2023 IB DP Mathematics - Analysis and Approaches Practice Paper with Answers

The general term in the expansion is given by \(T_{r+1} = \binom{6}{r} (2x^2)^{6-r} \left(-\frac{1}{ax}\right)^r = \binom{6}{r} 2^{6-r} (-1)^r a^{-r} x^{12-3r}\). For the constant term, the exponent of \(x\) must be 0, which gives \(12 - 3r = 0\), so \(r = 4\). Substituting \(r = 4\) into the general term gives the constant term: \(\binom{6}{4} 2^2 (-1)^4 a^{-4} = 15 \times 4 \times a^{-4} = \frac{60}{a^4}\). Setting this equal to the given constant term: \(\frac{60}{a^4} = \frac{15}{4}\). This simplifies to \(15a^4 = 240\), which gives \(a^4 = 16\). Since \(a > 0\), we find \(a = 2\).

M1 for expressing the general term in the binomial expansion. A1 for setting up the equation for the exponent of \(x\): \(12 - 3r = 0\) and finding \(r = 4\). M1 for substituting \(r = 4\) to find the constant term expression in terms of \(a\). A1 for obtaining \(\frac{60}{a^4}\). M1 for setting their expression equal to \(\frac{15}{4}\) and solving for \(a^4\). A1 for \(a = 2\) (rejecting \(a = -2\) since \(a > 0\)).

To find the inverse function, let \(y = \ln(x-2) + 3\). Rearranging to make \(x\) the subject: \(y - 3 = \ln(x-2)\). Take the exponential of both sides: \(e^{y-3} = x - 2\). Thus, \(x = e^{y-3} + 2\). Interchanging variables gives \(f^{-1}(x) = e^{x-3} + 2\). For part (b), we solve the inequality \(e^{x-3} + 2 < e^4 + 2\). Subtracting 2 from both sides gives \(e^{x-3} < e^4\). Taking the natural logarithm of both sides gives \(x - 3 < 4\), which simplifies to \(x < 7\).

M1 for attempting to make \(x\) (or \(y\)) the subject of the equation. A1 for obtaining \(e^{y-3} = x - 2\) or equivalent. A1 for the correct inverse function \(f^{-1}(x) = e^{x-3} + 2\). M1 for setting up the inequality \(e^{x-3} + 2 < e^4 + 2\). A1 for simplifying to \(e^{x-3} < e^4\) or \(x - 3 < 4\). A1 for the correct final answer \(x < 7\).

Using the identity \(\cos^2 x = 1 - \sin^2 x\), we substitute this into the equation: \(2(1 - \sin^2 x) - \sin x = 1\). Expanding and simplifying gives: \(2 - 2\sin^2 x - \sin x = 1 \Rightarrow 2\sin^2 x + \sin x - 1 = 0\). Let \(u = \sin x\). The quadratic equation becomes \(2u^2 + u - 1 = 0\). Factoring the quadratic gives \((2u - 1)(u + 1) = 0\), which yields \(u = \frac{1}{2}\) or \(u = -1\). Thus, \(\sin x = \frac{1}{2}\) or \(\sin x = -1\). For the interval \(0 \le x \le 2\pi\): \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\) for \(\sin x = \frac{1}{2}\), and \(x = \frac{3\pi}{2}\) for \(\sin x = -1\). Therefore, the solutions are \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\).

M1 for using the identity \(\cos^2 x = 1 - \sin^2 x\). A1 for obtaining the quadratic equation \(2\sin^2 x + \sin x - 1 = 0\). M1 for attempting to solve the quadratic equation in \(\sin x\). A1 for finding the correct values \(\sin x = \frac{1}{2}\) and \(\sin x = -1\). A1 for identifying both solutions for \(\sin x = \frac{1}{2}\): \(x = \frac{\pi}{6}, \frac{5\pi}{6}\). A1 for identifying the solution for \(\sin x = -1\): \(x = \frac{3\pi}{2}\).

First, find the \(y\)-coordinate of the point of tangency by substituting \(x = e\) into the equation of the curve: \(y = \frac{\ln e}{e^2} = \frac{1}{e^2} = e^{-2}\). Next, find the derivative \(\frac{dy}{dx}\) using the quotient rule: \(\frac{dy}{dx} = \frac{\left(\frac{1}{x}\right)x^2 - (\ln x)(2x)}{(x^2)^2} = \frac{x - 2x\ln x}{x^4} = \frac{1 - 2\ln x}{x^3}\). Substitute \(x = e\) to find the gradient of the tangent, \(m\): \(m = \frac{1 - 2\ln e}{e^3} = \frac{1 - 2}{e^3} = -\frac{1}{e^3} = -e^{-3}\). The equation of the tangent is given by \(y - y_1 = m(x - x_1)\). Substituting the point \((e, e^{-2})\) and gradient \(m = -e^{-3}\): \(y - e^{-2} = -e^{-3}(x - e)\). Expanding this gives \(y - e^{-2} = -e^{-3}x + e^{-2}\), which simplifies to \(y = -e^{-3}x + 2e^{-2}\).

