2025 IB DP Mathematics - Analysis and Approaches 模擬試題連答案詳解

Let the first term of the geometric series be \(u_1\). The sum of the first two terms is given by \(u_1 + u_1 r = 15 \implies u_1(1 + r) = 15\). The sum to infinity is given by \(S_\infty = \frac{u_1}{1 - r} = 27 \implies u_1 = 27(1 - r)\). Substituting the expression for \(u_1\) into the first equation yields \(27(1 - r)(1 + r) = 15\). This simplifies to \(27(1 - r^2) = 15 \implies 1 - r^2 = \frac{15}{27} = \frac{5}{9}\). Solving for \(r^2\) gives \(r^2 = 1 - \frac{5}{9} = \frac{4}{9}\). Since \(r > 0\), we take the positive square root to get \(r = \frac{2}{3}\).

M1: Attempt to write down expressions for the sum of the first two terms and the sum to infinity. A1: Correct equations \(u_1(1+r) = 15\) and \(\frac{u_1}{1-r} = 27\) (or equivalent). M1: Attempt to eliminate \(u_1\) to form an equation in terms of \(r\) only. A1: Correct simplified equation \(27(1-r^2) = 15\) (or equivalent). A1: Correct value \(r^2 = \frac{4}{9}\). A1: \(r = \frac{2}{3}\) (accepting only the positive root with a note of rejecting the negative root).

(a) To find the inverse, let \(y = \frac{kx + 5}{2x - 3}\). Swapping \(x\) and \(y\) gives \(x = \frac{ky + 5}{2y - 3}\). Rearranging to make \(y\) the subject: \(x(2y - 3) = ky + 5 \implies 2xy - 3x = ky + 5 \implies 2xy - ky = 3x + 5 \implies y(2x - k) = 3x + 5 \implies y = \frac{3x + 5}{2x - k}\). Thus, \(f^{-1}(x) = \frac{3x + 5}{2x - k}\). (b) Since \(f(x) = f^{-1}(x)\), we equate the two expressions: \rac{kx + 5}{2x - 3} = \frac{3x + 5}{2x - k}\). Comparing the coefficients of the terms in the denominators (or numerators), we find that \(k = 3\).

(a) M1: Setting up an equation \(y = f(x)\) and swapping \(x\) and \(y\) (or making \(x\) the subject). A1: Rearranging to get \(2xy - ky = 3x + 5\) (or equivalent intermediate step). A1: Correct final expression for the inverse \(f^{-1}(x) = \frac{3x + 5}{2x - k}\). (b) M1: Equating \(f(x)\) and \(f^{-1}(x)\). M1: Comparing the corresponding parts of the functions. A1: Correct value \(k = 3\).

We use the double angle identity \(\cos(2\theta) = 1 - 2\sin^2\theta\). Substituting this into the equation gives \((1 - 2\sin^2\theta) + 3\sin\theta = 2 \implies -2\sin^2\theta + 3\sin\theta - 1 = 0 \implies 2\sin^2\theta - 3\sin\theta + 1 = 0\). Factoring this quadratic equation gives \((2\sin\theta - 1)(\sin\theta - 1) = 0\). This yields two cases: \(\sin\theta = \frac{1}{2}\,\) or \(\,\sin\theta = 1\). For \(0 \le \theta \le 2\pi\), \(\sin\theta = \frac{1}{2}\) gives \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\). \(\sin\theta = 1\) gives \(\theta = \frac{\pi}{2}\). Thus, the solutions are \(\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\).

M1: Substituting \(\cos(2\theta) = 1 - 2\sin^2\theta\). A1: Correct quadratic equation \(2\sin^2\theta - 3\sin\theta + 1 = 0\). M1: Attempting to factor or solve the quadratic equation. A1: Correct values for \(\sin\theta\) (\(\sin\theta = \frac{1}{2}, \sin\theta = 1\)). A1: Finding \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\). A1: Finding \(\theta = \frac{\pi}{2}\).

First, find the \(y\)-coordinate of the point of tangency: \(f(\sqrt{3}) = (\sqrt{3})^2 e^{3 - (\sqrt{3})^2} = 3 e^{0} = 3\). Next, differentiate \(f(x)\) using the product rule and chain rule: \(f'(x) = 2x e^{3 - x^2} + x^2 \cdot e^{3 - x^2}(-2x) = e^{3 - x^2}(2x - 2x^3)\). Find the gradient of the tangent at \(x = \sqrt{3}\): \(f'(\sqrt{3}) = e^0(2\sqrt{3} - 2(3\sqrt{3})) = 2\sqrt{3} - 6\sqrt{3} = -4\sqrt{3}\). The equation of the tangent is \(y - 3 = -4\sqrt{3}(x - \sqrt{3}) \implies y - 3 = -4\sqrt{3}x + 12 \implies y = -4\sqrt{3}x + 15\).

A1: Correctly finding the \(y\)-coordinate, \(y = 3\). M1: Applying the product rule to differentiate \(f(x)\). A1: Correct derivative expression \(f'(x) = 2x e^{3-x^2} - 2x^3 e^{3-x^2}\) (or equivalent). M1: Substituting \(x = \sqrt{3}\) into their derivative. A1: Correct gradient of the tangent, \(m = -4\sqrt{3}\). A1: Correct final equation of the tangent, \(y = -4\sqrt{3}x + 15\).

We use integration by parts, \(\int u \, dv = uv - \int v \, du\). Let \(u = \ln(x) \implies du = \frac{1}{x} \, dx\). Let \(dv = x^3 \, dx \implies v = \frac{x^4}{4}\). Applying the integration by parts formula: \(\int x^3 \ln(x) \, dx = \frac{x^4}{4} \ln(x) - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{x^4}{4} \ln(x) - \frac{1}{4} \int x^3 \, dx = \frac{x^4}{4} \ln(x) - \frac{x^4}{16} + C\). Evaluating this from \(1\) to \(e\): \(\left[ \frac{x^4}{4} \ln(x) - \frac{x^4}{16} \right]_{1}^{e} = \left( \frac{e^4}{4} \ln(e) - \frac{e^4}{16} \right) - \left( \frac{1^4}{4} \ln(1) - \frac{1^4}{16} \right) = \left( \frac{e^4}{4} - \frac{e^4}{16} \right) - \left( 0 - \frac{1}{16} \right) = \frac{3e^4}{16} + \frac{1}{16} = \frac{3e^4 + 1}{16}\).

M1: Attempting integration by parts with appropriate choice of \(u = \ln(x)\) and \(dv = x^3 \, dx\). A1: Correct components \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^4}{4}\). A1: Correct substitution into the integration by parts formula: \(\frac{x^4}{4} \ln(x) - \int \frac{x^3}{4} \, dx\). A1: Correct indefinite integral \(\frac{x^4}{4} \ln(x) - \frac{x^4}{16}\). M1: Substituting limits \(1\) and \(e\) into their integrated expression. A1: Correct simplified final value: \rac{3e^4 + 1}{16}\).

(a) Using the law of total probability: \(\mathrm{P}(B) = \mathrm{P}(A) \mathrm{P}(B|A) + \mathrm{P}(A') \mathrm{P}(B|A')\). Since \(\mathrm{P}(A) = 0.6\), we have \(\mathrm{P}(A') = 0.4\). Thus, \(\mathrm{P}(B) = (0.6)(0.3) + (0.4)(0.8) = 0.18 + 0.32 = 0.5\). (b) Using the definition of conditional probability: \(\mathrm{P}(A|B') = \frac{\mathrm{P}(A \cap B')}{\mathrm{P}(B')}\). We have \(\mathrm{P}(B') = 1 - \mathrm{P}(B) = 1 - 0.5 = 0.5\). Also, \(\mathrm{P}(A \cap B') = \mathrm{P}(A) \mathrm{P}(B'|A) = 0.6(1 - 0.3) = 0.6(0.7) = 0.42\). Therefore, \(\mathrm{P}(A|B') = \frac{0.42}{0.5} = 0.84\) (or \(\frac{21}{25}\)).

