2024 IB DP Mathematics - Analysis and Approaches 模擬試題連答案詳解

Let the first term be \(u_1\) and the common difference be \(d\). Using the formula for the \(n\)-th term of an arithmetic sequence \(u_n = u_1 + (n-1)d\): \(u_3 = u_1 + 2d = 15\) and \(u_7 = u_1 + 6d = 31\). Subtracting the first equation from the second gives: \(4d = 16 \implies d = 4\). Substituting \(d = 4\) back into the first equation: \(u_1 + 2(4) = 15 \implies u_1 = 7\). The sum of the first \(n\) terms is given by: \(S_n = \frac{n}{2} [2u_1 + (n-1)d]\). Substituting \(u_1 = 7\) and \(d = 4\): \(S_n = \frac{n}{2} [2(7) + (n-1)(4)] = \frac{n}{2} [14 + 4n - 4] = \frac{n}{2} [4n + 10] = n(2n + 5)\). We are given \(S_n = 1950\): \(n(2n + 5) = 1950 \implies 2n^2 + 5n - 1950 = 0\). Factorizing the quadratic equation: \((2n + 65)(n - 30) = 0\). Since \(n\) must be a positive integer, we reject the negative solution \(n = -32.5\). Thus, \(n = 30\).

M1: Attempt to write two equations in terms of \(u_1\) and \(d\). A1: Correct values found for \(d = 4\) and \(u_1 = 7\). M1: Correct substitution of their \(u_1\) and \(d\) into the arithmetic series sum formula. A1: Formulating the quadratic equation \(2n^2 + 5n - 1950 = 0\). M1: Attempt to solve their quadratic equation (by factoring or using the quadratic formula). A1: Finding \(n = 30\) (with negative root explicitly rejected or ignored as a valid term index).

(a) To find the composite function, we substitute \(g(x)\) into \(f(x)\): \((f \circ g)(x) = f(g(x)) = \ln(3(e^{2x} + 1) - 2)\). Simplifying the expression: \(3(e^{2x} + 1) - 2 = 3e^{2x} + 3 - 2 = 3e^{2x} + 1\). Therefore, \((f \circ g)(x) = \ln(3e^{2x} + 1)\). (b) To solve \((f \circ g)(x) = \ln(13)\), we set \(\ln(3e^{2x} + 1) = \ln(13)\). Taking the exponential of both sides gives \(3e^{2x} + 1 = 13\). Subtracting 1 gives \(3e^{2x} = 12\). Dividing by 3 gives \(e^{2x} = 4\). Taking the natural logarithm of both sides gives \(2x = \ln(4)\). Since \(\ln(4) = \ln(2^2) = 2\ln(2)\), we have \(2x = 2\ln(2)\), which simplifies to \(x = \ln(2)\).

M1: Attempt to substitute \(g(x)\) into \(f(x)\). A1: Correctly simplifying to obtain \((f \circ g)(x) = \ln(3e^{2x} + 1)\). M1: Setting their composite function equal to \(\ln(13)\). A1: Correctly obtaining \(3e^{2x} = 12\) or equivalent. M1: Correct use of logarithms to solve for \(2x\) or \(x\). A1: Correct final value \(x = \ln(2)\) (accept \(x = \frac{1}{2}\ln(4)\)).

We start by using the trigonometric identity \(\cos^2(x) = 1 - \sin^2(x)\) to rewrite the equation in terms of \(\sin(x)\): \(2(1 - \sin^2(x)) + 3\sin(x) = 3\). Expanding the brackets: \(2 - 2\sin^2(x) + 3\sin(x) = 3\). Rearranging all terms to one side gives: \(2\sin^2(x) - 3\sin(x) + 1 = 0\). This is a quadratic in terms of \(\sin(x)\), which can be factored as: \((2\sin(x) - 1)(\sin(x) - 1) = 0\). This gives two possible cases: \(2\sin(x) - 1 = 0 \implies \sin(x) = \frac{1}{2}\) or \(\sin(x) - 1 = 0 \implies \sin(x) = 1\). For \(0 \le x \le 2\pi\): the solutions to \(\sin(x) = \frac{1}{2}\) are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). The solution to \(\sin(x) = 1\) is \(x = \frac{\pi}{2}\). Thus, the complete set of solutions is \(x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\).

M1: Attempt to use the identity \(\cos^2(x) = 1 - \sin^2(x)\). A1: Correctly forming the quadratic equation \(2\sin^2(x) - 3\sin(x) + 1 = 0\). M1: Attempt to factorize or solve the quadratic equation. A1: Correct roots \(\sin(x) = \frac{1}{2}\) and \(\sin(x) = 1\). A1: Correct solutions for \(\sin(x) = \frac{1}{2}\) (\(x = \frac{\pi}{6}, \frac{5\pi}{6}\)). A1: Correct solution for \(\sin(x) = 1\) (\(x = \frac{\pi}{2}\)).

Since the sum of the probabilities must equal 1, we have: \(a + a + b + 0.1 = 1 \implies 2a + b = 0.9\). The expectation of \(X\) is given by \(\mathrm{E}(X) = \sum x P(X=x)\): \(\mathrm{E}(X) = 1(a) + 2(a) + 3(b) + 4(0.1) = 3a + 3b + 0.4\). Since \(\mathrm{E}(X) = 2.4\), we set up the equation: \(3a + 3b + 0.4 = 2.4 \implies 3a + 3b = 2.0 \implies a + b = \frac{2}{3}\). We now have a system of two linear equations: (1) \(2a + b = \frac{9}{10}\) and (2) \(a + b = \frac{2}{3}\). Subtracting equation (2) from equation (1) gives: \(a = \frac{9}{10} - \frac{2}{3} = \frac{27 - 20}{30} = \frac{7}{30}\). Substituting this value of \(a\) back into equation (2): \(b = \frac{2}{3} - \frac{7}{30} = \frac{20 - 7}{30} = \frac{13}{30}\). Thus, \(a = \frac{7}{30}\) and \(b = \frac{13}{30}\).

M1: Using the property that the sum of probabilities is 1 to form \(2a + b = 0.9\) (or equivalent). M1: Attempt to write an expression for \(\mathrm{E}(X)\) in terms of \(a\) and \(b\). A1: Correct equation \(3a + 3b + 0.4 = 2.4\) (or equivalent). M1: Setting up and attempting to solve the system of two equations. A1: Finding the correct value \(a = \frac{7}{30}\) (or equivalent decimal \(0.233\)). A1: Finding the correct value \(b = \frac{13}{30}\) (or equivalent decimal \(0.433\)).

First, find the y-coordinate of the point of tangency by evaluating \(f(e)\): \(f(e) = e^2 \ln(e) = e^2\). Next, find the derivative of \(f(x)\) using the product rule: \(f'(x) = \frac{d}{dx}(x^2) \ln(x) + x^2 \frac{d}{dx}(\ln(x)) = 2x \ln(x) + x^2 \left(\frac{1}{x}\right) = 2x \ln(x) + x\). Evaluate \(f'(e)\) to find the gradient of the tangent: \(f'(e) = 2e \ln(e) + e = 2e(1) + e = 3e\). The equation of the tangent line is given by \(y - f(e) = f'(e)(x - e)\). Substituting the known values: \(y - e^2 = 3e(x - e)\). Expanding and rearranging into the form \(y = mx + c\): \(y - e^2 = 3ex - 3e^2 \implies y = 3ex - 2e^2\).

A1: Correct y-coordinate \(f(e) = e^2\). M1: Attempt to apply the product rule to differentiate \(f(x)\). A1: Correct derivative \(f'(x) = 2x \ln(x) + x\). A1: Correct gradient at \(x = e\), which is \(3e\). M1: Substituting their point and gradient into the equation of a line. A1: Correct final equation in the required form \(y = 3ex - 2e^2\).

The general term in the expansion is given by \(\binom{8}{r} (2x)^{8-r} \left(-\frac{3}{x}\right)^r\). Simplifying this expression: \(\binom{8}{r} 2^{8-r} x^{8-r} (-3)^r x^{-r} = \binom{8}{r} 2^{8-r} (-3)^r x^{8-2r}\). We want the term containing \(x^4\), so we set the exponent of \(x\) equal to 4: \(8 - 2r = 4 \implies 2r = 4 \implies r = 2\). Now, substitute \(r = 2\) into the general term: \(\binom{8}{2} 2^{8-2} (-3)^2 = \binom{8}{2} 2^6 (-3)^2\). Calculate each component: \(\binom{8}{2} = \frac{8 \times 7}{2} = 28\), \(2^6 = 64\), and \((-3)^2 = 9\). The coefficient is therefore: \(28 \times 64 \times 9 = 28 \times 576 = 16128\).

