2025 IB DP Mathematics - Analysis and Approaches Practice Paper with Answers

(a) Let the first term of the arithmetic sequence be \(u_1\). The first, third, and seventh terms are \(u_1\), \(u_1 + 2d\), and \(u_1 + 6d\). Since these form a geometric sequence, we have: \(\frac{u_1 + 2d}{u_1} = \frac{u_1 + 6d}{u_1 + 2d}\) which simplifies to \((u_1 + 2d)^2 = u_1(u_1 + 6d)\). Expanding both sides: \(u_1^2 + 4u_1d + 4d^2 = u_1^2 + 6u_1d\). Subtracting \(u_1^2\) and rearranging: \(4d^2 = 2u_1d\). Since \(d \neq 0\), we can divide by \(2d\) to get \(2d = u_1\). The common ratio \(r\) is \(r = \frac{u_1 + 2d}{u_1} = \frac{2d + 2d}{2d} = 2\). (b) Given \(u_1 = 3\), we find the common difference using \(2d = u_1 \implies d = 1.5\). The sum of the first \(10\) terms of the arithmetic sequence is: \(S_{10} = \frac{10}{2} [2u_1 + 9d] = 5 [2(3) + 9(1.5)] = 5 [6 + 13.5] = 5 [19.5] = 97.5\).

(a) M1 for setting up the geometric ratio equation: \(\frac{u_1 + 2d}{u_1} = \frac{u_1 + 6d}{u_1 + 2d}\). A1 for simplifying to find the relation between \(u_1\) and \(d\): \(2d = u_1\). AG for substituting back and showing that \(r = 2\). (b) M1 for finding \(d = 1.5\). M1 for substituting into the arithmetic sum formula. A1 for the correct answer 97.5 (or \(\frac{195}{2}\)).

(a) Let \(y = \frac{3x + 1}{x - 2}\). Interchanging \(x\) and \(y\) and rearranging to make \(y\) the subject: \(x(y - 2) = 3y + 1 \implies xy - 2x = 3y + 1 \implies y(x - 3) = 2x + 1 \implies y = \frac{2x + 1}{x - 3}\). Thus, \(f^{-1}(x) = \frac{2x + 1}{x - 3}\) for \(x \neq 3\). (b) Setting the two expressions equal: \(\frac{2x + 1}{x - 3} = \frac{3x + 1}{x - 2}\). Multiplying both sides: \((2x + 1)(x - 2) = (3x + 1)(x - 3) \implies 2x^2 - 3x - 2 = 3x^2 - 8x - 3 \implies x^2 - 5x - 1 = 0\). Using the quadratic formula, \(x = \frac{5 \pm \sqrt{29}}{2}\).

(a) M1 for attempting to interchange variables and isolate \(y\). M1 for gathering terms: \(y(x - 3) = 2x + 1\). A1 for \(f^{-1}(x) = \frac{2x + 1}{x - 3}\). (b) M1 for setting up and expanding the equation. A1 for simplifying to \(x^2 - 5x - 1 = 0\). A1 for the correct solutions \(x = \frac{5 \pm \sqrt{29}}{2}\).

Using the double-angle identity \(\sin(2\theta) = 2\sin\theta\cos\theta\), we can rewrite the equation as: \(2\sin\theta\cos\theta = \sqrt{3}\cos\theta\). Rearranging gives: \(\cos\theta(2\sin\theta - \sqrt{3}) = 0\). This gives two branches of equations: 1) \(\cos\theta = 0 \implies \theta = \frac{\pi}{2}, \frac{3\pi}{2}\) for the given domain. 2) \(\sin\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}, \frac{2\pi}{3}\) for the given domain. The solutions are \(\theta = \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{2}\).

M1 for applying the double-angle identity. M1 for factorizing to obtain \(\cos\theta(2\sin\theta - \sqrt{3}) = 0\). A1 for identifying both branches. A1 for finding the correct solutions for \(\cos\theta = 0\). A1 for finding the correct solutions for \(\sin\theta = \frac{\sqrt{3}}{2}\).

(a) Using the product rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 \cdot e^{-2x} + x \cdot (-2 e^{-2x}) = (1 - 2x)e^{-2x}\). (b) When \(x = 1\), the y-coordinate is \(y = 1 \cdot e^{-2} = e^{-2}\). The gradient of the tangent at \(x = 1\) is \(m = (1 - 2(1))e^{-2} = -e^{-2}\). Using the point-slope form: \(y - e^{-2} = -e^{-2}(x - 1) \implies y - e^{-2} = -e^{-2}x + e^{-2} \implies y = -e^{-2}x + 2e^{-2}\).

(a) M1 for using the product rule. A1 for obtaining \(\frac{\mathrm{d}y}{\mathrm{d}x} = (1 - 2x)e^{-2x}\). (b) A1 for finding the point \((1, e^{-2})\). M1 for substituting \(x = 1\) into the derivative to find the gradient \(-e^{-2}\). M1 for using the equation of a line. A1 for \(y = -e^{-2}x + 2e^{-2}\) (or equivalent).

