2023 IB DP Mathematics - Analysis and Approaches 模擬試題連答案詳解

First, find the common difference \(d\) and first term \(u_1\) of the arithmetic sequence. We have \(u_3 = u_1 + 2d = 8\) and \(u_8 = u_1 + 7d = 23\). Subtracting these equations gives \(5d = 15\), which means \(d = 3\). Substituting \(d = 3\) back into the first equation gives \(u_1 + 6 = 8\), so \(u_1 = 2\). The first three terms of the geometric sequence are \(t_1 = u_1 = 2\), \(t_2 = u_3 = 8\), and \(t_3 = u_k = u_1 + (k-1)d = 2 + 3(k-1) = 3k - 1\). Since these terms form a geometric sequence, the common ratio is constant, so \(\frac{t_2}{t_1} = \frac{t_3}{t_2}\). This gives \(\frac{8}{2} = \frac{3k-1}{8}\), which simplifies to \(4 = \frac{3k-1}{8}\). Multiplying both sides by 8 gives \(32 = 3k - 1\), so \(3k = 33\), yielding \(k = 11\).

M1 for setting up simultaneous equations for \(u_3\) and \(u_8\). A1 for finding \(d = 3\) and \(u_1 = 2\). M1 for expressing the terms of the geometric sequence as 2, 8, and \(3k-1\). M1 for setting up the geometric ratio equation \(\frac{8}{2} = \frac{3k-1}{8}\). A1 for solving to get \(3k = 33\). A1 for final answer \(k = 11\).

(a) To find \((g \circ f)(x)\), we substitute \(f(x)\) into \(g(x)\): \((g \circ f)(x) = g(f(x)) = e^{2\ln(x-2)} + 2\). Using the properties of logarithms, we write \(2\ln(x-2) = \ln((x-2)^2)\). Thus, \((g \circ f)(x) = e^{\ln((x-2)^2)} + 2 = (x-2)^2 + 2\). Here, \(a=1\), \(b=2\), and \(c=2\). (b) The domain of \((g \circ f)\) is the domain of \(f\), which is \(x > 2\). Since \(x > 2\), we have \(x - 2 > 0\), so \((x - 2)^2 > 0\). Therefore, the range of \((g \circ f)(x) = (x-2)^2 + 2\) is \(y > 2\).

M1 for substituting \(f(x)\) into \(g(x)\). M1 for applying log laws to write \(2\ln(x-2) = \ln((x-2)^2)\). A1 for simplifying to \((x-2)^2 + 2\). A1 for correct domain \(x > 2\). A2 for correct range \(y > 2\) (award A1 for \(y \ge 2\)).

Using the Pythagorean identity \(\cos^2(x) = 1 - \sin^2(x)\), we rewrite the equation as \(2(1 - \sin^2(x)) + 3 \sin(x) - 3 = 0\). Simplifying gives \(2 - 2 \sin^2(x) + 3 \sin(x) - 3 = 0\), which simplifies further to \(-2 \sin^2(x) + 3 \sin(x) - 1 = 0\). Multiplying by \(-1\), we get \(2 \sin^2(x) - 3 \sin(x) + 1 = 0\). Factoring this quadratic equation yields \((2\sin(x) - 1)(\sin(x) - 1) = 0\). This gives two possible cases: \(\sin(x) = \frac{1}{2}\) or \(\sin(x) = 1\). In the interval \(0 \le x \le 2\pi\), \(\sin(x) = \frac{1}{2}\) has solutions \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). In the same interval, \(\sin(x) = 1\) has solution \(x = \frac{\pi}{2}\). Thus, the full set of solutions is \(x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\).

M1 for using \(\cos^2(x) = 1 - \sin^2(x)\) to form a quadratic in \(\sin(x)\). A1 for obtaining \(2 \sin^2(x) - 3 \sin(x) + 1 = 0\). M1 for factoring to \((2\sin(x) - 1)(\sin(x) - 1) = 0\). A1 for finding \(\sin(x) = \frac{1}{2}\) and \(\sin(x) = 1\). A1 for \(x = \frac{\pi}{6}, \frac{5\pi}{6}\). A1 for \(x = \frac{\pi}{2}\).

Let \(u = x^2\). Then the derivative is \(\frac{du}{dx} = 2x\), which gives \(dx = \frac{du}{2x}\). We also change the integration limits: when \(x = 0\), \(u = 0^2 = 0\); when \(x = \sqrt{\pi}\), \(u = (\sqrt{\pi})^2 = \pi\). Substituting these into the integral, we get \(\int_{0}^{\pi} x \sin(u) \frac{du}{2x} = \frac{1}{2} \int_{0}^{\pi} \sin(u) \, du\). The antiderivative of \(\sin(u)\) is \(-\cos(u)\). Evaluating this from \(0\) to \(\pi\) yields \(\frac{1}{2} [-\cos(u)]_{0}^{\pi} = \frac{1}{2} (-\cos(\pi) - (-\cos(0))) = \frac{1}{2} (-(-1) + 1) = \frac{1}{2}(1 + 1) = 1\).

