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

First, find the sum of the first three terms of the arithmetic sequence: \(S_3 = \frac{3}{2}[2u_1 + (3-1)d] = \frac{3}{2}[2\ln k + 2\ln 2] = 3\ln k + 3\ln 2 = \ln(k^3) + \ln 8 = \ln(8k^3)\). Next, find the sum to infinity of the geometric sequence: \(S_\infty = \frac{v_1}{1 - r} = \frac{\ln k}{1 - 1/2} = 2\ln k = \ln(k^2)\). Equating the two sums gives \(\ln(8k^3) = \ln(k^2)\), which simplifies to \(8k^3 = k^2\). Since \(k > 0\), we divide both sides by \(k^2\) to obtain \(8k = 1\), which gives \(k = \frac{1}{8}\).

Award [1 mark] for expressing the sum of the first three terms of the arithmetic sequence, e.g., \(S_3 = 3\ln k + 3\ln 2\). Award [1 mark] for simplifying the arithmetic sum to \(\ln(8k^3)\). Award [1 mark] for expressing the sum to infinity of the geometric sequence as \(\ln(k^2)\). Award [1 mark] for equating the two expressions to obtain \(\ln(8k^3) = \ln(k^2)\). Award [1 mark] for obtaining the algebraic equation \(8k^3 = k^2\). Award [1 mark] for the final answer \(k = \frac{1}{8}\).

(a) To find \(f^{-1}(x)\), let \(y = \frac{2x+1}{x-3}\). Rearranging to make \(x\) the subject: \(y(x - 3) = 2x + 1\), which gives \(xy - 3y = 2x + 1\), then \(xy - 2x = 3y + 1\), and \(x(y - 2) = 3y + 1\). Thus, \(x = \frac{3y + 1}{y - 2}\), which means \(f^{-1}(x) = \frac{3x + 1}{x - 2}\) for \(x \neq 2\). (b) Setting \(f^{-1}(x) = f(x)\): \(\frac{3x + 1}{x - 2} = \frac{2x + 1}{x - 3}\). Cross-multiplying gives \((3x + 1)(x - 3) = (2x + 1)(x - 2)\). Expanding both sides: \(3x^2 - 8x - 3 = 2x^2 - 3x - 2\). Rearranging gives the quadratic equation \(x^2 - 5x - 1 = 0\). Using the quadratic formula, we obtain \(x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-1)}}{2(1)} = \frac{5 \pm \sqrt{29}}{2}\).

(a) Award [1 mark] for swapping variables or setting \(y = f(x)\). Award [1 mark] for making \(x\) the subject. Award [1 mark] for the correct expression for \(f^{-1}(x)\). (b) Award [1 mark] for equating the two expressions and cross-multiplying. Award [1 mark] for simplifying to obtain \(x^2 - 5x - 1 = 0\). Award [1 mark] for applying the quadratic formula correctly to obtain \(x = \frac{5 \pm \sqrt{29}}{2}\).

Dividing the entire equation by 2 gives: \(\frac{\sqrt{3}}{2} \sin(2x) - \frac{1}{2} \cos(2x) = \frac{1}{2}\). This can be recognized as a compound angle expansion of the form \(\sin(A - B) = \sin A \cos B - \cos A \sin B\). Let \(\cos\theta = \frac{\sqrt{3}}{2}\) and \(\sin\theta = \frac{1}{2}\), which gives \(\theta = \frac{\pi}{6}\). Thus, the equation can be written as \(\sin\left(2x - \frac{\pi}{6}\right) = \frac{1}{2}\). Since \(0 \le x \le \pi\), the domain for the argument is \(-\frac{\pi}{6} \le 2x - \frac{\pi}{6} \le \frac{11\pi}{6}\). Within this interval, \(\sin\theta = \frac{1}{2}\) has solutions \(2x - \frac{\pi}{6} = \frac{\pi}{6}\) and \(2x - \frac{\pi}{6} = \frac{5\pi}{6}\). Solving each: 1) \(2x = \frac{\pi}{3} \implies x = \frac{\pi}{6}\); 2) \(2x = \pi \implies x = \frac{\pi}{2}\). Both values lie in the given domain.

Award [1 mark] for dividing by 2 or attempting to use compound angle identities. Award [1 mark] for correctly expressing the left hand side as \(\sin\left(2x - \frac{\pi}{6}\right) = \frac{1}{2}\). Award [1 mark] for identifying the correct interval for the composite argument. Award [2 marks] (1 mark each) for finding the two correct angles \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\). Award [1 mark] for the correct final values of \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\).

Let \(u = x^2 + 1\), then \(\mathrm{d}u = 2x \\, \mathrm{d}x\), which means \(x \\, \mathrm{d}x = \frac{1}{2} \mathrm{d}u\). We rewrite the numerator as \(x^3 \\, \mathrm{d}x = x^2(x \\, \mathrm{d}x) = (u - 1) \left(\frac{1}{2} \mathrm{d}u\right)\). Next, we change the limits of integration: when \(x = 0\), \(u = 1\); when \(x = \sqrt{3}\), \(u = 4\). Substituting into the integral: \(\int_{1}^{4} \frac{u-1}{\sqrt{u}} \cdot \frac{1}{2} \\, \mathrm{d}u = \frac{1}{2} \int_{1}^{4} (u^{1/2} - u^{-1/2}) \\, \mathrm{d}u\). Integrating term by term gives \(\frac{1}{2} \left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right]_{1}^{4} = \left[ \frac{1}{3}u^{3/2} - u^{1/2} \right]_{1}^{4}\). Evaluating at the limits: at \(u = 4\), \(\frac{1}{3}(8) - 2 = \frac{2}{3}\); at \(u = 1\), \(\frac{1}{3}(1) - 1 = -\frac{2}{3}\). Subtracting the lower limit value from the upper limit value: \(\frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}\).

Award [1 mark] for finding \(\mathrm{d}u = 2x \\, \mathrm{d}x\) and substituting \(x^2 = u - 1\). Award [1 mark] for correctly changing the integration limits to \(1\) and \(4\). Award [1 mark] for writing the integral in terms of \(u\) as \(\frac{1}{2} \int_{1}^{4} (u^{1/2} - u^{-1/2}) \\, \mathrm{d}u\). Award [2 marks] (1 mark for each integrated term) for obtaining the antiderivative \(\frac{1}{3}u^{3/2} - u^{1/2}\). Award [1 mark] for substituting the limits correctly to get \(\frac{4}{3}\).

(a) The sum of probabilities must equal 1: \(\sum \mathrm{P}(X = x) = 1\). This gives \(k(1^2) + k(2^2) + k(3^2) + k(4^2) = 1\), which simplifies to \(k(1 + 4 + 9 + 16) = 1\), hence \(30k = 1\), so \(k = \frac{1}{30}\). (b) By definition, \(\mathrm{P}(X \ge 3 \mid X \ge 2) = \frac{\mathrm{P}(X \ge 3)}{\mathrm{P}(X \ge 2)}\). We calculate \(\mathrm{P}(X \ge 2) = \mathrm{P}(X=2) + \mathrm{P}(X=3) + \mathrm{P}(X=4) = \frac{4}{30} + \frac{9}{30} + \frac{16}{30} = \frac{29}{30}\) and \(\mathrm{P}(X \ge 3) = \mathrm{P}(X=3) + \mathrm{P}(X=4) = \frac{9}{30} + \frac{16}{30} = \frac{25}{30}\). Therefore, \(\mathrm{P}(X \ge 3 \mid X \ge 2) = \frac{25/30}{29/30} = \frac{25}{29}\).

