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We are given \(u_1 = v_1 = 3\). For the arithmetic sequence, the second term is \(u_2 = 3 + d\) and the fifth term is \(u_5 = 3 + 4d\). For the geometric sequence, the second term is \(v_2 = 3r\) and the third term is \(v_3 = 3r^2\). Since \(u_2 = v_2\), we have \(3 + d = 3r \implies d = 3r - 3 = 3(r-1)\). Since \(u_5 = v_3\), we have \(3 + 4d = 3r^2\). Substituting the expression for \(d\) into this equation yields: \(3 + 4(3r - 3) = 3r^2 \implies 3 + 12r - 12 = 3r^2 \implies 3r^2 - 12r + 9 = 0\). Dividing the entire equation by 3 gives: \(r^2 - 4r + 3 = 0\). Factoring the quadratic equation gives: \((r-1)(r-3) = 0\). Since we are given that \(r \neq 1\), we must have \(r = 3\). Substituting \(r = 3\) back into the equation for \(d\) gives: \(d = 3(3) - 3 = 6\).

[M1] for equating \(u_2 = v_2\) to express \(d\) in terms of \(r\) (or vice versa). [M1] for equating \(u_5 = v_3\) to set up a second equation. [M1] for substituting to obtain a quadratic equation in terms of \(r\) (or \(d\)). [A1] for solving the quadratic equation to find \(r = 3\) (rejecting \(r = 1\)). [A2.22] for correctly calculating \(d = 6\).

The general term in the binomial expansion of \((a + b)^n\) is given by \(\binom{n}{r} a^{n-r} b^r\). For the expansion of \(\left( 2x^2 - \frac{1}{x} \right)^6\), the general term is: \(\binom{6}{r} (2x^2)^{6-r} \left(-\frac{1}{x}\right)^r = \binom{6}{r} 2^{6-r} (-1)^r x^{2(6-r)} x^{-r} = \binom{6}{r} 2^{6-r} (-1)^r x^{12-3r}\). To find the term independent of \(x\), we set the exponent of \(x\) to zero: \(12 - 3r = 0 \implies r = 4\). Substituting \(r = 4\) into the coefficient expression: \(\binom{6}{4} 2^{6-4} (-1)^4 = 15 \times 2^2 \times 1 = 15 \times 4 = 60\).

[M1] for writing a correct general term using the binomial theorem. [M1] for simplifying the exponent of \(x\) to obtain \(12 - 3r\). [A1] for setting \(12 - 3r = 0\) and solving to find \(r = 4\). [M1] for evaluating the binomial coefficient \(\binom{6}{4} = 15\). [A2.22] for the final correct term of 60.

First, we find the \(y\)-coordinate of the point of tangency by substituting \(x = 0\) into the equation of the curve: \(y = e^{2(0)} \ln(0 + 1) = 1 \times 0 = 0\). Thus, the tangent line passes through the origin \((0, 0)\). Next, we differentiate \(y\) using the product rule: \(\frac{dy}{dx} = \frac{d}{dx}(e^{2x}) \ln(x + 1) + e^{2x} \frac{d}{dx}(\ln(x + 1)) = 2e^{2x} \ln(x + 1) + e^{2x} \frac{1}{x+1}\). Evaluating the derivative at \(x = 0\) to find the gradient of the tangent: \(m = 2e^0 \ln(1) + e^0 \frac{1}{1} = 0 + 1 = 1\). Using the point-slope form \(y - y_1 = m(x - x_1)\) with point \((0, 0)\) and gradient \(m = 1\), we obtain the equation: \(y - 0 = 1(x - 0) \implies y = x\).

[A1] for finding the point of contact \((0, 0)\). [M1] for applying the product rule to find the derivative. [A1] for the correct derivative expression \(\frac{dy}{dx} = 2e^{2x} \ln(x+1) + \frac{e^{2x}}{x+1}\). [M1] for substituting \(x = 0\) into their derivative to find the gradient \(m = 1\). [A2.22] for the correct final equation of the tangent line \(y = x\) (or equivalent).

We can evaluate this integral using the substitution method. Let \(u = \tan(x)\). Then the derivative is \(du = \sec^2(x) \, dx\). Next, we find the new limits of integration: when \(x = 0\), \(u = \tan(0) = 0\); when \(x = \frac{\pi}{4}\), \(u = \tan\left(\frac{\pi}{4}\right) = 1\). Substituting these into the integral gives: \(\int_{0}^{1} u^3 \, du = \left[ \frac{u^4}{4} \right]_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}\).

[M1] for using the substitution \(u = \tan(x)\) and finding \(du = \sec^2(x) \, dx\). [A1] for correctly changing the limits of integration to \(0\) and \(1\). [M1] for integrating \(u^3\) to obtain \(\frac{u^4}{4}\). [A2.22] for substituting the limits and obtaining the final correct answer of \(\frac{1}{4}\).

Using the double-angle identity \(\sin(2x) = 2 \sin(x) \cos(x)\), we can rewrite the equation as: \(2 \sin(x) \cos(x) = \cos(x) \implies 2 \sin(x) \cos(x) - \cos(x) = 0\). Factoring out \(\cos(x)\) gives: \(\cos(x) (2 \sin(x) - 1) = 0\). This yields two cases: 1) \(\cos(x) = 0\). In the interval \(0 \le x \le \pi\), the solution is \(x = \frac{\pi}{2}\). 2) \(2 \sin(x) - 1 = 0 \implies \sin(x) = \frac{1}{2}\). In the interval \(0 \le x \le \pi\), the solutions are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). The sum of all solutions is: \(\frac{\pi}{2} + \frac{\pi}{6} + \frac{5\pi}{6} = \frac{3\pi}{6} + \frac{\pi}{6} + \frac{5\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2}\).

[M1] for utilizing the double-angle identity to express the equation in terms of single-angle trig functions. [M1] for factoring the equation to obtain \(\cos(x)(2\sin(x)-1) = 0\). [A1] for finding the solution \(x = \frac{\pi}{2}\). [A1] for finding the solutions \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). [A2.22] for obtaining the correct sum of the solutions as \(\frac{3\pi}{2}\).

We use the addition rule for probability: \(\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)\). Substituting the given values: \(0.8 = 0.6 + 0.4 - \mathrm{P}(A \cap B) \implies 0.8 = 1.0 - \mathrm{P}(A \cap B) \implies \mathrm{P}(A \cap B) = 0.2\). We want to find the conditional probability \(\mathrm{P}(A' \mid B) = \frac{\mathrm{P}(A' \cap B)}{\mathrm{P}(B)}\rename\). Note that \(\mathrm{P}(A' \cap B) = \mathrm{P}(B) - \mathrm{P}(A \cap B) = 0.4 - 0.2 = 0.2\). Thus, \(\mathrm{P}(A' \mid B) = \frac{0.2}{0.4} = 0.5\).

[M1] for applying the probability addition formula. [A1] for obtaining \(\mathrm{P}(A \cap B) = 0.2\). [M1] for identifying or calculating \(\mathrm{P}(A' \cap B) = \mathrm{P}(B) - \mathrm{P}(A \cap B)\). [A1] for finding \(\mathrm{P}(A' \cap B) = 0.2\). [A1.22] for correctly calculating the conditional probability as \(0.5\) (or equivalent fraction).

To find the inverse function \(f^{-1}(x)\), let \(y = \frac{2x+1}{x-3}\). We then swap \(x\) and \(y\) and solve for \(y\): \(x = \frac{2y+1}{y-3} \implies x(y-3) = 2y+1 \implies xy - 3x = 2y+1\). Rearranging terms containing \(y\) to one side: \(xy - 2y = 3x + 1 \implies y(x-2) = 3x + 1 \implies y = \frac{3x+1}{x-2}\). Thus, the inverse function is \(f^{-1}(x) = \frac{3x+1}{x-2}\). Now, evaluating at \(x = 5\): \(f^{-1}(5) = \frac{3(5)+1}{5-2} = \frac{16}{3}\).

[M1] for attempting to swap \(x\) and \(y\) or make \(x\) the subject. [M1] for multiplying by the denominator and grouping terms. [A1] for correctly deriving the inverse function \(f^{-1}(x) = \frac{3x+1}{x-2}\). [M1] for substituting \(x = 5\) into their inverse expression. [A2.22] for obtaining the correct value of \(\frac{16}{3}\).

