IB DP · Thinka-original Practice Paper

2024 IB DP Mathematics - Analysis and Approaches Practice Paper with Answers

Thinka Nov 2024 SL (TZ1) IB Diploma Programme-Style Mock — Mathematics - Analysis and Approaches

160 marks180 mins2024

An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 SL (TZ1) IB Diploma Programme Mathematics - Analysis and Approaches paper. Not affiliated with or reproduced from IB.

Paper 1 Section A

Answer all questions. Answers must be written within the answer boxes provided.

6 Question · 35 marks

Question 1 · Short Answer

6 marks

Consider the expansion of $\left(x + \frac{2}{x^2}\right)^6 (x^3 - 1)$.

Find the term independent of $x$ in this expansion.

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Worked solution

The general term in the expansion of $\left(x + \frac{2}{x^2}\right)^6$ is:
$T_{r+1} = \binom{6}{r} x^{6-r} \left(\frac{2}{x^2}\right)^r = \binom{6}{r} 2^r x^{6-3r}$ for $r = 0, 1, \dots, 6$.

Multiplying this expansion by $(x^3 - 1)$:
$(x^3 - 1) \sum_{r=0}^{6} \binom{6}{r} 2^r x^{6-3r} = \sum_{r=0}^{6} \binom{6}{r} 2^r x^{9-3r} - \sum_{r=0}^{6} \binom{6}{r} 2^r x^{6-3r}$.

To find the term independent of $x$, we identify terms with an exponent on $x$ equal to 0:
- From the first summation: $9 - 3r = 0 \implies r = 3$.
The corresponding term is $\binom{6}{3} 2^3 = 20 \times 8 = 160$.
- From the second summation: $6 - 3r = 0 \implies r = 2$.
The corresponding term is $-\binom{6}{2} 2^2 = -15 \times 4 = -60$.

Combining these, the term independent of $x$ is:
$160 - 60 = 100$.

Marking scheme

M1 for attempting to write the general term of the expansion of $\left(x + \frac{2}{x^2}\right)^6$.
A1 for the correct general term $\binom{6}{r} 2^r x^{6-3r}$.
M1 for setting up equations for the powers of $x$ that lead to a constant (i.e., $9-3r=0$ and $6-3r=0$).
A1 for identifying $r = 3$ and $r = 2$.
M1 for substituting and evaluating the two constant terms to get $160$ and $-60$.
A1 for the correct final answer of $100$.

Question 2 · Short Answer

6 marks

The function $f$ is defined by $f(x) = \frac{ax + b}{x - 2}$ for $x \in \mathbb{R}, x \neq 2$, where $a, b \in \mathbb{R}$.

(a) Given that $f(3) = 5$, find an equation relating $a$ and $b$.

(b) Find an expression for $f^{-1}(x)$ in terms of $a$ and $b$.

(c) Given that $f^{-1}(x) = f(x)$ for all $x$ in the domain of $f$, find the value of $a$ and the value of $b$.

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Worked solution

(a) Substituting $x = 3$ and $f(3) = 5$ into the function definition:
$5 = \frac{3a + b}{3 - 2} \implies 3a + b = 5$.

(b) To find $f^{-1}(x)$, let $y = \frac{ax + b}{x - 2}$.
Rearranging to make $x$ the subject:
$y(x - 2) = ax + b$
$xy - 2y = ax + b$
$xy - ax = 2y + b$
$x(y - a) = 2y + b$
$x = \frac{2y + b}{y - a}$.
Thus, $f^{-1}(x) = \frac{2x + b}{x - a}$.

(c) Since $f^{-1}(x) = f(x)$ for all $x$ in the domain, we compare:
$\frac{2x + b}{x - a} = \frac{ax + b}{x - 2}$.
Comparing the denominators, we find $a = 2$.
Substituting $a = 2$ into our equation from part (a):
$3(2) + b = 5 \implies 6 + b = 5 \implies b = -1$.

Marking scheme

(a) A1 for $3a + b = 5$ (or any equivalent form).

(b) M1 for attempting to swap variables or rearrange to make $x$ the subject.
A1 for correct algebraic factorization to group $x$ terms, e.g., $x(y - a) = 2y + b$.
A1 for the correct expression $f^{-1}(x) = \frac{2x + b}{x - a}$.

(c) M1 for equating $f^{-1}(x) = f(x)$ to solve for $a = 2$.
A1 for both correct values: $a = 2$ and $b = -1$.

Question 3 · Short Answer

6 marks

Consider a triangle $ABC$ where $AB = 5\text{ cm}$, $BC = 7\text{ cm}$, and $\cos(\hat{B}) = \frac{1}{5}$.

(a) Find the length of $AC$.

(b) Find the exact value of $\sin(\hat{A})$.

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Worked solution

(a) Using the cosine rule on $\triangle ABC$:
$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\hat{B})$
$AC^2 = 5^2 + 7^2 - 2(5)(7)\left(\frac{1}{5}\right)$
$AC^2 = 25 + 49 - 14 = 60$.
Thus, $AC = \sqrt{60} = 2\sqrt{15}\text{ cm}$.

(b) Since $\cos(\hat{B}) = \frac{1}{5} > 0$, angle $\hat{B}$ is acute.
$\sin(\hat{B}) = \sqrt{1 - \cos^2(\hat{B})} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$.
Using the sine rule:
$\frac{\sin(\hat{A})}{BC} = \frac{\sin(\hat{B})}{AC} \implies \sin(\hat{A}) = \frac{BC \sin(\hat{B})}{AC}$
$\sin(\hat{A}) = \frac{7 \cdot \frac{2\sqrt{6}}{5}}{2\sqrt{15}} = \frac{7\sqrt{6}}{5\sqrt{15}} = \frac{7}{5}\sqrt{\frac{2}{5}} = \frac{7\sqrt{2}}{5\sqrt{5}} = \frac{7\sqrt{10}}{25}$.

Marking scheme

(a) M1 for substituting correctly into the cosine rule.
A1 for $AC^2 = 60$.
A1 for the exact length $AC = 2\sqrt{15}$ (accept $\sqrt{60}$).

(b) M1 for finding $\sin(\hat{B}) = \frac{2\sqrt{6}}{5}$ (or equivalent).
M1 for attempting to use the sine rule to express $\sin(\hat{A})$.
A1 for the exact simplified value $\sin(\hat{A}) = \frac{7\sqrt{10}}{25}$ (or any exact equivalent, e.g., $\frac{7\sqrt{2}}{5\sqrt{5}}$).