(a) Since the sum of all probabilities must equal 1, we have: \(10k^2 + 3k + 0.4 + 0.2 = 1 \implies 10k^2 + 3k - 0.4 = 0\). Multiplying the equation by 10 gives: \(100k^2 + 30k - 4 = 0 \implies 50k^2 + 15k - 2 = 0\). Factorizing the quadratic: \((10k - 1)(5k + 2) = 0\). This yields \(k = 0.1\) or \(k = -0.4\). Since \(k > 0\), we have \(k = 0.1\). (b) Substituting \(k = 0.1\) into the distribution: \(\mathrm{P}(X=1) = 0.1\), \(\mathrm{P}(X=2) = 0.3\), \(\mathrm{P}(X=3) = 0.4\), and \(\mathrm{P}(X=4) = 0.2\). The expectation is: \(\mathrm{E}(X) = \sum x \mathrm{P}(X=x) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7\).

(a) M1 for setting the sum of probabilities to 1. M1 for obtaining a standard quadratic form. A1 for factorizing or solving the quadratic. A1 for choosing the positive solution \(k = 0.1\). (b) M1 for using the expectation formula. A1 for finding the correct expected value \(2.7\).

(a) Substituting \(x = \frac{\pi}{6}\) into both functions: \(f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\), \(g\left(\frac{\pi}{6}\right) = \cos\left(2 \cdot \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\). Since both equal \(\frac{1}{2}\), they intersect at this point. (b) Since \(g(x) \ge f(x)\) on \([0, \frac{\pi}{6}]\), the area is: \(A = \int_{0}^{\frac{\pi}{6}} (\cos(2x) - \sin(x)) \mathrm{d}x = \left[ \frac{1}{2}\sin(2x) + \cos(x) \right]_{0}^{\frac{\pi}{6}}\). Evaluating at the upper limit: \(\frac{1}{2}\sin\left(\frac{\pi}{3}\right) + \cos\left(\frac{\pi}{6}\right) = \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}\). Evaluating at the lower limit: \(\frac{1}{2}\sin(0) + \cos(0) = 1\). Subtracting the lower limit value: \(A = \frac{3\sqrt{3}}{4} - 1\).

(a) M1 for evaluating both functions at the given point. AG for showing they both equal \(1/2\). (b) M1 for setting up the correct difference integral. M1 for finding the antiderivative. A1 for evaluating the upper limit correctly. A1 for evaluating the lower limit correctly. A1 for the correct final area \(\frac{3\sqrt{3}}{4} - 1\).

(a) (i) Using the product rule:
\(f'(x) = 1 \cdot e^{-2x} + x \cdot (-2 e^{-2x}) = (1 - 2x)e^{-2x}\)

(ii) For a local maximum, set \(f'(x) = 0\):
\((1 - 2x)e^{-2x} = 0 \implies 1 - 2x = 0 \implies x = \frac{1}{2}\)
Since \(x = \frac{1}{2}\), the corresponding \(y\)-coordinate is:
\(f\left(\frac{1}{2}\right) = \frac{1}{2} e^{-2(1/2)} = \frac{1}{2e}\)
Thus, the local maximum point is \(\left(\frac{1}{2}, \frac{1}{2e}\right)\).

(b) To find the point of inflexion, find \(f''(x)\) using the product rule:
\(f''(x) = -2e^{-2x} + (1-2x)(-2e^{-2x}) = (-2 - 2 + 4x)e^{-2x} = 4(x-1)e^{-2x}\)
Set \(f''(x) = 0\):
\(4(x-1)e^{-2x} = 0 \implies x = 1\)
Justification:
For \(x < 1\), \(f''(x) < 0\) (the curve is concave down).
For \(x > 1\), \(f''(x) > 0\) (the curve is concave up).
Since the second derivative changes sign at \(x = 1\), there is a point of inflexion at \(x = 1\).

(c) The area \(A\) is given by:
\(A = \int_{0}^{\ln 2} x e^{-2x} \, dx\)
Using integration by parts, let \(u = x \implies du = dx\) and \(dv = e^{-2x}\,dx \implies v = -\frac{1}{2}e^{-2x}\):
\(\int x e^{-2x} \, dx = -\frac{1}{2}x e^{-2x} - \int -\frac{1}{2}e^{-2x} \, dx = -\frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x}\)
Evaluating this from \(0\) to \(ln 2\):
\(A = \left[ -\frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} \right]_{0}^{\ln 2}\)
\(A = \left( -\frac{1}{2}(\ln 2)e^{-2\ln 2} - \frac{1}{4}e^{-2\ln 2} \right) - \left( 0 - \frac{1}{4}e^{0} \right)\)
Since \(e^{-2\ln 2} = (e^{\ln 2})^{-2} = 2^{-2} = \frac{1}{4}\):
\(A = \left( -\frac{\ln 2}{8} - \frac{1}{16} \right) + \frac{1}{4}\)
\(A = \frac{-2\ln 2 - 1 + 4}{16} = \frac{3 - 2\ln 2}{16}\)

(d) The volume \(V\) of the solid of revolution is:
\(V = \pi \int_{0}^{\ln 2} [f(x)]^2 \, dx = \pi \int_{0}^{\ln 2} x^2 e^{-4x} \, dx\)

(a) (i)
- M1: Attempt to apply product rule.
- A1: Correct derivative \(f'(x) = (1 - 2x)e^{-2x}\).
(ii)
- M1: Setting \(f'(x) = 0\) and solving for \(x\).
- A1: Correct coordinates \(\left(\frac{1}{2}, \frac{1}{2e}\right)\).

