MetadataPastPaper.sampleTitle

We use the substitution method. Let \(u = e^x\).
Then \(du = e^x \, dx\).

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

Substitute these into the integral:
\(\int_{1}^{2} \frac{1}{1 + u^2} \, du\)

Integrating, we get:
\(\left[ \arctan(u) \right]_{1}^{2} = \arctan(2) - \arctan(1)\)

Since \(\arctan(1) = \frac{\pi}{4}\), the final answer is:
\(\arctan(2) - \frac{\pi}{4}\)

M1: Attempt to use substitution with \(u = e^x\) and \(du = e^x \, dx\).
A1: Correctly changing the limits of integration to 1 and 2.
A1: Correct integration to get \(\arctan(u)\).
M1: Substituting the limits into the integrated expression.
A1: Evaluating \(\arctan(1) = \frac{\pi}{4}\).
A1.333: Correct final answer: \(\arctan(2) - \frac{\pi}{4}\).

The general term in the expansion of \((a + b)^n\) is given by \(\binom{n}{r} a^{n-r} b^r\).
For this binomial, the general term is:
\(\binom{8}{r} (2x)^{8-r} \left(-\frac{1}{x}\right)^r\)

Simplifying this expression:
\(\binom{8}{r} 2^{8-r} (-1)^r x^{8-r} x^{-r} = \binom{8}{r} 2^{8-r} (-1)^r x^{8-2r}\)

We want to find the coefficient of \(x^2\), so we set the exponent of \(x\) equal to 2:
\(8 - 2r = 2 \implies 2r = 6 \implies r = 3\)

Now we substitute \(r = 3\) back into the coefficient part of the term:
Coefficient \(= \binom{8}{3} 2^{8-3} (-1)^3\)
\(= \binom{8}{3} 2^5 (-1)\)

Calculate the components:
\(\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\)
\(2^5 = 32\)

Multiply them together:
\(Coefficient = 56 \times 32 \times (-1) = -1792\)

M1: Attempt to write the general term using the binomial expansion formula.
A1: Correct simplified power of \(x\), e.g., \(8 - 2r\).
M1: Setting \(8 - 2r = 2\) and solving for \(r = 3\).
A1: Correct evaluation of \(\binom{8}{3} = 56\).
A1: Correct evaluation of \(2^5 = 32\) and accounting for the negative sign.
A1.333: Correct final coefficient of \(-1792\).

Using the Pythagorean identity \(\cos^2(x) = 1 - \sin^2(x)\), we substitute this into the equation:
\(2(1 - \sin^2(x)) + 3 \sin(x) - 3 = 0\)
\(2 - 2 \sin^2(x) + 3 \sin(x) - 3 = 0\)
\(-2 \sin^2(x) + 3 \sin(x) - 1 = 0\)
\(2 \sin^2(x) - 3 \sin(x) + 1 = 0\)

Let \(u = \sin(x)\). The equation becomes a quadratic in terms of \(u\):
\(2u^2 - 3u + 1 = 0\)
\((2u - 1)(u - 1) = 0\)

This gives two possible solutions for \(u\):
\(u = \frac{1}{2}\) or \(u = 1\)

Therefore, we solve:
1) \(\sin(x) = \frac{1}{2}\) for \(0 \le x \le 2\pi\):
\(x = \frac{\pi}{6}, \frac{5\pi}{6}\)

2) \(\sin(x) = 1\) for \(0 \le x \le 2\pi\):
\(x = \frac{\pi}{2}\)

Combining these, the solutions are:
\(x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\)

M1: Attempt to write the equation in terms of \(\sin(x)\) using the Pythagorean identity.
A1: Correct quadratic equation \(2 \sin^2(x) - 3 \sin(x) + 1 = 0\).
M1: Factorising or solving the quadratic equation.
A1: Correct values \(\sin(x) = \frac{1}{2}\) and \(\sin(x) = 1\).
A1: Correct solutions for \(\sin(x) = \frac{1}{2}\), which are \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).
A1.333: Correct solution for \(\sin(x) = 1\), which is \(\frac{\pi}{2}\).

First, we find the composite function \((f \circ g)(x)\):
\((f \circ g)(x) = f(g(x)) = f(2x - 3)\)
\(= \frac{3(2x - 3) + 1}{(2x - 3) - 2}\)
\(= \frac{6x - 9 + 1}{2x - 5} = \frac{6x - 8}{2x - 5}\)

To find the inverse function, let \(y = \frac{6x - 8}{2x - 5}\). We swap \(x\) and \(y\) and solve for \(y\):
\(x = \frac{6y - 8}{2y - 5}\)
\(x(2y - 5) = 6y - 8\)
\(2xy - 5x = 6y - 8\)
\(2xy - 6y = 5x - 8\)
\(y(2x - 6) = 5x - 8\)
\(y = \frac{5x - 8}{2x - 6}\)

So, \((f \circ g)^{-1}(x) = \frac{5x - 8}{2x - 6}\).

To find the domain of the inverse function, we observe where its denominator is zero:
\(2x - 6 = 0 \implies x = 3\).
Thus, the domain is \(x \in \mathbb{R}, x \neq 3\).

M1: Attempt to find the composite function by substituting \(g(x)\) into \(f(x)\).
A1: Correct simplified composite function \((f \circ g)(x) = \frac{6x - 8}{2x - 5}\).
M1: Attempt to find the inverse by swapping variables and isolating \(y\).
A1: Correct algebraic manipulation to group \(y\) terms together, e.g., \(y(2x - 6) = 5x - 8\).
A1: Correct inverse function expression \((f \circ g)^{-1}(x) = \frac{5x - 8}{2x - 6}\) (or equivalent).
A1.333: Correctly stating the domain as \(x \in \mathbb{R}, x \neq 3\).

(a) Since the sum of probabilities must equal 1:
\(\sum_{x=1}^{4} P(X = x) = 1 \implies k(1) + k(2) + k(3) + k(4) = 1\)
\(10k = 1 \implies k = 0.1\)

(b) The expected value is:
\(\text{E}(X) = \sum x P(X = x) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4)\)
\(= 0.1 + 0.4 + 0.9 + 1.6 = 3\)

(c) First, we find \(\text{E}(X^2)\):
\(\text{E}(X^2) = \sum x^2 P(X = x) = 1^2(0.1) + 2^2(0.2) + 3^2(0.3) + 4^2(0.4)\)
\(= 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4)\)
\(= 0.1 + 0.8 + 2.7 + 6.4 = 10\)

Now, use the variance formula:
\(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\)
\(= 10 - 3^2 = 10 - 9 = 1\)

M1: Set sum of probabilities equal to 1.
A1: Correct value \(k = 0.1\).
M1: Correct method to calculate expectation \(\text{E}(X)\).
A1: Correct value \(\text{E}(X) = 3\).
M1: Correct method to calculate \(\text{E}(X^2)\) and find \(\text{Var}(X)\).
A1.333: Correct value \(\text{Var}(X) = 1\).

First, find the \(y\)-coordinate at \(x = e\):
\(y = e^2 \ln(e) = e^2\)
So, the point of contact is \((e, e^2)\).

Next, we find the derivative \(\frac{dy}{dx}\) using the product rule:
\(\frac{dy}{dx} = \frac{d}{dx}(x^2) \ln(x) + x^2 \frac{d}{dx}(\ln x)\)
\(\frac{dy}{dx} = 2x \ln(x) + x^2 \left(\frac{1}{x}\right) = 2x \ln(x) + x\)

At \(x = e\), the gradient of the tangent is:
\(m_{\text{tangent}} = 2e \ln(e) + e = 2e(1) + e = 3e\)

The gradient of the normal, \(m_{\text{normal}}\), is the negative reciprocal of the tangent gradient:
\(m_{\text{normal}} = -\frac{1}{3e}\)

The equation of the normal line is:
\(y - e^2 = -\frac{1}{3e}(x - e)\)

Multiply the entire equation by \(3e\) to remove the fraction:
\(3e(y - e^2) = -(x - e)\)
\(3ey - 3e^3 = -x + e\)

Rearrange into the form \(ax + by + d = 0\):
\(x + 3ey - 3e^3 - e = 0\)
\(x + 3ey - (3e^3 + e) = 0\)

A1: Correctly finding the \(y\)-coordinate of the point \((e, e^2)\).
M1: Applying the product rule to find the derivative \(\frac{dy}{dx}\).
A1: Correct derivative \(2x \ln(x) + x\).
A1: Evaluating the tangent gradient as \(3e\), and identifying the normal gradient as \(-\frac{1}{3e}\).
M1: Substituting the point and gradient into the equation of a straight line.
A1.333: Expressing the normal equation in the required form: \(x + 3ey - (3e^3 + e) = 0\) (or equivalent).