(a) Award [1 mark] for setting up the sum of probabilities equal to 1. Award [1 mark] for obtaining \(30k = 1\) and concluding \(k = \frac{1}{30}\). (b) Award [1 mark] for recognizing the conditional probability formula. Award [1 mark] for finding \(\mathrm{P}(X \ge 2) = \frac{29}{30}\). Award [1 mark] for finding \(\mathrm{P}(X \ge 3) = \frac{25}{30}\). Award [1 mark] for the correct final simplified fraction \(\frac{25}{29}\).

To find the total distance, we set \(v(t) = 0\): \(3t^2 - 12t + 9 = 0 \implies 3(t-1)(t-3) = 0\). The particle changes direction at \(t = 1\). The total distance is \(D = \int_{0}^{1} v(t) \\, \mathrm{d}t - \int_{1}^{3} v(t) \\, \mathrm{d}t\). The displacement function is \(s(t) = t^3 - 6t^2 + 9t\). Evaluating at key times: \(s(0) = 0\), \(s(1) = 4\), and \(s(3) = 0\). The distance from \(t=0\) to \(t=1\) is \(|4-0| = 4\) meters. The distance from \(t=1\) to \(t=3\) is \(|0-4| = 4\) meters. Total distance travelled is \(4 + 4 = 8\) meters.

Award [1 mark] for setting \(v(t) = 0\) and finding critical times \(t = 1\) and \(t = 3\). Award [1 mark] for recognizing that the total distance requires splitting the interval at \(t = 1\). Award [1 mark] for integrating \(v(t)\) to obtain \(s(t) = t^3 - 6t^2 + 9t\). Award [1 mark] for calculating \(s(1) = 4\). Award [1 mark] for calculating \(s(3) = 0\). Award [1 mark] for summing the absolute differences to obtain \(8\).

Using the properties of logarithms, we combine the terms on the left: \(\log_2((x - 3)(x - 1)) = 3\). Converting to exponential form: \((x - 3)(x - 1) = 2^3 \implies x^2 - 4x + 3 = 8\). This gives \(x^2 - 4x - 5 = 0\). Factorizing the quadratic equation: \((x - 5)(x + 1) = 0\), giving potential solutions \(x = 5\) or \(x = -1\). Checking the domain: we require \(x - 3 > 0 \implies x > 3\) and \(x - 1 > 0 \implies x > 1\). Since we must have \(x > 3\), the solution \(x = -1\) is rejected. The only valid solution is \(x = 5\).

Award [1 mark] for applying the product law: \(\log_2((x - 3)(x - 1))\). Award [1 mark] for converting to exponential form: \((x - 3)(x - 1) = 8\). Award [1 mark] for expanding and forming \(x^2 - 4x - 5 = 0\). Award [1 mark] for solving to find potential values \(x = 5\) and \(x = -1\). Award [1 mark] for identifying the domain restriction \(x > 3\). Award [1 mark] for rejecting \(x = -1\) to give the final answer \(x = 5\).

(a) First, \(\mathrm{P}(B') = 1 - \mathrm{P}(B) = 0.6\). Using the addition rule: \(\mathrm{P}(A \cup B') = \mathrm{P}(A) + \mathrm{P}(B') - \mathrm{P}(A \cap B')\). Substituting the values: \(0.8 = 0.6 + 0.6 - \mathrm{P}(A \cap B')\), which gives \(\mathrm{P}(A \cap B') = 0.4\). Since \(\mathrm{P}(A) = \mathrm{P}(A \cap B) + \mathrm{P}(A \cap B')\), we have \(0.6 = \mathrm{P}(A \cap B) + 0.4 \implies \mathrm{P}(A \cap B) = 0.2\). (b) By definition: \(\mathrm{P}(A' \mid B') = \frac{\mathrm{P}(A' \cap B')}{\mathrm{P}(B')}\). Using De Morgan's Law: \(\mathrm{P}(A' \cap B') = 1 - \mathrm{P}(A \cup B)\). We find \(\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) = 0.6 + 0.4 - 0.2 = 0.8\). Thus, \(\mathrm{P}(A' \cap B') = 1 - 0.8 = 0.2\). Finally, \(\mathrm{P}(A' \mid B') = \frac{0.2}{0.6} = \frac{1}{3}\).

(a) Award [1 mark] for finding \(\mathrm{P}(B') = 0.6\). Award [1 mark] for finding \(\mathrm{P}(A \cap B') = 0.4\). Award [1 mark] for calculating \(\mathrm{P}(A \cap B) = 0.2\). (b) Award [1 mark] for expressing \(\mathrm{P}(A' \mid B') = \frac{\mathrm{P}(A' \cap B')}{\mathrm{P}(B')}\). Award [1 mark] for finding \(\mathrm{P}(A' \cap B') = 0.2\). Award [1 mark] for the final simplified answer \(\frac{1}{3}\).

To find \(f(x)\), we integrate the gradient function: \(f(x) = \int \frac{6x}{\sqrt{3x^2 + 4}} \, dx\). Using integration by substitution with \(u = 3x^2 + 4\), we have \(\frac{du}{dx} = 6x\), which gives \(du = 6x \, dx\). Substituting these into the integral, we get \(f(x) = \int u^{-1/2} \, du = 2u^{1/2} + C = 2\sqrt{3x^2 + 4} + C\). To find the constant of integration \(C\), we substitute the coordinates of the point \((2, 7)\): \(2\sqrt{3(2)^2 + 4} + C = 7\), which simplifies to \(2\sqrt{16} + C = 7\), leading to \(8 + C = 7\) and thus \(C = -1\). The function is therefore \(f(x) = 2\sqrt{3x^2 + 4} - 1\). Evaluating this at \(x = 0\) gives \(f(0) = 2\sqrt{4} - 1 = 4 - 1 = 3\).

(M1) for attempting to integrate the gradient function. (M1) for a valid substitution, e.g., setting \(u = 3x^2 + 4\). (A1) for the correct antiderivative \(2\sqrt{3x^2 + 4} + C\). (M1) for substituting \((2, 7)\) to find \(C\). (A1) for finding \(C = -1\). (A1) for the final answer 3.

(a)
To find the \(y\)-intercept, evaluate \(f(0)\):
\(f(0) = (0-2)^2 e^{0} = 4 \times 1 = 4\)
So, the \(y\)-intercept is \((0, 4)\).

To find the \(x\)-intercept, solve \(f(x) = 0\):
\((x-2)^2 e^{-x} = 0\)
Since \(e^{-x} > 0\) for all \(x \in \mathbb{R}\), we must have:
\((x-2)^2 = 0 \implies x = 2\)
So, the \(x\)-intercept is \((2, 0)\).

(b)
Using the product rule on \(f(x) = (x-2)^2 e^{-x}\):
Let \(u(x) = (x-2)^2 \implies u'(x) = 2(x-2)\)
Let \(v(x) = e^{-x} \implies v'(x) = -e^{-x}\)

\(f'(x) = u'(x)v(x) + u(x)v'(x)\)
\(f'(x) = 2(x-2)e^{-x} + (x-2)^2(-e^{-x})\)
Factor out \((x-2)e^{-x}\):
\(f'(x) = (x-2)e^{-x} \left[ 2 - (x-2) \right]\)
\(f'(x) = (x-2)e^{-x}(4-x)\)
\(f'(x) = (x-2)(4-x)e^{-x}\).