(b)
- M1: Attempt to find second derivative \(f''(x)\).
- A1: Correct second derivative \(4(x-1)e^{-2x}\).
- A1: Finding \(x = 1\).
- R1: Valid justification of sign change of \(f''(x)\) around \(x = 1\).

(c)
- M1: Setting up correct definite integral.
- M1: Attempt at integration by parts (assigning \(u\) and \(v\)).
- A1: Correct antiderivative \(-\frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x}\).
- M1: Substituting limits \(0\) and \(\ln 2\) correctly.
- A1: Correct simplified answer \(\frac{3 - 2\ln 2}{16}\).

(d)
- M1: Use of volume of revolution formula \(V = \pi \int y^2 \, dx\) with limits.
- A1: Correct expression \(\pi \int_{0}^{\ln 2} x^2 e^{-4x} \, dx\).

(a) To show the lines intersect, we equate the vector equations component-by-component:
1) \(-1 + \lambda = 1 \implies \lambda = 2\)
2) \(\lambda = \mu \implies \mu = 2\)
3) \(3 = 1 + \mu\)
Substituting \(\mu = 2\) into the third equation yields \(3 = 1 + 2 = 3\), which is consistent.
Therefore, the lines intersect. The point of intersection \(P\) is found by substituting \(\lambda = 2\) into \(L_1\):
\(P = (-1+2, 2, 3) = (1, 2, 3)\).

(b) The direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\).
The angle \(\theta\) between them satisfies:
\(\cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|}\)
\(\mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(0) + (1)(1) + (0)(1) = 1\)
\(|\mathbf{d}_1| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}\)
\(|\mathbf{d}_2| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}\)
\(\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}\)
Since the angle between lines is acute, \(\theta = \frac{\pi}{3}\) (or \(60^\circ\)).

(c) (i) At \(\lambda = 4\), point \(A\) has coordinates:
\(A = (-1+4, 4, 3) = (3, 4, 3)\).

(ii) Method 1 (Geometric):
The point of intersection is \(P(1, 2, 3)\). The vector \(\vec{PA}\) is:
\(\vec{PA} = \mathbf{a} - \mathbf{p} = \begin{pmatrix} 3-1 \\ 4-2 \\ 3-3 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}\)
The length of \(\vec{PA}\) is:
\(|\vec{PA}| = \sqrt{2^2 + 2^2 + 0^2} = \sqrt{8} = 2\sqrt{2}\)
Since \(\theta = \frac{\pi}{3}\) is the angle between \(L_1\) and \(L_2\), the shortest distance \(d\) from \(A\) to \(L_2\) is:
\(d = |\vec{PA}| \sin \theta = 2\sqrt{2} \sin \left(\frac{\pi}{3}\right) = 2\sqrt{2} \cdot \frac{\sqrt{3}}{2} = \sqrt{6}\).

Method 2 (Vector projection):
Let \(Q\) be the foot of the perpendicular from \(A\) to \(L_2\). Then \(Q\) has coordinates \((1, \mu, 1+\mu)\).
The vector \(\vec{AQ} = \begin{pmatrix} 1-3 \\ \mu-4 \\ 1+\mu-3 \end{pmatrix} = \begin{pmatrix} -2 \\ \mu-4 \\ \mu-2 \end{pmatrix}\).
For \(\vec{AQ}\) to be perpendicular to the direction of \(L_2\), \(\mathbf{d}_2\):
\(\vec{AQ} \cdot \mathbf{d}_2 = 0 \implies 0(-2) + 1(\mu-4) + 1(\mu-2) = 0\)
\(2\mu - 6 = 0 \implies \mu = 3\).
Then \(\vec{AQ} = \begin{pmatrix} -2 \\ -1 \\ 1 \end{pmatrix}\).
Its magnitude is \(|\vec{AQ}| = \sqrt{(-2)^2 + (-1)^2 + 1^2} = \sqrt{6}\).

(a)
- M1: Setting up the system of equations by equating components.
- A1: Solving for \(\lambda\) and \(\mu\) (e.g. \(\lambda=2, \mu=2\)).
- R1: Showing consistency in the remaining equation.
- A1: Correct coordinates of the intersection point \(P(1, 2, 3)\).

(b)
- A1: Identifying the direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\).
- M1: Applying the scalar product formula \(\cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1| |\mathbf{d}_2|}\).
- A1: Correct value \(\cos \theta = \frac{1}{2}\).
- A1: Correct angle \(\theta = \frac{\pi}{3}\) (or \(60^\circ\)).

(c) (i)
- A1: Correct coordinates of \(A(3, 4, 3)\).
(ii)
- M1: Finding the vector \(\vec{PA}\) or setting up a generic vector \(\vec{AQ}\).
- A1: Correct vector \(\vec{PA} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}\) (magnitude \(\sqrt{8}\)) or setting up \(\vec{AQ} \cdot \mathbf{d}_2 = 0\).
- M1: Attempting to use trigonometry (\(d = |\vec{PA}|\sin\theta\)) or solving for \(\mu = 3\).
- A1: Correct value of \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\) or vector \(\vec{AQ} = \begin{pmatrix} -2 \\ -1 \\ 1 \end{pmatrix}\).
- A1: Correct shortest distance \(\sqrt{6}\).