Question 4 · Short Answer

6 marks

Consider the curve with equation $y = e^{-x} (x^2 - 3x + 1)$ for $x \in \mathbb{R}$.

(a) Find $\frac{dy}{dx}$.

(b) Find the coordinates of the points on the curve where the tangent is horizontal.

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Worked solution

(a) Using the product rule with $u = e^{-x}$ and $v = x^2 - 3x + 1$:
$u' = -e^{-x}$
$v' = 2x - 3$

$\frac{dy}{dx} = u v' + v u' = e^{-x}(2x - 3) + (x^2 - 3x + 1)(-e^{-x})$
$\frac{dy}{dx} = e^{-x}(2x - 3 - x^2 + 3x - 1) = e^{-x}(-x^2 + 5x - 4)$.

(b) A horizontal tangent occurs when $\frac{dy}{dx} = 0$.
Since $e^{-x} > 0$ for all real $x$, we solve:
$-x^2 + 5x - 4 = 0 \implies x^2 - 5x + 4 = 0$
$(x - 1)(x - 4) = 0$.
So, $x = 1$ or $x = 4$.

Substituting these $x$-values back into the original curve equation to find the $y$-coordinates:
- When $x = 1$: $y = e^{-1}(1^2 - 3(1) + 1) = -e^{-1}$.
- When $x = 4$: $y = e^{-4}(4^2 - 3(4) + 1) = 5e^{-4}$.

Thus, the coordinates are $(1, -e^{-1})$ and $(4, 5e^{-4})$.

Marking scheme

(a) M1 for an attempt to use the product rule.
A1 for correct derivatives of both components: $-e^{-x}$ and $2x - 3$.
A1 for simplifying to $e^{-x}(-x^2 + 5x - 4)$ (or equivalent).

(b) M1 for setting their derivative equal to 0 and solving the resulting quadratic equation.
A1 for identifying the correct $x$-values, $x = 1$ and $x = 4$.
A1 for obtaining both correct coordinates: $(1, -e^{-1})$ and $(4, 5e^{-4})$ (accept equivalent exact forms like $(1, -\frac{1}{e})$ and $(4, \frac{5}{e^4})$).

Question 5 · Short Answer

5 marks

A discrete random variable $X$ has the following probability distribution:

$P(X = x) = \begin{cases} 10k^2, & x = 1 \\ 8k, & x = 2 \\ k, & x = 3 \end{cases}$

where $k$ is a positive constant.

(a) Find the value of $k$.

(b) Find the expected value $\text{E}(X)$.

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Worked solution

(a) Since the sum of all probabilities in a probability distribution must equal 1:
$10k^2 + 8k + k = 1 \implies 10k^2 + 9k - 1 = 0$.

Factorizing the quadratic equation:
$(10k - 1)(k + 1) = 0$.
This gives the roots $k = 0.1$ or $k = -1$.
Since $k$ must be a positive constant, we reject $k = -1$. Thus, $k = 0.1$.

(b) Substituting $k = 0.1$ into the probability distribution table:
$P(X = 1) = 10(0.1)^2 = 0.1$
$P(X = 2) = 8(0.1) = 0.8$
$P(X = 3) = 0.1$

Now, compute $\text{E}(X)$:
$\text{E}(X) = \sum x P(X = x) = 1(0.1) + 2(0.8) + 3(0.1)$
$\text{E}(X) = 0.1 + 1.6 + 0.3 = 2.0$.

Marking scheme

(a) M1 for equating the sum of probabilities to 1.
A1 for factorizing the quadratic correctly to find the roots $k = 0.1$ and $k = -1$.
A1 for choosing $k = 0.1$ and explicitly rejecting $k = -1$ because $k > 0$.

(b) M1 for applying the expected value formula $\text{E}(X) = \sum x P(X=x)$.
A1 for the correct final answer of $2$.

Question 6 · Short Answer

6 marks

Let $f(x) = \frac{e^x}{e^x + 3}$ for $x \in \mathbb{R}$.

(a) Show that the derivative of $\ln(e^x + 3)$ is $f(x)$.

(b) Find the exact area of the region bounded by the curve $y = f(x)$, the $x$-axis, and the vertical lines $x = 0$ and $x = \ln 2$.

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Worked solution

(a) Let $g(x) = \ln(e^x + 3)$.
Using the chain rule:
$g'(x) = \frac{1}{e^x + 3} \cdot \frac{d}{dx}(e^x + 3) = \frac{e^x}{e^x + 3} = f(x)$.
Thus, the derivative of $\ln(e^x + 3)$ is indeed $f(x)$.

(b) The area of the region is given by the definite integral:
$\text{Area} = \int_{0}^{\ln 2} f(x) \, dx$.

Using the result from part (a), an antiderivative of $f(x)$ is $\ln(e^x + 3)$:
$\text{Area} = \left[ \ln(e^x + 3) \right]_{0}^{\ln 2}$
$\text{Area} = \ln(e^{\ln 2} + 3) - \ln(e^0 + 3)$
$\text{Area} = \ln(2 + 3) - \ln(1 + 3)$
$\text{Area} = \ln 5 - \ln 4 = \ln\left(\frac{5}{4}\right)$.

Marking scheme

(a) M1 for attempting to differentiate $\ln(e^x + 3)$ using the chain rule.
A1 for showing clearly that the derivative simplifies to $\frac{e^x}{e^x + 3}$.

(b) M1 for setting up the area as the definite integral $\int_{0}^{\ln 2} f(x) \, dx$.
A1 for stating the antiderivative is $\ln(e^x + 3)$.
M1 for correctly substituting the limits $\ln 2$ and $0$ into the antiderivative.
A1 for the exact value of $\ln\left(\frac{5}{4}\right)$ (accept exact equivalents like $\ln 1.25$ or $\ln 5 - \ln 4$).

Paper 1 Section B

Answer all questions in the answer booklet provided. Start each question on a new page.

3 Question · 45 marks

Question 1 · Extended Response

15 marks

Let $ f(x) = x e^{1-x^2} $, for $ x \in \mathbb{R} $.