Let \(a\) be the first term and \(d\) be the common difference of the arithmetic sequence.
We are given:
\(u_3 = a + 2d = 8\)
\(u_7 = a + 6d = 20\)

Subtract the first equation from the second:
\((a + 6d) - (a + 2d) = 20 - 8\)
\(4d = 12 \implies d = 3\)

Substitute \(d = 3\) back into the first equation:
\(a + 2(3) = 8 \implies a = 2\)

For the geometric sequence:
- The first term \(g_1 = d = 3\).
- The second term \(g_2 = a = 2\).
- The common ratio \(r = \frac{g_2}{g_1} = \frac{2}{3}\).

Now, we calculate the sum of the first 5 terms of this geometric sequence using the sum formula \(S_n = \frac{g_1(1 - r^n)}{1 - r}\):
\(S_5 = \frac{3 \left(1 - \left(\frac{2}{3}\right)^5\right)}{1 - \frac{2}{3}}\)
\(S_5 = \frac{3 \left(1 - \frac{32}{243}\right)}{\frac{1}{3}}\)
\(S_5 = 9 \left(\frac{243 - 32}{243}\right) = 9 \left(\frac{211}{243}\right)\)
\(S_5 = \frac{211}{27}\)

M1: Set up simultaneous equations for the arithmetic sequence.
A1: Correctly solve to find \(d = 3\) and \(a = 2\).
A1: Identify geometric sequence properties: first term \(g_1 = 3\) and common ratio \(r = \frac{2}{3}\).
M1: Substitute these parameters into the geometric series sum formula for \(n = 5\).
A1: Correctly evaluate the power term \(\left(\frac{2}{3}\right)^5 = \frac{32}{243}\).
A1.333: Correctly simplify the fraction to obtain \(\frac{211}{27}\).

(a) The dot product is:
\(\vec{u} \cdot \vec{v} = (1)(1) + (0)(p) + (1)(0) = 1\)

(b) First, find the magnitudes of both vectors:
\(|\vec{u}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}\)
\(|\vec{v}| = \sqrt{1^2 + p^2 + 0^2} = \sqrt{1 + p^2}\)

We use the angle formula \(\cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}\) where \(\theta = \frac{\pi}{3}\):
\(\cos\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{2} \sqrt{1+p^2}}\)

Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\):
\[\frac{1}{2} = \frac{1}{\sqrt{2}\sqrt{1+p^2}}\]
\(\sqrt{2}\sqrt{1+p^2} = 2\)

Square both sides:
\(2(1+p^2) = 4\)
\(1 + p^2 = 2\)
\(p^2 = 1\)
\(p = \pm 1\)

(a)
M1: Attempt to calculate dot product.
A1: Correct value of 1.

(b)
A1: Find magnitudes \(|\vec{u}| = \sqrt{2}\) and \(|\vec{v}| = \sqrt{1+p^2}\).
M1: Substitute into the cosine formula with \(\cos(\pi/3) = 1/2\).
A1: Solve the equation to get \(1+p^2 = 2\).
A1.333: Correct final values \(p = 1\) and \(p = -1\).

(a) We use the product rule with \(u = x\) and \(v = (4-x^2)^{\frac{1}{2}}\).
The derivatives are:
\(\frac{du}{dx} = 1\)
Using the chain rule:
\(\frac{dv}{dx} = \frac{1}{2}(4-x^2)^{-\frac{1}{2}} \cdot (-2x) = -\frac{x}{\sqrt{4-x^2}}\)

Applying the product rule:
\(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)
\(\frac{dy}{dx} = x\left(-\frac{x}{\sqrt{4-x^2}}\right) + \sqrt{4-x^2}\)
\(\frac{dy}{dx} = -\frac{x^2}{\sqrt{4-x^2}} + \frac{4-x^2}{\sqrt{4-x^2}}\)
\(\frac{dy}{dx} = \frac{4-2x^2}{\sqrt{4-x^2}}\)

(b) To find the local maximum, we set \(\frac{dy}{dx} = 0\):
\(\frac{4-2x^2}{\sqrt{4-x^2}} = 0 \implies 4-2x^2 = 0\)

\(2x^2 = 4 \implies x^2 = 2 \implies x = \pm\sqrt{2}\)

Since the denominator \(\sqrt{4-x^2}\) is positive for all \(x \in (-2, 2)\), we examine the sign of the numerator \(4-2x^2\):
- For \(-\sqrt{2} < x < \sqrt{2}\), \(\frac{dy}{dx} > 0\).
- For \(\sqrt{2} < x < 2\), \(\frac{dy}{dx} < 0\).
Thus, a local maximum occurs at \(x = \sqrt{2}\).

Substituting \(x = \sqrt{2}\) into the original curve equation:
\(y = \sqrt{2}\sqrt{4-(\sqrt{2})^2} = \sqrt{2}\sqrt{2} = 2\)

Thus, the coordinates of the local maximum point are \((\sqrt{2}, 2)\).

**(a)**
* **M1** for attempting to apply the product rule (or quotient rule by rewriting).
* **A1** for correct derivative of \(\sqrt{4-x^2}\), which is \(-\frac{x}{\sqrt{4-x^2}}\).
* **A1** for simplifying to the given rational expression. [3 marks]

**(b)**
* **M1** for setting \(\frac{dy}{dx} = 0\) and solving for \(x\).
* **A1** for finding \(x = \sqrt{2}\) (and identifying it as the local maximum, e.g., via the first derivative test).
* **A1** for finding the corresponding y-coordinate, \(y = 2\), and stating the final coordinates as \((\sqrt{2}, 2)\). [3 marks]

(a) To show the function is odd, evaluate \( f_k(-x) \):
\( f_k(-x) = (-x) \sqrt{k - (-x)^2} = -x \sqrt{k - x^2} = -f_k(x) \).
Since \( f_k(-x) = -f_k(x) \) for all \( x \) in the domain, \( f_k(x) \) is an odd function.

(b) Applying the product rule and chain rule:
\( f'_k(x) = 1 \cdot \sqrt{k-x^2} + x \cdot \frac{1}{2\sqrt{k-x^2}} \cdot (-2x) \)
\( f'_k(x) = \sqrt{k-x^2} - \frac{x^2}{\sqrt{k-x^2}} \)
\( f'_k(x) = \frac{(k-x^2) - x^2}{\sqrt{k-x^2}} = \frac{k-2x^2}{\sqrt{k-x^2}} \).

To find stationary points, set \( f'_k(x) = 0 \):
\( k - 2x^2 = 0 \Rightarrow x^2 = \frac{k}{2} \Rightarrow x = \pm\sqrt{\frac{k}{2}} \).

For \( x = \sqrt{\frac{k}{2}} \):
\( y = \sqrt{\frac{k}{2}} \sqrt{k - \frac{k}{2}} = \sqrt{\frac{k}{2}} \sqrt{\frac{k}{2}} = \frac{k}{2} \).
This is the local maximum point: \( \left(\sqrt{\frac{k}{2}}, \frac{k}{2}\right) \).

Since \( f_k(x) \) is odd, the local minimum point is at: \( \left(-\sqrt{\frac{k}{2}}, -\frac{k}{2}\right) \).

(c) For \( k = 4 \), \( f_4(x) = x \sqrt{4-x^2} \). The region in the first quadrant is bounded by \( x = 0 \) and the positive x-intercept \( x = 2 \).
\( A = \int_{0}^{2} x(4-x^2)^{1/2} dx \).
Let \( u = 4-x^2 \), hence \( du = -2x\,dx \).
When \( x = 0 \), \( u = 4 \); when \( x = 2 \), \( u = 0 \).
\( A = \int_{4}^{0} -\frac{1}{2} u^{1/2} du = \frac{1}{2} \int_{0}^{4} u^{1/2} du \)
\( A = \frac{1}{2} \left[ \frac{2}{3}u^{3/2} \right]_{0}^{4} = \frac{1}{3} (4^{3/2} - 0) = \frac{8}{3} \).