(a) Let \( S = 1 + \omega + \omega^2 + \omega^3 + \omega^4 \).
This is a finite geometric series with first term \( 1 \) and common ratio \( \omega \neq 1 \).
Using the sum formula:
\( S = \frac{\omega^5 - 1}{\omega - 1} \)
Since \( \omega = e^{i \frac{2\pi}{5}} \), we have:
\( \omega^5 = \left( e^{i \frac{2\pi}{5}} \right)^5 = e^{i 2\pi} = 1 \)
Therefore, \( S = \frac{1 - 1}{\omega - 1} = 0 \).

(b) By Euler's formula, \( \omega^k = \cos\left(\frac{2k\pi}{5}\right) + i \sin\left(\frac{2k\pi}{5}\right) \).
Since \( 1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0 \), the real part of the sum must also be equal to zero:
\( 1 + \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right) + \cos\left(\frac{8\pi}{5}\right) = 0 \)
Using the cosine identity \( \cos(2\pi - \theta) = \cos \theta \):
\( \cos\left(\frac{8\pi}{5}\right) = \cos\left(2\pi - \frac{2\pi}{5}\right) = \cos\left(\frac{2\pi}{5}\right) \)
\( \cos\left(\frac{6\pi}{5}\right) = \cos\left(2\pi - \frac{4\pi}{5}\right) = \cos\left(\frac{4\pi}{5}\right) \)
Substituting these back into the equation yields:
\( 1 + 2\cos\left(\frac{2\pi}{5}\right) + 2\cos\left(\frac{4\pi}{5}\right) = 0 \)
\( 2\left( \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) \right) = -1 \)
\( \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2} \).

(c) Let \( x = \cos\left(\frac{2\pi}{5}\right) \).
Using the double-angle identity for cosine:
\( \cos\left(\frac{4\pi}{5}\right) = \cos\left(2 \cdot \frac{2\pi}{5}\right) = 2\cos^2\left(\frac{2\pi}{5}\right) - 1 = 2x^2 - 1 \)
Substituting this into the result from part (b):
\( x + (2x^2 - 1) = -\frac{1}{2} \)
Multiply by 2 to clear fractions:
\( 4x^2 + 2x - 2 = -1 \implies 4x^2 + 2x - 1 = 0 \)
Solving this quadratic equation using the quadratic formula:
\( x = \frac{-2 \pm \sqrt{2^2 - 4(4)(-1)}}{2(4)} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4} \)
Since \( \frac{2\pi}{5} \) is in the first quadrant (i.e. \( 0 < \frac{2\pi}{5} < \frac{\pi}{2} \)), we must have \( \cos\left(\frac{2\pi}{5}\right) > 0 \).
Since \( \sqrt{5} > 1 \), the positive solution is selected:
\( \cos\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5} - 1}{4} \).

(a)
- M1: Attempting to use the sum of a geometric series formula.
- A1: Correct substitution of \(\omega^5\) and simplifying.
- A1: Recognizing \(\omega^5 = 1\) and concluding the result.

(b)
- M1: Taking the real part of both sides of the equation from part (a).
- A1: Correctly writing the expansion with cosines.
- M1: Applying the symmetry identity \(\cos(2\pi - \theta) = \cos\theta\).
- A1: Correctly simplifying the sum to \(1 + 2\cos\left(\frac{2\pi}{5}\right) + 2\cos\left(\frac{4\pi}{5}\right) = 0\).
- A1: Rearranging to show \(\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2}\).

(c)
- M1: Using the double-angle identity \(\cos(2\theta) = 2\cos^2\theta - 1\).
- A1: Correctly substituting to get the quadratic equation in terms of \(x = \cos\left(\frac{2\pi}{5}\right)\).
- M1: Rewriting in standard quadratic form \(4x^2 + 2x - 1 = 0\).
- M1: Applying the quadratic formula.
- A1: Obtaining \(\frac{-1 \pm \sqrt{5}}{4}\).
- R1: Reasoned choice of positive root because \(\frac{2\pi}{5}\) is in the first quadrant.
- A1: Final correct value \(\frac{\sqrt{5} - 1}{4}\).

Let \(t\) be the time in years.

Arthur's investment value after \(t\) years is given by:
\(A(t) = 5000 \left(1 + \frac{0.042}{12}\right)^{12t} = 5000 (1.0035)^{12t}\)

Beatrice's investment value after \(t\) years is given by:
\(B(t) = 5500 (1 - 0.015)^t = 5500 (0.985)^t\)

We want to find the minimum integer \(n\) such that \(A(n) > B(n)\).

Using a GDC to find the intersection of \(y = 5000 (1.0035)^{12t}\) and \(y = 5500 (0.985)^t\):
\(5000 (1.0035)^{12t} = 5500 (0.985)^t\)

Taking the natural logarithm of both sides:
\(\ln(5000) + 12t \ln(1.0035) = \ln(5500) + t \ln(0.985)\)
\(t \left(12 \ln(1.0035) - \ln(0.985)\right) = \ln(1.1)\)
\(t \approx 1.67\) years.

