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(a) Using a financial calculator (TVM solver) with:
\(N = 36\)
\(I\% = 4.8\)
\(PV = -15\,000\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)

We find the monthly payment \(PMT = \$447.60\) (correct to 2 decimal places).

(b) The total amount paid over the 3 years is:
\(447.603... \times 36 = \$16\,113.71\)

Total interest paid is:
\(16\,113.71 - 15\,000 = \$1113.71\) (or $1110 to 3 significant figures).

M1 for setting up the correct parameters in the TVM solver (N=36, I=4.8, PV=-15000).
A1 for a monthly payment of $447.60 (accept $448).
M1 for multiplying the monthly payment by 36.
A1 for finding the total interest paid as $1113.71 (accept $1110).

(a) To avoid making a loss, the bakery must break even or make a profit, so we solve \(P(x) \ge 0\).

Setting \(P(x) = 0\):
\(-0.25x^2 + 18x - 120 = 0\)

Using the quadratic formula or a GDC, we find the roots:
\(x \approx 7.43\) or \(x \approx 64.6\)

Since \(x\) must be an integer, we test surrounding values:
\(P(7) = -0.25(7)^2 + 18(7) - 120 = -6.25\) (loss)

\(P(8) = -0.25(8)^2 + 18(8) - 120 = 8\) (profit)

Thus, the minimum number of loaves to sell is 8.

(b) The maximum profit occurs at the vertex of the parabola where:
\(x = -\frac{b}{2a} = -\frac{18}{2(-0.25)} = 36\)

Evaluating \(P(36)\):
\(P(36) = -0.25(36)^2 + 18(36) - 120 = 204\)

So the maximum daily profit is $204.

M1 for setting \(P(x) = 0\) and solving.
A1 for finding the root \(x \approx 7.43\).
A1 for rounding up to 8 loaves.
M1 for finding the x-coordinate of the vertex (x=36).
A1 for calculating the maximum profit of $204.

(a) Triangle \(OAP\) is a right-angled triangle with \(\angle AOP = 90^\circ\).

Using basic trigonometry:
\(\tan(35^\circ) = \frac{h}{15}\)

\(h = 15 \tan(35^\circ) \approx 10.503\text{ m}\)

So \(h \approx 10.5\text{ m}\).

(b) The angle \(\angle AOB\) is the difference between the bearings:
\(\angle AOB = 130^\circ - 40^\circ = 90^\circ\)

Since triangle \(AOB\) is a right-angled triangle at \(O\), we apply Pythagoras' theorem:
\(AB = \sqrt{OA^2 + OB^2} = \sqrt{15^2 + 22^2} = \sqrt{225 + 484} = \sqrt{709} \approx 26.627\text{ m}\)

So the distance \(AB \approx 26.6\text{ m}\).

M1 for setting up the tangent ratio: \(\tan(35^\circ) = h/15\).
A1 for \(h \approx 10.5\text{ m}\).
M1 for finding \(\angle AOB = 90^\circ\).
M1 for using Pythagoras' theorem.
A1 for \(AB \approx 26.6\text{ m}\).

(a) Using the distance formula, \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\):

\(PA = \sqrt{(5-2)^2 + (6-8)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.61\text{ km}\)

\(PB = \sqrt{(5-8)^2 + (6-10)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.00\text{ km}\)

\(PC = \sqrt{(5-4)^2 + (6-2)^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.12\text{ km}\)

\(PD = \sqrt{(5-10)^2 + (6-4)^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39\text{ km}\)

Since \(PA\) is the shortest distance, Station A is the closest.

(b) To find the perpendicular bisector of \(AC\):

1. Find the midpoint of \(AC\):
\(M = \left(\frac{2+4}{2}, \frac{8+2}{2}\right) = (3, 5)\)

2. Find the gradient of \(AC\):
\(m = \frac{2-8}{4-2} = -3\)

3. The gradient of the perpendicular bisector is:
\(m_{\perp} = -\frac{1}{-3} = \frac{1}{3}\)

4. Use the point-slope form with \(M(3, 5)\):
\(y - 5 = \frac{1}{3}(x - 3) \implies y = \frac{1}{3}x + 4\) (or \(x - 3y + 12 = 0\)).

M1 for calculating at least two distances correctly.
A1 for identifying Station A as the closest.
M1 for finding the midpoint of \(AC\) as \((3, 5)\).
M1 for finding the gradient of \(AC\) as \(-3\).
M1 for finding the perpendicular gradient as \(1/3\).
A1 for the correct equation of the perpendicular bisector.