Using the properties of logarithms, we can combine the terms on the left side: \(\log_2(x(x-3)) = 2\). Converting this logarithmic equation into its equivalent exponential form: \(x(x-3) = 2^2 \implies x^2 - 3x = 4\). Rearranging this into a standard quadratic equation: \(x^2 - 3x - 4 = 0\). Factoring the quadratic: \((x-4)(x+1) = 0\). This gives two potential solutions: \(x = 4\) or \(x = -1\). However, we must check the domain of the original logarithmic expressions. We require \(x > 0\) and \(x-3 > 0 \implies x > 3\). Thus, \(x = -1\) is an extraneous solution and is rejected. The only valid solution is \(x = 4\).

[M1] for applying the log addition rule to combine terms. [M1] for converting the log equation to an exponential form. [A1] for setting up the correct quadratic equation \(x^2 - 3x - 4 = 0\). [M1] for finding the roots \(x = 4\) and \(x = -1\). [A1.22] for rejecting \(x = -1\) with a valid reason and stating the final solution \(x = 4\).

(a) For an infinite geometric series to converge, the common ratio \(r\) must satisfy \(|r| < 1\). Substituting the given ratio: \(\left| \frac{1}{2}\ln x \right| < 1\). This simplifies to: \(-1 < \frac{1}{2}\ln x < 1\) which gives \(-2 < \ln x < 2\). Exponentiating both sides with base \(e\), we get: \(e^{-2} < x < e^2\). (b) The sum to infinity is given by \(S_{\infty} = \frac{u_1}{1 - r}\). Substituting \(u_1 = \ln x\) and \(r = \frac{1}{2}\ln x\): \(S_{\infty} = \frac{\ln x}{1 - \frac{1}{2}\ln x} = \frac{2\ln x}{2 - \ln x}\). We are given that \(S_{\infty} = \frac{2}{3}\), so: \rac{2\ln x}{2 - \ln x} = \frac{2}{3}\. Dividing both sides by 2 gives: \(\frac{\ln x}{2 - \ln x} = \frac{1}{3}\). Cross-multiplying: \(3\ln x = 2 - \ln x\), which simplifies to \(4\ln x = 2\), hence \(\ln x = \frac{1}{2}\). Therefore, \(x = e^{1/2} = \sqrt{e}\). Since \(\ln(\sqrt{e}) = \frac{1}{2}\), the common ratio is \(r = \frac{1}{4}\), which satisfies the convergence condition \(|r| < 1\).

(a) M1 for using the convergence condition \(|r| < 1\). A1 for obtaining \(-2 < \ln x < 2\). A1 for the final interval \(e^{-2} < x < e^2\) (or equivalent). (b) M1 for substituting \(u_1\) and \(r\) into the sum to infinity formula and setting it equal to \(\frac{2}{3}\). A1 for simplifying the resulting equation to \(\ln x = \frac{1}{2}\) (or equivalent). A1 for obtaining \(x = e^{1/2}\) (or \(\sqrt{e}\)).

(a) To find the \( x \)-intercept, set \( f(x) = 0 \):
\( (x-2)e^{-x} = 0 \)
Since \( e^{-x} > 0 \) for all \( x \), we have:
\( x = 2 \)
So, the \( x \)-intercept is \( (2, 0) \).

To find the \( y \)-intercept, set \( x = 0 \):
\( f(0) = (0-2)e^{0} = -2 \)
So, the \( y \)-intercept is \( (0, -2) \).

(b) Using the product rule to find \( f'(x) \):
\( f'(x) = 1 \cdot e^{-x} + (x-2)(-e^{-x}) \)
\( f'(x) = e^{-x}(1 - x + 2) = (3-x)e^{-x} \)

To find the local maximum, set \( f'(x) = 0 \):
\( (3-x)e^{-x} = 0 \Rightarrow x = 3 \)

Substitute \( x = 3 \) back into \( f(x) \):
\( f(3) = (3-2)e^{-3} = e^{-3} \)

Since \( f'(x) > 0 \) for \( x < 3 \) and \( f'(x) < 0 \) for \( x > 3 \), the point \( (3, e^{-3}) \) is a local maximum.

(c) To find the point of inflection, differentiate \( f'(x) \) using the product rule to obtain \( f''(x) \):
\( f''(x) = -1 \cdot e^{-x} + (3-x)(-e^{-x}) \)
\( f''(x) = e^{-x}(-1 - 3 + x) = (x-4)e^{-x} \)

Set \( f''(x) = 0 \):
\( (x-4)e^{-x} = 0 \Rightarrow x = 4 \)

Substitute \( x = 4 \) back into \( f(x) \):
\( f(4) = (4-2)e^{-4} = 2e^{-4} \)

Since \( f''(x) \) changes sign around \( x = 4 \) (negative for \( x < 4 \), positive for \( x > 4 \)), the point of inflection is \( (4, 2e^{-4}) \).

(d) (i) The area \( A \) of the region \( R \) is given by:
\( A = \int_{2}^{k} (x-2)e^{-x} \, dx \)

Using integration by parts, let \( u = x-2 \) and \( dv = e^{-x} \, dx \).
Then \( du = dx \) and \( v = -e^{-x} \).

\( \int (x-2)e^{-x} \, dx = -(x-2)e^{-x} - \int (-e^{-x}) \, dx \)
\( = -(x-2)e^{-x} - e^{-x} = e^{-x}(-x + 2 - 1) = -(x-1)e^{-x} \)

Evaluating this from \( x = 2 \) to \( x = k \):
\( A = \left[ -(x-1)e^{-x} \right]_{2}^{k} = \left( -(k-1)e^{-k} \right) - \left( -(2-1)e^{-2} \right) \)
\( A = e^{-2} - (k-1)e^{-k} \) (as required).

(ii) As \( k \to \infty \):
\( \lim_{k\to\infty} e^{-k} = 0 \) and \( \lim_{k\to\infty} k e^{-k} = 0 \), which implies \( \lim_{k\to\infty} (k-1)e^{-k} = 0 \).

Therefore, the limiting value of the area is:
\( \lim_{k\to\infty} \left( e^{-2} - (k-1)e^{-k} \right) = e^{-2} \) (or \( \frac{1}{e^2} \)).

(a) [2 marks]
M1 for setting \( f(x) = 0 \) or \( x = 0 \).
A1 for both intercepts correct: \( (2, 0) \) and \( (0, -2) \).

(b) [4 marks]
M1 for attempting product rule to find \( f'(x) \).
A1 for \( f'(x) = (3-x)e^{-x} \).
M1 for setting \( f'(x) = 0 \) to find \( x = 3 \).
A1 for correct coordinates of local maximum: \( (3, e^{-3}) \).

(c) [4 marks]
M1 for attempting product rule to find \( f''(x) \).
A1 for \( f''(x) = (x-4)e^{-x} \).
M1 for setting \( f''(x) = 0 \) to find \( x = 4 \).
A1 for correct coordinates of point of inflection: \( (4, 2e^{-4}) \).

(d) [8 marks]
(i)
M1 for setting up the integral \( \int_{2}^{k} (x-2)e^{-x} \, dx \).
M1 for choosing correct parts \( u = x-2 \) and \( dv = e^{-x} \, dx \).
A1 for finding the antiderivative \( -(x-1)e^{-x} \).
M1 for applying the limits of integration \( 2 \) and \( k \).
A1 for obtaining the shown expression: \( e^{-2} - (k-1)e^{-k} \).

(ii)
M1 for recognizing the limit behavior of \( e^{-k} \) as \( k \to \infty \).
M1 for identifying that \( \lim_{k\to\infty} (k-1)e^{-k} = 0 \).
A1 for final limiting value \( e^{-2} \).

(a) We equate \( \sqrt{3} \sin(2x) - \cos(2x) = R \sin(2x - \alpha) \).
Using the identity \( R \sin(2x - \alpha) = R \sin(2x)\cos(\alpha) - R \cos(2x)\sin(\alpha) \), we compare coefficients:
\( R \cos(\alpha) = \sqrt{3} \)
\( R \sin(\alpha) = 1 \)

Squaring and adding these equations:
\( R^2(\cos^2(\alpha) + \sin^2(\alpha)) = (\sqrt{3})^2 + 1^2 \)
\( R^2 = 4 \Rightarrow R = 2 \) (since \( R > 0 \)).

