle \(\angle ACB\).
A1 for \(C \approx 32.1^\circ\) (or \(32.12^\circ\)).
A1 for bearing \(308^\circ\) (accept \(307.88...^\circ\)).

(a) Set \(f(x) = g(x)\):
\(\mathrm{e}^{0.5x} - 2 = \frac{4}{x-1}\).
Using GDC to find the intersection point for \(x > 1\):
\(x \approx 2.85\) (or \(2.8510...\)).

(b) A translation by \(\begin{pmatrix} 2 \\ -1 \end{pmatrix}\) shifts the graph \(2\) units right and \(1\) unit down.
\(h(x) = g(x-2) - 1 = \frac{4}{(x-2)-1} - 1 = \frac{4}{x-3} - 1\).

(c) Set \(f(x) = h(x)\) for \(x > 3\):
\(\mathrm{e}^{0.5x} - 2 = \frac{4}{x-3} - 1\).
Using GDC to solve this equation:
\(x \approx 3.73\) (or \(3.7323...\)).
Substitute back into \(f(x)\) to find \(y\):
\(y = \mathrm{e}^{0.5(3.7323)} - 2 \approx 4.46\).
So the coordinates are \((3.73, 4.46)\).

(a)
M1 for setting up the equation \(f(x) = g(x)\).
A1 for \(x \approx 2.85\).

(b)
M1 for substituting \(x-2\) into the definition of \(g(x)\).
A1 for \(h(x) = \frac{4}{x-3} - 1\).

(c)
M1 for setting \(f(x) = h(x)\) and attempting to solve (finding \(x \approx 3.73\) or \(y \approx 4.46\)).
A1 for coordinates \((3.73, 4.46)\) (accept \((3.73, 4.46)\)).

(a) The value \(V_n\) of the machine after \(n\) years is:
\(V_n = 25000 \times (1 - 0.12)^n = 25000 \times (0.88)^n\).
For \(n = 5\):
\(V_5 = 25000 \times (0.88)^5 \approx 13193.29\).
To the nearest dollar, this is \(\$13\,193\).

(b) We want to find the smallest integer \(n\) such that:
\(25000 \times (0.88)^n < 5000\)
\(0.88^n < 0.2\).
Taking the natural logarithm of both sides:
\(n \ln(0.88) < \ln(0.2)\).
Since \(\ln(0.88) < 0\), we reverse the inequality:
\(n > \frac{\ln(0.2)}{\ln(0.88)} \approx 12.59\).
Since \(n\) must be an integer, it will take \(13\) complete years.

(a)
M1 for substituting \(n = 5\) into a correct geometric progression / decay expression.
A1 for \(\$13\,193\).

(b)
M1 for setting up the inequality \(25000 \times (0.88)^n < 5000\).
M1 for a valid method of solving the inequality (e.g., using logarithms, or a GDC table/solver).
A1 for \(n > 12.59\) (or boundary \(12.59...\)).
A1 for \(13\) (must be an integer).

(a) The intersection point is found by setting \(4 - x^2 = \ln(x)\).
Using GDC to solve:
\(x \approx 1.8411 \approx 1.84\).

(b) The curve \(y = \ln(x)\) intersects the x-axis at \(x = 1\).
The curve \(y = 4 - x^2\) intersects the positive x-axis at \(x = 2\).
Let \(x_0 = 1.8411\) be the x-coordinate of the intersection point.
The area of the region bounded by the curves and the x-axis is split into two parts:
\(A = \int_{1}^{x_0} \ln(x) \text{ d}x + \int_{x_0}^{2} (4 - x^2) \text{ d}x\).
Using GDC to compute these definite integrals:
\(\int_{1}^{1.8411} \ln(x) \text{ d}x \approx 0.2827\)
\(\int_{1.8411}^{2} (4 - x^2) \text{ d}x \approx 0.0499\).
Adding these values:
\(A \approx 0.2827 + 0.0499 = 0.3326 \approx 0.333\).

(a)
M1 for setting \(4 - x^2 = \ln(x)\).
A1 for \(x \approx 1.84\).

(b)
M1 for finding the x-intercepts \(x=1\) and \(x=2\).
M1 for split integral expression: \(\int_{1}^{x_0} \ln(x) \text{ d}x + \int_{x_0}^{2} (4 - x^2) \text{ d}x\).
A1 for finding either individual integral correctly (\(0.283\) or \(0.050\)).
A1 for \(0.333\) (accept \(0.332\)).

(a) We want to find \(P(1400 \le T \le 1650)\) where \(T \sim N(1500, 120^2)\).
Using the GDC: \(\text{normalcdf}(1400, 1650, 1500, 120) \approx 0.690111\)
To 3 significant figures, \(P(1400 \le T \le 1650) = 0.690\).

(b) Let \(k\) be the minimum lifetime. We have \(P(T > k) = 0.15\), which means \(P(T \le k) = 0.85\).
Using the GDC inverse normal function: \(k = \text{invNorm}(0.85, 1500, 120) \approx 1624.37\).
To the nearest hour, the minimum lifetime of a premium bulb is 1624 hours.

