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

First, use the logarithm law \(\log_a(u) + \log_a(v) = \log_a(uv)\) to combine the terms on the left-hand side: \(\log_3((x-2)(x+4)) = 3\). Convert the logarithmic equation into its exponential form: \((x-2)(x+4) = 3^3\) which simplifies to \((x-2)(x+4) = 27\). Expand the left-hand side: \(x^2 + 2x - 8 = 27\). Rearrange to form a quadratic equation equal to zero: \(x^2 + 2x - 35 = 0\). Factorize the quadratic expression: \((x+7)(x-5) = 0\). This gives two possible solutions: \(x = -7\) or \(x = 5\). We must check the domain of the original logarithmic functions. For \(\log_3(x-2)\) to be defined, we require \(x - 2 > 0\) which means \(x > 2\). Therefore, \(x = -7\) is not a valid solution. The only valid solution is \(x = 5\).

M1 for applying the product rule of logarithms: \(\log_3((x-2)(x+4)) = 3\). A1 for converting to exponential form: \((x-2)(x+4) = 27\). A1 for forming the quadratic equation: \(x^2 + 2x - 35 = 0\). M1 for solving the quadratic equation (by factorizing or quadratic formula). A1 for identifying the roots: \(x = 5\) and \(x = -7\). R1 for rejecting \(x = -7\) with a valid reason (e.g. domain restriction \(x > 2\)), leading to the final answer \(x = 5\).

(a) To find the inverse function, set \(y = f(x)\) and rearrange to solve for \(x\): \(y = \frac{2x+3}{x-1}\), which leads to \(y(x-1) = 2x+3\), and \(yx - y = 2x+3\). Grouping \(x\) terms: \(yx - 2x = y + 3\), so \(x(y-2) = y + 3\), which gives \(x = \frac{y+3}{y-2}\). Thus, \(f^{-1}(x) = \frac{x+3}{x-2}\). The domain of \(f^{-1}(x)\) is all real numbers except where the denominator is zero: \(x \neq 2\). (b) We are given \((f \circ f)(a) = 5\), which means \(f(f(a)) = 5\). Taking the inverse on both sides: \(f(a) = f^{-1}(5)\). Substitute \(5\) into \(f^{-1}(x)\) to get \(f^{-1}(5) = \frac{5+3}{5-2} = \frac{8}{3}\). Now set \(f(a) = \frac{8}{3}\): \(\frac{2a+3}{a-1} = \frac{8}{3}\). This yields \(3(2a+3) = 8(a-1)\), so \(6a + 9 = 8a - 8\). Rearranging gives \(2a = 17\), so \(a = \frac{17}{2}\) (or \(8.5\)).

(a) M1 for attempting to swap variables and rearrange. A1 for obtaining the correct expression: \(f^{-1}(x) = \frac{x+3}{x-2}\). A1 for stating the correct domain: \(x \neq 2\) (accept \(x \in \mathbb{R}, x \neq 2\)). (b) M1 for recognizing the relationship \(f(a) = f^{-1}(5)\) or for substituting \(f(a)\) into \(f(x)\). A1 for finding \(f^{-1}(5) = \frac{8}{3}\). A1 for solving the linear equation to get \(a = \frac{17}{2}\) (or \(8.5\)).

Use the double-angle identity for cosine: \(\cos(2x) = 1 - 2\sin^2(x)\). Substitute this into the original equation: \((1 - 2\sin^2(x)) - 3\sin(x) = 1\). Subtract 1 from both sides: \(-2\sin^2(x) - 3\sin(x) = 0\). Multiply by \(-1\) to simplify: \(2\sin^2(x) + 3\sin(x) = 0\). Factorize the equation: \(\sin(x)(2\sin(x) + 3) = 0\). This gives two cases: 1) \(\sin(x) = 0\) and 2) \(2\sin(x) + 3 = 0\) which means \(\sin(x) = -1.5\). Since the range of the sine function is \([-1, 1]\), the equation \(\sin(x) = -1.5\) has no real solutions. For \(\sin(x) = 0\) in the interval \(0 \le x \le 2\pi\), the solutions are \(x = 0, \pi, 2\pi\).

M1 for applying the double-angle identity \(\cos(2x) = 1 - 2\sin^2(x)\). A1 for the correct simplified equation: \(2\sin^2(x) + 3\sin(x) = 0\). M1 for factorizing to find \(\sin(x)(2\sin(x) + 3) = 0\). A1 for stating that \(\sin(x) = -1.5\) has no solutions. A2 for the correct solutions \(x = 0, \pi, 2\pi\). Award A1 if only two of the three solutions are found.

A tangent is parallel to the \(x\)-axis if its gradient is zero. First, find the derivative of the curve with respect to \(x\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 6x - 9\). Set the derivative equal to zero to find the \(x\)-coordinates of these points: \(3x^2 - 6x - 9 = 0\). Divide the equation by 3: \(x^2 - 2x - 3 = 0\). Factorizing the quadratic expression gives \((x-3)(x+1) = 0\). This yields two \(x\)-coordinates: \(x = 3\) and \(x = -1\). Next, substitute these values back into the original curve equation to find the corresponding \(y\)-coordinates. For \(x = 3\): \(y = (3)^3 - 3(3)^2 - 9(3) + 5 = -22\), so one point is \((3, -22)\). For \(x = -1\): \(y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = 10\), so the other point is \((-1, 10)\). The coordinates of the points are \((-1, 10)\) and \((3, -22)\).

M1 for attempting to find the derivative of the cubic function. A1 for the correct derivative: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 6x - 9\). M1 for setting their derivative equal to 0. A1 for finding the correct \(x\)-values: \(x = 3\) and \(x = -1\). A1 for correctly calculating \(y = 10\) when \(x = -1\). A1 for correctly calculating \(y = -22\) when \(x = 3\).

