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

To find \(k\) and \(m\), we use two properties of probability distributions. First, the sum of all probabilities is 1: \(\sum \text{P}(X=x) = 1 \implies k(1) + k(2) + k(3) + k(4) + m = 1 \implies 10k + m = 1\). Second, the expected value is 3.8: \(\text{E}(X) = \sum x \cdot \text{P}(X=x) = 1(k) + 2(2k) + 3(3k) + 4(4k) + 5(m) = 30k + 5m = 3.8\). We now solve these simultaneous equations. From the first equation, \(m = 1 - 10k\). Substituting this into the second equation: \(30k + 5(1 - 10k) = 3.8 \implies 30k + 5 - 50k = 3.8 \implies -20k = -1.2 \implies k = 0.06\). Substituting \(k = 0.06\) back to find \(m\): \(m = 1 - 10(0.06) = 0.4\).

(M1) for setting up the sum of probabilities equation: \(10k + m = 1\). (A1) for the correct sum of probabilities equation. (M1) for setting up the expectation equation: \(30k + 5m = 3.8\). (A1) for the correct expectation equation. (M1) for attempting to solve the simultaneous equations. (A1) for correct values: \(k = 0.06\) and \(m = 0.4\).

Let \(u = \cos(x^2)\). The derivative with respect to \(x\) is \(\frac{\text{d}u}{\text{d}x} = -2x \sin(x^2)\), which gives \(x \sin(x^2) \text{d}x = -\frac{1}{2} \text{d}u\). Next, we find the new limits of integration: when \(x = 0\), \(u = \cos(0) = 1\), and when \(x = \sqrt{\pi}\), \(u = \cos(\pi) = -1\). Substituting these into the integral gives: \(\int_{1}^{-1} e^u \left(-\frac{1}{2} \text{d}u\right) = \frac{1}{2} \int_{-1}^{1} e^u \text{d}u\). Evaluating this integral yields: \(\frac{1}{2} [e^u]_{-1}^{1} = \frac{1}{2}(e^1 - e^{-1}) = \frac{1}{2}(e - e^{-1})\).

(M1) for attempting integration by substitution with \(u = \cos(x^2)\). (A1) for finding the correct differential relation: \(\text{d}u = -2x \sin(x^2) \text{d}x\). (A1) for finding the correct new limits of integration: 1 and -1. (M1) for writing the integral entirely in terms of \(u\) with correct signs. (A1) for correct integration to \(\frac{1}{2}[e^u]\). (A1) for the correct final exact value \(\frac{1}{2}(e - e^{-1})\) or equivalent.

We first simplify the terms using the properties of logarithms: \(u_1 = \ln x\), \(u_2 = \ln(x^{1/2}) = \frac{1}{2} \ln x\), and \(u_3 = \ln(x^{1/4}) = \frac{1}{4} \ln x\). (a) The common ratio \(r\) is \(r = \frac{u_2}{u_1} = \frac{\frac{1}{2} \ln x}{\ln x} = \frac{1}{2}\). (b) Since \(|r| < 1\), the sum to infinity \(S_\infty\) is defined and given by: \(S_\infty = \frac{u_1}{1 - r} = 8 \implies \frac{\ln x}{1 - 1/2} = 8 \implies 2 \ln x = 8 \implies \ln x = 4 \implies x = e^4\).

(a) (M1) for applying logarithm rules to simplify terms, e.g., \(\ln(\sqrt{x}) = \frac{1}{2}\ln x\). (A1) for finding \(r = \frac{1}{2}\). (b) (M1) for substituting into the sum to infinity formula: \(\frac{\ln x}{1 - 1/2} = 8\). (A1) for simplifying to \(2\ln x = 8\). (M1) for isolating \(\ln x = 4\). (A1) for the correct final answer \(x = e^4\).

Using the identity \(\cos^2(x) = 1 - \sin^2(x)\), we rewrite the equation in terms of \(\sin(x)\): \(2(1 - \sin^2(x)) + 3\sin(x) - 3 = 0 \implies 2 - 2\sin^2(x) + 3\sin(x) - 3 = 0 \implies 2\sin^2(x) - 3\sin(x) + 1 = 0\). Let \(y = \sin(x)\). This gives the quadratic equation \(2y^2 - 3y + 1 = 0\), which factors to \((2y - 1)(y - 1) = 0\). Therefore, we have \(\sin(x) = \frac{1}{2}\) or \(\sin(x) = 1\). Within the interval \(0 \le x \le 2\pi\): from \(\sin(x) = 1\), we get \(x = \frac{\pi}{2}\); from \(\sin(x) = \frac{1}{2}\), we get \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

(M1) for substituting \(\cos^2(x) = 1 - \sin^2(x)\). (A1) for obtaining the correct quadratic equation: \(2\sin^2(x) - 3\sin(x) + 1 = 0\). (M1) for factoring or solving the quadratic to get \(\sin(x) = \frac{1}{2}\) or \(\sin(x) = 1\). (A1) for \(x = \frac{\pi}{2}\). (A1)(A1) for \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