Since \(n\) must be an integer representing complete years:
For \(n = 1\):
\(A(1) = 5000 (1.0035)^{12} \approx \$5214.09\)
\(B(1) = 5500 (0.985)^1 = \$5417.50\)
Here, \(A(1) < B(1)\).

For \(n = 2\):
\(A(2) = 5000 (1.0035)^{24} \approx \$5437.34\)
\(B(2) = 5500 (0.985)^2 \approx \$5336.24\)
Here, \(A(2) > B(2)\).

Thus, the minimum number of complete years is \(2\).

M1: For setting up the compound interest formula for Arthur's investment, \(5000\left(1 + \frac{0.042}{12}\right)^{12t}\) or equivalent.
M1: For setting up the depreciation formula for Beatrice's investment, \(5500(0.985)^t\) or equivalent.
M1: For setting up the inequality or equation to find the intersection point, e.g., \(5000(1.0035)^{12t} = 5500(0.985)^t\).
A1: For finding the decimal value \(t \approx 1.67\).
R1: For comparing values at \(n = 1\) and \(n = 2\) or interpreting the inequality for integer values.
A1: For the final correct answer \(2\).

(a) The amplitude \(a\) is:
\(a = \frac{\text{max} - \text{min}}{2} = \frac{26 - 16}{2} = 5\)

The vertical shift \(d\) is:
\(d = \frac{\text{max} + \text{min}}{2} = \frac{26 + 16}{2} = 21\)

The period is 24 hours, so:
\(b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{24} = \frac{\pi}{12}\)

Since the maximum occurs at \(t = 14\), we can choose the horizontal shift \(c = 14\).
So the function is \(T(t) = 5 \cos\left(\frac{\pi}{12}(t - 14)\right) + 21\).

(b) We want to find the interval where \(T(t) > 23\):
\(5 \cos\left(\frac{\pi}{12}(t - 14)\right) + 21 > 23\)
\(\cos\left(\frac{\pi}{12}(t - 14)\right) > 0.4\)

Using a GDC to solve \(T(t) = 23\) on the interval \(0 \le t \le 24\), we find the intersection points:
\(t_1 \approx 9.572\)
\(t_2 \approx 18.428\)

The temperature is above \(23^\circ\text{C}\) between these two times.
Total duration = \(18.428 - 9.572 = 8.856\) hours.

To 3 significant figures, the total time is \(8.86\) hours.

(a)
A1: For \(a = 5\).
A1: For \(d = 21\).
A1: For \(b = \frac{\pi}{12}\) (accept standard decimal representation \(0.262\)).
A1: For \(c = 14\).

(b)
M1: For attempting to solve \(T(t) = 23\) (e.g., drawing line \(y = 23\) on GDC or setting up equation).
A1: For finding the two correct intersection points \(t_1 \approx 9.57\) and \(t_2 \approx 18.4\).
A1: For finding the correct difference of \(8.86\) hours (accept \(8.85\) or \(8.86\)).

(a) The total surface area of a closed rectangular box with a square base is:
\(A = 2x^2 + 4xh = 400\)

Expressing \(h\) in terms of \(x\):
\(4xh = 400 - 2x^2\)
\(h = \frac{400 - 2x^2}{4x} = \frac{100}{x} - \frac{x}{2}\)

The volume of the box is:
\(V = x^2 h = x^2 \left(\frac{100}{x} - \frac{x}{2}\right) = 100x - \frac{1}{2}x^3\) (as required).

(b) To find the maximum volume, we can use a GDC to find the local maximum of the volume function, or find where the derivative is zero:
\(V'(x) = 100 - \frac{3}{2}x^2 = 0\)
\(x^2 = \frac{200}{3} \approx 66.67\)
\(x = \sqrt{\frac{200}{3}} \approx 8.16497\text{ cm}\)

The maximum volume is:
\(V(8.16497) = 100(8.16497) - \frac{1}{2}(8.16497)^3 \approx 544.331\text{ cm}^3\)

To 3 significant figures:
\(x \approx 8.16\text{ cm}\)
Maximum volume \(\approx 544\text{ cm}^3\).

(a)
M1: For expressing the surface area in terms of \(x\) and \(h\): \(2x^2 + 4xh = 400\).
A1: For correctly making \(h\) the subject: \(h = \frac{100}{x} - \frac{x}{2}\) and substituting it into the volume formula \(V = x^2 h\) to obtain the required expression.

(b)
M1: For setting up the derivative \(V'(x) = 100 - 1.5x^2 = 0\) or utilizing the maximum-finding feature on a GDC.
A1: For \(x \approx 8.16\text{ cm}\) (accept \(\sqrt{\frac{200}{3}}\) or \(8.165\)).
A1: For maximum volume \(\approx 544\text{ cm}^3\) (accept \(544.33\)).

(a) Let us determine the angle \(\angle ABC\).
The bearing from A to B is \(060^\circ\).
If we draw a North-South line through B, the angle between the South-pointing ray and the line segment BA is \(60^\circ\) (alternate interior angles).
The bearing from B to C is \(150^\circ\), which is \(30^\circ\) East of South.
Therefore, the angle \(\angle ABC = 60^\circ + 30^\circ = 90^\circ\).