To find \( \alpha \):
\( \tan(\alpha) = \frac{1}{\sqrt{3}} \)
Since \( 0 < \alpha < \frac{\pi}{2} \), we have:
\( \alpha = \frac{\pi}{6} \)

Thus, \( g(x) = 2 \sin\left(2x - \frac{\pi}{6}\right) \).

(b) Since \( 0 \le x \le \pi \), we have \( -\frac{\pi}{6} \le 2x - \frac{\pi}{6} \le \frac{11\pi}{6} \).

The maximum value of \( g(x) \) is \( 2 \), occurring when:
\( 2x - \frac{\pi}{6} = \frac{\pi}{2} \Rightarrow 2x = \frac{2\pi}{3} \Rightarrow x = \frac{\pi}{3} \)
Thus, the local maximum point is \( \left(\frac{\pi}{3}, 2\right) \).

The minimum value of \( g(x) \) is \( -2 \), occurring when:
\( 2x - \frac{\pi}{6} = \frac{3\pi}{2} \Rightarrow 2x = \frac{5\pi}{3} \Rightarrow x = \frac{5\pi}{6} \)
Thus, the local minimum point is \( \left(\frac{5\pi}{6}, -2\right) \).

(c) To solve \( g(x) = 1 \):
\( 2 \sin\left(2x - \frac{\pi}{6}\right) = 1 \Rightarrow \sin\left(2x - \frac{\pi}{6}\right) = \frac{1}{2} \)
Within \( -\frac{\pi}{6} \le 2x - \frac{\pi}{6} \le \frac{11\pi}{6} \), the solutions are:
\( 2x - \frac{\pi}{6} = \frac{\pi}{6} \Rightarrow 2x = \frac{\pi}{3} \Rightarrow x = \frac{\pi}{6} \)
\( 2x - \frac{\pi}{6} = \frac{5\pi}{6} \Rightarrow 2x = \pi \Rightarrow x = \frac{\pi}{2} \)

So, the solutions are \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{2} \).

(d) The range of \( g(x) \) is \( [-2, 2] \).
Therefore, the values of \( g(x) + 3 \) lie in the interval:
\( -2 + 3 \le g(x) + 3 \le 2 + 3 \)
\( 1 \le g(x) + 3 \le 5 \)

Taking the reciprocal to find the range of \( h(x) = \frac{1}{g(x)+3} \):
\( \frac{1}{5} \le h(x) \le 1 \)

Thus, the range is \( \left[ \frac{1}{5}, 1 \right] \).

(a) [5 marks]
M1 for expanding \( R \sin(2x - \alpha) \).
A1 for equating coefficients to get \( R \cos(\alpha) = \sqrt{3} \) and \( R \sin(\alpha) = 1 \).
M1 for finding \( R \) using Pythagoras' identity.
A1 for \( R = 2 \).
A1 for \( \alpha = \frac{\pi}{6} \).

(b) [5 marks]
M1 for recognizing maximum is when \( \sin(\theta) = 1 \) and minimum is when \( \sin(\theta) = -1 \).
M1 for setting up \( 2x - \frac{\pi}{6} = \frac{\pi}{2} \) and solving for \( x \).
A1 for local maximum at \( \left(\frac{\pi}{3}, 2\right) \).
M1 for setting up \( 2x - \frac{\pi}{6} = \frac{3\pi}{2} \) and solving for \( x \).
A1 for local minimum at \( \left(\frac{5\pi}{6}, -2\right) \).

(c) [4 marks]
M1 for writing \( \sin\left(2x - \frac{\pi}{6}\right) = \frac{1}{2} \).
A1 for identifying either key angle \( \frac{\pi}{6} \) or \( \frac{5\pi}{6} \).
A1 for \( x = \frac{\pi}{6} \).
A1 for \( x = \frac{\pi}{2} \).

(d) [4 marks]
M1 for utilizing range of \( g(x) \) as \( [-2, 2] \).
A1 for finding bounds of the denominator: \( [1, 5] \).
M1 for applying reciprocals to the bounds.
A1 for final range \( \left[ \frac{1}{5}, 1 \right] \) (accept interval or inequality notation).

(a) Using standard sequence formulae:
\( u_2 = 4 + d \)
\( v_3 = 4r^2 \)
\( u_3 = 4 + 2d \)
\( v_2 = 4r \)

(b) From the given equations:
1) \( u_2 = v_3 \Rightarrow 4 + d = 4r^2 \Rightarrow d = 4r^2 - 4 \)
2) \( u_3 = v_2 \Rightarrow 4 + 2d = 4r \)

Substitute the expression for \( d \) from the first equation into the second equation:
\( 4 + 2(4r^2 - 4) = 4r \)
\( 4 + 8r^2 - 8 = 4r \)
\( 8r^2 - 4r - 4 = 0 \)

Divide the entire equation by 4:
\( 2r^2 - r - 1 = 0 \) (as required).

(c) Solve the quadratic equation \( 2r^2 - r - 1 = 0 \):
\( (2r + 1)(r - 1) = 0 \)
This gives \( r = -\frac{1}{2} \) or \( r = 1 \).

Since the geometric sequence has a sum to infinity, the common ratio must satisfy \( |r| < 1 \).
Therefore, we reject \( r = 1 \) and choose:
\( r = -\frac{1}{2} \)

Now substitute \( r = -\frac{1}{2} \) back to find \( d \):
\( d = 4\left(-\frac{1}{2}\right)^2 - 4 = 4\left(\frac{1}{4}\right) - 4 = 1 - 4 = -3 \).
So, \( r = -\frac{1}{2} \) and \( d = -3 \).

(d) The sum to infinity is given by:
\( S_{\infty} = \frac{v_1}{1-r} = \frac{4}{1 - \left(-\frac{1}{2}\right)} = \frac{4}{\frac{3}{2}} = \frac{8}{3} \).

(e) The sum of the first 10 terms of the arithmetic sequence is:
\( S_{10} = \frac{10}{2} [2u_1 + 9d] \)
\( S_{10} = 5 [2(4) + 9(-3)] \)
\( S_{10} = 5 [8 - 27] \)
\( S_{10} = 5 [-19] = -95 \).

(a) [2 marks]
A1 for \( u_2 = 4 + d \) and \( u_3 = 4 + 2d \).
A1 for \( v_3 = 4r^2 \) and \( v_2 = 4r \).

(b) [4 marks]
M1 for establishing the system of equations: \( 4+d=4r^2 \) and \( 4+2d=4r \).
M1 for expressing \( d \) as \( 4r^2 - 4 \).
M1 for substituting \( d \) into the second equation.
A1 for showing algebraic simplification to obtain \( 2r^2 - r - 1 = 0 \) clearly.

(c) [4 marks]
M1 for attempting to factorize or solve the quadratic equation.
A1 for finding roots \( r = -1/2 \) and \( r = 1 \).
M1 for choosing \( r = -1/2 \) with a valid reason related to convergence \( (|r| < 1) \).
A1 for finding \( d = -3 \).

(d) [3 marks]
M1 for using the infinite geometric sum formula.
A1 for substituting values correctly: \( \frac{4}{1 - (-0.5)} \).
A1 for \( S_{\infty} = \frac{8}{3} \).

(e) [5 marks]
M1 for recalling the arithmetic progression sum formula.
M1 for substituting \( n=10 \), \( u_1=4 \), and \( d=-3 \).
A1 for obtaining \( 5[8 - 27] \).
A1 for calculating \( 5 \times (-19) \).
A1 for final answer \( -95 \).

Using the formula for the \(n\)-th term of a geometric sequence, we have \(u_3 = a r^2 = 18\), which gives \(a = \frac{18}{r^2}\). Using the formula for the sum of the first \(n\) terms, we have \(S_6 = a \frac{1 - r^6}{1 - r} = 728\). Substituting the expression for \(a\) into the sum formula gives: \(\frac{18}{r^2} \frac{1 - r^6}{1 - r} = 728\). Since \(r \neq 1\), this equation can be solved using a GDC by finding the positive points of intersection of the functions \(y = \frac{18(1 - x^6)}{x^2(1 - x)}\) and \(y = 728\). The two positive solutions are \(r = 3\) and \(r \approx 0.173\) (correct to three significant figures).