Let \(X\) represent the mass of an apple, where \(X \sim N(150, 18^2)\).

(a) We seek \(P(140 < X < 165)\).

Using a graphics display calculator (GDC) with standard normal cumulative distribution function (normalcdf):
\(P(140 < X < 165) \approx 0.50842\)

Correct to 3 significant figures, the probability is \(0.508\).

(b) We are given that \(P(X < w) = 0.15\).

Using the inverse normal function on a GDC:
\(w \approx 131.34\)

Correct to 3 significant figures, \(w = 131\text{ g}\).

M1 for writing the probability statement \(P(140 < X < 165)\).
A1 for \(0.508\).
M1 for writing the probability statement \(P(X < w) = 0.15\) or showing standardisation step.
A1 for finding \(131.34\).
A1 for rounding to \(131\).

(a) The null hypothesis \(H_0\): Preferred study environment is independent of grade level (or there is no association between preferred study environment and grade level).

(b) Total for Grade 11 row = \(40 + 35 + 25 = 100\).
Total for Library column = \(40 + 30 = 70\).
Grand total \(N = 200\).

\(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{100 \times 70}{200} = 35\).

(c) Performing a chi-squared two-way test on a GDC using the given table, we obtain:
\(\chi^2 \approx 4.631\)
\(p\text{-value} \approx 0.09871\)

Correct to 3 significant figures, the \(p\)-value is \(0.0987\).

R1 for stating a correct null hypothesis mentioning independence or lack of association.
M1 for using the formula \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\).
A1 for 35.
M1 for setting up the matrix and running the chi-squared test on the GDC.
A1 for \(p = 0.0987\).

(a) We rewrite the expression as:
\(A(r) = 2\pi r^2 + 1000r^{-1}\)

Differentiating with respect to \(r\):
\(A'(r) = 4\pi r - 1000r^{-2} = 4\pi r - \frac{1000}{r^2}\)

(b) To minimize the surface area, we set \(A'(r) = 0\):

\(4\pi r - \frac{1000}{r^2} = 0\)

\(4\pi r^3 = 1000\)

\(r^3 = \frac{250}{\pi}\)

\(r = \left(\frac{250}{\pi}\right)^{1/3} \approx 4.30127\text{ cm}\)

Correct to 3 significant figures, \(r = 4.30\text{ cm}\).

M1 for writing \(1000/r\) as \(1000r^{-1}\) (or using the quotient rule).
A1 for \(4\pi r\).
A1 for \(-\frac{1000}{r^2}\).
M1 for setting their \(A'(r) = 0\).
A1 for isolating \(r^3 = 250/\pi\) or equivalent.
A1 for \(4.30\text{ cm}\).

(a) The acceleration \(a(t)\) is the derivative of the velocity with respect to time:
\(a(t) = v'(t) = -t + 4\)

At \(t = 3\):
\(a(3) = -3 + 4 = 1\text{ m s}^{-2}\)

(b) The total distance is given by \(\int_{0}^{6} |v(t)| dt\).
Since the velocity function is positive on the interval \([0, 6]\) (the roots of \(v(t) = 0\) are at \(t \approx -0.47\) and \(t \approx 8.47\)), we can evaluate the definite integral directly:

\(\int_{0}^{6} (-0.5t^2 + 4t + 2) dt = \left[ -\frac{0.5t^3}{3} + 2t^2 + 2t \right]_0^6\)

\(= \left( -\frac{216}{6} + 2(36) + 2(6) \right) - 0\)

\(= -36 + 72 + 12 = 48\text{ m}\).

M1 for differentiating \(v(t)\) to find \(a(t)\).
A1 for \(1\text{ m s}^{-2}\).
M1 for setting up the definite integral for distance: \(\int_{0}^{6} v(t) dt\).
M1 for finding the correct antiderivative \(-\frac{1}{6}t^3 + 2t^2 + 2t\).
A1 for \(48\text{ m}\).

(a) Using the compound interest formula, \(FV = PV\left(1 + \frac{r}{k}\right)^{kn}\). Here, \(PV = 5000\), \(r = 0.042\), \(k = 12\), and \(n = 5\). Therefore, \(FV = 5000\left(1 + \frac{0.042}{12}\right)^{60} = 5000(1.0035)^{60} \approx 6166.47\) USD. (b) For the investment to double, we need \(FV \ge 10000\). So, \(5000(1.0035)^{12t} \ge 10000\) which simplifies to \(1.0035^{12t} \ge 2\). Taking the natural logarithm of both sides: \(12t \ln(1.0035) \ge \ln(2)\), which gives \(t \ge \frac{\ln(2)}{12 \ln(1.0035)} \approx 16.53\) years. Since we need the minimum number of complete years, we round up to the next integer, which is \(17\) years.