M1 for using substitution with \(u = x^2\). A1 for finding \(du = 2x \, dx\) or equivalent. A1 for correctly changing the limits to \(0\) and \(\pi\). M1 for integrating to obtain \(-\frac{1}{2}\cos(u)\) (or equivalent in terms of \(x\)). A1 for substituting the limits. A1 for the final value 1.

Since the sum of the probabilities must equal 1, we have \(a + b + 0.3 + 0.1 = 1\), which simplifies to \(a + b = 0.6\). The expected value of \(X\) is defined by \(\mathbb{E}(X) = \sum x \cdot \mathbb{P}(X=x)\). Substituting the given values, we get \(1(a) + 2(b) + 3(0.3) + 4(0.1) = 2.4\), which simplifies to \(a + 2b + 0.9 + 0.4 = 2.4\). This further simplifies to \(a + 2b + 1.3 = 2.4\), so \(a + 2b = 1.1\). We now have a system of two linear equations: (1) \(a + b = 0.6\) and (2) \(a + 2b = 1.1\). Subtracting equation (1) from equation (2) gives \(b = 0.5\). Substituting \(b = 0.5\) back into equation (1) yields \(a + 0.5 = 0.6\), which gives \(a = 0.1\).

M1 for setting the sum of probabilities to 1: \(a + b + 0.4 = 1\). A1 for obtaining \(a + b = 0.6\). M1 for setting up the expectation equation: \(a + 2b + 1.3 = 2.4\). A1 for obtaining \(a + 2b = 1.1\). M1 for solving the simultaneous equations. A1 for correct values \(a = 0.1\) and \(b = 0.5\).

The curve crosses the \(x\)-axis when \(y = 0\). This means \(\frac{\ln(x)}{x} = 0\), which occurs when \(\ln(x) = 0\), so \(x = 1\). The point of contact is therefore \((1, 0)\). Next, we find the derivative \(\frac{dy}{dx}\) using the quotient rule. Let \(u = \ln(x)\) and \(v = x\), so \(u' = \frac{1}{x}\) and \(v' = 1\). The quotient rule gives \(\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{\frac{1}{x} \cdot x - \ln(x) \cdot 1}{x^2} = \frac{1 - \ln(x)}{x^2}\). At \(x = 1\), the gradient of the tangent is \(m = \frac{1 - \ln(1)}{1^2} = \frac{1 - 0}{1} = 1\). Using the point-slope formula \(y - y_1 = m(x - x_1)\) with point \((1, 0)\) and gradient \(m = 1\), we get \(y - 0 = 1(x - 1)\), which simplifies to \(y = x - 1\).

M1 for finding the \(x\)-intercept by setting \(y = 0\) to get \(x = 1\). A1 for the coordinates of the point \((1, 0)\). M1 for applying the quotient rule to find \(\frac{dy}{dx}\). A1 for obtaining \(\frac{dy}{dx} = \frac{1 - \ln(x)}{x^2}\). A1 for finding the gradient \(m = 1\). A1 for the final equation \(y = x - 1\).

**(a)** Using the product rule to find \(f'(x)\):
\(f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1\).
For a minimum point, set \(f'(x) = 0\):
\(\ln(x) + 1 = 0 \implies \ln(x) = -1 \implies x = e^{-1} = \frac{1}{e}\).
Substituting this back into \(f(x)\) to find the minimum value:
\(f\left(\frac{1}{e}\right) = \frac{1}{e} \ln\left(\frac{1}{e}\right) = \frac{1}{e}(-1) = -\frac{1}{e}\).

**(b)** At \(x = e\), the \(y\)-coordinate is:
\(y = f(e) = e \ln(e) = e\).
The gradient of the tangent at \(x = e\) is:
\(m = f'(e) = \ln(e) + 1 = 2\).
The equation of the tangent line is:
\(y - e = 2(x - e) \implies y = 2x - e\).

**(c)** Using integration by parts, let \(u = \ln(x)\) and \(dv = x \, dx\).
Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\).
\(\int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx\)
\(= \frac{x^2}{2} \ln(x) - \frac{1}{2} \int x \, dx\)
\(= \frac{x^2}{2} \ln(x) - \frac{x^2}{4} + C\).

**(d)** Find the \(x\)-intercept: \(x \ln(x) = 0 \implies x = 1\) (since \(x > 0\)).
For \(1 \le x \le e\), \(f(x) \ge 0\).
The enclosed area is given by:
\(A = \int_{1}^{e} x \ln(x) \, dx = \left[ \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \right]_{1}^{e}\)
\(= \left( \frac{e^2}{2} \ln(e) - \frac{e^2}{4} \right) - \left( \frac{1^2}{2} \ln(1) - \frac{1^2}{4} \right)\)
\(= \left( \frac{e^2}{2} - \frac{e^2}{4} \right) - \left( 0 - \frac{1}{4} \right) = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2+1}{4}\).