M1: Attempt to write the general term using the binomial expansion formula (with correct powers). A1: Correctly simplifying the power of \(x\) to find \(8 - 2r\). M1: Setting their exponent equal to 4 and solving for \(r\) to get \(r = 2\). A1: Correct calculation of the binomial coefficient \(\binom{8}{2} = 28\). M1: Combining all parts of the coefficient (including signs). A1: Correct final answer of \(16128\).

Two vectors are perpendicular if and only if their scalar (dot) product is equal to zero: \(\mathbf{v} \cdot \mathbf{w} = 0\). Calculating the scalar product: \(\mathbf{v} \cdot \mathbf{w} = p(p-4) + 2(p) + (-4)(2) = p^2 - 4p + 2p - 8 = p^2 - 2p - 8\). Setting this expression to zero gives the quadratic equation: \(p^2 - 2p - 8 = 0\). Factorizing the quadratic equation: \((p - 4)(p + 2) = 0\). Solving for \(p\) gives: \(p = 4\) or \(p = -2\).

M1: Recognizing that perpendicular vectors have a dot product of zero. M1: Attempting to calculate the dot product in terms of \(p\). A1: Correct expression for the dot product \(p^2 - 2p - 8\). M1: Setting their expression to zero and attempting to solve the quadratic equation. A1, A1: Correct values \(p = 4\) and \(p = -2\).

We use the integration method of substitution. Let \(u = x^2\). Differentiating both sides gives \(du = 2x \\, dx\), which means \(x \\, dx = \frac{1}{2} du\). Next, we find the new limits of integration: when \(x = 0\), \(u = 0^2 = 0\); when \(x = \sqrt{\pi}\), \(u = (\sqrt{\pi})^2 = \pi\). Substituting these into the integral gives: \(\int_0^{\pi} \sin(u) \cdot \frac{1}{2} \\, du = \frac{1}{2} \int_0^{\pi} \sin(u) \\, du\). Integrating \(\sin(u)\) gives \(-\cos(u)\): \(\frac{1}{2} [-\cos(u)]_0^{\pi} = \frac{1}{2} (-\cos(\pi) - (-\cos(0)))\). Since \(\cos(\pi) = -1\) and \(\cos(0) = 1\), we evaluate the limits: \(\frac{1}{2} (-(-1) + 1) = \frac{1}{2} (1 + 1) = 1\).

M1: Attempting to use substitution with \(u = x^2\) (or equivalent). A1: Finding correct differential relation \(x \\, dx = \frac{1}{2} du\) (or equivalent). A1: Correctly changing the limits of integration to \(0\) and \(\pi\) (or reversing substitution later). M1: Integrating to get \(-\cos(u)\) (or \(-\frac{1}{2}\cos(x^2)\)). M1: Correct substitution of limits into their integrated expression. A1: Correct final value of \(1\).

(a) To find the stationary point, we differentiate \(f(x)\) using the product rule: Let \(u = x \implies u' = 1\) and \(v = \text{e}^{-kx} \implies v' = -k\text{e}^{-kx}\). Thus, \(f'(x) = \text{e}^{-kx} - kx\text{e}^{-kx} = \text{e}^{-kx}(1 - kx)\). Setting \(f'(x) = 0\), we have \(1 - kx = 0 \implies x = \frac{1}{k}\). To show this is a local maximum, we can use the first derivative test: for \(x < \frac{1}{k}\), \(1 - kx > 0\) so \(f'(x) > 0\); for \(x > \frac{1}{k}\), \(1 - kx < 0\) so \(f'(x) < 0\). Since the derivative changes sign from positive to negative, \(x = \frac{1}{k}\) is a local maximum. (b) The \(y\)-coordinate of \(P\) is found by evaluating \(f\left(\frac{1}{k}\right)\): \(f\left(\frac{1}{k}\right) = \frac{1}{k}\text{e}^{-k\left(\frac{1}{k}\right)} = \frac{1}{k\text{e}}\). We are given that the \(y\)-coordinate is \(\frac{2}{\text{e}}\), so we set \(\frac{1}{k\text{e}} = \frac{2}{\text{e}}\). Multiplying both sides by \(\text{e}\) gives \(\frac{1}{k} = 2\), which yields \(k = \frac{1}{2}\).

(a) M1 for attempting product rule differentiation of \(f(x)\). A1 for correctly finding the derivative and setting it to 0 to obtain \(x = \frac{1}{k}\). R1 for justifying that \(x = \frac{1}{k}\) is a local maximum (either using a sign diagram/argument or the second derivative). (b) M1 for substituting their \(x\)-coordinate back into \(f(x)\). M1 for setting their expression for the \(y\)-coordinate equal to \(\frac{2}{\text{e}}\). A1 for correctly solving for \(k\) to obtain \(k = \frac{1}{2}\).

(a) To find the local maximum, we find the first derivative of \( f(x) \) using the product rule: \( f'(x) = 2x e^{-x} - x^2 e^{-x} = x(2 - x)e^{-x} \). Setting \( f'(x) = 0 \) gives \( x = 0 \) or \( x = 2 \). Since \( x \ge 0 \) and the derivative changes sign from positive to negative at \( x = 2 \), there is a local maximum at \( x = 2 \). The \( y \)-coordinate is \( f(2) = 2^2 e^{-2} = 4e^{-2} \). Thus, the coordinates are \( (2, 4e^{-2}) \). (b) To find the points of inflection, we find the second derivative: \( f''(x) = (2-2x)e^{-x} - (2x-x^2)e^{-x} = (x^2 - 4x + 2)e^{-x} \). Setting \( f''(x) = 0 \) yields \( x^2 - 4x + 2 = 0 \). Solving using the quadratic formula gives \( x = \frac{4 \pm \sqrt{16 - 8}}{2} = 2 \pm \sqrt{2} \). Both points are valid since \( 2 - \sqrt{2} > 0 \). Substituting these into \( f(x) \): \( f(2-\sqrt{2}) = (2-\sqrt{2})^2 e^{-(2-\sqrt{2})} = (6-4\sqrt{2})e^{-2+\sqrt{2}} \), and \( f(2+\sqrt{2}) = (2+\sqrt{2})^2 e^{-(2+\sqrt{2})} = (6+4\sqrt{2})e^{-2-\sqrt{2}} \). (c) Using integration by parts, let \( u = x^2 \) and \( dv = e^{-x} dx \). Then \( du = 2x dx \) and \( v = -e^{-x} \). Thus, \( \int x^2 e^{-x} dx = -x^2 e^{-x} + 2 \int x e^{-x} dx \). Applying integration by parts again to \( \int x e^{-x} dx \) with \( u = x \) and \( dv = e^{-x} dx \): \( du = dx \) and \( v = -e^{-x} \). This gives \( \int x e^{-x} dx = -xe^{-x} - e^{-x} \). Substituting back: \( \int x^2 e^{-x} dx = -x^2 e^{-x} + 2(-xe^{-x} - e^{-x}) + C = -e^{-x}(x^2 + 2x + 2) + C \). (d) The area is given by \( \int_0^2 x^2 e^{-x} dx = \left[ -e^{-x}(x^2 + 2x + 2) \right]_0^2 = -e^{-2}(4 + 4 + 2) - (-e^0(2)) = -10e^{-2} + 2 = 2 - 10e^{-2} \).

(a) M1 for applying product rule. A1 for correct derivative \( x(2-x)e^{-x} \). M1 for setting to 0 and solving. A1 for coordinates \( (2, 4e^{-2}) \). (b) M1 for finding second derivative. A1 for setting quadratic to 0 and obtaining \( x = 2 \pm \sqrt{2} \). A2 for both correct \( y \)-coordinates. (c) M1 for first application of integration by parts. A1 for correct intermediate expression. M1 for second application of integration by parts. A2 for showing the final expression rigorously. (d) M1 for setup of the definite integral. M1 for evaluating at limits. A1 for correct final area \( 2 - 10e^{-2} \).

