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

First, find the \(y\)-coordinate of the point of contact by substituting \(x = 2\) into the equation of the curve:
\(y = \ln(2^2 - 3(2) + 3) = \ln(4 - 6 + 3) = \ln(1) = 0\).
So the point of contact is \((2, 0)\).

Next, find the derivative using the chain rule:
\(\frac{dy}{dx} = \frac{2x - 3}{x^2 - 3x + 3}\).

Evaluate the gradient of the tangent at \(x = 2\):
\(m = \frac{2(2) - 3}{2^2 - 3(2) + 3} = \frac{1}{1} = 1\).

Using the point-slope formula \(y - y_1 = m(x - x_1)\):
\(y - 0 = 1(x - 2)\)
\(y = x - 2\).

- **M1**: Attempting to find the \(y\)-coordinate by substituting \(x = 2\) into the equation of the curve.
- **A1**: Finding the point of contact is \((2, 0)\) (or stating \(y = 0\)).
- **M1**: Attempting to differentiate using the chain rule.
- **A1**: Correct derivative expression \(\frac{dy}{dx} = \frac{2x - 3}{x^2 - 3x + 3}\).
- **A1**: Evaluating the gradient at \(x = 2\) to get \(m = 1\).
- **A0.83**: Correct equation of the tangent line: \(y = x - 2\).

(a)
Using the inverse function relationship, we have \(f^{-1}(4) = 5 \iff f(5) = 4\).
Substituting \(x = 5\) into the definition of \(f(x)\):
\(f(5) = \frac{3(5) + a}{5 - 2} = \frac{15 + a}{3}\).
Setting this expression equal to \(4\):
\(\frac{15 + a}{3} = 4 \implies 15 + a = 12 \implies a = -3\).
Thus, \(a = -3\) is shown.

(b)
With \(a = -3\), the function is \(f(x) = \frac{3x - 3}{x - 2}\).
To find the range, we examine the horizontal asymptote of this rational function.
As \(x \to \pm\infty\), \(f(x) \to \frac{3}{1} = 3\).
Since the function is a linear-over-linear rational function, it takes all real values except the value of its horizontal asymptote.
Hence, the range of \(f\) is \(y \in \mathbb{R}, y \neq 3\).

- **M1**: Applying the relationship \(f^{-1}(4) = 5 \iff f(5) = 4\) (or finding the algebraic inverse function first).
- **M1**: Substituting \(x = 5\) into \(f(x)\) and equating to \(4\).
- **A1**: Obtaining the correct value \(a = -3\) with clear algebraic steps.
- **M1**: Attempting to find the horizontal asymptote of \(f\) (e.g., taking the limit as \(x \to \infty\) or making \(x\) the subject of \(y = f(x)\)).
- **A1.83**: Stating the correct range: \(y \in \mathbb{R}, y \neq 3\) (accept equivalent notations like \(f(x) \neq 3\)).

We use the identity \(\cos^2 x = 1 - \sin^2 x\) to write the equation in terms of \(\sin x\) only:
\(2(1 - \sin^2 x) + 3 \sin x - 3 = 0\)
\(2 - 2 \sin^2 x + 3 \sin x - 3 = 0\)
\(-2 \sin^2 x + 3 \sin x - 1 = 0\)
\(2 \sin^2 x - 3 \sin x + 1 = 0\).

Now, we factorize this quadratic in terms of \(\sin x\):
\((2\sin x - 1)(\sin x - 1) = 0\).

This yields two possible values for \(\sin x\):
1) \(\sin x = \frac{1}{2}\)
2) \(\sin x = 1\).

Solving these equations within the interval \(0 \le x \le 2\pi\):
- For \(\sin x = \frac{1}{2}\), the solutions are \(x = \frac{\pi}{6}\) and \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\).
- For \(\sin x = 1\), the solution is \(x = \frac{\pi}{2}\).

Thus, the set of solutions is:
\(x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\).

- **M1**: Substituting the Pythagorean identity \(\cos^2 x = 1 - \sin^2 x\).
- **A1**: Obtaining the correct simplified quadratic equation: \(2 \sin^2 x - 3 \sin x + 1 = 0\).
- **M1**: Factorizing or solving the quadratic equation.
- **A1**: Finding both possible values for the trigonometric ratio: \(\sin x = \frac{1}{2}\) and \(\sin x = 1\).
- **A1.83**: Identifying all three correct angles in the given range: \(x = \frac{\pi}{6}\), \(x = \frac{\pi}{2}\), and \(x = \frac{5\pi}{6}\) (award partial accuracy marks if some are missing or if extra invalid values are present).

Because the terms are consecutive members of an arithmetic sequence, the difference between consecutive terms must be constant:
\(d = u_2 - u_1 = u_3 - u_2\).

Substituting the expressions for the terms:
\((k^2 - 1) - (2k - 1) = (5k - 7) - (k^2 - 1)\).

We simplify both sides of this equation:
\(k^2 - 2k = 5k - 6 - k^2\).

Rearranging all terms to one side yields the quadratic equation:
\(2k^2 - 7k + 6 = 0\).