(a) If \(A\) and \(B\) are mutually exclusive, then \(\mathrm{P}(A \cap B) = 0\). Therefore, \(\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) = 0.6 + 0.4 = 1.0\). (b) If \(A\) and \(B\) are independent, then \(\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B) = 0.6 \times 0.4 = 0.24\). 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) Using the general addition rule: \(\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)\), which means \(0.8 = 0.6 + 0.4 - \mathrm{P}(A \cap B)\). This simplifies to \(0.8 = 1.0 - \mathrm{P}(A \cap B)\), yielding \(\mathrm{P}(A \cap B) = 0.2\). Now use the conditional probability formula: \(\mathrm{P}(A \mid B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} = \frac{0.2}{0.4} = 0.5\).

(a) M1 for recognizing \(\mathrm{P}(A \cap B) = 0\) and adding the probabilities. A1 for \(1.0\). (b) M1 for calculating \(\mathrm{P}(A \cap B) = 0.6 \times 0.4 = 0.24\). A1 for \(0.76\). (c) M1 for calculating \(\mathrm{P}(A \cap B) = 0.2\) using the addition formula. A1 for \(\mathrm{P}(A \mid B) = 0.5\) (or \(\frac{1}{2}\)).

First, find the antiderivative of each term: \(\int e^{2x} \, \mathrm{d}x = \frac{1}{2}e^{2x}\) and \(\int e^{-x} \, \mathrm{d}x = -e^{-x}\). So the indefinite integral is: \(\int (e^{2x} + e^{-x}) \, \mathrm{d}x = \left[ \frac{1}{2}e^{2x} - e^{-x} \right]_{0}^{\ln 2}\). Now, substitute the upper limit \(\ln 2\): \(\frac{1}{2}e^{2\ln 2} - e^{-\ln 2} = \frac{1}{2}e^{\ln(4)} - e^{\ln(1/2)} = \frac{1}{2}(4) - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}\). Substitute the lower limit \(0\): \(\frac{1}{2}e^{2(0)} - e^{-(0)} = \frac{1}{2}(1) - 1 = -\frac{1}{2}\). Subtract the lower limit evaluation from the upper limit evaluation: \(\left(\frac{3}{2}\right) - \left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{2} = 2\).

M1 for attempting to integrate \(e^{2x}\) to obtain \(k e^{2x}\) (where \(k \neq 1\)). A1 for the correct term \(\frac{1}{2}e^{2x}\). M1 for integrating \(e^{-x}\) to obtain \(-e^{-x}\). A1 for the complete correct antiderivative: \(\frac{1}{2}e^{2x} - e^{-x}\). M1 for substituting limits \(\ln 2\) and \(0\) and using logarithm properties. A1 for the final value \(2\).

Let \(u_1\) be the first term and \(d\) be the common difference. Using the formula for the \(n\)-th term of an arithmetic sequence, \(u_n = u_1 + (n-1)d\): \(u_4 = u_1 + 3d = 11\) (Equation 1). Using the formula for the sum of the first \(n\) terms, \(S_n = \frac{n}{2}(2u_1 + (n-1)d)\): \(S_{10} = \frac{10}{2}(2u_1 + 9d) = 155\), which simplifies to \(5(2u_1 + 9d) = 155\), and thus \(2u_1 + 9d = 31\) (Equation 2). We now solve Equations 1 and 2 simultaneously. Multiply Equation 1 by 2: \(2u_1 + 6d = 22\) (Equation 3). Subtract Equation 3 from Equation 2: \((2u_1 + 9d) - (2u_1 + 6d) = 31 - 22\), which simplifies to \(3d = 9\), giving \(d = 3\). Substitute \(d = 3\) back into Equation 1: \(u_1 + 3(3) = 11\), so \(u_1 + 9 = 11\), which yields \(u_1 = 2\). Thus, the first term is \(u_1 = 2\) and the common difference is \(d = 3\).

M1 for writing a correct equation for \(u_4\): \(u_1 + 3d = 11\). M1 for writing a correct formula/equation for \(S_{10}\): \(\frac{10}{2}(2u_1 + 9d) = 155\). A1 for simplifying the sum equation to \(2u_1 + 9d = 31\). M1 for attempting to solve the simultaneous equations. A1 for finding \(d = 3\). A1 for finding \(u_1 = 2\).

(a) To find the length of \(AC\), we use the cosine rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\). Substitute the given values: \(AC^2 = 5^2 + 8^2 - 2(5)(8)\cos(60^\circ)\). This gives \(AC^2 = 25 + 64 - 80(0.5) = 89 - 40 = 49\). Taking the positive square root, we get \(AC = 7\text{ cm}\). (b) To find the area of triangle \(ABC\), we use the area formula: \(\text{Area} = \frac{1}{2}(AB)(BC)\sin(\angle ABC)\). Substitute the given values: \(\text{Area} = \frac{1}{2}(5)(8)\sin(60^\circ)\). This yields \(\text{Area} = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ cm}^2\).

(a) M1 for choosing the cosine rule with correct substitution. A1 for \(AC^2 = 5^2 + 8^2 - 2(5)(8)\cos(60^\circ)\). A1 for \(AC = 7\text{ cm}\). (b) M1 for choosing the sine-based area formula. A1 for correct substitution: \(\frac{1}{2}(5)(8)\sin(60^\circ)\). A1 for the final exact area: \(10\sqrt{3}\text{ cm}^2\).