(d) The volume of revolution is given by:
\( V = \pi \int_{0}^{2} [f_4(x)]^2 dx = \pi \int_{0}^{2} x^2(4-x^2) dx \)
\( V = \pi \int_{0}^{2} (4x^2 - x^4) dx \)
\( V = \pi \left[ \frac{4}{3}x^3 - \frac{1}{5}x^5 \right]_{0}^{2} \)
\( V = \pi \left( \left( \frac{32}{3} - \frac{32}{5} \right) - 0 \right) \)
\( V = 32\pi \left( \frac{1}{3} - \frac{1}{5} \right) = 32\pi \left( \frac{2}{15} \right) = \frac{64\pi}{15} \).

(a) M1 for attempting to substitute \(-x\) into the expression. A1 for correct simplification showing \(f_k(-x) = -f_k(x)\).

(b) M1 for using the product rule. A1 for correct differentiation of the second term. A1 for getting the correct numerator \(k-2x^2\). M1 for setting their numerator equal to 0. A1 for the correct coordinates of the local maximum. A1 for the correct coordinates of the local minimum.

(c) M1 for setting up the integral for the area. M1 for choosing and executing a valid integration technique (substitution). A1 for correct integration bounds/antiderivative. A1 for the final exact area \(\frac{8}{3}\).

(d) M1 for setting up the volume integral of \(\pi y^2\). A1 for correct expansion of the integrand. M1 for integrating term-by-term. A1 for substituting limits correctly. A2 for the final exact simplified volume \(\frac{64\pi}{15}\).

(a) We want to write \( \sqrt{3} \sin(2t) - \cos(2t) = R\sin(2t)\cos\phi - R\cos(2t)\sin\phi \).
Equating coefficients:
\( R \cos\phi = \sqrt{3} \)
\( R \sin\phi = 1 \)
Squaring and adding:
\( R^2 = (\sqrt{3})^2 + 1^2 = 4 \Rightarrow R = 2 \).

To find \( \phi \):
\( \tan\phi = \frac{1}{\sqrt{3}} \).
Since \( 0 < \phi < \pi \), we get \( \phi = \frac{\pi}{6} \).
Thus, \( s(t) = 2 \sin\left(2t - \frac{\pi}{6}\right) \).

(b) The particle is at rest when its velocity is zero, i.e., \( v(t) = s'(t) = 0 \).
\( s'(t) = 4 \cos\left(2t - \frac{\pi}{6}\right) \).
Setting \( v(t) = 0 \):
\( \cos\left(2t - \frac{\pi}{6}\right) = 0 \)
\( 2t - \frac{\pi}{6} = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \)

First value:
\( 2t - \frac{\pi}{6} = \frac{\pi}{2} \Rightarrow 2t = \frac{2\pi}{3} \Rightarrow t = \frac{\pi}{3} \).

Second value:
\( 2t - \frac{\pi}{6} = \frac{3\pi}{2} \Rightarrow 2t = \frac{10\pi}{6} = \frac{5\pi}{3} \Rightarrow t = \frac{5\pi}{6} \).

(c) To find the total distance traveled in \( 0 \le t \le \frac{\pi}{2} \), we identify any turning points in this interval.
The only time the particle is at rest within this interval is at \( t = \frac{\pi}{3} \).
We calculate the position at the boundaries and the turning point:
At \( t = 0 \):
\( s(0) = 2 \sin\left(-\frac{\pi}{6}\right) = -1 \).

At \( t = \frac{\pi}{3} \):
\( s\left(\frac{\pi}{3}\right) = 2 \sin\left(\frac{2\pi}{3} - \frac{\pi}{6}\right) = 2 \sin\left(\frac{\pi}{2}\right) = 2 \).

At \( t = \frac{\pi}{2} \):
\( s\left(\frac{\pi}{2}\right) = 2 \sin\left(\pi - \frac{\pi}{6}\right) = 2 \sin\left(\frac{5\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1 \).

Total distance is computed in two segments:
Distance 1: from \( t = 0 \) to \( t = \frac{\pi}{3} \):
\( |s\left(\frac{\pi}{3}\right) - s(0)| = |2 - (-1)| = 3 \).

Distance 2: from \( t = \frac{\pi}{3} \) to \( t = \frac{\pi}{2} \):
\( |s\left(\frac{\pi}{2}\right) - s\left(\frac{\pi}{3}\right)| = |1 - 2| = 1 \).

Total distance = \( 3 + 1 = 4 \) meters.

(a) M1 for attempting to find \(R\) by squaring and adding. A1 for \(R=2\). M1 for attempting to solve \(\tan\phi = \frac{1}{\sqrt{3}}\). A1 for \(\phi = \frac{\pi}{6}\).

(b) M1 for finding the derivative function \(s'(t)\). A1 for correct derivative. M1 for setting their derivative equal to 0. A1 for setting the argument to \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). A1 for \(t = \frac{\pi}{3}\). A1 for \(t = \frac{5\pi}{6}\).

(c) M1 for identifying the relevant turning point \(t = \frac{\pi}{3}\) within the interval. M1 for evaluating \(s(0)\). A1 for \(s(0) = -1\). M1 for evaluating \(s(\frac{\pi}{3})\). A1 for \(s(\frac{\pi}{3}) = 2\). M1 for evaluating \(s(\frac{\pi}{2})\). A1 for \(s(\frac{\pi}{2}) = 1\). A1 for calculating total distance as \(3 + 1 = 4\).

(a) Since \( f(x) \) is a probability density function, the total area under the curve is 1:
\( \int_{1}^{a} \frac{k}{x^2} dx = 1 \)
\( k \left[ -\frac{1}{x} \right]_{1}^{a} = 1 \)
\( k \left( 1 - \frac{1}{a} \right) = 1 \)
\( k \left( \frac{a-1}{a} \right) = 1 \Rightarrow k = \frac{a}{a-1} \).

(b) The expected value is given by:
\( E(X) = \int_{1}^{a} x f(x) dx = \int_{1}^{a} x \frac{k}{x^2} dx \)
\( E(X) = k \int_{1}^{a} \frac{1}{x} dx = k [ \ln x ]_{1}^{a} = k \ln a \).
Substituting \( k = \frac{a}{a-1} \):
\( E(X) = \frac{a}{a-1} \ln a \).
We are given \( E(X) = \frac{3}{2} \ln 3 \).
Comparing \( \frac{a}{a-1} \ln a \) with \( \frac{3}{2} \ln 3 \), if \( a = 3 \):
\( \frac{3}{3-1} \ln 3 = \frac{3}{2} \ln 3 \).
Since the function \( g(y) = \frac{y}{y-1} \ln y \) is strictly increasing for \( y > 1 \) (since its derivative is positive), \( a = 3 \) is the unique solution.

(c) For \( a = 3 \), we have \( k = \frac{3}{3-1} = \frac{3}{2} \).

(i) The cumulative distribution function is:
\( F(x) = \int_{1}^{x} \frac{3}{2t^2} dt = \frac{3}{2} \left[ -\frac{1}{t} \right]_{1}^{x} = \frac{3}{2} \left( 1 - \frac{1}{x} \right) \) for \( 1 \le x \le 3 \).

(ii) Let \( m \) be the median. Then \( F(m) = 0.5 \):
\( \frac{3}{2} \left( 1 - \frac{1}{m} \right) = \frac{1}{2} \)
\( 1 - \frac{1}{m} = \frac{1}{3} \)
\( \frac{1}{m} = \frac{2}{3} \Rightarrow m = 1.5 \).

(iii) The probability that a single observation is greater than the median is \( 1 - F(m) = 0.5 \).
Let the two independent observations be \( X_1 \) and \( X_2 \).
\( P(X_1 > m \cap X_2 > m) = P(X_1 > m) \cdot P(X_2 > m) = 0.5 \times 0.5 = 0.25 \).

(a) M1 for integrating \(\frac{k}{x^2}\) from 1 to \(a\) and setting to 1. A1 for finding correct antiderivative. A1 for obtaining the correct relation to show \(k = \frac{a}{a-1}\).

(b) M1 for writing the correct integral expression for \(E(X)\). A1 for finding the antiderivative as \(k \ln x\). A1 for obtaining \(k \ln a\). M1 for substituting the value of \(k\) to get \(\frac{a}{a-1} \ln a = \frac{3}{2} \ln 3\). A1 for evaluating both sides with \(a = 3\). R1 for reasoning that this solution is unique.