We factorize the quadratic expression:
\((2k - 3)(k - 2) = 0\).

Solving for \(k\) gives:
\(k = \frac{3}{2}\) or \(k = 2\).

- **M1**: Applying the property of arithmetic sequences that the difference between consecutive terms is constant.
- **A1**: Writing a correct algebraic equation representing this property: \((k^2 - 1) - (2k - 1) = (5k - 7) - (k^2 - 1)\).
- **M1**: Expanding and simplifying the equation.
- **A1**: Arriving at the correct quadratic equation: \(2k^2 - 7k + 6 = 0\).
- **M1**: Factoring the quadratic expression (or applying the quadratic formula).
- **A0.83**: Stating the correct final values for \(k\): \(k = \frac{3}{2}\) (or \(1.5\)) and \(k = 2\).

(a) If \(A\) and \(B\) are mutually exclusive, then \(\mathrm{P}(A \cap B) = 0\).
Using the addition rule:
\(\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) = 0.6 + 0.4 = 1.0\).

(b) If \(A\) and \(B\) are independent, then:
\(\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) = 0.6 \times 0.4 = 0.24\).
Using the addition rule:
\(\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) = 0.6 + 0.4 - 0.24 = 0.76\).

(c) We are given \(\mathrm{P}(A \mid B) = 0.3\).
From the definition of conditional probability:
\(\mathrm{P}(A \cap B) = \mathrm{P}(A \mid B) \times \mathrm{P}(B) = 0.3 \times 0.4 = 0.12\).
Using the addition rule:
\(\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) = 0.6 + 0.4 - 0.12 = 0.88\).

- **A1**: Correct probability for (a): \(1.0\) (or \(1\)).
- **M1**: Applying \(\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)\) for part (b).
- **A1**: Correct probability for (b): \(0.76\).
- **M1**: Applying \(\mathrm{P}(A \cap B) = \mathrm{P}(A \mid B)\mathrm{P}(B)\) for part (c).
- **A1.83**: Correct probability for (c): \(0.88\).

We use the integration by substitution method.
Let \(u = -x^2\).
Then, the derivative is:
\(\frac{du}{dx} = -2x \implies x \, dx = -\frac{1}{2} du\).

Now, change the limits of integration according to the substitution:
- When \(x = 0\), \(u = -(0)^2 = 0\).
- When \(x = 1\), \(u = -(1)^2 = -1\).

Substitute into the integral:
\(\int_{0}^{1} x e^{-x^2} \, dx = \int_{0}^{-1} e^u \left(-\frac{1}{2}\right) du\)
\(= \frac{1}{2} \int_{-1}^{0} e^u \, du\).

Integrate and evaluate:
\(= \frac{1}{2} \left[ e^u \right]_{-1}^{0}\)
\(= \frac{1}{2} \left( e^0 - e^{-1} \right)\)
\(= \frac{1}{2} \left( 1 - \frac{1}{e} \right)\)
\(= \frac{e - 1}{2e}\).

- **M1**: Choosing a suitable substitution, e.g., \(u = -x^2\) or \(u = x^2\).
- **A1**: Finding the correct differential relation, e.g., \(du = -2x \, dx\) or equivalent.
- **M1**: Attempting to change limits of integration (or returning to the original variable \(x\) after integrating).
- **A1**: Finding the correct antiderivative: \(-\frac{1}{2} e^{-x^2}\) (or \(-\frac{1}{2} e^u\)).
- **A1.83**: Substituting the limits correctly and simplifying to the exact value: \(\frac{e - 1}{2e}\) (or equivalent form like \(\frac{1}{2}(1 - e^{-1})\)).

(a) Set \(f(x) = 0\):
\(\sin(2x) - 2\sin(x) = 0\)
\(2\sin(x)\cos(x) - 2\sin(x) = 0\)
\(2\sin(x)(\cos(x) - 1) = 0\)
This gives \(\sin(x) = 0\) or \(\cos(x) = 1\).
In the interval \([0, 2\pi]\), the solutions are \(x = 0, \pi, 2\pi\).

(b)(i) Differentiating \(f(x)\):
\(f'(x) = 2\cos(2x) - 2\cos(x)\)
Using the double-angle identity \(\cos(2x) = 2\cos^2(x) - 1\):
\(f'(x) = 2(2\cos^2(x) - 1) - 2\cos(x) = 4\cos^2(x) - 2\cos(x) - 2\)

(b)(ii) Set \(f'(x) = 0\):
\(2(2\cos^2(x) - \cos(x) - 1) = 0\)
\(2(2\cos(x) + 1)(\cos(x) - 1) = 0\)
For \(0 < x < 2\pi\), \(\cos(x) = 1\) has no solutions.
So \(2\cos(x) + 1 = 0 \implies \cos(x) = -\frac{1}{2}\).
This yields \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\).
Testing the intervals for the sign of \(f'(x)\):
For \(0 < x < \frac{2\pi}{3}\), \(f'(x) < 0\) (decreasing).
For \(\frac{2\pi}{3} < x < \frac{4\pi}{3}\), \(f'(x) > 0\) (increasing).
For \(\frac{4\pi}{3} < x < 2\pi\), \(f'(x) < 0\) (decreasing).
Thus, \(x = \frac{2\pi}{3}\) is a local minimum, and \(x = \frac{4\pi}{3}\) is a local maximum.