To find the area of the region R, we evaluate the definite integral: Area = \int_{0}^{4} \frac{x}{\sqrt{2x + 1}} \, dx. We can use integration by substitution. Let u = 2x + 1, which gives du = 2 \, dx, or dx = \frac{du}{2}. From this, we also have x = \frac{u - 1}{2}. Next, we find the new limits of integration: When x = 0, u = 2(0) + 1 = 1. When x = 4, u = 2(4) + 1 = 9. Substituting these into the integral gives: \int_{1}^{9} \frac{\frac{u-1}{2}}{\sqrt{u}} \frac{du}{2} = \frac{1}{4} \int_{1}^{9} \frac{u-1}{u^{1/2}} \, du = \frac{1}{4} \int_{1}^{9} (u^{1/2} - u^{-1/2}) \, du. Now, we find the antiderivative of each term: \frac{1}{4} \left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right]_{1}^{9}. Evaluating at the upper limit u = 9: \frac{2}{3}(9)^{3/2} - 2(9)^{1/2} = \frac{2}{3}(27) - 2(3) = 18 - 6 = 12. Evaluating at the lower limit u = 1: \frac{2}{3}(1)^{3/2} - 2(1)^{1/2} = \frac{2}{3} - 2 = -\frac{4}{3}. Subtracting the lower limit value from the upper limit value: \frac{1}{4} \left( 12 - \left(-\frac{4}{3}\right) \right) = \frac{1}{4} \left( \frac{40}{3} \right) = \frac{10}{3}. Alternatively, using the substitution v = \sqrt{2x+1}: Then v^2 = 2x+1, which gives x = \frac{v^2-1}{2} and dx = v \, dv. The limits of integration become: When x = 0, v = 1. When x = 4, v = 3. This transforms the integral into: \int_{1}^{3} \frac{\frac{v^2-1}{2}}{v} \cdot v \, dv = \frac{1}{2} \int_{1}^{3} (v^2 - 1) \, dv = \frac{1}{2} \left[ \frac{v^3}{3} - v \right]_{1}^{3} = \frac{1}{2} \left( (9 - 3) - (\frac{1}{3} - 1) \right) = \frac{1}{2} \left( 6 + \frac{2}{3} \right) = \frac{1}{2} \left( \frac{20}{3} \right) = \frac{10}{3}.

M1: Writing down the correct definite integral representing the area. M1: Attempting an appropriate substitution (such as u = 2x + 1 or v = \sqrt{2x+1}) with a correct differential relationship. A1: Correctly transforming the integral, including the limits of integration. M1: Correctly integrating the transformed expression. A1: Correctly substituting the limits of integration. A1: Correct final exact value of \frac{10}{3}.

f(x) = x e^{-kx}

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

(b) To find the local maximum, set f'(x) = 0:
1 - kx = 0 \implies x = \frac{1}{k}.
Substituting this x-value into f(x) yields:
f\left(\frac{1}{k}\right) = \frac{1}{k}e^{-k(\frac{1}{k})} = \frac{1}{ke}.
Thus, the coordinates of the local maximum are \left(\frac{1}{k}, \frac{1}{ke}\right).

(c) Point of inflexion occurs when f''(x) = 0.
Using the product rule on f'(x):
f''(x) = -k e^{-kx} + (1-kx)(-k e^{-kx}) = -k(2 - kx)e^{-kx} = k(kx-2)e^{-kx}.
Set f''(x) = 0 \implies kx - 2 = 0 \implies x = \frac{2}{k}.

(d) Integrating by parts:
Let u = x \implies du = dx
dv = e^{-kx}dx \implies v = -\frac{1}{k}e^{-kx}
\int x e^{-kx} dx = -\frac{x}{k}e^{-kx} - \int -\frac{1}{k}e^{-kx} dx = -\frac{x}{k}e^{-kx} - \frac{1}{k^2}e^{-kx}
Evaluating from 0 to \frac{2}{k}:
\left[-\frac{x}{k}e^{-kx} - \frac{1}{k^2}e^{-kx}\right]_{0}^{\frac{2}{k}} = \left(-\frac{2}{k^2}e^{-2} - \frac{1}{k^2}e^{-2}\right) - \left(0 - \frac{1}{k^2}\right) = \frac{1}{k^2}(1 - 3e^{-2}).

(e) The area under the curve is given by \int_{0}^{\frac{2}{k}} f(x) \, dx = \frac{1}{k^2}(1 - 3e^{-2}).
Equating to the given area:
\frac{1}{k^2}(1 - 3e^{-2}) = 4 - 12e^{-2} = 4(1 - 3e^{-2})
Since 1 - 3e^{-2}
e 0, we can divide both sides by (1 - 3e^{-2}) to get:
\frac{1}{k^2} = 4 \implies k^2 = \frac{1}{4}
Since k > 0, we have k = \frac{1}{2}.

(a) M1 for applying the product rule, A1 for correct simplification to obtain the given expression.

(b) M1 for setting f'(x) = 0, A1 for x = \frac{1}{k}, A1 for y = \frac{1}{ke}.

(c) M1 for finding the second derivative f''(x), A1 for setting f''(x) = 0, A1 for the final coordinate x = \frac{2}{k}.

(d) M1 for setting up integration by parts (choosing u and dv), A1 for correct du and v, A1 for the correct unsimplified antiderivative, M1 for substituting the upper and lower limits, A1 for evaluating the expression at both limits, A1 for obtaining the shown result.

(e) M1 for equating the integration result to the given area, M1 for factoring out (1 - 3e^{-2}) on the right hand side, A1 for k^2 = \frac{1}{4}, A1 for k = \frac{1}{2} with justification that k > 0.

(a) By De Moivre's Theorem, z^n = \cos(n\theta) + \mathrm{i}\sin(n\theta) and z^{-n} = \cos(-n\theta) + \mathrm{i}\sin(-n\theta) = \cos(n\theta) - \mathrm{i}\sin(n\theta).
Summing these gives:
z^n + \frac{1}{z^n} = (\cos(n\theta) + \mathrm{i}\sin(n\theta)) + (\cos(n\theta) - \mathrm{i}\sin(n\theta)) = 2\cos(n\theta).
Subtracting these gives:
z^n - \frac{1}{z^n} = (\cos(n\theta) + \mathrm{i}\sin(n\theta)) - (\cos(n\theta) - \mathrm{i}\sin(n\theta)) = 2\mathrm{i}\sin(n\theta).

