2026 Edexcel IAL Mathematics (YMA01) Practice Paper with Answers

(a) Let \(u = x\) and \(dv = \sec^2(2x) \, dx\).
Then \(du = dx\) and \(v = \frac{1}{2}\tan(2x)\).

Applying the integration by parts formula \(\int u \, dv = uv - \int v \, du\):
\[ \int x \sec^2(2x) \, dx = \frac{1}{2}x \tan(2x) - \int \frac{1}{2}\tan(2x) \, dx \]

Since \(\int \tan(2x) \, dx = -\frac{1}{2}\ln|\cos(2x)|\), we have:
\[ \int x \sec^2(2x) \, dx = \frac{1}{2}x \tan(2x) - \frac{1}{2} \left( -\frac{1}{2}\ln|\cos(2x)| \right) + C \]
\[ = \frac{1}{2}x \tan(2x) + \frac{1}{4} \ln|\cos(2x)| + C \]

(b) The volume \(V\) of revolution is given by:
\[ V = \pi \int_{0}^{\pi/8} y^2 \, dx = \pi \int_{0}^{\pi/8} x \sec^2(2x) \, dx \]

Using the result from part (a):
\[ V = \pi \left[ \frac{1}{2}x \tan(2x) + \frac{1}{4} \ln|\cos(2x)| \right]_{0}^{\pi/8} \]

At the upper limit \(x = \frac{\pi}{8}\):
\[ \frac{1}{2}\left(\frac{\pi}{8}\right) \tan\left(\frac{\pi}{4}\right) + \frac{1}{4} \ln\left|\cos\left(\frac{\pi}{4}\right)\right| = \frac{\pi}{16}(1) + \frac{1}{4} \ln\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{16} - \frac{1}{8}\ln 2 \]

At the lower limit \(x = 0\):
\[ 0 + \frac{1}{4} \ln(1) = 0 \]

Thus, the exact volume is:
\[ V = \pi \left( \frac{\pi}{16} - \frac{1}{8}\ln 2 \right) = \frac{\pi^2}{16} - \frac{\pi}{8} \ln 2 \]

(a)
- M1: Applies integration by parts formula with correct assignments of \(u = x\) and \(dv = \sec^2(2x)\).
- A1: Obtains \(du = dx\) and \(v = \frac{1}{2}\tan(2x)\) correctly.
- M1: Integrates \(\int \tan(2x) \, dx\) to obtain a term of the form \(k\ln|\cos(2x)|\).
- A1: Obtains \(-\frac{1}{2}\ln|\cos(2x)|\) correctly.
- A1*: Fully completes the proof to show \(\frac{1}{2}x \tan(2x) + \frac{1}{4} \ln|\cos(2x)| + C\) with no errors.

(b)
- B1: States or uses the volume of revolution formula \(V = \pi \int y^2 \, dx\).
- M1: Applies the limits \(0\) and \(\frac{\pi}{8}\) to the integrated expression from part (a).
- A1: Evaluates the upper limit correctly to get \(\frac{\pi}{16} - \frac{1}{8}\ln 2\).
- A1: Evaluates the lower limit correctly to get \(0\).
- A1*: Fully simplifies the result to establish the given exact volume of \(\frac{\pi^2}{16} - \frac{\pi}{8} \ln 2\).

(a) Applying the quotient rule with \(u = e^{3x}\) and \(v = 2x+1\):
\[ \frac{du}{dx} = 3e^{3x}, \quad \frac{dv}{dx} = 2 \]
\[ \frac{dy}{dx} = \frac{3e^{3x}(2x+1) - 2e^{3x}}{(2x+1)^2} \]
Factorise \(e^{3x}\) out of the numerator:
\[ \frac{dy}{dx} = \frac{e^{3x}(3(2x+1) - 2)}{(2x+1)^2} = \frac{e^{3x}(6x+1)}{(2x+1)^2} \]
For a stationary point, we set \(\frac{dy}{dx} = 0\):
\[ e^{3x}(6x+1) = 0 \]
Since \(e^{3x} > 0\) for all real \(x\), we must have:
\[ 6x + 1 = 0 \implies x = -\frac{1}{6} \]