(c)(i) Differentiating \(f'(x)\):
\(f''(x) = \frac{d}{dx}(4\cos^2(x) - 2\cos(x) - 2) = 8\cos(x)(-\sin(x)) + 2\sin(x)\)
\(f''(x) = -8\sin(x)\cos(x) + 2\sin(x) = 2\sin(x)(1 - 4\cos(x))\)

(c)(ii) Set \(f''(x) = 0\) for \(0 < x < 2\pi\):
Either \[\sin(x) = 0 \implies x = \pi\\] or \[1 - 4\cos(x) = 0 \implies \cos(x) = \frac{1}{4} \implies x = \arccos\left(\frac{1}{4}\right)\text{ or } x = 2\pi - \arccos\left(\frac{1}{4}\right)\\]
Since \(f''(x)\) changes sign at each of these three points, the x-coordinates of the points of inflection are \(x = \arccos\left(\frac{1}{4}\right)\), \(x = \pi\), and \(x = 2\pi - \arccos\left(\frac{1}{4}\right)\).

(a)
M1 for attempting to use the identity \(\sin(2x) = 2\sin(x)\cos(x)\).
M1 for factoring \(2\sin(x)(\cos(x) - 1) = 0\).
A1 for identifying the correct x-intercepts: \(x = 0, \pi, 2\pi\).

(b)(i)
M1 for attempting to use the double-angle identity for \(\cos(2x)\).
A1 for the correct expression: \(4\cos^2(x) - 2\cos(x) - 2\).

(b)(ii)
M1 for setting \(f'(x) = 0\) and attempting to factorize.
A1 for identifying \(\cos(x) = -1/2\).
A1 for finding both critical points \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\).
M1 for a valid test (e.g., sign table or second derivative) to classify the points.
A1 for identifying \(x = \frac{2\pi}{3}\) as a local minimum.
A1 for identifying \(x = \frac{4\pi}{3}\) as a local maximum.

(c)(i)
M1 for attempting to differentiate \(f'(x)\) using the chain rule.
A1 for obtaining \(-8\sin(x)\cos(x) + 2\sin(x)\).
A1 for factorizing out \(2\sin(x)\) to show the given result.

(c)(ii)
M1 for setting \(f''(x) = 0\).
A1 for finding \(x = \pi\).
A2 for finding \(x = \arccos(1/4)\) and \(x = 2\pi - \arccos(1/4)\) (award A1 for one correct and A2 for both).
R1 for verifying that these are points of inflection (sign change occurs).

(a) The probability of drawing two red marbles is:
\(P(R = 2) = P(R_1) \times P(R_2 | R_1) = \frac{4}{n} \times \frac{3}{n-1} = \frac{12}{n(n-1)}\)

(b)(i) The event \(R \ge 1\) is the complement of \(R = 0\).
\(P(R = 0) = P(B_1) \times P(B_2 | B_1) = \frac{n-4}{n} \times \frac{n-5}{n-1} = \frac{n^2 - 9n + 20}{n(n-1)}\)
Thus,
\(P(R \ge 1) = 1 - \frac{n^2 - 9n + 20}{n(n-1)} = \frac{n(n-1) - (n^2 - 9n + 20)}{n(n-1)} = \frac{8n - 20}{n(n-1)}\)
Setting this equal to \(\frac{2}{3}\):
\(\frac{8n - 20}{n(n-1)} = \frac{2}{3} \implies 3(8n - 20) = 2n(n-1)\)
\(24n - 60 = 2n^2 - 2n \implies 2n^2 - 26n + 60 = 0 \implies n^2 - 13n + 30 = 0\)

(b)(ii) Factoring the quadratic:
\((n-10)(n-3) = 0\)
Since \(n \ge 5\), we must have \(n = 10\).

(c) For \(n = 10\):
\(P(R = 2) = \frac{12}{10 \times 9} = \frac{12}{90} = \frac{2}{15}\)
\(P(R = 0) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}\)
\(P(R = 1) = 1 - P(R = 0) - P(R = 2) = 1 - \frac{1}{3} - \frac{2}{15} = \frac{8}{15}\)
So the distribution of \(R\) is:
\(P(R=0) = \frac{1}{3}\), \(P(R=1) = \frac{8}{15}\), \(P(R=2) = \frac{2}{15}\)

(d) For a fair game, the entry cost \(k\) must equal the expected winnings \(E(X)\):
\(E(X) = 0 \times P(R=0) + 5 \times P(R=1) + 15 \times P(R=2)\)
\(E(X) = 0 + 5 \left(\frac{8}{15}\right) + 15 \left(\frac{2}{15}\right) = \frac{40 + 30}{15} = \frac{70}{15} = \frac{14}{3}\)
Thus, \(k = \frac{14}{3}\).

