2024 IB DP Mathematics - Analysis and Approaches Practice Paper with Answers

The common difference \(d\) of the arithmetic sequence is given by:
\(d = u_2 - u_1 = \ln(2x) - \ln x = \ln\left(\frac{2x}{x}\right) = \ln 2\).

The sum of the first 10 terms of an arithmetic sequence is given by the formula:
\(S_{10} = \frac{10}{2} [2u_1 + 9d]\).

Substituting \(u_1 = \ln x\) and \(d = \ln 2\) into the formula:
\(S_{10} = 5 [2 \ln x + 9 \ln 2] = 10 \ln x + 45 \ln 2\).

We are given that the sum is equal to \(10 \ln 5 + 25 \ln 2\). Equating the two expressions:
\(10 \ln x + 45 \ln 2 = 10 \ln 5 + 25 \ln 2\).

Subtracting \(45 \ln 2\) from both sides:
\(10 \ln x = 10 \ln 5 - 20 \ln 2\).

Dividing both sides by 10:
\(\ln x = \ln 5 - 2 \ln 2\).

Using the laws of logarithms:
\(\ln x = \ln 5 - \ln(2^2) = \ln 5 - \ln 4 = \ln\left(\frac{5}{4}\right)\).

Thus, \(x = \frac{5}{4}\) (or \(1.25\)).

M1 for finding the common difference \(d = \ln 2\).
M1 for substituting \(u_1\) and \(d\) into the arithmetic sum formula \(S_{10} = \frac{10}{2} [2u_1 + 9d]\).
A1 for simplifying the sum to \(10 \ln x + 45 \ln 2\).
M1 for equating this to the given sum: \(10 \ln x + 45 \ln 2 = 10 \ln 5 + 25 \ln 2\).
A1 for isolating \(\ln x\): \(\ln x = \ln 5 - 2 \ln 2\).
A1 for applying log laws to obtain \(x = \frac{5}{4}\) (or 1.25).

To find the inverse function, we let \(y = \frac{3e^x + 1}{e^x - 2}\) and rearrange the equation to make \(e^x\) the subject:
\(y(e^x - 2) = 3e^x + 1\)
\(ye^x - 2y = 3e^x + 1\)
\(ye^x - 3e^x = 2y + 1\)
\(e^x(y - 3) = 2y + 1\)
\(e^x = \frac{2y + 1}{y - 3}\).

Taking the natural logarithm of both sides:
\(x = \ln\left(\frac{2y + 1}{y - 3}\right)\).

Interchanging \(x\) and \(y\), we get the inverse function:
\(f^{-1}(x) = \ln\left(\frac{2x + 1}{x - 3}\right)\).

To find the domain of \(f^{-1}\), the argument of the natural logarithm must be strictly positive:
\(\frac{2x + 1}{x - 3} > 0\).

The critical values are \(x = -\frac{1}{2}\) and \(x = 3\). Testing intervals, the rational expression is positive when \(x < -\frac{1}{2}\) or \(x > 3\).

Thus, the domain of \(f^{-1}\) is \(x < -\frac{1}{2}\) or \(x > 3\).

M1 for setting \(y = \frac{3e^x + 1}{e^x - 2}\) and attempting to solve for \(e^x\).
A1 for obtaining \(ye^x - 3e^x = 2y + 1\) or equivalent.
A1 for finding \(e^x = \frac{2y + 1}{y - 3}\).
M1 for taking the natural logarithm to find the expression for \(f^{-1}(x)\).
M1 for setting up the inequality \(\frac{2x + 1}{x - 3} > 0\) to find the domain.
A1 for the correct domain: \(x < -\frac{1}{2}\) or \(x > 3\) (or equivalent interval notation).

We can divide both sides of the equation by 2:
\(\frac{\sqrt{3}}{2} \sin(2x) + \frac{1}{2} \cos(2x) = \frac{1}{2}\).

Using the compound angle identity \(\sin(A + B) = \sin A \cos B + \cos A \sin B\), we can let \(\cos B = \frac{\sqrt{3}}{2}\) and \(\sin B = \frac{1}{2}\), which gives \(B = \frac{\pi}{6}\).

Thus, the equation simplifies to:
\(\sin\left(2x + \frac{\pi}{6}\right) = \frac{1}{2}\).

Given that \(0 \le x \le \pi\), we find the corresponding interval for the angle \(2x + \frac{\pi}{6}\):
\(0 \le 2x \le 2\pi \implies \frac{\pi}{6} \le 2x + \frac{\pi}{6} \le \frac{13\pi}{6}\).

Within this interval, the values for which \(\sin\theta = \frac{1}{2}\) are:
\(2x + \frac{\pi}{6} = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}\).

Solving each of these for \(x\):
1) \(2x + \frac{\pi}{6} = \frac{\pi}{6} \implies 2x = 0 \implies x = 0\)
2) \(2x + \frac{\pi}{6} = \frac{5\pi}{6} \implies 2x = \frac{2\pi}{3} \implies x = \frac{\pi}{3}\)
3) \(2x + \frac{\pi}{6} = \frac{13\pi}{6} \implies 2x = 2\pi \implies x = \pi\).

Therefore, the solutions in the given domain are \(x = 0, \frac{\pi}{3}, \pi\).

