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

For an infinite geometric series to converge, we require \( |r| < 1 \). Here, the common ratio is \( r = \frac{1}{\ln(k^2)} = \frac{1}{2\ln(k)} \). Since \( k > 1 \), we have \( \ln(k) > 0 \), which implies that \( r > 0 \). Therefore, the convergence condition \( |r| < 1 \) simplifies to \( \frac{1}{2\ln(k)} < 1 \). Since \( \ln(k) > 0 \), we can multiply both sides of the inequality by \( 2\ln(k) \) to obtain \( 1 < 2\ln(k) \). Dividing by 2 yields \( \ln(k) > \frac{1}{2} \). Exponentiating both sides with base \( e \) gives \( k > e^{1/2} \). Since \( e^{1/2} > 1 \), this automatically satisfies the initial condition \( k > 1 \). Thus, the series converges for all \( k > e^{1/2} \) (or \( k > \sqrt{e} \)).

M1 for setting up the convergence condition \( |r| < 1 \). A1 for expressing the common ratio as \( r = \frac{1}{2\ln(k)} \). M1 for setting up the inequality \( \frac{1}{2\ln(k)} < 1 \). A1 for obtaining the simplified inequality \( \ln(k) > \frac{1}{2} \). A1.83 for the final correct set of values \( k > e^{1/2} \) (or \( k > \sqrt{e} \)).

Since \( (g \circ f)(x) = x \), the function \( g(x) \) is the inverse of \( f(x) \), which means \( g(x) = f^{-1}(x) \). Let \( y = \frac{2x+1}{x-3} \). We solve for \( x \) in terms of \( y \) by multiplying both sides by \( x-3 \): \( y(x-3) = 2x+1 \), which expands to \( xy - 3y = 2x + 1 \). Grouping the terms containing \( x \) on one side gives \( xy - 2x = 3y + 1 \). Factoring out \( x \) yields \( x(y-2) = 3y + 1 \). Thus, we find \( x = \frac{3y+1}{y-2} \). Interchanging the variables, we get the expression for the inverse function: \( g(x) = \frac{3x+1}{x-2} \). The domain of \( g(x) \) is the range of \( f(x) \). Since \( f(x) \) has a horizontal asymptote at \( y = 2 \), its range is all real numbers except 2. Thus, the domain of \( g(x) \) is \( x \in \mathbb{R}, x \neq 2 \).

M1 for realizing that \( g \) is the inverse function of \( f \). M1 for attempting to solve for \( x \) in terms of \( y \). A1 for obtaining \( x(y-2) = 3y+1 \) or equivalent. A1 for the correct expression \( g(x) = \frac{3x+1}{x-2} \). A1.83 for stating the correct domain \( x \neq 2 \) (or \( \mathbb{R} \setminus \{2\} \)).

We use the Pythagorean identity \( \cos^2(x) = 1 - \sin^2(x) \) to rewrite the equation in terms of sine: \( 2(1 - \sin^2(x)) + 3\sin(x) - 3 = 0 \). Expanding this gives \( 2 - 2\sin^2(x) + 3\sin(x) - 3 = 0 \), which simplifies to \( -2\sin^2(x) + 3\sin(x) - 1 = 0 \). Multiplying the entire equation by \(-1\) yields the quadratic equation \( 2\sin^2(x) - 3\sin(x) + 1 = 0 \). We factor this quadratic expression to obtain \( (2\sin(x) - 1)(\sin(x) - 1) = 0 \). This gives two possible cases: \( \sin(x) = \frac{1}{2} \) or \( \sin(x) = 1 \). Within the specified interval \( 0 \leq x \leq 2\pi \), the solutions for \( \sin(x) = \frac{1}{2} \) are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \). The solution for \( \sin(x) = 1 \) is \( x = \frac{\pi}{2} \). Combining these, the complete set of solutions is \( x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \).

M1 for substituting \( \cos^2(x) = 1 - \sin^2(x) \). A1 for writing the correct simplified quadratic equation \( 2\sin^2(x) - 3\sin(x) + 1 = 0 \). M1 for factoring the quadratic equation. A1 for finding the solutions \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \). A1.83 for finding the solution \( x = \frac{\pi}{2} \) and stating all three correct solutions with no extra incorrect values in the interval.