M1 for writing \(a r^2 = 18\) or equivalent. M1 for substituting into the sum formula to obtain \(\frac{18(1 - r^6)}{r^2(1 - r)} = 728\). A1 for setting up the equation on GDC. A1 for \(r = 3\). A2 for \(r = 0.173\) (or 0.1729).

(a) Since \(\mathrm{e}^{0.4x} > 0\) for all real \(x\), we have \(\mathrm{e}^{0.4x} - 3 > -3\). Therefore, the range of \(f\) is \(f(x) > -3\). (b) We set up the composite function equation \((f \circ g)(x) = 2.5 \implies f(g(x)) = 2.5\). Substituting \(g(x)\) into \(f\) gives \(\mathrm{e}^{0.4 \ln(x^2 + 2)} - 3 = 2.5 \implies \mathrm{e}^{0.4 \ln(x^2 + 2)} = 5.5\). Using a GDC to find the points of intersection or solving analytically: \((x^2 + 2)^{0.4} = 5.5 \implies x^2 + 2 = 5.5^{2.5} \implies x^2 \approx 68.940 \implies x \approx \pm 8.30\). Therefore, \(x \approx 8.30\) and \(x \approx -8.30\).

A1 for the range \(f(x) > -3\). M1 for writing the composite equation \(\mathrm{e}^{0.4 \ln(x^2 + 2)} - 3 = 2.5\). M1 for simplifying to \((x^2 + 2)^{0.4} = 5.5\) or equivalent. A1 for \(x^2 \approx 68.9\). A2 for obtaining both \(x \approx 8.30\) and \(x \approx -8.30\) (award A1 if only one value is given).

(a) The area of a triangle is given by \(\text{Area} = \frac{1}{2} a c \sin B\). Substituting the given values: \(16.5 = \frac{1}{2} (5.8) (7.2) \sin B \implies 16.5 = 20.88 \sin B \implies \sin B \approx 0.79023\). The first possible angle is \(B_1 = \arcsin(0.79023) \approx 52.209^\circ \approx 52.2^\circ\). The second possible angle is \(B_2 = 180^\circ - 52.209^\circ = 127.791^\circ \approx 128^\circ\). (b) Since \(\widehat{B}\) is obtuse, we use \(B \approx 127.791^\circ\). Using the cosine rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos B \implies AC^2 = 7.2^2 + 5.8^2 - 2(7.2)(5.8) \cos(127.791^\circ) \implies AC^2 = 51.84 + 33.64 - 83.52(-0.6128) \approx 136.66 \implies AC = \sqrt{136.66} \approx 11.7\text{ cm}\).

M1 for substituting into the area formula: \(16.5 = 0.5 \times 7.2 \times 5.8 \sin B\). A1 for \(\sin B \approx 0.790\). A1 for \(52.2^\circ\). A1 for \(128^\circ\). M1 for substituting the obtuse angle and given sides into the cosine rule. A1 for \(AC \approx 11.7\text{ cm}\).

(a) Let \(X\) be the weight of an apple, where \(X \sim N(\mu, 12^2)\). We are given \(\mathrm{P}(X > 145) = 0.15 \implies \mathrm{P}(X \le 145) = 0.85\). Under standardisation, \(z = \frac{145 - \mu}{12}\). From the inverse normal distribution on a GDC, the \(z\)-score corresponding to a left-tail area of \(0.85\) is \(z \approx 1.0364\). Thus, \(1.0364 = \frac{145 - \mu}{12} \implies 145 - \mu \approx 12.437 \implies \mu \approx 132.563 \approx 133\text{ grams}\). (b) Let \(Y\) be the number of apples weighing more than \(145\text{ grams}\) in a sample of 8. Since each apple is selected independently, \(Y \sim \mathrm{B}(8, 0.15)\). We need to find \(\mathrm{P}(Y \ge 2)\): \(\mathrm{P}(Y \ge 2) = 1 - \mathrm{P}(Y \le 1)\). Using the binomial cumulative distribution function on a GDC: \(\mathrm{P}(Y \le 1) \approx 0.6572\). Therefore, \(\mathrm{P}(Y \ge 2) = 1 - 0.6572 = 0.3428 \approx 0.343\).

M1 for setting up the standardisation equation \(z = \frac{145 - \mu}{12}\). A1 for finding the correct \(z\)-score of \(1.04\) or \(1.036\). A1 for \(\mu \approx 133\). M1 for identifying the binomial distribution \(Y \sim \mathrm{B}(8, 0.15)\). M1 for the probability expression \(1 - \mathrm{P}(Y \le 1)\). A1 for the correct probability \(0.343\) (or 0.3428).

(a) To find the points of intersection, we solve \(\ln(x^2 + 1) = 4 - x^2\). Using the GDC solver, we find the two roots: \(x_1 \approx -1.6412 \approx -1.64\) and \(x_2 \approx 1.6412 \approx 1.64\). (b) The enclosed area \(A\) is given by the integral of the upper curve minus the lower curve between these intersection limits: \(A = \int_{-1.6412}^{1.6412} \left( (4 - x^2) - \ln(x^2 + 1) \right) \mathrm{d}x\). Evaluating this definite integral on a GDC yields \(A \approx 8.3648 \approx 8.36\).

M1 for equating the curve equations: \(\ln(x^2+1) = 4-x^2\). A1 for \(x \approx -1.64\) and A1 for \(x \approx 1.64\). M1 for setting up the definite integral for the area with correct boundaries. A2 for \(8.36\) (accept 8.36 or 8.37 depending on intermediate rounding).

(a) We construct the augmented matrix: \(\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 2 & 5 & 1 & | & 10 \\ 3 & 7 & k & | & 12 \end{pmatrix}\). Performing row operations: \(R_2 \to R_2 - 2R_1\) and \(R_3 \to R_3 - 3R_1\) yields: \(\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 3 & | & 4 \\ 0 & 1 & k+3 & | & 3 \end{pmatrix}\). Performing \(R_3 \to R_3 - R_2\) yields: \(\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 3 & | & 4 \\ 0 & 0 & k & | & -1 \end{pmatrix}\). The system fails to have a unique solution when the leading coefficient in the third row is zero, which gives \(k = 0\). (b) Substituting \(k = 0\) into the third row gives the equation \(0z = -1 \implies 0 = -1\), which is a contradiction. Therefore, the system is inconsistent and has no solution.

M1 for representing the system as an augmented matrix or system of equations. M1 for a correct row reduction step to eliminate \(x\) from the second and third rows. A1 for the reduced system showing a relation like \(kz = -1\). A1 for concluding \(k = 0\). R1 for substituting \(k = 0\) to show the contradiction \(0 = -1\). R1 for concluding that this contradiction means there is no solution.

(a) Equating the components of \(L_1\) and \(L_2\): \(1 + 2\lambda = 5 - \mu \implies 2\lambda + \mu = 4\) (Equation 1), \(-2 + \lambda = 3\mu \implies \lambda - 3\mu = 2\) (Equation 2), and \(3 - \lambda = 1 + 2\mu \implies \lambda + 2\mu = 2\) (Equation 3). From Equation 1, \(\mu = 4 - 2\lambda\). Substituting into Equation 2: \(\lambda - 3(4 - 2\lambda) = 2 \implies 7\lambda = 14 \implies \lambda = 2\). This gives \(\mu = 4 - 2(2) = 0\). Checking in Equation 3: \(2 + 2(0) = 2\), which is consistent. Substituting \(\mu = 0\) into \(L_2\), we find the intersection point coordinates are \((5, 0, 1)\). (b) The direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}\). The angle \(\theta\) is found using the dot product formula: \(\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|}\). Here, \(\mathbf{d}_1 \cdot \mathbf{d}_2 = 2(-1) + 1(3) + (-1)(2) = -1\). The magnitudes are \(|\mathbf{d}_1| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6}\) and \(|\mathbf{d}_2| = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{14}\). Thus, \(\cos \theta = \frac{|-1|}{\sqrt{6}\sqrt{14}} = \frac{1}{\sqrt{84}} \approx 0.1091\). This gives \(\theta = \arccos(0.1091) \approx 83.7^\circ\).