A1 for finding the correct \(y\)-coordinate, \(y = e^{-2}\). M1 for an attempt to use the quotient rule (or product rule) to differentiate \(y\). A1 for the correct derivative expression \(\frac{1 - 2\ln x}{x^3}\). A1 for finding the gradient \(m = -e^{-3}\) (or \(-\frac{1}{e^3}\)). M1 for substituting their point and gradient into the tangent equation formula. A1 for the correct equation in the requested form: \(y = -e^{-3}x + 2e^{-2}\) (or \(y = -\frac{1}{e^3}x + \frac{2}{e^2}\)).

We use substitution. Let \(u = x^2 + 1\), which gives \(\mathrm{d}u = 2x \, \mathrm{d}x\), or \(x \, \mathrm{d}x = \frac{1}{2} \mathrm{d}u\). We also change the integration limits: when \(x = 0\), \(u = 0^2 + 1 = 1\); when \(x = \sqrt{3}\), \(u = (\sqrt{3})^2 + 1 = 4\). Substituting these into the integral gives: \(\int_{1}^{4} \frac{1}{2\sqrt{u}} \, \mathrm{d}u = \frac{1}{2} \int_{1}^{4} u^{-1/2} \, \mathrm{d}u\). Integrating gives: \(\frac{1}{2} \left[ 2u^{1/2} \right]_{1}^{4} = \left[ \sqrt{u} \right]_{1}^{4}\). Evaluating this at the limits gives: \(\sqrt{4} - \sqrt{1} = 2 - 1 = 1\).

M1 for an attempt at integration by substitution, setting \(u = x^2 + 1\) and finding \(\mathrm{d}u = 2x \, \mathrm{d}x\). A1 for correct integration of the function, obtaining \(\sqrt{x^2+1}\) (or \(u^{1/2}\)). A1 for finding the correct new limits of integration (1 and 4) or showing correct substitution back. M1 for evaluating the integrated expression at the limits. A1 for the correct exact value \(1\).

Since the sum of the probabilities must equal 1, we have: \(p + 0.3 + q + 0.2 = 1\), which simplifies to \(p + q = 0.5\) (Equation 1). Using the formula for expectation, \(\mathrm{E}(X) = \sum x \cdot \mathrm{P}(X=x)\), we have: \(1(p) + 2(0.3) + 3(q) + 4(0.2) = 2.5\). This simplifies to \(p + 0.6 + 3q + 0.8 = 2.5\), which gives \(p + 3q = 1.1\) (Equation 2). Subtracting Equation 1 from Equation 2 gives: \(2q = 0.6 \Rightarrow q = 0.3\). Substituting \(q = 0.3\) back into Equation 1: \(p + 0.3 = 0.5 \Rightarrow p = 0.2\). Thus, the values are \(p = 0.2\) and \(q = 0.3\).

M1 for using the fact that the sum of probabilities is 1. A1 for the equation \(p + q = 0.5\) (or equivalent). M1 for writing an expression for \(\mathrm{E}(X)\) in terms of \(p\) and \(q\). A1 for the equation \(p + 3q = 1.1\) (or equivalent). M1 for a valid method to solve the simultaneous equations. A1 for finding \(p = 0.2\) and \(q = 0.3\).

a. Using the product rule to find the derivative of \( f(x) = x^2 e^{-x} \):
\( f'(x) = 2x e^{-x} - x^2 e^{-x} = x(2 - x) e^{-x} \).
To find the stationary points, set \( f'(x) = 0 \).
Since \( e^{-x} > 0 \), this gives \( x = 0 \) or \( x = 2 \).
For \( 0 < x < 2 \), \( f'(x) > 0 \), and for \( x > 2 \), \( f'(x) < 0 \). Thus, a local maximum occurs at \( x = 2 \).
Evaluating \( f(2) \):
\( f(2) = 2^2 e^{-2} = 4e^{-2} \).
So, the coordinates of the local maximum are \( (2, 4e^{-2}) \).

b. Using the product rule on \( f'(x) = (2x - x^2)e^{-x} \) to find the second derivative:
\( f''(x) = (2 - 2x)e^{-x} - (2x - x^2)e^{-x} = (x^2 - 4x + 2)e^{-x} \).
For points of inflexion, set \( f''(x) = 0 \).
Since \( e^{-x} \neq 0 \), we solve \( x^2 - 4x + 2 = 0 \).
Using the quadratic formula:
\( x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(2)}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2} \).
Both values are positive, so they lie in the domain \( x \ge 0 \). Since \( f''(x) \) changes sign at these points, the \( x \)-coordinates of the points of inflexion are \( x = 2 - \sqrt{2} \) and \( x = 2 + \sqrt{2} \).

c. The area of the region is given by \( I = \int_0^2 x^2 e^{-x} \, dx \).
We use integration by parts: \( \int u \, dv = uv - \int v \, du \).
Let \( u = x^2 \) and \( dv = e^{-x} \, dx \), which gives \( du = 2x \, dx \) and \( v = -e^{-x} \).
\( \int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2 \int x e^{-x} \, dx \).
Apply integration by parts again to the remaining integral:
Let \( u = x \) and \( dv = e^{-x} \, dx \), which gives \( du = dx \) and \( v = -e^{-x} \).
\( \int x e^{-x} \, dx = -x e^{-x} - \int -e^{-x} \, dx = -x e^{-x} - e^{-x} \).
Substituting this back, the antiderivative of \( f(x) \) is:
\( -x^2 e^{-x} + 2(-x e^{-x} - e^{-x}) = -e^{-x}(x^2 + 2x + 2) \).
Now apply the limits from 0 to 2:
\( \left[ -e^{-x}(x^2 + 2x + 2) \right]_0^2 = -e^{-2}(2^2 + 2(2) + 2) - (-e^0(0^2 + 2(0) + 2)) \)
\( = -e^{-2}(10) - (-2) = 2 - 10e^{-2} \).