(e) Winning exactly $5 corresponds to the event \(R = 1\), which has probability \(p = \frac{8}{15}\).
Let \(Y\) be the number of times this occurs in 3 games. Then \(Y \sim \text{B}\left(3, \frac{8}{15}\right)\).
\(P(Y \ge 2) = P(Y=2) + P(Y=3)\)
\(P(Y=2) = \binom{3}{2} \left(\frac{8}{15}\right)^2 \left(\frac{7}{15}\right) = 3 \times \frac{64 \times 7}{3375} = \frac{1344}{3375}\)
\(P(Y=3) = \left(\frac{8}{15}\right)^3 = \frac{512}{3375}\)
\(P(Y \ge 2) = \frac{1344 + 512}{3375} = \frac{1856}{3375}\)

(a)
M1 for calculating \(P(R = 2)\) as a product of two probabilities.
A1 for showing \(\frac{4}{n} \times \frac{3}{n-1} = \frac{12}{n(n-1)}\).

(b)(i)
M1 for using the complement method: \(P(R \ge 1) = 1 - P(R = 0)\).
M1 for expressing \(P(R = 0)\) in terms of \(n\).
A1 for setting up the equation and correctly performing algebra to get the quadratic equation.

(b)(ii)
A1 for correctly stating \(n = 10\) and rejecting \(n = 3\).

(c)
A1 for \(P(R=0) = 1/3\).
A1 for \(P(R=2) = 2/15\).
A1 for \(P(R=1) = 8/15\).

(d)
M1 for writing down the expected value formula.
A1 for substituting the correct probabilities and winnings.
A1 for concluding \(k = 14/3\).

(e)
M1 for identifying the binomial setup with \(p = 8/15\) and \(n = 3\).
M1 for correctly substituting into the formula for \(P(Y=2)\) and \(P(Y=3)\).
A1 for simplifying to \(1856/3375\).

(a)(i) Substituting \(k = 4\) and setting \(f(x) = 0\):
\(e^{2x} - 4e^x + 3 = 0\)
Let \(u = e^x\). This yields the quadratic:
\(u^2 - 4u + 3 = 0\)
\((u - 3)(u - 1) = 0 \implies e^x = 3 \text{ or } e^x = 1\)
Taking the natural logarithm of both sides gives the solutions \(x = \ln 3\) and \(x = 0\).

(a)(ii) Finding the derivative of \(f\):
\(f'(x) = 2e^{2x} - 4e^x\)
Setting \(f'(x) = 0\):
\(2e^x(e^x - 2) = 0 \implies e^x = 2 \implies x = \ln 2\)
Since \(f'(x) < 0\) for \(x < \ln 2\) and \(f'(x) > 0\) for \(x > \ln 2\), this is indeed a local minimum.
The y-coordinate is:
\(f(\ln 2) = e^{2\ln 2} - 4e^{\ln 2} + 3 = 4 - 8 + 3 = -1\)
So the coordinates are \((\ln 2, -1)\).

(b)(i) Let \(u = e^x\). Since \(x \in \mathbb{R}\), \(u\) must be strictly positive (\(u > 0\)).
The equation becomes \(u^2 - ku + 3 = 0\).
For the equation \(f(x) = 0\) to have two distinct real solutions, the quadratic must have two distinct positive real roots for \(u\).
The discriminant of the quadratic must be strictly positive:
\(\Delta = k^2 - 12 > 0 \implies k > 2\sqrt{3} \text{ or } k < -2\sqrt{3}\)
Furthermore, for both roots to be positive, their product must be positive (which is \(u_1 u_2 = 3 > 0\), always true) and their sum must be positive:
\(u_1 + u_2 = k > 0\)
Combining the conditions \(k > 0\) and \((k > 2\sqrt{3} \text{ or } k < -2\sqrt{3})\) yields:
\(k > 2\sqrt{3}\).

(b)(ii) Since the product of the roots is \(u_1 u_2 = 3 > 0\), both roots must have the same sign. Thus, we cannot have one positive and one negative root.
Therefore, exactly one positive root can only occur if the quadratic has a single repeated positive real root (a double root).
This occurs when the discriminant is zero:
\(\Delta = k^2 - 12 = 0 \implies k = \pm 2\sqrt{3}\)
For the repeated root to be positive, we need the sum of the roots to be positive, so \(k > 0\).
Hence, \(k = 2\sqrt{3}\).

(a)(i)
M1 for substituting \(k = 4\) and writing the equation in terms of \(u = e^x\).
M1 for factorizing the quadratic expression.
A1 for obtaining \(x = 0\) and \(x = \ln 3\).

(a)(ii)
M1 for finding the derivative \(f'(x) = 2e^{2x} - 4e^x\).
M1 for setting \(f'(x) = 0\) and solving to get \(x = \ln 2\).
A1 for the correct y-coordinate of \(-1\).
A1 for stating the coordinates \((\ln 2, -1)\).