**(a)**
M1: Attempt to apply the product rule to find \(f'(x)\).
A1: Correct derivative \(f'(x) = \ln(x) + 1\).
M1: Setting their \(f'(x) = 0\) and solving for \(x\).
A1: Correctly showing that the minimum value is \(-\frac{1}{e}\).

**(b)**
A1: Correctly finding the coordinate \((e, e)\).
A1: Correct gradient of the tangent \(m = 2\).
A1: Correct equation of the tangent \(y = 2x - e\) (any equivalent form).

**(c)**
M1: Attempting integration by parts with correct assignment of \(u\) and \(dv\).
A1: Correct \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\).
M1: Applying the integration by parts formula correctly.
A1: Correct final expression \(\frac{x^2}{2} \ln(x) - \frac{x^2}{4} + C\) (condone missing \(+ C\)).

**(d)**
A1: Correctly identifying the lower boundary of integration as \(x = 1\).
M1: Setting up the definite integral with correct limits from \(1\) to \(e\).
A1: Correct substitution of limits into the integrated function.
A1: Correctly simplifying to obtain \(\frac{e^2+1}{4}\).

**(a)** (i) The vertical asymptote occurs where the denominator is zero: \(x - b = 0 \implies x = b\). Since the vertical asymptote is given as \(x = 3\), it follows that \(b = 3\).
(ii) Substituting \((4, 11)\) into \(f(x) = \frac{2x + a}{x - 3}\):
\(11 = \frac{2(4) + a}{4 - 3} \implies 11 = 8 + a \implies a = 3\).

**(b)** (i) To find the inverse, let \(y = f(x)\):
\(y = \frac{2x + 3}{x - 3}\)
\(y(x - 3) = 2x + 3 \implies yx - 3y = 2x + 3\)
\(yx - 2x = 3y + 3 \implies x(y - 2) = 3y + 3\)
\(x = \frac{3y + 3}{y - 2}\).
Thus, \(f^{-1}(x) = \frac{3x + 3}{x - 2}\).
The domain of \(f^{-1}\) is the range of \(f\). The horizontal asymptote of \(f(x)\) is \(y = 2\), so the domain of \(f^{-1}(x)\) is \(x \in \mathbb{R}, x \neq 2\).

(ii) The graphs of a function and its inverse intersect along the line \(y = x\). Setting \(f(x) = x\):
\(\frac{2x + 3}{x - 3} = x \implies 2x + 3 = x^2 - 3x \implies x^2 - 5x - 3 = 0\).
Applying the quadratic formula:
\(x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-3)}}{2} = \frac{5 \pm \sqrt{25 + 12}}{2} = \frac{5 \pm \sqrt{37}}{2}\).
Since these points lie on \(y = x\), the coordinates of intersection are:
\(\left(\frac{5 + \sqrt{37}}{2}, \frac{5 + \sqrt{37}}{2}\right)\) and \(\left(\frac{5 - \sqrt{37}}{2}, \frac{5 - \sqrt{37}}{2}\right)\).

**(c)** (i) \((f \circ g)(x) = f(g(x)) = f(\ln(x)) = \frac{2\ln(x) + 3}{\ln(x) - 3}\).
(ii) Setting \((f \circ g)(x) = 3\):
\(\frac{2\ln(x) + 3}{\ln(x) - 3} = 3\)
\(2\ln(x) + 3 = 3(\ln(x) - 3)\)
\(2\ln(x) + 3 = 3\ln(x) - 9\)
\(\ln(x) = 12\)
\(x = e^{12}\).

**(a)**
(i) R1: Reasoning that the vertical asymptote occurs when the denominator is zero to show \(b = 3\).
(ii) M1: Substituting \(x = 4, y = 11\) and \(b = 3\) into the function. A1: Correctly solving to get \(a = 3\).

**(b)**
(i) M1: Writing an equation swapping \(x\) and \(y\) (or rearranging \(y = f(x)\) to make \(x\) the subject). M1: Correct algebraic manipulation to isolate \(x\) or \(y\). A1: Correct inverse expression \(f^{-1}(x) = \frac{3x + 3}{x - 2}\). A1: Correctly stating the domain as \(x \neq 2\) (accept \(\mathbb{R} \setminus \{2\}\)).
(ii) M1: Setting \(f(x) = x\) (or \(f(x) = f^{-1}(x)\)). A1: Correct quadratic equation \(x^2 - 5x - 3 = 0\). A1: Correctly stating both coordinates of intersection \(\left(\frac{5 \pm \sqrt{37}}{2}, \frac{5 \pm \sqrt{37}}{2}\right)\).

**(c)**
(i) A1: Correct composite expression \(\frac{2\ln(x) + 3}{\ln(x) - 3}\).
(ii) M1: Setting their composite function equal to \(3\). M1: Clearing the fraction correctly to obtain \(2\ln(x) + 3 = 3\ln(x) - 9\). A1: Correctly isolating \(\ln(x) = 12\). A1: Correct solution \(x = e^{12}\).