(b) Expanding \left(z - \frac{1}{z}\right)^5 using the binomial theorem:
\left(z - \frac{1}{z}\right)^5 = z^5 - 5z^3 + 10z - \frac{10}{z} + \frac{5}{z^3} - \frac{1}{z^5}
= \left(z^5 - \frac{1}{z^5}\right) - 5\left(z^3 - \frac{1}{z^3}\right) + 10\left(z - \frac{1}{z}\right).
Using part (a), we substitute the corresponding identities:
(2\mathrm{i}\sin\theta)^5 = 2\mathrm{i}\sin(5\theta) - 5(2\mathrm{i}\sin(3\theta)) + 10(2\mathrm{i}\sin\theta)
32\mathrm{i}^5\sin^5\theta = 2\mathrm{i}(\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta).
Since \mathrm{i}^5 = \mathrm{i}, this simplifies to:
32\mathrm{i}\sin^5\theta = 2\mathrm{i}(\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta).
Dividing both sides by 32\mathrm{i} gives:
\sin^5\theta = \frac{1}{16}(\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta).

(c) Integrating the identity:
\int_{0}^{\frac{\pi}{2}} \sin^5\theta \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1}{16}(\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta) \, d\theta
= \frac{1}{16} \left[ -\frac{1}{5}\cos(5\theta) + \frac{5}{3}\cos(3\theta) - 10\cos\theta \right]_{0}^{\frac{\pi}{2}}.
At \theta = \frac{\pi}{2}:
-\frac{1}{5}\cos\left(\frac{5\pi}{2}\right) + \frac{5}{3}\cos\left(\frac{3\pi}{2}\right) - 10\cos\left(\frac{\pi}{2}\right) = 0 + 0 - 0 = 0.
At \theta = 0:
-\frac{1}{5}\cos(0) + \frac{5}{3}\cos(0) - 10\cos(0) = -\frac{1}{5} + \frac{5}{3} - 10 = \frac{-3 + 25 - 150}{15} = -\frac{128}{15}.
Thus:
\int_{0}^{\frac{\pi}{2}} \sin^5\theta \, d\theta = \frac{1}{16} \left(0 - \left(-\frac{128}{15}\right)\right) = \frac{128}{240} = \frac{8}{15}.

(a) M1 for applying De Moivre's theorem to z^n, A1 for correct expression for z^{-n}, A1 for showing the sum simplifies to 2\cos(n\theta), A1 for showing the difference simplifies to 2\mathrm{i}\sin(n\theta).

(b) M1 for correct binomial expansion of \left(z - \frac{1}{z}\right)^5, A1 for correct grouping of terms with their corresponding coefficients, M1 for substituting 2\mathrm{i}\sin(k\theta) for each group, A1 for LHS simplification to 32\mathrm{i}\sin^5\theta, M1 for factoring out 2\mathrm{i} on the RHS, A1 for dividing and showing the required identity.

(c) M1 for substituting the identity into the integral, A1 for correct antiderivatives of the three terms, M1 for substituting the upper limit to get 0, M1 for substituting the lower limit and evaluating, A1 for correct evaluation of fractions to -\frac{128}{15}, A1 for the final value \frac{8}{15}.

(a) 1. Direction vectors are \mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} and \mathbf{d}_2 = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}. Since \mathbf{d}_1 is not a scalar multiple of \mathbf{d}_2, the lines are not parallel.
2. To find if they intersect, we set up a system of linear equations:
1 + 2\lambda = 2 - \mu \implies 2\lambda + \mu = 1 (Eq 1)
\lambda = 5 + 2\mu \implies \lambda - 2\mu = 5 (Eq 2)
-1 + 2\lambda = \mu \implies 2\lambda - \mu = 1 (Eq 3)
Solving Eq 1 and Eq 2 simultaneously:
From Eq 1, \mu = 1 - 2\lambda.
Substitute into Eq 2: \lambda - 2(1 - 2\lambda) = 5 \implies 5\lambda = 7 \implies \lambda = \frac{7}{5}.
Then \mu = 1 - 2\left(\frac{7}{5}\right) = -\frac{9}{5}.
We test these values in Eq 3:
LHS = 2\left(\frac{7}{5}\right) - \left(-\frac{9}{5}\right) = \frac{23}{5}
eq 1 (RHS).
Since there is no common solution, the lines do not intersect. Since they are not parallel and do not intersect, they are skew.

(b) The perpendicular vector \mathbf{n} is the cross product of \mathbf{d}_1 and \mathbf{d}_2:
\mathbf{n} = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} \times \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(1) - 2(2) \\ 2(-1) - 2(1) \\ 2(2) - 1(-1) \end{pmatrix} = \begin{pmatrix} -3 \\ -4 \\ 5 \end{pmatrix}.
Any non-zero scalar multiple is acceptable, e.g., \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix}.

(c) Let A(1, 0, -1) be a point on L_1 and B(2, 5, 0) be a point on L_2.
The vector \mathbf{AB} = \begin{pmatrix} 2-1 \\ 5-0 \\ 0-(-1) \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ 1 \end{pmatrix}.
The shortest distance d is the projection of \mathbf{AB} onto \mathbf{n}:
d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left|\begin{pmatrix} 1 \\ 5 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ -4 \\ 5 \end{pmatrix}\right|}{\sqrt{(-3)^2 + (-4)^2 + 5^2}} = \frac{|-3 - 20 + 5|}{\sqrt{50}} = \frac{18}{5\sqrt{2}} = \frac{9\sqrt{2}}{5}.