a.
- M1 for attempting product rule to find \( f'(x) \)
- A1 for correct derivative \( f'(x) = x(2 - x)e^{-x} \)
- M1 for setting \( f'(x) = 0 \) and identifying \( x = 2 \) as the local maximum
- A1 for \( (2, 4e^{-2}) \)

b.
- M1 for attempting to find \( f''(x) \) using product rule
- A1 for correct \( f''(x) = (x^2 - 4x + 2)e^{-x} \)
- M1 for setting \( f''(x) = 0 \) and attempting to solve the quadratic equation
- A1 for \( x = 2 - \sqrt{2} \) and \( x = 2 + \sqrt{2} \)

c.
- M1 for setting up the area integral \( \int_0^2 x^2 e^{-x} \, dx \)
- M1 for first integration by parts step, finding \( -x^2 e^{-x} + 2 \int x e^{-x} \, dx \)
- A1 for correct terms
- M1 for second integration by parts step, finding \( \int x e^{-x} \, dx = -e^{-x}(x + 1) \)
- A1 for correct complete antiderivative \( -e^{-x}(x^2 + 2x + 2) \)
- M1 for substituting limits \( 0 \) and \( 2 \)
- A1 for \( 2 - 10e^{-2} \)

a. Since \( f(x) \) is a probability density function, the total area under the curve is 1:
\( \int_{0}^{6} f(x) \, dx = 1 \)
\( \int_{0}^{2} k x^2 \, dx + \int_{2}^{6} k(6 - x) \, dx = 1 \)
\( k \left( \left[ \frac{x^3}{3} \right]_0^2 + \left[ 6x - \frac{x^2}{2} \right]_2^6 \right) = 1 \)
Evaluating the first integral:
\( \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \)
Evaluating the second integral:
\( \left[ 6x - \frac{x^2}{2} \right]_2^6 = (36 - 18) - (12 - 2) = 18 - 10 = 8 \)
Combining the integrals:
\( k \left( \frac{8}{3} + 8 \right) = 1 \Rightarrow k \left( \frac{32}{3} \right) = 1 \Rightarrow k = \frac{3}{32} \).

b. We want to find \( \text{P}(1 \le X \le 4) = \int_{1}^{4} f(x) \, dx \):
\( \text{P}(1 \le X \le 4) = \frac{3}{32} \left( \int_{1}^{2} x^2 \, dx + \int_{2}^{4} (6 - x) \, dx \right) \)
Evaluating the first integral:
\( \int_{1}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \)
Evaluating the second integral:
\( \int_{2}^{4} (6 - x) \, dx = \left[ 6x - \frac{x^2}{2} \right]_2^4 = (24 - 8) - (12 - 2) = 16 - 10 = 6 \)
Summing these values:
\( \text{P}(1 \le X \le 4) = \frac{3}{32} \left( \frac{7}{3} + 6 \right) = \frac{3}{32} \left( \frac{25}{3} \right) = \frac{25}{32} \).

c. Let \( m \) be the median. Since \( \int_{0}^{2} f(x) \, dx = \frac{3}{32} \cdot \frac{8}{3} = \frac{1}{4} = 0.25 < 0.5 \), the median must lie in the interval \( 2 < m \le 6 \).
Thus, \( \int_{m}^{6} f(x) \, dx = 0.5 \).
\( \frac{3}{32} \int_{m}^{6} (6 - x) \, dx = \frac{1}{2} \Rightarrow \int_{m}^{6} (6 - x) \, dx = \frac{16}{3} \).
We can integrate this directly using substitution or geometry:
Let \( u = 6 - x \), then \( du = -dx \). When \( x = m \), \( u = 6 - m \). When \( x = 6 \), \( u = 0 \).
\( \int_{0}^{6 - m} u \, du = \left[ \frac{u^2}{2} \right]_0^{6 - m} = \frac{(6 - m)^2}{2} \).
Set this equal to \( \frac{16}{3} \):
\( \frac{(6 - m)^2}{2} = \frac{16}{3} \Rightarrow (6 - m)^2 = \frac{32}{3} \).
Taking the square root (noting that \( m < 6 \), so \( 6 - m > 0 \)):
\( 6 - m = \sqrt{\frac{32}{3}} = \frac{4\sqrt{2}}{\sqrt{3}} = \frac{4\sqrt{6}}{3} \).
Thus, \( m = 6 - \frac{4}{3}\sqrt{6} \).