**(a)** The sum of all probabilities in a probability distribution must equal 1:
\(\sum P(X = x) = 1\)
\(k + 2k + 3k + \frac{1}{4} + \frac{1}{12} = 1\)
\(6k + \frac{3}{12} + \frac{1}{12} = 1\)
\(6k + \frac{4}{12} = 1 \implies 6k + \frac{1}{3} = 1\)
\(6k = \frac{2}{3} \implies k = \frac{2}{18} = \frac{1}{9}\).

**(b)** Using the formula for expectation \(E(X) = \sum x \cdot P(X = x)\):
\(E(X) = 1(k) + 2(2k) + 3(3k) + 4\left(\frac{1}{4}\right) + 5\left(\frac{1}{12}\right)\)
\(= k + 4k + 9k + 1 + \frac{5}{12}\)
\(= 14k + \frac{17}{12}\).
Substituting \(k = \frac{1}{9}\):
\(E(X) = 14\left(\frac{1}{9}\right) + \frac{17}{12} = \frac{14}{9} + \frac{17}{12}\).
Using a common denominator of 36:
\(E(X) = \frac{14 \times 4}{36} + \frac{17 \times 3}{36} = \frac{56 + 51}{36} = \frac{107}{36}\).

**(c)** The possible independent pairs \((X_1, X_2)\) that sum to 5 are:
\((1, 4)\), \((2, 3)\), \((3, 2)\), and \((4, 1)\).
The probabilities of each outcome with \(k = \frac{1}{9}\) are:
\(P(X = 1) = \frac{1}{9}\)
\(P(X = 2) = \frac{2}{9}\)
\(P(X = 3) = \frac{3}{9}\)
\(P(X = 4) = \frac{1}{4}\)
Calculating the individual joint probabilities:
\(P(1, 4) = \frac{1}{9} \times \frac{1}{4} = \frac{1}{36}\)
\(P(2, 3) = \frac{2}{9} \times \frac{3}{9} = \frac{6}{81} = \frac{2}{27}\)
\(P(3, 2) = \frac{3}{9} \times \frac{2}{9} = \frac{6}{81} = \frac{2}{27}\)
\(P(4, 1) = \frac{1}{4} \times \frac{1}{9} = \frac{1}{36}\).
Summing these probabilities:
\(P(X_1 + X_2 = 5) = 2 \times \frac{1}{36} + 2 \times \frac{2}{27} = \frac{1}{18} + \frac{4}{27}\).
Using a common denominator of 54:
\(P(X_1 + X_2 = 5) = \frac{3}{54} + \frac{8}{54} = \frac{11}{54}\).

**(d)** We seek the conditional probability:
\(P(X_1 \ge 3 \mid X_1 + X_2 = 5) = \frac{P(X_1 \ge 3 \cap X_1 + X_2 = 5)}{P(X_1 + X_2 = 5)}\).
The pairs from part (c) where \(X_1 \ge 3\) are \((3, 2)\) and \((4, 1)\).
The sum of their probabilities is:
\(P((3, 2) \cup (4, 1)) = P(3, 2) + P(4, 1) = \frac{2}{27} + \frac{1}{36}\).
Using a common denominator of 108:
\(P(3, 2) + P(4, 1) = \frac{8}{108} + \frac{3}{108} = \frac{11}{108}\).
Now calculate the conditional probability:
\(P(X_1 \ge 3 \mid X_1 + X_2 = 5) = \frac{\frac{11}{108}}{\frac{11}{54}} = \frac{11}{108} \times \frac{54}{11} = \frac{54}{108} = \frac{1}{2}\).

**(a)**
M1: Setting up equation where sum of all probabilities is equal to 1.
A1: Correct simplification to \(6k + \frac{1}{3} = 1\) (or equivalent form).
A1: Showing step-by-step resolution leading to \(k = \frac{1}{9}\).

**(b)**
M1: Attempting to apply formula for expected value \(\sum x P(X = x)\).
A1: Correct expression in terms of \(k\) (e.g., \(14k + \frac{17}{12}\)).
A1: Correct final fraction \(\frac{107}{36}\).

**(c)**
M1: Identifying the four valid pairs: \((1, 4)\), \((2, 3)\), \((3, 2)\), and \((4, 1)\).
M1: Correct calculation of at least two individual joint probabilities (e.g. \(P(1, 4) = \frac{1}{36}\) and \(P(2, 3) = \frac{2}{27}\)).
A1: Correctly adding the four joint probabilities.
A1: Correct final simplified fraction \(\frac{11}{54}\).

**(d)**
M1: Writing the correct formula for conditional probability.
M1: Correctly identifying the favorable outcomes where \(X_1 \ge 3\) given the sum is 5, i.e., \((3, 2)\) and \((4, 1)\).
A1: Calculating the joint probability for these favorable outcomes, \(\frac{11}{108}\) (or equivalent).
M1: Substituting values into the conditional probability formula.
A1: Correct final answer of \(\frac{1}{2}\) (or 0.5).