(d) The normal to the plane \Pi is parallel to \mathbf{n} = \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix}.
The Cartesian equation is of the form 3x + 4y - 5z = D.
Using the point A(1, 0, -1) on L_1:
3(1) + 4(0) - 5(-1) = 8 \implies D = 8.
Thus, the equation of the plane is 3x + 4y - 5z = 8.

(a) R1 for stating skew lines are not parallel and do not intersect, A1 for verifying they are not parallel, M1 for establishing a system of three linear equations for intersection, A1 for finding correct values for \lambda and \mu from two equations, A1 for substituting into the third equation and showing inconsistency, R1 for concluding the lines are skew.

(b) M1 for using the cross product of the direction vectors, A1 for setting up the determinant or component-wise products, A2 for the correct vector \begin{pmatrix} -3 \\ -4 \\ 5 \end{pmatrix} (or scalar multiple).

(c) M1 for finding a vector \mathbf{AB} between L_1 and L_2, A1 for correct \mathbf{AB}, M1 for using the projection formula, A1 for the dot product calculation, A1 for the correct final distance \frac{9\sqrt{2}}{5}.

(d) M1 for recognizing that the normal to the plane is parallel to the cross product vector \mathbf{n}, M1 for substituting a point on L_1 to find the constant, A1 for the correct Cartesian equation 3x + 4y - 5z = 8.

The sum of the first \(n\) terms of the arithmetic sequence is given by: \(T_n = \frac{n}{2}[2(5) + (n-1)(1.5)] = \frac{n}{2}(1.5n + 8.5)\). The sum of the first \(n\) terms of the geometric sequence is given by: \(S_n = \frac{2(1.15^n - 1)}{1.15 - 1} = \frac{40}{3}(1.15^n - 1)\). We wish to find the smallest integer \(n\) such that \(S_n > T_n\), which is equivalent to finding the smallest integer \(n\) where \(\frac{40}{3}(1.15^n - 1) > \frac{n}{2}(1.5n + 8.5)\). Using a GDC to tabulate or solve this inequality: For \(n = 28\), \(T_{28} = 707\) and \(S_{28} \approx 654.12\). For \(n = 29\), \(T_{29} = 754\) and \(S_{29} \approx 754.24\). Thus, the least value of \(n\) is \(29\).

M1 for writing down a correct expression for \(T_n\). M1 for writing down a correct expression for \(S_n\). M1 for setting up the inequality \(S_n > T_n\). M1 for using GDC to evaluate values or find the intersection. A1 for identifying the values of both sums at \(n = 28\) or \(n = 29\). A1 for the correct final answer \(n = 29\).

(a) Substituting \(t = 0\) and \(P(0) = 150\): \(\frac{1200}{1 + a e^0} = 150 \implies 1 + a = 8 \implies a = 7\). (b) Substituting \(t = 3\), \(P(3) = 450\), and \(a = 7\): \(\frac{1200}{1 + 7 e^{-3b}} = 450 \implies 1 + 7 e^{-3b} = \frac{8}{3} \implies e^{-3b} = \frac{5}{21} \implies b = -\frac{1}{3}\ln\left(\frac{5}{21}\right) \approx 0.47833\), which rounds to \(0.478\) to three significant figures. (c) The rate of change at \(t = 5\) is given by \(P'(5)\). Differentiating \(P(t)\) using a GDC at \(t = 5\) yields \(P'(5) \approx 137\) rabbits per year (or more precisely \(136.6\)).

(a) M1 for substituting \(t=0\) into the formula. A1 for \(a = 7\). (b) M1 for substituting \(t=3\) and \(P=450\) to set up the equation. A1 for showing the algebraic steps leading to \(b \approx 0.478\). (c) M1 for recognizing that the rate of change is \(P'(5)\). A1 for correct GDC numerical differentiation setup. A1 for the correct value \(137\) (or \(136.6\)).

(a) Using parallel lines, the angle from BA to the South direction at B is \(50^\circ\). The bearing of C from B is \(160^\circ\), which is \(20^\circ\) past South. Therefore, \(\angle ABC = 180^\circ - (160^\circ - 50^\circ) = 70^\circ\). Applying the cosine rule on triangle ABC: \(AC^2 = 12^2 + 18^2 - 2(12)(18)\cos(70^\circ) \approx 320.25 \implies AC \approx 17.9\text{ km}\). (b) Using the sine rule to find the angle \(\angle BAC\): \(\frac{\sin(\angle BAC)}{18} = \frac{\sin(70^\circ)}{17.895} \implies \sin(\angle BAC) \approx 0.9452 \implies \angle BAC \approx 70.93^\circ\). The bearing of C from A is \(50^\circ + 70.93^\circ = 120.93^\circ \approx 121^\circ\).

(a) M1 for finding \(\angle ABC = 70^\circ\). M1 for a correct application of the cosine rule. A1 for \(17.9\text{ km}\). (b) M1 for a correct application of the sine rule to find \(\angle BAC\). A1 for \(\angle BAC \approx 70.9^\circ\). A1 for the bearing \(121^\circ\) (or \(120.9^\circ\)).

(a) Setting the two curves equal: \(3\cos(x) = x^2 - 2\). Using a GDC, the x-coordinates of the intersection points are \(x \approx -1.493\) and \(x \approx 1.493\) (or \(x = \pm 1.49\) to three significant figures). (b) The enclosed area is given by the integral \(A = \int_{-1.493}^{1.493} (3\cos(x) - (x^2 - 2)) \text{ d}x\). Evaluating this using a GDC gives \(A \approx 9.74\). (c) The volume of the solid of revolution is \(V = \pi \int_{-1.493}^{1.493} ((3\cos(x))^2 - (x^2 - 2)^2) \text{ d}x\). Evaluating this using a GDC gives \(V \approx 25.4\).

(a) M1 for setting up the equation \(3\cos(x) = x^2 - 2\). A1 for \(x = \pm 1.49\). (b) M1 for setting up the area integral with correct limits. M1 for integrating the difference of the functions. A1 for \(9.74\). (c) M1 for setting up the volume integral with \(\pi\) and squared functions. A1 for \(25.4\).