M1 for equating at least two components of the line equations. A1 for obtaining \(\lambda = 2\) or \(\mu = 0\). A1 for coordinates \((5, 0, 1)\). M1 for calculating the dot product and magnitudes of the direction vectors. A1 for \(\cos \theta = \frac{1}{\sqrt{84}}\) or equivalent. A1 for \(\theta \approx 83.7^\circ\) (or \(1.46\text{ radians}\)).

Let \(V\) be the volume of water, \(h\) be the depth, and \(r\) be the radius of the water surface. The volume of a cone is given by \(V = \frac{1}{3} \pi r^2 h\). By similar triangles, the ratio of the radius to the height is \(\frac{r}{h} = \frac{3}{9} \implies r = \frac{1}{3}h\). Substituting this into the volume formula yields: \(V = \frac{1}{3} \pi \left(\frac{1}{3}h\right)^2 h = \frac{\pi}{27} h^3\). Differentiating both sides with respect to time \(t\) using the chain rule: \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}h} \cdot \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\pi}{9} h^2 \frac{\mathrm{d}h}{\mathrm{d}t}\). We are given \(\frac{\mathrm{d}V}{\mathrm{d}t} = 0.5\). Substituting \(h = 4\): \(0.5 = \frac{\pi}{9} (4)^2 \frac{\mathrm{d}h}{\mathrm{d}t} \implies 0.5 = \frac{16\pi}{9} \frac{\mathrm{d}h}{\mathrm{d}t}\). Solving for \(\frac{\mathrm{d}h}{\mathrm{d}t}\): \(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{0.5 \times 9}{16\pi} = \frac{9}{32\pi} \approx 0.0895\text{ m min}^{-1}\) (to three significant figures).

M1 for using similar triangles to find \(r = \frac{1}{3}h\). A1 for obtaining \(V = \frac{\pi}{27}h^3\). M1 for the chain rule differentiation: \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\pi}{9}h^2 \frac{\mathrm{d}h}{\mathrm{d}t}\). A1 for evaluating the derivative at \(h = 4\) to get \(\frac{16\pi}{9}\). M1 for substituting the known rate of \(0.5\) and solving. A1 for \(\frac{9}{32\pi}\) or \(0.0895\text{ m min}^{-1}\).

**(a)** Let \(X\) be the weight of a chocolate bar. We are given that \(X \sim N(150, \sigma^2)\) and \(P(X < 142) = 0.08\). Standardizing this value gives \(P\left(Z < \frac{142 - 150}{\sigma}\right) = 0.08\). Using the inverse normal function on a GDC, we find the \(z\)-score corresponding to a left-tail probability of \(0.08\): \(z \approx -1.40507\). Setting up the equation: \(\frac{142 - 150}{\sigma} = -1.40507 \Rightarrow \frac{-8}{\sigma} = -1.40507\). Solving for \(\sigma\): \(\sigma = \frac{-8}{-1.40507} \approx 5.69366\). Thus, \(\sigma \approx 5.69\text{ g}\) (to 3 significant figures). **(b)** First, we find the probability, \(p\), that a single randomly chosen chocolate bar weighs more than \(153\text{ g}\): \(p = P(X > 153)\). Using a GDC with \(\mu = 150\) and \(\sigma = 5.69366\): \(p \approx 0.299132\). Now, let \(Y\) be the number of chocolate bars weighing more than \(153\text{ g}\) out of a sample of 5. This follows a binomial distribution: \(Y \sim B(5, 0.299132)\). We need to find the probability that at least two bars weigh more than \(153\text{ g}\): \(P(Y \ge 2) = 1 - P(Y \le 1) = 1 - [P(Y = 0) + P(Y = 1)]\). Using the binomial cumulative distribution function on a GDC: \(P(Y \le 1) \approx 0.529977\). Therefore, \(P(Y \ge 2) = 1 - 0.529977 \approx 0.470023\). Thus, the probability is \(0.470\) (to 3 significant figures).

**(a)**
(M1) for an attempt to standardize or use the inverse normal function on a GDC.
(A1) for \(\frac{-8}{\sigma} = -1.40507\) (or equivalent equation).
(A1) for \(\sigma \approx 5.69\) (accept 5.69366...).

**(b)**
(M1) for an attempt to calculate \(P(X > 153)\) using their standard deviation.
(A1) for \(p \approx 0.299\) (accept 0.299132...).
(M1) for recognizing a binomial distribution \(Y \sim B(5, p)\) and attempting to find \(P(Y \ge 2)\).
(A1) for \(0.470\) (accept 0.470023...).

(a) Let \(X \sim N(150, 12^2)\).
Using GDC: \(P(140 < X < 165) \approx 0.692\).

(b) We are given \(P(X > w) = 0.15\), which means \(P(X \le w) = 0.85\).
Using the inverse normal function on a GDC:
\(w = \text{invNorm}(0.85, 150, 12) \approx 162.437\).
Thus, \(w \approx 162\) g (to 3 significant figures).

(c) Let \(Y\) be the number of Premium apples in a box of 20. Then \(Y \sim B(20, 0.15)\).
(i) \(P(Y = 4) = \binom{20}{4} (0.15)^4 (0.85)^{16} \approx 0.182\).
(ii) \(P(Y \ge 2) = 1 - P(Y \le 1) = 1 - (P(Y = 0) + P(Y = 1))\)
\(P(Y \le 1) \approx 0.17556\).
\(P(Y \ge 2) \approx 1 - 0.17556 = 0.824\).

(d) For the 3rd Premium apple to be the 10th apple selected, there must be exactly 2 Premium apples in the first 9 selections, and the 10th selection must be a Premium apple.
Let \(W \sim B(9, 0.15)\).
\(P(W = 2) = \binom{9}{2} (0.15)^2 (0.85)^7 \approx 0.25967\).
Required probability = \(P(W = 2) \times 0.15 = 0.25967 \times 0.15 \approx 0.0390\) (or 0.03895 to 4 s.f.).

(e) Let \(Y_p \sim N(\mu_p, 10^2)\).
\(P(Y_p < 120) = 0.10\).
Standardizing gives: \(P\left(Z < \frac{120 - \mu_p}{10}\right) = 0.10\).
Using inverse normal, \(z_{0.10} \approx -1.28155\).
\(\frac{120 - \mu_p}{10} = -1.28155 \Rightarrow 120 - \mu_p = -12.8155 \Rightarrow \mu_p \approx 132.8155\).
Thus, \(\mu_p \approx 133\) g.

(a) M1 for attempting to find normal probability with correct limits and parameters.
A2 for 0.692 (accept 0.6920).

(b) M1 for writing a correct probability statement (e.g., \(P(X > w) = 0.15\) or \(P(X < w) = 0.85\)).
A1 for identifying the correct z-score of approximately 1.036 (implied by GDC use).
A1 for \(w = 162\) (accept 162.4).

(c) (i) M1 for recognizing binomial model \(B(20, 0.15)\).
A1 for 0.182.
(ii) M1 for expressing the probability as \(1 - P(Y \le 1)\) or \(P(Y \ge 2)\).
M1 for calculating \(P(Y \le 1)\).
A1 for 0.824.

(d) M1 for recognizing the negative binomial scenario (2 Premium in first 9 and Premium on the 10th).
M1 for calculating binomial probability \(\binom{9}{2} (0.15)^2 (0.85)^7\).
A1 for 0.2597.
A1 for 0.0390 (or 0.03895).

(e) M1 for standardizing to get \(\frac{120 - \mu_p}{10}\).
M1 for equating to \(-1.282\) (or \(-1.28\)).
A1 for \(\mu_p = 133\) (accept 132.8).

(a) Initial velocity is \(v(0)\):
\(v(0) = 0^2 \cos(0) - 2(0) + 5 = 5 \text{ m s}^{-1}\).

(b) Acceleration is the derivative of velocity:
\(a(t) = v'(t)\).
Using GDC numerical differentiation at \(t = 3\):
\(v'(t) = 2t \cos\left(\frac{\pi t}{4}\right) - \frac{\pi t^2}{4} \sin\left(\frac{\pi t}{4}\right) - 2\).
At \(t = 3\):
\(a(3) = 6 \cos\left(\frac{3\pi}{4}\right) - \frac{9\pi}{4} \sin\left(\frac{3\pi}{4}\right) - 2 \approx -11.2 \text{ m s}^{-2}\).