[M1] for substituting correct values into the compound interest formula. [A1] for 6166.47. [M1] for setting up the equation or inequality to double the investment. [M1] for solving for t using logarithms or a GDC. [A1] for t approximately 16.5. [A1] for 17 years.

(a) The midpoint of \(AB\) is \(\left(\frac{2+8}{2}, \frac{8+6}{2}\right) = (5, 7)\). The gradient of \(AB\) is \(m = \frac{6-8}{8-2} = -\frac{1}{3}\). The gradient of the perpendicular bisector is the negative reciprocal, which is \(3\). The equation is \(y - 7 = 3(x - 5)\), which simplifies to \(y = 3x - 8\). (b) To find the circumcenter, we find the intersection of the two perpendicular bisectors: \(y = 3x - 8\) and \(y = x - 1\). Equating them: \(3x - 8 = x - 1 \Rightarrow 2x = 7 \Rightarrow x = 3.5\). Substituting \(x = 3.5\) into the second equation: \(y = 3.5 - 1 = 2.5\). Thus, the coordinates are \((3.5, 2.5)\).

[M1] for finding the midpoint (5, 7) and the gradient of AB. [M1] for finding the perpendicular gradient of 3. [A1] for the equation y = 3x - 8. [M1] for setting up the system of linear equations. [M1] for solving for x. [A1] for the correct coordinates (3.5, 2.5).

(a) Let \(X\) be the weight of a chocolate bar. \(X \sim N(150, 4^2)\). Using a graphic display calculator (GDC), we find \(P(X > 155) \approx 0.10565\). Correct to three significant figures, the probability is \(0.106\). (b) Let \(Y\) be the number of chocolate bars in a pack of \(6\) that weigh more than \(155\) grams. \(Y \sim B(6, 0.10565)\). We want to find \(P(Y \ge 2) = 1 - P(Y \le 1)\). Using a GDC for binomial cumulative distribution, we find \(P(Y \le 1) \approx 0.87373\). Thus, \(P(Y \ge 2) = 1 - 0.87373 = 0.12627\), which is \(0.126\) correct to three significant figures.

[M1] for setting up the normal distribution parameters. [A1] for 0.106. [M1] for identifying the binomial distribution model B(6, 0.10565). [M1] for setting up the calculation 1 - P(Y <= 1). [M1] for finding P(Y <= 1) = 0.874. [A1] for the final probability 0.126.

(a) The volume of a cylinder is given by \(V = \pi r^2 h = 500\), which gives \(h = \frac{500}{\pi r^2}\). The total surface area of a closed cylinder is \(A = 2\pi r^2 + 2\pi r h\). Substituting the expression for \(h\): \(A = 2\pi r^2 + 2\pi r\left(\frac{500}{\pi r^2}\right) = 2\pi r^2 + \frac{1000}{r}\). (b) To find the minimum surface area, we differentiate \(A\) with respect to \(r\): \(\frac{dA}{dr} = 4\pi r - \frac{1000}{r^2}\). Setting the derivative equal to zero: \(4\pi r - \frac{1000}{r^2} = 0 \Rightarrow 4\pi r^3 = 1000 \Rightarrow r^3 = \frac{250}{\pi}\). Solving for \(r\) gives \(r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30\) cm. (c) Substituting \(r \approx 4.3012\) back into the surface area equation: \(A = 2\pi(4.3012)^2 + \frac{1000}{4.3012} \approx 349\text{ cm}^2\).

[M1] for expressing h in terms of r. [A1] for showing the given formula for A. [M1] for finding the derivative dA/dr. [M1] for setting the derivative to zero. [A1] for r = 4.30 cm. [A1] for 349 cm^2.