(c)
Set \(f'(x) = 0\):
\((x-2)(4-x)e^{-x} = 0 \implies x = 2\) or \(x = 4\).

To classify the stationary points, analyze the sign of \(f'(x)\):
- For \(x < 2\), \(f'(x) < 0\) (decreasing)
- For \(2 < x < 4\), \(f'(x) > 0\) (increasing)
- For \(x > 4\), \(f'(x) < 0\) (decreasing)

Thus:
At \(x = 2\), there is a local minimum.
\(f(2) = (2-2)^2 e^{-2} = 0\)
Coordinate of local minimum: \((2, 0)\).

At \(x = 4\), there is a local maximum.
\(f(4) = (4-2)^2 e^{-4} = 4e^{-4}\)
Coordinate of local maximum: \((4, 4e^{-4})\).

(d)
The region is bounded by \(y = f(x)\), the \(x\)-axis, and the \(y\)-axis, which means the limits of integration are from \(x = 0\) to the \(x\)-intercept at \(x = 2\).

\(\text{Area} = \int_{0}^{2} (x-2)^2 e^{-x} \, dx\)

Use integration by parts: \(\int u \, dv = uv - \int v \, du\)
Let \(u = (x-2)^2 \implies du = 2(x-2) \, dx\)
Let \(dv = e^{-x} \, dx \implies v = -e^{-x}\)

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

Apply integration by parts again to the remaining integral \(\int (x-2)e^{-x} \, dx\):
Let \(u = x-2 \implies du = dx\)
Let \(dv = e^{-x} \, dx \implies v = -e^{-x}\)

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

Substitute this back into the main equation:
\(\int (x-2)^2 e^{-x} \, dx = -(x-2)^2 e^{-x} + 2e^{-x}(1-x)\)
\(= e^{-x} [ -(x^2 - 4x + 4) + 2 - 2x ]\)
\(= e^{-x} [ -x^2 + 2x - 2 ] = -e^{-x}(x^2 - 2x + 2)\)

Now, evaluate from 0 to 2:
\(\left[ -e^{-x}(x^2 - 2x + 2) \right]_{0}^{2} = \left( -e^{-2}(2^2 - 2(2) + 2) \right) - \left( -e^{0}(0^2 - 2(0) + 2) \right)\)
\(= -e^{-2}(4 - 4 + 2) - (-1(2))\)
\(= -2e^{-2} + 2\)
\(= 2 - 2e^{-2}\).

(a)
- **M1** for attempting to find the \(y\)-intercept by setting \(x=0\) or \(x\)-intercept by setting \(y=0\).
- **A1** for correct \(y\)-intercept coordinates: \((0,4)\).
- **A1** for correct \(x\)-intercept coordinates: \((2,0)\).

(b)
- **M1** for attempting to use the product rule.
- **A1** for obtaining correct individual terms \(2(x-2)e^{-x}\) and \(-(x-2)^2 e^{-x}\).
- **M1** for factoring out common terms such as \((x-2)e^{-x}\).
- **AG** for showing the final structured form clearly.

(c)
- **M1** for setting \(f'(x) = 0\) and identifying critical values \(x = 2, 4\).
- **A1** for verifying the nature of the stationary points (e.g., using a sign table or second derivative).
- **A1** for correct coordinates of the minimum: \((2, 0)\).
- **A1** for correct coordinates of the maximum: \((4, 4e^{-4})\).

(d)
- **M1** for setting up the definite integral with correct limits from \(0\) to \(2\).
- **M1** for attempting integration by parts once.
- **A1** for obtaining \(-(x-2)^2 e^{-x} + 2 \int (x-2)e^{-x} \, dx\) or equivalent.
- **M1** for attempting integration by parts a second time.
- **A1** for obtaining the correct antiderivative: \(-e^{-x}(x^2 - 2x + 2)\) or equivalent.
- **M1** for substituting limits \(0\) and \(2\) into their antiderivative.
- **A1** for obtaining the exact area: \(2 - 2e^{-2}\).

(a)
For \(P(X=x)\) to be a valid probability distribution, the sum of all probabilities must equal 1:
\(\sum_{x=2}^{4} P(X=x) = 1\)

Evaluate each probability:
\(P(X=2) = k(2-1)(5-2) = 3k\)
\(P(X=3) = k(3-1)(5-3) = 4k\)
\(P(X=4) = k(4-1)(5-4) = 3k\)

Sum of probabilities:
\(3k + 4k + 3k = 10k\)
Set equal to 1:
\(10k = 1 \implies k = 0.1\).

(b)
Using \(k = 0.1\), the probability distribution is:
\(P(X=2) = 0.3\)
\(P(X=3) = 0.4\)
\(P(X=4) = 0.3\)

Expected Value \(\text{E}(X)\):
\(\text{E}(X) = \sum x \cdot P(X=x)\)
\(\text{E}(X) = 2(0.3) + 3(0.4) + 4(0.3) = 0.6 + 1.2 + 1.2 = 3\).

To find \(\text{Var}(X)\), first calculate \(\text{E}(X^2)\):
\(\text{E}(X^2) = \sum x^2 \cdot P(X=x)\)
\(\text{E}(X^2) = 2^2(0.3) + 3^2(0.4) + 4^2(0.3) = 4(0.3) + 9(0.4) + 16(0.3)\)
\(\text{E}(X^2) = 1.2 + 3.6 + 4.8 = 9.6\).

Now, calculate the variance:
\(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\)
\(\text{Var}(X) = 9.6 - 3^2 = 9.6 - 9 = 0.6\).

(c)(i)
Let \(S = X_1 + X_2\). Since \(X_1\) and \(X_2\) are independent and can only take values \(2, 3, 4\), the ways to get a sum of 6 are:
\((X_1=2 \cap X_2=4)\), \((X_1=3 \cap X_2=3)\), and \((X_1=4 \cap X_2=2)\).

Using independence:
\(P(X_1=2 \cap X_2=4) = 0.3 \times 0.3 = 0.09\)
\(P(X_1=3 \cap X_2=3) = 0.4 \times 0.4 = 0.16\)
\(P(X_1=4 \cap X_2=2) = 0.3 \times 0.3 = 0.09\)

Summing these probabilities:
\(P(X_1+X_2=6) = 0.09 + 0.16 + 0.09 = 0.34\).

(c)(ii)
We want to find \(P(S \ge 6 \mid S \text{ is even})\).
By definition of conditional probability:
\(P(S \ge 6 \mid S \text{ is even}) = \frac{P(S \ge 6 \cap S \text{ is even})}{P(S \text{ is even})}\)

The possible values of \(S = X_1+X_2\) range from \(2+2=4\) to \(4+4=8\).
The even sums possible are \(S \in \{4, 6, 8\}\).

Find the probability of each even sum:
\(P(S=4) = P(X_1=2 \cap X_2=2) = 0.3 \times 0.3 = 0.09\)
\(P(S=6) = 0.34\) (from part c(i))
\(P(S=8) = P(X_1=4 \cap X_2=4) = 0.3 \times 0.3 = 0.09\)

So,
\(P(S \text{ is even}) = P(S=4) + P(S=6) + P(S=8) = 0.09 + 0.34 + 0.09 = 0.52\).

The event \((S \ge 6 \cap S \text{ is even})\) corresponds to \(S \in \{6, 8\}\):
\(P(S \ge 6 \cap S \text{ is even}) = P(S=6) + P(S=8) = 0.34 + 0.09 = 0.43\).