(a) M1: Applying the law of total probability. A1: Correct numerical substitution: \(0.6 \times 0.3 + 0.4 \times 0.8\). A1: Correct calculation \(\mathrm{P}(B) = 0.5\). (b) M1: Writing down the conditional probability formula \(\mathrm{P}(A|B') = \frac{\mathrm{P}(A \cap B')}{\mathrm{P}(B')}\). A1: Correctly finding both \(\mathrm{P}(A \cap B') = 0.42\) and \(\mathrm{P}(B') = 0.5\) (or equivalent). A1: Correct final answer \(0.84\) (or \(\frac{21}{25}\)).

Using log laws, we simplify the equations: \(\log_3(x) + 2\log_3(y) = 5\) and \(2\log_3(x) - \log_3(y) = 5\). Let \(u = \log_3(x)\) and \(v = \log_3(y)\). The system becomes: (1) \(u + 2v = 5\) and (2) \(2u - v = 5\). From (2), we have \(v = 2u - 5\). Substituting this into (1) gives \(u + 2(2u - 5) = 5 \implies u + 4u - 10 = 5 \implies 5u = 15 \implies u = 3\). Substituting \(u = 3\) back into the expression for \(v\) gives \(v = 2(3) - 5 = 1\). Since \(u = \log_3(x) = 3\), we have \(x = 3^3 = 27\). Since \(v = \log_3(y) = 1\), we have \(y = 3^1 = 3\).

M1: Applying log laws to simplify power terms in logs (e.g. \(\log_3(y^2) = 2\log_3(y)\) or \(\log_3(x^2) = 2\log_3(x)\)). A1: Correct system of linear equations in terms of \(\log_3(x)\) and \(\log_3(y)\). M1: Attempting to solve the simultaneous equations. A1: Correct values for \(\log_3(x) = 3\) and \(\log_3(y) = 1\). A1: Correct value \(x = 27\). A1: Correct value \(y = 3\).

The displacement \(s(t)\) is the integral of the velocity function \(v(t)\): \(s(t) = \int v(t) \, dt = \int (6\sin(2t) - 4\cos(t)) \, dt\). Finding the indefinite integral gives \(s(t) = -3\cos(2t) - 4\sin(t) + C\), where \(C\) is the constant of integration. Using the initial condition \(s(0) = 3\): \(3 = -3\cos(0) - 4\sin(0) + C \implies 3 = -3(1) - 0 + C \implies C = 6\). Thus, the expression for the displacement is \(s(t) = -3\cos(2t) - 4\sin(t) + 6\).

M1: Recognizing that displacement is the integral of velocity, \(s(t) = \int v(t) \, dt\). M1: Attempting to integrate \(6\sin(2t) - 4\cos(t)\). A1: Correct integration of the sine term: \(-3\cos(2t)\). A1: Correct integration of the cosine term: \(-4\sin(t)\). M1: Substituting \(t = 0\) and \(s = 3\) to find the constant \(C\). A1: Correct final expression: \(s(t) = -3\cos(2t) - 4\sin(t) + 6\).

To find the coordinates of the local maximum point, we first find the derivative of \(f(x)\) using the quotient rule. Let \(u = \ln(2x)\) and \(v = x^2\). Then \(u' = \frac{1}{x}\) and \(v' = 2x\). Applying the quotient rule: \(f'(x) = \frac{u'v - uv'}{v^2} = \frac{\frac{1}{x}(x^2) - \ln(2x)(2x)}{(x^2)^2} = \frac{x - 2x\ln(2x)}{x^4} = \frac{1 - 2\ln(2x)}{x^3}\). To find the local maximum, we set \(f'(x) = 0\): \(\frac{1 - 2\ln(2x)}{x^3} = 0 \implies 1 - 2\ln(2x) = 0 \implies \ln(2x) = \frac{1}{2}\). Solving for \(x\): \(2x = e^{1/2} \implies x = \frac{\sqrt{e}}{2}\). To find the corresponding \(y\)-coordinate, we substitute this \(x\)-value back into the original function: \(y = f\left(\frac{\sqrt{e}}{2}\right) = \frac{\ln\left(2 \cdot \frac{\sqrt{e}}{2}\right)}{\left(\frac{\sqrt{e}}{2}\right)^2} = \frac{\ln(\sqrt{e})}{\frac{e}{4}} = \frac{\frac{1}{2}}{\frac{e}{4}} = \frac{2}{e}\). Thus, the coordinates of the local maximum point are \(\left(\frac{\sqrt{e}}{2}, \frac{2}{e}\right)\).

**M1** for attempting to use the quotient rule (or product rule); **A1** for correct derivative \(f'(x) = \frac{1 - 2\ln(2x)}{x^3}\); **M1** for setting their derivative equal to zero, \(f'(x) = 0\); **A1** for obtaining the correct \(x\)-coordinate \(x = \frac{\sqrt{e}}{2}\); **M1** for substituting their \(x\)-value into \(f(x)\); **A1** for the correct final coordinates \(\left(\frac{\sqrt{e}}{2}, \frac{2}{e}\right)\).

(a) Using the product rule on the first term, we have f'(x) = 1 * \ln x + x * (1/x) - k = \ln x + 1 - k. (b) For a local minimum, set f'(x) = 0, which gives \ln x + 1 - k = 0, so \ln x = k - 1, yielding x = e^{k-1}. Since the second derivative is f''(x) = 1/x, which is positive for all x > 0, this value of x corresponds to a local minimum. The y-coordinate is f(e^{k-1}) = e^{k-1} \ln(e^{k-1}) - k e^{k-1} = e^{k-1}(k - 1) - k e^{k-1} = -e^{k-1}. Thus, the coordinates are (e^{k-1}, -e^{k-1}). (c) At x = e^k, the y-coordinate is f(e^k) = e^k \ln(e^k) - k e^k = k e^k - k e^k = 0. The gradient of the tangent is f'(e^k) = \ln(e^k) + 1 - k = k + 1 - k = 1. Using the point-slope formula, the equation of the tangent line is y - 0 = 1 * (x - e^k), which simplifies to y = x - e^k. (d) Since k > 1, f(x) = x(\ln x - k) is negative for all 1 <= x < e^k. Thus, the area of the enclosed region is given by the integral from 1 to e^k of -f(x) dx, which is the integral from 1 to e^k of (k x - x \ln x) dx. Using integration by parts for the second term, where u = \ln x and dv = x dx, we get the integral of x \ln x dx = (1/2) x^2 \ln x - (1/4) x^2. Thus, the antiderivative of (k x - x \ln x) is (1/2) k x^2 - (1/2) x^2 \ln x + (1/4) x^2. Evaluating this expression from 1 to e^k: At x = e^k, we have (1/2) k e^{2k} - (1/2) e^{2k} (k) + (1/4) e^{2k} = (1/4) e^{2k}. At x = 1, we have (1/2) k (1)^2 - 0 + (1/4) (1)^2 = (1/2) k + 1/4. Subtracting the lower limit from the upper limit, the total area is (1/4) e^{2k} - (1/2) k - 1/4.