(b)(i)
M1 for making the substitution \(u = e^x\) and noting that \(u > 0\).
M1 for finding the discriminant \(\Delta = k^2 - 12\).
R1 for explaining that two distinct solutions for \(x\) require two distinct positive roots for \(u\).
M1 for stating the conditions for positive roots (discriminant and sum of roots are positive).
A1 for concluding \(k > 2\sqrt{3}\).

(b)(ii)
R1 for recognizing that because the product of the roots is positive, the roots must have the same sign.
M1 for recognizing that exactly one positive root occurs when there is a repeated positive root.
A1 for finding \(k = 2\sqrt{3}\) (rejecting \(k = -2\sqrt{3}\)).

(a) Let \(W\) be the weight of a pear, so \(W \sim N(150, \sigma^2)\). We are given \(P(W > 180) = 0.15\), which implies \(P(W \le 180) = 0.85\). Standardizing the value, we have \(P\left(Z \le \frac{180 - 150}{\sigma}\right) = 0.85\). Using a GDC, the z-score corresponding to a cumulative probability of 0.85 is \(z \approx 1.03643\). Thus, \(\frac{30}{\sigma} = 1.03643\), and solving for \(\sigma\) yields \(\sigma \approx 28.9455\) grams, which is \(28.9\) to three significant figures. (b) We want to find \(P(W < 130)\). Using \(\mu = 150\) and \(\sigma \approx 28.9455\), we compute \(P(W < 130) = \text{normalcdf}(-\infty, 130, 150, 28.9455) \approx 0.24479\), which is \(0.245\) to three significant figures.

(a) M1 for setting up the probability equation \(P(W > 180) = 0.15\) or standardizing. A1 for finding the z-score \(1.036\). A1 for the correct value of \(\sigma = 28.9\). (b) M1 for setting up the normal cumulative distribution with mean 150 and standard deviation found in part (a). A1 for the correct probability \(0.245\).

(a) The particle is at rest when its velocity is zero, so \(4\ln(t+1) - t^2 = 0\). Using a GDC to find the positive root in the interval \(0 \le t \le 3\), we find \(t \approx 2.13926\) seconds. To three significant figures, \(t \approx 2.14\) s. (b) The total distance travelled is given by \(\int_{0}^{3} |v(t)| dt = \int_{0}^{3} |4\ln(t+1) - t^2| dt\). Splitting the integral where the velocity changes sign: \(\int_{0}^{2.139} (4\ln(t+1) - t^2) dt - \int_{2.139}^{3} (4\ln(t+1) - t^2) dt\). Evaluating this numerically on a GDC yields \(2.5453 + 1.3646 \approx 3.9099\) metres, which rounds to \(3.91\) m to three significant figures.

(a) M1 for setting \(v(t) = 0\). A1 for the correct root \(t \approx 2.14\). (b) M1 for setting up the integral of the absolute value of velocity. M1 for splitting the integral at the boundary \(t \approx 2.14\). A1 for evaluating each part correctly. A1 for the final distance \(3.91\).

(a) The bearing of B from A is \(060^\circ\), which means the direction from A to B makes an angle of \(60^\circ\) with the North line. The South-bound line from B makes an angle of \(60^\circ\) with BA (alternate interior angles). Since the bearing of C from B is \(130^\circ\), the angle between the South line from B and BC is \(180^\circ - 130^\circ = 50^\circ\). Hence, \(\angle ABC = 60^\circ + 50^\circ = 110^\circ\). Applying the Cosine Rule to triangle ABC: \(AC^2 = 15^2 + 22^2 - 2(15)(22)\cos(110^\circ) \approx 934.73\). Taking the square root, \(AC \approx 30.573\) km, which is \(30.6\) km to three significant figures. (b) To find the bearing of A from C, we find \(\angle ACB\) using the Sine Rule: \(\frac{\sin(\angle ACB)}{15} = \frac{\sin(110^\circ)}{30.573} \Rightarrow \sin(\angle ACB) \approx 0.4611\), giving \(\angle ACB \approx 27.46^\circ\). The bearing of B from C is \(130^\circ + 180^\circ = 310^\circ\). Since A lies West of BC, the bearing of A from C is \(310^\circ - 27.46^\circ = 282.54^\circ\), which is \(283^\circ\) to the nearest degree.

(a) M1 for finding \(\angle ABC = 110^\circ\). M1 for applying the Cosine Rule correctly. A1 for \(30.6\) km. (b) M1 for using the Sine Rule to find \(\angle ACB\). A1 for finding \(\angle ACB \approx 27.5^\circ\). A1 for calculating the correct bearing \(283^\circ\).