(a) Find $ f'(x) $. [3 marks]

(b) Find the coordinates of the local minimum and the local maximum of the graph of $ f $. [4 marks]

(c) Find the equation of the tangent to the curve of $ f $ at the point where $ x = 1 $. [3 marks]

(d) Find the area of the region enclosed by the curve of $ f $, the $ x $-axis, and the vertical line $ x = 1 $. [5 marks]

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Worked solution

(a) Using the product rule and chain rule:

$ f'(x) = (1)(e^{1-x^2}) + (x)(-2x e^{1-x^2}) $

$ f'(x) = e^{1-x^2}(1 - 2x^2) $

(b) Set $ f'(x) = 0 $:

$ e^{1-x^2}(1 - 2x^2) = 0 \implies 1 - 2x^2 = 0 \implies x = \pm\frac{1}{\sqrt{2}} $

Substitute $ x $ back into $ f(x) $:

$ f\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} e^{1 - 1/2} = \sqrt{\frac{e}{2}} $

$ f\left(-\frac{1}{\sqrt{2}}\right) = -\frac{1}{\sqrt{2}} e^{1 - 1/2} = -\sqrt{\frac{e}{2}} $

Using a sign diagram for $ f'(x) $:

For $ x < -\frac{1}{\sqrt{2}} $, $ f'(x) < 0 $; for $ -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} $, $ f'(x) > 0 $; for $ x > \frac{1}{\sqrt{2}} $, $ f'(x) < 0 $.

Thus, the local minimum is at $ \left(-\frac{1}{\sqrt{2}}, -\sqrt{\frac{e}{2}}\right) $ and the local maximum is at $ \left(\frac{1}{\sqrt{2}}, \sqrt{\frac{e}{2}}\right) $.

(c) At $ x = 1 $:

$ f(1) = 1 \cdot e^0 = 1 $

$ f'(1) = e^0(1 - 2) = -1 $

The equation of the tangent is: $ y - 1 = -1(x - 1) \implies y = -x + 2 $.

(d) The region is enclosed between $ x = 0 $, $ x = 1 $, the curve, and the x-axis.

$ \text{Area} = \int_{0}^{1} x e^{1-x^2} \mathrm{d}x $

Let $ u = 1 - x^2 $, then $ \mathrm{d}u = -2x \mathrm{d}x \implies x \mathrm{d}x = -\frac{1}{2} \mathrm{d}u $.

When $ x = 0 $, $ u = 1 $. When $ x = 1 $, $ u = 0 $.

$ \text{Area} = \int_{1}^{0} -\frac{1}{2} e^u \mathrm{d}u = \frac{1}{2} \int_{0}^{1} e^u \mathrm{d}u = \frac{1}{2} [e^u]_0^1 = \frac{1}{2}(e - 1) $.

Marking scheme

(a) [3 marks]

M1: Attempt to use the product rule.

A1: Correct derivative of the exponential part, $ -2x e^{1-x^2} $.

A1: Correct simplified derivative: $ e^{1-x^2}(1 - 2x^2) $.

(b) [4 marks]

M1: Setting $ f'(x) = 0 $.

A1: Correct $ x $-coordinates: $ x = \pm\frac{1}{\sqrt{2}} $ (or equivalent).

A1: Correct $ y $-coordinates: $ y = \pm\sqrt{\frac{e}{2}} $ (or equivalent).

R1: Clear justification of local minimum and maximum (e.g., via first derivative test sign diagram or second derivative test).

(c) [3 marks]

A1: Finding $ f(1) = 1 $ and $ f'(1) = -1 $.

M1: Correct substitution of their point and gradient into a linear equation form.

A1: Correct equation: $ y = -x + 2 $ (or equivalent).

(d) [5 marks]

M1: Setting up the correct definite integral with limits $ 0 $ and $ 1 $.

M1: Attempting integration by substitution.

A1: Correct antiderivative: $ -\frac{1}{2} e^{1-x^2} $.

M1: Correct application of fundamental theorem of calculus (substituting limits $ 0 $ and $ 1 $).

A1: Correct final exact area: $ \frac{e-1}{2} $ (or equivalent).

Question 2 · Extended Response

15 marks

Let $ f(x) = \frac{2x + 1}{x - 3} $, for $ x \in \mathbb{R}, x \neq 3 $, and $ g(x) = 2\ln(x - 1) $, for $ x \in \mathbb{R}, x > 1 $.

(a) Show that $ f^{-1}(x) = \frac{3x + 1}{x - 2} $, stating its domain. [4 marks]

(b) Find an expression for $ (f \circ g)(x) $, stating its domain. [4 marks]

(c) Solve the equation $ (f \circ g)(x) = 3 $. [4 marks]

(d) Describe a sequence of two transformations that maps the graph of $ y = \ln(x) $ to the graph of $ y = g(x) $. [3 marks]

Show answer & marking scheme

Worked solution

(a) Let $ y = \frac{2x+1}{x-3} $.

Interchange $ x $ and $ y $: $ x = \frac{2y+1}{y-3} $

$ x(y-3) = 2y + 1 \implies xy - 3x = 2y + 1 $

$ xy - 2y = 3x + 1 \implies y(x-2) = 3x + 1 $

$ y = \frac{3x+1}{x-2} $

Therefore, $ f^{-1}(x) = \frac{3x + 1}{x - 2} $.

The domain of $ f^{-1} $ is the range of $ f $. Since the horizontal asymptote of $ f $ is $ y = 2 $, the range of $ f $ is $ \mathbb{R} \setminus \{2\} $.

So, the domain of $ f^{-1} $ is $ x \in \mathbb{R}, x \neq 2 $.

(b) $ (f \circ g)(x) = f(g(x)) = \frac{2(2\ln(x-1)) + 1}{2\ln(x-1) - 3} = \frac{4\ln(x-1) + 1}{2\ln(x-1) - 3} $.

For the domain: we must have $ x - 1 > 0 \implies x > 1 $.

Also, the denominator cannot be zero: $ 2\ln(x-1) - 3 \neq 0 \implies \ln(x-1) \neq 1.5 \implies x - 1 \neq e^{1.5} \implies x \neq e^{1.5} + 1 $.

Thus, the domain is $ x > 1, x \neq e^{1.5} + 1 $.

(c) Solve $ (f \circ g)(x) = 3 $:

$ \frac{4\ln(x-1) + 1}{2\ln(x-1) - 3} = 3 $

$ 4\ln(x-1) + 1 = 3(2\ln(x-1) - 3) $

$ 4\ln(x-1) + 1 = 6\ln(x-1) - 9 $

$ 10 = 2\ln(x-1) \implies \ln(x-1) = 5 $

$ x - 1 = e^5 \implies x = e^5 + 1 $.