(c)(i) M1 for setting up the integral for \(F(x)\). A1 for correct integration. A1 for the final form of \(F(x)\).
(c)(ii) M1 for setting \(F(m) = 0.5\). A1 for getting \(1 - \frac{1}{m} = \frac{1}{3}\). A1 for showing \(m = 1.5\).
(c)(iii) M1 for stating the probability of being greater than the median is 0.5. M1 for multiplying independent probabilities. A1 for finding the final answer \(0.25\) (or \(\frac{1}{4}\)).

The sum of the first \(n\) terms of the arithmetic sequence is given by: \(S_n(\text{arith}) = \frac{n}{2}[2(12) + (n - 1)(4.5)] = 9.75n + 2.25n^2\). The sum of the first \(n\) terms of the geometric sequence is given by: \(S_n(\text{geom}) = \frac{3(1.15^n - 1)}{1.15 - 1} = 20(1.15^n - 1)\). We use a GDC to find the smallest integer \(n\) such that \(20(1.15^n - 1) > 9.75n + 2.25n^2\). Testing values of \(n\): For \(n = 36\): \(S_{36}(\text{geom}) \approx 3043.08\) and \(S_{36}(\text{arith}) = 3267\), so \(S_{36}(\text{geom}) < S_{36}(\text{arith})\). For \(n = 37\): \(S_{37}(\text{geom}) \approx 3502.54\) and \(S_{37}(\text{arith}) = 3441\), so \(S_{37}(\text{geom}) > S_{37}(\text{arith})\). Thus, the least value of \(n\) is \(37\).

M1: Setting up the formula for \(S_n(\text{arith})\).
M1: Setting up the formula for \(S_n(\text{geom})\).
M1: Attempting to solve the inequality or comparing terms near the crossover point using a table or graph on a GDC.
A1: Finding correct values of \(S_{36}\) for both sequences.
A1: Finding correct values of \(S_{37}\) for both sequences.
A1: Correct final answer \(n = 37\).

(a) First find \(g(3) = \ln(3^2 + 1) = \ln(10)\). Then, \((f \circ g)(3) = f(\ln(10)) = e^{0.5 \ln(10)} - 2 = e^{\ln(\sqrt{10})} - 2 = \sqrt{10} - 2 \approx 1.16\). (b) The composite function is \((f \circ g)(x) = e^{0.5 \ln(x^2 + 1)} - 2 = \sqrt{x^2 + 1} - 2\). We solve the equation \(\sqrt{x^2 + 1} - 2 = \ln(x^2 + 1)\). Let \(u = x^2 + 1\). The equation becomes \(\sqrt{u} - 2 = \ln(u)\). Using a GDC to solve for \(u \ge 1\) yields \(u \approx 28.694\). Then \(x^2 + 1 = 28.694 \implies x^2 = 27.694 \implies x \approx \pm 5.26\).

(a) M1: Attempting to find \(g(3)\).
A1: Correct exact value \(\sqrt{10} - 2\) or 3 s.f. value \(1.16\).
(b) M1: Correct expression for the composite function \(\sqrt{x^2+1}-2\).
M1: Equating the composite function to \(g(x)\).
A1: Finding \(x \approx 5.26\).
A1: Finding \(x \approx -5.26\).

(a) In triangle \(PAB\), the angle \(\angle PAB = 95^\circ\) (since the angle between the line \(AP\) and the North line at \(A\) is \(75^\circ\), and the angle of the line \(AB\) with the South-pointing line at \(A\) is \(180^\circ - 160^\circ = 20^\circ\); thus, \(\angle PAB = 75^\circ + 20^\circ = 95^\circ\)). Using the Cosine Rule: \(PB^2 = 12^2 + 18^2 - 2(12)(18)\cos(95^\circ)\). \(PB^2 \approx 144 + 324 - 432(-0.08716) \approx 505.65\). \(PB \approx 22.486 \approx 22.5\text{ km}\). (b) Using the Sine Rule to find the angle \(\angle APB\): \(\frac{\sin(\angle APB)}{18} = \frac{\sin(95^\circ)}{22.486}\). \(\sin(\angle APB) \approx \frac{18 \times 0.99619}{22.486} \approx 0.7974\). \(\angle APB \approx \arcsin(0.7974) \approx 52.88^\circ\). The bearing of \(B\) from \(P\) is \(75^\circ + 52.88^\circ = 127.88^\circ \approx 128^\circ\).

(a) M1: Finding the interior angle \(\angle PAB = 95^\circ\).
M1: Substituting values correctly into the Cosine Rule.
A1: Distance \(PB \approx 22.5\text{ km}\) (or \(22.49\text{ km}\)).
(b) M1: Applying the Sine Rule (or Cosine Rule) to find angle \(\angle APB\).
A1: Finding angle \(\angle APB \approx 52.9^\circ\).
A1: Correct bearing of \(128^\circ\) (or \(127.9^\circ\)).

(a) The particle is at rest when \(v(t) = 0\). We solve \(3t \cos(0.8t) - 2 = 0\) for \(0 \le t \le 6\) using a GDC. This yields two roots: \(t_1 \approx 0.867\text{ s}\) and \(t_2 \approx 1.28\text{ s}\). (b) The total distance travelled is given by \(\int_{0}^{5} |v(t)| dt = \int_{0}^{5} |3t \cos(0.8t) - 2| dt\). Using a GDC for numerical integration, we calculate the value of this integral: \(\int_{0}^{5} |3t \cos(0.8t) - 2| dt \approx 32.0\text{ m}\).

(a) M1: Equating the velocity function to 0.
A1: First rest time \(t \approx 0.867\text{ s}\).
A1: Second rest time \(t \approx 1.28\text{ s}\).
(b) M1: Setting up the correct integral \(\int_{0}^{5} |v(t)| dt\) for total distance.
M1: Attempting to integrate numerically using GDC.
A1: Correct total distance \(32.0\text{ m}\).

(a) Let \(X\) be the mass of an apple, where \(X \sim N(150, \sigma^2)\). We are given \(P(X > 185) = 0.08\). Standardizing, we get \(P\left(Z > \frac{185 - 150}{\sigma}\right) = 0.08\). Using the inverse normal function on a GDC: \(z_{0.92} \approx 1.40507\). Therefore, \(\frac{35}{\sigma} = 1.40507 \implies \sigma = \frac{35}{1.40507} \approx 24.91 \approx 24.9\text{ g}\). (b) Let \(Y\) be the number of apples in the sample of 5 with a mass greater than 185 g. Here, \(Y \sim B(5, 0.08)\). We need to find \(P(Y \ge 2)\): \(P(Y \ge 2) = 1 - P(Y \le 1)\). Using the binomial cumulative distribution function on a GDC: \(P(Y \le 1) \approx 0.94564\). Thus, \(P(Y \ge 2) = 1 - 0.94564 \approx 0.0544\).

(a) M1: Writing down the standardized equation with \(z\).
M1: Finding the correct z-score \(z \approx 1.405\) from a GDC.
A1: Standard deviation \(\sigma \approx 24.9\text{ g}\).
(b) M1: Recognizing the binomial distribution with parameters \(n = 5\) and \(p = 0.08\).
M1: Writing down the expression \(1 - P(Y \le 1)\) or equivalent sum.
A1: Correct probability \(0.0544\).

(a) The total area of the circle is \(\pi r^2 = 64\pi\). The area of the minor segment is \(\frac{1}{2}r^2(\theta - \sin\theta) = 32(\theta - \sin\theta)\). The area of the major region is the total area minus the minor segment: \(64\pi - 32(\theta - \sin\theta) = 180\). (b) Solving the equation using a GDC: \(32(\theta - \sin\theta) = 64\pi - 180 \approx 21.062\). \(\theta - \sin\theta \approx 0.6582\). Using the solver on a GDC yields \(\theta \approx 1.655 \approx 1.66\text{ rad}\). (c) The perimeter of the minor segment is the sum of the minor arc length \(L\) and the chord length \(C\). \(L = r\theta = 8(1.6548) \approx 13.238\text{ cm}\). \(C = 2r\sin(\theta/2) = 16\sin(0.8274) \approx 11.776\text{ cm}\). Perimeter \(P = 13.238 + 11.776 = 25.014 \approx 25.0\text{ cm}\).

(a) M1: Formulating the area of the major region as \(\text{Area of circle} - \text{Area of minor segment}\).
A1: Correct equation \(64\pi - 32(\theta - \sin\theta) = 180\).
(b) A1: Correct angle \(\theta \approx 1.66\) (accept \(1.65\)).
(c) M1: Calculating arc length \(L\) and chord length \(C\).
M1: Summing the two lengths to find the perimeter.
A1: Correct perimeter \(25.0\text{ cm}\).