Therefore,
\(P(S \ge 6 \mid S \text{ is even}) = \frac{0.43}{0.52} = \frac{43}{52}\).

(a)
- **M1** for attempting to sum the probabilities in terms of \(k\) and set the sum equal to 1.
- **A1** for obtaining \(3k + 4k + 3k = 1\) (or equivalent expression).
- **AG** for showing clearly that \(k = 0.1\).

(b)
- **M1** for attempting to use the expectation formula \(\sum x P(X=x)\).
- **A1** for \(\text{E}(X) = 3\).
- **M1** for attempting to find \(\text{E}(X^2)\).
- **M1** for attempting to use \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\).
- **A1** for \(\text{Var}(X) = 0.6\).

(c)(i)
- **M1** for identifying the three possible outcomes: \((2,4)\), \((3,3)\), \((4,2)\).
- **A1** for calculating the probability of individual outcomes (e.g., \(0.09\), \(0.16\), \(0.09\)).
- **A1** for the correct sum: \(0.34\).

(c)(ii)
- **M1** for stating or using the conditional probability formula.
- **M1** for identifying the even sums as \(\{4, 6, 8\}\).
- **A1** for calculating \(P(S=4) = 0.09\) or \(P(S=8) = 0.09\).
- **A1** for finding \(P(S \text{ is even}) = 0.52\).
- **A1** for identifying \(P(S \ge 6 \cap S \text{ is even}) = P(S=6) + P(S=8) = 0.43\).
- **A2** for the final simplified fraction \(\frac{43}{52}\) (accept equivalent exact decimals \(\approx 0.827\)).

(a)
Using the area formula for a triangle:
\(\text{Area} = \frac{1}{2} \times AB \times AC \times \sin \theta\)
\(\text{Area} = \frac{1}{2} \times 6 \times 4 \times \sin \theta = 12 \sin \theta\).

(b)
Using the cosine rule in \(\triangle ABC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cos \theta\)
\(BC^2 = 6^2 + 4^2 - 2(6)(4) \cos \theta\)
\(BC^2 = 36 + 16 - 48 \cos \theta = 52 - 48 \cos \theta\).

(c)
**Method 1: Supplementary Angles & Cosine Rule**
Let \(BD = CD = x = \frac{BC}{2}\).
Let \(\angle ADB = \alpha\), which means \(\angle ADC = 180^\circ - \alpha\).

Apply the cosine rule in \(\triangle ABD\):
\(AB^2 = AD^2 + BD^2 - 2(AD)(BD) \cos \alpha\)
\(36 = AD^2 + x^2 - 2(AD)x \cos \alpha\) --- (Equation 1)

Apply the cosine rule in \(\triangle ACD\):
\(AC^2 = AD^2 + CD^2 - 2(AD)(CD) \cos(180^\circ - \alpha)\)
Since \(CD = x\) and \
\cos(180^\circ - \alpha) = -\cos \alpha\):
\(16 = AD^2 + x^2 + 2(AD)x \cos \alpha\) --- (Equation 2)

Add Equation 1 and Equation 2:
\(36 + 16 = (AD^2 + x^2 - 2(AD)x \cos \alpha) + (AD^2 + x^2 + 2(AD)x \cos \alpha)\)
\(52 = 2AD^2 + 2x^2\)
\(26 = AD^2 + x^2\)

Since \(x = \frac{BC}{2}\), we have \(x^2 = \frac{BC^2}{4}\):
\(26 = AD^2 + \frac{BC^2}{4}\)
\(AD^2 = 26 - \frac{52 - 48 \cos \theta}{4}\)
\(AD^2 = 26 - (13 - 12 \cos \theta)\)
\(AD^2 = 13 + 12 \cos \theta\).

**Method 2: Vectors**
Let \(\vec{AB} = \mathbf{u}\) and \(\vec{AC} = \mathbf{v}\).
Then \(|\mathbf{u}| = 6\), \(|\mathbf{v}| = 4\), and \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta = 24 \cos \theta\).
Since \(D\) is the midpoint of \(BC\):
\(\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC}) = \frac{1}{2}(\mathbf{u} + \mathbf{v})\)

Thus, the squared length is:
\(AD^2 = \vec{AD} \cdot \vec{AD} = \frac{1}{4}(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})\)
\(AD^2 = \frac{1}{4}(|\mathbf{u}|^2 + 2\mathbf{u} \cdot \mathbf{v} + |\mathbf{v}|^2)\)
\(AD^2 = \frac{1}{4}(36 + 48 \cos \theta + 16)\)
\(AD^2 = \frac{1}{4}(52 + 48 \cos \theta) = 13 + 12 \cos \theta\).

(d)(i)
Given \(AD = \sqrt{19} \implies AD^2 = 19\).
\(13 + 12 \cos \theta = 19\)
\(12 \cos \theta = 6 \implies \cos \theta = \frac{1}{2}\)
Since \(0 < \theta < \pi\), we have:
\(\theta = \frac{\pi}{3}\) (or \(60^\circ\)).

(d)(ii)
Substitute \(\theta = \frac{\pi}{3}\) into the equation from part (b):
\(BC^2 = 52 - 48 \cos\left(\frac{\pi}{3}\right)\)
\(BC^2 = 52 - 48(0.5) = 52 - 24 = 28\)
\(BC = \sqrt{28} = 2\sqrt{7}\).

(d)(iii)
Using the sine rule in \(\triangle ABC\):
\(\frac{BC}{\sin \theta} = \frac{AC}{\sin(A\hat{B}C)}\)
\(\frac{2\sqrt{7}}{\sin(\pi/3)} = \frac{4}{\sin(A\hat{B}C)}\)

Since \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\):
\(\frac{2\sqrt{7}}{\frac{\sqrt{3}}{2}} = \frac{4}{\sin(A\hat{B}C)}\)
\(\frac{4\sqrt{7}}{\sqrt{3}} = \frac{4}{\sin(A\hat{B}C)}\)
\(\sin(A\hat{B}C) = \frac{\sqrt{3}}{\sqrt{7}} = \frac{\sqrt{21}}{7}\).

(a)
- **M1** for attempting to use the triangle area formula \(\frac{1}{2}ab\sin C\).
- **A1** for obtaining \(12 \sin \theta\).

(b)
- **M1** for attempting to use the cosine rule on \(\triangle ABC\).
- **A1** for substitution of the correct values: \(6^2 + 4^2 - 2(6)(4)\cos\theta\).
- **AG** for showing the final algebraic step to get \(BC^2 = 52 - 48 \cos \theta\).

(c)
- **M1** for identifying a valid strategy (either supplementary angles with the cosine rule or vector addition).
- **A1** for correctly writing down two cosine rule equations for the sub-triangles OR expressing \(\vec{AD}\) in terms of vectors.
- **A1** for utilizing supplementary angle relation \(\cos(180^\circ - \alpha) = -\cos\alpha\) OR expanding the vector dot product.
- **M1** for substituting the expression for \(BC^2\) from part (b).
- **AG** for obtaining \(AD^2 = 13 + 12 \cos \theta\) through clear working.

(d)(i)
- **M1** for setting \(13 + 12\cos\theta = 19\) and solving for \(\cos\theta\).
- **A1** for \(\theta = \frac{\pi}{3}\) (or \(60^\circ\)).