(a) Let \(X\) be the mass of an orange, so \(X \sim N(\mu, \sigma^2)\). We are given \(P(X < 115) = 0.08\) and \(P(X > 160) = 0.15 \implies P(X < 160) = 0.85\). Standardizing these: \(\frac{115 - \mu}{\sigma} = \Phi^{-1}(0.08) \approx -1.4051 \implies \mu - 1.405\sigma = 115\). Similarly, \(\frac{160 - \mu}{\sigma} = \Phi^{-1}(0.85) \approx 1.0364 \implies \mu + 1.036\sigma = 160\). (b) Solving the two simultaneous equations: Subtracting the first equation from the second gives \(2.4415\sigma = 45 \implies \sigma \approx 18.4\text{ g}\). Substituting this back gives \(\mu = 160 - 1.0364(18.43) \approx 141\text{ g}\).

(a) M1 for attempting to standardize \(P(X < 115) = 0.08\). A1 for \(\mu - 1.405\sigma = 115\). M1 for attempting to standardize \(P(X < 160) = 0.15\). A1 for \(\mu + 1.036\sigma = 160\). (b) A1 for \(\sigma = 18.4\). A1 for \(\mu = 141\).

(a) The three roots are \(3\), \(p + q\mathrm{i}\), and \(p - q\mathrm{i}\). The sum of these roots is \(3 + (p + q\mathrm{i}) + (p - q\mathrm{i}) = 3 + 2p\). Since the sum is given as 1, we have \(3 + 2p = 1 \implies 2p = -2 \implies p = -1\). (b) The sum of the squares of the roots is \(3^2 + (p + q\mathrm{i})^2 + (p - q\mathrm{i})^2 = 9 + (p^2 - q^2 + 2pq\mathrm{i}) + (p^2 - q^2 - 2pq\mathrm{i}) = 9 + 2p^2 - 2q^2\). Since this sum is 5, and substituting \(p = -1\): \(9 + 2(-1)^2 - 2q^2 = 5 \implies 11 - 2q^2 = 5 \implies 2q^2 = 6 \implies q^2 = 3 \implies q = \sqrt{3}\) (since \(q > 0\)). (c) Using Vieta's formulas: The sum of the roots is \(-a = 1 \implies a = -1\). The product of the roots taken pairwise is \(b = r_1 r_2 + r_2 r_3 + r_3 r_1\). With \(r_2, r_3 = -1 \pm \sqrt{3}\mathrm{i}\), we have \(r_2 r_3 = (-1)^2 + 3 = 4\) and \(r_1(r_2+r_3) = 3(-2) = -6\), so \(b = 4 - 6 = -2\).

(a) M1 for setting up the sum of roots in terms of \(p\). A1 for obtaining \(p = -1\). (b) M1 for expanding the sum of squares of the complex roots. A1 for simplifying the expression to \(9 + 2p^2 - 2q^2 = 5\). A1 for obtaining \(q = \sqrt{3}\). (c) A1 for \(a = -1\). A1 for \(b = -2\).

(a) The direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\). The angle \(\theta\) satisfies \(\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{ |\mathbf{d}_1| |\mathbf{d}_2| } = \frac{|1(2) + 2(-1) - 2(4)|}{\sqrt{9}\sqrt{21}} = \frac{8}{3\sqrt{21}} \approx 0.5819 \implies \theta \approx 54.4^\circ\) (or \(0.950\) radians). (b) For intersection, we equate the components: \(2 + \lambda = 5 + 2\mu\) (1), \(-1 + 2\lambda = 1 - \mu\) (2), \(3 - 2\lambda = 2 + 4\mu\) (3). Solving (1) and (2) simultaneously gives \(\lambda = 1.4\) and \(\mu = -0.8\). Substituting these into the third equation: LHS \(= 3 - 2(1.4) = 0.2\) and RHS \(= 2 + 4(-0.8) = -1.2\). Since LHS \(\neq\) RHS, the lines do not intersect.

(a) M1 for identifying the correct direction vectors. M1 for setting up the dot product equation. A1 for \(54.4^\circ\) (or \(0.950\) radians). (b) M1 for setting up three linear equations. A1 for solving any two equations to find \(\lambda\) and \(\mu\). A1 for demonstrating a contradiction in the third equation to complete the proof.

(a) The particle is at rest when \(v(t) = 0\). Using a GDC to solve \(3t - 2\sin(1.5t) - 4 = 0\) yields \(t \approx 1.70\text{ s}\). (b) Acceleration is the derivative of velocity: \(a(t) = v'(t) = 3 - 3\cos(1.5t)\). At \(t = 4\), \(a(4) = v'(4) \approx 0.120\text{ m s}^{-2}\) (or \(0.119\)). (c) The total distance traveled in the first 6 seconds is given by \(\int_{0}^{6} |v(t)| \text{ d}t = \int_{0}^{6} |3t - 2\sin(1.5t) - 4| \text{ d}t\). Evaluating this using a GDC gives \(\text{Distance} \approx 37.3\text{ m}\) (or \(37.2\)).

(a) M1 for setting \(v(t) = 0\). A1 for \(t \approx 1.70\text{ s}\). (b) M1 for recognizing that \(a(t) = v'(t)\). A1 for \(0.120\text{ m s}^{-2}\). (c) M1 for setting up the integral of the absolute value of velocity. A1 for the correct integral expression with limits. A1 for \(37.3\text{ m}\).