First, we find the \( y \)-coordinate of the point of contact by substituting \( x = e \) into the curve equation: \( y = e^2 \ln(e) = e^2(1) = e^2 \). So the point is \( (e, e^2) \). Next, we find the derivative of \( y \) with respect to \( x \) using the product rule: \( \frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \ln(x) + x^2 \cdot \frac{d}{dx}(\ln(x)) = 2x \ln(x) + x^2 \left(\frac{1}{x}\right) = 2x \ln(x) + x \). To find the gradient of the tangent, \( m \), we evaluate the derivative at \( x = e \): \( m = 2e \ln(e) + e = 2e(1) + e = 3e \). Using the point-slope formula for a straight line: \( y - y_1 = m(x - x_1) \), we substitute the point \( (e, e^2) \) and gradient \( m = 3e \): \( y - e^2 = 3e(x - e) \). Expanding the right-hand side gives \( y - e^2 = 3ex - 3e^2 \). Adding \( e^2 \) to both sides yields the final equation: \( y = 3ex - 2e^2 \).

A1 for finding the correct \( y \)-coordinate \( e^2 \). M1 for applying the product rule to differentiate \( x^2 \ln(x) \). A1 for obtaining the correct derivative \( 2x \ln(x) + x \). A1 for calculating the correct gradient of the tangent \( m = 3e \). A1.83 for writing the final tangent equation in the form \( y = mx + c \) as \( y = 3ex - 2e^2 \).

We use the definition of the expected value of a discrete random variable, \( \text{E}(X) = \sum x \cdot \text{P}(X=x) \). Substituting the values from the table, we get: \( \text{E}(X) = 1(k) + 2(2k) + 3(1 - 3k) \). Expanding this expression yields: \( \text{E}(X) = k + 4k + 3 - 9k \), which simplifies to: \( \text{E}(X) = 3 - 4k \). We are given that \( \text{E}(X) = 2.1 \). Therefore, we can set up the equation: \( 3 - 4k = 2.1 \). Rearranging the terms to solve for \( k \) gives: \( 4k = 3 - 2.1 \), which simplifies to \( 4k = 0.9 \). Solving for \( k \) yields: \( k = \frac{0.9}{4} = 0.225 \) (or \( \frac{9}{40} \)).

M1 for writing down the correct expected value expression in terms of \( k \). A1 for simplifying the expression to \( 3 - 4k \). M1 for setting up the equation \( 3 - 4k = 2.1 \). A2.83 for solving the equation correctly to get \( k = 0.225 \) (or equivalent fraction \( \frac{9}{40} \)).

We use the method of integration by substitution. Let \( u = \sin(2x) \). The differential is \( du = 2\cos(2x) \, dx \), which implies that \( \cos(2x) \, dx = \frac{1}{2} du \). Next, we change the limits of integration to be in terms of \( u \): when \( x = 0 \), \( u = \sin(0) = 0 \); when \( x = \frac{\pi}{4} \), \( u = \sin\left(2 \cdot \frac{\pi}{4}\right) =  \sin\left(\frac{\pi}{2}\right) = 1 \). Substituting these into the integral gives: \( \int_{0}^{\pi/4} \cos(2x) e^{\sin(2x)} \, dx = \int_{0}^{1} e^u \cdot \frac{1}{2} du \). This simplifies to: \( \frac{1}{2} \int_{0}^{1} e^u \, du = \frac{1}{2} [e^u]_0^1 = \frac{1}{2}(e^1 - e^0) = \frac{1}{2}(e - 1) \). Thus, the exact value of the integral is \( \frac{e-1}{2} \).

M1 for using integration by substitution. A1 for finding the correct differential \( du = 2\cos(2x)\,dx \). A1 for finding the correct new limits of integration, \( u = 0 \) and \( u = 1 \) (or for returning to the original variable with correct limits). A1 for finding the antiderivative \( \frac{1}{2}e^{\sin(2x)} \) or \( \frac{1}{2}e^u \). A1.83 for evaluating correctly to obtain the exact value \( \frac{e-1}{2} \) (or equivalent expression such as \( \frac{1}{2}(e-1) \)).