Since $ e^5 + 1 > 1 $ and $ e^5 + 1 \neq e^{1.5} + 1 $, this solution is valid.

(d) Starting with $ y = \ln(x) $:

1. Translate horizontally by 1 unit to the right to get $ y = \ln(x - 1) $.

2. Stretch vertically with a scale factor of 2 to get $ y = 2\ln(x - 1) $.

(Note: The order of these two transformations does not matter).

Marking scheme

(a) [4 marks]

M1: Attempt to swap $ x $ and $ y $ or make $ x $ the subject of $ y = f(x) $.

M1: Correct algebraic rearrangement to isolate the variable.

A1: Correct expression for $ f^{-1}(x) = \frac{3x+1}{x-2} $.

A1: Correct domain: $ x \in \mathbb{R}, x \neq 2 $ (or equivalent).

(b) [4 marks]

M1: Correct substitution of $ g(x) $ into $ f(x) $.

A1: Correct expression $ \frac{4\ln(x-1)+1}{2\ln(x-1)-3} $.

M1: Identifying both restrictions: $ x - 1 > 0 $ and denominator $ \neq 0 $.

A1: Correct domain: $ x > 1 $ and $ x \neq e^{1.5} + 1 $ (or equivalent).

(c) [4 marks]

M1: Setting their composite function equal to 3.

A1: Correct linear equation in terms of $ \ln(x-1) $: $ 4\ln(x-1) + 1 = 6\ln(x-1) - 9 $.

M1: Solving for $ \ln(x-1) $ to find $ \ln(x-1) = 5 $.

A1: Correct exact value: $ x = e^5 + 1 $.

(d) [3 marks]

A1: Translation of 1 unit to the right (or translation vector $ \begin{pmatrix} 1 \\ 0 \end{pmatrix} $).

A1: Vertical stretch with scale factor of 2.

A1: Stating both transformations clearly (any order is acceptable).

Question 3 · Extended Response

15 marks

(a) Prove the identity: $ \frac{\sin(2\theta)}{1 + \cos(2\theta)} = \tan\theta $. [4 marks]

(b) Hence or otherwise, solve the equation $ \sin(2\theta) = \sqrt{3}(1 + \cos(2\theta)) $ for $ 0 \le \theta \le \pi $. [4 marks]

(c) Show that $ \cos^4\theta - \sin^4\theta = \cos(2\theta) $. [3 marks]

(d) Given that $ \tan\theta = -2 $ and $ \frac{\pi}{2} < \theta < \pi $, find the exact value of:

(i) $ \sin(2\theta) $

(ii) $ \cos(2\theta) $. [4 marks]

Show answer & marking scheme

Worked solution

(a) LHS $ = \frac{\sin(2\theta)}{1 + \cos(2\theta)} $

Using the double angle identities $ \sin(2\theta) = 2\sin\theta\cos\theta $ and $ \cos(2\theta) = 2\cos^2\theta - 1 $:

LHS $ = \frac{2\sin\theta\cos\theta}{1 + (2\cos^2\theta - 1)} $

LHS $ = \frac{2\sin\theta\cos\theta}{2\cos^2\theta} $

LHS $ = \frac{\sin\theta}{\cos\theta} = \tan\theta = $ RHS. Thus proven.

(b) Rewrite the equation: $ \sin(2\theta) - \sqrt{3}(1+\cos(2\theta)) = 0 $.

Using double angle formulas:

$ 2\sin\theta\cos\theta - \sqrt{3}(2\cos^2\theta) = 0 $

$ 2\cos\theta(\sin\theta - \sqrt{3}\cos\theta) = 0 $

This gives two possible equations:

1) $ \cos\theta = 0 \implies \theta = \frac{\pi}{2} $ (since $ 0 \le \theta \le \pi $).

2) $ \sin\theta - \sqrt{3}\cos\theta = 0 \implies \tan\theta = \sqrt{3} \implies \theta = \frac{\pi}{3} $ (since $ 0 \le \theta \le \pi $).

So the solutions are $ \theta = \frac{\pi}{3} $ and $ \theta = \frac{\pi}{2} $.

(c) LHS $ = \cos^4\theta - \sin^4\theta $

Using the difference of squares: $ \cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta) $

Using the Pythagorean identity $ \cos^2\theta + \sin^2\theta = 1 $:

LHS $ = (\cos^2\theta - \sin^2\theta)(1) $

LHS $ = \cos(2\theta) = $ RHS. Thus proven.

(d) Given $ \tan\theta = -2 $ and $ \frac{\pi}{2} < \theta < \pi $ (Quadrant II):

Using the identity $ \sec^2\theta = 1 + \tan^2\theta $:

$ \sec^2\theta = 1 + (-2)^2 = 5 \implies \cos^2\theta = \frac{1}{5} \implies \cos\theta = -\frac{1}{\sqrt{5}} $ (since $ \theta $ is in QII).

Then $ \sin\theta = \tan\theta\cos\theta = (-2)\left(-\frac{1}{\sqrt{5}}\right) = \frac{2}{\sqrt{5}} $.

(i) $ \sin(2\theta) = 2\sin\theta\cos\theta = 2 \left(\frac{2}{\sqrt{5}}\right) \left(-\frac{1}{\sqrt{5}}\right) = -\frac{4}{5} $.

(ii) $ \cos(2\theta) = \cos^2\theta - \sin^2\theta = \left(-\frac{1}{\sqrt{5}}\right)^2 - \left(\frac{2}{\sqrt{5}}\right)^2 = \frac{1}{5} - \frac{4}{5} = -\frac{3}{5} $.

Marking scheme

(a) [4 marks]

M1: Correct substitution of $ \frac{\sin(2\theta)}{1 + \cos(2\theta)} $ with relevant identity templates.

M1: Correct substitution of $ \cos(2\theta) = 2\cos^2\theta - 1 $.

A1: Correct simplification of the denominator to $ 2\cos^2\theta $.

A1: Cancelling terms correctly to obtain $ \tan\theta $.

(b) [4 marks]

M1: Expressing the equation as $ 2\sin\theta\cos\theta = 2\sqrt{3}\cos^2\theta $ or equivalent factorization.

A1: Correctly factoring to $ 2\cos\theta(\sin\theta - \sqrt{3}\cos\theta) = 0 $.

A1: Finding $ \theta = \frac{\pi}{2} $.