(b) At the point where \(x = 0\):
\[ y = \frac{e^0}{2(0)+1} = 1 \]
Substitute \(x = 0\) into \(\frac{dy}{dx}\) to find the gradient of the tangent:
\[ m_t = \frac{e^0(6(0)+1)}{(2(0)+1)^2} = \frac{1(1)}{1} = 1 \]
The gradient of the normal, \(m_n\), is:
\[ m_n = -\frac{1}{m_t} = -1 \]
The equation of the normal at \((0, 1)\) is:
\[ y - 1 = -1(x - 0) \]
\[ y - 1 = -x \]
Rearranging into the requested form:
\[ x + y - 1 = 0 \]

(a)
- M1: Attempts quotient rule (or product rule on \(e^{3x}(2x+1)^{-1}\)) with clear structure of \(\frac{u'v - v'u}{v^2}\).
- A1: Correctly differentiates numerator and denominator to find \(u' = 3e^{3x}\) and \(v' = 2\).
- A1: Obtains a correct unsimplified expression for \(\frac{dy}{dx}\).
- M1: Sets their \(\frac{dy}{dx} = 0\) and attempts to solve for \(x\).
- A1*: Completes calculations to show \(x = -\frac{1}{6}\) cleanly.

(b)
- B1: Finds the correct y-coordinate of the point of interest, \(y = 1\).
- M1: Substitutes \(x = 0\) into \(\frac{dy}{dx}\) to find the gradient of the tangent.
- A1: Identifies the gradient of the tangent as \(1\).
- M1: Applies the perpendicular gradient rule \(m_n = -\frac{1}{m_t}\) and attempts to write the straight-line equation of the normal.
- A1: Expresses the normal in the form \(x + y - 1 = 0\) (or any integer multiple thereof).

(a) Differentiating the equation with respect to \(x\) implicitly:
\[ 2x - 2\left(y + x\frac{dy}{dx}\right) + 4y\frac{dy}{dx} + 2 = 0 \]
\[ 2x - 2y - 2x\frac{dy}{dx} + 4y\frac{dy}{dx} + 2 = 0 \]
Divide all terms by 2:
\[ x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} + 1 = 0 \]
Group the terms containing \(\frac{dy}{dx}\):
\[ (2y - x)\frac{dy}{dx} = y - x - 1 \]
Solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{y - x - 1}{2y - x} \]

(b) The tangent is parallel to the y-axis where the gradient \(\frac{dy}{dx}\) is undefined, meaning the denominator is zero:
\[ 2y - x = 0 \implies x = 2y \]
Substitute \(x = 2y\) back into the original curve equation:
\[ (2y)^2 - 2(2y)y + 2y^2 + 2(2y) = 6 \]
\[ 4y^2 - 4y^2 + 2y^2 + 4y = 6 \]
\[ 2y^2 + 4y - 6 = 0 \]
Divide by 2:
\[ y^2 + 2y - 3 = 0 \]
\[ (y + 3)(y - 1) = 0 \implies y = 1 \text{ or } y = -3 \]
Now find the corresponding values of \(x\) using \(x = 2y\):
- When \(y = 1\), \(x = 2(1) = 2\). Point: \((2, 1)\).
- When \(y = -3\), \(x = 2(-3) = -6\). Point: \((-6, -3)\).

Thus, the coordinates of the points are \((2, 1)\) and \((-6, -3)\).

(a)
- M1: Attempts to differentiate \(2xy\) implicitly using the product rule.
- A1: Obtains \(2y + 2x\frac{dy}{dx}\) for the product rule term.
- A1: Fully differentiates the remaining terms to get \(2x + 4y\frac{dy}{dx} + 2 = 0\).
- M1: Collects like terms to group all \(\frac{dy}{dx}\) terms together on one side.
- A1*: Successfully factorises and rearranges to produce the given expression cleanly.

(b)
- M1: Sets the denominator of \(\frac{dy}{dx}\) equal to zero, obtaining \(x = 2y\).
- M1: Substitutes \(x = 2y\) (or equivalent) back into the original curve equation to get a quadratic in one variable.
- A1: Forms the correct simplified quadratic equation, e.g., \(2y^2 + 4y - 6 = 0\).
- M1: Solves their quadratic to find two values for \(y\).
- A1: Obtains the correct coordinates \((2, 1)\) and \((-6, -3)\).