(a) Using the quotient rule: Let \( u = \ln(x) \) and \( v = x^2 \). Then \( u' = \frac{1}{x} \) and \( v' = 2x \). \( f'(x) = \frac{\frac{1}{x} \cdot x^2 - \ln(x) \cdot 2x}{(x^2)^2} = \frac{x - 2x\ln(x)}{x^4} = \frac{1 - 2\ln(x)}{x^3} \). (b) For the local maximum, set \( f'(x) = 0 \): \( 1 - 2\ln(x) = 0 \Rightarrow \ln(x) = \frac{1}{2} \Rightarrow x = e^{1/2} = \sqrt{e} \). Substituting \( x = \sqrt{e} \) back into \( f(x) \): \( y = \frac{\ln(\sqrt{e})}{(\sqrt{e})^2} = \frac{1/2}{e} = \frac{1}{2e} \). The coordinates are \( \left(\sqrt{e}, \frac{1}{2e}\right) \). (c) For the point of inflection, find \( f''(x) \) using the quotient rule: \( f''(x) = \frac{-\frac{2}{x} \cdot x^3 - (1 - 2\ln(x)) \cdot 3x^2}{x^6} = \frac{-2x^2 - 3x^2 + 6x^2\ln(x)}{x^6} = \frac{6\ln(x) - 5}{x^4} \). Set \( f''(x) = 0 \): \( 6\ln(x) - 5 = 0 \Rightarrow \ln(x) = \frac{5}{6} \Rightarrow x = e^{5/6} \). Substituting \( x = e^{5/6} \) back into \( f(x) \): \( y = \frac{\ln(e^{5/6})}{(e^{5/6})^2} = \frac{5/6}{e^{5/3}} = \frac{5}{6e^{5/3}} \). The coordinates are \( \left(e^{5/6}, \frac{5}{6e^{5/3}}\right) \). (d) The curve intersects the x-axis when \( f(x) = 0 \Rightarrow \ln(x) = 0 \Rightarrow x = 1 \). The vertical line passing through the local maximum is \( x = \sqrt{e} \). The area \( R \) is given by \( \int_{1}^{\sqrt{e}} \frac{\ln(x)}{x^2} dx \). Using integration by parts: let \( u = \ln(x) \), \( dv = x^{-2} dx \), then \( du = \frac{1}{x} dx \), \( v = -\frac{1}{x} \). \( \int \frac{\ln(x)}{x^2} dx = -\frac{\ln(x)}{x} - \int -\frac{1}{x^2} dx = -\frac{\ln(x)}{x} - \frac{1}{x} = -\frac{\ln(x) + 1}{x} \). Evaluating from 1 to \( \sqrt{e} \): \( \left[ -\frac{\ln(x) + 1}{x} \right]_{1}^{\sqrt{e}} = \left( -\frac{\ln(\sqrt{e}) + 1}{\sqrt{e}} \right) - \left( -\frac{\ln(1) + 1}{1} \right) = -\frac{1.5}{\sqrt{e}} + 1 = 1 - \frac{3}{2\sqrt{e}} \).

(a) M1 for attempting quotient rule, A1 for correct numerator expression, A1 for simplified derivative. (b) M1 for setting \( f'(x) = 0 \), A1 for correct x-coordinate, A1 for correct y-coordinate. (c) M1 for derivative of \( f'(x) \), A1 for correct second derivative, A1 for finding \( x = e^{5/6} \), A1 for correct y-coordinate. (d) M1 for identifying lower limit \( x = 1 \), M1 for setting up integral, M1 for integration by parts process, A1 for correct antiderivative, A1 for correct final exact value.