(a) The vertical asymptote occurs when the denominator is zero: \(2x + b = 0 \implies x = -\frac{b}{2}\). Since the vertical asymptote is at \(x = -3\), we have \(-\frac{b}{2} = -3 \implies b = 6\). The horizontal asymptote is determined by the ratio of the leading coefficients: \(y = \frac{a}{2}\). Since the horizontal asymptote is at \(y = 2\), we have \(\frac{a}{2} = 2 \implies a = 4\). Thus, \(f(x) = \frac{4x + 3}{2x + 6}\). (b) To find the inverse, let \(x = \frac{4y + 3}{2y + 6}\). Multiplying both sides by the denominator: \(x(2y + 6) = 4y + 3 \implies 2xy + 6x = 4y + 3\). Grouping terms with \(y\): \(2xy - 4y = 3 - 6x \implies y(2x - 4) = 3 - 6x \implies y = \frac{3 - 6x}{2x - 4}\). Thus, \(f^{-1}(x) = \frac{3 - 6x}{2x - 4}\).

(a) (M1) for equating the denominator to zero at \(x = -3\) to find \(b\). (A1) for \(b = 6\). (A1) for \(a = 4\) from horizontal asymptote analysis. (b) (M1) for swapping \(x\) and \(y\) (or attempting to make \(x\) the subject). (M1) for correct algebraic steps to group the variable of interest: \(2xy - 4y = 3 - 6x\). (A1) for the correct expression \(f^{-1}(x) = \frac{3 - 6x}{2x - 4}\) (or equivalent, such as \(\frac{6x - 3}{4 - 2x}\)).

First, find the gradient function of the curve by differentiating \(y\) with respect to \(x\): \(\frac{\text{d}y}{\text{d}x} = 3x^2 - 6x - 9\). Since the tangent is parallel to the line \(y = 15x - 7\), their gradients must be equal, so we set the derivative to 15: \(3x^2 - 6x - 9 = 15 \implies 3x^2 - 6x - 24 = 0\). Dividing by 3 gives \(x^2 - 2x - 8 = 0\), which factors to \((x - 4)(x + 2) = 0\). This yields \(x = 4\) and \(x = -2\). Substituting these back into the original curve equation to find the \(y\)-coordinates: for \(x = 4\), \(y = 4^3 - 3(4^2) - 9(4) + 5 = 64 - 48 - 36 + 5 = -15\); for \(x = -2\), \(y = (-2)^3 - 3(-2)^2 - 9(-2) + 5 = -8 - 12 + 18 + 5 = 3\). The coordinates of the points are \((4, -15)\) and \((-2, 3)\).

(M1)(A1) for finding the correct derivative: \(\frac{\text{d}y}{\text{d}x} = 3x^2 - 6x - 9\). (M1) for equating the derivative to 15: \(3x^2 - 6x - 9 = 15\). (A1) for finding both correct \(x\) values: \(x = 4\) and \(x = -2\). (A1) for finding the point \((4, -15)\). (A1) for finding the point \((-2, 3)\).

(a) Using the properties of logarithms, for x > 0 we have f(x) = x \ln(x^2) = 2x \ln x. Applying the product rule: f'(x) = \frac{d}{dx}(2x) \cdot \ln x + 2x \cdot \frac{d}{dx}(\ln x) = 2 \ln x + 2x \left(\frac{1}{x}\right) = 2 \ln x + 2 = 2(1 + \ln x).

(b) For a stationary point, we set f'(x) = 0. This gives 2(1 + \ln x) = 0 \implies \ln x = -1 \implies x = e^{-1} = \frac{1}{e}. Substituting this value into f(x) gives the y-coordinate: f(e^{-1}) = e^{-1} \ln((e^{-1})^2) = e^{-1} \ln(e^{-2}) = -2e^{-1} = -\frac{2}{e}. Since f'(x) < 0 for 0 < x < e^{-1} and f'(x) > 0 for x > e^{-1}, this point is a local minimum. The exact coordinates of the local minimum are (e^{-1}, -2e^{-1}).

(c) Since f(x) \ge 0 for x \ge 1, the area of R is given by Area = \int_{1}^{e} 2x \ln x \, dx. We use integration by parts with u = \ln x and dv = 2x \, dx. This gives du = \frac{1}{x} \, dx and v = x^2. Thus, \int 2x \ln x \, dx = x^2 \ln x - \int x \, dx = x^2 \ln x - \frac{x^2}{2} + C. Evaluating this from 1 to e: [x^2 \ln x - \frac{x^2}{2}]_{1}^{e} = (e^2 \ln e - \frac{e^2}{2}) - (1^2 \ln 1 - \frac{1}{2}) = (e^2 - \frac{e^2}{2}) + \frac{1}{2} = \frac{e^2 + 1}{2}.

(d) The volume of revolution V is given by V = \pi \int_{1}^{e^2} [g(x)]^2 \, dx = \pi \int_{1}^{e^2} \frac{(\ln x)^2}{x} \, dx. We use the substitution u = \ln x, which gives du = \frac{1}{x} \, dx. Changing the limits of integration: when x = 1, u = 0; when x = e^2, u = 2. This yields V = \pi \int_{0}^{2} u^2 \, du = \pi [\frac{u^3}{3}]_{0}^{2} = \frac{8\pi}{3}.