(a) The general term in the expansion is: \(T_{r+1} = \binom{9}{r} (2x^2)^{9-r} \left(\frac{k}{x}\right)^r = \binom{9}{r} 2^{9-r} x^{18-2r} k^r x^{-r} = \binom{9}{r} 2^{9-r} k^r x^{18-3r}\). (b) To find the term in \(x^3\), set the power of \(x\) to 3: \(18 - 3r = 3 \implies 3r = 15 \implies r = 5\). (c) The coefficient is \(\binom{9}{5} 2^{9-5} k^5 = 126 \times 16 \times k^5 = 2016 k^5\). Set this equal to the given coefficient: \(2016 k^5 = 2064384 \implies k^5 = 1024\). Since \(1024 = 2^{10} = 4^5\), we find \(k = 4\).

(a) M1: Attempting to write down the general term of the expansion.
A1: Correct simplified general term \(\binom{9}{r} 2^{9-r} k^r x^{18-3r}\).
(b) M1: Setting the exponent of \(x\) equal to 3.
A1: Correct value \(r = 5\).
(c) M1: Equating the binomial term coefficient to \(2064384\).
A1: Correct value \(k = 4\).

(a) To find the point of intersection, we set \(f(x) = g(x) \implies \sin(x^2) = \cos(x)\). Using a GDC to find the root in the interval \([0, \frac{\pi}{2}]\) yields \(x \approx 0.84937 \approx 0.849\). (b) The region is bounded by \(x = 0\) and the intersection point \(x \approx 0.84937\). In this interval, \(\cos(x) \ge \sin(x^2)\). The area is given by: \(A = \int_{0}^{0.84937} (\cos(x) - \sin(x^2)) dx\). Using a GDC to compute this numerical integral gives: \(A \approx 0.554\).

(a) M1: Setting up the equation \(\sin(x^2) = \cos(x)\).
A1: Correct x-coordinate \(x \approx 0.849\).
(b) M1: Setting up the integral with the correct limits \(0\) and \(0.849\).
M1: Correct integrand \(\cos(x) - \sin(x^2)\).
A1: Correct area value \(0.554\).

Let \(X\) be the mass of a box of chocolates, where \(X \sim N(\mu, \sigma^2)\).

We are given:
\(P(X < 245) = 0.10\) and \(P(X > 255) = 0.15 \implies P(X < 255) = 0.85\)

Using the standard normal distribution \(Z \sim N(0, 1)\), we find the \(z\)-scores corresponding to these probabilities using a GDC:
\(z_1 = \Phi^{-1}(0.10) \approx -1.28155\)
\(z_2 = \Phi^{-1}(0.85) \approx 1.03643\)

Using the standardization formula \(z = \frac{x - \mu}{\sigma}\), we set up a system of two equations:
\(\frac{245 - \mu}{\sigma} = -1.28155 \implies 245 - \mu = -1.28155\sigma\) (Equation 1)
\(\frac{255 - \mu}{\sigma} = 1.03643 \implies 255 - \mu = 1.03643\sigma\) (Equation 2)

Subtracting Equation 1 from Equation 2 gives:
\(10 = 2.31798\sigma\)
\(\sigma \approx 4.31410...\)

Substituting \(\sigma\) back into Equation 2:
\(255 - \mu = 1.03643(4.31410...)\)

\(\mu \approx 250.5289...\)

To 3 significant figures:
\(\mu \approx 251\text{ g}\)
\(\sigma \approx 4.31\text{ g}\)

**[1 mark]** for finding the correct \(z\)-score for \(10\%\): \(z_1 \approx -1.28\) (or a standardizing equation set to a \(z\)-value in the range \([-1.29, -1.28]\)).

**[1 mark]** for finding the correct \(z\)-score for \(85\%\): \(z_2 \approx 1.04\) (or a standardizing equation set to a \(z\)-value in the range \([1.03, 1.04]\)).

**[1 mark]** for setting up Equation 1: \(245 - \mu = -1.28155\sigma\) (or equivalent).

**[1 mark]** for setting up Equation 2: \(255 - \mu = 1.03643\sigma\) (or equivalent).

**[1 mark]** for a valid method to solve the simultaneous equations (e.g., elimination, substitution, or GDC solver showing intermediate working).

**[1 mark]** for both correct final values rounded to 3 significant figures: \(\mu \approx 251\) and \(\sigma \approx 4.31\). (Accept \(\mu = 251\) and \(\sigma = 4.31\)).

(a) To find the local minimum, we find where \(f'(x) = 0\):
\(f'(x) = e^{0.2x}(0.2\cos(x) - \sin(x)) = 0\).
Since \(e^{0.2x} \ne 0\), we have \(0.2\cos(x) - \sin(x) = 0 \implies \tan(x) = 0.2\).
For \(0 \le x \le 4\), the solution giving a minimum is \(x = \arctan(0.2) + \pi \approx 3.33898\).
Substituting back into the function: \(y = f(3.33898) \approx 1.0879\).
To 3 significant figures, the coordinates of the local minimum are \((3.34, 1.09)\).

(b) (i) The area of \(R\) is given by:
\(\text{Area} = \int_{0}^{4} f(x) \, dx = \int_{0}^{4} (e^{0.2x} \cos(x) + 3) \, dx\).
(ii) Using a GDC to evaluate the integral:
\(\text{Area} \approx 9.9084 \approx 9.91\).

(c) (i) The volume of revolution is given by:
\(V = \pi \int_{0}^{4} [f(x)]^2 \, dx = \pi \int_{0}^{4} (e^{0.2x} \cos(x) + 3)^2 \, dx\).
(ii) Using a GDC to evaluate the integral:
\(V \approx \pi \times 29.4869 \approx 92.636 \approx 92.6\).

(d) (i) Evaluating the function at the required points:
\(f(0) = e^0 \cos(0) + 3 = 4.000\) (exactly 4)
\(f(2) = e^{0.4} \cos(2) + 3 \approx 2.37918 \approx 2.379\)
\(f(4) = e^{0.8} \cos(4) + 3 \approx 1.54529 \approx 1.545\).
(ii) The quadratic curve \(g(x) = ax^2 + bx + c\) passes through these points:
At \(x = 0\): \(c = 4\).
At \(x = 2\): \(4a + 2b + 4 = 2.37918 \implies 4a + 2b = -1.62082\).
At \(x = 4\): \(16a + 4b + 4 = 1.54529 \implies 16a + 4b = -2.45471\).
Solving this system of linear equations:
From the first equation, \(2b = -1.62082 - 4a \implies 4b = -3.24164 - 8a\).
Substituting into the second equation:
\(16a - 3.24164 - 8a = -2.45471 \implies 8a = 0.78693 \implies a \approx 0.098366\).
Then, \(2b = -1.62082 - 4(0.098366) = -2.01428 \implies b \approx -1.00714\).
Thus, to 3 significant figures, \(a \approx 0.0984\), \(b \approx -1.01\), \(c = 4\).
(iii) The volume of the second model is:
\(V_{\text{new}} = \pi \int_{0}^{4} (0.098366x^2 - 1.00714x + 4)^2 \, dx\).
Evaluating this integral using a GDC:
\(V_{\text{new}} \approx \pi \times 27.2708 \approx 85.674\).
Percentage Error \(= \frac{|V_{\text{new}} - V|}{V} \times 100\% = \frac{|85.674 - 92.636|}{92.636} \times 100\% \approx 7.5155\% \approx 7.52\%\).

(a)
- M1 for attempting to find \(f'(x) = 0\) (e.g., using GDC or by differentiating).
- A1 for \(x \approx 3.34\).
- A1 for \(y \approx 1.09\).

(b)
- (i) A1 for the correct integral expression (including limits and \(dx\)).
- (ii) A2 for \(9.91\).

(c)
- (i) A1 for the correct integral expression including \(\pi\) and the function squared.
- (ii) A2 for \(92.6\).

(d)
- (i) A2 for all three values correct (allow A1 if 2 are correct).
- (ii) M1 for setting up a system of equations, A1 for \(a \approx 0.0984\) and \(b \approx -1.01\), A1 for \(c = 4\).
- (iii) M1 for attempting to set up the volume integral for the quadratic model, A1 for finding \(V_{\text{new}} \approx 85.7\) (or \(27.3\pi\)), M1 for substituting into the percentage error formula, A1 for \(7.52\%\) (accept answers in range \(7.50\% - 7.55\%\) depending on intermediate rounding).