a.
- M1 for setting up \( \int_0^6 f(x) \, dx = 1 \)
- M1 for splitting the integral at \( x = 2 \)
- A1 for finding \( \int_0^2 x^2 \, dx = \frac{8}{3} \)
- A1 for finding \( \int_2^6 (6 - x) \, dx = 8 \)
- A1 for obtaining \( k = \frac{3}{32} \) (since this is a 'show that' question, sufficient intermediate working must be shown)

b.
- M1 for setting up \( \int_1^4 f(x) \, dx \) split into two integrals
- A1 for \( \int_1^2 x^2 \, dx = \frac{7}{3} \)
- A1 for \( \int_2^4 (6 - x) \, dx = 6 \)
- M1 for adding the evaluated parts multiplied by \( \frac{3}{32} \)
- A1 for \( \frac{25}{32} \)

c.
- M1 for reasoning that \( m > 2 \) (e.g., finding the area up to 2 is 0.25)
- M1 for setting up \( \int_m^6 f(x) \, dx = 0.5 \) or \( \int_2^m f(x) \, dx = 0.25 \)
- M1 for performing integration on the second piece with limit \( m \)
- M1 for obtaining a quadratic equation in terms of \( m \) or \( 6 - m \), such as \( (6 - m)^2 = \frac{32}{3} \)
- A1 for \( m = 6 - \frac{4}{3}\sqrt{6} \)

a. Let \( -8\text{i} \) be written in exponential form:
\( -8\text{i} = 8 e^{\text{i}\left(-\frac{\pi}{2} + 2k\pi\right)} \) for \( k \in \mathbb{Z} \).
Let \( z = r e^{\text{i}\theta} \). Then \( z^3 = r^3 e^{3\text{i}\theta} = 8 e^{\text{i}\left(-\frac{\pi}{2} + 2k\pi\right)} \).
Comparing magnitudes:
\( r^3 = 8 \Rightarrow r = 2 \).
Comparing arguments:
\( 3\theta = -\frac{\pi}{2} + 2k\pi \Rightarrow \theta = -\frac{\pi}{6} + \frac{2k\pi}{3} \).
We choose \( k = 0, 1, -1 \) to keep \( -\pi < \theta \le \pi \):
For \( k = 0 \): \( \theta = -\frac{\pi}{6} \)
For \( k = 1 \): \( \theta = -\frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{2} \)
For \( k = -1 \): \( \theta = -\frac{\pi}{6} - \frac{2\pi}{3} = -\frac{5\pi}{6} \).
So, the three roots are:
\( z_1 = 2 e^{\text{i}\frac{\pi}{2}} \), \( z_2 = 2 e^{-\text{i}\frac{\pi}{6}} \), \( z_3 = 2 e^{-\text{i}\frac{5\pi}{6}} \).

b. All three roots have a modulus of 2, so the points \( A \), \( B \), and \( C \) lie on a circle of radius 2 centered at the origin.
The difference in the arguments between any two adjacent roots is:
\( \frac{\pi}{2} - \left(-\frac{\pi}{6}\right) = \frac{2\pi}{3} \) and \( -\frac{\pi}{6} - \left(-\frac{5\pi}{6}\right) = \frac{2\pi}{3} \).
Since the angle subtended at the origin between any two adjacent vertices is \( \frac{2\pi}{3} \) and the distances from the origin are equal, the three points form an equilateral triangle.
To find the area, we can sum the areas of the three congruent triangles formed by the origin and the vertices:
Area = \( 3 \times \left(\frac{1}{2} a b \sin C\right) = 3 \times \left(\frac{1}{2} \times 2 \times 2 \times \sin\frac{2\pi}{3}\right) \)
\( = 6 \sin\frac{2\pi}{3} = 6 \left(\frac{\sqrt{3}}{2}\right) = 3\sqrt{3} \).

c. First, write the roots in Cartesian form to determine their real parts:
\( z_1 = 2 e^{\text{i}\frac{\pi}{2}} = 2\text{i} \Rightarrow \text{Re}(z_1) = 0 \)
\( z_2 = 2 e^{-\text{i}\frac{\pi}{6}} = 2\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\text{i}\right) = \sqrt{3} - \text{i} \Rightarrow \text{Re}(z_2) = \sqrt{3} \)
\( z_3 = 2 e^{-\text{i}\frac{5\pi}{6}} = 2\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}\text{i}\right) = -\sqrt{3} - \text{i} \Rightarrow \text{Re}(z_3) = -\sqrt{3} \).
The root with the largest real part is \( z = 2 e^{-\text{i}\frac{\pi}{6}} \).
Now compute \( w \):
\( w = z \cdot e^{\text{i}\frac{5\pi}{12}} = 2 e^{-\text{i}\frac{\pi}{6}} \cdot e^{\text{i}\frac{5\pi}{12}} \)
\( w = 2 e^{\text{i}\left(-\frac{2\pi}{12} + \frac{5\pi}{12}\right)} = 2 e^{\text{i}\frac{3\pi}{12}} = 2 e^{\text{i}\frac{\pi}{4}} \).
Now calculate \( w^4 \):
\( w^4 = \left(2 e^{\text{i}\frac{\pi}{4}}\right)^4 = 2^4 e^{\text{i}\pi} = 16 (-1) = -16 \).
In the form \( a + b\text{i} \), this is \( -16 + 0\text{i} \), so \( w^4 = -16 \).