(c) The particle is at rest when \(v(t) = 0\).
Using the GDC root-finding tool for \(t^2 \cos\left(\frac{\pi t}{4}\right) - 2t + 5 = 0\) on \(0 \le t \le 10\):
\(t_1 \approx 2.18\) s,
\(t_2 \approx 6.24\) s,
\(t_3 \approx 9.81\) s.

(d) Total distance traveled is given by \(\int_{0}^{6} |v(t)| \mathrm{d}t\).
Using GDC numerical integration:
\(\int_{0}^{6} \left|t^2 \cos\left(\frac{\pi t}{4}\right) - 2t + 5\right| \mathrm{d}t \approx 57.5\) m.

(e) (i) Displacement \(s(t)\) from the starting position at time \(t\) is:
\(s(t) = \int_{0}^{t} v(x) \mathrm{d}x\) (or \(s(t) = \int_{0}^{t} \left(x^2 \cos\left(\frac{\pi x}{4}\right) - 2x + 5\right) \mathrm{d}x\)).
(ii) The particle returns to its starting position when \(s(T) = 0\) for \(T > 0\).
Therefore, we solve \(\int_{0}^{T} v(t) \mathrm{d}t = 0\).
Using GDC numerical integration and solver on \(T > 0\):
Since the velocity is positive on \([0, 2.18]\) and negative on \([2.18, 6.24]\), the displacement increases to a maximum at \(t \approx 2.18\) and then decreases, passing through zero before \(t = 6.24\).
Solving \(\int_{0}^{T} \left(t^2 \cos\left(\frac{\pi t}{4}\right) - 2t + 5\right) \mathrm{d}t = 0\) yields:
\(T \approx 3.28\) s.

(a) A1 for 5 (units not required).

(b) M1 for recognizing \(a(t) = v'(t)\).
A1 for expressing derivative (either analytically or implied by GDC).
A1 for \(-11.2\) (accept \(-11.24\)).

(c) M1 for setting \(v(t) = 0\).
A2 for all three correct values: \(2.18, 6.24, 9.81\) (award A1 if only two are correct).

(d) M1 for setting up the correct integral \(\int_{0}^{6} |v(t)| \mathrm{d}t\).
M2 for showing splitting of intervals or GDC entry: \(\int_{0}^{2.18} v(t) \mathrm{d}t - \int_{2.18}^{6} v(t) \mathrm{d}t\).
A1 for \(57.5\).

(e) (i) A2 for \(s(t) = \int_{0}^{t} v(x) \mathrm{d}x\) (accept equivalent expressions).
(ii) M1 for setting \(s(T) = 0\).
M2 for establishing the integral equation \(\int_{0}^{T} v(t) \mathrm{d}t = 0\) on GDC.
A2 for \(T \approx 3.28\).

(a) (i) Amplitude \(a = \frac{\text{Max} - \text{Min}}{2} = \frac{14.6 - 8.2}{2} = 3.2\).
(ii) Vertical shift \(d = \frac{\text{Max} + \text{Min}}{2} = \frac{14.6 + 8.2}{2} = 11.4\).

(b) The period of the tide is 12 hours.
\(b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{12} = \frac{\pi}{6}\).

(c) The maximum of a cosine function occurs when the argument is zero (or multiples of \(2\pi\)).
Since high tide (maximum depth) occurs at \(t = 3\):
\(b(3 - c) = 0 \Rightarrow c = 3\).

(d) Substituting the values into the equation:
\(D(t) = 3.2 \cos\left(\frac{\pi}{6}(t - 3)\right) + 11.4\).

(e) At 10:30, \(t = 10.5\).
\(D(10.5) = 3.2 \cos\left(\frac{\pi}{6}(10.5 - 3)\right) + 11.4\)
\(= 3.2 \cos\left(\frac{7.5\pi}{6}\right) + 11.4 = 3.2 \cos(1.25\pi) + 11.4\)
\(= 3.2\left(-\frac{\sqrt{2}}{2}\right) + 11.4 \approx 9.14\) m.

(f) We need \(D(t) \ge 12.5\) for \(0 \le t \le 24\).
Set \(3.2 \cos\left(\frac{\pi}{6}(t - 3)\right) + 11.4 = 12.5\).
\(\cos\left(\frac{\pi}{6}(t - 3)\right) = \frac{1.1}{3.2} = 0.34375\).
\(\frac{\pi}{6}(t - 3) = \pm \arccos(0.34375) + 2k\pi\).
\(\arccos(0.34375) \approx 1.2201\) rad.
\(t - 3 \approx \pm \frac{6}{\pi}(1.2201) + 12k \approx \pm 2.3302 + 12k\).
\(t \approx 3 \pm 2.3302 + 12k\).
Within the range \(0 \le t \le 24\), the critical times are:
\(t_1 = 3 - 2.3302 = 0.670\) (from \(k=0\))
\(t_2 = 3 + 2.3302 = 5.33\) (from \(k=0\))
\(t_3 = 15 - 2.3302 = 12.67\) (from \(k=1\))
\(t_4 = 15 + 2.3302 = 17.33\) (from \(k=1\))

The ship can safely enter during the intervals \([0.670, 5.33]\) and \([12.67, 17.33]\).
Total safe duration = \((5.33 - 0.670) + (17.33 - 12.67) = 4.66 + 4.66 = 9.32\) hours (or 9 hours 19 minutes).

(a) (i) M1 for using the formula \(a = \frac{\text{Max} - \text{Min}}{2}\).
A1 for showing clearly \(a = 3.2\).
(ii) M1 for using the formula \(d = \frac{\text{Max} + \text{Min}}{2}\).
A1 for showing clearly \(d = 11.4\).

(b) M1 for setting up \(b = \frac{2\pi}{\text{period}}\).
A1 for \(b = \frac{\pi}{6}\).

(c) M1 for equating argument to 0 at \(t = 3\).
A1 for \(c = 3\).

(d) A1 for the correct equation.

(e) M1 for substituting \(t = 10.5\) into the model.
A1 for \(9.14\) (or \(9.14\) m).

(f) M1 for setting up the inequality \(D(t) \ge 12.5\).
M2 for finding the critical boundaries \(t \approx 0.670, 5.33, 12.67, 17.33\) (award A1 for two correct values, A2 for all four).
M2 for identifying the two correct intervals where \(D(t) \ge 12.5\).
A2 for adding the durations to find \(9.32\) hours (accept \(9.3\) or \(9.33\) hours depending on intermediate rounding).

(a) Since \(x = \cos\theta\), for \(n = 0\), we have \(T_0(\cos\theta) = \cos(0) = 1\), which gives \(T_0(x) = 1\). For \(n = 1\), we have \(T_1(\cos\theta) = \cos\theta\), which gives \(T_1(x) = x\).

(b) Starting with the trigonometric identity:
\(\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos\theta\cos(n\theta)\)
By definition, \(T_k(\cos\theta) = \cos(k\theta)\), so we can substitute this to get:
\(T_{n+1}(\cos\theta) + T_{n-1}(\cos\theta) = 2\cos\theta T_n(\cos\theta)\)
Substituting \(x = \cos\theta\) yields:
\(T_{n+1}(x) + T_{n-1}(x) = 2xT_n(x)\)
Rearranging this gives:
\(T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x)\).

(c) Using the recurrence relation:
\(T_2(x) = 2xT_1(x) - T_0(x) = 2x(x) - 1 = 2x^2 - 1\)
\(T_3(x) = 2xT_2(x) - T_1(x) = 2x(2x^2 - 1) - x = 4x^3 - 3x\)
\(T_4(x) = 2xT_3(x) - T_2(x) = 2x(4x^3 - 3x) - (2x^2 - 1) = 8x^4 - 8x^2 + 1\).