(a) Let \(L \sim N(50.2, 0.8^2)\).
Using GDC normal CDF with lower limit \(49.5\), upper limit \(51.0\), \(\mu = 50.2\), and \(\sigma = 0.8\):
\(P(49.5 < L < 51.0) \approx 0.65056 \approx 0.651\).

(b) (i) A rod is discarded if \(L < 49.0\) or \(L > 51.8\).
Using GDC:
\(P(L < 49.0) \approx 0.066807\)
\(P(L > 51.8) \approx 0.022750\)
Sum of probabilities \(= 0.066807 + 0.022750 = 0.089557\)
This rounds to \(0.0896\) correct to 3 significant figures.
(ii) Let \(X\) be the number of discarded rods in a batch of 150.
Then \(X \sim B(150, 0.089557)\) (or using \(0.0896\)).
We need to find \(P(10 \le X \le 20) = P(X \le 20) - P(X \le 9)\).
Using GDC binomial CDF:
If using \(p = 0.089557\):
\(P(X \le 20) \approx 0.97047\)
\(P(X \le 9) \approx 0.13430\)
\(P(10 \le X \le 20) = 0.97047 - 0.13430 \approx 0.83617 \approx 0.836\).
If using \(p = 0.0896\):
\(P(X \le 20) \approx 0.97025\)
\(P(X \le 9) \approx 0.13346\)
\(P(10 \le X \le 20) \approx 0.83679 \approx 0.837\).

(c) (i) Under the new distribution, \(L_{\text{new}} \sim N(50.2, \sigma_{\text{new}}^2)\).
We are given \(P(L_{\text{new}} > 51.5) = 0.05\).
Thus, \(P\left(Z > \frac{51.5 - 50.2}{\sigma_{\text{new}}}\right) = 0.05\).
From the inverse normal distribution, \(z_{0.05} \approx 1.64485\).
\(\frac{1.3}{\sigma_{\text{new}}} = 1.64485 \implies \sigma_{\text{new}} = \frac{1.3}{1.64485} \approx 0.79034 \approx 0.790\text{ cm}\).
(ii) We need to find \(P(L_{\text{new}} < 51.0 \mid L_{\text{new}} > 49.5) = \frac{P(49.5 < L_{\text{new}} < 51.0)}{P(L_{\text{new}} > 49.5)}\).
Using \(\mu = 50.2\) and \(\sigma_{\text{new}} = 0.79034\):
\(P(49.5 < L_{\text{new}} < 51.0) \approx 0.65640\)
\(P(L_{\text{new}} > 49.5) \approx 0.81211\)
Conditional probability \(= \frac{0.65640}{0.81211} \approx 0.80826 \approx 0.808\).

(d) Let \(Y\) be the number of defective rods in a sample of 10. \(Y \sim B(10, 0.0896)\).
The machine is stopped if \(Y > 1\).
\(P(Y > 1) = 1 - P(Y \le 1) = 1 - [P(Y = 0) + P(Y = 1)]\).
Using GDC binomial cumulative or probability density functions:
\(P(Y = 0) \approx 0.39037\)
\(P(Y = 1) \approx 0.38415\)
\(P(Y \le 1) \approx 0.77452\)
\(P(Y > 1) = 1 - 0.77452 \approx 0.22548 \approx 0.225\).

(a)
- M1 for attempting to use normal CDF on GDC.
- A1 for \(0.651\).

(b)
- (i) M1 for recognizing both tails are needed: \(P(L < 49.0) + P(L > 51.8)\).
- A1 for \(0.066807 + 0.022750\) (or equivalent values seen).
- AG for showing this sum equals \(0.0896\) to 3 s.f.
- (ii) M1 for identifying binomial distribution with parameters \(n = 150\), \(p = 0.0896\).
- M1 for attempting to calculate \(P(10 \le X \le 20)\).
- A1 for \(0.836\) (using more accurate \(p\)) or \(0.837\) (using \(0.0896\)).

(c)
- (i) M1 for setting up the standardized equation: \(\frac{1.3}{\sigma} = z\).
- A1 for \(z = 1.64\) (or more accurate \(1.64485\)).
- A1 for \(\sigma_{\text{new}} \approx 0.790\).
- (ii) M1 for using conditional probability formula: \(\frac{P(49.5 < L < 51.0)}{P(L > 49.5)}\).
- A1 for numerator \(0.656\) and denominator \(0.812\).
- A1 for \(0.808\).

(d)
- M1 for defining \(Y \sim B(10, 0.0896)\).
- M1 for writing down \(1 - P(Y \le 1)\) or equivalent sum.
- A1 for \(0.225\).

(a) If the drone passes through \(H\), there must exist a value \(t\) such that:
\(20 - 2t = 10 \implies 2t = 10 \implies t = 5\).
Substituting \(t = 5\) into the other components:
\(y = 10 + 3(5) = 25\) (which matches \(H\)'s y-coordinate).
\(z = 5 + 1(5) = 10 \ne 12\) (which does not match \(H\)'s z-coordinate).
Since no single value of \(t\) satisfies all three coordinate equations, the drone does not pass directly through the hub \(H\).

(b) Let \(D(t)\) be the position of the drone at time \(t\): \(D(t) = (20 - 2t, 10 + 3t, 5 + t)\).
The vector \(\overrightarrow{HD} = \begin{pmatrix} 20 - 2t - 10 \\ 10 + 3t - 25 \\ 5 + t - 12 \end{pmatrix} = \begin{pmatrix} 10 - 2t \\ -15 + 3t \\ -7 + t \end{pmatrix}\).
For the closest approach, \(\overrightarrow{HD}\) must be perpendicular to the direction vector of the line, \(\mathbf{d} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\).
Thus, \(\overrightarrow{HD} \cdot \mathbf{d} = 0\):
\((10 - 2t)(-2) + (-15 + 3t)(3) + (-7 + t)(1) = 0\)
\(-20 + 4t - 45 + 9t - 7 + t = 0\)
\(14t - 72 = 0 \implies t = \frac{72}{14} = \frac{36}{7} \approx 5.14\text{ seconds}\).

(c) To find the minimum distance, evaluate the magnitude of \(\overrightarrow{HD}\) at \(t = \frac{36}{7}\):
\(\overrightarrow{HD} = \begin{pmatrix} 10 - 2(36/7) \\ -15 + 3(36/7) \\ -7 + (36/7) \end{pmatrix} = \begin{pmatrix} -2/7 \\ 3/7 \\ -13/7
\end{pmatrix}\).
\(\text{Distance} = \sqrt{\left(-\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{13}{7}\right)^2} = \sqrt{\frac{4 + 9 + 169}{49}} = \sqrt{\frac{182}{49}} = \sqrt{\frac{26}{7}} \approx 1.93\text{ m}\).

(d) (i) The plane \(\Pi\) has equation \(3x + 2y - z = D\).
Substitute \(H(10, 25, 12)\):
\(3(10) + 2(25) - 12 = 30 + 50 - 12 = 68\).
So the Cartesian equation is \(3x + 2y - z = 68\).
(ii) Substitute the parametric equations of the line into the plane equation:
\(3(20 - 2t) + 2(10 + 3t) - (5 + t) = 68\)
\(60 - 6t + 20 + 6t - 5 - t = 68\)
\(75 - t = 68 \implies t = 7\).
Substitute \(t = 7\) back into the drone's position equation:
\(x = 20 - 2(7) = 6\)
\(y = 10 + 3(7) = 31\)
\(z = 5 + 7 = 12\).
So the coordinates of the intersection point \(P\) are \((6, 31, 12)\).
(iii) The distance from the intersection point \(P\) to the hub \(H\) is:
\(HP = \sqrt{(6-10)^2 + (31-25)^2 + (12-12)^2} = \sqrt{(-4)^2 + 6^2 + 0^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21\text{ m}\).
Since the distance \(HP \approx 7.21\text{ m}\) is less than the blade length of \(8\text{ m}\), the drone passes through the circular region swept out by the rotating blades of the turbine.

(a)
- M1 for attempting to solve \(20-2t = 10\).
- A1 for showing the resulting \(t = 5\) does not yield \(z = 12\) (showing \(10 \ne 12\)).