M1 for dividing by 2 or attempting to use a compound angle identity.
A1 for expressing the equation in the form \(\sin\left(2x + \frac{\pi}{6}\right) = \frac{1}{2}\) or \(\cos\left(2x - \frac{\pi}{3}\right) = \frac{1}{2}\).
M1 for establishing the correct domain for the angle: \(\frac{\pi}{6} \le 2x + \frac{\pi}{6} \le \frac{13\pi}{6}\).
A1 for finding at least two correct values for the compound angle (e.g., \(\frac{\pi}{6}, \frac{5\pi}{6}\)).
A2 for all three correct solutions: \(x = 0, \frac{\pi}{3}, \pi\) (A1 if only two correct solutions are found).

The sum of the probabilities in a probability distribution must equal 1:
\(P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1 \implies 3k + a + b = 1\).

We are given that \(P(X \ge 3) = 0.7\), which means:
\(P(X = 3) + P(X = 4) = a + b = 0.7\).

Substituting \(a + b = 0.7\) into the first equation:
\(3k + 0.7 = 1 \implies 3k = 0.3 \implies k = 0.1\).

Using \(k = 0.1\), we have \(P(X = 1) = 0.1\) and \(P(X = 2) = 0.2\).

Now we use the expected value \(E(X) = 3\):
\(E(X) = 1(k) + 2(2k) + 3(a) + 4(b) = 3\)
\(5k + 3a + 4b = 3\).

Substitute \(k = 0.1\) into this equation:
\(5(0.1) + 3a + 4b = 3 \implies 0.5 + 3a + 4b = 3 \implies 3a + 4b = 2.5\).

We also have the system of two equations:
1) \(a + b = 0.7 \implies 3a + 3b = 2.1\)
2) \(3a + 4b = 2.5\).

Subtracting the first from the second gives:
\(b = 0.4\).

Substituting \(b = 0.4\) back into \(a + b = 0.7\) gives:
\(a = 0.3\).

Thus, the values are \(a = 0.3\), \(b = 0.4\), and \(k = 0.1\).

M1 for the sum of probabilities equation: \(3k + a + b = 1\).
M1 for identifying \(a + b = 0.7\) from \(P(X \ge 3) = 0.7\).
A1 for finding \(k = 0.1\).
M1 for formulating the expectation equation: \(5k + 3a + 4b = 3\) and substituting \(k = 0.1\).
A1 for obtaining \(b = 0.4\).
A1 for obtaining \(a = 0.3\).

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

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

Since \(\frac{d}{dx}(\ln(x^2 - 3)) = \frac{2x}{x^2 - 3}\), we have:
\(\frac{dy}{dx} = \ln(x^2 - 3) + x \cdot \frac{2x}{x^2 - 3} = \ln(x^2 - 3) + \frac{2x^2}{x^2 - 3}\).

Substitute \(x = 2\) to find the gradient of the tangent, \(m\):
\(m = \ln(2^2 - 3) + \frac{2(2^2)}{2^2 - 3} = \ln(1) + \frac{8}{1} = 8\).

Using the point-slope form of a linear equation, the equation of the tangent line is:
\(y - y_1 = m(x - x_1) \implies y - 0 = 8(x - 2) \implies y = 8x - 16\).

A1 for finding the correct \(y\)-coordinate \(y = 0\).
M1 for applying the product rule to differentiate \(x \ln(x^2 - 3)\).
M1 for applying the chain rule to obtain \(\frac{d}{dx}(\ln(x^2 - 3)) = \frac{2x}{x^2 - 3}\).
A1 for the correct derivative expression: \(\frac{dy}{dx} = \ln(x^2 - 3) + \frac{2x^2}{x^2 - 3}\).
A1 for substituting \(x = 2\) to find the tangent gradient \(m = 8\).
A1 for the correct equation of the tangent line: \(y = 8x - 16\).

(a) For \(\vec{u}\) and \(\vec{v}\) to be perpendicular, their scalar (dot) product must equal zero:
\(\vec{u} \cdot \vec{v} = 0\)
\(1(p) + 2(p^2) - 1(p+2) = 0\)
\(p + 2p^2 - p - 2 = 0 \implies 2p^2 - 2 = 0\).

Solving for \(p\):
\(2p^2 = 2 \implies p^2 = 1 \implies p = 1\) or \(p = -1\).

(b) For \(\vec{v}\) to be parallel to \(\vec{w} = \begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix}\), there must exist a non-zero scalar \(k\) such that \(\vec{v} = k\vec{w}\):
\(\begin{pmatrix} p \\ p^2 \\ p+2 \end{pmatrix} = k \begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix}\).

This results in three equations:
1) \(p = 2k\)
2) \(p^2 = 4k\)
3) \(p+2 = 4k\).

From equations (2) and (3), we can equate the expressions for \(4k\):
\(p^2 = p+2 \implies p^2 - p - 2 = 0\).

Factoring the quadratic equation:
\((p-2)(p+1) = 0\), which gives \(p = 2\) or \(p = -1\).

Now we test both values of \(p\) in equation (1) and check for consistency:
- If \(p = 2\): From (1), \(2 = 2k \implies k = 1\). Substituting \(p=2\) and \(k=1\) into (2) and (3) gives consistent results (\(4 = 4\)).
- If \(p = -1\): From (1), \(-1 = 2k \implies k = -0.5\). Substituting into (2) gives \((-1)^2 = 4(-0.5) \implies 1 = -2\), which is a contradiction.

Therefore, the only valid solution is \(p = 2\).

(a)
M1 for setting up the scalar product \(\vec{u} \cdot \vec{v} = 0\).
A1 for simplifying to \(2p^2 - 2 = 0\).
A1 for finding the correct values \(p = 1\) or \(p = -1\).