(d) Let \(P(n)\) be the statement: 'The leading coefficient of \(T_n(x)\) is \(2^{n-1}\)'.
For \(n=1\), \(T_1(x) = x\), which has leading coefficient \(1 = 2^0\). So \(P(1)\) is true.
For \(n=2\), \(T_2(x) = 2x^2 - 1\), which has leading coefficient \(2 = 2^1\). So \(P(2)\) is true.
Assume \(P(k-1)\) and \(P(k)\) are true for some integer \(k \ge 2\). This means the leading term of \(T_k(x)\) is \(2^{k-1}x^k\) and the leading term of \(T_{k-1}(x)\) is \(2^{k-2}x^{k-1}\).
Using the recurrence relation:
\(T_{k+1}(x) = 2xT_k(x) - T_{k-1}(x)\)
The leading term of \(2xT_k(x)\) is \(2x \cdot (2^{k-1}x^k) = 2^k x^{k+1}\). Since the degree of \(T_{k-1}(x)\) is \(k-1\), which is strictly less than \(k+1\), it does not affect the highest degree term of \(T_{k+1}(x)\).
Thus, the leading term of \(T_{k+1}(x)\) is \(2^k x^{k+1}\), which means its leading coefficient is \(2^k = 2^{(k+1)-1}\). Therefore, \(P(k+1)\) is true.
By the principle of mathematical induction, \(P(n)\) is true for all integers \(n \ge 1\).

(e) Set \(T_n(x) = 0\), which means \(\cos(n\theta) = 0\).
Thus, \(n\theta = \frac{\pi}{2} + k\pi = \frac{(2k+1)\pi}{2}\) for \(k \in \mathbb{Z}\).
So \(\theta_k = \frac{(2k+1)\pi}{2n}\).
To obtain unique values of \(x = \cos\theta\) in \([-1, 1]\), we choose \(\theta \in [0, \pi]\), which corresponds to \(k = 0, 1, \dots, n-1\).
Thus, the roots are \(x_k = \cos\left(\frac{(2k+1)\pi}{2n}\right)\) for \(k = 0, 1, \dots, n-1\).

(f) The absolute maximum value of \(|T_n(x)|\) on \([-1, 1]\) is \(1\). These occur when \(\cos(n\theta) = \pm 1\).
This gives \(n\theta = k\pi\), so \(\theta_k = \frac{k\pi}{n}\).
To restrict \(\theta \in [0, \pi]\), we choose \(k = 0, 1, \dots, n\).
The \(x\)-coordinates of the extrema are \(x_k = \cos\left(\frac{k\pi}{n}\right)\).
The corresponding \(y\)-coordinates are \(T_n(x_k) = \cos(k\pi) = (-1)^k\).
Thus, the coordinates of the local extrema on \([-1, 1]\) are \(\left(\cos\left(\frac{k\pi}{n}\right), (-1)^k\right)\) for \(k = 0, 1, \dots, n\).

(g) (i) The leading coefficient of \(T_n(x)\) is \(2^{n-1}\). Thus, the leading coefficient of \(P_n(x) = 2^{1-n}T_n(x)\) is \(2^{1-n} \cdot 2^{n-1} = 2^0 = 1\). Since its leading coefficient is \(1\), \(P_n(x)\) is monic.
(ii) Since the maximum value of \(|T_n(x)|\) on \([-1, 1]\) is \(1\), the maximum value of \(|P_n(x)|\) is \(2^{1-n} \cdot 1 = 2^{1-n}\).

(h) (i) Since both \(P_n(x)\) and \(Q_n(x)\) are monic polynomials of degree \(n\), their leading terms \(x^n\) cancel when subtracted. Thus, the maximum possible degree of \(F(x) = P_n(x) - Q_n(x)\) is \(n-1\).
(ii) At the points \(x_k = \cos\left(\frac{k\pi}{n}\right)\) (where \(k = 0, 1, \dots, n\)), we have \(P_n(x_k) = 2^{1-n}(-1)^k\). By assumption, \(|Q_n(x_k)| < 2^{1-n}\).
If \(k\) is even, \(P_n(x_k) = 2^{1-n}\), so \(F(x_k) = 2^{1-n} - Q_n(x_k) > 0\).
If \(k\) is odd, \(P_n(x_k) = -2^{1-n}\), so \(F(x_k) = -2^{1-n} - Q_n(x_k) < 0\).
Thus, the sign of \(F(x_k)\) is positive for even \(k\) and negative for odd \(k\). Since \(F(x)\) is continuous and changes sign between each consecutive pair of points \(x_k\) and \(x_{k+1}\), the Intermediate Value Theorem guarantees that \(F(x)\) must have at least one root in each of the \(n\) intervals between these points. Hence, \(F(x)\) has at least \(n\) distinct roots.
(iii) A non-zero polynomial of degree at most \(n-1\) can have at most \(n-1\) roots. Since \(F(x)\) has at least \(n\) roots, \(F(x)\) must be the zero polynomial. However, \(F(x_k) \neq 0\) for all \(k\), so \(F(x)\) is not the zero polynomial, which is a contradiction. Thus, no such monic polynomial \(Q_n(x)\) exists, proving that the maximum absolute value of any monic polynomial of degree \(n\) on \([-1, 1]\) is at least \(2^{1-n}\).

(a) [2 marks]
M1 for substituting \(n=0, 1\) into the definition.
A1 for correctly showing \(T_0(x) = 1\) and \(T_1(x) = x\).

(b) [3 marks]
M1 for substituting \(T_k(\cos\theta) = \cos(k\theta)\) into the trigonometric identity.
M1 for replacing \(\cos\theta\) with \(x\).
A1 for completing the proof of the recurrence relation.

(c) [3 marks]
A1 for \(T_2(x) = 2x^2 - 1\).
A1 for \(T_3(x) = 4x^3 - 3x\).
A1 for \(T_4(x) = 8x^4 - 8x^2 + 1\).

(d) [3 marks]
M1 for establishing base cases \(n=1, 2\).
M1 for stating induction hypothesis and substituting into recurrence relation.
A1 for demonstrating the leading coefficient is \(2^k\) for \(T_{k+1}(x)\) and completing the induction step.

(e) [3 marks]
M1 for setting \(\cos(n\theta) = 0\).
M1 for finding \(\theta_k = \frac{(2k+1)\pi}{2n}\).
A1 for writing \(x_k = \cos\left(\frac{(2k+1)\pi}{2n}\right)\) for \(k=0, 1, \dots, n-1\).

(f) [4 marks]
M1 for identifying that extrema occur where \(T_n(x) = \pm 1\).
A1 for finding \(\theta_k = \frac{k\pi}{n}\).
A1 for obtaining \(x_k = \cos\left(\frac{k\pi}{n}\right)\).
A1 for obtaining \(y_k = (-1)^k\) and writing the complete coordinates.

(g) [3 marks]
(i) A1 for showing that the leading coefficient of \(P_n(x)\) is \(2^{1-n} \cdot 2^{n-1} = 1\).
(ii) M1 for relating the maximum of \(|P_n(x)|\) to that of \(|T_n(x)|\).
A1 for concluding that the maximum absolute value is \(2^{1-n}\).

(h) [6.5 marks]
(i) A1 for stating the maximum possible degree of \(F(x)\) is \(n-1\).
(ii) M1 for finding the value of \(P_n(x_k)\).
M1 for determining that \(F(x_k) > 0\) for even \(k\) and \(F(x_k) < 0\) for odd \(k\).
M1 for applying the Intermediate Value Theorem to show a root exists in each interval.
A1.5 for concluding there are at least \(n\) distinct roots.
(iii) A1 for explaining the contradiction: a non-zero polynomial of degree at most \(n-1\) cannot have \(n\) roots unless it is identically zero, which contradicts \(F(x_k) \neq 0\).

(a) The distance of any point \((x(\theta), y(\theta))\) from the origin is:
\(r(\theta) = \sqrt{x(\theta)^2 + y(\theta)^2}\)
\(r(\theta) = \sqrt{(a e^{b\theta} \cos\theta)^2 + (a e^{b\theta} \sin\theta)^2}\)
\(r(\theta) = \sqrt{a^2 e^{2b\theta}(\cos^2\theta + \sin^2\theta)}\)
Since \(\cos^2\theta + \sin^2\theta = 1\), we have:
\(r(\theta) = \sqrt{a^2 e^{2b\theta}} = a e^{b\theta}\) (since \(a > 0\) and \(e^{b\theta} > 0\)).

(b) Using the product rule for differentiation:
\(\frac{dx}{d\theta} = \frac{d}{d\theta}(a e^{b\theta}) \cos\theta + a e^{b\theta} \frac{d}{d\theta}(\cos\theta)\)
\(\frac{dx}{d\theta} = a b e^{b\theta}\cos\theta - a e^{b\theta}\sin\theta = a e^{b\theta}(b\cos\theta - \sin\theta)\)

\(\frac{dy}{d\theta} = \frac{d}{d\theta}(a e^{b\theta}) \sin\theta + a e^{b\theta} \frac{d}{d\theta}(\sin\theta)\)
\(\frac{dy}{d\theta} = a b e^{b\theta}\sin\theta + a e^{b\theta}\cos\theta = a e^{b\theta}(b\sin\theta + \cos\theta)\).