A1: Finding $ \theta = \frac{\pi}{3} $.

(c) [3 marks]

M1: Factorising using difference of squares to get $ (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta) $.

A1: Recognizing and substituting $ \cos^2\theta + \sin^2\theta = 1 $.

A1: Recognizing and substituting $ \cos^2\theta - \sin^2\theta = \cos(2\theta) $ to complete the proof.

(d) [4 marks]

M1: Attempt to find $ \sin\theta $ and $ \cos\theta $ using trigonometric identities or reference triangle, including correct signs for Quadrant II ($ \cos\theta = -\frac{1}{\sqrt{5}} $, $ \sin\theta = \frac{2}{\sqrt{5}} $).

A1: Finding $ \sin(2\theta) = -\frac{4}{5} $.

M1: Attempt to calculate $ \cos(2\theta) $ using a double-angle formula.

A1: Finding $ \cos(2\theta) = -\frac{3}{5} $.

Paper 2 Section A

Answer all questions. Answers must be written within the answer boxes provided. GDC required.

6 Question · 34.98 marks

Question 1 · Short Answer

5.83 marks

The weights of bags of flour are normally distributed with mean $\mu$ and standard deviation $\sigma = 12$ grams. It is known that 15% of the bags weigh less than 485 grams.

(a) Find the value of $\mu$.

(b) A bag is selected at random. Given that it weighs more than 495 grams, find the probability that it weighs less than 510 grams.

Show answer & marking scheme

Worked solution

(a) Let $X$ be the weight of a bag of flour, so $X \sim N(\mu, 12^2)$.
Given $P(X < 485) = 0.15$.
Using the standard normal distribution, we find $z$ such that $P(Z < z) = 0.15$.
From the GDC (inverse normal), $z \approx -1.03643$.
Then, $\frac{485 - \mu}{12} = -1.03643 \implies 485 - \mu = -12.437 \implies \mu \approx 497.437\text{ g}$.
Thus, $\mu = 497\text{ g}$ (to 3 significant figures).

(b) We want to find the conditional probability $P(X < 510 \mid X > 495)$.
By the conditional probability formula:
$P(X < 510 \mid X > 495) = \frac{P(495 < X < 510)}{P(X > 495)}$.
Using $\mu = 497.437$ and $\sigma = 12$ in the GDC:
$P(495 < X < 510) \approx 0.43163$
$P(X > 495) \approx 0.58062$
Now, calculate the ratio:
$\frac{0.43163}{0.58062} \approx 0.74339$.
Thus, the probability is $0.743$ (to 3 significant figures).

Marking scheme

(a) [3 Marks]
M1 for setting up the standardized equation: $\frac{485 - \mu}{12} = z$.
A1 for finding $z \approx -1.036$.
A1 for $\mu = 497$ (accept 497.4).

(b) [3 Marks]
M1 for writing down the correct conditional probability expression: $\frac{P(495 < X < 510)}{P(X > 495)}$.
A1 for finding both probabilities correctly (e.g. numerator $\approx 0.432$ or $0.421$, denominator $\approx 0.581$ or $0.566$).
A1 for the final answer $0.743$.

Question 2 · Short Answer

5.83 marks

Let $f(x) = e^{0.5x} - 2$ and $g(x) = \ln(x + 3)$ for $x > -3$.

(a) Find the $x$-coordinates of the points of intersection of the graphs of $f$ and $g$.

(b) Hence, solve the inequality $f(x) > g(x)$.

Show answer & marking scheme

Worked solution

(a) To find the points of intersection, we set $f(x) = g(x)$:
$e^{0.5x} - 2 = \ln(x + 3)$.
Using a GDC to find the roots of this equation on the interval $x > -3$:
$x_1 \approx -2.8315$
$x_2 \approx 2.6225$
So, the $x$-coordinates are $x = -2.83$ and $x = 2.62$ (to 3 significant figures).

(b) The inequality $f(x) > g(x)$ holds where the graph of $f$ lies strictly above the graph of $g$.
By observing the graphs on the GDC, this occurs for values of $x$ to the left of the first intersection point and to the right of the second intersection point.
Considering the domain $x > -3$, the solution is:
$-3 < x < -2.83$ or $x > 2.62$.

Marking scheme

(a) [3 Marks]
M1 for attempting to solve $f(x) = g(x)$ (e.g., sketching graphs or setting up the equation).
A1 for $x \approx -2.83$.
A1 for $x \approx 2.62$.

(b) [3 Marks]
M1 for identifying the correct intervals from the graph.
A1 for $-3 < x < -2.83$ (must include the lower bound $-3$).
A1 for $x > 2.62$.

Question 3 · Short Answer

5.83 marks

A particle moves along a straight line. Its velocity $v(t)$ in $\text{m s}^{-1}$ at time $t$ seconds, for $0 \le t \le 10$, is given by $v(t) = 5 - 3t\cos(0.2t^2)$.

(a) Find the value of $t$ when the particle first comes to rest.

(b) Find the total distance travelled by the particle in the first 6 seconds.

Show answer & marking scheme

Worked solution

(a) The particle is at rest when its velocity is zero, so we solve:
$5 - 3t\cos(0.2t^2) = 0$.
Using a GDC to find the smallest positive root of this equation:
$t \approx 5.0381\dots$
Thus, the particle first comes to rest at $t = 5.04\text{ s}$ (to 3 significant figures).

(b) The total distance travelled in the first 6 seconds is given by the integral of the speed:
$\text{Distance} = \int_{0}^{6} |v(t)| \, dt = \int_{0}^{6} |5 - 3t\cos(0.2t^2)| \, dt$.
Using the GDC's numerical integration function, we evaluate this expression:
$\text{Distance} \approx 40.499\dots\text{ m}$.
Thus, the total distance travelled is $40.5\text{ m}$ (to 3 significant figures).

Marking scheme

(a) [2 Marks]
M1 for setting $v(t) = 0$.
A1 for $t \approx 5.04$.

(b) [4 Marks]
M1 for recognizing that total distance is the integral of absolute velocity.
A1 for writing the correct integral expression: $\int_{0}^{6} |5 - 3t\cos(0.2t^2)| \, dt$.
M1 for using GDC numerical integration.
A1 for the final answer $40.5$ (accept $40.5\text{ m}$).

Question 4 · Short Answer

5.83 marks

A boat leaves port $P$ and sails 12 km on a bearing of $075^\circ$ to point $A$. It then changes direction and sails 18 km to point $B$. The bearing of $B$ from $P$ is $115^\circ$.