(b)
M1 for setting up the parallel vector relation \(\vec{v} = k\vec{w}\) or equivalent ratio of coordinates.
A1 for obtaining the quadratic equation \(p^2 - p - 2 = 0\) and finding the potential solutions \(p=2\) and \(p=-1\).
R1 for testing consistency and correctly rejecting \(p = -1\) to conclude \(p = 2\).

(a) To find the local minimum, we find the first derivative: \( f'(x) = e^{2x} + 2x e^{2x} = (1 + 2x)e^{2x} \). Setting \( f'(x) = 0 \) gives \( 1 + 2x = 0 \implies x = -1/2 \). The corresponding \( y \)-coordinate is \( f(-1/2) = -1/2 e^{-1} = -1/(2e) \). To confirm it is a local minimum, we check the second derivative: \( f''(x) = 2e^{2x} + 2(1+2x)e^{2x} = (4x + 4)e^{2x} \). Since \( f''(-1/2) = 2e^{-1} > 0 \), the point is indeed a local minimum. Thus, the coordinates are \( (-1/2, -1/(2e)) \). (b) To find the point of inflection, we set \( f''(x) = 0 \implies (4x + 4)e^{2x} = 0 \implies x = -1 \). Since the second derivative changes sign at \( x = -1 \), this is a point of inflection. The \( y \)-coordinate is \( f(-1) = -e^{-2} \). Thus, the coordinates are \( (-1, -e^{-2}) \). (c) Let \( u = x \) and \( \mathrm{d}v = e^{2x} \,\mathrm{d}x \). Then \( \mathrm{d}u = \mathrm{d}x \) and \( v = \frac{1}{2}e^{2x} \). Using the integration by parts formula: \( \int u \,\mathrm{d}v = u v - \int v \,\mathrm{d}u \), we get \( \int x e^{2x} \,\mathrm{d}x = \frac{1}{2}x e^{2x} - \int \frac{1}{2}e^{2x} \,\mathrm{d}x = \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C \). (d) The curve intersects the \( x \)-axis where \( x e^{2x} = 0 \implies x = 0 \). For \( 0 \le x \le 2 \), \( f(x) \ge 0 \). Thus, the area of the region is given by \( \int_0^2 x e^{2x} \,\mathrm{d}x = [\frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x}]_0^2 = (\frac{1}{2}(2)e^4 - \frac{1}{4}e^4) - (0 - \frac{1}{4}e^0) = e^4 - \frac{1}{4}e^4 + \frac{1}{4} = \frac{3}{4}e^4 + \frac{1}{4} \).

(a) M1 for finding the derivative using product rule. A1 for setting derivative to 0 and obtaining \( x = -1/2 \). A1 for correct y-coordinate \( -1/(2e) \). R1 for validating the local minimum (e.g., using second derivative test or first derivative sign chart). (b) M1 for setting second derivative to 0. A1 for \( x = -1 \). A1 for correct y-coordinate \( -e^{-2} \). (c) M1 for correctly identifying \( u \) and \( \mathrm{d}v \). A1 for correct integration of \( v = \frac{1}{2}e^{2x} \). M1 for applying the integration by parts formula. AG for showing the given expression. (d) M1 for recognizing the limits are \( 0 \) and \( 2 \). M1 for substituting the limits into the integrated expression. A1 for substituting upper limit correctly. A1 for substituting lower limit correctly and obtaining the final exact area: \( \frac{3}{4}e^4 + \frac{1}{4} \).

(a) Using the double angle identity \( \cos(2x) = 2 \cos^2 x - 1 \), we have \( 2 \cos^2 x = \cos(2x) + 1 \). Substituting this into the function gives \( f(x) = (\cos(2x) + 1) + \sin(2x) = \cos(2x) + \sin(2x) + 1 \). (b) Expanding the right-hand side: \( \sqrt{2} \sin(2x + \frac{\pi}{4}) = \sqrt{2} (\sin(2x)\cos(\frac{\pi}{4}) + \cos(2x)\sin(\frac{\pi}{4})) \). Since \( \cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \), this becomes \( \sqrt{2} (\sin(2x) \cdot \frac{1}{\sqrt{2}} + \cos(2x) \cdot \frac{1}{\sqrt{2}}) = \sin(2x) + \cos(2x) \). Thus, the identity is shown. (c) From parts (a) and (b), we can express the function as \( f(x) = \sqrt{2} \sin(2x + \frac{\pi}{4}) + 1 \). The maximum value of the sine function is 1, so the maximum value of \( f(x) \) is \( \sqrt{2}(1) + 1 = \sqrt{2} + 1 \). This maximum occurs when \( \sin(2x + \frac{\pi}{4}) = 1 \implies 2x + \frac{\pi}{4} = \frac{\pi}{2} \implies 2x = \frac{\pi}{4} \implies x = \frac{\pi}{8} \). Since \( 0 \le x \le \pi \), the value \( x = \frac{\pi}{8} \) lies within the interval. (d) To solve \( f(x) = 1 \), we set \( \sqrt{2} \sin(2x + \frac{\pi}{4}) + 1 = 1 \implies \sin(2x + \frac{\pi}{4}) = 0 \). For \( 0 \le x \le \pi \), we have \( \frac{\pi}{4} \le 2x + \frac{\pi}{4} \le \frac{9\pi}{4} \). The values where the sine function equals 0 in this domain are \( 2x + \frac{\pi}{4} = \pi \) and \( 2x + \frac{\pi}{4} = 2\pi \). Solving these equations: \( 2x = \frac{3\pi}{4} \implies x = \frac{3\pi}{8} \) and \( 2x = \frac{7\pi}{4} \implies x = \frac{7\pi}{8} \). Both solutions lie within the interval \( [0, \pi] \).