(a) Using the identity \( \sin^2(x) + \cos^2(x) = 1 \), we have \( \sin^2(x) = 1 - \cos^2(x) \). Substituting this into the function yields \( f(x) = 1 - \cos^2(x) + \cos(x) \). (b) Setting \( f(x) = 1 \) gives \( 1 - \cos^2(x) + \cos(x) = 1 \Rightarrow \cos(x)(1 - \cos(x)) = 0 \). Thus, either \( \cos(x) = 0 \) or \( \cos(x) = 1 \). In the interval \( 0 \le x \le 2\pi \), this gives \( x = 0, \frac{\pi}{2}, \frac{3\pi}{2}, 2\pi \). (c) To find the stationary points, we set the derivative to zero: \( f'(x) = 2\sin(x)\cos(x) - \sin(x) = \sin(x)(2\cos(x) - 1) = 0 \). This gives \( \sin(x) = 0 \Rightarrow x = 0, \pi, 2\pi \) or \( \cos(x) = \frac{1}{2} \Rightarrow x = \frac{\pi}{3}, \frac{5\pi}{3} \). Evaluating \( f(x) \) at these values: \( f(0) = 1 \), \( f(\pi) = -1 \), \( f(2\pi) = 1 \), \( f(\frac{\pi}{3}) = \frac{5}{4} \), \( f(\frac{5\pi}{3}) = \frac{5}{4} \). Using the second derivative \( f''(x) = 4\cos^2(x) - \cos(x) - 2 \): at \( x = \pi \), \( f''(\pi) = 3 > 0 \) (local minimum); at \( x = 0, 2\pi \), \( f''(x) = 1 > 0 \) (local minima); at \( x = \frac{\pi}{3}, \frac{5\pi}{3} \), \( f''(x) = -1.5 < 0 \) (local maxima). (d) The area enclosed is given by integrating the absolute difference \( |f(x) - 1| = |\cos(x) - \cos^2(x)| \). The curve intersects \( y = 1 \) at \( x = 0, \frac{\pi}{2}, \frac{3\pi}{2}, 2\pi \). Thus, the area is split into three integrals: \( A = \int_{0}^{\pi/2} (\cos(x) - \cos^2(x)) dx - \int_{\pi/2}^{3\pi/2} (\cos(x) - \cos^2(x)) dx + \int_{3\pi/2}^{2\pi} (\cos(x) - \cos^2(x)) dx \). Using the identity \( \cos^2(x) = \frac{1+\cos(2x)}{2} \), the antiderivative is \( F(x) = \sin(x) - \frac{x}{2} - \frac{\sin(2x)}{4} \). Evaluating at the boundaries: \( F(0) = 0 \), \( F(\frac{\pi}{2}) = 1 - \frac{\pi}{4} \), \( F(\frac{3\pi}{2}) = -1 - \frac{3\pi}{4} \), \( F(2\pi) = -\pi \). This yields \( I_1 = 1 - \frac{\pi}{4} \), \( I_2 = -(-2 - \frac{\pi}{2}) = 2 + \frac{\pi}{2} \), and \( I_3 = 1 - \frac{\pi}{4} \). Summing these gives the total area \( A = 1 - \frac{\pi}{4} + 2 + \frac{\pi}{2} + 1 - \frac{\pi}{4} = 4 \).

(a) M1 for using Pythagorean identity to rewrite the equation. (b) M1 for setting up the equation and factoring out \( \cos(x) \). A2 for all four correct solutions. (c) M1 for setting first derivative to 0. A2 for locating all five candidate points. M1 for evaluating second derivative or using a sign diagram. A1 for identifying local maxima at \( (\frac{\pi}{3}, \frac{5}{4}) \) and \( (\frac{5\pi}{3}, \frac{5}{4}) \). A1 for identifying local minimum at \( (\pi, -1) \). (d) M1 for identifying correct integration intervals. M1 for using double-angle identity to integrate \( \cos^2(x) \). A1 for correct antiderivative. M1 for correct substitution of limits. A1 for final simplified answer of \( 4 \).

(a) Since the total probability must equal 1, we set \( \int_{-\infty}^{\infty} f(x) dx = 1 \). This gives \( k \int_{0}^{2} x^2 dx + k \int_{2}^{6} (6-x) dx = 1 \). Evaluating the integrals: \( \int_{0}^{2} x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \), and \( \int_{2}^{6} (6-x) dx = \left[ 6x - \frac{x^2}{2} \right]_2^6 = (36 - 18) - (12 - 2) = 18 - 10 = 8 \). Thus, \( k \left( \frac{8}{3} + 8 \right) = 1 \Rightarrow k \left( \frac{32}{3} \right) = 1 \Rightarrow k = \frac{3}{32} \). (b) For \( 2 \le x \le 6 \), the cumulative distribution is \( F(x) = \int_{0}^{2} f(t) dt + \int_{2}^{x} f(t) dt = \frac{3}{32} \left( \frac{8}{3} \right) + \int_{2}^{x} \frac{3}{32}(6-t) dt = \frac{1}{4} + \frac{3}{32} \left[ 6t - \frac{t^2}{2} \right]_2^x = \frac{1}{4} + \frac{3}{32} \left( 6x - \frac{x^2}{2} - 10 \right) = \frac{8}{32} + \frac{18x}{32} - \frac{3x^2}{64} - \frac{30}{32} = \frac{-3x^2 + 36x - 44}{64} \). (c) The median \( m \) satisfies \( F(m) = 0.5 \). Since \( F(2) = 0.25 < 0.5 \), the median lies in the interval \( [2, 6] \). Setting \( \frac{-3m^2 + 36m - 44}{64} = \frac{1}{2} \Rightarrow -3m^2 + 36m - 44 = 32 \Rightarrow 3m^2 - 36m + 76 = 0 \). Applying the quadratic formula: \( m = \frac{36 \pm \sqrt{(-36)^2 - 4(3)(76)}}{6} = \frac{36 \pm \sqrt{1296 - 912}}{6} = \frac{36 \pm \sqrt{384}}{6} = 6 \pm \frac{8\sqrt{6}}{6} = 6 \pm \frac{4}{3}\sqrt{6} \). Since \( m \le 6 \), we choose \( m = 6 - \frac{4}{3}\sqrt{6} \). (d) The expected value is \( E(X) = \int_{0}^{6} x f(x) dx = \frac{3}{32} \int_{0}^{2} x^3 dx + \frac{3}{32} \int_{2}^{6} (6x-x^2) dx \). Integrating both parts: \( \int_{0}^{2} x^3 dx = \left[ \frac{x^4}{4} \right]_0^2 = 4 \), and \( \int_{2}^{6} (6x-x^2) dx = \left[ 3x^2 - \frac{x^3}{3} \right]_2^6 = (108 - 72) - (12 - \frac{8}{3}) = 36 - \frac{28}{3} = \frac{80}{3} \). Thus, \( E(X) = \frac{3}{32} (4) + \frac{3}{32} \left( \frac{80}{3} \right) = \frac{3}{8} + \frac{5}{2} = \frac{23}{8} \).

(a) M1 for setting up the equation where the sum of integrals equals 1. A1 for correct integration of both parts. M1 for solving for \( k \) successfully to show \( k = 3/32 \). (b) M1 for splitting the CDF integral at \( x=2 \). A1 for correct evaluation of the constant part \( F(2) = 1/4 \). M1 for integrating the second part with variable limit \( x \). A1 for algebraic simplification leading to the final expression. (c) M1 for establishing that the median lies in the interval \( [2, 6] \) and setting \( F(m) = 0.5 \). A1 for translating to a quadratic equation. A1 for choosing the correct sign in the quadratic formula. (d) M1 for setup of the expected value integral. A1 for integration of both terms. M1 for evaluating limits correctly. A1 for final correct expectation of \( 23/8 \).

Let \(X\) be the weight of an apple. We have \(X \sim N(150, 12^2)\).

(a) We want to find \(P(X > 165)\).
Using a GDC normal cumulative distribution function:
\(P(X > 165) \approx 0.105649\)
So, the probability is \(0.106\) (to 3 significant figures).