(a) Using the identity \( \sin^2(x) = 1 - \cos^2(x) \): \( f(x) = 2(1 - \cos^2(x)) + 3\cos(x) = -2\cos^2(x) + 3\cos(x) + 2 \). So \( a = -2 \), \( b = 3 \), \( c = 2 \). (b) Set \( f(x) = 0 \): \( -2\cos^2(x) + 3\cos(x) + 2 = 0 \Rightarrow 2\cos^2(x) - 3\cos(x) - 2 = 0 \). Let \( t = \cos(x) \). The quadratic is \( 2t^2 - 3t - 2 = 0 \Rightarrow (2t + 1)(t - 2) = 0 \). This gives \( t = -\frac{1}{2} \) or \( t = 2 \). Since \( -1 \le \cos(x) \le 1 \), there are no solutions for \( \cos(x) = 2 \). For \( \cos(x) = -\frac{1}{2} \) in the interval \( 0 \le x \le 2\pi \), the solutions are \( x = \frac{2\pi}{3} \) and \( x = \frac{4\pi}{3} \). (c) Let \( t = \cos(x) \) where \( -1 \le t \le 1 \). We look for the maximum and minimum of the quadratic function \( g(t) = -2t^2 + 3t + 2 \). The vertex of the parabola is at \( t = -\frac{3}{2(-2)} = \frac{3}{4} \). Since \( \frac{3}{4} \in [-1, 1] \), the maximum value is \( g\left(\frac{3}{4}\right) = -2\left(\frac{9}{16}\right) + 3\left(\frac{3}{4}\right) + 2 = -\frac{9}{8} + \frac{18}{8} + 2 = \frac{25}{8} \). Testing endpoints: \( g(-1) = -2(-1)^2 + 3(-1) + 2 = -3 \) and \( g(1) = -2(1)^2 + 3(1) + 2 = 3 \). The minimum value is \( -3 \). Thus, the range of \( f(x) \) is \( \left[-3, \frac{25}{8}\right] \). (d) Differentiating \( f(x) \): \( f'(x) = 4\sin(x)\cos(x) - 3\sin(x) \). At \( x = \frac{\pi}{3} \): \( f'\left(\frac{\pi}{3}\right) = 4\sin\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{3}\right) - 3\sin\left(\frac{\pi}{3}\right) = 4\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - 3\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} - \frac{3\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} \).

(a) M1 for identity substitution, A1 for correct form. (b) M1 for forming quadratic in cosine, M1 for factoring quadratic, R1 for rejecting \( \cos(x) = 2 \), A1 for \( x = \frac{2\pi}{3} \), A1 for \( x = \frac{4\pi}{3} \). (c) M1 for recognizing vertex location, A1 for finding vertex y-value, M1 for checking boundary values, A1 for minimum value, A1 for correct range notation. (d) M1 for correct derivative, M1 for substituting \( \frac{\pi}{3} \) with exact trig ratios, A1 for final answer.

(a) The sum of the probabilities must equal 1: \( k + 2k^2 + 3k + 0.12 = 1 \Rightarrow 2k^2 + 4k - 0.88 = 0 \Rightarrow k^2 + 2k - 0.44 = 0 \). Multiplying by 100 and dividing by 4 gives: \( 25k^2 + 50k - 11 = 0 \). Factorizing the quadratic: \( (5k - 1)(5k + 11) = 0 \). Since probabilities must be positive and \( P(X=1) = k \), \( k > 0 \). Thus, we reject \( k = -2.2 \), leaving the only valid solution as \( k = \frac{1}{5} = 0.2 \). (b) Substitute \( k = 0.2 \) into the table: \( P(X=1) = 0.2 \), \( P(X=2) = 2(0.2)^2 = 0.08 \), \( P(X=3) = 3(0.2) = 0.6 \), \( P(X=4) = 0.12 \). \( \text{E}(X) = 1(0.2) + 2(0.08) + 3(0.6) + 4(0.12) = 0.2 + 0.16 + 1.8 + 0.48 = 2.64 \). (c) The probability of obtaining an odd score is \( P(\text{Odd}) = P(X=1) + P(X=3) = 0.2 + 0.6 = 0.8 \). The probability of losing is \( 1 - 0.8 = 0.2 \). Expected gain \( \text{E}(G) = 5(0.8) - 10(0.2) = 4 - 2 = 2 \). (d) Let \( Y \) be the number of rounds won. \( Y \sim B(3, 0.8) \). We want to find \( P(Y \ge 2) = P(Y = 2) + P(Y = 3) \). \( P(Y=2) = \binom{3}{2} (0.8)^2 (0.2)^1 = 3(0.64)(0.2) = 0.384 \). \( P(Y=3) = (0.8)^3 = 0.512 \). \( P(Y \ge 2) = 0.384 + 0.512 = 0.896 \).