(a) Find the distance from $P$ to $B$.

(b) Find the bearing of $B$ from $A$.

Show answer & marking scheme

Worked solution

(a) In $\triangle PAB$, we are given:
- $PA = 12$ km
- $AB = 18$ km
- The angle $\angle APB$ is the difference between the two bearings: $115^\circ - 75^\circ = 40^\circ$.

Using the Sine Rule to find angle $\angle PBA$:
$\frac{\sin \angle PBA}{PA} = \frac{\sin \angle APB}{AB} \implies \frac{\sin \angle PBA}{12} = \frac{\sin 40^\circ}{18}$
$\sin \angle PBA = \frac{12 \sin 40^\circ}{18} \approx 0.4285$
$\angle PBA \approx 25.37^\circ$ (since $\angle PBA$ must be acute in this configuration).

Now find the angle $\angle PAB$:
$\angle PAB = 180^\circ - 40^\circ - 25.37^\circ = 114.63^\circ$.

Using the Sine Rule to find $PB$:
$\frac{PB}{\sin 114.63^\circ} = \frac{18}{\sin 40^\circ} \implies PB = \frac{18 \sin 114.63^\circ}{\sin 40^\circ} \approx 25.46\text{ km}$.
So, the distance from $P$ to $B$ is $25.5\text{ km}$ (to 3 significant figures).

(b) To find the bearing of $B$ from $A$:
Draw a North-South line at $A$. Since the bearing of $A$ from $P$ is $075^\circ$, the angle from the South direction at $A$ to $AP$ is $75^\circ$ (alternate interior angles).
Thus, the bearing of $P$ from $A$ is $180^\circ + 75^\circ = 255^\circ$.

Since $\angle PAB = 114.63^\circ$ and $B$ lies to the east of the line $AP$, the bearing of $B$ from $A$ is obtained by turning counter-clockwise from the line $AP$:
$\text{Bearing} = 255^\circ - 114.63^\circ = 140.37^\circ$.
Thus, the bearing is $140^\circ$ (to 3 significant figures).

Marking scheme

(a) [4 Marks]
M1 for finding $\angle APB = 40^\circ$.
M1 for applying the Sine Rule to find $\angle PBA \approx 25.37^\circ$.
A1 for finding $\angle PAB \approx 114.63^\circ$.
A1 for $PB \approx 25.5\text{ km}$ (accept 25.46).

(b) [2 Marks]
M1 for attempting to find the bearing using angles at parallel lines (e.g. finding bearing of $P$ from $A$ as $255^\circ$).
A1 for the final bearing $140^\circ$ (accept 140).

Question 5 · Short Answer

5.83 marks

Mia deposits $500 into a savings account at the beginning of each year. The account pays an annual interest rate of 4.2%, compounded annually.

(a) Find the total value of her savings at the end of 10 years.

(b) Find the minimum number of years it will take for the total value of her savings to exceed $15,000.

Show answer & marking scheme

Worked solution

(a) At the end of 10 years, each annual deposit of $500 will have accumulated compound interest.
The first deposit earns interest for 10 years: $500 \times 1.042^{10}$.
The tenth deposit earns interest for 1 year: $500 \times 1.042^1$.
The total value is the sum of a geometric series:
$S_{10} = 500(1.042)^1 + 500(1.042)^2 + \dots + 500(1.042)^{10}$.
This is a geometric series with first term $u_1 = 521$, common ratio $r = 1.042$, and number of terms $n = 10$.
Using the sum formula $S_n = \frac{u_1(r^n - 1)}{r - 1}$:
$S_{10} = \frac{521(1.042^{10} - 1)}{1.042 - 1} \approx \frac{521(1.508958 - 1)}{0.042} \approx 6313.43$.
Thus, the total value is $\$6310$ (to 3 significant figures, or $\$6313.43$).

(b) We want to find the minimum integer $n$ such that:
$\frac{521(1.042^n - 1)}{0.042} > 15000$.
Simplify the inequality:
$1.042^n - 1 > \frac{15000 \times 0.042}{521}$
$1.042^n - 1 > 1.20921$
$1.042^n > 2.20921$
Taking natural logarithms on both sides:
$n \ln(1.042) > \ln(2.20921)$
$n > \frac{\ln(2.20921)}{\ln(1.042)} \approx 19.26$.
Since $n$ must be an integer, the minimum number of years required is 20.

Marking scheme

(a) [3 Marks]
M1 for identifying a geometric series with $u_1 = 521$ (or $500 \times 1.042$) and $r = 1.042$.
M1 for substituting values into the sum of a GP formula.
A1 for $6310$ (or $6313.43$).

(b) [3 Marks]
M1 for setting up the inequality or equation: $S_n > 15000$.
M1 for attempting to solve the inequality (using GDC table, solver, or logs).
A1 for $20$ (years).

Question 6 · Short Answer

5.83 marks

Let $f(x) = \cos(x^2)$ and $g(x) = x^2 - 1$.

(a) Find the $x$-coordinates of the points of intersection of the graphs of $f$ and $g$.

(b) Find the area of the region completely enclosed by the graphs of $f$ and $g$.

Show answer & marking scheme

Worked solution

(a) To find the points of intersection, we set $f(x) = g(x)$:
$\cos(x^2) = x^2 - 1$.
Using a GDC to find the roots of $\cos(x^2) - x^2 + 1 = 0$:
$x_1 \approx -1.13287$
$x_2 \approx 1.13287$
So, the $x$-coordinates of the points of intersection are $x \approx -1.13$ and $x \approx 1.13$ (to 3 significant figures).

(b) The region is enclosed between $x = -1.13287$ and $x = 1.13287$. In this interval, the graph of $f(x)$ lies above the graph of $g(x)$.
The area $A$ is given by:
$A = \int_{-1.13287}^{1.13287} (\cos(x^2) - (x^2 - 1)) \, dx$.
Using the GDC to compute this definite integral numerically:
$A \approx 3.2173$.
Thus, the area of the enclosed region is $3.22$ (to 3 significant figures).

Marking scheme

(a) [2 Marks]
M1 for attempting to solve $f(x) = g(x)$ (e.g. writing $\cos(x^2) = x^2 - 1$).
A1 for both $x \approx -1.13$ and $x \approx 1.13$.