a.
- M1 for writing \( -8\text{i} \) in exponential/polar form
- M1 for using De Moivre's theorem / finding the cube roots
- A1 for modulus \( r = 2 \)
- M1 for finding the arguments \( \theta = \frac{\pi}{2}, -\frac{\pi}{6}, -\frac{5\pi}{6} \)
- A1 for the final three roots

b.
- R1 for explaining that vertices lie on a circle of radius 2 centered at the origin
- R1 for showing the angle between any adjacent roots at the origin is \( \frac{2\pi}{3} \) (hence establishing it is equilateral)
- M1 for attempting to find the area using trigonometry (e.g., \( 3 \times \frac{1}{2} \times 2 \times 2 \times \sin\frac{2\pi}{3} \) or using side length \( s = \sqrt{12} \) with \( \frac{\sqrt{3}}{4}s^2 \))
- A2 for the exact area \( 3\sqrt{3} \)

c.
- M1 for identifying \( z = 2 e^{-\text{i}\frac{\pi}{6}} \) (or \( \sqrt{3} - \text{i} \)) as the root with the largest real part
- M1 for multiplying by \( e^{\text{i}\frac{5\pi}{12}} \)
- A1 for obtaining \( w = 2 e^{\text{i}\frac{\pi}{4}} \)
- M1 for applying De Moivre's theorem to find \( w^4 \)
- A1 for \( -16 \)

Let \(X\) be the weight of a bag of rice, so \(X \sim N(\mu, \sigma^2)\).

We are given:
\(P(X < 45) = 0.22\)
\(P(X > 65) = 0.15 \implies P(X < 65) = 0.85\)

Using the standard normal distribution \(Z \sim N(0, 1)\):
\(P\left(Z < \frac{45 - \mu}{\sigma}\right) = 0.22 \implies \frac{45 - \mu}{\sigma} = \text{invNorm}(0.22) \approx -0.77219\)
\(P\left(Z < \frac{65 - \mu}{\sigma}\right) = 0.85 \implies \frac{65 - \mu}{\sigma} = \text{invNorm}(0.85) \approx 1.03643\)

We now have a system of two linear equations:
1) \(45 - \mu = -0.77219\sigma\)
2) \(65 - \mu = 1.03643\sigma\)

Subtracting equation (1) from equation (2):
\(20 = 1.80862\sigma\)
\(\sigma \approx 11.058\)

Substituting \(\sigma\) back into equation (1):
\(45 - \mu = -0.77219(11.058)\)
\(45 - \mu \approx -8.539\)
\
\mu \approx 53.539\)

To 3 significant figures:
\(\mu \approx 53.5\) and \(\sigma \approx 11.1\).

M1: Set up the standardized equations using the inverse normal function.
A1: Obtain \(z_1 \approx -0.772\) and \(z_2 \approx 1.04\).
M1: Attempt to solve the simultaneous equations for \(\mu\) and \(\sigma\).
A1: Correct value of \(\sigma \approx 11.1\) (or \(11.1\) to 3 s.f., accept \(11.058\)).
A1: Correct value of \(\mu \approx 53.5\) (or \(53.5\) to 3 s.f., accept \(53.539\)).

(a) To find the points of intersection, we set \(\cos x = x^2 - 2\).
Using a graphic display calculator (GDC) to solve this equation in the domain \([-\pi, \pi]\), we obtain:
\(x \approx -1.45435\) and \(x \approx 1.45435\)
Thus, the \(x\)-coordinates to 3 significant figures are \(x \approx -1.45\) and \(x \approx 1.45\).

(b) The area of the region enclosed by the curves is given by:
\(\text{Area} = \int_{-1.45435}^{1.45435} (\cos x - (x^2 - 2)) \, dx\)

Evaluating this integral using the GDC:
\(\text{Area} \approx 5.7528\)

To 3 significant figures, the area is \(5.75\).

(a)
M1: Setting up the equation \(\cos x = x^2 - 2\) (can be implied by a sketch or GDC usage).
A1A1: Correct intersection coordinates \(x \approx -1.45\) and \(x \approx 1.45\) (accept more precise values like \(\pm 1.454\)).

(b)
M1: Writing a correct integral expression representing the area.
A1: Correct limits and correct integrand.
A1: Correct final area of \(5.75\) (accept \(5.753\)).

(a) The maximum depth is \(14.6\) and the minimum depth is \(8.2\).
The amplitude \(a\) is:
\(a = \frac{14.6 - 8.2}{2} = 3.2\)

The vertical shift \(c\) is:
\(c = \frac{14.6 + 8.2}{2} = 11.4\)

The minimum depth occurs \(6.2\) hours after the maximum. This represents half of the period, so the full period \(T\) is:
\(T = 2 \times 6.2 = 12.4\) hours.

The frequency coefficient \(b\) is:
\(b = \frac{2\pi}{12.4} = \frac{\pi}{6.2} \approx 0.5067\)
To 3 significant figures, \(b \approx 0.507\).

(b) To find the first time after midnight when the water depth is \(10.0\) metres, we solve:
\(3.2 \cos(0.5067 t) + 11.4 = 10.0\)
\(3.2 \cos(0.5067 t) = -1.4\)
\(\cos(0.5067 t) = -0.4375\)

Using the GDC to find the first positive solution:
\(0.5067 t = \arccos(-0.4375) \approx 2.0236\) radians
\(t \approx \frac{2.0236}{0.5067} \approx 3.9935\)

Thus, the first time after midnight is approximately \(3.99\) hours (or 3 hours and 60 \(\times 0.99 \approx 59\) minutes, i.e., 03:59).