(a) M1 for attempting to use product rule, A1 for product rule derivative term \ln x + 1, A1 for final derivative f'(x) = \ln x + 1 - k. (b) M1 for setting f'(x) = 0, A1 for x = e^{k-1}, M1 for substituting x back into f(x) to find y, A1 for y = -e^{k-1} leading to (e^{k-1}, -e^{k-1}). (c) M1 for finding f(e^k) = 0, M1 for finding gradient f'(e^k) = 1, A1 for setting up the line equation, A1 for showing clearly that y = x - e^k. (d) M1 for setting up the correct area integral with limits and negative sign, M1 for attempting integration by parts on x \ln x, A1 for correct integration of x \ln x, A1 for the full correct antiderivative, M1 for substituting the limits e^k and 1, A1 for evaluating each boundary correctly, A1 for the final simplified expression (1/4)e^{2k} - (1/2)k - 1/4.

(a) The general term of an arithmetic sequence is given by u_n = u_1 + (n-1)d. Substituting the given values: u_n = \ln a + (n-1)\ln 2 = \ln a + \ln(2^{n-1}) = \ln(a \cdot 2^{n-1}). (b) The sum of the first three terms of the arithmetic sequence is S_3 = u_1 + u_2 + u_3 = \ln a + \ln(2a) + \ln(4a) = \ln(a * 2a * 4a) = \ln(8a^3). We are given S_3 = 3 \ln 4 = \ln(4^3) = \ln 64. Thus, \ln(8a^3) = \ln 64, which implies 8a^3 = 64, so a^3 = 8, and since a > 0, we have a = 2. (c) The first three terms of the geometric sequence are v_1 = a = 2, v_2 = 2r, and v_3 = 2r^2. Their sum is 2 + 2r + 2r^2 = 26. Dividing by 2 gives 1 + r + r^2 = 13, which rearranges to r^2 + r - 12 = 0. Factoring gives (r + 4)(r - 3) = 0. Since r > 0, we find r = 3. (d) For these values of a and r, we have v_n = 2 * 3^{n-1} and e^{u_n} = e^{\ln(2 * 2^{n-1})} = 2 * 2^{n-1}. We want to solve the inequality v_n > 10 e^{u_n}, which is 2 * 3^{n-1} > 10 * (2 * 2^{n-1}). Dividing both sides by 2 yields 3^{n-1} > 10 * 2^{n-1}. Dividing by 2^{n-1} gives (3/2)^{n-1} > 10. Taking the natural logarithm of both sides gives (n - 1) \ln(1.5) > \ln 10. Since 1.5 > 1, \ln(1.5) is positive, so we can divide by it without changing the inequality sign: n - 1 > \ln 10 / \ln 1.5, which gives n > 1 + \ln 10 / \ln 1.5. Thus, P = 10 and Q = 1.5.

(a) M1 for using the formula u_n = u_1 + (n-1)d, A1 for substituting the given terms, A1 for combining the terms using log laws to show the required result. (b) M1 for writing an expression for S_3, A1 for simplifying it to \ln(8a^3), M1 for expressing 3 \ln 4 as \ln 64, A1 for solving 8a^3 = 64 to obtain a = 2. (c) M1 for writing the sum of the first three terms as 2 + 2r + 2r^2 = 26, A1 for forming the quadratic equation r^2 + r - 12 = 0, M1 for factoring or solving the quadratic, A1 for rejecting the negative root and stating r = 3. (d) M1 for expressing v_n as 2 * 3^{n-1}, M1 for expressing e^{u_n} as 2 * 2^{n-1}, A1 for writing the correct inequality 2 * 3^{n-1} > 20 * 2^{n-1}, M1 for simplifying this to (1.5)^{n-1} > 10, M1 for taking natural logs on both sides, A1 for correct algebraic manipulation to isolate n, A1 for final answer in the required form with P = 10 and Q = 1.5 (or Q = 3/2).

(a) To show they intersect, we set r_1 = r_2 to obtain a system of three equations: 1 + 2\lambda = 4 + \mu (Eq 1), 2 - \lambda = 2 + \mu (Eq 2), and -1 + \lambda = -1 - \mu (Eq 3). From Eq 2, we have \mu = -\lambda. Substituting this into Eq 1 gives 1 + 2\lambda = 4 - \lambda, which simplifies to 3\lambda = 3, so \lambda = 1. Consequently, \mu = -1. Checking these parameters in Eq 3: -1 + 1 = 0 and -1 - (-1) = 0, which is consistent. Substituting \lambda = 1 into r_1 gives the point P(1+2, 2-1, -1+1) = (3, 1, 0). (b) The direction vectors of L_1 and L_2 are d_1 = (2, -1, 1)^T and d_2 = (1, 1, -1)^T. Their dot product is d_1 . d_2 = 2*1 + (-1)*1 + 1*(-1) = 2 - 1 - 1 = 0. Since the dot product of their non-zero direction vectors is 0, L_1 and L_2 are perpendicular. (c) For \lambda = 3, point A is A(1+6, 2-3, -1+3) = (7, -1, 2). The vector PA is OA - OP = (7-3, -1-1, 2-0)^T = (4, -2, 2)^T. The distance AP is the magnitude of this vector: |PA| = \sqrt{4^2 + (-2)^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6}. (d) Since the lines L_1 and L_2 are perpendicular and intersect at P, the triangle APB is a right-angled triangle at P. The area of the triangle is Area = (1/2) * AP * PB. We are given Area = 6\sqrt{2}, so (1/2) * (2\sqrt{6}) * PB = 6\sqrt{2}, which simplifies to \sqrt{6} * PB = 6\sqrt{2}. Thus, PB = 6\sqrt{2}/\sqrt{6} = 6/\sqrt{3} = 2\sqrt{3}. Since B lies on L_2, its position vector is of the form b = (4+\mu, 2+\mu, -1-\mu)^T. The vector PB is b - OP = (4+\mu-3, 2+\mu-1, -1-\mu-0)^T = (1+\mu, 1+\mu, -1-\mu)^T = (\mu+1)(1, 1, -1)^T. Its magnitude is |PB| = |\mu+1| * \sqrt{1^2 + 1^2 + (-1)^2} = |\mu+1|\sqrt{3}. Setting this equal to 2\sqrt{3} yields |\mu+1|\sqrt{3} = 2\sqrt{3}, so |\mu+1| = 2. This gives two cases: Case 1: \mu+1 = 2, so \mu = 1. The position vector of B is (5, 3, -2)^T. Case 2: \mu+1 = -2, so \mu = -3. The position vector of B is (1, -1, 2)^T.

(a) M1 for equating the components of r_1 and r_2, A1 for finding a consistent system of equations, A1 for solving the system to get \lambda = 1 and \mu = -1, A1 for verifying with the third component, A1 for clearly stating the coordinates of P as (3, 1, 0). (b) M1 for identifying the correct direction vectors, M1 for evaluating the dot product, A1 for stating that the product is 0 and concluding they are perpendicular. (c) A1 for finding the coordinates of A as (7, -1, 2), M1 for finding the vector PA, A1 for calculating the magnitude AP = 2\sqrt{6}. (d) M1 for identifying that the triangle is right-angled at P, M1 for setting up the area equation (1/2) * AP * PB = 6\sqrt{2}, A1 for finding the correct distance PB = 2\sqrt{3}, M1 for expressing PB in terms of \mu, A1 for setting up the equation |\mu+1| = 2, A1 for finding \mu = 1 and \mu = -3, A1 for providing both correct position vectors: (5, 3, -2)^T and (1, -1, 2)^T.

(a) Let \(X\) be the mass of a puppy, where \(X \sim N(3.5, 0.4^2)\).
Using a GDC to find the normal cumulative probability:
\(P(X > 4.1) \approx 0.066807 \approx 0.0668\).

(b) We are given \(P(X < w) = 0.15\).
Using the inverse normal function on a GDC:
\(w = \text{invNorm}(0.15, 3.5, 0.4) \approx 3.0854 \approx 3.09\).