**(a)**
Let \(W\) be the weight of an apple, where \(W \sim N(150, \sigma^2)\).
We are given that \(\text{P}(W > 175) = 0.15\).
This means \(\text{P}(W \le 175) = 0.85\). (M1)

Standardizing the boundary value:
\(\text{P}\left(Z \le \frac{175 - 150}{\sigma}\right) = 0.85\)

Using the inverse normal cumulative distribution function on a GDC:
\(\frac{25}{\sigma} \approx 1.03643...\) (A1)

Solving for \(\sigma\):
\(\sigma = \frac{25}{1.03643...} \approx 24.1202...\)
\(\sigma \approx 24.1\text{ g}\) (to 3 significant figures). (A1)

**(b)**
We want to find \(\text{P}(130 < W < 160)\) where \(W \sim N(150, 24.1202^2)\). (M1)

Using the normal cumulative distribution function on a GDC: (M1)
\(\text{P}(130 < W < 160) \approx 0.457\) (to 3 significant figures). (A1)

*(Note: If using the rounded value \(\sigma = 24.1\), the probability is \(0.458\).)*

**(a)**
**M1** for attempting to use the symmetry or complement property of the normal distribution (e.g., writing \(\text{P}(W \le 175) = 0.85\) or standardizing \(\frac{175-150}{\sigma}\)).
**A1** for finding the correct z-critical value of \(1.03643...\) (or setting up the equation \(\frac{25}{\sigma} = 1.036...\)).
**A1** for \(\sigma \approx 24.1\) (accept \(24.1202...\)).

**(b)**
**M1** for identifying the required probability interval \(\text{P}(130 < W < 160)\) (can be shown in a sketch or standardizing formula).
**M1** for evidence of using the normal CDF function on a GDC with their \(\sigma\) and \(\mu = 150\).
**A1** for \(0.457\) (accept \(0.458\) if \(\sigma = 24.1\) is used).

(a) (i) Using the cosine rule:
\(AC^2 = 120^2 + 150^2 - 2(120)(150)\cos(75^\circ)\)
\(AC^2 = 14400 + 22500 - 36000(0.258819) = 27582.52\)
\(AC \approx 166.08 \approx 166\text{ m}\)

(ii) Using the sine rule:
\(\frac{\sin(BCA)}{120} =
\frac{\sin(75^\circ)}{166.08}\)
\
\sin(BCA) = \frac{120 \sin(75^\circ)}{166.08} \approx 0.69792\)
\(BCA \approx 44.26^\circ \approx 44.3^\circ\) (or \(0.773\text{ rad}\))

(b) The area of \(\triangle ABC = \frac{1}{2} \times 120 \times 150 \times \sin(75^\circ) \approx 8693.33\text{ m}^2\).
Also, \(\text{Area} = \frac{1}{2} \times AC \times BD\).
\(8693.33 = \frac{1}{2} \times 166.08 \times BD \implies BD \approx 104.7\text{ m}\) (or \(105\text{ m}\)).

(c) In right-angled triangle \(BD\text{Top}\):
\(\tan(22^\circ) = \frac{h}{BD}\)
\(h = 104.69 \times \tan(22^\circ) \approx 42.3\text{ m}\).

(d) (i) Angle of the sector in radians is \(\theta = 75^\circ \times \frac{\pi}{180} = \frac{5\pi}{12} \approx 1.309\text{ rad}\).
\(\text{Area of sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} r^2 \left(\frac{5\pi}{12}\right) = \frac{5\pi}{24} r^2\).
Setting area equal to half of park area:
\(\frac{5\pi}{24} r^2 = \frac{8693.33}{2} = 4346.67\)
\(r^2 = \frac{4346.67 \times 24}{5\pi} \approx 6641.17 \implies r \approx 81.5\text{ m}\).

(ii) Since the radius of the sprinkler is \(r \approx 81.5\text{ m}\) and the minimum perpendicular distance from \(B\) to the boundary \(AC\) is \(BD \approx 104.7\text{ m}\), the sprinkler's water never reaches any point on the boundary \(AC\).
Therefore, the percentage of the boundary watered is \(0\%\).

(a) (i) M1 for applying the Cosine Rule. A1 for correct substitution. A1 for \(166\text{ m}\) (or \(166.08\text{ m}\)). [3 marks]
(ii) M1 for applying the Sine Rule. A1 for \(44.3^\circ\) (or \(0.773\text{ rad}\)). [2 marks]

(b) M1 for finding the area of the triangle. M1 for setting up the area equation \(\frac{1}{2} \times AC \times BD = \text{Area}\). A1 for \(105\text{ m}\) (or \(104.7\text{ m}\)). [3 marks]

(c) M1 for identifying the correct trigonometric ratio \(\tan(22^\circ) = \frac{h}{BD}\). A1 for substitution of their \(BD\). A1 for \(42.3\text{ m}\). [3 marks]

(d) (i) M1 for calculating area of the sector. M1 for setting up equation \(\text{Area of sector} = 0.5 \times 8693.33\). A1 for \(r^2 \approx 6641.17\). A1 for \(81.5\text{ m}\). [4 marks]
(ii) R1 for comparing \(r \approx 81.5\) with the perpendicular distance \(BD \approx 104.7\). A1 for stating \(0\%\). [2 marks]

(a) The particle is at rest when \(v(t) = 0\).
Since \(3t e^{-0.2t} = 0 \implies t = 0\),
and \
\sin(0.8t) = 0 \implies 0.8t = k\pi\) for \(k \in \mathbb{Z}\).
For \(0 \le t \le 10\):
\(t = 0\)
\(t = \frac{\pi}{0.8} \approx 3.927 \approx 3.93\text{ s}\)
\(t = \frac{2\pi}{0.8} \approx 7.854 \approx 7.85\text{ s}\)

(b) Speed is \(|v(t)|\). Using GDC to locate extrema:
Local maximum at \(t \approx 2.12\), where \(v \approx 4.12\).
Local minimum at \(t \approx 5.76\), where \(v \approx -5.43\).
Local maximum at \(t \approx 9.40\), where \(v \approx 4.09\).
Comparing magnitudes, the maximum speed is \(5.43\text{ m s}^{-1}\).