(a)
M1: Attempt to find the amplitude \(a\) or vertical shift \(c\).
A1: Correct values \(a = 3.2\) and \(c = 11.4\).
M1: Find the period \(12.4\) and use it to find \(b\).
A1: Correct value \(b \approx 0.507\) (or \(\frac{\pi}{6.2}\)).

(b)
M1: Set up the equation \(D(t) = 10\).
A1: Correct solution for the first time \(t \approx 3.99\) hours.

(a) Under Plan A, interest is compounded monthly. The formula for the value of the investment after \(t\) years is:
\(V_A(t) = P \left(1 + \frac{i}{12}\right)^{12t}\)

For \(t = 5\) years:
\(V_A(5) = 10000 \left(1 + \frac{0.045}{12}\right)^{60} = 10000 (1.00375)^{60}\)
Using a calculator:
\(V_A(5) \approx 12517.96\) dollars.
To the nearest dollar, the value is \(\$12\,518\).

(b) The value of the investment under Plan A after \(10\) years is:
\(V_A(10) = 10000 (1.00375)^{120} \approx 15669.93\) dollars.

Under Plan B, the investment earns simple interest. The value after \(10\) years is given by:
\(V_B(10) = 11500 \left(1 + 10 \times \frac{r}{100}\right) = 11500 (1 + 0.1r)\)

We set \(V_B(10) = V_A(10)\):
\(11500 (1 + 0.1r) = 15669.93\)
\(1 + 0.1r = \frac{15669.93}{11500} \approx 1.36260\)
\(0.1r \approx 0.36260\)
\(r \approx 3.6260\)

Thus, to 3 significant figures, \(r \approx 3.63\).

(a)
M1: Use of compound interest formula with correct parameters (\(12\) compounds per year, \(t=5\)).
A1: Correct value \(\$12\,518\) (accept \(\$12517.96\) or \(\$12\,500\) to 3 s.f.).

(b)
M1: Calculate the value under Plan A at \(10\) years (\(15669.93\)).
M1: Set up simple interest equation for Plan B equated to the value from Plan A.
A1: Correct value of \(r \approx 3.63\) (accept \(3.62\) due to rounding of intermediate steps).

Since the tower \(TA\) is vertical, the triangles \(TAB\) and \(TAC\) are right-angled at \(A\).
We can express the distances on the ground, \(AB\) and \(AC\), in terms of \(h\):
\(\tan(32^\circ) = \frac{h}{AB} \implies AB = \frac{h}{\tan(32^\circ)}\)
\(\tan(25^\circ) = \frac{h}{AC} \implies AC = \frac{h}{\tan(25^\circ)}\)

In triangle \(BAC\) on the ground, we are given \(BC = 45\) m and \(\angle BAC = 110^\circ\).
Applying the Cosine Rule in triangle \(BAC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(\angle BAC)\)

Substitute the expressions in terms of \(h\):
\(45^2 = \left(\frac{h}{\tan(32^\circ)}\right)^2 + \left(\frac{h}{\tan(25^\circ)}\right)^2 - 2 \left(\frac{h}{\tan(32^\circ)}\right) \left(\frac{h}{\tan(25^\circ)}\right) \cos(110^\circ)\)

\(2025 = h^2 \left[ \frac{1}{\tan^2(32^\circ)} + \frac{1}{\tan^2(25^\circ)} - \frac{2 \cos(110^\circ)}{\tan(32^\circ) \tan(25^\circ)} \right]\)

Using GDC to evaluate the coefficients:
\(\tan(32^\circ) \approx 0.62487\)
\(\tan(25^\circ) \approx 0.46631\)
\(\cos(110^\circ) \approx -0.34202\)

This gives:
\(\frac{1}{\tan^2(32^\circ)} \approx 2.56106\)
\(\frac{1}{\tan^2(25^\circ)} \approx 4.59892\)
\(-\frac{2 \cos(110^\circ)}{\tan(32^\circ) \tan(25^\circ)} \approx -\frac{2(-0.34202)}{0.62487 \times 0.46631} \approx 2.34758\)

Sum of terms inside the bracket:
\(2.56106 + 4.59892 + 2.34758 \approx 9.50756\)

Therefore:
\(2025 = 9.50756 h^2\)
\(h^2 \approx 212.988\)
\(h \approx 14.594\)

To 3 significant figures, the height of the tower is \(14.6\) metres.

M1: Expressing ground distances \(AB\) and \(AC\) in terms of \(h\).
A1: Correct expressions \(AB = \frac{h}{\tan(32^\circ)}\) and \(AC = \frac{h}{\tan(25^\circ)}\).
M1: Applying the Cosine Rule to triangle \(BAC\).
A1: Correct substitution of values into the Cosine Rule equation.
A1: Correct final answer \(h \approx 14.6\) metres (accept \(14.59\)).

(a) Entering the data into a graphic display calculator (GDC):
\(x = [2, 5, 8, 10, 12, 15]\)
\(y = [45, 56, 68, 75, 80, 92]\)

The GDC gives the linear correlation coefficient:
\(r \approx 0.99868\)

To 3 significant figures, \(r \approx 0.999\).