(d)(ii)
- **M1** for substituting \(\theta = \frac{\pi}{3}\) back into the formula for \(BC^2\).
- **A1** for \(BC = 2\sqrt{7}\) (accept \(\sqrt{28}\)).

(d)(iii)
- **M1** for attempting to use the sine rule on \(\triangle ABC\).
- **A1** for substituting the correct values: \(\frac{2\sqrt{7}}{\sin(\pi/3)} = \frac{4}{\sin(A\hat{B}C)}\).
- **M1** for solving for \\sin(A\hat{B}C)\) using \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\).
- **A1** for the exact rationalized answer \(\frac{\sqrt{21}}{7}\) (or equivalent exact form \(\frac{\sqrt{3}}{\sqrt{7}}\)).

(a) The total value at the end of \(n\) years is given by the sum of a geometric series: \(S_n = 1500 \times 1.042^n + 1500 \times 1.042^{n-1} + \dots + 1500 \times 1.042^1 = 1500 \times \frac{1.042(1.042^n - 1)}{1.042 - 1}\). For \(n = 5\), \(S_5 = 1500 \times \frac{1.042(1.042^5 - 1)}{0.042} \approx 8499.63\). (b) We set \(S_n > 25000\), which gives \(1500 \times \frac{1.042(1.042^n - 1)}{0.042} > 25000\). Simplifying, \(1.042^n - 1 > 0.6718\), so \(1.042^n > 1.6718\). Solving for \(n\) using logarithms or a GDC solver gives \(n > 12.49\). Since \(n\) must be an integer, the minimum number of years is 13.

(a) M1 for setting up the geometric series sum formula with correct parameters. A1 for correct substitution of \(n=5\). A1 for correct calculation of 8500 dollars (or 8499.63 dollars). (b) M1 for setting up the inequality \(S_n > 25000\). A1 for obtaining \(n > 12.49\) (or equivalent). A1 for the final answer of 13.

(a) To find the point of intersection, we set \(f(x) = g(x)\), which gives \(e^{0.5x} - 2 = \ln(x^2 + 1) + 1\). Using a GDC to find the intersection of these two curves within the domain \([-3, 5]\), we obtain \(x \approx 3.4251\). To 3 significant figures, \(x = 3.43\). (b) By examining the graphs on the GDC, the curve of \(f(x)\) lies above \(g(x)\) for values of \(x\) greater than the intersection point. Thus, the solution to \(f(x) > g(x)\) within the domain is \(3.43 < x \le 5\).

(a) M1 for attempting to solve \(f(x) = g(x)\) (e.g., sketching graphs or writing the equation). A2 for \(x \approx 3.43\). (b) M1 for identifying the correct region where \(f(x)\) is above \(g(x)\). A2 for the correct interval \(3.43 < x \le 5\) (accept open interval at 3.43, closed at 5).

(a) The area of a triangle is given by \(\text{Area} = \frac{1}{2} a c \sin B\). Substituting the known values: \(28 = \frac{1}{2} \times 9 \times 7 \sin B \implies 28 = 31.5 \sin B\). Thus, \(\sin B = \frac{28}{31.5} = \frac{8}{9}\). Solving for \(B\), we find \(B_1 = \arcsin(8/9) \approx 62.729^\circ\) and \(B_2 = 180^\circ - 62.729^\circ \approx 117.271^\circ\). To 3 significant figures, the possible values are \(62.7^\circ\) and \(117^\circ\). (b) For an obtuse angle, \(B \approx 117.271^\circ\). Using the cosine rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos B = 7^2 + 9^2 - 2(7)(9) \cos(117.271^\circ) = 49 + 81 - 126 \cos(117.271^\circ) \approx 130 - 126(-0.4581) \approx 187.72\). Taking the square root, we obtain \(AC \approx 13.701\text{ cm}\). To 3 significant figures, \(AC = 13.7\text{ cm}\).

(a) M1 for using the area formula with correct substitution. A1 for \(\sin B = 8/9\) (or equivalent). A1 for both angles correct (62.7 and 117). (b) M1 for selecting the obtuse angle \(117.271^\circ\) and substituting into the cosine rule. A1 for showing the correct substitution: \(AC^2 = 7^2 + 9^2 - 2(7)(9) \cos(117.271^\circ)\). A1 for \(AC \approx 13.7\text{ cm}\).

(a) Let the center of the circular cross-section be \(O\). A perpendicular line from \(O\) to the water surface bisects the chord of width \(24\text{ cm}\) into two segments of \(12\text{ cm}\). The distance \(d\) from the center \(O\) to the chord is \(d = \sqrt{15^2 - 12^2} = \sqrt{225 - 144} = 9\text{ cm}\). If the water level is below the center, the depth is \(h = 15 - 9 = 6\text{ cm}\). If the water level is above the center, the depth is \(h = 15 + 9 = 24\text{ cm}\). (b) When the pipe is less than half full, \(h = 6\text{ cm}\). The angle \(\theta\) subtended by the water surface at the center satisfies \(\cos(\theta/2) = 9/15 = 0.6\), which gives \(\theta = 2 \arccos(0.6) \approx 1.8546\text{ radians}\). The cross-sectional area of the water is the area of the circular segment: \(A = A_{\text{sector}} - A_{\text{triangle}} = \frac{1}{2} r^2 (\theta - \sin \theta) = \frac{1}{2} (15)^2 (1.8546 - \sin 1.8546) \approx 112.5(1.8546 - 0.96) = 112.5(0.8946) \approx 100.64\text{ cm}^2\). To 3 significant figures, the area is \(101\text{ cm}^2\).

(a) M1 for finding the distance from the center to the chord using Pythagoras' theorem. A1 for 6 cm. A1 for 24 cm. (b) M1 for finding the central angle \(\theta\) in radians (or degrees). M1 for calculating the area of the segment as the difference between the sector and the triangle. A1 for \(101\text{ cm}^2\).

(a) We are given that \(P(X > 180) = 0.15\), which is equivalent to \(P(Z > \frac{180-150}{\sigma}) = 0.15\). Using a GDC for the standard normal distribution, we find the z-score such that \(P(Z > z) = 0.15\) is \(z \approx 1.0364\). Setting \(\frac{30}{\sigma} = 1.0364\), we get \(\sigma = \frac{30}{1.0364} \approx 28.945\text{ grams}\). To 3 significant figures, \(\sigma = 28.9\text{ grams}\). (b) We want to find the mass \(x\) such that \(P(X > x) = 0.05\), or \(P(X \le x) = 0.95\). Using the GDC's inverse normal function with \(\mu = 150\) and \(\sigma \approx 28.945\), we obtain \(x \approx 197.61\text{ grams}\). To 3 significant figures, the minimum mass is \(198\text{ grams}\).

(a) M1 for writing a correct probability statement \(P(X > 180) = 0.15\) or standardizing. A1 for finding the correct z-score of \(1.0364\). A1 for \(\sigma \approx 28.9\text{ grams}\). (b) M1 for setting up \(P(X > x) = 0.05\) or \(P(X \le x) = 0.95\). A1 for showing the use of inverse normal with the calculated \(\sigma\). A1 for \(198\text{ grams}\) (or \(197.6\text{ grams}\)).