(a) M1 for attempting to use the double angle identity for cos. A1 for correctly substituting \( 2\cos^2 x = \cos(2x) + 1 \). AG for obtaining the exact form. (b) M1 for using compound angle formula on \( \sin(2x + \frac{\pi}{4}) \). A1 for correct exact values of \( \cos(\frac{\pi}{4}) \) and \( \sin(\frac{\pi}{4}) \). AG for showing the identity. (c) A1 for identifying max value of \( \sqrt{2} + 1 \). M1 for setting the sine argument to \( \frac{\pi}{2} \). A1 for \( x = \frac{\pi}{8} \). (d) M1 for setting equation to 0: \( \sin(2x + \frac{\pi}{4}) = 0 \). M1 for finding the adjusted domain \( [\frac{\pi}{4}, \frac{9\pi}{4}] \) or solving for argument. A1 for \( 2x + \frac{\pi}{4} = \pi \implies x = \frac{3\pi}{8} \). A1 for \( 2x + \frac{\pi}{4} = 2\pi \implies x = \frac{7\pi}{8} \).

(a) For \( f(x) \) to be a valid probability density function, the total integral must equal 1: \( \int_0^2 k(4x - x^3) \,\mathrm{d}x = 1 \). Evaluating the integral: \( k \left[ 2x^2 - \frac{x^4}{4} \right]_0^2 = 1 \implies k \left( (2(4) - \frac{16}{4}) - 0 \right) = 1 \implies k(8 - 4) = 1 \implies 4k = 1 \implies k = \frac{1}{4} \). (b) The expected value is given by: \( \mathrm{E}(X) = \int_0^2 x f(x) \,\mathrm{d}x = \int_0^2 \frac{1}{4}(4x^2 - x^4) \,\mathrm{d}x \). Evaluating the integral: \( \mathrm{E}(X) = \frac{1}{4} \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_0^2 = \frac{1}{4} \left( \frac{32}{3} - \frac{32}{5} - 0 \right) = \frac{8}{3} - \frac{8}{5} = \frac{40 - 24}{15} = \frac{16}{15} \). (c) The probability \( \mathrm{P}(X > 1) \) is given by: \( \int_1^2 \frac{1}{4}(4x - x^3) \,\mathrm{d}x = \frac{1}{4} \left[ 2x^2 - \frac{x^4}{4} \right]_1^2 = \frac{1}{4} \left( (8 - 4) - (2 - \frac{1}{4}) \right) = \frac{1}{4} \left( 4 - \frac{7}{4} \right) = \frac{1}{4} \left( \frac{9}{4} \right) = \frac{9}{16} \). (d) To find \( \mathrm{P}(X < 1 \mid X < \sqrt{2}) \), we use conditional probability: \( \mathrm{P}(X < 1 \mid X < \sqrt{2}) = \frac{\mathrm{P}(X < 1 \cap X < \sqrt{2})}{\mathrm{P}(X < \sqrt{2})} = \frac{\mathrm{P}(X < 1)}{\mathrm{P}(X < \sqrt{2})} \). First, find \( \mathrm{P}(X < \sqrt{2}) = \int_0^{\sqrt{2}} \frac{1}{4}(4x - x^3) \,\mathrm{d}x = \frac{1}{4} \left[ 2x^2 - \frac{x^4}{4} \right]_0^{\sqrt{2}} = \frac{1}{4} \left( 2(2) - \frac{4}{4} - 0 \right) = \frac{3}{4} \). Since \( \mathrm{P}(X < 1) = 1 - \mathrm{P}(X > 1) = 1 - \frac{9}{16} = \frac{7}{16} \). Thus, the conditional probability is \( \frac{7/16}{3/4} = \frac{7}{16} \times \frac{4}{3} = \frac{7}{12} \).

(a) M1 for setting total integral to 1. A1 for correct integration of the polynomial. M1 for substituting limits of 0 and 2. AG for showing \( k = \frac{1}{4} \). (b) M1 for correct definition of \( \mathrm{E}(X) \). A1 for correct integration. M1 for substituting limits. A1 for exact value \( \frac{16}{15} \). (c) M1 for setting up the correct integral with limits 1 and 2. A1 for correct substitution of limits. A1 for exact value \( \frac{9}{16} \). (d) M1 for writing the conditional probability formula. M1 for calculating \( \mathrm{P}(X < \sqrt{2}) = \frac{3}{4} \). A1 for \( \mathrm{P}(X < 1) = \frac{7}{16} \). A1 for exact value \( \frac{7}{12} \).

(a) Let \(X\) be the mass of an apple, so \(X \sim N(150, \sigma^2)\).
We are given that \(P(X > 180) = 0.15\).
This can be standardized as:
\(P\left(Z > \frac{180 - 150}{\sigma}\right) = 0.15\)
\(P\left(Z < \frac{30}{\sigma}\right) = 0.85\)
Using a GDC or standard normal tables, we find the \(z\)-score for which the cumulative probability is 0.85:
\(z_{0.85} \approx 1.03643\)
Therefore:
\(\frac{30}{\sigma} = 1.03643 \Rightarrow \sigma = \frac{30}{1.03643} \approx 28.945\text{ g}\)
So, \(\sigma \approx 28.9\text{ g}\) (to 3 significant figures).

(b) We want to find \(P(130 < X < 160)\) using \(\sigma \approx 28.945\).
Using the normal cumulative distribution function (normalcdf) on a GDC with mean \(\mu = 150\) and standard deviation \(\sigma = 28.945\):
\(P(130 < X < 160) \approx 0.39036\)
So, the probability is approximately \(0.390\) (to 3 significant figures).