(b) Let \(Y\) be the number of apples in the sample of 10 that weigh more than \(165\text{ g}\).
\(Y \sim B(10, p)\) where \(p \approx 0.105649\).
We want to find \(P(Y \ge 3) = 1 - P(Y \le 2)\).
Using a GDC binomial cumulative distribution function:
\(P(Y \le 2) \approx 0.916578\)
Therefore, \(P(Y \ge 3) = 1 - 0.916578 \approx 0.083422\)
So, the probability is \(0.0834\) (to 3 significant figures).

(a)
[M1] for setting up the normal distribution model \(X \sim N(150, 12^2)\).
[A1] for \(0.106\) (accept \(0.105649...\)).

(b)
[M1] for identifying the binomial distribution model \(Y \sim B(10, 0.105649)\).
[M1] for recognizing the need to calculate \(P(Y \ge 3)\) or \(1 - P(Y \le 2)\).
[A1] for \(P(Y \le 2) \approx 0.917\).
[A1] for \(0.0834\) (accept \(0.083422...\)).

Let \(C\) be the base of the tower on the ground, so that \(T\), \(B\), and \(A\) lie in the same vertical plane.
In triangle \(ABT\):
\(\angle TAB = 25^\circ\)
\(\angle TBC = 40^\circ \implies \angle TBA = 180^\circ - 40^\circ = 140^\circ\)
\(\angle ATB = 180^\circ - (25^\circ + 140^\circ) = 15^\circ\)

(a) Applying the sine rule in triangle \(ABT\):
\(\frac{BT}{\sin(25^\circ)} = \frac{AB}{\sin(15^\circ)}\)
Given \(AB = 50\text{ m}\):
\(BT = \frac{50 \sin(25^\circ)}{\sin(15^\circ)}\)
Using GDC:
\(BT \approx \frac{50 \times 0.422618}{0.258819} \approx 81.638\text{ m}\)
So, the distance is \(81.6\text{ m}\) (to 3 significant figures).

(b) In the right-angled triangle \(BCT\):
\(\sin(40^\circ) = \frac{\text{height}}{BT}\)
\(\text{height} = BT \sin(40^\circ) = 81.638 \times \sin(40^\circ) \approx 52.476\text{ m}\)
So, the height of the tower is \(52.5\text{ m}\) (to 3 significant figures).

(a)
[M1] for finding \(\angle TBA = 140^\circ\) or \(\angle ATB = 15^\circ\).
[M1] for a correct application of the sine rule: \(\frac{BT}{\sin(25^\circ)} = \frac{50}{\sin(15^\circ)}\).
[A1] for \(81.6\text{ m}\) (accept \(81.638...\)).

(b)
[M1] for attempting to use right-angled triangle trigonometry, e.g., \(\sin(40^\circ) = \frac{h}{BT}\).
[A1] for substituting their value of \(BT\).
[A1] for \(52.5\text{ m}\) (accept \(52.476...\)).

(a) The intersection points occur when \(f(x) = g(x) \implies \ln(x) = x^2 - 4x + 2\).
Using a GDC to find the roots of \(\ln(x) - x^2 + 4x - 2 = 0\):
\(x_1 \approx 0.712\)
\(x_2 \approx 3.83\)

(b) The area \(A\) of the enclosed region is given by:
\(A = \int_{x_1}^{x_2} (f(x) - g(x)) \, \mathrm{d}x = \int_{0.712}^{3.83} (\ln(x) - (x^2 - 4x + 2)) \, \mathrm{d}x\)
Using a GDC to evaluate this definite integral:
\(A \approx 5.746\)
So, the area is \(5.75\) (to 3 significant figures).

(a)
[M1] for setting up the equation \(\ln(x) = x^2 - 4x + 2\).
[A1] for \(x \approx 0.712\) (accept \(0.7116...\)).
[A1] for \(x \approx 3.83\) (accept \(3.828...\)).

(b)
[M1] for setting up the integral \(\int_{0.712}^{3.83} (\ln(x) - (x^2 - 4x + 2)) \, \mathrm{d}x\).
[A1] for correct integrand and limits.
[A1] for \(5.75\) (accept \(5.746...\)).

Let \(P = 15000\) and \(r = 1.045\).
The deposits are made at the beginning of each year.
At the end of year 10, the total value \(S_{10}\) is:
\(S_{10} = 15000(1.045)^{10} + 15000(1.045)^9 + \dots + 15000(1.045)\)
This is a geometric series with first term \(a = 15000 \times 1.045 = 15675\), common ratio \(R = 1.045\), and \(n = 10\) terms.

(a) \(S_{10} = 15675 \frac{1.045^{10} - 1}{1.045 - 1}\)
Using a GDC:
\(S_{10} \approx 192617.55\)
So, the total value is \(\$192,618\) (or \(\$193,000\) to 3 significant figures).

(b) We want to find the smallest integer \(n\) such that:
\(15675 \frac{1.045^n - 1}{0.045} > 300000\)
Using a GDC to solve or by trial and error:
For \(n = 14\): \(S_{14} \approx 296068.51\)
For \(n = 15\): \(S_{15} \approx 325066.59\)
Thus, the minimum number of years required is 15.

(a)
[M1] for identifying the geometric series parameters, e.g., \(a = 15675\) and \(R = 1.045\) (or equivalent TVM solver setup).
[A1] for a correct expression or setup: \(15675 \frac{1.045^{10} - 1}{0.045}\).
[A1] for \(\$192,618\) (accept \(\$193,000\)).

(b)
[M1] for setting up the inequality \(15675 \frac{1.045^n - 1}{0.045} > 300000\).
[A1] for finding \(n \approx 14.1\).
[A1] for rounding up to the next integer, \(15\) (years).

(a) As \(t \to \infty\), \(\mathrm{e}^{-kt} \to 0\), so \(T(t) \to 22\). The room temperature is \(22^\circ\text{C}\).

(b) At \(t = 0\), \(T(0) = 22 + a \mathrm{e}^{0} = 22 + a\).
Given \(T(0) = 85\):
\(22 + a = 85 \implies a = 63\).

(c) (i) At \(t = 5\), \(T(5) = 22 + 63 \mathrm{e}^{-5k} = 58\).
\(63 \mathrm{e}^{-5k} = 36 \implies \mathrm{e}^{-5k} = \frac{36}{63} = \frac{4}{7}\)
\(-5k = \ln\left(\frac{4}{7}\right) \implies k = -\frac{1}{5}\ln\left(\frac{4}{7}\right) \approx 0.111923\)
So, \(k \approx 0.112\) (to 3 significant figures).

(ii) The rate of change of temperature is given by \(T'(t) = -a k \mathrm{e}^{-kt} = -63 \times 0.111923 \mathrm{e}^{-0.111923 t}\).
At \(t = 10\):
\(T'(10) = -63 \times 0.111923 \mathrm{e}^{-1.11923} \approx -2.3024^\circ\text{C min}^{-1}\).
The cooling rate is \(2.30^\circ\text{C min}^{-1}\) (or the rate of change is \(-2.30^\circ\text{C min}^{-1}\)).

(a)
[A1] for \(22\) (or \(22^\circ\text{C}\)).

(b)
[A1] for \(a = 63\).

(c)
[M1] for setting up the equation \(22 + 63 \mathrm{e}^{-5k} = 58\) (or using their \(a\)).
[A1] for \(k \approx 0.112\) (accept \(0.111923...\)).
[M1] for attempting to find the derivative \(T'(t)\) and evaluating it at \(t = 10\).
[A1] for \(-2.30\) (or \(2.30\)) (accept \(-2.3024...\)).

(a) The number of observations is \(n = 8\).
\(\bar{x} = \t\t\frac{\sum x}{n} = \frac{48}{8} = 6\)
\(\bar{y} = \t\t\frac{\sum y}{n} = \frac{520}{8} = 65\)

(b) The slope \(m\) of the regression line of \(y\) on \(x\) is:
\(m = \frac{n \sum xy - \sum x \sum y}{n \sum x^2 - (\sum x)^2}\)
\(m = \frac{8(3240) - (48)(520)}{8(312) - 48^2} = \frac{25920 - 24960}{2496 - 2304} = \frac{960}{192} = 5\)
The \(y\)-intercept \(c\) is:
\(c = \bar{y} - m\bar{x} = 65 - 5(6) = 35\)
So, the regression line equation is \(y = 5x + 35\).