(a) M1 for attempting to use the product rule. A1 for correct derivatives of both terms. AG for clearly showing the steps leading to the given expression 2(1 + \ln x).

(b) M1 for setting f'(x) = 0. A1 for solving x = e^{-1}. M1 for substituting their x-value into f(x). A1 for correct coordinates (e^{-1}, -2e^{-1}).

(c) M1 for setting up the correct definite integral. M1 for attempting integration by parts. A1 for the correct antiderivative. A1 for the correct final area \frac{e^2 + 1}{2}.

(d) M1 for setting up the volume integral with \pi. M1 for attempting substitution u = \ln x. A1 for the correct antiderivative in terms of u (or x). A1 for the exact volume \frac{8\pi}{3}.

(a) Since the sum of all probabilities in a probability distribution is 1, we have: 0.1 + a + b + 0.2 = 1 \implies a + b = 0.7. Using the expectation formula E(X) = \sum x \cdot P(X=x) = 2.6: 1(0.1) + 2a + 3b + 4(0.2) = 2.6 \implies 0.1 + 2a + 3b + 0.8 = 2.6 \implies 2a + 3b = 1.7. We can solve these simultaneous equations by expressing a = 0.7 - b. Substituting into the second equation: 2(0.7 - b) + 3b = 1.7 \implies 1.4 + b = 1.7 \implies b = 0.3. This gives a = 0.7 - 0.3 = 0.4.

(b) To find the variance, we first calculate E(X^2): E(X^2) = 1^2(0.1) + 2^2(0.4) + 3^2(0.3) + 4^2(0.2) = 0.1 + 1.6 + 2.7 + 3.2 = 7.6. Using the formula Var(X) = E(X^2) - [E(X)]^2: Var(X) = 7.6 - (2.6)^2 = 7.6 - 6.76 = 0.84.

(c) The possible outcomes (X_1, X_2) where X_1 + X_2 \ge 6 are: (2, 4), (3, 3), (4, 2), (3, 4), (4, 3), and (4, 4). Since the two observations are independent, the probabilities are: P(2, 4) = 0.4 \times 0.2 = 0.08, P(3, 3) = 0.3 \times 0.3 = 0.09, P(4, 2) = 0.2 \times 0.4 = 0.08, P(3, 4) = 0.3 \times 0.2 = 0.06, P(4, 3) = 0.2 \times 0.3 = 0.06, P(4, 4) = 0.2 \times 0.2 = 0.04. Summing these probabilities: P(X_1 + X_2 \ge 6) = 0.08 + 0.09 + 0.08 + 0.06 + 0.06 + 0.04 = 0.41.

(d) We are seeking the conditional probability P(X_1 = 4 | X_1 + X_2 \ge 6) = \frac{P(X_1 = 4 \cap (X_1 + X_2 \ge 6))}{P(X_1 + X_2 \ge 6)}. The outcomes where X_1 = 4 and X_1 + X_2 \ge 6 are (4, 2), (4, 3), and (4, 4). The combined probability of these outcomes is P(4, 2) + P(4, 3) + P(4, 4) = 0.08 + 0.06 + 0.04 = 0.18. Thus, the conditional probability is \frac{0.18}{0.41} = \frac{18}{41}.

(a) M1 for using the fact that probabilities sum to 1. A1 for setting up the equation a + b = 0.7. M1 for setting up the expectation equation. A1 for obtaining 2a + 3b = 1.7. A1 for correctly solving the simultaneous equations to obtain the given values.

(b) M1 for attempting to find E(X^2). A1 for finding E(X^2) = 7.6. A1 for calculating Var(X) = 0.84.

(c) M1 for identifying the correct combinations for a sum of at least 6. M1 for calculating individual probabilities. A2 for summing the correct values to get 0.41.

(d) M1 for stating the correct conditional probability expression. A1 for identifying and calculating the numerator probability of 0.18. A1 for the final answer of 18/41 (or 0.439).

(a) Using the double-angle identity \cos(2x) = 1 - 2\sin^2(x), we substitute this into the expression for f(x): f(x) = 2(1 - 2\sin^2(x)) - 4\sin(x) + 1 = 2 - 4\sin^2(x) - 4\sin(x) + 1 = -4\sin^2(x) - 4\sin(x) + 3.

(b) To solve f(x) = 0, we solve -4\sin^2(x) - 4\sin(x) + 3 = 0. Let u = \sin(x). The quadratic equation becomes 4u^2 + 4u - 3 = 0. Factoring this yields (2u - 1)(2u + 3) = 0. Therefore, u = \frac{1}{2} or u = -\frac{3}{2}. Since the range of the sine function is [-1, 1], we reject u = -\frac{3}{2}. Solving \sin(x) = \frac{1}{2} on the interval 0 \le x \le 2\pi gives the solutions x = \frac{\pi}{6} and x = \frac{5\pi}{6}.

(c) To find the stationary points, we differentiate f(x): f'(x) = -8\sin(x)\cos(x) - 4\cos(x) = -4\cos(x)(2\sin(x) + 1). Setting f'(x) = 0 gives two possibilities:
1. \cos(x) = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}
2. 2\sin(x) + 1 = 0 \implies \sin(x) = -\frac{1}{2} \implies x = \frac{7\pi}{6}, \frac{11\pi}{6}.