(a) As \(t \to \infty\), \(e^{-kt} \to 0\), so \(P(t) \to A\). Since the limit is 5000, we have \(A = 5000\). At \(t = 0\), \(P(0) = \frac{5000}{1 + B} = 500\), which yields \(B = 9\). At \(t = 4\), \(P(4) = \frac{5000}{1 + 9e^{-4k}} = 1500\). Solving this equation: \(1 + 9e^{-4k} = \frac{10}{3} \Rightarrow 9e^{-4k} = \frac{7}{3} \Rightarrow e^{-4k} = \frac{7}{27}\). Taking natural logarithms: \(k = -\frac{1}{4}\ln\left(\frac{7}{27}\right) \approx 0.33753\). So, \(k \approx 0.338\). (b) The rate of growth at \(t = 6\) is \(P'(6)\). Using a GDC to differentiate \(P(t) = \frac{5000}{1 + 9e^{-0.33753t}}\right) at \)t=6\) yields \(P'(6) \approx 418.8\). To the nearest integer, this is \(419\) bacteria per hour.

(a) A1 for \(A = 5000\). A1 for \(B = 9\). M1 for substituting \(t = 4, P = 1500\) to set up an equation. A1 for \(k \approx 0.338\). (b) M1 for finding or setting up the derivative at \(t = 6\). A1 for the correct rate of growth \(419\) (accept 418 to 420).

(a) Compounded monthly for 5 years means \(60\) periods. Substituting into the future value formula: \(13488.50 = 10000 \left(1 + \frac{r}{1200}\right)^{60}\). Dividing by 10000: \(1.34885 = \left(1 + \frac{r}{1200}\right)^{60}\). Taking the 60th root: \(1 + \frac{r}{1200} = (1.34885)^{1/60} \approx 1.00500\). Therefore, \(\frac{r}{1200} = 0.00500 \Rightarrow r = 6.00\). (b) We want to solve \(10000 (1.005)^n > 20000\), which simplifies to \(1.005^n > 2\). Taking natural logarithms: \(n \ln(1.005) > \ln(2) \Rightarrow n > \frac{\ln(2)}{\ln(1.005)} \approx 138.98\). Rounding up to the next integer, the minimum number of complete months is \(139\).

(a) M1 for substituting into the compound interest formula with correct parameters. A1 for setting up the equation. A1 for \(r = 6.00\). (b) M1 for setting up the inequality \(10000(1.005)^n > 20000\). A1 for finding the boundary \(n \approx 138.98\). A1 for rounding up to \(139\).

(a) Entering the values into a GDC, we perform a linear regression analysis and obtain the Pearson's product-moment correlation coefficient \(r \approx 0.9953\), which is \(0.995\) to three significant figures. (b) From the GDC linear regression output, we get the gradient \(a \approx 4.7908\) and the y-intercept \(b \approx 34.1939\). Thus, the regression line equation is \(y = 4.79x + 34.2\). (c) Substituting \(x = 6\) into the regression equation: \(y = 4.7908(6) + 34.1939 \approx 62.9387\). To three significant figures, the estimated exam score is \(62.9\).

(a) G1 for finding \(r = 0.995\) (accept 0.9953). (b) G1 for gradient \(a \approx 4.79\), G1 for y-intercept \(b \approx 34.2\). (c) M1 for substituting \(x = 6\) into their regression equation. A1 for finding \(62.9\).

(a) We want to find \(P(V > 758)\) where \(V \sim N(750, 6^2)\).
Using a GDC (normalcdf with lower = 758, upper = \(10^{99}\), \(\mu = 750\), \(\sigma = 6\)):
\(P(V > 758) \approx 0.0912112\)
\(P(V > 758) \approx 0.0912\) (to 3 sig. figs.)

(b) We are given \(P(V < k) = 0.025\).
Using the inverse normal function on a GDC (invNorm with area = 0.025, \(\mu = 750\), \(\sigma = 6\)):
\(k \approx 738.240\)
\(k \approx 738\) mL (to 3 sig. figs.)

(c) First, let us find the probability, \(p\), that a single bottle from the upgraded machine contains more than 745 mL, where \(V_{\text{new}} \sim N(750, 4.2^2)\).
\(p = P(V_{\text{new}} > 745)\)
Using a GDC:
\(p \approx 0.883049\)

Let \(X\) be the number of bottles containing more than 745 mL out of 12. Then \(X \sim B(12, 0.883049)\).
We want to find \(P(X \ge 10)\).
\(P(X \ge 10) = 1 - P(X \le 9)\)
Using a GDC (binomcdf with \(n = 12\), \(p = 0.883049\), \(x = 9\)):
\(P(X \le 9) \approx 0.169335\)
\(P(X \ge 10) \approx 1 - 0.169335 = 0.830665\)
\(P(X \ge 10) \approx 0.831\) (to 3 sig. figs.)