(c) For \(x = 7.5\):
\(y = 5(7.5) + 35 = 37.5 + 35 = 72.5\)
The estimated test score is \(72.5\text{ points}\).

(a)
[A1] for \(\bar{x} = 6\).
[A1] for \(\bar{y} = 65\).

(b)
[M1] for a correct method to find the slope \(m\) or intercept \(c\).
[A1] for \(y = 5x + 35\).

(c)
[M1] for substituting \(x = 7.5\) into their regression equation.
[A1] for \(72.5\).

(a) Acceleration \(a(t) = v'(t)\).
Using GDC numerical derivative at \(t = 2.5\):
\(a(2.5) \approx -25.238\text{ m s}^{-2}\)
So, the acceleration is \(-25.2\text{ m s}^{-2}\) (to 3 significant figures).

(b) The particle is at rest when \(v(t) = 0 \implies 3t^2 \cos(t) - 2t = 0\).
Since \(t \ne 0\), we have \(3t \cos(t) - 2 = 0\).
Using a GDC to find the first positive root of this equation in the interval \([0, 5]\):
\(t \approx 4.85\text{ seconds}\) (to 3 significant figures).

(c) The total distance travelled in the first 4 seconds is:
\(\text{Distance} = \int_{0}^{4} |v(t)| \, \mathrm{d}t = \int_{0}^{4} |3t^2 \cos(t) - 2t| \, \mathrm{d}t\)
Using a GDC to evaluate this definite integral:
\(\text{Distance} \approx 63.473\text{ m}\)
So, the total distance travelled is \(63.5\text{ m}\) (to 3 significant figures).

(a)
[M1] for recognizing that acceleration is the derivative of velocity.
[A1] for \(-25.2\text{ m s}^{-2}\) (accept \(-25.238...\)).

(b)
[M1] for setting up the equation \(3t^2 \cos(t) - 2t = 0\).
[A1] for \(t \approx 4.85\) (accept \(4.846...\)).

(c)
[M1] for setting up the integral \(\int_{0}^{4} |v(t)| \, \mathrm{d}t\).
[A1] for \(63.5\text{ m}\) (accept \(63.473...\)).

(a) Substituting \(k = 2\):
\(2 \sin^2(x) + 3 \cos(x) = 2\)
Using the identity \(\sin^2(x) = 1 - \cos^2(x)\):
\(2(1 - \cos^2(x)) + 3 \cos(x) = 2\)
\(2 - 2 \cos^2(x) + 3 \cos(x) = 2\)
\(-2 \cos^2(x) + 3 \cos(x) = 0\)
\(\cos(x)(3 - 2 \cos(x)) = 0\)
Since \(3 - 2 \cos(x) = 0 \implies \cos(x) = 1.5\) has no real solutions, we must have:
\(\cos(x) = 0 \implies x = \frac{\pi}{2}\text{ or } x = \frac{3\pi}{2}\).

(b) Let \(u = \cos(x)\). For \(x \in [0, 2\pi]\), \(u \in [-1, 1]\).
The equation becomes:
\(2(1 - u^2) + 3u = k \implies -2u^2 + 3u + 2 = k\).
Let \(g(u) = -2u^2 + 3u + 2\).
For the original equation in \(x\) to have exactly four distinct real solutions, the quadratic equation \(g(u) = k\) must have two distinct real roots \(u_1, u_2\) both lying strictly inside the interval \((-1, 1)\) (since each \(u \in (-1, 1)\) corresponds to exactly two distinct values of \(x\)).
The vertex of the parabola \(y = g(u)\) is at \(u = \frac{-3}{2(-2)} = 0.75\).
The maximum value at this vertex is \(g(0.75) = -2(0.75)^2 + 3(0.75) + 2 = 3.125\).
At the boundary \(u = 1\):
\(g(1) = -2(1)^2 + 3(1) + 2 = 3\).
At the boundary \(u = -1\):
\(g(-1) = -2(-1)^2 + 3(-1) + 2 = -3\).
For \(k\) to yield two distinct roots in \((-1, 1)\), the line \(y = k\) must intersect the curve \(y = g(u)\) twice in the interval \-1 < u < 1.
Looking at the graph of \(g(u)\), this occurs when \(3 < k < 3.125\).
Thus, the range of values is \(3 < k < 3.125\) (or \(3 < k < \frac{25}{8}\)).

(a)
[M1] for substituting \(\sin^2(x) = 1 - \cos^2(x)\) to obtain a quadratic in terms of \(\cos(x)\).
[A1] for \(\cos(x) = 0\).
[A1] for \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\) (accept \(1.57\) and \(4.71\)).

(b)
[M1] for expressing the equation as a quadratic \(g(u) = -2u^2 + 3u + 2 = k\) and identifying that we need two distinct roots \(u \in (-1, 1)\).
[M1] for finding the vertex maximum value of \(3.125\) and/or the boundary value \(g(1) = 3\).
[A1] for the correct range \(3 < k < 3.125\) (or equivalent).

(a) Let \(W \sim N(150, \sigma^2)\). We are given that \(P(W > 180) = 0.15\), which is equivalent to \(P(W \le 180) = 0.85\). Standardizing this gives: \(P(Z \le \frac{180 - 150}{\sigma}) = 0.85\). Using a graphic display calculator, we find the z-score corresponding to a cumulative probability of \(0.85\) is \(z \approx 1.03643...\). Thus, \(\frac{30}{\sigma} = 1.03643...\), which gives \(\sigma \approx 28.9\text{ g}\) (or more accurately \(28.9454...\)). (b) Let \(Y\) be the number of apples in the sample of \(4\) that weigh more than \(180\text{ g}\). Since each apple is chosen at random, \(Y\) follows a binomial distribution with \(Y \sim B(4, 0.15)\). We want to find \(P(Y \le 2)\). Using a graphic display calculator for the binomial cumulative distribution, we get \(P(Y \le 2) \approx 0.988\) (or more accurately \(0.988018...\)).

(a) M1 for standardizing or setting up a probability equation, e.g. \(P(Z < \frac{30}{\sigma}) = 0.85\) or \(\frac{180 - 150}{\sigma} = \text{invNorm}(0.85)\). A1 for the correct z-score of \(1.0364...\) (accept \(1.04\)). A1 for \(\sigma \approx 28.9\) (accept \(28.8\) if \(z = 1.04\) is used, or more accurate \(28.945...\)). (b) M1 for recognizing a binomial distribution model, \(Y \sim B(4, 0.15)\). M1 for calculating the cumulative probability \(P(Y \le 2)\) (either summing the terms or using binomial CDF on GDC). A1 for \(0.988\) (accept \(0.988018...\)).

(a) Let \(X \sim N(150, 18^2)\). We want to find \(P(130 \le X \le 170)\).
Using a graphic display calculator (GDC):
\(P(130 \le X \le 170) \approx 0.7385...\)
Rounding to 3 significant figures gives \(0.739\).

(b) (i) Using GDC for the interval \([125, 165]\):
\(P(\text{Medium}) = P(125 \le X \le 165) \approx 0.7152... \approx 0.715\).
(ii) Using GDC for the interval \([165, \infty)\):
\(P(\text{Large}) = P(X > 165) \approx 0.2023... \approx 0.202\).

(c) Let \(Y\) be the number of Large apples in a sample of 8. Then \(Y \sim B(8, p)\) where \(p = 0.20233...\).
We want to find \(P(Y \ge 3) = 1 - P(Y \le 2)\).
Using the binomial cumulative distribution function on a GDC:
\(P(Y \le 2) \approx 0.79374...\)
Therefore, \(P(Y \ge 3) = 1 - 0.79374... \approx 0.20626... \approx 0.206\).

(d) Let \(W\) be the selling price of an apple. First, find the probability of a Small apple:
\(P(\text{Small}) = P(X < 125) \approx 0.08245...\) (or \(1 - 0.71522... - 0.20233... = 0.08245...\)).

The expected selling price is:
\(E(W) = 0.20 \times P(\text{Small}) + 0.50 \times P(\text{Medium}) + 0.90 \times P(\text{Large})\)
\(E(W) = 0.20(0.08245...) + 0.50(0.71522...) + 0.90(0.20233...)\)
\(E(W) \approx 0.01649 + 0.35761 + 0.18210 = 0.55620... \approx \$0.556\).