Now, we evaluate the y-coordinates:
- For x = \frac{\pi}{2}: f(\frac{\pi}{2}) = -4(1)^2 - 4(1) + 3 = -5.
- For x = \frac{3\pi}{2}: f(\frac{3\pi}{2}) = -4(-1)^2 - 4(-1) + 3 = 3.
- For x = \frac{7\pi}{6}: f(\frac{7\pi}{6}) = -4(-\frac{1}{2})^2 - 4(-\frac{1}{2}) + 3 = 4.
- For x = \frac{11\pi}{6}: f(\frac{11\pi}{6}) = -4(-\frac{1}{2})^2 - 4(-\frac{1}{2}) + 3 = 4.

To classify the stationary points, we find the second derivative f''(x) = -8\cos(2x) + 4\sin(x):
- At x = \frac{\pi}{2}: f''(\frac{\pi}{2}) = -8\cos(\pi) + 4\sin(\frac{\pi}{2}) = 8 + 4 = 12 > 0 \implies (\frac{\pi}{2}, -5) is a local minimum.
- At x = \frac{3\pi}{2}: f''(\frac{3\pi}{2}) = -8\cos(3\pi) + 4\sin(\frac{3\pi}{2}) = 8 - 4 = 4 > 0 \implies (\frac{3\pi}{2}, 3) is a local minimum.
- At x = \frac{7\pi}{6}: f''(\frac{7\pi}{6}) = -8\cos(\frac{7\pi}{3}) + 4\sin(\frac{7\pi}{6}) = -4 - 2 = -6 < 0 \implies (\frac{7\pi}{6}, 4) is a local maximum.
- At x = \frac{11\pi}{6}: f''(\frac{11\pi}{6}) = -8\cos(\frac{11\pi}{3}) + 4\sin(\frac{11\pi}{6}) = -4 - 2 = -6 < 0 \implies (\frac{11\pi}{6}, 4) is a local maximum.

(a) M1 for substituting \cos(2x) = 1 - 2\sin^2(x). A1 for correct algebraic simplification. AG for clearly showing the steps leading to the given expression.

(b) M1 for attempting to factorize the quadratic equation. A1 for finding \sin(x) = 1/2. R1 for rejecting \sin(x) = -3/2. A1 for x = \pi/6. A1 for x = 5\pi/6.

(c) M1 for finding the derivative f'(x). A1 for setting f'(x) = 0 and finding the four critical values. A1 for finding the correct y-values for each critical value. M1 for finding the second derivative or analyzing first derivative signs. A1 for classifying (\pi/2, -5) and (3\pi/2, 3) as local minimums. A2 for classifying (7\pi/6, 4) and (11\pi/6, 4) as local maximums.

Let \(X\) be the weight of a randomly chosen chocolate bar. We have \(X \sim N(150, 6^2)\).

(a) Using a graphic display calculator (GDC):
\(P(X > 155) \approx 0.202328\)
To three significant figures, \(P(X > 155) = 0.202\).

(b) Let \(Y\) be the number of chocolate bars in the packet that weigh more than 155 grams. Then \(Y \sim B(10, p)\), where \(p = 0.202328...\).
We want to find \(P(Y \ge 3)\):
\(P(Y \ge 3) = 1 - P(Y \le 2)\)
Using a GDC with binomial cumulative probability:
\(P(Y \le 2) \approx 0.67228...\)
\(P(Y \ge 3) = 1 - 0.67228... \approx 0.3277...\)
To three significant figures, \(P(Y \ge 3) = 0.328\).

(a)
(M1) for setting up the normal distribution calculation, i.e., \(P(X > 155)\) with mean 150 and standard deviation 6.
(A1) for \(0.202\) (accept \(0.202328...\))

(b)
(M1) for identifying the binomial distribution model, \(Y \sim B(10, 0.202...)\).
(M1) for attempting to calculate \(1 - P(Y \le 2)\) or summing individual binomial probabilities from 3 to 10.
(A1) for \(P(Y \le 2) \approx 0.672\).
(A1) for \(0.328\) (accept \(0.3277...\) or \(0.328\) based on rounded \(p\)).

(a) To find the points of intersection, we set \(f(x) = g(x)\):
\(3 \cos(x) = e^{0.2x}\)
Using a GDC to find the intersection points in the interval \([-\pi, \pi]\), we obtain:
\(x_1 \approx -1.311\) and \(x_2 \approx 1.137\)
To three significant figures, \(x = -1.31\) and \(x = 1.14\).

(b) The area \(A\) of the enclosed region is given by:
\(A = \int_{-1.3112...}^{1.1374...} (3 \cos(x) - e^{0.2x}) \, dx\)
Using the integration function on a GDC:
\(A \approx 3.192\)
To three significant figures, \(A = 3.19\).

(a)
(M1) for setting up the equation \(3 \cos(x) = e^{0.2x}\) or sketching both curves.
(A1) for both \(x\)-coordinates correct to 3 sf: \(x = -1.31\) and \(x = 1.14\) (accept \(-1.311\) and \(1.137\)).