Since triangle ABC is a right-angled triangle at B:
\(AC^2 = AB^2 + BC^2 = 8^2 + 12^2 = 64 + 144 = 208\)
\(AC = \sqrt{208} \approx 14.4222\text{ km}\)

To 3 significant figures, the distance is \(14.4\text{ km}\).

(b) Let \(\theta = \angle BAC\).
Since \(\angle ABC = 90^\circ\):
\(\tan(\theta) = \frac{BC}{AB} = \frac{12}{8} = 1.5\)
\(\theta = \arctan(1.5) \approx 56.31^\circ\)

The bearing of B from A is \(060^\circ\).
Since the path from B to C moves clockwise from the direction of AB, the bearing of C from A is:
\(\text{Bearing} = 60^\circ + 56.31^\circ = 116.31^\circ\)

To 3 significant figures, the bearing is \(116^\circ\).

(a)
M1: For deducing that \(\angle ABC = 90^\circ\) (or for setting up the cosine rule with \(\cos(90^\circ)\)).
A1: For a correct expression for the distance, e.g., \(AC = \sqrt{8^2 + 12^2}\).
A1: For \(AC \approx 14.4\text{ km}\) (accept \(\sqrt{208}\) or \(14.42\)).

(b)
M1: For attempting to find the angle \(\angle BAC\) using a correct trigonometric ratio (e.g., \(\tan(\theta) = \frac{12}{8}\) or sine/cosine rule).
A1: For finding \(\angle BAC \approx 56.3^\circ\).
A1: For the final bearing of \(116^\circ\) (accept \(116.3^\circ\)).

(a) Let \(X\) be the mass of an orange, where \(X \sim N(180, \sigma^2)\).
We are given \(P(X < 150) = 0.15\).

Standardizing this gives:
\(P\left(Z < \frac{150 - 180}{\sigma}\right) = 0.15\)
\(P\left(Z < \frac{-30}{\sigma}\right) = 0.15\)

Using the inverse normal function on a GDC:
\(z_{0.15} \approx -1.03643\)

Therefore:
\(\frac{-30}{\sigma} = -1.03643\)
\(\sigma = \frac{30}{1.03643} \approx 28.9454\text{ g}\)

To 3 significant figures, \(\sigma \approx 28.9\text{ g}\).

(b) We want to find \(P(X > 210)\).
Since the normal distribution is symmetric about the mean \(\mu = 180\):
The distance from \(150\) to the mean is \(180 - 150 = 30\).
The distance from \(210\) to the mean is \(210 - 180 = 30\).

By symmetry:
\(P(X > 210) = P(X < 150) = 0.15\)

Alternatively, using \(\sigma = 28.9454\) on a GDC:
\(P(X > 210) \approx 0.150\).

(a)
M1: For standardizing the variable: \(\frac{150 - 180}{\sigma}\).
A1: For obtaining \(z \approx -1.036\) from GDC.
A1: For \(\sigma \approx 28.9\text{ g}\) (accept \(28.95\) or \(29.0\)).

(b)
M1: For recognizing the symmetry about the mean \(\mu = 180\) or setting up the GDC normal cumulative distribution with \(\text{lower} = 210\), \(\text{mean} = 180\), and \(\text{SD} = 28.945\).
A2: For finding the correct probability of \(0.150\) (accept \(0.15\)).

(a) The particle is at rest when \(v(t) = 0\).
Using a GDC to find the roots of \(t^2 \sin(2t) - t + 3 = 0\) in the interval \(0 \le t \le 6\):
\(t_1 \approx 1.77\text{ s}\)
\(t_2 \approx 3.15\text{ s}\)
\(t_3 \approx 4.68\text{ s}\)

(b) The acceleration is the derivative of the velocity function:
\(a(t) = v'(t)\)

Using the numerical derivative feature on a GDC at \(t = 3\):
\(a(3) \approx 14.6\text{ m s}^{-2}\)

Alternatively, differentiating by hand:
\(v'(t) = 2t \sin(2t) + 2t^2 \cos(2t) - 1\)
\(v'(3) = 6 \sin(6) + 18 \cos(6) - 1 \approx 14.607\text{ m s}^{-2}\)

To 3 significant figures, \(a(3) \approx 14.6\text{ m s}^{-2}\).

(c) The total distance traveled is given by:
\(\text{Distance} = \int_{0}^{6} |v(t)| dt\)

Using the GDC numerical integration feature to evaluate \(\int_{0}^{6} |t^2 \sin(2t) - t + 3| dt\):
\(\text{Distance} \approx 53.8\text{ m}\).

(a)
M1: For setting \(v(t) = 0\) (can be shown by sketching the graph of velocity).
A1: For all three correct values of \(t\): \(1.77\), \(3.15\), \(4.68\) (accept \(1.771\), \(3.148\), \(4.675\)).

(b)
M1: For recognizing that \(a(t) = v'(t)\).
A1: For \(14.6\text{ m s}^{-2}\) (accept \(14.61\)).