(a) Option A has a nominal annual rate of 4.2% compounded monthly, so the monthly rate is \(0.042 / 12 = 0.0035\). The value after 5 years (60 months) is \(50000 \times (1.0035)^{60} \approx 61651.81\). To 3 significant figures, this is $61,700. (b)(i) The common ratio is \(r = \frac{53045}{51500} = 1.03\). (ii) The value after 5 years is \(u_5 = u_1 \times r^4 = 51500 \times (1.03)^4 \approx 57963.70\). To 3 significant figures, this is $58,000. (c)(i) The monthly interest rate is 0.35%, so each month the balance increases by a factor of 1.0035. After withdrawing $200, the new balance is \(A_n = A_{n-1}(1.0035) - 200\), with \(A_0 = 50000\). (ii) Using the recurrence relation or a financial calculator (TVM Solver with \(N=60, I\%=4.2, PV=-50000, PMT=200, P/Y=12, C/Y=12\)), we find \(A_{60} = 50000(1.0035)^{60} - 200 \times \frac{1.0035^{60}-1}{0.0035} \approx 48335.45\). To 3 significant figures, this is $48,300. (iii) Set \(A_n = 40000\). Thus, \(40000 = 50000(1.0035)^n - 200 \times \frac{1.0035^n-1}{0.0035}\). Simplifying this equation: \(40000 = 50000(1.0035)^n - \frac{200}{0.0035}(1.0035^n) + \frac{200}{0.0035}\). Multiplying by 0.0035: \(140 = 175(1.0035)^n - 200(1.0035)^n + 200 \implies -60 = -25(1.0035)^n \implies 1.0035^n = 2.4\). Taking the natural logarithm: \(n = \frac{\ln(2.4)}{\ln(1.0035)} \approx 250.57\). Since we want the balance to first fall below $40,000, we round up to 251 months.

(a) M1 for setting up the compounding formula: \(50000 \times (1 + 0.042/12)^{60}\), A1 for 61651.81, A1 for $61,700. (b)(i) M1 for \(53045 / 51500\), A1 for 1.03. (ii) M1 for \(51500 \times (1.03)^4\), A1 for $58,000. (c)(i) R1 for explaining the factor 1.0035 (from 1 + 0.042/12), R1 for explaining the subtraction of 200. (ii) M1 for substituting \(n=60\) into the formula or setting up TVM solver, A1 for 48335.45, A1 for $48,300. (iii) M1 for setting up the equation \(A_n = 40000\), M1 for simplifying to \(1.0035^n = 2.4\), A1 for \(n \approx 250.57\), A1 for 251 months.

(a) The volume of a cylinder is \(V = \pi r^2 h\). Setting \(V = 150\), we have \(150 = \pi r^2 h \implies h = \frac{150}{\pi r^2}\). (b) The total cost \(C\) is the cost of the two circular bases plus the cost of the curved side. The area of the two bases is \(2\pi r^2\), and the area of the curved side is \(2\pi r h\). Therefore, \(C(r) = 80(2\pi r^2) + 50(2\pi r h) = 160\pi r^2 + 100\pi r h\). Substituting \(h = \frac{150}{\pi r^2}\) gives \(C(r) = 160\pi r^2 + 100\pi r \left(\frac{150}{\pi r^2}\right) = 160\pi r^2 + \frac{15000}{r}\). (c) Differentiating \(C(r)\) with respect to \(r\): \(C'(r) = 320\pi r - \frac{15000}{r^2}\). (d)(i) Set \(C'(r) = 0 \implies 320\pi r - \frac{15000}{r^2} = 0 \implies 320\pi r^3 = 15000 \implies r^3 = \frac{15000}{320\pi} = \frac{375}{8\pi} \approx 14.9208\). Taking the cube root, \(r \approx 2.4619\text{ m}\). To 3 significant figures, \(r = 2.46\text{ m}\). (ii) The minimum cost is \(C(2.4619) = 160\pi (2.4619)^2 + \frac{15000}{2.4619} \approx 3046.46 + 6092.93 = 9139.39\). To the nearest dollar, this is $9,139. (e) The volume of the liquid in the tank is \(V_{\text{liquid}} = \pi r^2 y\), where \(y\) is the depth of the liquid. Since \(r\) is constant, differentiating both sides with respect to time \(t\) gives \(\frac{dV}{dt} = \pi r^2 \frac{dy}{dt}\). We are given \(\frac{dV}{dt} = 5\). Using \(r \approx 2.4619\), we have \(5 = \pi (2.4619)^2 \frac{dy}{dt} \implies \frac{dy}{dt} = \frac{5}{\pi (2.4619)^2} \approx 0.2626\text{ m min}^{-1}\). To 3 significant figures, the rate of change of depth is \(0.263\text{ m min}^{-1}\).