The sum of the first \(n\) terms of the arithmetic sequence is: \(S_{n,\text{arith}} = \frac{n}{2} [2(12) + (n-1)(3.5)] = \frac{n}{2} [3.5n + 20.5]\). The sum of the first \(n\) terms of the geometric sequence is: \(S_{n,\text{geom}} = \frac{4(1.15^n - 1)}{1.15 - 1} = \frac{4(1.15^n - 1)}{0.15}\). We use a GDC to find the smallest integer \(n\) satisfying the inequality \(S_{n,\text{geom}} > S_{n,\text{arith}}\). For \(n = 30\): \(S_{30,\text{arith}} = 1882.5\) and \(S_{30,\text{geom}} \approx 1738.98\) (the condition is not met). For \(n = 31\): \(S_{31,\text{arith}} = 1999.5\) and \(S_{31,\text{geom}} \approx 2003.83\) (the condition is met). Thus, the least value of \(n\) is \(31\).

M1: Attempt to write down formulas for \(S_{n,\text{arith}}\) and \(S_{n,\text{geom}}\). A1: Correct expression for \(S_{n,\text{arith}}\). A1: Correct expression for \(S_{n,\text{geom}}\). M1: Attempt to solve inequality using GDC (e.g., table or intersection search). A1: Correct comparison of values for \(n=30\) and \(n=31\). A1: Final answer \(n=31\).

(a) Since \(T(0) = 85\), we have \(a e^{0} + 18 = 85 \implies a + 18 = 85 \implies a = 67\). (b) Since \(T(15) = 41\), we have \(67 e^{-15b} + 18 = 41 \implies 67 e^{-15b} = 23 \implies e^{-15b} = \frac{23}{67} \implies -15b = \ln\left(\frac{23}{67}\right) \implies b \approx 0.0713\) (or \(0.07128...\)). (c) The rate of change of temperature is given by \(T'(t) = -67b e^{-bt}\). We set the rate of decrease to \(1.5\), which means \(-T'(t) = 1.5 \implies 67b e^{-bt} = 1.5\). Substituting \(b \approx 0.07128\) gives: \(67(0.07128) e^{-0.07128t} = 1.5 \implies e^{-0.07128t} \approx 0.3141 \implies -0.07128t \approx \ln(0.3141) \implies t \approx 16.2\text{ minutes}\).

(a) A1: Correct value of \(a = 67\). (b) M1: Substitution of values to set up equation for \(b\). A1: Correct value of \(b \approx 0.0713\) (accept \(0.07128...\)). (c) M1: Differentiating to obtain \(T'(t)\). M1: Setting their \(-T'(t) = 1.5\) (or equivalent). A1: Correct value of \(t \approx 16.2\) (accept \(16.24...\)).

(a) The bearings from \(A\) are measured clockwise from North. \(\angle TAB = 110^\circ - 35^\circ = 75^\circ\). (b) The bearing of \(A\) from \(B\) is \(110^\circ + 180^\circ = 290^\circ\). The bearing of \(T\) from \(B\) is \(340^\circ\). Thus, the interior angle \(\angle TBA = 340^\circ - 290^\circ = 50^\circ\). The third angle in the triangle is \(\angle ATB = 180^\circ - (75^\circ + 50^\circ) = 55^\circ\). Using the Law of Sines: \(\frac{AT}{\sin(50^\circ)} = \frac{120}{\sin(55^\circ)} \implies AT = 120 \times \frac{\sin(50^\circ)}{\sin(55^\circ)} \approx 112.22\text{ m}\). So \(AT \approx 112\text{ m}\). (c) In the right-angled triangle representing the height, \(\tan(18^\circ) = \frac{h}{AT} \implies h = 112.22 \times \tan(18^\circ) \approx 36.46\text{ m}\). So \(h \approx 36.5\text{ m}\).

(a) M1: Attempt to use given bearings. A1: Correct angle \(75^\circ\). (b) M1: Attempt to calculate interior angle at \(B\) (e.g. \(50^\circ\)) and third angle \(55^\circ\). A1: Correct application of Sine Rule leading to \(AT \approx 112\text{ m}\) (accept \(112.22...\)). (c) M1: Using right-angled trigonometry with \(AT\) and \(18^\circ\). A1: Correct height \(h \approx 36.5\text{ m}\) (accept \(36.46...\)).

(a) Let \(X \sim N(\mu, 12^2)\). We are given \(P(X < 980) = 0.08\). Standardizing gives \(P\left(Z < \frac{980 - \mu}{12}\right) = 0.08\). Using inverse normal: \(\frac{980 - \mu}{12} \approx -1.4051 \implies 980 - \mu \approx -16.86 \implies \mu \approx 996.86\text{ grams}\). To 3 significant figures, \(\mu \approx 997\text{ grams}\). (b) We seek \(P(995 < X < 1015)\) with \(\mu = 996.86\) and \(\sigma = 12\). Using the GDC normalCDF tool: \(P(995 < X < 1015) \approx 0.496\) (or \(0.4963\)).