(c)
M1: For setting up the integral of the absolute value of the velocity function: \(\int_{0}^{6} |v(t)| dt\).
A1: For \(53.8\text{ m}\) (accept \(53.81\)).

(a) Let \(X\) be the weight of a chocolate bar. We have \(X \sim N(150, 3.5^2)\). Using a GDC to find \(P(X < 145)\): \(z = \frac{145-150}{3.5} = -1.42857\). Thus, \(P(X < 145) \approx 0.0765637\). To 3 significant figures, this is 0.0766. (b) Let \(Y\) be the number of chocolate bars weighing less than 145 grams in a sample of 10. Then \(Y \sim B(10, 0.0765637)\). We need to find \(P(Y \ge 2) = 1 - P(Y \le 1)\). Using the binomial cumulative distribution on a GDC: \(P(Y \le 1) \approx 0.824490\). Therefore, \(P(Y \ge 2) = 1 - 0.824490 = 0.175510 \approx 0.176\). (c) We are given \(P(X > k) = 0.15\), which is equivalent to \(P(X \le k) = 0.85\). Using the inverse normal function on a GDC: \(k \approx 153.6275\). To 3 significant figures, \(k = 154\). (d) We want to find the probability that a premium chocolate bar weighs more than 155 grams. This is a conditional probability: \(P(X > 155 \mid X > k)\). Since \(155 > 153.6275\), the intersection \(P(X > 155 \cap X > 153.6275) = P(X > 155)\). We calculate \(P(X > 155)\) using a GDC: \(P(X > 155) \approx 0.0765637\). Then, \(P(X > 155 \mid X > k) = \frac{P(X > 155)}{P(X > k)} = \frac{0.0765637}{0.15} \approx 0.510425\). To 3 significant figures, the probability is 0.510.

(a) M1 for recognizing normal distribution and setting up \(P(X < 145)\). A1 for finding the correct z-value of -1.43 (or equivalent GDC parameters). A1 for 0.0766. (b) M1 for identifying the binomial distribution parameters \(n=10\) and \(p=0.07656...\). M1 for the correct probability expression \(1 - P(Y \le 1)\) or equivalent. A1 for finding \(P(Y \le 1) \approx 0.8245\). A1 for 0.176. (c) M1 for setting up \(P(X \le k) = 0.85\) or \(P(X > k) = 0.15\). A1 for GDC inverse normal setup or finding \(z \approx 1.036\). A1 for \(k = 154\) (accept 153.6). (d) M1 for realizing a conditional probability is needed: \(P(X > 155 \mid X > k)\). M1 for identifying that \(P(X > 155 \cap X > k) = P(X > 155)\). A1 for calculating \(P(X > 155) \approx 0.0766\). M1 for calculating the quotient \(\frac{0.07656...}{0.15}\). A1 for 0.510.

(a) The volume of a cylinder is given by \(V = \pi r^2 h = 500\), which gives \(h = \frac{500}{\pi r^2}\). The total surface area of a closed cylinder is \(A = 2\pi r^2 + 2\pi r h\). Substituting \(h\) into the surface area equation: \(A = 2\pi r^2 + 2\pi r \left(\frac{500}{\pi r^2}\right) = 2\pi r^2 + \frac{1000}{r}\). (b) To find the minimum surface area, we differentiate \(A\) with respect to \(r\): \(\frac{dA}{dr} = 4\pi r - \frac{1000}{r^2}\). Setting \(\frac{dA}{dr} = 0\) gives \(4\pi r^3 = 1000 \implies r^3 = \frac{250}{\pi}\). Thus, \(r = \left(\frac{250}{\pi}\right)^{1/3} \approx 4.30127\text{ cm}\). The minimum surface area is \(A(4.30127) = 2\pi (4.30127)^2 + \frac{1000}{4.30127} \approx 116.24 + 232.49 = 348.73\text{ cm}^2\). To 3 significant figures, \(r = 4.30\text{ cm}\) and \(A = 349\text{ cm}^2\). (c) (i) The cost is given by the sum of the top/bottom cost and the curved side cost: \(C = 0.05(2\pi r^2) + 0.03\left(\frac{1000}{r}\right) = 0.10\pi r^2 + \frac{30}{r}\). (ii) To minimize cost, we differentiate \(C\) with respect to \(r\): \(\frac{dC}{dr} = 0.20\pi r - \frac{30}{r^2}\). Setting \(\frac{dC}{dr} = 0\) gives \(0.20\pi r^3 = 30 \implies r^3 = \frac{150}{\pi}\). Thus, \(r = \left(\frac{150}{\pi}\right)^{1/3} \approx 3.6278\text{ cm}\). The minimum cost is \(C(3.6278) = 0.10\pi (3.6278)^2 + \frac{30}{3.6278} \approx 4.135 + 8.269 = 12.404\) dollars. To 3 significant figures, \(r = 3.63\text{ cm}\) and the minimum cost is $12.40. (d) Since the local minimum of the cost function occurs at \(r \approx 3.63\text{ cm}\) (which is within the interval \([3, 6]\)), the cost function decreases on \([3, 3.63]\) and increases on \([3.63, 6]\). The maximum cost must therefore occur at one of the boundaries. We check the boundary values: At \(r = 3\): \(C(3) = 0.10\pi(3)^2 + \frac{30}{3} = 0.9\pi + 10 \approx 12.827\) dollars. At \(r = 6\): \(C(6) = 0.10\pi(6)^2 + \frac{30}{6} = 3.6\pi + 5 \approx 16.310\) dollars. Since \(16.310 > 12.827\), the maximum possible cost is $16.31 (occurring when \(r = 6\)).