(a)
- (M1) for attempting to standardize the normal variable: \(\frac{180 - 150}{\sigma}\)
- (A1) for finding the correct \(z\)-value of \(1.03643\dots\)
- (A1) for \(\sigma \approx 28.9\text{ g}\) (accept 28.95 or 29)

(b)
- (M1) for setting up the correct probability interval: \(P(130 < X < 160)\)
- (M1) for using their standard deviation from (a) in a GDC normal cumulative distribution calculation
- (A1) for \(0.390\) (accept 0.39)

(a) The particle is at rest when \(v(t) = 0\).
\(5 e^{-0.4t} \sin(t) = 0\)
Since \(5 e^{-0.4t} > 0\) for all \(t\), we must have \(\sin(t) = 0\).
For \(0 \le t \le 6\), the non-zero solution is:
\(t = \pi \approx 3.14\text{ seconds}\).

(b) The speed is given by \(|v(t)|\).
Using a GDC to find the maximum value of \(|5 e^{-0.4t} \sin(t)|\) on \([0, 6]\):
By sketching the graph, the maximum occurs at \(t \approx 1.19\text{ s}\) in the first interval \([0, \pi]\).
At this point, the speed is:
\(v(1.19) \approx 5 e^{-0.4(1.19)} \sin(1.19) \approx 2.8837\text{ m s}^{-1}\).
(The local maximum speed in the interval \([\pi, 6]\) is much smaller, occurring at \(t \approx 4.33\text{ s}\) with speed \(|-0.824| = 0.824\text{ m s}^{-1}\)).
Thus, the maximum speed is \(2.88\text{ m s}^{-1}\).

(c) The total distance travelled is given by \(\int_0^6 |v(t)| \, dt\).
Using a GDC to evaluate this definite integral:
\(\int_0^6 |5 e^{-0.4t} \sin(t)| \, dt \approx 7.089\text{ m}\).
So, the total distance travelled is approximately \(7.09\text{ m}\).

(a)
- (A1) for \(t = \pi\) (or \(3.14\))

(b)
- (M1) for identifying that the maximum speed occurs when the derivative of velocity is zero (or by locating the peak of the absolute velocity graph using a GDC)
- (A1) for \(2.88\text{ m s}^{-1}\) (accept \(2.8837\dots\))

(c)
- (M1) for writing the integral expression for distance: \(\int_0^6 |v(t)| \, dt\) (or splitting it into \(\int_0^{\pi} v(t) \, dt - \int_{\pi}^6 v(t) \, dt\))
- (A1) for showing the GDC calculation of either of the components or setting up the correct integral
- (A1) for the final distance \(7.09\text{ m}\) (accept \(7.089\dots\))

(a) After \(1.5\) hours, the distances travelled by the boats are:
\(PA = 15 \times 1.5 = 22.5\text{ km}\)
\(PB = 22 \times 1.5 = 33\text{ km}\)

The angle between the two bearings is \(\angle APB = 115^\circ - 35^\circ = 80^\circ\).

Using the cosine rule to find the distance \(AB\):
\(AB^2 = PA^2 + PB^2 - 2(PA)(PB)\cos(\angle APB)\)
\(AB^2 = 22.5^2 + 33^2 - 2(22.5)(33)\cos(80^\circ)\)
\(AB^2 = 506.25 + 1089 - 1485 \cos(80^\circ)\)
\(AB^2 \approx 1595.25 - 257.87 = 1337.38\)
\(AB \approx \sqrt{1337.38} \approx 36.570\text{ km}\)

So the distance is \(36.6\text{ km}\) (to 3 significant figures).

(b) To find the bearing of \(B\) from \(A\), we first find \(\angle PAB\) using the sine rule:
\(\frac{\sin(\angle PAB)}{PB} = \frac{\sin(\angle APB)}{AB}\)
\(\frac{\sin(\angle PAB)}{33} = \frac{\sin(80^\circ)}{36.570}\)
\(\sin(\angle PAB) = \frac{33 \sin(80^\circ)}{36.570} \approx 0.88856\)
Since \(PB^2 < PA^2 + AB^2\) (i.e. \(1089 < 506.25 + 1337.38\)), \(\angle PAB\) is acute:
\(\angle PAB \approx \arcsin(0.88856) \approx 62.69^\circ\)

The bearing of \(P\) from \(A\) is \(35^\circ + 180^\circ = 215^\circ\).
Since boat \(B\) lies to the south-east of boat \(A\), we subtract \(\angle PAB\) from the bearing of \(P\) from \(A\) to find the bearing of \(B\) from \(A\):
\(\text{Bearing} = 215^\circ - 62.69^\circ = 152.31^\circ\)

So the bearing is \(152^\circ\) (to 3 significant figures).

(a)
- (A1) for calculating the lengths \(PA = 22.5\) and \(PB = 33\)
- (M1) for substituting correctly into the cosine rule: \(AB^2 = 22.5^2 + 33^2 - 2(22.5)(33)\cos(80^\circ)\)
- (A1) for \(AB \approx 36.6\text{ km}\) (accept \(36.570\dots\))

(b)
- (M1) for using the sine rule (or cosine rule) to find \(\angle PAB\): e.g. \(\frac{\sin(\angle PAB)}{33} = \frac{\sin(80^\circ)}{36.570}\)
- (A1) for \(\angle PAB \approx 62.7^\circ\) (accept \(62.69^\circ\dots\))
- (A1) for a final bearing of \(152^\circ\) (accept \(152.31^\circ\dots\))

(a) At \(t=0\), \(T_1(0) = 20 + 75 e^0 = 20 + 75 = 95\, ^\circ\text{C}\).