(a) M1 for sum of probabilities equal to 1, A1 for setting up quadratic equation, M1 for factorizing/solving, A1 for both roots, R1 for rejecting the negative root and concluding \( k = 0.2 \). (b) M1 for evaluating individual probabilities, M1 for applying the formula of expectation, A1 for \( 2.64 \). (c) M1 for finding winning probability \( 0.8 \), M1 for expected value formula, A1 for correct answer of $2 (or 2). (d) M1 for identifying binomial model, M1 for identifying correct terms to sum, A1 for calculations of both terms, A1 for final probability \( 0.896 \) (or equivalent fraction).

On 31 December of the tenth year, the first deposit has earned interest for 10 years, the second for 9 years, and so on, up to the tenth deposit which has earned interest for 1 year. The total amount \(S_{10}\) is given by the sum of a geometric series: \(S_{10} = 1000(1.04)^1 + 1000(1.04)^2 + \dots + 1000(1.04)^{10}\). This is a geometric series with first term \(u_1 = 1000(1.04) = 1040\), common ratio \(r = 1.04\), and number of terms \(n = 10\). Using the sum formula: \(S_{n} = \frac{u_1(r^n - 1)}{r - 1}\), we get \(S_{10} = \frac{1040(1.04^{10} - 1)}{1.04 - 1}\). Using a GDC to calculate this value: \(S_{10} \approx 12486.35\). To the nearest USD, the total amount is \(12486\).

M1 for identifying the series as geometric. A1 for writing the correct first term \(u_1 = 1040\) and ratio \(r = 1.04\). M1 for substituting correct values into the geometric series sum formula. A1 for a correct intermediate step or unrounded value \(12486.35\). A1 for the final answer of \(12486\) (accept \(12500\) from 3 s.f. rounding).

We need to solve the equation \(P(t) = 400\) for \(t\): \(\frac{600}{1 + 5e^{-0.4t}} = 400\). Multiply both sides by \(1 + 5e^{-0.4t}\): \(600 = 400(1 + 5e^{-0.4t})\). Divide by 400: \(1.5 = 1 + 5e^{-0.4t}\). Subtract 1: \(0.5 = 5e^{-0.4t}\). Divide by 5: \(e^{-0.4t} = 0.1\). Take the natural logarithm of both sides: \(-0.4t = \ln(0.1)\). Solving for \(t\) gives \(t = \frac{\ln(0.1)}{-0.4} \approx 5.75646\). To three significant figures, \(t = 5.76\) years.

M1 for setting up the equation \(P(t) = 400\). M1 for algebraic simplification or setting up GDC intersection. A1 for simplifying to \(e^{-0.4t} = 0.1\) or equivalent. M1 for taking natural logarithms to solve for \(t\). A1 for \(5.76\).

The area of a triangle is given by the formula: \(\text{Area} = \frac{1}{2} a c \sin B\). Substituting the given values: \(24 = \frac{1}{2} (9)(7) \sin B\), which simplifies to \(24 = 31.5 \sin B\). Thus, \(\sin B = \frac{24}{31.5} = \frac{16}{21} \approx 0.761905\). We find the acute angle using the inverse sine function on a GDC: \(B_1 = \arcsin\left(\frac{16}{21}\right) \approx 0.86595\text{ radians}\). Since \(\sin(\pi - B) = \sin B\), there is a second possible obtuse angle: \(B_2 = \pi - 0.86595 \approx 2.27564\text{ radians}\). To three significant figures, the two possible values for the angle \(B\) are \(0.866\) and \(2.28\).

M1 for using the area of a triangle formula. A1 for writing the correct equation \(24 = 31.5 \sin B\) or equivalent. M1 for solving for \(\sin B\) to get \(\sin B \approx 0.762\). A1 for finding the first angle \(B_1 \approx 0.866\). M1 for calculating the second angle \(B_2 = \pi - B_1\). A1 for finding the second angle \(B_2 \approx 2.28\).