(c) Let \(Y\) be the number of puppies weighing more than \(4.1\text{ kg}\).
Then \(Y \sim B(3, 0.066807)\).
Using binomial probability distribution on a GDC:
\(P(Y = 2) = \binom{3}{2} (0.066807)^2 (1 - 0.066807) \approx 0.012497 \approx 0.0125\).

(a) M1 for setting up the normal model, A1 for the correct probability of 0.0668. [2 marks]
(b) M1 for set-up with inverse normal, A1 for 3.09. [2 marks]
(c) M1 for recognizing binomial distribution, A1 for 0.0125. [2 marks]

(a) The particle is at rest when \(v(t) = 0\).
\(3t^2 e^{-0.8t} - 1 = 0\)
Using a GDC to solve this equation for \(0 \le t \le 5\):
\(t \approx 0.6482\) and \(t \approx 3.791\)
So, \(t \approx 0.648\) seconds and \(t \approx 3.79\) seconds.

(b) The maximum velocity occurs when \(v'(t) = 0\).
Using a GDC to find the local maximum of \(v(t)\):
This occurs at \(t = 2.5\).
\(v(2.5) = 3(2.5)^2 e^{-0.8(2.5)} - 1 = 18.75 e^{-2} - 1 \approx 1.538 \approx 1.54\text{ m s}^{-1}\).

(c) The total distance travelled is given by \\int_{0}^{5} |v(t)| dt.
Using GDC integration:
\\int_{0}^{5} |3t^2 e^{-0.8t} - 1| dt \\approx 3.1235 \\approx 3.12\\text{ m}\).

(a) M1 for setting \(v(t) = 0\), A1 for obtaining both solutions: 0.648 and 3.79. [2 marks]
(b) M1 for finding the maximum point on GDC, A1 for 1.54. [2 marks]
(c) M1 for writing the correct definite integral with absolute value, A1 for 3.12. [2 marks]

(a) Let us determine the internal angle \(\angle ABC\).
The bearing of \(B\) from \(A\) is \(065^\circ\), so the reverse direction \(BA\) has a bearing of \(65^\circ + 180^\circ = 245^\circ\).
The bearing of \(C\) from \(B\) is \(145^\circ\).
Therefore, the angle \(\angle ABC = 245^\circ - 145^\circ = 100^\circ\).
Using the cosine rule in \(\triangle ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 12^2 + 18^2 - 2(12)(18)\cos(100^\circ)\)
\(AC^2 = 144 + 324 - 432\cos(100^\circ)\)
\(AC^2 \approx 543.016 \implies AC \approx 23.303 \approx 23.3\text{ km}\).

(b) To find the bearing of \(C\) from \(A\), we first find \(\angle BAC\) using the sine rule:
\\frac{\\sin(\\angle BAC)}{BC} = \\frac{\\sin(\\angle ABC)}{AC}
\\frac{\\sin(\\angle BAC)}{18} = \\frac{\\sin(100^\\circ)}{23.303}
\\sin(\\angle BAC) \\approx 0.7607 \\implies \\angle BAC \\approx 49.53^\\circ\).
The bearing of \(C\) from \(A\) is \(65^\circ + 49.53^\circ = 114.53^\circ \approx 115^\circ\) (to 3 s.f.).

(a) M1 for calculating \(\angle ABC = 100^\circ\), M1 for correct substitution into cosine rule, A1 for 23.3. [3 marks]
(b) M1 for correct substitution into sine rule, A1 for \(\angle BAC \approx 49.5^\circ\), A1 for the bearing 115 (or 114.5). [3 marks]

(a) Let \(v_n\) represent the annual profit of Company B in year \(n\).
\(v_n\) is a geometric sequence with \(v_1 = 100000\) and \(r = 1.08\).
\(v_{10} = 100000 \times (1.08)^9 \approx 199900.46 \approx \\\$199,900\) (to 3 s.f.).

(b) Let \(u_n\) represent the annual profit of Company A in year \(n\).
\(u_n\) is an arithmetic sequence with \(u_1 = 150000\) and \(d = 12000\).
The sum over 15 years is:
\(S_{15} = \frac{15}{2} \left(2(150000) + 14(12000)\right) = 7.5 \times (300000 + 168000) = \\\$3,510,000\).

(c) We want to find the smallest integer \(n\) such that \(v_n > u_n\).
\(100000 \times 1.08^{n-1} > 150000 + 12000(n-1)\)
Using the GDC to compare the values for different integers \(n\):
For \(n = 16\):
\(v_{16} \approx 317,217\) and \(u_{16} = 330,000\) (since \(v_{16} < u_{16}\))
For \(n = 17\):
\(v_{17} \approx 342,594\) and \(u_{17} = 342,000\) (since \(v_{17} > u_{17}\))
Thus, the first year is the 17th year.

(a) M1 for setting up the geometric sequence formula, A1 for \\$199,900. [2 marks]
(b) M1 for using the arithmetic series sum formula, A1 for \\$3,510,000. [2 marks]
(c) M1 for formulating the inequality or setting up GDC table comparison, A1 for Year 17. [2 marks]

(a) As \(x \to \infty\), \(e^{-0.5x} \to 0\), meaning \(g(x) \to 3\).
The horizontal asymptote is \(y = 3\).

(b) Intersection points occur where \(f(x) = g(x)\):
\(\ln(x^2 - 3x + 4) = 3 - e^{-0.5x}\)
Using the GDC equation solver or graphing features to find the points of intersection:
\(x \approx -0.475\) and \(x \approx 5.64\).

(c) Comparing the graphs on GDC, the curve of \(f(x)\) lies above the curve of \(g(x)\) when \(x\) is to the left of the first intersection or to the right of the second.
Therefore, the solution to \(f(x) > g(x)\) is \(x < -0.475\) or \(x > 5.64\).

(a) A1 for \(y = 3\). [1 mark]
(b) M1 for equating \(f(x) = g(x)\), A1 for \(x \approx -0.475\), A1 for \(x \approx 5.64\). [3 marks]
(c) M1 for interpreting the inequality graphically, A1 for \(x < -0.475\) or \(x > 5.64\). [2 marks]

(a) At points of intersection, \(\cos(2x) = x^2 - 2\).
Using GDC solver or graph intersection feature:
\(x \approx -1.153\) and \(x \approx 1.153\).
To 3 s.f., the x-coordinates are \(x = -1.15\) and \(x = 1.15\).

(b) The area of the region is given by the integral:
\(A = \int_{-1.153}^{1.153} \left( \cos(2x) - (x^2 - 2) \right) dx\)
Using GDC integration directly:
\(A \approx 4.3314 \approx 4.33\).

(a) M1 for setting up the equation, A1 for obtaining \(x \approx \pm 1.15\). [2 marks]
(b) M1 for the correct integral expression, M1 for using the limits from part (a), A1 for correct GDC numerical integration, A1 for 4.33. [4 marks]

(a) Entering the data in the statistics list of the GDC and performing a linear regression analysis:
\(r \approx 0.9951 \approx 0.995\).

(b) From the GDC linear regression output:
\(a \approx 2.5739 \approx 2.57\)
\(b \approx 40.544 \approx 40.5\)
So, the equation is \(y = 2.57x + 40.5\).

(c) Substitute \(x = 14\) into the regression equation:
\(y = 2.5739(14) + 40.544 \approx 76.5786 \approx 76.6\).

(a) M1 for entering data into GDC list, A1 for 0.995. [2 marks]
(b) M1 for accessing linear regression values, A1 for \(y = 2.57x + 40.5\). [2 marks]
(c) M1 for substituting 14 into their regression equation, A1 for 76.6. [2 marks]

(a) The area of a sector is given by \(A = \frac{1}{2}r^2 \theta\).
\(60 = \frac{1}{2}r^2 \theta \implies r^2 \theta = 120\).
The perimeter of a sector is given by \(P = 2r + r\theta\).
So, \(2r + r\theta = 32\).