(b)
- M1 for writing an expression for \(\overrightarrow{HD}\) in terms of \(t\).
- M1 for setting up the scalar product \(\overrightarrow{HD} \cdot \mathbf{d} = 0\) (or attempting to minimize the distance expression squared using calculus/GDC).
- A1 for a correct linear equation in \(t\) (e.g. \(14t - 72 = 0\)).
- A2 for \(t \approx 5.14\) (or \(\frac{36}{7}\)).

(c)
- M1 for substituting their value of \(t\) back into the distance or vector expression.
- A1 for obtaining the vector \(\begin{pmatrix} -0.286 \\ 0.429 \\ -1.86 \end{pmatrix}\) (or exact fractional form).
- A1 for \(1.93\text{ m}\) (or \(\sqrt{\frac{26}{7}}\text{ m}\)).

(d)
- (i) M1 for attempting \(\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}\).
- A1 for \(3x + 2y - z = 68\).
- (ii) M1 for substituting the line equation into their plane equation.
- A1 for finding \(t = 7\).
- A1 for coordinates \((6, 31, 12)\).
- (iii) M1 for finding the distance between \(P\) and \(H\).
- A1 for \(HP = \sqrt{52} \approx 7.21\text{ m}\).
- R1 for stating that since \(7.21 < 8\), the drone does pass through the swept region.

**Part A**

(a) Differentiating using the product rule:
\[f_n'(x) = n x^{n-1} e^{-x} - x^n e^{-x} = x^{n-1} e^{-x} (n - x)\]
Setting \(f_n'(x) = 0\), and since \(x > 0\) and \(e^{-x} > 0\), we get \(x = n\).
For \(0 < x < n\), \(f_n'(x) > 0\), and for \(x > n\), \(f_n'(x) < 0\).
Thus, by the first derivative test, \(x = n\) is indeed a local maximum.

(b) Finding the second derivative:
\[f_n''(x) = \frac{d}{dx} [e^{-x}(n x^{n-1} - x^n)] = -e^{-x}(n x^{n-1} - x^n) + e^{-x}(n(n-1)x^{n-2} - n x^{n-1})\]
\[f_n''(x) = e^{-x} x^{n-2} [ -n x + x^2 + n(n-1) - n x ] = e^{-x} x^{n-2} [ x^2 - 2n x + n(n-1) ]\]
Setting \(f_n''(x) = 0\), since \(e^{-x} x^{n-2} \neq 0\) for \(x > 0\), we solve:
\[x^2 - 2n x + n(n-1) = 0\]
Using the quadratic formula:
\[x = \frac{2n \pm \sqrt{4n^2 - 4n(n-1)}}{2} = n \pm \sqrt{n}\]
Since \(f_n''(x)\) changes sign at these roots, they are indeed points of inflection.

(c) The maximum value is \(y_n = f_n(n) = n^n e^{-n}\).
The \(y\)-coordinates of the points of inflection are:
\[y_I = f_n(n \pm \sqrt{n}) = (n \pm \sqrt{n})^n e^{-(n \pm \sqrt{n})}\]
Taking the ratio:
\[\frac{y_I}{y_n} = \frac{(n \pm \sqrt{n})^n e^{-(n \pm \sqrt{n})}}{n^n e^{-n}} = \left(\frac{n \pm \sqrt{n}}{n}\right)^n e^{\mp \sqrt{n}} = \left(1 \pm \frac{1}{\sqrt{n}}\right)^n e^{\mp \sqrt{n}}\]
Taking the natural logarithm yields the desired result:
\[\ln\left(\frac{y_I}{y_n}\right) = n \ln\left(1 \pm \frac{1}{\sqrt{n}}\right) \mp \sqrt{n}\]

**Part B**

(d) (i) Let \(u = \frac{1}{\sqrt{n}}\). Substituting into the Taylor expansion:
\[\ln\left(1 - \frac{1}{\sqrt{n}}\right) = -\frac{1}{\sqrt{n}} - \frac{1}{2n} - \frac{1}{3n\sqrt{n}} - \dots\]
Multiplying by \(n\) and adding \(\sqrt{n}\):
\[n \ln\left(1 - \frac{1}{\sqrt{n}}\right) + \sqrt{n} = n\left(-\frac{1}{\sqrt{n}} - \frac{1}{2n} - \frac{1}{3n\sqrt{n}} - \dots\right) + \sqrt{n}\]
\[= -\sqrt{n} - \frac{1}{2} - \frac{1}{3\sqrt{n}} - \dots + \sqrt{n} = -\frac{1}{2} - \frac{1}{3\sqrt{n}} - \dots\]
For large \(n\), the terms with \(\sqrt{n}\) in the denominator approach 0, leaving \(-\frac{1}{2}\).

(ii) Using the limit:
\[\lim_{n \to \infty} \frac{f_n(n - \sqrt{n})}{f_n(n)} = e^{-1/2} = \frac{1}{\sqrt{e}}\]

**Part C**

(e) (i) Using integration by parts:
Let \(u = x^n \implies du = n x^{n-1} \, dx\)
Let \(dv = e^{-x} \, dx \implies v = -e^{-x}\)
\[I_n = [-x^n e^{-x}]_0^\infty - \int_0^\infty (-e^{-x})(n x^{n-1}) \, dx\]
Since \(\lim_{x \to \infty} x^n e^{-x} = 0\) and the expression is 0 at \(x = 0\) for \(n \ge 1\), the boundary term vanishes.
Thus, \(I_n = n \int_0^\infty x^{n-1} e^{-x} \, dx = n I_{n-1}\).

(ii) Given \(I_0 = 1\), we deduce that \(I_n = n!\).

(f) The function \(g_n(x) = \frac{x^n e^{-x}}{n!}\) is the PDF of a Gamma distribution. By the Central Limit Theorem, for large \(n\), this distribution approaches a normal distribution with mean \(\mu = n\) and standard deviation \(\sigma = \sqrt{n}\).
The inflection points \(x = n \pm \sqrt{n}\) correspond to \(\mu \pm \sigma\).
The area under a normal curve within one standard deviation of the mean is approximately \(0.683\) (or \(68.3\%\)).

**(a)**
- M1: For attempting to find the first derivative of \(f_n(x)\).
- A1: For obtaining \(f_n'(x) = x^{n-1} e^{-x}(n-x)\).
- R1: For showing the sign change of the first derivative around \(x=n\).
- A1: For concluding that \(x=n\) is a local maximum.

**(b)**
- M1: For attempting to find the second derivative.
- A1: For obtaining \(f_n''(x) = e^{-x} x^{n-2} [x^2 - 2nx + n(n-1)]\).
- M1: For setting \(f_n''(x) = 0\).
- A1: For solving the quadratic equation correctly.
- A1: For concluding that the points of inflection are at \(x = n \pm \sqrt{n}\).

**(c)**
- A1: For stating \(y_n = n^n e^{-n}\).
- A1: For stating \(y_I = (n \pm \sqrt{n})^n e^{-(n \pm \sqrt{n})}\).
- M1: For simplifying the ratio \(\frac{y_I}{y_n}\).
- A1: For taking the natural logarithm to reach the final form.

**(d)(i)**
- M1: For writing the expansion of \ (\ln(1 - 1/\sqrt{n})\).
- A1: For multiplying the expansion by \(n\).
- M1: For adding \(\sqrt{n}\) and simplifying.
- A1: For showing the limiting behavior as \(n \to \infty\).

**(d)(ii)**
- M1: For exponentiating the result from d(i).
- A1: For obtaining the final limit as \(e^{-1/2}\).

**(e)(i)**
- M1: For setting up the integration by parts formula with correct choices of \(u\) and \(dv\).
- A1: For finding \(du\) and \(v\).
- M1: For writing down the integration by parts step.
- R1: For showing that the boundary term \([-x^n e^{-x}]_0^\infty\) is zero.
- A1: For showing \(I_n = n I_{n-1}\).

**(e)(ii)**
- A1: For concluding \(I_n = n!\).

**(f)**
- M1: For identifying that \(g_n(x)\) represents a probability density function.
- M1: For identifying that \(x = n \pm \sqrt{n}\) corresponds to \(\mu \pm \sigma\) of a normal distribution.
- A1: For giving the final estimate of \(0.683\) (accept \(68\%\) or \(68.3\%\)).