(e) We want the conditional probability \(P(\text{Large} \mid \text{Not Small}) = \frac{P(\text{Large} \cap \text{Not Small})}{P(\text{Not Small})}\).
Since Large apples are already "Not Small", \(P(\text{Large} \cap \text{Not Small}) = P(\text{Large}) = 0.20233...\).
\(P(\text{Not Small}) = 1 - P(\text{Small}) = 1 - 0.08245... = 0.91755...\)
Thus, \(P(\text{Large} \mid \text{Not Small}) = \frac{0.20233...}{0.91755...} \approx 0.22051... \approx 0.221\).

(a)
M1: For attempting to use the normal cumulative distribution function on a GDC.
A1: Correct parameters shown or implied (e.g., lower = 130, upper = 170, \(\mu = 150\), \(\sigma = 18\)).
A1: \(0.739\) (accept 0.738).

(b)
(i) A1: \(0.715\) (or 0.7152).
(ii) A1: \(0.202\) (or 0.2023).

(c)
M1: Recognizing binomial distribution \(Y \sim B(8, p)\) with \(p = 0.202\).
A1: Correct setup for finding the probability, such as \(1 - P(Y \le 2)\).
A1: \(0.206\).

(d)
A1: Correct probability for Small apples \(P(\text{Small}) \approx 0.0824\).
M1: Correct expected value expression substituting their probabilities: \(0.20 P(S) + 0.50 P(M) + 0.90 P(L)\).
A1: \(\$0.556\) (accept \(0.56\)).

(e)
M1: Recognizing conditional probability structure \(P(L \mid S')\).
A1: Correct formula or ratio: \(\frac{P(L)}{1 - P(S)}\) or \(\frac{0.20233...}{0.91755...}\).
M1: Correct denominator value of \(0.918\) (or \(1 - 0.0824\)).
A1: Correct numerator value of \(0.202\).
A1: \(0.221\) (accept 0.220).

(a) The volume of revolution \(V\) about the \(x\)-axis is given by:
\(V = \pi \int_{0}^{5} y^2 \, dx = \pi \int_{0}^{5} \left(\sqrt{x} e^{0.1x}\right)^2 \, dx = \pi \int_{0}^{5} x e^{0.2x} \, dx\)
Using integration by parts:
Let \(u = x \implies du = dx\)
Let \(dv = e^{0.2x} \, dx \implies v = 5 e^{0.2x}\)
\(\int x e^{0.2x} \, dx = 5x e^{0.2x} - \int 5e^{0.2x} \, dx = 5x e^{0.2x} - 25e^{0.2x}\)
Evaluating this from 0 to 5:
\(\left[5(5) e^{1} - 25 e^{1}\right] - \left[0 - 25 e^{0}\right] = [25e - 25e] - [-25] = 25\)
Thus, \(V = 25\pi \approx 78.5398... \approx 78.5\text{ dm}^3\).

(b) Let \(h\) be the depth of the water at any time \(t\). The volume of water at depth \(h\) is:
\(V(h) = \pi \int_0^h x e^{0.2x} \, dx\)
We are given that \(\frac{dV}{dt} = 2.5\). We want to find \(\frac{dh}{dt}\) when \(h = 3\).
By the Fundamental Theorem of Calculus:
\(\frac{dV}{dh} = \pi h e^{0.2h}\)
At \(h = 3\):
\(\frac{dV}{dh} = 3\pi e^{0.6}\)
By the chain rule:
\(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \implies 2.5 = 3\pi e^{0.6} \cdot \frac{dh}{dt}\)
\(\frac{dh}{dt} = \frac{2.5}{3\pi e^{0.6}} \approx 0.14557... \approx 0.146\text{ dm min}^{-1}\).

(c) The area \(A\) of the circular surface of the water when the depth is \(h\) is:
\(A(h) = \pi [y(h)]^2 = \pi h e^{0.2h}\)
We want to find \(\frac{dA}{dt}\) when \(h = 3\).
Differentiating \(A\) with respect to \(h\) using the product rule:
\(\frac{dA}{dh} = \pi \left(1 \cdot e^{0.2h} + h \cdot 0.2 e^{0.2h}\right) = \pi e^{0.2h} (1 + 0.2h)\)
At \(h = 3\):
\(\frac{dA}{dh} = \pi e^{0.6} (1 + 0.6) = 1.6 \pi e^{0.6}\)
By the chain rule:
\(\frac{dA}{dt} = \frac{dA}{dh} \cdot \frac{dh}{dt} = \left(1.6 \pi e^{0.6}\right) \cdot \left(\frac{2.5}{3\pi e^{0.6}}\right)\)
The terms \(\pi e^{0.6}\) cancel out:
\(\frac{dA}{dt} = \frac{1.6 \times 2.5}{3} = \frac{4}{3} \approx 1.33\text{ dm}^2\text{ min}^{-1}\).

(d) Since the water is poured in at a constant rate:
\(t = \frac{\text{Total Volume}}{\text{Rate}}\)
\(t = \frac{25\pi}{2.5} = 10\pi \approx 31.4159... \approx 31.4\text{ minutes}\).

(a)
M1: For attempting to write down the volume of revolution integral \(\pi \int y^2 \, dx\).
A1: Correct integrand and limits: \(\pi \int_{0}^{5} x e^{0.2x} \, dx\).
A1: Analytical integration or GDC evaluation of the integral to 25.
A1: \(25\pi\) (or \(78.5\)).

(b)
M1: Recognizing the relationship \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\).
M1: Finding \(\frac{dV}{dh} = \pi h e^{0.2h}\) using the Fundamental Theorem of Calculus.
A1: Correct value of \(\frac{dV}{dh}\) at \(h = 3\) (either \(3\pi e^{0.6}\) or \(17.17\)).
M1: Substituting \(\frac{dV}{dt} = 2.5\) into their relation.
A1: Correct expression for \(\frac{dh}{dt} = \frac{2.5}{3\pi e^{0.6}}\).
A1: \(0.146\).

(c)
M1: Formulating the area equation \(A = \pi y^2 = \pi h e^{0.2h}\).
M1: Attempting to differentiate \(A\) with respect to \(h\) (using product rule).
A1: \(\frac{dA}{dh} = \pi e^{0.2h}(1 + 0.2h)\).
M1: Correct substitution into the chain rule \(\frac{dA}{dt} = \frac{dA}{dh} \cdot \frac{dh}{dt}\).
A1: \(\frac{4}{3}\) (or \(1.33\)).

(d)
M1: Formulating time as \(\frac{\text{Volume}}{\text{Rate}}\).
A1: \(\frac{25\pi}{2.5}\) seen.
A1: \(31.4\) (or \(10\pi\)).

(a) (i) Since the maximum value of \(\cos(0.5t)\) is 1, the maximum height of wave \(A\) is \(1 + 2 = 3\text{ m}\).
(ii) Period of \(B = \frac{2\pi}{0.25} = 8\pi \approx 25.1\text{ s}\).

(b) To find the times when the waves are at the same height, set \(y_A(t) = y_B(t)\):
\(\cos(0.5t) + 2 = 1.5\sin(0.25t) + 1.5\)
Using the double angle identity \(\cos(2\theta) = 1 - 2\sin^2(\theta)\) with \(\theta = 0.25t\):
\(\cos(0.5t) = 1 - 2\sin^2(0.25t)\)
Substitute this identity into the wave equation:
\(1 - 2\sin^2(0.25t) + 2 = 1.5\sin(0.25t) + 1.5\)
\(3 - 2\sin^2(0.25t) = 1.5\sin(0.25t) + 1.5\)
Rearranging all terms to one side:
\(2\sin^2(0.25t) + 1.5\sin(0.25t) - 1.5 = 0\)
Multiplying the entire equation by 2:
\(4\sin^2(0.25t) + 3\sin(0.25t) - 3 = 0\) (as required).