(b)
(M1) for setting up the correct integral with their limits from part (a).
(M1) for the correct integrand \(3 \cos(x) - e^{0.2x}\).
(A1) for showing the antiderivative \(3 \sin(x) - 5e^{0.2x}\) (if worked analytically, though not strictly required) OR for indicating GDC usage.
(A1) for the correct area of \(3.19\) (accept \(3.192...\)).

(a) To find the length \(AC\), we use the cosine rule:
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 120^2 + 85^2 - 2(120)(85)\cos(72^\circ)\)
\(AC^2 = 14400 + 7225 - 20400 \cos(72^\circ)\)
\(AC^2 = 21625 - 6303.95...\)
\(AC^2 = 15321.04...\)
\(AC = \sqrt{15321.04...} \approx 123.78\)
To three significant figures, \(AC = 124\text{ m}\).

(b) The area \(A\) of the triangle is given by:
\(A = \frac{1}{2} a c \sin(B)\)
\(A = \frac{1}{2} (120)(85) \sin(72^\circ)\)
\(A = 5100 \sin(72^\circ)\)
\(A \approx 4850.39...\)
To three significant figures, the area is \(4850\text{ m}^2\).

(a)
(M1) for substituting correctly into the cosine rule: \(120^2 + 85^2 - 2(120)(85)\cos(72^\circ)\).
(A1) for obtaining \(AC^2 \approx 15321.04...\).
(A1) for \(124\) (or \(123.78...\)).

(b)
(M1) for substituting correctly into the area formula: \(\frac{1}{2} (120)(85) \sin(72^\circ)\).
(A1) for obtaining \(4850.39...\).
(A1) for \(4850\) (accept \(4850.4\)).

(a) Using the formula for the \(n\)-th term of an arithmetic sequence:
\(u_{20} = u_1 + (20 - 1)d\)
\(u_{20} = 12 + 19(4.5) = 12 + 85.5 = 97.5\).

(b) The sum of the first \(n\) terms of the arithmetic sequence is:
\(A_n = \frac{n}{2}(2u_1 + (n-1)d) = \frac{n}{2}(24 + (n-1)4.5)\)
\(A_n = \frac{n}{2}(4.5n + 19.5) = 2.25n^2 + 9.75n\).

The sum of the first \(n\) terms of the geometric sequence is:
\(S_n = \frac{v_1(r^n - 1)}{r - 1} = \frac{3(1.15^n - 1)}{1.15 - 1} = 20(1.15^n - 1)\).

We require \(S_n > A_n\):
\(20(1.15^n - 1) > 2.25n^2 + 9.75n\).
Using a GDC to find the values of \(S_n\) and \(A_n\) for integer values of \(n\):
For \(n = 36\):
\(A_{36} = 2.25(36)^2 + 9.75(36) = 3267\)
\(S_{36} = 20(1.15^{36} - 1) \approx 3042.92\)
(Here, \(S_{36} < A_{36}\))

For \(n = 37\):
\(A_{37} = 2.25(37)^2 + 9.75(37) = 3441\)
\(S_{37} = 20(1.15^{37} - 1) \approx 3502.50\)
(Here, \(S_{37} > A_{37}\))

Thus, the smallest value of \(n\) is \(37\).

(a)
(M1) for substituting correctly into the \(n\)-th term formula: \(12 + 19(4.5)\).
(A1) for \(97.5\).

(b)
(M1) for setting up correct expressions for \(A_n\) and \(S_n\) in terms of \(n\).
(M1) for writing the inequality or setting them equal: \(20(1.15^n - 1) > 2.25n^2 + 9.75n\).
(A1) for finding or showing evidence of evaluating values near \(n = 37\) (e.g., table values for \(n = 36\) and \(n = 37\)).
(A1) for \(n = 37\).

(a) The maximum depth (high tide) is \(a + c = 14.6\).
The minimum depth (low tide) is \(-a + c = 8.2\).
Solving these simultaneously:
\(2c = 22.8 \implies c = 11.4\)
\(a = 14.6 - 11.4 = 3.2\)

The time between high tide and low tide is half the period:
\(\frac{\text{Period}}{2} = 6.2 \implies \text{Period} = 12.4\text{ hours}\)
\(b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{12.4} = \frac{5\pi}{31} \approx 0.507\) (to 3 sf).

(b) We require \(d(t) \ge 10\) for \(0 \le t \le 12\):
\(3.2 \cos\left(\frac{5\pi}{31} t\right) + 11.4 \ge 10\)
Using a GDC to find the intersections of \(y = d(t)\) and \(y = 10\) on the interval \([0, 12]\):
\(t_1 \approx 3.994\text{ hours}\)
\(t_2 \approx 8.406\text{ hours}\)

Since \(d(0) = 14.6 > 10\), the depth is at least 10 meters during the intervals:
\(0 \le t \le 3.994\) and \(8.406 \le t \le 12\)

The total duration of safety is:
\(T = 3.994 + (12 - 8.406) = 3.994 + 3.594 = 7.588\text{ hours}\)
To three significant figures, the total time is \(7.59\) hours.