(b) [4 Marks]
M1 for seting up the integral of difference: $\int (f(x) - g(x)) \, dx$.
A1 for writing the correct integral with correct limits: \[ \int_{-1.13287}^{1.13287} (\cos(x^2) - x^2 + 1) \, dx \]
M1 for using GDC numerical integration.
A1 for $3.22$ (accept 3.2173).

Paper 2 Section B

Answer all questions in the answer booklet provided. Start each question on a new page. GDC required.

3 Question · 45 marks

Question 1 · Extended Response

15 marks

The masses of coffee bags, $X$ grams, are normally distributed with mean $\mu$ and standard deviation $\sigma$.

It is known that $5\%$ of the bags have a mass less than $248\text{ g}$ and $10\%$ of the bags have a mass greater than $254\text{ g}$.

(a) Show that $\mu = 251$ and $\sigma = 2.05$, correct to 3 significant figures. [5 marks]

(b) A bag is selected at random. Find the probability that its mass is between $250\text{ g}$ and $255\text{ g}$. [2 marks]

(c) A customer buys 8 randomly selected bags. Find the probability that exactly 3 of these bags have a mass between $250\text{ g}$ and $255\text{ g}$. [3 marks]

(d) The coffee shop owner packs bags into boxes of $n$ bags. The probability that at least one bag in a box has a mass less than $248\text{ g}$ is greater than $0.75$. Find the minimum value of $n$. [5 marks]

Show answer & marking scheme

Worked solution

(a) Let $X \sim N(\mu, \sigma^2)$. From the given information:
$P(X < 248) = 0.05 \implies \frac{248 - \mu}{\sigma} = z_1$
Using a GDC, $z_1 \approx -1.64485$.
So, $248 - \mu = -1.64485\sigma$ (Equation 1).

$P(X > 254) = 0.10 \implies P(X < 254) = 0.90 \implies \frac{254 - \mu}{\sigma} = z_2$
Using a GDC, $z_2 \approx 1.28155$.
So, $254 - \mu = 1.28155\sigma$ (Equation 2).

Subtracting Equation 1 from Equation 2 gives:
$6 = 2.92640\sigma \implies \sigma \approx 2.05029$
Thus, $\sigma = 2.05$ (to 3 s.f.).

Substituting $\sigma = 2.05029$ back into Equation 2:
$\mu = 254 - 1.28155(2.05029) \approx 251.372$
Thus, $\mu = 251$ (to 3 s.f.).

(b) If using the 3 s.f. values $\mu = 251$ and $\sigma = 2.05$:
$P(250 < X < 255) = \text{normalcdf}(250, 255, 251, 2.05) \approx 0.665$.
If using the unrounded values $\mu = 251.372$ and $\sigma = 2.05029$:
$P(250 < X < 255) \approx 0.713$.

(c) Let $Y$ be the number of bags with mass between 250 g and 255 g. $Y \sim B(8, p)$.
If $p = 0.665$:
$P(Y = 3) = \binom{8}{3}(0.665)^3(0.335)^5 \approx 0.0691$.
If $p = 0.713$:
$P(Y = 3) = \binom{8}{3}(0.713)^3(0.287)^5 \approx 0.0401$.

(d) Let $W$ be the number of underweight bags in a box of $n$ bags. The probability of a bag being underweight is $0.05$.
$W \sim B(n, 0.05)$.
We require $P(W \ge 1) > 0.75 \implies 1 - P(W = 0) > 0.75 \implies 1 - (0.95)^n > 0.75 \implies (0.95)^n < 0.25$.
Taking logarithms:
$n \ln(0.95) < \ln(0.25) \implies n > \frac{\ln(0.25)}{\ln(0.95)} \approx 27.03$.
Since $n$ must be an integer, the minimum value is $n = 28$.

Marking scheme

(a) [5 marks]
M1: For attempting to write down probability statements in terms of z-values.
A1: Finding $z_1 \approx -1.64$ and $z_2 \approx 1.28$.
M1: Setting up two simultaneous equations in terms of $\mu$ and $\sigma$.
A1: Finding $\sigma \approx 2.0503$, which rounds to $2.05$.
A1: Finding $\mu \approx 251.37$, which rounds to $251$.

(b) [2 marks]
M1: Setting up the normal cumulative probability.
A1: For $0.665$ (using 3 s.f. values) or $0.713$ (using unrounded values).

(c) [3 marks]
M1: Identifying binomial distribution with $n = 8$ and their probability $p$ from part (b).
M1: Attempting to calculate $P(Y = 3)$.
A1: For $0.0691$ (using $0.665$) or $0.0401$ (using $0.713$).

(d) [5 marks]
M1: Setting up binomial distribution $W \sim B(n, 0.05)$.
M1: Identifying the condition $P(W \ge 1) > 0.75$ as $1 - (0.95)^n > 0.75$.
A1: Simplifying to $(0.95)^n < 0.25$.
M1: Solving the inequality using logs or GDC.
A1: Correct minimum value $n = 28$.

Question 2 · Extended Response

15 marks

A particle moves along a straight line so that its velocity, $v\text{ m s}^{-1}$, at time $t$ seconds, is given by $v(t) = 3\text{e}^{-0.2t}\sin(t) - 0.5$, for $0 \le t \le 6$.

(a) Find the initial velocity of the particle. [1 mark]

(b) Find the acceleration of the particle at $t = 2$. [3 marks]

(c) Find the values of $t$ in the interval $0 \le t \le 6$ when the particle is at rest. [3 marks]

(d) Find the total distance travelled by the particle between $t = 0$ and $t = 6$. [4 marks]

(e) Given that the particle starts at the origin, find its maximum displacement from the origin during the interval $0 \le t \le 6$. [4 marks]

Show answer & marking scheme

Worked solution

(a) At $t = 0$, $v(0) = 3\text{e}^0\sin(0) - 0.5 = -0.5\text{ m s}^{-1}$.

(b) Acceleration is the derivative of velocity, $a(t) = v'(t)$.
By product rule:
$a(t) = 3(-0.2\text{e}^{-0.2t}\sin(t) + \text{e}^{-0.2t}\cos(t))$.
At $t = 2$:
$a(2) = 3\text{e}^{-0.4}(\cos(2) - 0.2\sin(2)) \approx -1.20\text{ m s}^{-2}$ (to 3 s.f.).
Note: Finding the numerical derivative using GDC directly is also correct.