(b) Set \(T_1(t) = T_2(t)\):
\(20 + 75 e^{-0.08t} = 20 + 65 e^{-0.05t}\)
\(75 e^{-0.08t} = 65 e^{-0.05t}\)
Divide both sides by \(65 e^{-0.08t}\):
\(\frac{75}{65} = \frac{e^{-0.05t}}{e^{-0.08t}}\)
\(e^{0.03t} = \frac{15}{13}\)
Taking the natural logarithm of both sides:
\(0.03t = \ln\left(\frac{15}{13}\right)\)
\(t = \frac{\ln(15/13)}{0.03} \approx 4.7701\text{ minutes}\)
So, \(t \approx 4.77\text{ minutes}\).

(c) The rate of change of the coffee's temperature is given by \(T_1'(t)\).
\(T_1'(t) = 75 \times (-0.08) e^{-0.08t} = -6 e^{-0.08t}\)
At \(t = 4.7701\):
\(T_1'(4.7701) = -6 e^{-0.08(4.7701)} \approx -4.0967\, ^\circ\text{C min}^{-1}\)
So, the rate of change of the coffee's temperature is approximately \(-4.10\, ^\circ\text{C min}^{-1}\) (or cooling at a rate of \(4.10\, ^\circ\text{C min}^{-1}\)).

(a)
- (A1) for \(95\, ^\circ\text{C}\)

(b)
- (M1) for equating \(T_1(t) = T_2(t)\)
- (M1) for showing a valid method to solve the equation (either algebraically using logs or showing GDC solver setup)
- (A1) for \(t \approx 4.77\text{ minutes}\)

(c)
- (M1) for attempting to differentiate \(T_1(t)\) with respect to \(t\) (or evaluating numerical derivative on GDC at their value of \(t\))
- (A1) for \(-4.10\, ^\circ\text{C min}^{-1}\) (accept \(-4.1\) or a positive value if framed as 'cooling at a rate of 4.10')

(a) Using the sum formula for an arithmetic sequence:
\(S_n = \frac{n}{2} [2u_1 + (n-1)d]\)
\(S_{20} = \frac{20}{2} [2(5) + 19(3)]\)
\(S_{20} = 10 [10 + 57] = 670\).

(b) Write expressions for \(S_n\) and \(G_n\):
\(S_n = \frac{n}{2} [10 + 3(n-1)] = 1.5n^2 + 3.5n\)
\(G_n = \frac{v_1(r^n - 1)}{r - 1} = \frac{3(1.2^n - 1)}{1.2 - 1} = 15(1.2^n - 1)\)

We need to find the least integer \(n\) such that:
\(15(1.2^n - 1) > 1.5n^2 + 3.5n\)

Using a GDC to create a table of values or find the intersection of the continuous curves:
For \(n = 21\):
\(S_{21} = 1.5(21)^2 + 3.5(21) = 661.5 + 73.5 = 735\)
\(G_{21} = 15(1.2^{21} - 1) \approx 675.08\)
Since \(G_{21} < S_{21}\), the condition is not met.

For \(n = 22\):
\(S_{22} = 1.5(22)^2 + 3.5(22) = 726 + 77 = 803\)
\(G_{22} = 15(1.2^{22} - 1) \approx 813.09\)
Since \(G_{22} > S_{22}\), the condition is met.

Thus, the least value of \(n\) is \(22\).

(a)
- (M1) for substituting into the arithmetic sum formula
- (A1) for \(670\)

(b)
- (M1) for setting up the inequality \(15(1.2^n - 1) > 1.5n^2 + 3.5n\) (or equivalent)
- (M1) for attempting to use a GDC table or graph to compare values of \(S_n\) and \(G_n\)
- (A1) for showing values for \(n = 21\) and \(n = 22\) (e.g. \(G_{21} \approx 675\), \(S_{21} = 735\) and \(G_{22} \approx 813\), \(S_{22} = 803\))
- (A1) for \(n = 22\)

(a) Entering the data into the GDC lists and performing linear regression (LinReg \(ax+b\)):
\(r \approx 0.98767\)
So, \(r \approx 0.988\) (to 3 significant figures).

(b) The equation of the regression line of \(y\) on \(x\) is in the form \(y = mx + c\).
From the GDC:
\(m \approx 2.89036\)
\(c \approx 40.4988\)
So, the regression equation is:
\(y = 2.89x + 40.5\) (with coefficients given to 3 significant figures).

(c) To estimate the score for \(x = 10\), substitute \(x = 10\) into the regression equation:
\(y = 2.89036(10) + 40.4988 = 28.9036 + 40.4988 = 69.4024\)
So, the estimated exam score is approximately \(69.4\) (to 3 significant figures).

(a)
- (M1) for an attempt to use the bivariate statistics menu on a GDC with the given data
- (A1) for \(r \approx 0.988\) (accept \(0.987\dots\))

(b)
- (M1) for attempting to find the regression line parameters on a GDC
- (A1) for \(y = 2.89x + 40.5\) (accept \(y = 2.89036\dots x + 40.4988\dots\))

(c)
- (M1) for substituting \(x = 10\) into their regression equation
- (A1) for \(69.4\) (accept \(69\) if using rounded coefficients, but \(69.4\) is preferred from unrounded GDC values)

Let \(X\) be the weight of a randomly chosen apple, where \(X \sim N(150, 12^2)\).