(c) \(\text{Total distance} = \int_{0}^{10} |v(t)| dt = \int_{0}^{10} |3t e^{-0.2t} \sin(0.8t)| dt\).
Using GDC integration: \(\text{Total distance} \approx 25.8\text{ m}\).

(d) \(a(4) = v'(4)\).
Using GDC numerical derivative: \(a(4) \approx -4.32\text{ m s}^{-2}\).

(e) (i) \(s(8) = s(0) + \int_{0}^{8} v(t) dt = -3 + \int_{0}^{8} 3t e^{-0.2t} \sin(0.8t) dt\).
Using GDC to compute \(\int_{0}^{8} v(t) dt \approx -5.15\).
\(s(8) = -3 - 5.15 = -8.15\text{ m}\).

(ii) Furthest from origin means maximizing \(|s(t)|\).
Testing turning points of position where \(s'(t) = v(t) = 0\) (i.e., at \(t = 0, 3.93, 7.85\)) and the endpoint \(t = 10\):
- At \(t = 0\), \(s = -3 \implies |s| = 3\)
- At \(t \approx 3.93\), \(s = -3 + \int_0^{3.93} v(t) dt \approx -3 + 8.24 = 5.24 \implies |s| = 5.24\)
- At \(t \approx 7.85\), \(s = 5.24 - 13.52 = -8.28 \implies |s| = 8.28\)
- At \(t = 10\), \(s = -8.28 + 4.10 = -4.18 \implies |s| = 4.18\)
Thus, the particle is furthest from the origin at \(t \approx 7.85\text{ s}\) (where distance is \(8.28\text{ m}\)).

(a) M1 for setting \(v(t) = 0\). A1 for identifying correct boundary/solutions. A1 for \(t = 0\), \(t \approx 3.93\text{ s}\), and \(t \approx 7.85\text{ s}\). [3 marks]

(b) M1 for realizing speed is \(|v(t)|\). M1 for finding critical points/sketching. A1 for \(5.43\text{ m s}^{-1}\). [3 marks]

(c) M1 for writing integral expression \(\int_{0}^{10} |v(t)| dt\). A1 for correct bounds. A1 for \(25.8\text{ m}\). [3 marks]

(d) M1 for identifying \(a(t) = v'(t)\). M1 for using GDC derivative function. A1 for \(-4.32\text{ m s}^{-2}\). [3 marks]

(e) (i) M1 for using the fundamental theorem of calculus \(s(8) = s(0) + \int_0^8 v(t) dt\). A1 for \(-8.15\text{ m}\). [2 marks]
(ii) M1 for identifying turning points of \(s(t)\). M1 for checking magnitudes at these points. A1 for \(t = 7.85\text{ s}\). [3 marks]

(a) Let \(X\) be the mass of an apple. \(X \sim N(150, 12^2)\).
(i) \(P(140 < X < 165)\)
Using normalcdf on GDC with lower = 140, upper = 165, \(\mu = 150\), \(\sigma = 12\):
\(P(140 < X < 165) \approx 0.6912\) (or \(0.691\) to 3 s.f.).

(ii) \(P(X > 160)\)
Using normalcdf with lower = 160, upper = \(\infty\):
\(P(X > 160) \approx 0.2023\) (or \(0.202\) to 3 s.f.).

(b) We require \(P(X < d) = 0.08\).
Using invNorm on GDC with area = 0.08, \(\mu = 150\), \(\sigma = 12\):
\(d \approx 133.14 \approx 133\text{ g}\).

(c) Let \(Y\) be the number of apples in the sample of 15 that have a mass greater than \(160\text{ g}\).
\(Y \sim B(15, p)\), where \(p = 0.2023\).
(i) We want \(P(Y = 4)\).
Using binompdf on GDC with \(n = 15\), \(p = 0.2023\), \(x = 4\):
\(P(Y = 4) \approx 0.1944 \approx 0.194\).

(ii) We want \(P(Y \ge 2) = 1 - P(Y \le 1)\).
Using binomcdf on GDC with \(n = 15\), \(p = 0.2023\), \(x = 1\):
\(P(Y \le 1) \approx 0.1444\).
\(P(Y \ge 2) \approx 1 - 0.1444 = 0.8556 \approx 0.856\).

(d) Let \(W\) be the number of rejected crates. \(W \sim B(N, 0.05)\).
We require \(P(W \ge 1) > 0.99\).
\(1 - P(W = 0) > 0.99\)
\(1 - (0.95)^N > 0.99\)
\((0.95)^N < 0.01\)
Taking natural logarithms on both sides:
\(N \ln(0.95) < \ln(0.01)\)
Since \(\ln(0.95)\) is negative, reverse the inequality:
\(N > \frac{\ln(0.01)}{\ln(0.95)} \approx 89.78\).
Since \(N\) must be an integer, the minimum value is \(N = 90\).

(a) (i) M1 for correct setup. A1 for \(0.691\). [2 marks]
(ii) M1 for correct setup. A1 for \(0.202\). [2 marks]

(b) M1 for setting up the equation \(P(X < d) = 0.08\). M1 for inverse normal method. A1 for \(133\text{ g}\). [3 marks]

(c) (i) M1 for identifying binomial model with parameters \(n = 15\) and \(p = 0.2023\). A1 for correct substitution. A1 for \(0.194\). [3 marks]
(ii) M1 for recognizing \(P(Y \ge 2) = 1 - P(Y \le 1)\). A1 for finding \(P(Y \le 1) \approx 0.144\). A1 for \(0.856\). [3 marks]

(d) M1 for setting up the inequality \(1 - (0.95)^N > 0.99\). M1 for attempting to solve using logarithms or GDC table/graph. A1 for \(N > 89.78\). A1 for concluding \(N = 90\). [4 marks]