(d) We have the cost relationship \(C = 0.05 V_{\text{new}} + 1.20\).
Mean of cost:
\(E(C) = 0.05 E(V_{\text{new}}) + 1.20\)
\(E(C) = 0.05(750) + 1.20 = 37.5 + 1.20 = 38.7\) USD

Variance of cost:
\(\text{Var}(C) = (0.05)^2 \text{Var}(V_{\text{new}})\)
Since \(\sigma_{\text{new}} = 4.2\), \(\text{Var}(V_{\text{new}}) = 4.2^2 = 17.64\).
\(\text{Var}(C) = 0.0025 \times 17.64 = 0.0441\) USD\(^2\)

(a)
(M1) for attempting to use normal cumulative distribution function on GDC.
(A1) for correct setup/parameters.
(A1) for \(0.0912\). [3 marks]

(b)
(M1) for setting up the equation \(P(V < k) = 0.025\).
(A1) for attempting to use inverse normal function on GDC.
(A1) for \(738\) (accept 738.24). [3 marks]

(c)
(M1) for finding \(P(V_{\text{new}} > 745) = 0.883049\).
(M1) for recognizing a binomial distribution \(X \sim B(12, p)\).
(M1) for attempting to find \(P(X \ge 10)\) (e.g., writing \(1 - P(X \le 9)\)).
(A1) for correct binomial term or cumulative calculation.
(A1) for \(0.831\) (accept 0.830). [5 marks]

(d)
(M1) for calculating \(E(C) = 0.05(750) + 1.20\).
(A1) for \(38.7\).
(M1) for using \(\text{Var}(aX+b) = a^2\text{Var}(X)\) to write \(0.05^2 \times 4.2^2\).
(A1) for \(0.0441\). [3 marks]

(a) Initial temperature is when \(t = 0\):
\(T(0) = 20 + 60e^{0} \cos(0) = 20 + 60(1)(1) = 80^\circ\text{C}\).

(b) The rate of change of temperature is given by \(T'(t)\).
Using the product rule:
\(T'(t) = 60 \left( -0.05e^{-0.05t} \cos(0.1t) - 0.1e^{-0.05t} \sin(0.1t) \right)\)
\(T'(t) = -3e^{-0.05t} \cos(0.1t) - 6e^{-0.05t} \sin(0.1t)\).
At \(t = 10\):
\(T'(10) = -3e^{-0.5} \cos(1) - 6e^{-0.5} \sin(1)\)
Using a GDC (in radians mode):
\(T'(10) \approx 0.60653 \times (-3(0.54030) - 6(0.84147)) \approx -4.04539\)
So the rate of change is \(-4.05^\circ\text{C}\)/minute (to 3 sig. figs.).

(c) The temperature is decreasing at its maximum rate when the rate of change, \(T'(t)\), is at its minimum value (most negative).
This occurs when the derivative of the rate of change is zero, \(T''(t) = 0\), and \(T'(t) < 0\).
Using the GDC to find the first local minimum of the graph of \(y = T'(x)\) (or finding the first positive root of \(T''(t) = 0\)):
\(T''(t) = e^{-0.05t} [ -0.45\cos(0.1t) + 0.6\sin(0.1t) ] = 0\)
\(0.6\sin(0.1t) = 0.45\cos(0.1t) \Rightarrow \tan(0.1t) = 0.75\)
\(0.1t = \arctan(0.75) \approx 0.643501\)
\(t \approx 6.43501\)
So \(t \approx 6.44\) minutes (to 3 sig. figs.).

(d) The average temperature is given by:
\(T_{\text{avg}} = \frac{1}{30 - 0} \int_{0}^{30} T(t) \, dt = \frac{1}{30} \int_{0}^{30} \left(20 + 60e^{-0.05t} \cos(0.1t)\right) \, dt\)
Using a GDC to evaluate the definite integral:
\(\int_{0}^{30} \left(20 + 60e^{-0.05t} \cos(0.1t)\right) \, dt \approx 908.130\)
\(T_{\text{avg}} = \frac{908.130}{30} \approx 30.271\)
So the average temperature is \(30.3^\circ\text{C}\) (to 3 sig. figs.).

(a)
(M1) for substituting \(t = 0\) into \(T(t)\).
(A1) for \(80\) (or \(80^\circ\text{C}\)). [2 marks]

(b)
(M1) for attempting to differentiate \(T(t)\) (use of product/chain rule).
(A1) for correct expression of \(T'(t)\).
(A1) for \(-4.05\) (or \(-4.05^\circ\text{C}\)/minute). [3 marks]

(c)
(M1) for recognizing that the maximum rate of decrease occurs at a minimum of \(T'(t)\) (or setting \(T''(t) = 0\)).
(M1) for attempting to solve \(T''(t) = 0\) or using GDC to find the minimum of the derivative.
(A1) for obtaining \(\tan(0.1t) = 0.75\) (or equivalent equation).
(A1) for \(6.44\) (or \(6.44\) minutes). [4 marks]

(d)
(M1) for setting up the average value integral formula: \(\frac{1}{30} \int_{0}^{30} T(t) \, dt\).
(A2) for writing the correct integrand and limits.
(A1) for finding the value of the integral (\(908.130\) or similar).
(A1) for \(30.3\) (or \(30.3^\circ\text{C}\)). [5 marks]

(a) Since the mast has a height of 12 meters, the coordinates of \(P\) are \((0, 0, 12)\).
The vector \(\vec{PA}\) is:
\(\vec{PA} = \vec{OA} - \vec{OP} = \begin{pmatrix} 8 \\ 2 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 12 \end{pmatrix} = \begin{pmatrix} 8 \\ 2 \\ -12 \end{pmatrix}\).
The length of the cable \(PA\) is the magnitude of \(\vec{PA}\):
\(|\vec{PA}| = \sqrt{8^2 + 2^2 + (-12)^2} = \sqrt{64 + 4 + 144} = \sqrt{212} \approx 14.5602\)
So the length of the cable \(PA\) is \(14.6\) m (to 3 sig. figs.).