(a) M1 for \(V = \pi r^2 h\), A1 for rearranging to \(h = \frac{150}{\pi r^2}\). (b) M1 for expressing cost as \(80(2\pi r^2) + 50(2\pi r h)\), M1 for substituting \(h\), A1 for showing the final expression. (c) M1 for power rule on \(r^2\) term, M1 for power rule on \(r^{-1}\) term, A1 for \(320\pi r - \frac{15000}{r^2}\). (d)(i) M1 for setting \(C'(r) = 0\), A1 for \(r^3 \approx 14.92\), A1 for \(r \approx 2.46\text{ m}\). (ii) M1 for substituting \(r\) back into \(C(r)\), A1 for $9,139. (e) M1 for relating rates: \(\frac{dV}{dt} = \pi r^2 \frac{dy}{dt}\), M1 for substituting \(r\) and \(\frac{dV}{dt} = 5\), A1 for \(0.263\text{ m min}^{-1}\).

(a)(i) Entering the data into a GDC, the Pearson's product-moment correlation coefficient is \(r \approx -0.979507\). To 3 significant figures, \(r = -0.980\). (ii) The correlation is strong and negative. (b)(i) Using the GDC, the equation of the regression line of \(y\) on \(x\) is \(y = mx + c\), where \(m \approx -9.18174\) and \(c \approx 87.4800\). To 3 significant figures, \(y = -9.18x + 87.5\). (ii) Substituting \(x = 5.0\) into the regression equation: \(y = -9.18174(5.0) + 87.4800 \approx 41.5713\). To 3 significant figures, the estimated concentration is 41.6 ppm. (c)(i) Let \(X\) be the pollutant concentration. \(X \sim N(35, 8^2)\). We want to find \(P(X > 45)\). Using a GDC, \(P(X > 45) \approx 0.10565\). To 3 significant figures, \(P(X > 45) = 0.106\). (ii) Let \(Y\) be the number of highly contaminated soil samples. \(Y \sim B(3, 0.10565)\). We want \(P(Y \ge 1) = 1 - P(Y = 0) = 1 - (1 - 0.10565)^3 \approx 1 - (0.89435)^3 \approx 1 - 0.71536 = 0.28464\). To 3 significant figures, the probability is 0.285. (iii) We want \(P(X < 50 | X > 45) = \frac{P(45 < X < 50)}{P(X > 45)}\). Using GDC, \(P(45 < X < 50) \approx 0.075254\). Thus, the conditional probability is \(\frac{0.075254}{0.10565} \approx 0.71229\). To 3 significant figures, the probability is 0.712.

(a)(i) A2 for -0.980 (Award A1 for -0.98). (ii) A1 for "strong", A1 for "negative". (b)(i) A1 for gradient -9.18, A1 for y-intercept 87.5. (ii) M1 for substituting x = 5.0 into their equation, A1 for 41.6. (c)(i) M1 for setting up normal distribution parameters, A2 for 0.106. (ii) M1 for setting up binomial distribution or finding 1 - (1-p)^3, A1 for substituting their p, A1 for 0.285. (iii) M1 for using conditional probability formula, A1 for 0.712.

(a) Distance \(d(A, B) = \sqrt{(8-2)^2 + (11-9)^2} = \sqrt{6^2 + 2^2} = \sqrt{40} \approx 6.3245\text{ km}\). To 3 significant figures, the distance is 6.32 km. (b) Midpoint of \(AB = \left(\frac{2+8}{2}, \frac{9+11}{2}\right) = (5, 10)\). The gradient of \(AB\) is \(m = \frac{11-9}{8-2} = \frac{2}{6} = \frac{1}{3}\). The gradient of the perpendicular bisector is \(-\frac{1}{m} = -3\). The equation is \(y - 10 = -3(x - 5) \implies y = -3x + 25\). (c)(i) Solve the system: \(y = -3x + 25\) and \(y = x - 2\). We get \(-3x + 25 = x - 2 \implies 4x = 27 \implies x = 6.75\). Substituting \(x\): \(y = 6.75 - 2 = 4.75\). So the coordinates of \(P\) are \((6.75, 4.75)\). (ii) Any point on the perpendicular bisector of \(AB\) is equidistant from \(A\) and \(B\). Any point on the perpendicular bisector of \(BC\) is equidistant from \(B\) and \(C\). Since \(P\) lies on both, it must be equidistant from \(A\), \(B\), and \(C\). (d)(i) Distance from \(T(6.75, 4.75)\) to \(B(8, 11)\) is \(d(T, B) = \sqrt{(8-6.75)^2 + (11-4.75)^2} = \sqrt{1.25^2 + 6.25^2} = \sqrt{40.625} \approx 6.3738\text{ km}\). To 3 significant figures, this is 6.37 km. (ii) The circle centered at \(P(6.75, 4.75)\) passing through \(A, B, C\) has radius squared \(R^2 = 40.625\). Its equation is \((x - 6.75)^2 + (y - 4.75)^2 = 40.625\). (e) Distance from \(T(6.75, 4.75)\) to \(D(4, 1)\) is \(d(T, D) = \sqrt{(4-6.75)^2 + (1-4.75)^2} = \sqrt{(-2.75)^2 + (-3.75)^2} = \sqrt{21.625} \approx 4.6503\text{ km}\). Since \(4.65\text{ km} < 6.37\text{ km}\), the new patrol tower at \(T\) is closer to station \(D\) than it is to station \(B\).