(b) Rearranging the perimeter equation:
\(r\theta = 32 - 2r \implies \theta = \frac{32 - 2r}{r}\).
Substitute into the area equation:
\(r^2 \left(\frac{32 - 2r}{r}\right) = 120\)
\(r(32 - 2r) = 120\)
\(32r - 2r^2 = 120 \implies 2r^2 - 32r + 120 = 0\)
\(r^2 - 16r + 60 = 0\)
\((r - 6)(r - 10) = 0\)
Therefore, the possible values are \(r = 6\text{ cm}\) or \(r = 10\text{ cm}\).

(c) For the larger radius, \(r = 10\).
\(10\theta = 32 - 2(10) \implies 10\theta = 12 \implies \theta = 1.2\text{ radians}\).
The area of the segment is:
\(A_{\text{segment}} = \frac{1}{2}r^2(\theta - \sin\theta)\)
\(A_{\text{segment}} = \frac{1}{2}(10^2)(1.2 - \sin(1.2)) \approx 50 \times 0.26796 \approx 13.398 \approx 13.4\text{ cm}^2\).

(a) M1 for correct sector area formula step, A1 for writing down the perimeter equation. [2 marks]
(b) M1 for substitution to get a single quadratic in \(r\), M1 for factorizing or solving, A1 for \(r = 6\) and \(r = 10\). [3 marks]
(c) A1 for 13.4. [1 mark]

(a) Let \(Z \sim N(0, 1)\).

Using the inverse normal cumulative distribution function on a GDC:
For \(P(X < 35) = 0.12\):
\(P\left(Z < \frac{35 - \mu}{\sigma}\right) = 0.12 \implies \frac{35 - \mu}{\sigma} \approx -1.174986\)
\(\mu - 1.174986\sigma = 35\) (Equation 1)

For \(P(X > 80) = 0.05 \implies P(X \le 80) = 0.95\):
\(P\left(Z < \frac{80 - \mu}{\sigma}\right) = 0.95 \implies \frac{80 - \mu}{\sigma} \approx 1.644853\)
\(\mu + 1.644853\sigma = 80\) (Equation 2)

Subtracting Equation 1 from Equation 2:
\(2.819839\sigma = 45 \implies \sigma \approx 15.9583\)
So \(\sigma \approx 16.0\) (to 3 significant figures).

Substituting \(\sigma\) back into Equation 2:
\(\mu = 80 - 1.644853(15.9583) \approx 53.7510\)
So \(\mu \approx 53.8\) (to 3 significant figures).

(b) We want to find \(P(50 < X < 75)\) where \(X \sim N(53.7510, 15.9583^2)\).
Using a GDC to find the normal cumulative probability:
\(P(50 < X < 75) \approx 0.50141\)
To 3 significant figures, this probability is \(0.501\) (or \(0.501\) if using the 3 s.f. values \(\mu = 53.8\) and \(\sigma = 16.0\)).

**(a)**
* **M1** for attempting to write at least one standardization equation or finding a \(z\)-value.
* **A1** for finding both correct \(z\)-values: \(z_1 \approx -1.17\) and \(z_2 \approx 1.64\) (accept \(z_1 \in [-1.18, -1.17]\) and \(z_2 \in [1.64, 1.65]\)).
* **A1** for obtaining \(\sigma \approx 16.0\) (accept \(15.96\) or \(15.9583...\)).
* **A1** for obtaining \(\mu \approx 53.8\) (accept \(53.75\) or \(53.7510...\)).

**(b)**
* **M1** for attempting to find \(P(50 < X < 75)\) with their mean and standard deviation (e.g., writing the probability statement or a sketch or calculator inputs).
* **A1** for correct probability \(0.501\) (accept \(0.50141...\) or \(0.50131...\) if using 3 s.f. rounded values from part a).

(a)
At \(t = 0\):
\(v(0) = 3 - e^0 \cos(0) = 3 - 1 = 2 \text{ m s}^{-1}\).

(b)
The particle is at rest when \(v(t) = 0\).
Using a GDC to solve \(3 - e^{0.2t} \cos\left(\frac{\pi t}{3}\right) = 0\) on \(0 \le t \le 10\):
\(t \approx 5.71\) and \(t \approx 6.62\).

(c)
\(a(t) = v'(t)\).
Using a GDC to evaluate the numerical derivative at \(t = 4\):
\(a(4) \approx -1.80 \text{ m s}^{-2}\).

(d)
Total distance \(= \int_0^6 |v(t)| \mathrm{d}t\).
Using numerical integration on a GDC:
Total distance \(\approx 17.6\) m.

(e)
(i) \(s(t) = -2 + \int_0^t v(u) \mathrm{d}u\).

(ii) The critical points of \(s(t)\) occur where \(s'(t) = v(t) = 0\), which are at \(t_1 \approx 5.7126\) and \(t_2 \approx 6.6180\).
Evaluating the displacement at the endpoints and critical points:
- At \(t = 0\), \(s(0) = -2\).
- At \(t \approx 5.71\), \(s(5.7126) = -2 + \int_0^{5.7126} v(u) \mathrm{d}u \approx 15.6\) m.
- At \(t \approx 6.62\), \(s(6.6180) = -2 + \int_0^{6.6180} v(u) \mathrm{d}u \approx 15.5\) m.
- At \(t = 10\), \(s(10) = -2 + \int_0^{10} v(u) \mathrm{d}u \approx 34.7\) m.
Comparing the magnitudes, the maximum displacement from the origin is \(34.7\) m (to 3 s.f.) which occurs at \(t = 10\).

(iii) The particle passes through the origin when \(s(t) = 0\).
Using a GDC to find the root of \(-2 + \int_0^t v(u) \mathrm{d}u = 0\):
\(t \approx 0.960\) s.

(a)
M1 for substituting \(t=0\) into the equation.
A1 for \(2\).

(b)
M1 for setting \(v(t) = 0\).
A1 for \(5.71\), A1 for \(6.62\).

(c)
M1 for recognizing \(a(t) = v'(t)\).
A1 for \(-1.80\).

(d)
M1 for writing the integral for total distance \(\int_0^6 |v(t)| \mathrm{d}t\).
M1 for splitting the integral or using GDC absolute value function.
A1 for \(17.6\).

(e)
(i) A1 for correct integral expression including the constant \(-2\).
(ii) M1 for finding or listing candidates (endpoints and roots of \(v(t)\)).
A1 for checking \(s(5.71) \approx 15.6\) or \(s(6.62) \approx 15.5\).
A1 for checking \(s(10) \approx 34.7\).
A1 for identifying \(34.7\) m as the maximum displacement.
A1 for identifying \(t = 10\).
(iii) M1 for setting \(s(t) = 0\).
A1 for \(0.960\).

(a)
(i) We need to find \(P(48.5 < L < 51.5)\).
Using GDC: \(P(48.5 < L < 51.5) \approx 0.7887 = 0.789\).

(ii) We need to find \(P(L > 52.0)\).
Using GDC: \(P(L > 52.0) \approx 0.04779 = 0.0478\).

(b)
(i) A bolt is defective if \(L < 48.0\) or \(L > 52.0\).
Since the distribution is symmetric about the mean \(\mu = 50\):
\(P(\text{Defective}) = P(L < 48.0) + P(L > 52.0) = 2 \times P(L > 52.0)\)
From part (a)(ii), \(P(L > 52.0) \approx 0.04779\).
\(P(\text{Defective}) = 2 \times 0.04779 = 0.09558 \approx 0.0956\).