(b) The vector \(\vec{PB}\) is:
\(\vec{PB} = \vec{OB} - \vec{OP} = \begin{pmatrix} -4 \\ 10 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 12 \end{pmatrix} = \begin{pmatrix} -4 \\ 10 \\ -12 \end{pmatrix}\).
Its magnitude is:
\(|\vec{PB}| = \sqrt{(-4)^2 + 10^2 + (-12)^2} = \sqrt{16 + 100 + 144} = \sqrt{260} \approx 16.1245\).
To find the angle \(\theta\) between the cables, we use the dot product of \(\vec{PA}\) and \(\vec{PB}\):
\(\vec{PA} \cdot \vec{PB} = 8(-4) + 2(10) + (-12)(-12) = -32 + 20 + 144 = 132\).
\(\cos\theta = \frac{\vec{PA} \cdot \vec{PB}}{|\vec{PA}||\vec{PB}|} = \frac{132}{\sqrt{212} \times \sqrt{260}} \approx 0.562237\).
\(\theta = \arccos(0.562237) \approx 0.97368\) radians (or \(55.787^\circ\)).
So the angle is \(0.974\) radians or \(55.8^\circ\).

(c) The midpoint \(M\) of \(PA\) is:
\(M = \left(\frac{0+8}{2}, \frac{0+2}{2}, \frac{12+0}{2}\right) = (4, 1, 6)\).
The direction vector of the line \(BM\) is:
\(\vec{BM} = \vec{OM} - \vec{OB} = \begin{pmatrix} 4 \\ 1 \\ 6 \end{pmatrix} - \begin{pmatrix} -4 \\ 10 \\ 0 \end{pmatrix} = \begin{pmatrix} 8 \\ -9 \\ 6 \end{pmatrix}\).
Using the point \(B\) as the position vector, a vector equation of the line \(BM\) is:
\(\mathbf{r} = \begin{pmatrix} -4 \\ 10 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 8 \\ -9 \\ 6 \end{pmatrix}\), where \(\lambda \in \mathbb{R}\).

(d) Let \(Q\) be a general point on the line \(BM\). Its position vector is \(\vec{OQ} = \begin{pmatrix} -4 + 8\lambda \\ 10 - 9\lambda \\ 6\lambda \end{pmatrix}\).
The vector \(\vec{OQ}\) is perpendicular to the direction vector of the line \(\vec{d} = \begin{pmatrix} 8 \\ -9 \\ 6 \end{pmatrix}\) when \(Q\) is the closest point on the line to \(O\):
\(\vec{OQ} \cdot \vec{d} = 0\)
\(8(-4 + 8\lambda) - 9(10 - 9\lambda) + 6(6\lambda) = 0\)
\(-32 + 64\lambda - 90 + 81\lambda + 36\lambda = 0\)
\(181\lambda - 122 = 0 \Rightarrow \lambda = \frac{122}{181} \approx 0.674033\).
Substitute \(\lambda\) back to find the shortest distance (the magnitude of \(\vec{OQ}\)):
\(\vec{OQ} = \begin{pmatrix} -4 + 8(0.674033) \\ 10 - 9(0.674033) \\ 6(0.674033) \end{pmatrix} \approx \begin{pmatrix} 1.39226 \\ 3.93370 \\ 4.04420 \end{pmatrix}\).
\(|\vec{OQ}| = \sqrt{(1.39226)^2 + (3.93370)^2 + (4.04420)^2} = \sqrt{33.7679} \approx 5.8110\) m.
So the shortest distance is \(5.81\) m (to 3 sig. figs.).

(a)
(A1) for coordinates \(P(0, 0, 12)\).
(M1) for attempting to find the magnitude of the vector \(\vec{PA}\).
(A1) for \(14.6\) (accept \(\sqrt{212}\)). [3 marks]

(b)
(M1) for finding the vector \(\vec{PB}\).
(M1) for finding the dot product \(\vec{PA} \cdot \vec{PB} = 132\).
(M1) for attempting to use the scalar product formula \(\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\).
(A1) for \(0.974\) rad (or \(55.8^\circ\)). [4 marks]

(c)
(M1) for finding midpoint \(M(4, 1, 6)\).
(M1) for finding the direction vector \(\vec{BM}\) (or \(\vec{MB}\)).
(A1) for a correct vector equation of the line, including "\(\mathbf{r} = \dots\)" or equivalent. [3 marks]

(d)
(M1) for writing a parametric vector representing a general point on the line.
(M1) for setting the dot product of the general vector and the direction vector to zero.
(A1) for finding the correct parameter value \(\lambda \approx 0.674\).
(A1) for \(5.81\) m (or \(5.8110...\)). [4 marks]