(c) (i) Since each bulb has a constant probability of 0.15 of being premium, \(X\) follows a binomial distribution:
\(X \sim B(12, 0.15)\).

(ii) We need to find \(P(X \ge 3) = 1 - P(X \le 2)\).
Using GDC binomial cumulative distribution: \(P(X \le 2) = \text{binomcdf}(12, 0.15, 2) \approx 0.73582\).
\(P(X \ge 3) = 1 - 0.73582 \approx 0.26418\).
To 3 significant figures, the probability is 0.264.

(d) Let \(Y\) be the number of boxes containing at least 3 premium bulbs. \(Y\) follows a binomial distribution:
\(Y \sim B(50, p)\) where \(p = 0.26418\).
We need to find \(P(Y \ge 15) = 1 - P(Y \le 14)\).
Using GDC: \(P(Y \le 14) = \text{binomcdf}(50, 0.26418, 14) \approx 0.65538\).
\(P(Y \ge 15) = 1 - 0.65538 \approx 0.34462\).
To 3 significant figures, the probability is 0.345.

(a)
(M1) for setting up the normal probability equation \(P(1400 \le T \le 1650)\).
(A1) for correct parameters used in GDC.
(A1) for \(0.690\). [3 marks]

(b)
(M1) for stating \(P(T \le k) = 0.85\) or \(P(T > k) = 0.15\).
(A1) for \(1624.37...\)
(A1) for rounding to the nearest hour: \(1624\). [3 marks]

(c)
(i) (A1) for identifying Binomial, (A1) for parameters: \(X \sim B(12, 0.15)\). [2 marks]
(ii) (M1) for writing \(P(X \ge 3) = 1 - P(X \le 2)\).
(A1) for \(P(X \le 2) \approx 0.736\).
(A1) for \(0.264\). [3 marks]

(d)
(M1) for identifying new binomial model \(Y \sim B(50, 0.26418)\).
(A1) for finding/using the correct parameter \(p \approx 0.264\).
(M1) for calculating \(1 - P(Y \le 14)\).
(A1) for \(0.345\). [4 marks]

(a) At 04:00, \(t = 4\).
\(R_{\text{in}}(4) = 20 + 10\sin\left(\frac{4\pi}{6}\right) = 20 + 10\sin\left(\frac{2\pi}{3}\right) = 20 + 10\left(\frac{\sqrt{3}}{2}\right) = 20 + 5\sqrt{3} \approx 28.7\) liters per hour.

(b) Set \(R_{\text{in}}(t) = R_{\text{out}}(t)\):
\(20 + 10\sin\left(\frac{\pi t}{6}\right) = 15 + 0.5t\)
\(10\sin\left(\frac{\pi t}{6}\right) - 0.5t + 5 = 0\)
Using a GDC to solve this equation for \(0 \le t \le 24\), we find three intersection points:
\(t \approx 6.35\), \(t \approx 12.2\), and \(t \approx 17.3\) hours (to 3 significant figures).

(c) The total amount of water entering the tank is:
\(\int_{0}^{24} R_{\text{in}}(t) \, dt = \int_{0}^{24} \left(20 + 10\sin\left(\frac{\pi t}{6}\right)\right) \, dt\)
\(= \left[ 20t - \frac{60}{\pi}\cos\left(\frac{\pi t}{6}\right) \right]_0^{24}\)
\(= \left(480 - \frac{60}{\pi}\cos(4\pi)\right) - \left(0 - \frac{60}{\pi}\cos(0)\right)\)
\(= 480 - \frac{60}{\pi} + \frac{60}{\pi} = 480\) liters.

(d) (i) \(W(t) = 150 + \int_{0}^{t} (R_{\text{in}}(x) - R_{\text{out}}(x)) \, dx\)
\(W(t) = 150 + \int_{0}^{t} \left(5 + 10\sin\left(\frac{\pi x}{6}\right) - 0.5x\right) \, dx\)
\(W(t) = 150 + \left[ 5x - \frac{60}{\pi}\cos\left(\frac{\pi x}{6}\right) - 0.25x^2 \right]_0^t\)
\(W(t) = 150 + 5t - \frac{60}{\pi}\cos\left(\frac{\pi t}{6}\right) - 0.25t^2 + \frac{60}{\pi}\).

(ii) The maximum water occurs either at critical points where \(W'(t) = 0\) (from part (b)) or at the endpoints \(t = 0\) and \(t = 24\).
Evaluating \(W(t)\) at these points using GDC:
- \(W(0) = 150\) liters
- \(W(6.35) \approx 209.55\) liters (local max since \(W'(t)\) changes from positive to negative)
- \(W(12.2) \approx 141.2\) liters (local min)
- \(W(17.3) \approx 198.6\) liters (local max)
- \(W(24) = 126\) liters

Thus, the maximum amount of water in the tank is approximately 210 liters (or more precisely 210 to 3 s.f.), which occurs at \(t \approx 6.35\) hours (approximately 06:21).