(a)
(M1) for setting up equations to find \(a\) and \(c\): \(a = \frac{14.6 - 8.2}{2}\) or \(c = \frac{14.6 + 8.2}{2}\).
(A1) for \(a = 3.2\) and \(c = 11.4\).
(A1) for \(b = \frac{5\pi}{31}\) (accept \(0.507\) or \(0.5067...\)).

(b)
(M1) for setting up the inequality \(d(t) \ge 10\) or finding critical values \(t \approx 3.99\) and \(t \approx 8.41\).
(M1) for identifying the correct intervals: \([0, 3.994]\) and \([8.406, 12]\).
(A1) for \(7.59\) (accept \(7.588...\)).

(a) Entering the data into a GDC, the product-moment correlation coefficient is:
\(r \approx -0.99833...\)
To three significant figures, \(r = -0.998\).

(b) The equation of the regression line is of the form \(D = aT + b\).
From the GDC:
\(a \approx -0.28709...\)
\(b \approx 14.136...\)
Thus, the equation of the regression line is:
\(D = -0.287T + 14.1\) (or \(D = -0.28709...T + 14.136...\))

(c) Substituting \(T = 17.0\) into the regression equation:
\(D = -0.28709(17.0) + 14.136\)
\(D \approx -4.8805 + 14.136 \approx 9.2555\)
To three significant figures, the dissolved oxygen concentration is \(9.26\text{ mg/L}\).

(a)
(M1) for attempting to use a GDC to find the correlation coefficient.
(A1) for \(-0.998\) (accept \(-0.99833...\)).

(b)
(A1) for gradient \(a \approx -0.287\) (accept \(-0.28709...\)).
(A1) for intercept \(b \approx 14.1\) (accept \(14.136...\)).
(Award max 1 mark if variables other than \(D\) and \(T\) are used, e.g., \(y = -0.287x + 14.1\)).

(c)
(M1) for substituting \(T = 17.0\) into their regression equation.
(A1) for \(9.26\) (accept \(9.255...\) or answers matching their equation from part (b)).

(a)
The particle is at rest when \(v(t) = 0\).
Using a GDC to solve \(8 \cos(t) \ln(t + 1) - 3 = 0\) in the interval \(0 \le t \le 6\), we find:
\(t \approx 0.554\text{ s}\)
\(t \approx 0.998\text{ s}\)
\(t \approx 4.92\text{ s}\) (to 3 significant figures).

(b)
Acceleration is the rate of change of velocity, \(a(t) = v'(t)\).
We can find this value using the numerical derivative feature of the GDC at \(t = 2\):
\(a(2) = v'(2) \approx -9.10\text{ m s}^{-2}\) (to 3 significant figures).
Alternatively, differentiating analytically:
\(v'(t) = 8 \left( -\sin(t)\ln(t+1) + \frac{\cos(t)}{t+1} \right)\)
At \(t = 2\):
\(v'(2) = 8 \left( -\sin(2)\ln(3) + \frac{\cos(2)}{3} \right) \approx 8(-0.9093 \times 1.0986 - 0.1387) \approx -9.10\text{ m s}^{-2}\).

(c)
The total distance traveled is given by the integral of the speed:
\(D = \int_0^6 |v(t)| \text{ d}t\)
Using the GDC to evaluate \(\int_0^6 |8 \cos(t) \ln(t + 1) - 3| \text{ d}t\), we obtain:
\(\text{Distance} \approx 34.2\text{ m}\) (to 3 significant figures).

(d)
(i) The displacement \(s(t)\) at time \(t\) is given by:
\(s(t) = s(0) + \int_0^t v(u) \text{ d}u = 4 + \int_0^t (8 \cos(u) \ln(u + 1) - 3) \text{ d}u\).
Using the GDC to evaluate \(s(6)\):
\(s(6) = 4 + \int_0^6 v(t) \text{ d}t \approx 4 - 18.77 \approx -14.8\text{ m}\) (to 3 significant figures).

(ii) The distance from the origin is given by \(|s(t)|\).
The critical points of \(s(t)\) occur when \(s'(t) = v(t) = 0\), which are at \(t \approx 0.554\), \(t \approx 0.998\), and \(t \approx 4.92\).
We evaluate the displacement at the boundaries and these critical points:
- \(s(0) = 4\)
- \(s(0.554) = 4 + \int_0^{0.554} v(t) \text{ d}t \approx 3.17\)
- \(s(0.998) = 4 + \int_0^{0.998} v(t) \text{ d}t \approx 3.25\)
- \(s(4.92) = 4 + \int_0^{4.92} v(t) \text{ d}t \approx -21.8\)
- \(s(6) \approx -14.8\)

Comparing the absolute values:
- \(|s(0)| = 4\)
- \(|s(0.554)| \approx 3.17\)
- \(|s(0.998)| \approx 3.25\)
- \(|s(4.92)| \approx 21.8\)
- \(|s(6)| \approx 14.8\)

Therefore, the maximum distance of the particle from the origin in the interval is \(21.8\text{ m}\) (at \(t \approx 4.92\text{ s}\)).

(a)
M1 for attempting to solve \(v(t) = 0\) (e.g., sketch or equation).
A2 for all three correct values to 3 s.f. (award A1 if only two are correct).