(a) M1 for expressing \(h\) in terms of \(r\) using the volume formula. M1 for writing down the total surface area formula for a closed cylinder. A1 for substituting and simplifying to obtain the given expression. (b) M1 for differentiating \(A\) with respect to \(r\) (or using GDC to find the minimum point). A1 for finding \(r \approx 4.30\text{ cm}\). M1 for substituting the value of \(r\) back into the area equation. A1 for finding the area \(349\text{ cm}^2\). (c) (i) M1 for expressing the cost equation using the areas of the two parts. A1 for simplifying to \(C = 0.1\pi r^2 + \frac{30}{r}\). (ii) M1 for differentiating \(C\) (or setting up GDC minimization). A1 for \(r \approx 3.63\text{ cm}\). A1 for finding minimum cost $12.40. (d) M1 for evaluating the cost function at the endpoints \(r=3\) and \(r=6\). A1 for finding \(C(3) \approx 12.83\) and \(C(6) \approx 16.31\). A1 for concluding that the maximum cost is $16.31.

(a) The midline of the wave is \(d = \frac{\text{Max} + \text{Min}}{2} = \frac{12 + 4}{2} = 8\). The amplitude of the wave is \(a = \frac{\text{Max} - \text{Min}}{2} = \frac{12 - 4}{2} = 4\). The time between high tide and low tide is 6 hours, which represents half of the full cycle. Thus, the period of the function is \(2 \times 6 = 12\) hours. This gives \(b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{12} = \frac{\pi}{6}\). Since high tide (the maximum) occurs at \(t = 3\), the cosine function is shifted to the right by 3 units, so \(c = 3\). Thus, \(H(t) = 4 \cos\left(\frac{\pi}{6}(t - 3)\right) + 8\). (b) Maxima occur at intervals of 12 hours starting at \(t = 3\). Within the domain \([0, 24]\), these occur at \(t = 3\) and \(t = 15\). The coordinates of the local maxima are \((3, 12)\) and \((15, 12)\). Minima occur at intervals of 12 hours starting at \(t = 9\). Within the domain \([0, 24]\), these occur at \(t = 9\) and \(t = 21\). The coordinates of the local minima are \((9, 4)\) and \((21, 4)\). (c) We solve the inequality \(H(t) \ge 9.5\) for \(0 \le t \le 24\). We first find the intersection points by solving \(4 \cos\left(\frac{\pi}{6}(t - 3)\right) + 8 = 9.5\). Using a GDC to find the solutions in the domain \([0, 24]\): \(t_1 \approx 0.734\) hours, \(t_2 \approx 5.266\) hours, \(t_3 \approx 12.734\) hours, \(t_4 \approx 17.266\) hours. The boat can enter safely when \(t \in [0.734, 5.266]\) and \(t \in [12.734, 17.266]\). The duration of the first interval is \(5.266 - 0.734 = 4.532\) hours. The duration of the second interval is \(17.266 - 12.734 = 4.532\) hours. Total safe time \(= 4.532 + 4.532 = 9.064\) hours. To 3 significant figures, this is 9.06 hours. (d) To find the rate of change of the height at 12:00, we evaluate the derivative \(H'(t)\) at \(t = 12\). Differentiating \(H(t)\): \(H'(t) = -4 \left(\frac{\pi}{6}\right) \sin\left(\frac{\pi}{6}(t - 3)\right) = -\frac{2\pi}{3} \sin\left(\frac{\pi}{6}(t - 3)\right)\). At \(t = 12\): \(H'(12) = -\frac{2\text{\pi}}{3} \sin\left(\frac{9\pi}{6}\right) = -\frac{2\text{\pi}}{3} \sin\left(\frac{3\pi}{2}\right) = -\frac{2\pi}{3}(-1) = \frac{2\pi}{3} \approx 2.09\text{ meters per hour}\).

(a) M1 for \(d = 8\). M1 for \(a = 4\). M1 for determining the period is 12 hours. A1 for \(b = \frac{\pi}{6}\). A1 for \(c = 3\). (b) A1 for identifying both maxima at \((3, 12)\) and \((15, 12)\). A1 for identifying both minima at \((9, 4)\) and \((21, 4)\). A1 for correct formatting of coordinates and respecting the domain boundary. (c) M1 for setting up \(H(t) \ge 9.5\). A1 for finding the first two intersection points \(t \approx 0.734\) and \(t \approx 5.266\). A1 for finding the next two intersection points \(t \approx 12.734\) and \(t \approx 17.266\). M1 for computing the difference to find the durations of safe windows. A1 for the total safe time of 9.06 hours (accept 9.06 to 9.07). (d) M1 for finding the derivative function \(H'(t)\) or utilizing GDC numerical derivative at \(t=12\). A1 for \(2.09\text{ m/h}\) (or \(\frac{2\pi}{3}\)).