(a)
(M1) for substituting \(t = 4\) into \(R_{\text{in}}(t)\).
(A1) for \(28.7\) (accept \(20 + 5\sqrt{3}\)). [2 marks]

(b)
(M1) for setting up the equation \(R_{\text{in}}(t) = R_{\text{out}}(t)\).
(A2) for all three correct values: \(t = 6.35, 12.2, 17.3\) (award (A1) if only one or two are correct). [3 marks]

(c)
(M1) for setting up the integral \(\int_{0}^{24} R_{\text{in}}(t) \, dt\).
(A1) for correct integration.
(A1) for \(480\). [3 marks]

(d)
(i)
(M1) for setting up \(W(t) = 150 + \int (R_{\text{in}} - R_{\text{out}}) \, dt\).
(A1) for correct simplified formula: \(W(t) = 150 + 5t - \frac{60}{\pi}\cos\left(\frac{\pi t}{6}\right) - 0.25t^2 + \frac{60}{\pi}\).
[2 marks]

(ii)
(M1) for realizing the maximum occurs at one of the critical points or endpoints.
(M1) for evaluating \(W(t)\) at \(t \approx 6.35\) and \(t \approx 17.3\).
(A1) for \(W(6.35) \approx 210\) (or 209.55).
(A1) for \(W(17.3) \approx 199\).
(A1) for identifying the maximum is 210 liters at \(t = 6.35\) hours. [5 marks]

(a) Let us find the angles in the triangle \(ABY\).
- The bearing of \(B\) from \(A\) is \(110^\circ\), and the bearing of \(Y\) from \(A\) is \(160^\circ\).
Thus, \(\angle BAY = 160^\circ - 110^\circ = 50^\circ\).
- The bearing of \(A\) from \(B\) is \(110^\circ + 180^\circ = 290^\circ\).
The bearing of \(Y\) from \(B\) is \(220^\circ\).
Thus, \(\angle ABY = 290^\circ - 220^\circ = 70^\circ\).
- The angle at \(Y\) is \(\angle AYB = 180^\circ - (50^\circ + 70^\circ) = 60^\circ\).

Using the Sine Rule in \(\triangle ABY\):
\(\frac{AY}{\sin(70^\circ)} = \frac{AB}{\sin(60^\circ)}\)
\(AY = \frac{12 \sin(70^\circ)}{\sin(60^\circ)} \approx 13.018\text{ km}\).
To 3 significant figures, the distance is 13.0 km.

(b) Using the Sine Rule to find \(BY\):
\(\frac{BY}{\sin(50^\circ)} = \frac{12}{\sin(60^\circ)}\)
\(BY = \frac{12 \sin(50^\circ)}{\sin(60^\circ)} \approx 10.613\text{ km}\).
To 3 significant figures, the distance is 10.6 km.

(c) The yacht travels on a bearing of \(280^\circ\).
Let's determine the angle between the yacht's direction of travel and the line \(YA\).
The bearing of \(A\) from \(Y\) is \(160^\circ + 180^\circ = 340^\circ\).
The yacht's bearing is \(280^\circ\).
The angle between the line \(YA\) and the path is \(340^\circ - 280^\circ = 60^\circ\).

Let the closest point of approach to \(A\) on the path be \(D\). In the right-angled triangle \(ADY\), \(YA\) is the hypotenuse and \(\angle AYD = 60^\circ\).
The shortest distance is the perpendicular distance \(AD\):
\(AD = AY \sin(60^\circ) = 13.018 \times \sin(60^\circ) = 11.273\text{ km}\).
To 3 significant figures, the shortest distance is 11.3 km.

(d) The distance travelled along the path from \(Y\) to \(D\) is:
\(YD = AY \cos(60^\circ) = 13.018 \times 0.5 = 6.509\text{ km}\).
With a speed of 15 km/h, the time taken \(t\) is:
\(t = \frac{6.509}{15} \approx 0.43393\text{ hours}\).
Converting to minutes: \(0.43393 \times 60 \approx 26.036\text{ minutes}\).
Rounded to the nearest minute, the time taken is 26 minutes.
So the yacht is closest at 12:26.

(a)
(M1) for finding \(\angle BAY = 50^\circ\).
(M1) for finding \(\angle ABY = 70^\circ\) and thus \(\angle AYB = 60^\circ\).
(M1) for applying the Sine Rule correctly: \(\frac{AY}{\sin(70^\circ)} = \frac{12}{\sin(60^\circ)}\).
(A1) for \(13.0\text{ km}\). [4 marks]

(b)
(M1) for applying the Sine Rule for \(BY\).
(A1) for \(10.6\text{ km}\). [2 marks]

(c)
(M1) for finding the bearing of \(A\) from \(Y\) is \(340^\circ\).
(M1) for finding the angle between the path and the line \(YA\) is \(60^\circ\).
(M1) for recognizing that the shortest distance involves \(AY \sin(60^\circ)\).
(A1) for correct calculation setup.
(A1) for \(11.3\text{ km}\). [5 marks]

(d)
(M1) for finding the distance travelled \(YD = AY \cos(60^\circ)\).
(A1) for \(YD \approx 6.51\text{ km}\).
(M1) for using \(t = \frac{\text{distance}}{\text{speed}}\).
(A1) for \(26\text{ minutes}\) (giving 12:26). [4 marks]