(a) Since \(X\) follows a binomial distribution \(X \sim B(12, p)\), the mean is given by \(\mu = n p\). Substituting the known values: \(12 p = 3 \implies p = 0.25\). (b) (i) We require \(P(X = 4)\). Using the binomial probability formula or GDC (binompdf): \(P(X = 4) = \binom{12}{4} (0.25)^4 (0.75)^8 \approx 0.19357\). To 3 significant figures, the probability is \(0.194\). (ii) We require \(P(X \ge 5) = 1 - P(X \le 4)\). Using the cumulative binomial distribution on a GDC (binomcdf): \(P(X \le 4) \approx 0.84236\). Thus, \(P(X \ge 5) = 1 - 0.84236 \approx 0.15764\). To 3 significant figures, the probability is \(0.158\).

(a) M1 for using the mean formula \(np = 3\). A1 for \(p = 0.25\). (b) (i) M1 for identifying the binomial distribution formula or showing binompdf usage on GDC. A1 for \(0.194\). (ii) M1 for setting up \(1 - P(X \le 4)\) or equivalent sum. A1 for \(0.158\).

(a) The graphs of \(y = \cos(x)\) (a cosine wave symmetric about the y-axis) and \(y = x^2 - 1\) (a parabola opening upwards with vertex at \((0,-1)\)) intersect at two points. Using the solver or intersection function on a GDC, we find the coordinates of these points to be \((-1.177, 0.384)\) and \((1.177, 0.384)\). To 3 significant figures, the coordinates are \((-1.18, 0.384)\) and \((1.18, 0.384)\). (b) The area of the enclosed region is given by the integral of the upper curve minus the lower curve between the intersection points: \(A = \int_{-1.177}^{1.177} (\cos(x) - (x^2 - 1)) \, dx\). Using the GDC's numerical integration function, we evaluate this integral to obtain \(A \approx 3.1137\). To 3 significant figures, the area is \(3.11\).

(a) M1 for sketching both curves showing correct shapes and relative positions. A2 for coordinates of both intersection points correct (accept either 1.18 or 1.177 for x-coordinate, and 0.384 for y-coordinate). (b) M1 for writing a correct integral expression for the area, including correct limits. A2 for the correct evaluation of the area as \(3.11\).

(a) The particle is at rest when its velocity is zero, i.e., \(v(t) = 0\). We solve \(t^2 \cos(t) - t + 3 = 0\) for \(0 \le t \le 6\) using a GDC. This yields two solutions: \(t \approx 1.8899\text{ s}\) and \(t \approx 4.7895\text{ s}\). To 3 significant figures, the times are \(t = 1.89\text{ s}\) and \(t = 4.79\text{ s}\). (b) The total distance travelled is the integral of the speed: \(d = \int_{0}^{6} |v(t)| \, dt\). Using the numerical integration function on a GDC with the absolute value of the velocity function, we obtain \(d \approx 44.88\text{ m}\). To 3 significant figures, the total distance travelled is \(44.9\text{ m}\).

(a) M1 for setting \(v(t) = 0\). A1 for \(t \approx 1.89\text{ s}\). A1 for \(t \approx 4.79\text{ s}\). (b) M1 for expressing the total distance as the integral of the absolute value of velocity \(\int_{0}^{6} |v(t)| \, dt\) (or splitting the integral into three parts). A2 for the correct numerical evaluation of \(44.9\text{ m}\).

Let \(X\) represent the mass of a bird, where \(X \sim N(\mu, \sigma^2)\).

We are given:
\(P(X > 220) = 0.15 \implies P(X \le 220) = 0.85\)
\(P(X < 160) = 0.30\)

Standardizing the variables using \(Z \sim N(0, 1)\):
For \(P(Z \le z_1) = 0.85\):
\(z_1 = \Phi^{-1}(0.85) \approx 1.03643\)
So, \(\frac{220 - \mu}{\sigma} = 1.03643 \implies 220 - \mu = 1.03643\sigma\) (Equation 1)

For \(P(Z < z_2) = 0.30\):
\(z_2 = \Phi^{-1}(0.30) \approx -0.52440\)
So, \(\frac{160 - \mu}{\sigma} = -0.52440 \implies 160 - \mu = -0.52440\sigma\) (Equation 2)

Subtracting Equation 2 from Equation 1:
\(60 = 1.56083\sigma\)
\(\sigma = \frac{60}{1.56083} \approx 38.4410\)

Substituting \(\sigma\) back into Equation 2:
\(160 - \mu = -0.52440(38.4410)\)
\(\mu = 160 + 20.1585 \approx 180.159\)

To 3 significant figures:
\(\mu \approx 180\)
\(\sigma \approx 38.4\)

**[1 Mark]** for standardizing and attempting to find at least one z-score.
**[1 Mark]** for \(z_1 \approx 1.0364\).
**[1 Mark]** for \(z_2 \approx -0.5244\).
**[1 Mark]** for setting up a correct pair of simultaneous equations in terms of \(\mu\) and \(\sigma\).
**[1 Mark]** for solving to find \(\sigma \approx 38.4\) (accept 38.44).
**[1 Mark]** for solving to find \(\mu \approx 180\) (accept 180.16 or 180.2).

**Part A**

(a) We can write \( f_1(x) = \frac{\ln x}{x^{-1}} \). As \( x \to 0^+ \), \( \ln x \to -\infty \) and \( x^{-1} \to \infty \), which is an indeterminate form of type \( \frac{\infty}{\infty} \).
Applying L'HDpital's rule:
\( \lim_{x \to 0^+} \frac{\ln x}{x^{-1}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-x^{-2}} = \lim_{x \to 0^+} (-x) = 0 \). [3 marks]

(b) (i) For \( k = 2 \), \( f_2(x) = x(\ln x)^2 \).
Applying the product rule:
\( f'_2(x) = 1 \cdot (\ln x)^2 + x \cdot 2(\ln x) \cdot \frac{1}{x} = (\ln x)^2 + 2\ln x = \ln x (\ln x + 2) \).
To find the stationary points, set \( f'_2(x) = 0 \):
\( \ln x = 0 \implies x = 1 \)
\( \ln x + 2 = 0 \implies \ln x = -2 \implies x = e^{-2} \).
The corresponding y-coordinates are:
\( f_2(1) = 1(\ln 1)^2 = 0 \), giving the point \( (1, 0) \).
\( f_2(e^{-2}) = e^{-2} (\ln e^{-2})^2 = e^{-2} (-2)^2 = 4e^{-2} \), giving the point \( (e^{-2}, 4e^{-2}) \).

(ii) The second derivative is:
\( f''_2(x) = \frac{d}{dx}[(\ln x)^2 + 2\ln x] = \frac{2\ln x}{x} + \frac{2}{x} = \frac{2(\ln x + 1)}{x} \).
At \( x = 1 \):
\( f''_2(1) = \frac{2(0 + 1)}{1} = 2 > 0 \), so \( (1, 0) \) is a local minimum.
At \( x = e^{-2} \):
\( f''_2(e^{-2}) = \frac{2(-2 + 1)}{e^{-2}} = -2e^2 < 0 \), so \( (e^{-2}, 4e^{-2}) \) is a local maximum. [5 marks]

**Part B**

(c) (i) For \( f_k(x) = x(\ln x)^k \):
Using the product rule and chain rule:
\( f'_k(x) = 1 \cdot (\ln x)^k + x \cdot k(\ln x)^{k-1} \cdot \frac{1}{x} = (\ln x)^k + k(\ln x)^{k-1} = (\ln x)^{k-1}(\ln x + k) \).
(ii) Setting \( f'_k(x) = 0 \) gives:
\( (\ln x)^{k-1} = 0 \implies \ln x = 0 \implies x = 1 \)
\( \ln x + k = 0 \implies \ln x = -k \implies x = e^{-k} \). [5 marks]

(d) (i) Differentiating \( f'_k(x) = (\ln x)^k + k(\ln x)^{k-1} \):
\( f''_k(x) = k(\ln x)^{k-1} \frac{1}{x} + k(k-1)(\ln x)^{k-2} \frac{1}{x} = \frac{k(\ln x)^{k-2}}{x}(\ln x + k - 1) \).