(c) Let \(u = \sin(0.25t)\). The equation is \(4u^2 + 3u - 3 = 0\).
Using the quadratic formula:
\(u = \frac{-3 \pm \sqrt{3^2 - 4(4)(-3)}}{2(4)} = \frac{-3 \pm \sqrt{57}}{8}\)
Evaluating the positive root (since the negative root gives \(u \approx -1.32\), which is outside the range of sine):
\(\sin(0.25t) \approx 0.5687\)
Now solve for \(0.25t\) within the interval \(0 \le 0.25t \le 7.5\) (since \(0 \le t \le 30\)):
\(0.25t = \arcsin(0.5687) \approx 0.6049\) radians
\(0.25t = \pi - 0.6049 \approx 2.5367\) radians
\(0.25t = 2\pi + 0.6049 \approx 6.8881\) radians

Multiplying by 4 to get \(t\):
\(t_1 \approx 2.419... \approx 2.42\text{ s}\)
\(t_2 \approx 10.146... \approx 10.1\text{ s}\)
\(t_3 \approx 27.552... \approx 27.6\text{ s}\)

(d) Wave \(B\) is strictly higher than wave \(A\) when \(y_B(t) > y_A(t)\).
Looking at the graph or analyzing test points:
- On \([0, 2.42)\), \(y_A(t) > y_B(t)\)
- On \((2.42, 10.15)\), \(y_B(t) > y_A(t)\)
- On \((10.15, 27.55)\), \(y_A(t) > y_B(t)\)
- On \((27.55, 30]\), \(y_B(t) > y_A(t)\)

Total time = \((10.1468 - 2.4195) + (30 - 27.5524)\)
Total time = \(7.7273 + 2.4476 = 10.1749... \approx 10.2\text{ seconds}\).

(e) The rate of change of the height of wave \(A\) is given by the derivative:
\(y'_A(t) = -0.5\sin(0.5t)\)
At \(t = 5\):
\(y'_A(5) = -0.5\sin(2.5) \approx -0.2992... \approx -0.299\text{ m s}^{-1}\).

(a)
(i) A1: 3.
(ii) A1: Correct method (e.g., \(\frac{2\pi}{0.25}\)).
A1: \(8\pi\) (or \(25.1\)).

(b)
M1: Setting the two equations equal to each other.
M1: Attempting to use the double-angle formula for \(\cos(0.5t)\).
A1: Correct substitution of \(1 - 2\sin^2(0.25t)\).
M1: Collecting terms to form a quadratic equal to 0.
AG: Showing algebraic steps clearly to reach \(4\sin^2(0.25t) + 3\sin(0.25t) - 3 = 0\).

(c)
M1: Attempt to solve the quadratic equation (using formula or GDC solver).
A1: Finding the valid value for sine: \(\sin(0.25t) \approx 0.569\).
M1: Finding multiple angles (at least two) for \(0.25t\).
A1: Correct three values for \(t\): \(2.42\), \(10.1\), and \(27.6\).

(d)
M1: Identifying the correct intervals of \(t\) where \(y_B > y_A\).
A1: Setting up the interval lengths: \(10.15 - 2.42\) and \(30 - 27.55\).
A1: \(10.2\) (accept \(10.18\)).

(e)
M1: Attempting to differentiate \(y_A(t)\).
A1: \(y'_A(t) = -0.5\sin(0.5t)\).
A1: \(-0.299\).

(a)
(i) \(T_0(x) = \cos(0) = 1\) and \(T_1(x) = \cos(\arccos x) = x\).
(ii) Let \(\theta = \arccos x\), so \(x = \cos\theta\). Then \(T_2(x) = \cos(2\theta) = 2\cos^2\theta - 1 = 2x^2 - 1\).
(iii) Using \(T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)\) with \(n=2\), we get \(T_3(x) = 2x T_2(x) - T_1(x) = 2x(2x^2 - 1) - x = 4x^3 - 3x\).

(b)
(i) Adding \(\cos(A+B) = \cos A \cos B - \sin A \sin B\) and \(\cos(A-B) = \cos A \cos B + \sin A \sin B\) yields \(\cos(A+B) + \cos(A-B) = 2\cos A\cos B\).
(ii) Let \(A = n\theta\) and \(B = \theta\), where \(\theta = \arccos x\). Substituting into the identity from (b)(i) gives \(\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos(n\theta)\cos\theta\). Since \(x = \cos\theta\) and \(T_k(x) = \cos(k\theta)\), we obtain \(T_{n+1}(x) + T_{n-1}(x) = 2x T_n(x)\), which rearranges to \(T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)\).

(c)
Let \(P(n)\) be the proposition that \(T_n(x)\) is a polynomial of degree \(n\) with leading coefficient \(2^{n-1}\) for \(n \ge 1\).
Base cases:
For \(n=1\), \(T_1(x) = x\), which has degree 1 and leading coefficient \(1 = 2^0\). Thus \(P(1)\) is true.
For \(n=2\), \(T_2(x) = 2x^2-1\), which has degree 2 and leading coefficient \(2 = 2^1\). Thus \(P(2)\) is true.
Inductive step:
Assume \(P(k-1)\) and \(P(k)\) are true for some integer \(k \ge 2\). Then \(T_k(x) = 2^{k-1}x^k + \dots\) and \(T_{k-1}(x)\) has degree \(k-1\).
By the recurrence relation: \(T_{k+1}(x) = 2x T_k(x) - T_{k-1}(x) = 2x(2^{k-1}x^k + \dots) - T_{k-1}(x) = 2^k x^{k+1} + \dots - T_{k-1}(x)\).
Since \(T_{k-1}(x)\) has degree \(k-1\), the term of highest degree in \(T_{k+1}(x)\) is \(2^k x^{k+1}\). Thus \(T_{k+1}(x)\) is a polynomial of degree \(k+1\) with leading coefficient \(2^k\).
Hence, \(P(k) \land P(k-1) \implies P(k+1)\). Since both base cases are true, \(P(n)\) is true for all \(n \ge 1\) by mathematical induction.

(d)
(i) Setting \(T_n(x) = 0 \implies \cos(n \arccos x) = 0\). Let \(\theta = \arccos x\), where \(\theta \in [0, \pi]\). Then \(\cos(n\theta) = 0 \implies n\theta = \frac{\pi}{2} + m\pi\) for \(m \in \mathbb{Z}\). This gives \(\theta = \frac{(2m+1)\pi}{2n}\). Since \(0 \le \theta \le \pi\), we require \(0 \le m \le n-1\). Letting \(k = m+1\), we have \(k \in \{1, 2, \dots, n\}\) and \(\theta_k = \frac{(2k-1)\pi}{2n}\). Thus \(x_k = \cos\left(\frac{(2k-1)\pi}{2n}\right)\).
(ii) There are exactly \(n\) distinct roots because the \(n\) values of \(\theta_k\) are distinct and lie in the interval \((0, \pi)\). Since the cosine function is strictly decreasing (and thus injective) on \([0, \pi]\), each distinct \(\theta_k\) maps to a unique root \(x_k\).

(e)
(i) The local extrema occur when \(\cos(n\theta) = \pm 1\) for \(\theta \in (0, \pi)\), which gives \(n\theta = p\pi \implies \theta = \frac{p\pi}{n}\) for \(p \in \{1, 2, \dots, n-1\}\). The coordinates of these extrema are \(\left(\cos\left(\frac{p\pi}{n}\right), (-1)^p\right)\).
(ii) For \(y = T_4(x)\), the roots are \(x_k = \cos\left(\frac{(2k-1)\pi}{8}\right)\) (approx. \(\pm 0.924\), \(\pm 0.383\)). The local extrema in \((-1, 1)\) are \(\left(\pm\frac{\sqrt{2}}{2}, -1\right)\) and \((0, 1)\). The boundaries are \((-1, 1)\) and \((1, 1)\). The graph is a symmetric 'W' shape.

(f)
Let \(x = \cos\theta \implies dx = -\sin\theta d\theta\). The interval \([-1, 1]\) maps to \([\pi, 0]\).
\(\int_{-1}^{1} \frac{T_n(x)T_m(x)}{\sqrt{1-x^2}} dx = \int_{\pi}^{0} \frac{\cos(n\theta)\cos(m\theta)}{\sin\theta} (-\sin\theta d\theta) = \int_{0}^{\pi} \cos(n\theta)\cos(m\theta) d\theta\).
Using \(\cos(n\theta)\cos(m\theta) = \frac{1}{2}(\cos((n+m)\theta) + \cos((n-m)\theta))\), since \(n \neq m\), the integral becomes \(\frac{1}{2} \left[ \frac{\sin((n+m)\theta)}{n+m} + \frac{\sin((n-m)\theta)}{n-m} \right]_0^\pi = 0\).