(a) We want to find \(P(X > 165)\).
Using a standard normal distribution or a GDC:
\(P(X > 165) = P\left(Z > \frac{165 - 150}{12}\right) = P(Z > 1.25) \approx 0.10565\)
To 3 significant figures, \(P(X > 165) \approx 0.106\).

(b) We are given \(P(X < w) = 0.08\).
Using the inverse normal function on a GDC:
\(w \approx 133.136\)
To 3 significant figures, \(w \approx 133\) grams.

(c) Let \(Y\) be the number of apples weighing more than \(165\) grams in a box of \(20\).
\(Y \sim B(20, p)\) where \(p = 0.10565...\)
We want to find \(P(Y \ge 3) = 1 - P(Y \le 2)\).
Using a GDC with binomial cumulative probability:
\(P(Y \le 2) \approx 0.64373\)
\(P(Y \ge 3) = 1 - 0.64373 = 0.35627\)
To 3 significant figures, \(P(Y \ge 3) \approx 0.356\).

(d) Let \(q\) be the probability that an individual apple weighs more than \(d\) grams.
Since the selections are independent, the probability that all \(6\) apples weigh more than \(d\) grams is \(q^6 = 0.05\).
\(q = (0.05)^{1/6} \approx 0.60696\)
We require \(P(X > d) = 0.60696\), which means \(P(X < d) = 1 - 0.60696 = 0.39304\).
Using the inverse normal function on a GDC:
\(d \approx 146.743\)
To 3 significant figures, \(d \approx 147\) grams.

(e) We want to find the conditional probability \(P(140 < X < 160 \mid X > 140)\).
Using the formula for conditional probability:
\(P(140 < X < 160 \mid X > 140) = \frac{P(140 < X < 160)}{P(X > 140)}\)
Using GDC:
\(P(140 < X < 160) \approx 0.79767 - 0.20233 = 0.59534\)
\(P(X > 140) \approx 1 - 0.20233 = 0.79767\)
\(\text{Ratio} = \frac{0.59534}{0.79767} \approx 0.74635\)
To 3 significant figures, the probability is \(0.746\).

(a)
(M1) for attempting to standardize or write down standard normal probability expression.
(A1) for \(0.106\) (accept \(0.10565...\)).

(b)
(M1) for set-up equation \(P(X < w) = 0.08\).
(A1) for a graphical representation or standardizing formula.
(A1) for \(133\) (accept \(133.136...\)).

(c)
(M1) for identifying the distribution as binomial \(Y \sim B(20, 0.10565)\).
(M1) for attempting to find \(1 - P(Y \le 2)\).
(A1) for \(0.356\) (accept \(0.35627...\)).

(d)
(M1) for setting up the equation \(q^6 = 0.05\).
(A1) for finding \(q \approx 0.607\).
(M1) for setting up the equation \(P(X > d) = 0.607\) or \(P(X < d) = 0.393\).
(A1) for \(147\) (accept \(146.743...\)).

(e)
(M1) for writing down the conditional probability expression.
(A1) for finding \(P(140 < X < 160) \approx 0.595\) and \(P(X > 140) \approx 0.798\).
(A1) for \(0.746\) (accept \(0.74635...\)).

(a) The initial rate of flow is when \(t = 0\).
\(f(0) = 4\cos(0) - 0.5(0)^2 + 3(0) + 10 = 4(1) + 10 = 14\) cubic meters per hour.

(b) To find the maximum rate of flow, we find where the derivative \(f'(t) = 0\).
\(f'(t) = -3.2\sin(0.8t) - t + 3\)
Setting \(f'(t) = 0\):
\(-3.2\sin(0.8t) - t + 3 = 0\)
Using a GDC to solve this equation in the interval \(0 \le t \le 6\):
\(t \approx 0.8966\) hours.
To confirm it is a maximum, we can examine the second derivative \(f''(t) = -2.56\cos(0.8t) - 1\).
At \(t = 0.8966\), \(f''(0.8966) \approx -2.92 < 0\), confirming a local maximum.
To 3 significant figures, \(t \approx 0.897\) hours.

(c) The total volume of water is the definite integral of the flow rate function from \(t = 0\) to \(t = 4\).
\(V = \int_{0}^{4} (4\cos(0.8t) - 0.5t^2 + 3t + 10) \, dt\)
Using a GDC to evaluate the definite integral:
\(V \approx 53.041\)
To 3 significant figures, the total volume is \(53.0\) \(\text{m}^3\).

(d) The rate of change of the water flow rate is given by the derivative \(f'(t)\).
We want to find \(f'(3)\).
\(f'(3) = -3.2\sin(0.8 \times 3) - 3 + 3 = -3.2\sin(2.4)\)
Using a GDC in radian mode:
\(f'(3) \approx -3.2(0.67546) = -2.161\)
To 3 significant figures, the rate of change is \(-2.16\) \(\text{m}^3\text{h}^{-2}\).

(e) The rate of flow is decreasing most rapidly when its derivative \(f'(t)\) is at a local minimum.
This occurs when \(f''(t) = 0\) and changes from negative to positive.
\(f''(t) = -2.56\cos(0.8t) - 1 = 0\)
\(\cos(0.8t) = -\frac{1}{2.56} = -0.390625\)
In the interval \(0 \le t \le 6\), we solve for \(0.8t\):
\(0.8t = \arccos(-0.390625) \approx 1.9722\)
\(t = \frac{1.9722}{0.8} \approx 2.465\) hours.
Testing the second derivative behavior around \(t \approx 2.47\):
For \(t < 2.465\), \(f''(t) < 0\) (flow rate is concave down).
For \(t > 2.465\), \(f''(t) > 0\) (flow rate is concave up).
Thus, \(t \approx 2.465\) is indeed a point of inflection representing the minimum rate of change (most rapid decrease).
To 3 significant figures, \(t \approx 2.47\) hours.