(c) Squaring the derivatives:
\(\left(\frac{dx}{d\theta}\right)^2 = a^2 e^{2b\theta}(b^2\cos^2\theta - 2b\cos\theta\sin\theta + \sin^2\theta)\)
\(\left(\frac{dy}{d\theta}\right)^2 = a^2 e^{2b\theta}(b^2\sin^2\theta + 2b\sin\theta\cos\theta + \cos^2\theta)\)
Adding them together:
\(\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = a^2 e^{2b\theta} \left[ b^2(\cos^2\theta + \sin^2\theta) + (\sin^2\theta + \cos^2\theta) \right]\)
\(\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = a^2 e^{2b\theta}(b^2 + 1)\)
Taking the square root:
\(\frac{ds}{d\theta} = \sqrt{a^2(1+b^2)e^{2b\theta}} = a\sqrt{1+b^2}e^{b\theta}\).

(d) (i) The radial vector is:
\(\vec{OP} = \begin{pmatrix} a e^{b\theta}\cos\theta \\ a e^{b\theta}\sin\theta \end{pmatrix} = a e^{b\theta} \begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix}\)
The tangent vector is:
\(\vec{T} = \begin{pmatrix} a e^{b\theta}(b\cos\theta - \sin\theta) \\ a e^{b\theta}(b\sin\theta + \cos\theta) \end{pmatrix} = a e^{b\theta} \begin{pmatrix} b\cos\theta - \sin\theta \\ b\sin\theta + \cos\theta \end{pmatrix}\).

(ii) The cosine of the angle \(\phi\) is given by:
\(\cos\phi = \frac{\vec{OP} \cdot \vec{T}}{|\vec{OP}| |\vec{T}|}\)
The scalar product is:
\(\vec{OP} \cdot \vec{T} = a^2 e^{2b\theta} [\cos\theta(b\cos\theta - \sin\theta) + \sin\theta(b\sin\theta + \cos\theta)]\)
\(\vec{OP} \cdot \vec{T} = a^2 e^{2b\theta} [b\cos^2\theta - \cos\theta\sin\theta + b\sin^2\theta + \sin\theta\cos\theta] = a^2 e^{2b\theta} b\)
The magnitudes are:
\(|\vec{OP}| = a e^{b\theta}\)
\(|\vec{T}| = a \sqrt{1+b^2} e^{b\theta}\)
Thus:
\(\cos\phi = \frac{a^2 e^{2b\theta} b}{(a e^{b\theta})(a \sqrt{1+b^2} e^{b\theta})} = \frac{b}{\sqrt{1+b^2}}\).

(iii) Since \(b\) is a positive constant, the expression \(\frac{b}{\sqrt{1+b^2}}\) is a constant. Therefore, \(\cos\phi\) is constant, which means the angle \(\phi\) is constant and independent of \(\theta\) along the entire spiral.

(e) (i) The integral expression is:
\(L = \int_{-\infty}^{\theta_0} a\sqrt{1+b^2}e^{b\theta} d\theta\).

(ii) To evaluate the improper integral, we use limits:
\(L = \lim_{k \to -\infty} \int_{k}^{\theta_0} a\sqrt{1+b^2}e^{b\theta} d\theta\)
\(L = a\sqrt{1+b^2} \lim_{k \to -\infty} \left[ \frac{1}{b}e^{b\theta} \right]_k^{\theta_0}\)
\(L = \frac{a\sqrt{1+b^2}}{b} \lim_{k \to -\infty} (e^{b\theta_0} - e^{bk})\)
Since \(b > 0\), as \(k \to -\infty\), \(bk \to -\infty\), so \(e^{bk} \to 0\).
Therefore, the limit converges and:
\(L = \frac{a\sqrt{1+b^2}}{b} e^{b\theta_0}\).

(iii) We are given \(r(\theta_0) = a e^{b\theta_0}\) and \(\cos\phi = \frac{b}{\sqrt{1+b^2}}\).
Thus:
\(\frac{r(\theta_0)}{\cos\phi} = \frac{a e^{b\theta_0}}{\frac{b}{\sqrt{1+b^2}}} = \frac{a\sqrt{1+b^2}}{b} e^{b\theta_0}\)
This is exactly equal to the total arc length \(L\), hence \(L = \frac{r(\theta_0)}{\cos\phi}\).

(f) For \(a = 1\) and \(b = 0.1\), the distance crawled from \(\theta = 0\) to \(\theta = 2\pi\) is:
\(D = \int_{0}^{2\pi} \sqrt{1 + 0.1^2} e^{0.1\theta} d\theta = \sqrt{1.01} \left[ 10 e^{0.1\theta} \right]_0^{2

\pi}\)
\(D = 10\sqrt{1.01} (e^{0.2\pi} - 1)\)
Using a GDC, we evaluate:
\(D \approx 10 \times 1.004987 \times (1.87435 - 1) \approx 8.79\) (to 3 significant figures).

(a) [2 marks]
M1 for writing down the correct distance formula \(r(\theta) = \sqrt{x^2 + y^2}\) and substituting.
A1 for simplifying to \(a e^{b\theta}\).

(b) [4 marks]
M1 for product rule on \(x(\theta)\).
A1 for correct derivative \(\frac{dx}{d\theta} = a e^{b\theta}(b\cos\theta - \sin\theta)\).
M1 for product rule on \(y(\theta)\).
A1 for correct derivative \(\frac{dy}{d\theta} = a e^{b\theta}(b\sin\theta + \cos\theta)\).

(c) [4 marks]
M1 for writing \(\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2\) and substituting.
M1 for expanding both squares correctly.
M1 for factoring \(a^2 e^{2b\theta}\) and using trigonometric identity.
A1 for obtaining \(\frac{ds}{d\theta} = a\sqrt{1+b^2}e^{b\theta}\).

(d) [7 marks]
(i) [2 marks]
A1 for \(\vec{OP} = a e^{b\theta} \begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix}\).
A1 for \(\vec{T} = a e^{b\theta} \begin{pmatrix} b\cos\theta - \sin\theta \\ b\sin\theta + \cos\theta \end{pmatrix}\).
(ii) [4 marks]
M1 for using the formula \(\cos\phi = \frac{\vec{OP} \cdot \vec{T}}{|\vec{OP}| |\vec{T}|}\).
M1 for calculating the scalar product and simplifying to \(a^2 e^{2b\theta} b\).
M1 for dividing by the magnitudes.
A1 for concluding that \(\cos\phi = \frac{b}{\sqrt{1+b^2}}\).
(iii) [1 mark]
R1 for explaining that because \(b\) is a constant, \(\cos\phi\) is constant, hence the angle \(\phi\) is constant along the entire spiral.

(e) [8.5 marks]
(i) [2 marks]
A2 for writing down the correct integral expression \(L = \int_{-\infty}^{\theta_0} a\sqrt{1+b^2}e^{b\theta} d\theta\) (A1 for integrand, A1 for limits).
(ii) [3.5 marks]
M1 for setting up the limit \(\lim_{k \to -\infty}\).
A1 for the integration \(\left[ \frac{1}{b}e^{b\theta} \right]\).
M1 for taking the limit of \(e^{bk}\) as \(0\).
A0.5 for obtaining the final expression \(\frac{a\sqrt{1+b^2}}{b} e^{b\theta_0}\).
(iii) [3 marks]
M1 for substituting \(r(\theta_0) = a e^{b\theta_0}\) and \(\cos\phi\) into \(\frac{r(\theta_0)}{\cos\phi}\).
A1 for showing algebra leading to \(\frac{a\sqrt{1+b^2}}{b} e^{b\theta_0}\).
A1 for equating this result to \(L\).

(f) [2 marks]
M1 for setting up the integral \(\int_{0}^{2\pi} \sqrt{1.01}e^{0.1\theta}d\theta\) (or equivalent algebraic expression).
A1 for the final GDC-calculated value of \(8.79\) (accept \(8.78\) to \(8.80\)).