(a) Let \(X\) be the mass of a randomly selected apple, where \(X \sim N(150, 12^2)\). Using a GDC to find the normal cumulative probability: \(P(X > 165) \approx 0.10565\). To three significant figures, \(P(X > 165) = 0.106\). (b) Let \(Y\) be the number of large apples in a sample of 20. \(Y\) follows a binomial distribution, \(Y \sim B(20, 0.10565)\). We want to find the probability that at least 5 apples are large: \(P(Y \ge 5) = 1 - P(Y \le 4)\). Using the binomial cumulative distribution function on a GDC with \(n = 20\), \(p = 0.10565\), and \(x = 4\): \(P(Y \le 4) \approx 0.94783\). Therefore, \(P(Y \ge 5) = 1 - 0.94783 \approx 0.05217\). To three significant figures, the probability is \(0.0522\).

M1 for defining the normal model with correct mean and standard deviation. A1 for \(P(X > 165) \approx 0.106\). M1 for identifying the binomial model with parameters \(n = 20\) and \(p = 0.10565\) (or their answer from part a). M1 for expressing the required probability as \(1 - P(Y \le 4)\). A1 for \(0.0522\).

First, we find the x-coordinates of the points of intersection of the two graphs by solving \(\cos(x) = x^2 - 1\). Using a GDC, we find the intersection points: \(x_1 \approx -1.1765\) and \(x_2 \approx 1.1765\). In the interval \([-1.1765, 1.1765]\), the graph of \(f(x) = \cos(x)\) lies above the graph of \(g(x) = x^2 - 1\). The area \(A\) of the enclosed region is given by the definite integral: \(A = \int_{-1.1765}^{1.1765} (\cos(x) - (x^2 - 1)) \, dx\). Using a GDC to evaluate this definite integral, we get: \(A \approx 3.1136\). To three significant figures, the area is \(3.11\).

M1 for setting up the equation \(\cos(x) = x^2 - 1\). A1 for finding the intersection limits \(x \approx \pm 1.18\) (or more precisely \(\pm 1.1765\)). M1 for setting up the integral of the difference of the functions with the correct order. M1 for using GDC numerical integration. A1 for the correct area of \(3.11\).

(a) Inputting the given data into a GDC, we obtain the Pearson's product-moment correlation coefficient: \(r \approx 0.991\) (specifically \(0.99052\)). (b) The GDC provides the linear regression parameters for the line of best fit \(y = ax + b\): \(a \approx 3.78\) (specifically \(3.7803\)) and \(b \approx 37.9\) (specifically \(37.883\)). So, the equation of the regression line is \(y = 3.78x + 37.9\). (c) Substitute \(x = 8\) into the regression equation: \(y = 3.7803(8) + 37.883 = 30.2424 + 37.883 = 68.125\). To three significant figures, the estimated score is \(68.1\).

A1 for the correct value of \(r \approx 0.991\). M1 for attempting linear regression on GDC. A1 for \(a \approx 3.78\). A1 for \(b \approx 37.9\). M1 for substituting \(x = 8\) into their regression equation. A1 for \(68.1\).

(a) Let \(M\) be the mass of the apple, where \(M \sim N(150, 18^2)\).

(i) Using a GDC to find \(P(130 < M < 160)\):
\(P(130 < M < 160) \approx 0.578\) (3 s.f.)

(ii) Using a GDC to find \(P(M > 175)\):
\(P(M > 175) \approx 0.0824\) (3 s.f.)

(b) We are given \(P(M < w) = 0.08\).
Using the inverse normal function on a GDC:
\(w \approx 124.708 \approx 125\text{ g}\) (3 s.f.)

(c) Let \(X\) be the number of small apples in a box of 12.
Since the probability of any one apple being small is 0.08, \(X\) follows a binomial distribution:
\(X \sim B(12, 0.08)\).

(i) To find the probability that exactly 2 apples are small:
\(P(X = 2) = \binom{12}{2} (0.08)^2 (0.92)^{10}\)
Using a GDC:
\(P(X = 2) \approx 0.183\) (3 s.f.)

(ii) To find the probability of at least 2 small apples:
\(P(X \ge 2) = 1 - P(X \le 1) = 1 - (P(X = 0) + P(X = 1))\)
Using a GDC:
\(P(X \ge 2) \approx 0.249\) (3 s.f.)

**(a) (i)**
M1: For attempting to write a probability statement or showing the standardisation formula.
A1: Correct values or parameter inputs in GDC (e.g., lower=130, upper=160, mean=150, sd=18).
A1: \(0.578\) (accept 0.577902...)