(a) M1 for using distance formula, A1 for 6.32. (b) M1 for finding the midpoint (5, 10), M1 for finding perpendicular gradient -3, A1 for \(y = -3x + 25\). (c)(i) M1 for setting equations equal, A1 for x = 6.75, A1 for y = 4.75. (ii) R1 for stating that P lies on both perpendicular bisectors. (d)(i) M1 for distance formula, A1 for 6.37. (ii) M1 for using circle formula, A1 for \((x-6.75)^2 + (y-4.75)^2 = 40.6\). (e) M1 for calculating distance from T to D, A1 for finding \(d(T, D) \approx 4.65\text{ km}\), R1 for explicit comparison and correct conclusion.

(a)(i) The amplitude \(a\) is half the difference between maximum and minimum temperatures: \(a = \frac{28 - 12}{2} = 8\). (ii) The vertical shift \(d\) is the average of the maximum and minimum temperatures: \(d = \frac{28 + 12}{2} = 20\). (iii) The period is 24 hours. Therefore, \(b = \frac{2\pi}{24} = \frac{\pi}{12}\). (iv) A cosine curve with \(a > 0\) has its maximum when the cosine argument is zero. Since the maximum occurs at \(t = 14\), we set \(c = 14\). (b) Combining these: \(T(t) = 8 \cos\left(\frac{\pi}{12}(t - 14)\right) + 20\). (c)(i) At 08:00, \(t = 8\). Substituting: \(T(8) = 8 \cos\left(\frac{\pi}{12}(8 - 14)\right) + 20 = 8 \cos(-\pi/2) + 20 = 20^\circ\text{C}\). (ii) Solve \(8 \cos\left(\frac{\pi}{12}(t - 14)\right) + 20 = 24 \implies \cos\left(\frac{\pi}{12}(t - 14)\right) = 0.5\). The solutions in the interval \(0 \le t \le 24\) are given by: \(\frac{\pi}{12}(t - 14) = -\frac{\pi}{3} \implies t = 10\), and \(\frac{\pi}{12}(t - 14) = \frac{\pi}{3} \implies t = 18\). So the temperatures are exactly \(24^\circ\text{C}\) at \(t = 10\) and \(t = 18\). (d)(i) Differentiating \(T(t)\) using the chain rule: \(T'(t) = -8 \sin\left(\frac{\pi}{12}(t - 14)\right) \times \frac{\pi}{12} = -\frac{2\pi}{3} \sin\left(\frac{\pi}{12}(t - 14)\right)\). (ii) At 10:00, \(t = 10\). Substituting: \(T'(10) = -\frac{2\pi}{3} \sin\left(\frac{\pi}{12}(10 - 14)\right) = -\frac{2\pi}{3} \sin(-\pi/3) = \frac{\pi\sqrt{3}}{3} \approx 1.8138\). To 3 significant figures, \(T'(10) \approx 1.81^\circ\text{C}\text{ hour}^{-1}\). This means that at 10:00, the temperature in the greenhouse is increasing at a rate of \(1.81^\circ\text{C}\) per hour.

(a)(i) M1 for \(\frac{28 - 12}{2}\), A1 for showing it equals 8. (ii) A1 for 20. (iii) M1 for \(\frac{2\pi}{24}\), A1 for \(\frac{\pi}{12}\). (iv) A1 for 14. (b) A1 for writing the correct formula. (c)(i) M1 for substituting \(t = 8\), A1 for 20. (ii) M1 for setting equation equal to 24, A1 for \(t = 10\), A1 for \(t = 18\). (d)(i) M1 for using chain rule, A1 for \(-\frac{2\pi}{3} \sin\left(\frac{\pi}{12}(t - 14)\right)\). (ii) A1 for 1.81, R1 for explaining that the temperature is increasing at this rate at 10:00.