(b)
M1 for recognizing that \(a(t) = v'(t)\).
M1 for utilizing GDC numerical derivative or attempting analytical product rule.
A1 for \(-9.10\text{ m s}^{-2}\) (accept \(-9.1\)).

(c)
M1 for writing the integral of the absolute value of velocity: \(\int_0^6 |v(t)| \text{ d}t\).
A1 for setting up the correct limits.
A1 for \(34.2\text{ m}\) (accept values in the range \([34.1, 34.3]\)).

(d)(i)
M1 for writing \(s(t) = 4 + \int_0^t v(u) \text{ d}u\).
M1 for attempting to evaluate \(s(6)\) using GDC.
A1 for \(-14.8\text{ m}\) (accept values in the range \([-15.0, -14.6]\)).

(d)(ii)
M1 for identifying that the maximum distance from the origin occurs at a boundary or at a critical point where \(v(t) = 0\).
M1 for finding the value of \(s(t)\) at \(t \approx 4.92\) (or evaluating other candidates).
A1 for \(21.8\text{ m}\) (accept values in the range \([21.6, 22.0]\)).

(a)
(i) Let \(W\) be the weight of a randomly selected apple. We are given:
\(P(W < 110) = 0.12\) and \(P(W > 180) = 0.15 \implies P(W < 180) = 0.85\).
Standardizing these probabilities using \(Z \sim N(0, 1)\):
For \(P(Z < z_1) = 0.12\), the GDC gives \(z_1 \approx -1.17498\).
Hence: \(\frac{110 - \mu}{\sigma} = -1.175\) (or \(\mu - 1.175\sigma = 110\)).
For \(P(Z < z_2) = 0.85\), the GDC gives \(z_2 \approx 1.03643\).
Hence: \(\frac{180 - \mu}{\sigma} = 1.036\) (or \(\mu + 1.036\sigma = 180\)).

(ii) Solving the two equations simultaneously:
Subtracting the first equation from the second:
\(1.03643\sigma - (-1.17498\sigma) = 180 - 110\)
\(2.21141\sigma = 70 \implies \sigma \approx 31.654\text{ g} \approx 31.7\text{ g}\).
Substituting \(\sigma\) back to find \(\mu\):
\(\mu = 180 - 1.03643(31.654) \approx 147.19\text{ g} \approx 147\text{ g}\).

(b)
Using the normal parameters \(\mu = 147.19\) and \(\sigma = 31.654\):
\(P(W > 170) = P\left(Z > \frac{170 - 147.19}{31.654}\right) = P(Z > 0.7206)\).
Using GDC normalcdf(170, E99, 147.19, 31.654):
\(P(W > 170) \approx 0.23558 \approx 0.236\).

(c)
Let \(X\) be the number of Premium apples in a box of 12. \(X\) follows a binomial distribution \(X \sim B(12, 0.23558)\).
(i) \(P(X = 3)\):
Using GDC binompdf(12, 0.23558, 3):
\(P(X = 3) = \binom{12}{3} (0.23558)^3 (1-0.23558)^9 \approx 0.262\).

(ii) \(P(X \ge 2)\):
\(P(X \ge 2) = 1 - P(X \le 1) = 1 - [P(X=0) + P(X=1)]\).
Using GDC 1 - binomcdf(12, 0.23558, 1):
\(P(X \ge 2) \approx 1 - 0.1810 = 0.819\).

(d)
This is a binomial process with \(N = 100\) trials and success probability \(p = P(X \ge 2) \approx 0.819\).
Expected number of boxes \(E(Y) = Np = 100 \times 0.819 = 81.9\).

(e)
We want to find the conditional probability:
\(P(W < 160 \mid W > 130) = \frac{P(130 < W < 160)}{P(W > 130)}\).
Using GDC:
\(P(130 < W < 160) = \text{normalcdf}(130, 160, 147.19, 31.654) \approx 0.3637\).
\(P(W > 130) = \text{normalcdf}(130, E99, 147.19, 31.654) \approx 0.7065\).
Thus, the conditional probability is:
\(\frac{0.3637}{0.7065} \approx 0.515\) (to 3 significant figures).

(a)(i)
M1 for standardizing with \(z\)-values.
A1 for both equations correct (accept equations written with 3 s.f. \(z\)-values, i.e., \(-1.17\) and \(1.04\)).

(a)(ii)
M1 for attempting to solve the simultaneous equations.
A1 for \(\sigma \approx 31.7\text{ g}\) (or \(31.6\)).
A1 for \(\mu \approx 147\text{ g}\).

(b)
M1 for setting up the normal cumulative probability \(P(W > 170)\).
A1 for \(0.236\) (accept \(0.235\) or \(0.237\)).

(c)(i)
M1 for setting up the binomial distribution \(B(12, p)\).
A1 for \(0.262\).

(c)(ii)
M1 for calculating \(1 - P(X \le 1)\) or adding probabilities from \(X = 2\) to \(12\).
A1 for \(0.819\).

(d)
M1 for multiplying \(100\) by their answer from (c)(ii).
A1 for \(81.9\).