(ii) Let \(X\) be the number of defective bolts in a batch of 150.
\(X \sim B(150, 0.09558)\).
We want to find \(P(X \ge 15) = 1 - P(X \le 14)\).
Using GDC: \(P(X \le 14) \approx 0.5283\).
\(P(X \ge 15) = 1 - 0.5283 = 0.4717 \approx 0.472\).

(c)
(i) We want \(P(L' > 55.0) = 0.01\) where \(L' \sim N(\mu', 1.2^2)\).
Standardizing this gives:
\(P\left(Z > \frac{55.0 - \mu'}{1.2}\right) = 0.01\)
Using GDC inverse normal function: \(\frac{55.0 - \mu'}{1.2} = 2.3263\)
\(55.0 - \mu' = 2.7916\)
\(\mu' = 52.208 \approx 52.2\) mm.

(ii) We want the probability of defect to be \(0.02\) with \(\mu = 50\) and standard deviation \(\sigma'\).
By symmetry:
\(P(\text{Defective}) = 2 \times P(L < 48.0) = 0.02 \implies P(L < 48.0) = 0.01\).
Standardizing gives:
\(P\left(Z < \frac{48.0 - 50}{\sigma'}\right) = 0.01\)
Using GDC inverse normal function:
\(\frac{-2.0}{\sigma'} = -2.3263\)
\(\sigma' = \frac{2.0}{2.3263} \approx 0.8597 \approx 0.860\) mm.

(a)
(i) M1 for a correct normal probability expression.
A1 for \(0.789\).
(ii) A1 for \(0.0478\).

(b)
(i) M1 for using symmetry: \(2 \times P(L > 52.0)\).
A1 for showing \(2 \times 0.04779...\).
AG for obtaining \(0.0956\) (must show unrounded value first).
(ii) M1 for identifying Binomial distribution with parameters \(n = 150, p = 0.0956\).
M1 for writing or calculating \(1 - P(X \le 14)\).
A1 for \(0.472\).

(c)
(i) M1 for writing standardizing equation: \(\frac{55.0 - \mu'}{1.2} = z\).
A1 for \(z = 2.3263\).
A1 for \(52.2\).
(ii) M1 for dividing defect rate by 2 to get \(0.01\).
M1 for standardizing equation: \(\frac{-2}{\sigma'} = -z\).
A1 for \(z = 2.3263\).
M1 for solving for \(\sigma'\).
A1 for \(0.860\).

(a)
Since the bearing of \(B\) from \(C\) is \(040^\circ\) and the bearing of \(D\) from \(C\) is \(110^\circ\), the angle between them is:
\(\widehat{BCD} = 110^\circ - 40^\circ = 70^\circ\).

To find \(\widehat{CBD}\), first find \(\widehat{BDC}\):
The back-bearing of \(C\) from \(D\) is \(110^\circ + 180^\circ = 290^\circ\).
The bearing of \(B\) from \(D\) is \(320^\circ\).
Therefore, the angle \(\widehat{BDC} = 320^\circ - 290^\circ = 30^\circ\).
Using the angle sum of triangle \(BCD\):
\(\widehat{CBD} = 180^\circ - 70^\circ - 30^\circ = 80^\circ\).

(b)
In triangle \(BCD\), we can use the Sine Rule:
\(\frac{BC}{\sin(\widehat{BDC})} = \frac{CD}{\sin(\widehat{CBD})}\)
\(\frac{BC}{\sin(30^\circ)} = \frac{500}{\sin(80^\circ)}\)
\(BC = \frac{500 \sin(30^\circ)}{\sin(80^\circ)}\)
\(BC \approx 253.86 \approx 254\) m.

(c)
(i) Triangle \(ABC\) is right-angled at \(B\). Let the vertical height of \(A\) above sea level be \(AB\).
\(\tan(15^\circ) = \frac{AB}{BC}\)
\(AB = BC \tan(15^\circ) = 253.86 \times \tan(15^\circ) \approx 68.02 \approx 68.0\) m.

(ii) The height of the lighthouse \(h\) is:
\(h = AB - 45 = 68.02 - 45 = 23.02 \approx 23.0\) m.

(d)
(i) The path of the boat is the line segment \(CD\). The closest point to \(B\) on this path is the perpendicular projection \(N\) of \(B\) onto \(CD\).
In the right-angled triangle \(BNC\):
\(BN = BC \sin(\widehat{BCD}) = 253.86 \times \sin(70^\circ)\)
\(BN \approx 238.55 \approx 239\) m.

(ii) The angle of elevation \(\theta\) from any point on the path to \(A\) is given by \(\tan(\theta) = \frac{AB}{\text{distance}}\).
This angle is maximized when the horizontal distance to the base \(B\) is minimized (which is the closest distance \(BN\)).
\(\tan(\theta_{\max}) = \frac{AB}{BN} = \frac{68.02}{238.55} \approx 0.2851\)
\(\theta_{\max} = \arctan(0.2851) \approx 15.9^\circ\) (or \(0.278\) radians).

(a)
M1 for subtracting bearings: \(110 - 40 = 70^\circ\).
M1 for finding the angle \(\widehat{BDC} = 30^\circ\) (using back-bearings or alternate angles).
A1 for \(80^\circ\).

(b)
M1 for substituting correct values into the Sine Rule.
M1 for rearranging to make \(BC\) the subject.
A1 for intermediate value \(253.86\).
A1 for \(254\).

(c)
(i) M1 for using \(\tan\) ratio with \(BC\) and angle \(15^\circ\).
A1 for \(68.0\) (accept \(68.1\) if using rounded \(254\)).
(ii) A1 for \(23.0\).

(d)
(i) M1 for recognizing that the closest distance is perpendicular to \(CD\).
M1 for setting up the sine ratio in right-angled triangle \(BNC\).
A1 for expression \(253.86 \sin(70^\circ)\).
A1 for \(239\) (accept \(238.7\) if using rounded \(254\)).
(ii) M1 for recognizing that maximum elevation corresponds to minimum distance.
M1 for setting up \(\tan(\theta) = \frac{\text{height}}{\text{closest distance}}\).
A1 for \(\tan(\theta) = \frac{68.02}{238.55}\).
A1 for \(15.9^\circ\) or \(0.278\) rad.

**Part A: Introduction and basic properties**
**(a) (i)** For k = 0, f_0(x) = \frac{x^2}{x^2+1}.
As x \to \pm\infty, f_0(x) \to 1, so there is a horizontal asymptote at y = 1.
The derivative is f'_0(x) = \frac{2x(x^2+1) - x^2(2x)}{(x^2+1)^2} = \frac{2x}{(x^2+1)^2}. Setting this to 0 gives x = 0.
At x = 0, y = 0. Since f_0(x) \ge 0 for all x, (0,0) is a local (and global) minimum.
The graph is symmetric about the y-axis, starting near y = 1 at -\infty, dipping down to a minimum at (0,0), and rising back to approach y = 1 as x \to \infty.
**(ii)** The range of f_0(x) is [0, 1).
**(b)** Using the quotient rule for f_k(x) = \frac{x^2+kx}{x^2+1}:
f'_k(x) = \frac{(2x+k)(x^2+1) - (x^2+kx)(2x)}{(x^2+1)^2}
f'_k(x) = \frac{2x^3 + 2x + kx^2 + k - 2x^3 - 2kx^2}{(x^2+1)^2}
f'_k(x) = \frac{-kx^2 + 2x + k}{(x^2+1)^2}.