(a)
(A1) for \(14\).

(b)
(M1) for differentiating \(f(t)\) correctly: \(f'(t) = -3.2\sin(0.8t) - t + 3\).
(M1) for setting \(f'(t) = 0\).
(M1) for an attempt to solve on GDC.
(A1) for \(0.897\) (accept \(0.8966...\)).

(c)
(M1) for writing down the correct definite integral \(\int_{0}^{4} f(t) \, dt\).
(A1) for attempting to integrate or use GDC numerical integration.
(A1) for \(53.0\) (accept \(53.041...\)).

(d)
(M1) for recognizing that the rate of change is \(f'(3)\).
(A1) for substituting \(3\) into the derivative formula.
(A1) for \(-2.16\) (accept \(-2.161...\)).

(e)
(M1) for recognizing that the flow is decreasing most rapidly when \(f''(t) = 0\) or at the minimum of \(f'(t)\).
(M1) for \(-2.56\cos(0.8t) - 1 = 0\).
(A1) for solving to find the critical point \(2.465...\).
(A1) for justifying that this is a minimum of \(f'(t)\) (e.g., via table, derivative sign test, or graph analysis) and giving \(2.47\) (accept \(2.465...\)).

(a) Using the Cosine Rule in \(\triangle ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\widehat{ABC})\)
\(AC^2 = 80^2 + 110^2 - 2(80)(110)\cos(115^\circ)\)
\(AC^2 = 6400 + 12100 - 17600\cos(115^\circ)\)
\(AC^2 \approx 18500 - 17600(-0.422618) \approx 25938.08\)
\(AC \approx \sqrt{25938.08} \approx 161.053\) meters.
To 3 significant figures, \(AC \approx 161\) meters.

(b) Using the area formula for a triangle:
\(\text{Area} = \frac{1}{2}(AB)(BC)\sin(\widehat{ABC})\)
\(\text{Area} = \frac{1}{2}(80)(110)\sin(115^\circ) = 4400\sin(115^\circ) \approx 3987.75\) square meters.
To 3 significant figures, \(\text{Area} \approx 3990\) \(\text{m}^2\).

(c) Since the tower \(TB\) is vertical, \(\triangle TBA\) is right-angled at \(B\).
\(\tan(18^\circ) = \frac{TB}{AB} = \frac{h}{80}\)
\(h = 80\tan(18^\circ) \approx 80(0.32492) \approx 25.9936\) meters.
To 3 significant figures, the height is \(26.0\) meters.

(d) Since \(\triangle TBC\) is right-angled at \(B\), let \(\theta\) be the angle of elevation from \(C\) to \(T\).
\(\tan\theta = \frac{TB}{BC} = \frac{25.9936}{110} \approx 0.236305\)
\(\theta = \arctan(0.236305) \approx 13.296^\circ\) (or \(0.232\) radians).
To 3 significant figures, the angle of elevation is \(13.3^\circ\) (or \(0.232\) radians).

(e) Let \(P\) be a point on \(AC\). In the right-angled triangle \(\triangle TBP\), by Pythagoras' Theorem:
\(TP = \sqrt{TB^2 + BP^2} = \sqrt{h^2 + BP^2}\)
To minimize \(TP\), we must minimize the distance \(BP\) on the horizontal ground. The minimum distance from \(B\) to the line \(AC\) is the perpendicular distance from \(B\) to \(AC\).
Using the area of \(\triangle ABC\):
\(\text{Area} = \frac{1}{2} \times AC \times BP\)
\(3987.75 = \frac{1}{2} \times 161.053 \times BP\)
\(BP = \frac{2 \times 3987.75}{161.053} \approx 49.521\) meters.
Now we find the minimum length of the cable:
\(TP_{\min} = \sqrt{h^2 + BP^2} = \sqrt{25.9936^2 + 49.521^2} = \sqrt{675.667 + 2452.33} = \sqrt{3128.00} \approx 55.928\) meters.
To 3 significant figures, the minimum length of the cable is \(55.9\) meters.

(a)
(M1) for attempting to use the Cosine Rule.
(A1) for correct substitution: \(80^2 + 110^2 - 2(80)(110)\cos(115^\circ)\).
(A1) for \(161\) (accept \(161.053...\)).

(b)
(M1) for attempting to use the area formula \(\frac{1}{2}ab\sin C\).
(A1) for \(3990\) (accept \(3987.75...\)).

(c)
(M1) for recognizing that \(\triangle TBA\) is right-angled and writing a correct trigonometric ratio.
(A1) for substituting correctly: \(h = 80\tan(18^\circ)\).
(A1) for \(26.0\) (accept \(25.9936...\)).

(d)
(M1) for recognizing that \(\triangle TBC\) is right-angled and writing a correct trigonometric ratio.
(A1) for substituting correctly: \(\tan\theta = \frac{25.9936}{110}\).
(A1) for \(13.3^\circ\) or \(0.232\) radians (accept \(13.296...^\circ\) or \(0.23207...\) radians).

(e)
(M1) for recognizing that the minimum cable length occurs when \(BP\) is perpendicular to \(AC\).
(M1) for calculating \(BP\) using the area of the triangle: \(BP = \frac{2 \times \text{Area}}{AC}\).
(A1) for finding \(BP \approx 49.521\) meters.
(A1) for finding the minimum cable length using Pythagoras: \(\sqrt{25.9936^2 + 49.521^2} \approx 55.9\) (accept \(55.928...\)).