(e)
M1 for setting up the conditional probability formula \(\frac{P(130 < W < 160)}{P(W > 130)}\).
A1 for correct numerator and denominator values (e.g., \(0.364\) and \(0.706\)).
A1 for \(0.515\).

(a)
From the given bearings, the angle between \(AB\) and \(AC\) is:
\(\widehat{BAC} = 110^\circ - 40^\circ = 70^\circ\).
Using the Cosine Rule in triangle \(ABC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(\widehat{BAC})\)
\(BC^2 = 800^2 + 1200^2 - 2(800)(1200) \cos(70^\circ)\)
\(BC^2 = 640000 + 1440000 - 1920000 \cos(70^\circ)\)
\(BC^2 \approx 2080000 - 656678.4 = 1423321.6\)
\(BC \approx 1193.03\text{ m} \approx 1190\text{ m}\) (to 3 significant figures).

(b)
Let \(\widehat{ABC}\) be the angle inside the triangle at vertex \(B\).
Using the Sine Rule to find \(\widehat{ABC}\):
\(\frac{\sin(\widehat{ABC})}{AC} = \frac{\sin(\widehat{BAC})}{BC}\)
\(\frac{\sin(\widehat{ABC})}{1200} = \frac{\sin(70^\circ)}{1193.03}\)
\(\sin(\widehat{ABC}) = 1200 \times \frac{\sin(70^\circ)}{1193.03} \approx 0.94518\)
Since \(AC > BC\), \(\widehat{ABC}\) is the largest angle. Looking at the geometry, \(\widehat{ABC}\) is acute:
\(\widehat{ABC} \approx 70.93^\circ\).
Now we determine the bearing of \(C\) from \(B\).
Draw a north line at \(B\). The bearing of \(A\) from \(B\) is \(40^\circ + 180^\circ = 220^\circ\).
Since Tower \(C\) lies to the south-east and to the "left" of the vector \(\vec{BA}\), the bearing of \(C\) from \(B\) is:
\(\text{Bearing} = 220^\circ - \widehat{ABC} = 220^\circ - 70.93^\circ = 149.07^\circ \approx 149^\circ\).

(c)
(i) Since the first drone starts at \(B\) and flies towards \(C\) at \(5\text{ m s}^{-1}\), its distance from \(B\) after \(t\) seconds is:
\(d_1(t) = 5t\).

(ii) The second drone starts at \(A\) and flies towards \(B\) at \(4\text{ m s}^{-1}\).
The initial distance \(AB\) is 800 meters. After \(t\) seconds, the drone has traveled \(4t\) meters from \(A\).
Thus, its distance from \(B\) is:
\(d_2(t) = 800 - 4t\).

(d)
At any time \(t\), the positions of the two drones, \(D_1\) and \(D_2\), form a triangle with vertex \(B\), where the angle \(\widehat{D_1 B D_2}\) is \(\widehat{ABC} \approx 70.93^\circ\).
Applying the Cosine Rule to triangle \(B D_1 D_2\), the distance \(D(t)\) between the drones is:
\(D(t)^2 = (5t)^2 + (800 - 4t)^2 - 2(5t)(800 - 4t) \cos(70.93^\circ)\)
\(D(t) = \sqrt{25t^2 + (800 - 4t)^2 - 10t(800 - 4t) \cos(70.93^\circ)}\)

We enter this function into our GDC and find its minimum.
The quadratic inside the square root is:
\(f(t) = 25t^2 + 640000 - 6400t + 16t^2 - (8000t - 40t^2) \cos(70.93^\circ)\)
Using GDC to plot and find the minimum of \(D(t)\):
The coordinates of the minimum point are \((83.4, 514)\).
Therefore, the distance between the two drones is minimized at \(t \approx 83.4\text{ s}\) (or \(83.3\text{ s}\)), and the minimum distance is \(514\text{ m}\) (or \(515\text{ m}\)).

(a)
M1 for writing a correct Cosine Rule formula to find \(BC\).
A1 for substituting correct values into the formula: \(BC^2 = 800^2 + 1200^2 - 2(800)(1200)\cos(70^\circ)\).
A1 for \(1190\text{ m}\) (accept \(1193\text{ m}\)).

(b)
M1 for a correct Sine Rule setup to find angle \(\widehat{ABC}\).
A1 for \(\widehat{ABC} \approx 70.9^\circ\) (or \(71.0^\circ\)).
M1 for identifying relation of bearing to angle \(\widehat{ABC}\) (e.g., \(220^\circ - 70.9^\circ\) or equivalent diagrammatic argument).
A1 for \(149^\circ\).

(c)(i)
A1 for \(5t\).

(c)(ii)
M1 for recognizing the starting distance is \(800\) and the drone moves towards \(B\).
A1 for \(800 - 4t\).

(d)
M1 for writing down a distance expression between the two drones using Cosine Rule.
A1 for the correct formula: \(D(t) = \sqrt{(5t)^2 + (800 - 4t)^2 - 2(5t)(800 - 4t)\cos(70.93^\circ)}\).
M1 for using GDC to locate the minimum of the function.
A1 for \(t \approx 83.4\text{ s}\) (accept \(83.3\text{ s}\)).
A1 for minimum distance \(\approx 514\text{ m}\) (accept \(515\text{ m}\)).