(a) Let \(u = 1 + e^x\), which means \(e^x = u - 1\).
Differentiating gives \(du = e^x \, dx\), so \(dx = \frac{du}{e^x} = \frac{du}{u - 1}\).

The integrand term becomes:
\[ \frac{e^{2x}}{1+e^x} \, dx = \frac{e^x \cdot e^x}{1+e^x} \, dx = \frac{(u - 1) e^x}{u} \cdot \frac{du}{e^x} = \frac{u - 1}{u} \, du \]

Now we change the limits of integration:
- When \(x = 0\), \(u = 1 + e^0 = 2\).
- When \(x = \ln 3\), \(u = 1 + e^{\ln 3} = 4\).

The integral is rewritten as:
\[ \int_{2}^{4} \frac{u-1}{u} \, du = \int_{2}^{4} \left(1 - \frac{1}{u}\right) \, du \]
\[ = \left[ u - \ln u \right]_{2}^{4} \]
\[ = (4 - \ln 4) - (2 - \ln 2) \]
\[ = 2 - \ln 4 + \ln 2 = 2 - 2\ln 2 + \ln 2 = 2 - \ln 2 \]

(b) We can write \(\tan^3(\theta)\) as \(\tan(\theta)\tan^2(\theta) = \tan(\theta)(\sec^2(\theta) - 1)\).
Thus:
\[ \int_{0}^{\pi/4} \tan^3(\theta) \, d\theta = \int_{0}^{\pi/4} \left( \tan(\theta)\sec^2(\theta) - \tan(\theta) \right) \, d\theta \]

We split this into two integrals:
1) For \(I_1 = \int_{0}^{\pi/4} \tan(\theta)\sec^2(\theta) \, d\theta\), use substitution \(w = \tan(\theta)\), so \(dw = \sec^2(\theta) \, d\theta\).
- Limit at \(\theta = 0 \implies w = 0\).
- Limit at \(\theta = \pi/4 \implies w = 1\).
\[ I_1 = \int_{0}^{1} w \, dw = \left[ \frac{1}{2}w^2 \right]_{0}^{1} = \frac{1}{2} \]

2) For \(I_2 = \int_{0}^{\pi/4} \tan(\theta) \, d\theta\):
\[ I_2 = \left[ \ln|\sec(\theta)| \right]_{0}^{\pi/4} = \ln|\sec(\pi/4)| - \ln|\sec(0)| = \ln(\sqrt{2}) - \ln(1) = \frac{1}{2}\ln 2 \]

Combining the results:
\[ \int_{0}^{\pi/4} \tan^3(\theta) \, d\theta = I_1 - I_2 = \frac{1}{2} - \frac{1}{2}\ln 2 \]

(a)
- M1: Uses the substitution \(u = 1 + e^x\) to express \(dx\) in terms of \(du\).
- A1: Obtains the correct simplified integral \(\int \frac{u-1}{u} \, du\).
- B1: Changes the integration limits correctly from \([0, \ln 3]\) to \([2, 4]\).
- M1: Integrates \(1 - \frac{1}{u}\) to get \(u - \ln u\) and applies the new limits.
- A1: Fully simplifies to get the exact answer \(2 - \ln 2\).

(b)
- M1: Uses trigonometric identity \(\tan^2(\theta) = \sec^2(\theta) - 1\) to rewrite the integrand.
- A1: Splits into \(\tan(\theta)\sec^2(\theta)\) and \(\tan(\theta)\) correctly.
- M1: Uses a valid method to integrate \(\tan(\theta)\sec^2(\theta)\) (e.g. substitution or inspection).
- A1: Integrates both parts correctly to obtain \(\frac{1}{2}\tan^2(\theta) - \ln|\sec(\theta)|\) (or equivalent).
- A1*: Evaluates between limits \(0\) and \(\pi/4\) correctly to yield the shown result \(\frac{1}{2} - \frac{1}{2}\ln 2\).