**(a) (ii)**
M1: For standardising or setting up the correct inequality.
A1: \(0.0824\) (accept 0.082448...)

**(b)**
M1: For writing a correct probability statement \(P(M < w) = 0.08\).
M1: For a correct inverse normal setup or equation.
A1: \(125\) (accept 124.708...)

**(c) (i)**
M1: For recognising a binomial distribution with \(n = 12\) and \(p = 0.08\).
A1: For a correct probability formula or GDC input.
A1: \(0.183\) (accept 0.183472...)

**(c) (ii)**
M1: For identifying the need to calculate \(1 - P(X \le 1)\) or summing individual probabilities for \(X \ge 2\).
A1: For correct GDC input or individual probabilities (e.g., \(P(X \le 1) \approx 0.75133\)).
A2: \(0.249\) (accept 0.248671...)

(a) The initial velocity is when \(t = 0\):
\(v(0) = 4\ln(0 - 0 + 2) - 0.5(0)^2 + 3(0) - 2 = 4\ln(2) - 2 \approx 0.773\text{ m s}^{-1}\) (3 s.f.)

(b) The particle is at rest when \(v(t) = 0\):
\(4\ln(t^2 - 2t + 2) - 0.5t^2 + 3t - 2 = 0\)
Using a GDC to find the root in the interval \(0 \le t \le 10\):
\(t \approx 9.23\) seconds (3 s.f.)

(c) To find the maximum velocity, we use a GDC to find the coordinates of the maximum point of \(y = v(t)\) in the interval \(0 \le t \le 10\).
The maximum point is at \((4.96, 11.8)\) (to 3 s.f.).
Therefore, the maximum velocity is \(11.8\text{ m s}^{-1}\) and it occurs at \(t \approx 4.96\) seconds.

(d) The total distance travelled is given by \(\int_{0}^{8} |v(t)| \mathrm{d}t\).
Since the velocity \(v(t) > 0\) for all \(0 \le t \le 8\), this is equal to \(\int_{0}^{8} v(t) \mathrm{d}t\).
Using a GDC to compute this definite integral:
\(\int_{0}^{8} (4\ln(t^2 - 2t + 2) - 0.5t^2 + 3t - 2) \mathrm{d}t \approx 60.7\text{ m}\) (3 s.f.)

(e) The acceleration of \(P\) is \(a(t) = v'(t)\).
Differentiating \(v(t)\):
\(a(t) = 4 \cdot \frac{2t - 2}{t^2 - 2t + 2} - t + 3 = \frac{8t - 8}{t^2 - 2t + 2} - t + 3\)
At \(t = 5\):
\(a(5) = \frac{8(5) - 8}{5^2 - 2(5) + 2} - 5 + 3 = \frac{32}{17} - 2 = -\frac{2}{17} \approx -0.118\text{ m s}^{-2}\) (3 s.f.)

**(a)**
M1: For substituting \(t = 0\) into \(v(t)\).
A1: \(4\ln(2) - 2\) or \(0.773\) (accept 0.77258...)

**(b)**
M1: For setting \(v(t) = 0\).
M1: Attempting to solve using GDC solver or graphing.
A1: \(t \approx 9.23\) (accept 9.2327...)

**(c)**
M1: Recognising that maximum velocity occurs when \(v'(t) = 0\) or locating maximum on graph.
A1: \(t \approx 4.96\) (accept 4.958...)
A1: \(v_{\text{max}} \approx 11.8\) (accept 11.836...)

**(d)**
M1: For setting up the integral for distance: \(\int_{0}^{8} |v(t)| \mathrm{d}t\) or \(\int_{0}^{8} v(t) \mathrm{d}t\).
A2: Accurate GDC computation.
A1: \(60.7\) (accept 60.690...)