**Part B: Stationary points and their locus**
**(c) (i)** Stationary points occur when f'_k(x) = 0, which requires the numerator to be zero:
-kx^2 + 2x + k = 0.
For k
e 0, this is a quadratic equation kx^2 - 2x - k = 0.
The discriminant is \Delta = (-2)^2 - 4(k)(-k) = 4 + 4k^2.
Since k
e 0, k^2 > 0, which means \Delta = 4(1+k^2) > 4 > 0 for all k
e 0.
Thus, there are always exactly two distinct real roots, corresponding to two distinct stationary points.
**(ii)** By Vieta's formulas for the quadratic equation kx^2 - 2x - k = 0, the product of the roots is x_1 x_2 = \frac{-k}{k} = -1.
**(d)** From the stationary equation, we have k(x^2 - 1) = 2x.
Since x_1 x_2 = -1, the roots cannot be \pm 1. Thus x
e \pm 1, so we can express k as:
k = \frac{2x}{x^2 - 1}.
Substitute this expression for k back into the function y = f_k(x):
y = \frac{x^2 + \left(\frac{2x}{x^2-1}\right)x}{x^2+1} = \frac{x^2 + \frac{2x^2}{x^2-1}}{x^2+1}
Multiply the numerator and denominator by x^2 - 1:
y = \frac{x^2(x^2-1) + 2x^2}{(x^2-1)(x^2+1)} = \frac{x^4 - x^2 + 2x^2}{(x^2-1)(x^2+1)} = \frac{x^4 + x^2}{(x^2-1)(x^2+1)}
Factor out x^2 in the numerator:
y = \frac{x^2(x^2+1)}{(x^2-1)(x^2+1)} = \frac{x^2}{x^2-1}.
Thus, the stationary points always lie on the curve C: y = \frac{x^2}{x^2-1}.
**(e) (i)** Since A(x_1, y_1) and B(x_2, y_2) lie on C, we have:
y_1 = \frac{x_1^2}{x_1^2 - 1} and y_2 = \frac{x_2^2}{x_2^2 - 1}.
Since x_1 x_2 = -1 \implies x_2 = -\frac{1}{x_1}, we substitute this into y_2:
y_2 = \frac{(-1/x_1)^2}{(-1/x_1)^2 - 1} = \frac{1/x_1^2}{1/x_1^2 - 1} = \frac{1}{1 - x_1^2} = -\frac{1}{x_1^2 - 1}.
The y-coordinate of the midpoint M is:
y_M = \frac{y_1 + y_2}{2} = \frac{1}{2} \left( \frac{x_1^2}{x_1^2-1} - \frac{1}{x_1^2-1} \right) = \frac{1}{2} \left( \frac{x_1^2 - 1}{x_1^2 - 1} \right) = \frac{1}{2}.
Thus, the y-coordinate of M is constant and its value is \frac{1}{2}.
**(ii)** The x-coordinate of M is x_M = \frac{x_1 + x_2}{2}.
By Vieta's formulas, x_1 + x_2 = \frac{2}{k}.
Thus, x_M = \frac{2/k}{2} = \frac{1}{k}.

**Part C: Distance between stationary points**
**(f)** We have y_2 - y_1 = -\frac{1}{x_1^2-1} - \frac{x_1^2}{x_1^2-1} = -\frac{x_1^2+1}{x_1^2-1}.
We also know that x_2 - x_1 = -\frac{1}{x_1} - x_1 = -\frac{x_1^2+1}{x_1}.
Multiplying this by \frac{k}{2} = \frac{x_1}{x_1^2-1}:
\frac{k}{2}(x_2 - x_1) = \left(\frac{x_1}{x_1^2-1}\right) \left(-\frac{x_1^2+1}{x_1}\right) = -\frac{x_1^2+1}{x_1^2-1}.
Thus, y_2 - y_1 = \frac{k}{2}(x_2 - x_1).
**(g)** The square of the distance is d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.
Using the result from (f):
d^2 = (x_2 - x_1)^2 + \left(\frac{k}{2}(x_2 - x_1)\right)^2 = (x_2 - x_1)^2 \left(1 + \frac{k^2}{4}\right).
Since x_1, x_2 are the roots of kx^2 - 2x - k = 0:
(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2 = \left(\frac{2}{k}\right)^2 - 4(-1) = \frac{4}{k^2} + 4 = 4\left(1 + \frac{1}{k^2}\right).
Substitute this back:
d^2 = 4\left(1 + \frac{1}{k^2}\right)\left(1 + \frac{k^2}{4}\right) = 4\left(1 + \frac{k^2}{4} + \frac{1}{k^2} + \frac{1}{4}\right) = 4 + k^2 + \frac{4}{k^2} + 1 = k^2 + 5 + \frac{4}{k^2}.
**(h)** To minimize d^2 = k^2 + 5 + \frac{4}{k^2}, we can apply the AM-GM inequality to k^2 and \frac{4}{k^2}:
k^2 + \frac{4}{k^2} \ge 2\sqrt{k^2 \cdot \frac{4}{k^2}} = 2(2) = 4.
Thus, the minimum value of d^2 is 4 + 5 = 9, which means the minimum distance is d = 3.
Equality occurs when k^2 = \frac{4}{k^2} \implies k^4 = 4 \implies k^2 = 2 \implies k = \pm\sqrt{2}.

**(a)(i)**
M1 for identifying the horizontal asymptote y = 1.
A1 for identifying the stationary point at (0,0).
A1 for a correct sketch showing symmetry, approaching y = 1 on both sides, and having a minimum at (0,0).

**(a)(ii)**
A1 for the correct range: [0, 1) or 0 \le y < 1.

**(b)**
M1 for correctly applying the quotient rule.
A1 for correct expansion of the numerator: 2x^3 + 2x + kx^2 + k - 2x^3 - 2kx^2.
A1 for simplifying to show f'_k(x) = \frac{-kx^2+2x+k}{(x^2+1)^2}.

**(c)(i)**
M1 for equating numerator to 0 and identifying the discriminant \Delta = 4 + 4k^2.
A1 for stating that since k
e 0, k^2 > 0 \implies \Delta > 4 > 0, hence there are always two distinct real roots.

**(c)(ii)**
A1 for showing x_1 x_2 = -1 using Vieta's formulas.

**(d)**
M1 for rearranging the stationary point equation to k = \frac{2x}{x^2-1}.
M1 for substituting this into the expression for f_k(x).
A1 for correct algebraic simplification of the complex fraction.
A1 for obtaining y = \frac{x^2}{x^2-1}.

**(e)(i)**
M1 for expressing y_1 and y_2 in terms of x_1 and x_2.
M1 for substituting x_2 = -1/x_1 into y_2 to get y_2 = \frac{1}{1-x_1^2}.
A1 for showing y_M = \frac{y_1 + y_2}{2} = \frac{1}{2}.

**(e)(ii)**
A1 for x_M = \frac{1}{k}.

**(f)**
M1 for showing y_2 - y_1 = -\frac{x_1^2+1}{x_1^2-1}.
M1 for showing x_2 - x_1 = -\frac{x_1^2+1}{x_1}.
M1 for multiplying x_2 - x_1 by \frac{k}{2} = \frac{x_1}{x_1^2-1}.
A1 for confirming the two expressions are equal.

**(g)**
M1 for using d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.
M1 for substituting y_2 - y_1 = \frac{k}{2}(x_2 - x_1) to get d^2 = (x_2 - x_1)^2(1 + k^2/4).
M1 for finding (x_2 - x_1)^2 = 4 + 4/k^2 using Vieta's or quadratic formula.
A0.5 for expanding and simplifying to d^2 = k^2 + 5 + 4/k^2.

**(h)**
M1 for using AM-GM or setting derivative of d^2 to 0 to find the minimum.
A1 for minimum distance d = 3 occurring at k = \pm\sqrt{2}.