(a)
(i) \(T_0(x) = 1\) [A1], \(T_1(x) = x\) [A1]
(ii) Let \(\theta = \arccos x\) [M1], \(T_2(x) = \cos(2\theta) = 2\cos^2\theta - 1 = 2x^2-1\) [AG] [A1]
(iii) Using recurrence: \(T_3(x) = 2x T_2(x) - T_1(x) = 2x(2x^2-1) - x\) [M1] \(= 4x^3 - 3x\) [A1]
(b)
(i) Expanding both cosine terms [M1], and adding them yields \(2\cos A\cos B\) [A1]
(ii) Substituting \(A = n\theta\) and \(B = \theta\) [M1], rewriting in terms of \(T_n(x)\) [M1], showing the algebraic step to get the recurrence [AG] [A1]
(c)
Showing base cases: \(P(1)\) is true [A1], \(P(2)\) is true [A1].
Stating the induction hypothesis for \(k-1\) and \(k\) [M1].
Writing \(T_{k+1}(x) = 2x T_k(x) - T_{k-1}(x)\) [M1] and analyzing leading term: \(2x(2^{k-1}x^k) = 2^k x^{k+1}\) [A1].
Explaining that because \(T_{k-1}(x)\) has lower degree, the leading coefficient is unchanged [R1] and concluding the proof [R1].
(d)
(i) Setting \(\cos(n\theta) = 0\) [M1], finding general solutions for \(\theta\) in \([0, \pi]\) [M1], showing \(x_k = \cos\left(\frac{(2k-1)\pi}{2n}\right)\) [A1].
(ii) Reasoning that there are exactly \(n\) distinct roots because \(\theta_k\) are distinct in \((0, \pi)\) and \(\cos\) is injective on this interval [R1].
(e)
(i) Setting \(\cos(n\theta) = \pm 1\) [M1], finding coordinates: \(\left(\cos\left(\frac{p\pi}{n}\right), (-1)^p\right)\) [A1].
(ii) Correct 'W' shape symmetric about y-axis [A1], correct coordinates of intercepts and extrema shown on sketch [A1].
(f)
Performing substitution \(x = \cos\theta\) and correctly transforming differential and limits [M1].
Simplifying integrand to \(\cos(n\theta)\cos(m\theta)\) [A1].
Applying product-to-sum identity [M1].
Integrating to show result is 0 [A1].

(a)
The gradient of the tangent to \(y = x^2\) at \(P(t, t^2)\) is \(\frac{dy}{dx} = 2x \implies m_T = 2t\).
The gradient of the normal line is the negative reciprocal: \(m_N = -\frac{1}{2t}\) (for \(t \neq 0\)).
The equation of the normal line is: \(y - t^2 = -\frac{1}{2t}(x - t)\).
Multiplying both sides by \(2t\): \(2ty - 2t^3 = -x + t \implies x + 2ty - 2t^3 - t = 0\).
This equation is also valid for \(t = 0\), where the normal line is the y-axis, \(x = 0\).

(b)
The normal lines at \(P\) and \(Q\) are:
(1) \(x + 2t_1 y = 2t_1^3 + t_1\)
(2) \(x + 2t_2 y = 2t_2^3 + t_2\)
Subtracting (2) from (1) gives: \(2(t_1 - t_2)y = 2(t_1^3 - t_2^3) + (t_1 - t_2)\).
Since \(t_1 \neq t_2\), divide by \(2(t_1 - t_2)\): \(y_I = \frac{2(t_1 - t_2)(t_1^2 + t_1 t_2 + t_2^2)}{2(t_1 - t_2)} + \frac{t_1 - t_2}{2(t_1 - t_2)} = t_1^2 + t_1 t_2 + t_2^2 + \frac{1}{2}\).
Substituting this into (1) to solve for \(x_I\): \(x_I = 2t_1^3 + t_1 - 2t_1\left(t_1^2 + t_1 t_2 + t_2^2 + \frac{1}{2}\right) = 2t_1^3 + t_1 - 2t_1^3 - 2t_1^2 t_2 - 2t_1 t_2^2 - t_1 = -2t_1 t_2(t_1 + t_2)\).

(c)
Taking the limit as \(t_2 \to t_1\), let both \(t_1, t_2 \to t\).
Then \(x_I \to -2t^2(2t) = -4t^3\), and \(y_I \to t^2 + t^2 + t^2 + \frac{1}{2} = 3t^2 + \frac{1}{2}\).
Thus, the limiting coordinates of the point \(C(t)\) are \(\left(-4t^3, 3t^2 + \frac{1}{2}\right)\).

(d)
We have \(x = -4t^3\) and \(y = 3t^2 + \frac{1}{2}\).
From the second equation, \(t^2 = \frac{y - 1/2}{3}\).
Squaring the first equation gives: \(x^2 = 16t^6 = 16(t^2)^3\).
Substituting the expression for \(t^2\): \(x^2 = 16 \left(\frac{y - 1/2}{3}\right)^3 = \frac{16}{27}\left(y - \frac{1}{2}\right)^3 \implies 27x^2 = 16\left(y - \frac{1}{2}\right)^3\).

(e)
(i) \(\frac{\partial g}{\partial t} = 2y - 6t^2 - 1 = 0 \implies y = 3t^2 + \frac{1}{2}\).
(ii) Substituting \(y\) into \(g(x, y, t) = 0\): \(x + 2t\left(3t^2 + \frac{1}{2}\right) - 2t^3 - t = 0 \implies x + 6t^3 + t - 2t^3 - t = 0 \implies x = -4t^3\).
(iii) The parametric equations obtained are indeed \(x = -4t^3\) and \(y = 3t^2 + \frac{1}{2}\). Eliminating \(t\) as in part (d) yields the identical Cartesian equation \(27x^2 = 16\left(y - \frac{1}{2}\right)^3\).

(f)
(i) The sketch should show:
- The parabola \(y = x^2\) with its vertex at \((0, 0)\).
- The evolute with a cusp at \((0, 1/2)\), opening upwards and symmetric about the y-axis, crossing the parabola.
(ii) The boundary curve where the number of real roots of the cubic changes is exactly the evolute. Within the region above the evolute (inside the cusp), the discriminant of the cubic is positive, meaning there are 3 real roots for \(t\). Therefore, exactly 3 normal lines can be drawn to the parabola from any point in this region.

(a)
Finding gradient of tangent \(m_T = 2t\) [M1]
Stating gradient of normal \(m_N = -\frac{1}{2t}\) [A1]
Setting up equation of normal line [M1]
Simplifying to \(x + 2ty - 2t^3 - t = 0\) [AG] [A1]
(b)
Writing equations for both normals [M1]
Subtracting to eliminate \(x\) [M1]
Factoring the difference of cubes to find \(y_I\) [AG] [A1]
Substituting \(y_I\) back to solve for \(x\) [M1]
Simplifying to show \(x_I\) [AG] [A1]
(c)
Explicitly stating the limit process \(t_2 \to t_1\) [M1]
Taking limit for \(x_I\) to get \(-4t^3\) [A1]
Taking limit for \(y_I\) to get \(3t^2 + 1/2\) [A1]
Expressing coordinates as \(C(t)\) [A1]
(d)
Solving for \(t^2\) from \(y\) [M1]
Squaring \(x\) to express in terms of \(t^6\) [M1]
Substituting to obtain \(27x^2 = 16\left(y - \frac{1}{2}\right)^3\) [AG] [A1]
(e)
(i) Finding partial derivative and setting to 0 [M1], solving for \(y\) [A1]
(ii) Substituting into the normal equation [M1] and solving for \(x\) [A1]
(iii) Confirming both methods lead to the same parametric set [M1] and Cartesian equation [A1]
(f)
(i) Correct sketch of parabola [A1]
Correct sketch of evolute with cusp at \((0, 0.5)\) [A1]
Correct relative orientation and intersection shown [A1]
(ii) Explaining that the evolute represents the boundary of changing root count [R1]
Concluding that 3 normal lines can be drawn since there are 3 real roots in the cusp region [R1]