**(e)**
M1: Knowing that acceleration is the derivative of velocity, \(a(t) = v'(t)\).
M1: Correct derivative formula or calculation using GDC numerical derivative.
A1: \(-0.118\) (accept \(-0.117647...\))

(a) Using the area formula for a triangle, \(\text{Area} = \frac{1}{2}ab\sin C\):
\(A = \frac{1}{2} \cdot AB \cdot AC \cdot \sin\theta\)
\(A = \frac{1}{2} \cdot 120 \cdot 85 \cdot \sin\theta\)
\(A = 60 \cdot 85 \cdot \sin\theta = 5100\sin\theta\) (Shown)

(b) We are given \(A = 3500\):
\(5100\sin\theta = 3500 \implies \sin\theta = \frac{35}{51}\)
Using a GDC to find \(\arcsin\left(\frac{35}{51}\right)\):
The acute angle is \(\theta \approx 0.7565\) radians.
Since \(\theta\) is obtuse:
\(\theta = \pi - 0.7565 \approx 2.39\) radians (3 s.f.)

(c) (i) Using the cosine rule on triangle \(ABC\):
\(BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos\theta\)
\(BC^2 = 120^2 + 85^2 - 2 \cdot 120 \cdot 85 \cdot \cos(2.38524...)\)
Using \(\theta \approx 2.38524\):
\(BC^2 = 14400 + 7225 - 20400 \cdot \cos(2.38524)\)
\(BC^2 \approx 21625 - 20400 \cdot (-0.72734) \approx 36462.7\)
\(BC = \sqrt{36462.7} \approx 191\text{ m}\) (3 s.f.)

(ii) Using the sine rule:
\(\frac{\sin(A\hat{B}C)}{AC} = \frac{\sin\theta}{BC}\)
\(\sin(A\hat{B}C) = \frac{85 \sin(2.38524)}{190.952}\)
Using \(\sin(2.38524) = \frac{35}{51}\):
\(\sin(A\hat{B}C) \approx \frac{58.333}{190.952} \approx 0.30549\)
\(A\hat{B}C = \arcsin(0.30549) \approx 0.310\) radians (3 s.f.)

(d) The angle bisector \(AD\) divides triangle \(ABC\) into triangles \(ABD\) and \(ACD\).
The sum of their areas equals the area of triangle \(ABC\):
\(\text{Area}(ABD) + \text{Area}(ACD) = \text{Area}(ABC)\)
\(\frac{1}{2} \cdot AB \cdot AD \cdot \sin\left(\frac{\theta}{2}\right) + \frac{1}{2} \cdot AC \cdot AD \cdot \sin\left(\frac{\theta}{2}\right) = 3500\)
Substitute \(AB = 120\) and \(AC = 85\):
\(\frac{1}{2} \cdot 120 \cdot AD \cdot \sin\left(\frac{2.38524}{2}\right) + \frac{1}{2} \cdot 85 \cdot AD \cdot \sin\left(\frac{2.38524}{2}\right) = 3500\)
\(\left(60 + 42.5\right) \cdot AD \cdot \sin(1.19262) = 3500\)
\(102.5 \cdot AD \cdot 0.92934 \approx 3500\)
\(95.257 \cdot AD \approx 3500\)
\(AD \approx 36.7\text{ m}\) (3 s.f.)

**(a)**
M1: For substituting correct side lengths into the area formula \(\frac{1}{2}ab\sin C\).
A1: Showing clearly that \(\frac{1}{2} \times 120 \times 85 = 5100\) leading to the given expression.

**(b)**
M1: Setting \(5100\sin\theta = 3500\).
A1: Finding the reference acute angle \(\approx 0.757\) radians or \(43.3^\circ\).
A1: Finding the correct obtuse angle \(\approx 2.39\) (accept 2.38524...).

**(c) (i)**
M1: Setting up the Cosine Rule with correct values.
A1: Correct substitution of \(\cos(2.39)\) or similar.
A1: \(BC \approx 191\) (accept 190.95...)

**(c) (ii)**
M1: Setting up the Sine Rule: \(\frac{\sin(A\hat{B}C)}{85} = \frac{\sin(2.39)}{BC}\).
A1: Correct substitution of values.
A1: \(A\hat{B}C \approx 0.310\) (accept 0.31008... radians).

**(d)**
M1: Realising that the total area is the sum of the areas of \(\triangle ABD\) and \(\triangle ACD\).
M1: Setting up the equation: \(\frac{1}{2}(120)(AD)\sin(\theta/2) + \frac{1}{2}(85)(AD)\sin(\theta/2) = 3500\).
A1: Substituting the correct half-angle value (\(\theta/2 \approx 1.193\) rad).
A1: \(AD \approx 36.7\) (accept 36.74...)