(a) M1: Attempt to standardize. M1: Finding the standard z-score \(-1.4051\). A1: Correct mean \(\mu \approx 997\) (accept \(996.86...\)). (b) M1: Formulating correct probability statement. M1: Use of GDC normalCDF function with correct parameters. A1: Correct probability \(\approx 0.496\) (accept answers in range \([0.495, 0.497]\) depending on rounding).

(a) Setting \(f(x) = g(x)\) gives \(\ln(x^2 + 1) = \cos(x) + 0.5x\). Using GDC solver in the interval \([0, 5]\), we obtain the intersection point \(x \approx 1.2492\) (or \(1.25\) to 3 s.f.). (b) The region is bounded on the left by the \(y\)-axis (\(x = 0\)) and on the right by \(x \approx 1.2492\). Since \(g(x) \ge f(x)\) on \([0, 1.2492]\), the area is given by \(A = \int_{0}^{1.2492} (g(x) - f(x)) \, dx = \int_{0}^{1.2492} (\cos(x) + 0.5x - \ln(x^2 + 1)) \, dx\). Evaluating this using GDC numerical integration yields \(A \approx 0.871\).

(a) M1: Writing equation \(f(x) = g(x)\). A1: Correct \(x\)-coordinate of \(1.25\) (accept \(1.2492...\)). (b) M1: Formulating correct integral for the area. A1: Correct limits and integrand. M1: Using numerical integration on a GDC. A1: Correct area of \(0.871\) (accept \(0.8708...\)).

(a) We first compute the natural logarithm values for both variables: \(\ln d \in \{0, 0.4055, 0.6931, 0.9163, 1.0986, 1.2528, 1.3863\}\) and \(\ln I \in \{6.0403, 5.2204, 4.7005, 4.2195, 3.8067, 3.5264, 3.2581\}\). Performing linear regression of \(\ln I\) on \(\ln d\) on a GDC yields the equation: \(\ln I = -2.0135 \ln d + 6.0507\). Rounding coefficients to 3 s.f.: \(\ln I = -2.01 \ln d + 6.05\). (b) Comparing to the theoretical linearized formula: \(b = m \approx -2.01\) and \(\ln a = c = 6.0507 \implies a = e^{6.0507} \approx 424\). (c) When \(d = 2.8\), \(I = 424 \times 2.8^{-2.01} \approx 53.4\text{ lux}\).

(a) M1: Attempt to compute logarithm values. M1: Linear regression setup on GDC. A1: Correct equation with 3 s.f. coefficients: \(\ln I = -2.01 \ln d + 6.05\). (b) A1: Correct value of \(b \approx -2.01\). A1: Correct value of \(a \approx 424\) (accept range \([422, 426]\)). (c) A1: Correct predicted value \(I \approx 53.4\text{ lux}\) (accept range \([53.0, 54.0]\)).

(a)
The maximum height is \(80 + 45 = 125\text{ m}\).
The minimum height is \(80 - 45 = 35\text{ m}\).
Thus, the amplitude \(a = 45\) and the vertical shift \(d = 80\).

Since the blades rotate at \(12\) revolutions per minute, the time for one revolution (the period \(T\)) is:
\[T = \frac{60}{12} = 5\text{ seconds}\]

Thus,
\[b = \frac{2\pi}{T} = \frac{2\pi}{5} \approx 1.26\]

(b)
We solve \(h(t) = 100\):
\[45 \cos\left(\frac{2\pi}{5}t\right) + 80 = 100\]
\[\cos\left(\frac{2\pi}{5}t\right) = \frac{20}{45} = \frac{4}{9}\]
\[\frac{2\pi}{5}t = \arccos\left(\frac{4}{9}\right) \approx 1.1102\]
\[t_1 \approx 0.883\text{ seconds}\]

Using the symmetry of the cosine function (or the period \(T = 5\)):
\[t_2 = 5 - t_1 \approx 4.12\text{ seconds}\]

(c)
The difference in height is given by \(D(t) = |h(t) - g(t)|\).
Using a graphic display calculator to plot \(y = |h(t) - g(t)|\) over the interval \(t \in [0, 10]\), we locate the maximum points.