(b) Using the GDC's linear regression feature (\(y = ax + b\)):
\(a \approx 3.5808\)
\(b \approx 38.299\)

To 3 significant figures, the regression line of \(y\) on \(x\) is:
\(y = 3.58x + 38.3\)

(c) To estimate the exam score for \(x = 7\) hours of study:
\(y = 3.5808(7) + 38.299 \approx 25.0656 + 38.299 \approx 63.36\)

To 3 significant figures, the estimated exam score is \(63.4\).

(a)
A1: Correct value of \(r \approx 0.999\) (or \(0.99868...\)).

(b)
M1: Use of linear regression model on GDC.
A1A1: Correct parameters \(a \approx 3.58\) and \(b \approx 38.3\) (accept \(y = 3.58x + 38.3\)).

(c)
M1: Substituting \(x = 7\) into the regression equation.
A1: Correct estimate of \(63.4\) (accept \(63.3\) or \(63.36\) depending on rounding precision of equation).

(a) Let \(X\) be the mass of a randomly chosen apple, where \(X \sim N(150, 18^2)\). Using a GDC to find the normal cumulative probability: \(P(130 \le X \le 165) \approx 0.66415...\) to 3 significant figures is \(0.664\).

(b) Let \(k\) be the minimum mass. We require \(P(X > k) = 0.15\), which is equivalent to \(P(X \le k) = 0.85\). Using the inverse normal function on a GDC: \(k \approx 168.655...\). To the nearest gram, \(k = 169\text{ g}\).

(c) Let \(Y\) be the number of Premium apples in a batch of 10. \(Y\) follows a binomial distribution \(Y \sim B(10, 0.15)\). We need to find \(P(Y \ge 3) = 1 - P(Y \le 2)\). Using the binomial cumulative distribution function on a GDC: \(P(Y \le 2) \approx 0.82020...\). Therefore, \(P(Y \ge 3) = 1 - 0.82020... \approx 0.17980...\) which is \(0.180\) to 3 significant figures.

(d) The total production cost for 100 apples is \(100 \times 0.20 = 20.00\text{ USD}\). Let \(R\) be the revenue from a single apple. An apple is Premium with probability \(0.15\) and non-Premium with probability \(0.85\). The expected revenue per apple is: \(E(R) = (1.50 \times 0.15) + (0.60 \times 0.85) = 0.225 + 0.51 = 0.735\text{ USD}\). For a batch of 100 apples, the expected total revenue is: \(100 \times 0.735 = 73.50\text{ USD}\). Therefore, the expected profit is: \(\text{Expected Revenue} - \text{Total Cost} = 73.50 - 20.00 = 53.50\text{ USD}\).

(a)
M1 for setting up the normal probability equation \(P(130 \le X \le 165)\).
A1 for consistent GDC inputs/z-scores.
A1 for final answer 0.664.

(b)
M1 for writing \(P(X > k) = 0.15\) or \(P(X \le k) = 0.85\).
M1 for evidence of using inverse normal on a GDC.
A1 for 169 (must be to the nearest integer as requested).

(c)
M1 for identifying the binomial distribution \(B(10, 0.15)\).
M1 for setting up the calculation \(1 - P(Y \le 2)\).
A1 for 0.180.

(d)
A1 for finding total production cost of 20 USD.
M1 for setting up expected revenue per apple equation.
A1 for expected revenue per apple of 0.735 USD.
A1 for total expected revenue of 73.50 USD.
M1 for subtracting total cost from expected revenue.
A1 for 53.50 USD (accept 53.5).

(a) The particle is at rest when \(v(t) = 0\). Setting \(3 + 2t - e^{0.5t} \cos(t) = 0\) and solving on a GDC for \(0 \le t \le 8\) yields \(t \approx 5.7013...\) and \(t \approx 7.4024...\). To 3 significant figures, the times are \(t = 5.70\text{ s}\) and \(t = 7.40\text{ s}\).

(b) Acceleration is given by \(a(t) = v'(t)\). Using numerical differentiation on a GDC at \(t = 3\): \(a(3) \approx 4.8533...\text{ m s}^{-2}\). To 3 significant figures, \(a(3) = 4.85\text{ m s}^{-2}\).

(c) The total distance traveled is given by \(\int_{0}^{6} |v(t)| \, dt\). Since \(v(t)\) changes sign at \(t \approx 5.70\), we evaluate the integral on a GDC: \(\int_{0}^{6} |3 + 2t - e^{0.5t} \cos(t)| \, dt \approx 52.454...\text{ m}\). To 3 significant figures, the total distance is \(52.5\text{ m}\).

(d) (i) Since \(s(t) = s(0) + \int_{0}^{t} v(u) \, du\), we have \(s(t) = -2 + \int_{0}^{t} (3 + 2u - e^{0.5u} \cos(u)) \, du\).