**Part A: Finding the first few polynomials and recurrence relation**
**(a)** Since T_2(\cos\theta) = \cos(2\theta) = 2\cos^2\theta - 1, substituting x = \cos\theta gives T_2(x) = 2x^2 - 1.
**(b)** Since T_3(\cos\theta) = \cos(3\theta) = 4\cos^3\theta - 3\cos\theta, substituting x = \cos\theta gives T_3(x) = 4x^3 - 3x.
**(c)** Using the sum and difference formulas for cosine:
\cos((n+1)\theta) = \cos(n\theta)\cos\theta - \sin(n\theta)\sin\theta
\cos((n-1)\theta) = \cos(n\theta)\cos\theta + \sin(n\theta)\sin\theta
Adding these two equations:
\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos(n\theta)\cos\theta.
**(d)** Let x = \cos\theta, which means T_n(x) = \cos(n\theta), T_{n+1}(x) = \cos((n+1)\theta), and T_{n-1}(x) = \cos((n-1)\theta).
Substitute these into the identity from part (c):
T_{n+1}(x) + T_{n-1}(x) = 2x T_n(x)
Rearranging gives:
T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x).
**(e)** For n = 3:
T_4(x) = 2x T_3(x) - T_2(x) = 2x(4x^3 - 3x) - (2x^2 - 1) = 8x^4 - 6x^2 - 2x^2 + 1 = 8x^4 - 8x^2 + 1.
For n = 4:
T_5(x) = 2x T_4(x) - T_3(x) = 2x(8x^4 - 8x^2 + 1) - (4x^3 - 3x) = 16x^5 - 16x^3 + 2x - 4x^3 + 3x = 16x^5 - 20x^3 + 5x.

**Part B: Roots and Extrema of Chebyshev Polynomials**
**(f)** We want to solve T_n(x) = 0 for x \in [-1, 1].
Let x = \cos\theta with \theta \in [0, \pi].
T_n(\cos\theta) = 0 \implies \cos(n\theta) = 0.
The general solution is n\theta = \frac{\pi}{2} + m\pi for m \in \mathbb{Z}.
\theta = \frac{(2m+1)\pi}{2n}.
To get n distinct values of \theta in the interval [0, \pi], we choose m = 0, 1, \dots, n-1.
Equivalately, letting k = m+1, we have:
x_k = \cos\left( \frac{(2k-1)\pi}{2n} \right) for k = 1, 2, \dots, n.
These n values are all distinct and lie in [-1, 1]. Since T_n(x) is a polynomial of degree n, these are all its roots.
**(g)** For n = 5, the roots are:
x_k = \cos\left( \frac{(2k-1)\pi}{10} \right) for k = 1, 2, 3, 4, 5.
These are:
x_1 = \cos(\pi/10)
x_2 = \cos(3\pi/10)
x_3 = \cos(5\pi/10) = 0
x_4 = \cos(7\pi/10) = -\cos(3\pi/10)
x_5 = \cos(9\pi/10) = -\cos(\pi/10).
**(h)** The range of g(\theta) = \cos(n\theta) is [-1, 1], so the maximum value is 1 and the minimum value is -1.
These extreme values occur when \cos(n\theta) = \pm 1.
n\theta = m\pi \implies \theta = \frac{m\pi}{n} for m \in \mathbb{Z}.
For \theta \in [0, \pi], the values are \theta = \frac{m\pi}{n} for m = 0, 1, \dots, n.
Thus, the extrema occur at x = \cos\left( \frac{m\pi}{n} \right) for m = 0, 1, \dots, n.

**Part C: Relations to derivatives and Differential Equations**
**(i)** Let y = T_n(x) where x = \cos\theta.
Then y = \cos(n\theta).
Differentiating with respect to \theta:
\frac{dy}{d\theta} = -n\sin(n\theta).
Using the chain rule, \frac{dy}{d\theta} = \frac{dy}{dx} \frac{dx}{d\theta} = \frac{dy}{dx} (-\sin\theta).
Equating the two expressions:
\sin\theta \frac{dy}{dx} = n\sin(n\theta).
Differentiate both sides with respect to \theta again:
\frac{d}{d\theta} \left(
\sin\theta \frac{dy}{dx} \right) = \frac{d}{d\theta} (n\sin(n\theta))
\cos\theta \frac{dy}{dx} +
\sin\theta \frac{d}{d\theta}\left(\frac{dy}{dx}\right) = n^2 \cos(n\theta).
Since \frac{d}{d\theta}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2} \frac{dx}{d\theta} = -\sin\theta \frac{d^2y}{dx^2}, we substitute this back:
\cos\theta \frac{dy}{dx} -
\sin^2\theta \frac{d^2y}{dx^2} = n^2 \cos(n\theta).
Substitute x = \cos\theta, \sin^2\theta = 1-x^2, and y = \cos(n\theta):
x \frac{dy}{dx} - (1-x^2) \frac{d^2y}{dx^2} = n^2 y.
Rearranging terms:
(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + n^2 y = 0.
**(j)** For T_3(x) = 4x^3 - 3x, we have:
y = 4x^3 - 3x
\frac{dy}{dx} = 12x^2 - 3
\frac{d^2y}{dx^2} = 24x.
Substitute these into the differential equation with n = 3:
(1-x^2)(24x) - x(12x^2 - 3) + 9(4x^3 - 3x)
= 24x - 24x^3 - 12x^3 + 3x + 36x^3 - 27x
= (-24 - 12 + 36)x^3 + (24 + 3 - 27)x
= 0.
Since LHS = RHS = 0, the differential equation is verified for T_3(x).

**(a)**
A1 for T_2(x) = 2x^2 - 1.

**(b)**
A1 for T_3(x) = 4x^3 - 3x.

**(c)**
M1 for writing out both \cos((n+1)\theta) and \cos((n-1)\theta) using compound angle formulas.
A1 for correctly adding them to obtain 2\cos\theta \cos(n\theta).

**(d)**
M1 for substituting x = \cos\theta and T_k(x) = \cos(k\theta) into the identity.
A1 for obtaining the recurrence relation T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x).

**(e)**
A1 for T_4(x) = 8x^4 - 8x^2 + 1.
A1 for T_5(x) = 16x^5 - 20x^3 + 5x.

**(f)**
M1 for setting \cos(n\theta) = 0.
A1 for solving to get \theta = \frac{(2k-1)\pi}{2n} for k = 1, \dots, n.
A1 for stating that x_k = \cos\theta_k e.g. gives n distinct roots.

**(g)**
A1 for identifying the roots in terms of cosine: \cos(\pi/10), \cos(3\pi/10), 0, -\cos(3\pi/10), -\cos(\pi/10).
A2 for fully correct values (award 1 mark for any 3 correct).

**(h)**
M1 for recognizing that the extrema of cosine are \pm 1.
A1 for setting \cos(n\theta) = \pm 1 \implies n\theta = m\pi.
A1 for finding \theta = \frac{m\pi}{n} for m = 0, \dots, n.
A1 for stating the coordinates x = \cos(m\pi/n).

**(i)**
M1 for finding \frac{dy}{d\theta} = -n\sin(n\theta).
M1 for correctly applying the chain rule to relate \frac{dy}{d\theta} and \frac{dy}{dx}.
M1 for differentiating with respect to \theta again using the product rule.
A1 for correctly substituting the second derivative with respect to x.
A1 for substituting x = \cos\theta, \sin^2\theta = 1-x^2 and rearranging to the final form.

**(j)**
M1 for finding \frac{dy}{dx} = 12x^2 - 3.
M1 for finding \frac{d^2y}{dx^2} = 24x.
M1 for substituting into the differential equation.
A1.5 for full algebraic simplification showing that it reduces to 0.