The first maximum occurs at \(t \approx 0.389\text{ seconds}\).
Since both models have a period of \(5\) seconds, the difference function also has a period of \(5\) seconds.
Therefore, the second maximum occurs at:
\[t \approx 0.389 + 5 = 5.39\text{ seconds}\]

(d)
We want to find the total time in \([0, 10]\) for which \(h(t) > g(t)\).
We first find the intersection points where \(h(t) = g(t)\) in the first cycle \([0, 5]\):
\[t \approx 1.693\text{ s}\quad \text{and} \quad t \approx 4.085\text{ s}\]

Thus, during the first cycle of \(5\) seconds, \(h(t) > g(t)\) on the intervals:
\[[0, 1.693) \quad \text{and} \quad (4.085, 5]\]

The duration in the first cycle is:
\[1.693 + (5 - 4.085) = 2.608\text{ seconds}\]

Since the behavior repeats every \(5\) seconds, the total time in the first \(10\) seconds is:
\[2 \times 2.608 = 5.216 \approx 5.22\text{ seconds}\]

(a)
- M1 for identifying amplitude \(a = 45\)
- M1 for identifying vertical translation \(d = 80\)
- A1 for identifying period \(T = 5\)
- A1 for obtaining \(b = \frac{2\pi}{5}\) (or \(1.26\))

(b)
- M1 for setting up the equation \(45 \cos\left(\frac{2\pi}{5}t\right) + 80 = 100\)
- A1 for finding the first solution \(t_1 \approx 0.883\)
- M1 for using symmetry or period (e.g., \(5 - t_1\))
- A1 for finding the second solution \(t_2 \approx 4.12\)

(c)
- M1 for defining the difference function \(D(t) = |h(t) - g(t)|\)
- M1 for using GDC to find a maximum of the function
- A1 for the first time \(t_1 \approx 0.389\)
- A1 for the second time \(t_2 \approx 5.39\)

(d)
- M1 for finding the critical boundary intersection values \(t \approx 1.69\) and \(t \approx 4.09\)
- M1 for determining the time interval length in one period: \(1.693 + (5 - 4.085) = 2.608\)
- A1 for the final answer \(5.22\) (accept \(5.21\) to \(5.23\))

(a)
Let \(W \sim N(150, 12^2)\).
(i) Using the GDC:
\[P(W < 135) \approx 0.1056
\approx 0.106\]
(ii) Using the GDC:
\[P(140 < W < 165) \approx 0.6919 \approx 0.692\]

(b)
We are given \(P(W > k) = 0.15\), which means \(P(W \le k) = 0.85\).
Using the inverse normal function on the GDC with \(\mu = 150\) and \(\sigma = 12\):
\[k \approx 162.437 \approx 162\text{ grams}\]

(c)
First, we find the probability of a small apple, \(p_{\text{small}}\):
\[p_{\text{small}} = P(W < 130) \approx 0.04779\]

Let \(X\) be the number of small apples in a box of \(20\).
\[X \sim B(20, 0.04779)\]

(i) Using the binomial probability formula or GDC:
\[P(X = 3) = \binom{20}{3} (0.04779)^3 (1-0.04779)^{17} \approx 0.05393 \approx 0.0539\]

(ii) We need \(P(X \ge 2) = 1 - P(X \le 1)\):
Using the binomial cumulative distribution function on the GDC:
\[P(X \le 1) \approx 0.7533\]
\[P(X \ge 2) = 1 - 0.7533 = 0.2467 \approx 0.247\]

(d)
For the 8th tested apple to be the 5th large apple, there must be exactly 4 large apples in the first 7 tested apples, and the 8th tested apple must be large.

Let \(Y\) be the number of large apples in the first 7 trials.
\[Y \sim B(7, 0.15)\]
\[P(Y = 4) = \binom{7}{4} (0.15)^4 (0.85)^3 \approx 0.01088\]

Multiplying by the probability that the 8th apple is large:
\[P(\text{exactly 8 tested}) = 0.01088 \times 0.15 \approx 0.001632 \approx 0.00163\]

(e)
We want the conditional probability \(P(W > 160 \mid W \ge 130)\):
\[P(W > 160 \mid W \ge 130) = \frac{P(W > 160 \cap W \ge 130)}{P(W \ge 130)} = \frac{P(W > 160)}{P(W \ge 130)}\]

Using the GDC:
\[P(W > 160) \approx 0.20233\]
\[P(W \ge 130) = 1 - 0.04779 = 0.95221\]

Thus,
\[P(W > 160 \mid W \ge 130) = \frac{0.20233}{0.95221} \approx 0.21248 \approx 0.212\]

(a)
- A1 for \(P(W < 135) \approx 0.106\)
- M1 for setting up \(P(140 < W < 165)\)
- A1 for \(P(140 < W < 165) \approx 0.692\)

(b)
- M1 for writing \(P(W > k) = 0.15\) (or equivalent)
- M1 for setting up the inverse normal equation
- A1 for \(k \approx 162\) (accept \(162.4\))

(c)
- A1 for finding \(p_{\text{small}} \approx 0.0478\)
- A1 for \(P(X = 3) \approx 0.0539\)
- M1 for recognizing the complement rule: \(1 - P(X \le 1)\)
- A1 for \(P(X \ge 2) \approx 0.247\)

(d)
- M1 for recognizing a negative binomial situation (requiring 4 large in 7 trials)
- M1 for calculating \(\binom{7}{4} (0.15)^4 (0.85)^3 \times 0.15\) (or equivalent)
- A1 for \(0.00163\)