(ii) The displacement from the origin is given by \(|s(t)|\). Critical points of \(s(t)\) occur when \(s'(t) = v(t) = 0\), which are at \(t \approx 5.70\) and \(t \approx 7.40\). We compare displacement values at the endpoints and critical points:
- At \(t = 0\): \(s(0) = -2\text{ m}\)
- At \(t \approx 5.70\): \(s(5.70) = -2 + \int_{0}^{5.70} v(t) \, dt \approx 49.815...\text{ m}\)
- At \(t \approx 7.40\): \(s(7.40) = -2 + \int_{0}^{7.40} v(t) \, dt \approx 39.168...\text{ m}\)
- At \(t = 8\): \(s(8) = -2 + \int_{0}^{8} v(t) \, dt \approx 46.368...\text{ m}\)

Comparing the magnitudes, the maximum displacement of the particle from the origin is \(49.8\text{ m}\) (to 3 significant figures), which occurs at \(t \approx 5.70\text{ s}\).

(a)
M1 for setting \(v(t) = 0\).
A1 for \(t \approx 5.70\).
A1 for \(t \approx 7.40\).

(b)
M1 for recognizing that acceleration is the derivative of velocity.
A1 for \(4.85\) (accept \(4.85\text{ m s}^{-2}\)).

(c)
M1 for setting up the integral of the absolute value of velocity.
A1 for correct limits \(0\) and \(6\).
A1 for \(52.5\) (accept \(52.5\text{ m}\)).

(d) (i)
A1 for a correct expression for \(s(t)\).

(d) (ii)
M1 for identifying that maximum displacement occurs at critical points \(v(t) = 0\) or endpoints.
M1 for testing values of \(s(t)\) at critical points.
A1 for calculating \(s(5.70) \approx 49.8\).
A1 for calculating \(s(7.40) \approx 39.2\) or \(s(8) \approx 46.4\).
A1 for selecting the maximum value of \(49.8\) (accept \(49.8\text{ m}\)).
A1 for identifying the time \(t \approx 5.70\text{ s}\).

(a) The radius of the Ferris wheel is \(30\text{ m}\), which represents the amplitude of the motion, so \(A = 30\). The midline is at the average of the minimum and maximum heights: \(D = \frac{4 + (4+60)}{2} = 34\text{ m}\). The period is \(12\) minutes, so the frequency is \(B = \frac{2\pi}{12} = \frac{\pi}{6}\). Since the passenger starts at the lowest point at \(t=0\), we use a negative cosine function: \(h(t) = -30 \cos\left(\frac{\pi}{6}t\right) + 34\).

(b) Substituting \(t = 5\) into the function: \(h(5) = -30 \cos\left(\frac{5\pi}{6}\right) + 34 = -30\left(-\frac{\sqrt{3}}{2}\right) + 34 = 15\sqrt{3} + 34 \approx 59.9807...\text{ m}\). To 3 significant figures, the height is \(60.0\text{ m}\).

(c) The rate of change is given by \(h'(t)\). Differentiating \(h(t)\): \(h'(t) = 30\left(\frac{\pi}{6}\right) \sin\left(\frac{\pi}{6}t\right) = 5\pi \sin\left(\frac{\pi}{6}t\right)\). At \(t = 5\): \(h'(5) = 5\pi \sin\left(\frac{5\pi}{6}\right) = 5\pi (0.5) = 2.5\pi \approx 7.85398...\text{ m min}^{-1}\). To 3 significant figures, the rate of change is \(7.85\text{ m min}^{-1}\).

(d) We set \(h(t) = 45\) to find when the passenger is at this height: \(-30 \cos\left(\frac{\pi}{6}t\right) + 34 = 45 \Rightarrow \cos\left(\frac{\pi}{6}t\right) = -\frac{11}{30}\). Solving on a GDC within the first period \(0 \le t \le 12\) gives: \(t_1 \approx 3.7180\) and \(t_2 \approx 8.2820\). The total time spent at or above 45 meters is \(t_2 - t_1 = 8.2820 - 3.7180 = 4.564\text{ minutes}\). To 3 significant figures, this is \(4.56\text{ minutes}\).

(e) We set \(h(t) = g(t)\): \(-30 \cos\left(\frac{\pi}{6}t\right) + 34 = -25 \cos\left(\frac{\pi}{5}t - \frac{\pi}{3}\right) + 28\). Using a GDC to find the first intersection point for \(t > 0\) yields \(t \approx 0.80112...\text{ minutes}\). To 3 significant figures, the time is \(0.801\text{ minutes}\).

(a)
A1 for identifying the amplitude is 30.
A1 for identifying the midline is 34.
A1 for calculating the period coefficient \(B = \frac{\pi}{6}\) and correctly justifying the negative cosine structure.

(b)
M1 for substituting \(t = 5\) into the model.
A1 for \(60.0\) (or \(15\sqrt{3} + 34\)).

(c)
M1 for taking the derivative using the chain rule.
A1 for correct derivative formula \(h'(t) = 5\pi \sin\left(\frac{\pi}{6}t\right)\).
A1 for \(7.85\).

(d)
M1 for setting up the equation \(h(t) = 45\).
A1 for finding \(t_1 \approx 3.72\).
A1 for finding \(t_2 \approx 8.28\).
A1 for the final duration of \(4.56\) (accept \(4.56\text{ minutes}\)).

(e)
M1 for setting \(h(t) = g(t)\).
M1 for attempting to find the intersection using GDC solver or graph.
A1 for \(0.801\).