(c) The particle is at rest when $v(t) = 0$.
$3\text{e}^{-0.2t}\sin(t) - 0.5 = 0$.
Using the GDC solver in the interval $0 \le t \le 6$:
$t \approx 0.174\text{ s}$ and $t \approx 2.84\text{ s}$.

(d) Total distance is given by $\int_0^6 |v(t)| \text{ d}t$.
$\text{Distance} = \int_0^6 |3\text{e}^{-0.2t}\sin(t) - 0.5| \text{ d}t$.
Using numerical integration on the GDC:
$\text{Distance} \approx 6.85\text{ m}$.

(e) Displacement is $s(t) = \int_0^t v(u) \text{ d}u$.
The maximum displacement occurs at a local maximum or boundary point.
Evaluating displacement at key times:
- At $t=0$, $s(0) = 0$.
- At $t=0.174$, $s(0.174) = \int_0^{0.174} v(t)\text{ d}t \approx -0.0432\text{ m}$.
- At $t=2.84$, $s(2.84) = \int_0^{2.84} v(t)\text{ d}t \approx 2.93\text{ m}$.
- At $t=6$, $s(6) = \int_0^6 v(t)\text{ d}t \approx -0.901\text{ m}$.
Therefore, the maximum displacement of the particle from the origin is $2.93\text{ m}$ (at $t \approx 2.84\text{ s}$).

Marking scheme

(a) [1 mark]
A1: For $-0.5\text{ m s}^{-1}$.

(b) [3 marks]
M1: Knowing that acceleration is the derivative of velocity.
M1: Attempting to differentiate $v(t)$ (or setting up numerical derivative on GDC).
A1: For $-1.20\text{ m s}^{-2}$.

(c) [3 marks]
M1: Setting $v(t) = 0$.
A1: Finding $t \approx 0.174$.
A1: Finding $t \approx 2.84$.

(d) [4 marks]
M1: Setting up the integral of absolute velocity.
A1: Correct limits $0$ and $6$.
M1: Evaluating the integral on GDC.
A1: For $6.85\text{ m}$.

(e) [4 marks]
M1: Recognizing that maximum displacement occurs at turning points or endpoints.
M1: Finding displacement at $t \approx 2.84$ by evaluating $\int_0^{2.84} v(t)\text{ d}t$.
A1: Showing that other values are smaller (e.g., at $t=6$ or $t=0.174$).
A1: For $2.93\text{ m}$.

Question 3 · Extended Response

15 marks

An offshore oil rig $R$ is located $5\text{ km}$ from the nearest point $A$ on a straight shoreline. A refinery $B$ is located on the shoreline, $12\text{ km}$ from $A$.

A pipeline is to be laid underwater from $R$ to a point $P$ on the shoreline (between $A$ and $B$), and then along the shoreline from $P$ to $B$. The distance $AP = x\text{ km}$, where $0 \le x \le 12$.

(a) Write down an expression for the distance $RP$ in terms of $x$. [1 mark]

(b) Write down an expression for the distance $PB$ in terms of $x$. [1 mark]

The cost of laying the pipeline underwater is $\$80\,000$ per kilometer, and along the shoreline is $\$50\,000$ per kilometer. Let $C(x)$ represent the total cost of laying the pipeline, in thousands of dollars.

(c) Find an expression for $C(x)$. [2 marks]

(d) Sketch the graph of $C(x)$ for $0 \le x \le 12$, clearly showing the coordinates of the endpoints. [3 marks]

(e) Find the value of $x$ that minimizes the total cost, and find this minimum cost. [4 marks]

(f) Due to a marine reserve near $A$, the pipeline must connect to the shoreline at least $8\text{ km}$ from $A$ (so $x \ge 8$). Find the percentage increase in the minimum cost of the pipeline under this restriction. [4 marks]

Show answer & marking scheme

Worked solution

(a) Using Pythagoras' theorem in right-angled triangle $RAP$:
$RP = \sqrt{x^2 + 5^2} = \sqrt{x^2 + 25}$.

(b) Since the total distance from $A$ to $B$ is $12\text{ km}$ and $AP = x$:
$PB = 12 - x$.

(c) The cost is $\$80$ thousand per km underwater and $\$50$ thousand per km along land:
$C(x) = 80\sqrt{x^2 + 25} + 50(12 - x)$.

(d) At $x = 0$: $C(0) = 80(5) + 50(12) = 1000$.
At $x = 12$: $C(12) = 80\sqrt{144 + 25} = 1040$.
Minimum point occurs between $x=0$ and $x=12$.
The sketch should show a curve starting at $(0, 1000)$, dipping to a minimum, and rising to $(12, 1040)$.

(e) Using GDC to find the minimum of $C(x)$:
Minimum occurs at $x \approx 4.003 \approx 4.00\text{ km}$.
The minimum cost is $C(4.003) \approx 912.25\text{ thousand dollars}$ (or $\$912\,251$).

(f) Since the function $C(x)$ is strictly increasing for $x > 4.00$, the minimum cost on the domain $x \ge 8$ must occur at the boundary $x = 8$.
$C(8) = 80\sqrt{64 + 25} + 50(12 - 8) = 80\sqrt{89} + 200 \approx 954.721\text{ thousand dollars}$.
Percentage increase = $\frac{954.721 - 912.251}{912.251} \times 100 \approx 4.66\%$.

Marking scheme

(a) [1 mark]
A1: Correct expression $\sqrt{x^2 + 25}$.

(b) [1 mark]
A1: Correct expression $12 - x$.

(c) [2 marks]
M1: Correctly scaling distances by costs.
A1: Correct expression $C(x) = 80\sqrt{x^2 + 25} + 50(12 - x)$.

(d) [3 marks]
A1: Correct endpoints $(0, 1000)$ and $(12, 1040)$.
M1: Sketch showing a local minimum between $x=0$ and $x=12$.
A1: Correct overall shape of the curve.

(e) [4 marks]
M1: Realizing the minimum occurs when $C'(x) = 0$ or using GDC minimum tool.
A1: Finding $x \approx 4.00\text{ km}$.
M1: Evaluating $C(x)$ at this value.
A1: For $912$ thousand dollars (or $\$912\,251$).

(f) [4 marks]
M1: Identifying that the minimum cost under the restriction $x \ge 8$ occurs at $x = 8$.
A1: Calculating $C(8) \approx 954.72\text{ thousand dollars}$.
M1: Setting up percentage increase formula.
A1: For $4.66\%$.

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