(e)
- M1 for setting up the conditional probability fraction: \(\frac{P(W > 160)}{P(W \ge 130)}\)
- A1 for \(0.212\)

(a)
The volume \(V\) of the container is the sum of the volumes of the cylinder and the hemisphere:
\[V = \pi r^2 h + \frac{2}{3} \pi r^3\]

Setting \(V = 500\):
\[\pi r^2 h + \frac{2}{3} \pi r^3 = 500\]
\[\pi r^2 h = 500 - \frac{2}{3} \pi r^3\]
\[h = \frac{500}{\pi r^2} - \frac{2}{3} r\]

(b)
The total surface area is given by:
\[S = \text{Area of base} + \text{Area of cylinder wall} + \text{Area of hemisphere}\]
\[S = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h\]

Substitute the expression for \(h\) from part (a):
\[S = 3\pi r^2 + 2\pi r \left(\frac{500}{\pi r^2} - \frac{2}{3} r\right)\]
\[S = 3\pi r^2 + \frac{1000}{r} - \frac{4}{3}\pi r^2\]
\[S = \left(3 - \frac{4}{3}\right)\pi r^2 + \frac{1000}{r}\]
\[S = \frac{5}{3}\pi r^2 + \frac{1000}{r}\]

(c)
\[S'(r) = \frac{d}{dr}\left(\frac{5}{3}\pi r^2 + 1000 r^{-1}\right) = \frac{10}{3}\pi r - \frac{1000}{r^2}\]

(d)
To minimize surface area, we set \(S'(r) = 0\):
\[\frac{10}{3}\pi r = \frac{1000}{r^2}\]
\[r^3 = \frac{300}{\pi} \approx 95.493\]
\[r = \sqrt[3]{\frac{300}{\pi}} \approx 4.57\text{ cm}\]

To show this is a minimum, we find the second derivative:
\[S''(r) = \frac{10}{3}\pi + \frac{2000}{r^3}\]
For \(r > 0\), both terms in \(S''(r)\) are positive, hence \(S''(4.57) > 0\), which confirms a minimum.

(e)
The total cost \(C(r)\) in dollars is given by:
\[C(r) = 0.05 \left(\text{Area of base} + \text{Area of cylinder wall}\right) + 0.08 \left(\text{Area of hemisphere}\right)\]
\[C(r) = 0.05 \left(\pi r^2 + 2\pi r h\right) + 0.08 \left(2\pi r^2\right)\]
\[C(r) = 0.05 \left(\pi r^2 + \frac{1000}{r} - \frac{4}{3}\pi r^2\right) + 0.16\pi r^2\]
\[C(r) = 0.05 \left(-\frac{1}{3}\pi r^2 + \frac{1000}{r}\right) + 0.16\pi r^2\]
\[C(r) = \left(0.16 - \frac{0.05}{3}\right)\pi r^2 + \frac{50}{r}\]
\[C(r) = \frac{0.43}{3}\pi r^2 + \frac{50}{r}\]

Using GDC to find the minimum of \(C(r)\):
\[C'(r) = \frac{0.86}{3}\pi r - \frac{50}{r^2} = 0\]
\[r = \sqrt[3]{\frac{150}{0.86\pi}} \approx 3.81\text{ cm}\]

The minimum cost is:
\[C(3.81) \approx \frac{0.43}{3}\pi(3.8149)^2 +
\frac{50}{3.8149} \approx 19.7\text{ dollars}\]

(a)
- M1 for setting up volume equation \(V = \pi r^2 h + \frac{2}{3} \pi r^3 = 500\)
- A1 for making \(h\) the subject: \(h = \frac{500}{\pi r^2} -
\frac{2}{3} r\)

(b)
- M1 for writing down correct components of area: \(S = \pi r^2 + 2
\pi r h + 2\pi r^2\)
- M1 for substituting the expression for \(h\)
- A1 for correct algebraic expansion \(S = 3\pi r^2 + \frac{1000}{r} - \frac{4}{3}\pi r^2\)
- AG for clearly reaching the given expression

(c)
- A1 for \(\frac{10}{3}\pi r\)
- A1 for \(-\frac{1000}{r^2}\)

(d)
- M1 for setting \(S'(r) = 0\)
- A1 for \(r \approx 4.57\text{ cm}\)
- M1 for finding second derivative \(S''(r) = \frac{10}{3}\pi + \frac{2000}{r^3}\)
- A1 for stating \(S''(r) > 0\) (or evaluating at \(r \approx 4.57\)) and concluding it is a minimum

(e)
- M1 for setting up the correct cost function \(C(r) = \frac{0.43}{3}\pi r^2 + \frac{50}{r}\) (or unsimplified form)
- A1 for finding the radius \(r \approx 3.81\text{ cm}\)
- A1 for the minimum cost of \(19.7\) (accept \(19.66\))