(ii) For an inflection point, the second derivative must equal zero and change sign. \( f''_k(x) = 0 \) occurs when \( \ln x = 1-k \) (so \( x = e^{1-k} \)) or \( \ln x = 0 \) (so \( x = 1 \)).
Since \( k \ge 4 \) is even, \( k-2 \) is also even. Thus, the term \( (\ln x)^{k-2} \) is always non-negative, and does not change sign as \( x \) passes through \( 1 \). Also, the factor \( \ln x + k - 1 \) at \( x=1 \) is \( k-1 > 0 \), which does not change sign.
Therefore, \( f''_k(x) \) does not change sign at \( x = 1 \), meaning \( x = 1 \) is not an inflection point.
However, at \( x = e^{1-k} \), the factor \( \ln x + k - 1 \) changes sign from negative to positive as \( x \) increases through \( e^{1-k} \), while all other factors in \( f''_k(x) \) remain strictly positive. Hence, \( f''_k(x) \) changes sign at \( x = e^{1-k} \), so \( x = e^{1-k} \) is an inflection point. [4 marks]

**Part C**

(e) Let \( I_k = \int_0^1 x(\ln x)^k \, dx \).
Using integration by parts with \( u = (\ln x)^k \) and \( dv = x \, dx \):
\( du = k(\ln x)^{k-1} \frac{1}{x} \, dx \) and \( v = \frac{1}{2} x^2 \).
\( I_k = \left[ \frac{1}{2} x^2 (\ln x)^k \right]_0^1 - \int_0^1 \frac{1}{2} x^2 \cdot k(\ln x)^{k-1} \frac{1}{x} \, dx \).
For the boundary term, \( \lim_{x \to 1^-} \frac{1}{2} x^2 (\ln x)^k = 0 \).
As \( x \to 0^+ \), \( \lim_{x \to 0^+} x^2 (\ln x)^k = 0 \) (by repeated applications of L'HDpital's rule).
So the boundary term vanishes.
\( I_k = 0 - \frac{k}{2} \int_0^1 x (\ln x)^{k-1} \, dx = -\frac{k}{2} I_{k-1} \). [5 marks]

(f) Let the statement be \( P(k): I_k = (-1)^k \frac{k!}{2^{k+1}} \).
**Base Case (k = 0):**
\( I_0 = \int_0^1 x \, dx = \left[ \frac{1}{2} x^2 \right]_0^1 = \frac{1}{2} \).
Using the formula: \( (-1)^0 \frac{0!}{2^1} = \frac{1}{2} \).
Since \( I_0 = \frac{1}{2} \), \( P(0) \) is true.

**Inductive Step:**
Assume \( P(m) \) is true for some \( m \in \mathbb{N} \), i.e., \( I_m = (-1)^m \frac{m!}{2^{m+1}} \).
We must show that \( P(m+1) \) is true, i.e., \( I_{m+1} = (-1)^{m+1} \frac{(m+1)!}{2^{m+2}} \).
Using the reduction formula from part (e):
\( I_{m+1} = -\frac{m+1}{2} I_m \).
Substituting the inductive hypothesis:
\( I_{m+1} = -\frac{m+1}{2} \left( (-1)^m \frac{m!}{2^{m+1}} \right) \)
\( I_{m+1} = (-1)^{m+1} \frac{(m+1) \cdot m!}{2 \cdot 2^{m+1}} \)
\( I_{m+1} = (-1)^{m+1} \frac{(m+1)!}{2^{m+2}} \).
Thus, \( P(m) \implies P(m+1) \).
Since the base case is true and the inductive step holds, by mathematical induction, \( P(k) \) is true for all \( k \in \mathbb{N} \). [5 marks]

**Part A**

(a) [3 marks]
- M1: for rewriting \( f_1(x) = \frac{\ln x}{x^{-1}} \) to apply L'HDpital's rule.
- M1: for differentiating numerator and denominator correctly to get \( \frac{1/x}{-1/x^2} \).
- A1: for simplifying to \( -x \) and taking the limit to show 0 (AG).

(b) [5 marks]
- M1: for applying the product rule correctly to find \( f'_2(x) = \ln x (\ln x + 2) \).
- A1: for finding correct x-coordinates \( x = 1 \) and \( x = e^{-2} \).
- A1: for finding correct y-coordinates \( (1, 0) \) and \( (e^{-2}, 4e^{-2}) \).
- M1: for checking the second derivative at these points.
- A1: for correctly identifying \( (1, 0) \) as a minimum and \( (e^{-2}, 4e^{-2}) \) as a maximum.

**Part B**

(c) [5 marks]
- M1: for using product and chain rules to differentiate \( f_k(x) \).
- A1: for obtaining \( f'_k(x) = (\ln x)^k + k(\ln x)^{k-1} \) and factoring it to show the given result (AG).
- M1: for setting \( f'_k(x) = 0 \).
- A1: for \( x = 1 \).
- A1: for \( x = e^{-k} \).

(d) [4 marks]
- M1: for differentiating \( f'_k(x) \) again using product/chain rule.
- A1: for simplifying to show \( f''_k(x) = \frac{k(\ln x)^{k-2}}{x}(\ln x + k - 1) \) (AG).
- R1: for explaining that since \( k \ge 4 \) is even, \( (\ln x)^{k-2} \ge 0 \) does not change sign at \( x=1 \), so \( x=1 \) is not an inflection point.
- R1: for explaining that \( \ln x + k - 1 \) changes sign at \( x = e^{1-k} \), so \( x = e^{1-k} \) is an inflection point.

**Part C**

(e) [5 marks]
- M1: for choosing appropriate \( u \) and \( dv \) for integration by parts.
- A1: for correct differentiation of \( u \) and integration of \( dv \).
- A1: for establishing the correct boundary term limits as 0.
- M1: for writing the integration by parts formula.
- A1: for obtaining the recurrence relation \( I_k = -\frac{k}{2} I_{k-1} \) (AG).

(f) [5 marks]
- A1: for checking the base case \( k = 0 \) correctly on both sides.
- M1: for stating the inductive assumption \( P(m) \) is true.
- M1: for substituting \( I_m \) into the recurrence relation for \( I_{m+1} \).
- A1: for correctly simplifying the expression to the form \( (-1)^{m+1} \frac{(m+1)!}{2^{m+2}} \).
- R1: for a complete conclusion statement including both base case and induction step logic.