**Part A**

(a) Using the double angle identity \(\sin(2t) = 2 \sin t \cos t\):
\[y(t) = 2 \sin t \cos t + a \sin t = \sin t (2 \cos t + a)\]

(b) (i) The \(x\)-intercepts occur when \(y(t) = 0\):
\[\sin t (2 \cos t + a) = 0\]
This gives \(\sin t = 0\) or \(\cos t = -\frac{a}{2}\).
For \(\sin t = 0\), \(t = 0, \pi, 2\pi\).
At \(t = 0, 2\pi\), \(x = \cos(0) = 1 \implies (1, 0)\).
At \(t = \pi\), \(x = \cos(\pi) = -1 \implies (-1, 0)\).
For \(\cos t = -\frac{a}{2}\) (valid only if \(a \le 2\)), the \(x\)-coordinate is \(x = \cos t = -\frac{a}{2}\), giving the point \(\left(-\frac{a}{2}, 0\right)\).

(ii) If \(a > 2\), then \(-\frac{a}{2} < -1\). Since \(-1 \le \cos t \le 1\) for all real \(t\), the equation \(\cos t = -\frac{a}{2}\) has no solutions. Therefore, the only \(x\)-intercepts are the two points \((1, 0)\) and \((-1, 0)\).

(iii) A self-intersection occurs when two distinct parameter values \(t_1, t_2 \in [0, 2\pi)\) yield the same coordinates.
\[x(t_1) = x(t_2) \implies \cos t_1 = \cos t_2 \implies t_2 = 2\pi - t_1\]
Then, \(y(t_2) = y(2\pi - t_1) = -y(t_1)\).
For \(y(t_1) = y(t_2)\), we must have \(y(t_1) = 0\), which placing the intersection point on the \(x\)-axis.
Since \(0 \le a < 2\), \(\cos t = -\frac{a}{2}\) has two distinct solutions \(t_1 \in (0, \pi)\) and \(t_2 \in (\pi, 2\pi)\). These two parameters yield the single physical point \(\left(-\frac{a}{2}, 0\right)\), which is the unique self-intersection point.

**Part B**

(c) (i) For \(a = 1\), the intercepts are at \((1, 0)\), \((-1, 0)\) and the self-intersection is at \((-1/2, 0)\).
The curve has a figure-eight loop on the left.

(ii) The self-intersection occurs when \(\cos t = -1/2\), which corresponds to \(t = 2\pi/3\) and \(t = 4\pi/3\). The loop to the left of the intersection point is traversed for \(t \in [2\pi/3, 4\pi/3]\).
The area \(A\) is given by:
\[A = \int_{2\pi/3}^{4
\pi/3} y(t) (-x'(t)) \, dt = \int_{2\pi/3}^{4
\pi/3} \sin t (2 \cos t + 1) (\sin t) \, dt = \int_{2\pi/3}^{4\pi/3} \sin^2 t (2 \cos t + 1) \, dt\]

(iii) Evaluating the integral:
\[\int_{2\pi/3}^{4\pi/3} (2 \sin^2 t \cos t + \sin^2 t) \, dt = \left[ \frac{2}{3} \sin^3 t + \frac{t}{2} - \frac{\sin(2t)}{4} \right]_{2\pi/3}^{4\pi/3}\]
At \(t = 4\pi/3\):
\[\frac{2}{3} \left(-\frac{\sqrt{3}}{2}\right)^3 + \frac{2\pi}{3} - \frac{\sin(8\pi/3)}{4} = -\frac{\sqrt{3}}{4} + \frac{2\pi}{3} - \frac{\sqrt{3}}{8} = -\frac{3\sqrt{3}}{8} + \frac{2\pi}{3}\]
At \(t = 2\pi/3\):
\[\frac{2}{3} \left(\frac{\sqrt{3}}{2}\right)^3 + \frac{\pi}{3} - \frac{\sin(4\pi/3)}{4} = \frac{\sqrt{3}}{4} + \frac{\pi}{3} + \frac{\sqrt{3}}{8} = \frac{3\sqrt{3}}{8} + \frac{\pi}{3}\]
Subtracting the lower limit from the upper limit:
\[\left(-\frac{3\sqrt{3}}{8} + \frac{2\pi}{3}\right) - \left(\frac{3\sqrt{3}}{8} + \frac{\pi}{3}\right) = \frac{\pi}{3} - \frac{3\sqrt{3}}{4}\]
Taking the absolute value to represent the area:
\[A = \frac{3\sqrt{3}}{4} - \frac{\pi}{3}\]

**Part C**

(d) For \(a \ge 2\), the curve forms a single closed loop for \(t \in [0, 2
\pi]\). The total area is:
\[A = \int_0^{2\pi} y(t) (-x'(t)) \, dt = \int_0^{2\pi} \sin^2 t (2 \cos t + a) \, dt\]
\[= \int_0^{2\pi} 2 \sin^2 t \cos t \, dt + a \int_0^{2\pi} \sin^2 t \, dt\]
For the first integral:
\[\int_0^{2\pi} 2 \sin^2 t \cos t \, dt = \left[ \frac{2}{3} \sin^3 t \right]_0^{2\pi} = 0\]
For the second integral:
\[a \int_0^{2\pi} \frac{1 - \cos(2t)}{2} \, dt = a \left[ \frac{t}{2} - \frac{\sin(2t)}{4} \right]_0^{2\pi} = a\pi\]
Thus, the area is \(a\pi\).

(e) At the self-intersection point \((-1/2, 0)\), \(t = 2\pi/3\) and \(t = 4\pi/3\).
The derivative is:
\[\frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{2 \cos(2t) + \cos t}{-\sin t}\]
At \(t = 2\pi/3\):
\[\frac{dy}{dx} = \frac{2(-1/2) - 1/2}{-\sqrt{3}/2} = \frac{-3/2}{-\sqrt{3}/2} = \sqrt{3}\]
Tangent line: \(y = \sqrt{3}\left(x + \frac{1}{2}\right)\).

At \(t = 4\pi/3\):
\[\frac{dy}{dx} = \frac{2(-1/2) - 1/2}{\sqrt{3}/2} = \frac{-3/2}{\sqrt{3}/2} = -\sqrt{3}\]
Tangent line: \(y = -\sqrt{3}\left(x + \frac{1}{2}\right)\).

**(a)**
- M1: For utilizing the double-angle formula \(\sin(2t) = 2\sin t\cos t\).
- A1: For factoring out \(\sin t\) to obtain \(\sin t(2\cos t + a)\).

**(b)(i)**
- M1: For setting \(y(t) = 0\).
- A1: For obtaining \(t = 0, \pi, 2\pi\) and intercepts \((1,0)\), \((-1,0)\).
- M1: For setting \(\cos t = -a/2\).
- A1: For getting the intercept \((-a/2, 0)\).

**(b)(ii)**
- R1: For showing that if \(a > 2\), then \(|-a/2| > 1\).
- R1: For concluding \(\cos t = -a/2\) has no real solutions, hence only 2 intercepts.

**(b)(iii)**
- M1: For setting \(x(t_1) = x(t_2)\) and showing \(t_2 = 2\pi - t_1\).
- R1: For arguing that \(y(t_1) = y(t_2)\) requires \(y = 0\).
- A1: For finding the coordinate \((-a/2, 0)\).

**(c)(i)**
- A1: For correct loop shape resembling a figure-eight or teardrop loop on left.
- A1: For showing intercepts at \((1,0)\) and \((-1,0)\).
- A1: For showing the self-intersection at \((-1/2, 0)\).

**(c)(ii)**
- M1: For identifying the integration limits as \(t = 2\pi/3\) and \(t = 4\pi/3\).
- A1: For writing a correct integral expression for the area.

**(c)(iii)**
- M1: For attempting the integration of \(2 \sin^2 t \cos t + \sin^2 t\).
- A1: For finding the antiderivative \(\frac{2}{3} \sin^3 t + \frac{t}{2} - \frac{\sin 2t}{4}\).
- A1: For substitute the limits \(2\pi/3\) and \(4\pi/3\) correctly.
- A1: For simplifying to show the area is \(\frac{3\sqrt{3}}{4} - \frac{\pi}{3}\).

**(d)**
- M1: For setting up the area integral \(\int_0^{2\pi} y(t)(-x'(t)) \, dt\).
- M1: For splitting the integral into two parts.
- A1: For showing the first integral is 0.
- A1: For showing the second integral yields \(a\pi\).

**(e)**
- M1: For finding the derivative expression \(\frac{dy}{dx} = \frac{2\cos 2t + \cos t}{-\sin t}\).
- A1: For finding slopes \(\pm\sqrt{3}\).
- A1: For writing the equations of both tangent lines.