**Part A**

(a) For \( n = 0 \):
\( T_0(x) = \cos(0 \cdot \arccos x) = \cos(0) = 1 \).
For \( n = 1 \):
\( T_1(x) = \cos(1 \cdot \arccos x) = x \). [3 marks]

(b) Let \( \theta = \arccos x \), so \( x = \cos \theta \) and \( T_n(x) = \cos(n\theta) \).
Letting \( A = n\theta \) and \( B = \theta \) in the identity \( \cos(A+B) + \cos(A-B) = 2 \cos A \cos B \):
\( \cos((n+1)\theta) + \cos((n-1)\theta) = 2 \cos(n\theta) \cos\theta \).
Substituting \( T_{n+1}(x) = \cos((n+1)\theta) \), \( T_{n-1}(x) = \cos((n-1)\theta) \), \( T_n(x) = \cos(n\theta) \), and \( x = \cos\theta \):
\( T_{n+1}(x) + T_{n-1}(x) = 2x T_n(x) \).
Rearranging this gives:
\( T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x) \). [4 marks]

(c) Using the recurrence relation:
(i) \( T_2(x) = 2x T_1(x) - T_0(x) = 2x(x) - 1 = 2x^2 - 1 \).
(ii) \( T_3(x) = 2x T_2(x) - T_1(x) = 2x(2x^2 - 1) - x = 4x^3 - 3x \).
(iii) \( T_4(x) = 2x T_3(x) - T_2(x) = 2x(4x^3 - 3x) - (2x^2 - 1) = 8x^4 - 8x^2 + 1 \). [3 marks]

**Part B**

(d) Setting \( T_n(x) = 0 \) gives \( \cos(n \arccos x) = 0 \).
Let \( \theta = \arccos x \). Since \( x \in [-1, 1] \), \( \theta \in [0, \pi] \), so \( n\theta \in [0, n\pi] \).
\( \cos(n\theta) = 0 \implies n\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \dots, \frac{(2n-1)\pi}{2} \).
Thus, \( n\theta = \frac{2k-1}{2} \pi \) for \( k = 1, 2, \dots, n \).
Dividing by \( n \):
\( \theta = \frac{2k-1}{2n} \pi \).
Since \( x = \cos\theta \):
\( x_k = \cos\left( \frac{2k-1}{2n} \pi \right) \) for \( k = 1, 2, \dots, n \).
Since the cosine function is strictly decreasing on \( [0, \pi] \), each of these angles yields a distinct value of \( x_k \). Thus there are exactly \( n \) distinct roots. [4 marks]

(e) The local extrema of \( T_n(x) = \cos(n\theta) \) occur when the function reaches its absolute maximum or minimum value, which are \( 1 \) and \( -1 \).
This occurs when \( \cos(n\theta) = \pm 1 \implies n\theta = m\pi \) for \( m \in \mathbb{Z} \).
Since \( \theta \in [0, \pi] \), we have \( m = 0, 1, 2, \dots, n \).
Thus, the x-coordinates are \( x_m = \cos\left( \frac{m\pi}{n} \right) \) for \( m = 0, 1, 2, \dots, n \).
At these coordinates, \( T_n(x_m) = \cos(m\pi) = (-1)^m \), which is either \( 1 \) or \( -1 \). [4 marks]

**Part C**

(f) Let \( y = \cos(n \arccos x) \).
Using the chain rule with \( u = n \arccos x \):
\( \frac{dy}{dx} = -\sin(n \arccos x) \cdot \frac{d}{dx}[n \arccos x] = -\sin(n \arccos x) \cdot \left( -\frac{n}{\sqrt{1-x^2}} \right) = \frac{n \sin(n \arccos x)}{\sqrt{1-x^2}} \).
Multiplying both sides by \( \sqrt{1-x^2} \):
\( \sqrt{1-x^2} \frac{dy}{dx} = n \sin(n \arccos x) \).
Squaring both sides:
\( (1-x^2) \left(\frac{dy}{dx}\right)^2 = n^2 \sin^2(n \arccos x) \).
Since \( \sin^2(n \arccos x) = 1 - \cos^2(n \arccos x) = 1 - y^2 \), we obtain:
\( (1-x^2) \left(\frac{dy}{dx}\right)^2 = n^2 (1 - y^2) \). [5 marks]

(g) Differentiating both sides of \( (1-x^2) \left(\frac{dy}{dx}\right)^2 = n^2 (1 - y^2) \) with respect to \( x \):
Left-hand side (using the product and chain rules):
\( \frac{d}{dx}\left[ (1-x^2) \left(\frac{dy}{dx}\right)^2 \right] = -2x \left(\frac{dy}{dx}\right)^2 + (1-x^2) \cdot 2 \frac{dy}{dx} \frac{d^2 y}{dx^2} \).
Right-hand side (using the chain rule):
\( \frac{d}{dx}\left[ n^2 (1 - y^2) \right] = n^2 (-2y) \frac{dy}{dx} = -2n^2 y \frac{dy}{dx} \).
Equating the two sides:
\( -2x \left(\frac{dy}{dx}\right)^2 + 2(1-x^2) \frac{dy}{dx} \frac{d^2 y}{dx^2} = -2n^2 y \frac{dy}{dx} \).
Dividing both sides by \( 2 \frac{dy}{dx} \) (assuming \( \frac{dy}{dx} \ne 0 \)):
\( -x \frac{dy}{dx} + (1-x^2) \frac{d^2 y}{dx^2} = -n^2 y \).
Rearranging this gives:
\( (1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + n^2 y = 0 \). [5 marks]

**Part A**

(a) [3 marks]
- M1: for substituting \( n = 0 \) into the definition of \( T_n(x) \) and simplifying.
- A1: for showing \( T_0(x) = 1 \).
- A1: for showing \( T_1(x) = x \).

(b) [4 marks]
- M1: for setting \( \theta = \arccos x \) and relating \( T_n(x) \) to \( \cos(n\theta) \).
- M1: for substituting \( A = n\theta \) and \( B = \theta \) into the given identity.
- A1: for obtaining \( \cos((n+1)\theta) + \cos((n-1)\theta) = 2 \cos(n\theta) \cos\theta \).
- A1: for substituting \( T_k(x) \) back and showing the recurrence relation (AG).

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

**Part B**

(d) [4 marks]
- M1: for setting \( T_n(x) = 0 \) and stating \( n\theta = \arccos(0) \) within appropriate range.
- M1: for finding the general roots for \( n\theta \) as \( \frac{2k-1}{2}\pi \).
- A1: for \( x_k = \cos\left( \frac{2k-1}{2n}\pi \right) \).
- R1: for explaining why there are exactly \( n \) distinct roots (since cosine is strictly decreasing on \( [0, \pi] \)).

(e) [4 marks]
- M1: for recognizing that extrema occur when \( \cos(n\theta) = \pm 1 \).
- A1: for solving to find \( n\theta = m\pi \).
- M1: for identifying the domain of \( m \) as \( 0 \le m \le n \).
- A1: for concluding \( x_m = \cos\left(\frac{m\pi}{n}\right) \) and stating that the y-value is \( (-1)^m \) (which is \( 1 \) or \( -1 \)).

**Part C**

(f) [5 marks]
- M1: for using the chain rule to differentiate \( \cos(n \arccos x) \).
- A1: for getting the correct derivative \( \frac{n \sin(n \arccos x)}{\sqrt{1-x^2}} \).
- M1: for rearranging to group terms as \( \sqrt{1-x^2} y' = n \sin(n\theta) \).
- A1: for squaring both sides correctly.
- A1: for using Pythagorean identity to write \( \sin^2(n\theta) = 1 - y^2 \) and obtaining the given relation (AG).

(g) [5 marks]
- M1: for differentiating LHS using product and chain rules.
- A1: for correct LHS derivative: \( -2x (y')^2 + 2(1-x^2)y'y'' \).
- M1: for differentiating RHS using chain rule.
- A1: for correct RHS derivative: \( -2n^2 y y' \).
- A1: for dividing by \( 2y' \) and rearranging to get the final second-order ODE (AG).