(a) The perimeter \(P\) of a sector with radius \(r\) and angle \(\theta\) is:
\[ P = 2r + r\theta \]
We are given \(P = 40\), so:
\[ 2r + r\theta = 40 \implies r\theta = 40 - 2r \implies \theta = \frac{40 - 2r}{r} \]

The area \(A\) of the sector is given by:
\[ A = \frac{1}{2}r^2\theta \]
Substitute \(\theta = \frac{40 - 2r}{r}\) into this equation:
\[ A = \frac{1}{2}r^2 \left( \frac{40 - 2r}{r} \right) = \frac{1}{2}r(40 - 2r) = 20r - r^2 \quad \text{(as required)} \]

(b) Differentiating \(A\) with respect to \(r\):
\[ \frac{dA}{dr} = 20 - 2r \]
To find the stationary point, set \(\frac{dA}{dr} = 0\):
\[ 20 - 2r = 0 \implies r = 10 \]

To justify that this value of \(r\) yields a maximum area, find the second derivative:
\[ \frac{d^2A}{dr^2} = -2 \]
Since \(\frac{d^2A}{dr^2} < 0\) for all values of \(r\), the stationary point represents a maximum.

The maximum possible area is:
\[ A_{\text{max}} = 20(10) - (10)^2 = 200 - 100 = 100 \, \text{cm}^2 \]

(c) The corresponding angle \(\theta\) is obtained by substituting \(r = 10\) into the expression for \(\theta\):
\[ \theta = \frac{40 - 2(10)}{10} = \frac{20}{10} = 2 \, \text{radians} \]

(a)
- B1: Writes a correct formula for the perimeter: \(P = 2r + r\theta = 40\).
- M1: Rearranges the perimeter equation to express \(\theta\) in terms of \(r\).
- M1: Substitutes their expression for \(\theta\) into the sector area formula \(A = \frac{1}{2}r^2\theta\).
- A1*: Simplifies correctly to show the target formula \(A = 20r - r^2\).

(b)
- M1: Differentiates \(A = 20r - r^2\) with respect to \(r\).
- A1: Obtains the derivative \(\frac{dA}{dr} = 20 - 2r\), sets it to \(0\), and finds \(r = 10\).
- M1: Finds the second derivative \(\frac{d^2A}{dr^2} = -2\) and notes that it is negative to show it is a maximum.
- A1: Evaluates the maximum area as \(100\).

(c)
- M1: Uses their value of \(r\) in a correct equation relating \(r\) and \(\theta\).
- A1: Obtains \(\theta = 2\) radians.

(a) To find the points of intersection, set the equations of \(C\) and \(L\) equal to each other:
\[ 9 + 2x - x^2 = x + 3 \]
Rearrange into standard quadratic form:
\[ x^2 - x - 6 = 0 \]
Factorise the quadratic:
\[ (x - 3)(x + 2) = 0 \]
Thus, the x-coordinates of the intersection points are \(x = -2\) and \(x = 3\).

(b) The curve \(C\) lies above the line \(L\) within the interval \([-2, 3]\). The area \(A\) of the bounded region is given by:
\[ A = \int_{-2}^{3} (y_{\text{curve}} - y_{\text{line}}) \, dx \]
\[ A = \int_{-2}^{3} \left( (9 + 2x - x^2) - (x + 3) \right) \, dx \]
\[ A = \int_{-2}^{3} (6 + x - x^2) \, dx \]

Now we integrate term by term:
\[ A = \left[ 6x + \frac{1}{2}x^2 - \frac{1}{3}x^3 \right]_{-2}^{3} \]

Substitute the upper limit \(x = 3\):
\[ 6(3) + \frac{1}{2}(3)^2 - \frac{1}{3}(3)^3 = 18 + \frac{9}{2} - 9 = 9 + 4.5 = \frac{27}{2} \]

Substitute the lower limit \(x = -2\):
\[ 6(-2) + \frac{1}{2}(-2)^2 - \frac{1}{3}(-2)^3 = -12 + 2 + \frac{8}{3} = -10 + \frac{8}{3} = -\frac{22}{3} \]

Find the difference:
\[ A = \frac{27}{2} - \left( -\frac{22}{3} \right) = \frac{27}{2} + \frac{22}{3} = \frac{81 + 44}{6} = \frac{125}{6} \]

(a)
- M1: Equates the curve and the line to establish a quadratic equation.
- A1: Obtains the correct quadratic equation \(x^2 - x - 6 = 0\) (or equivalent).
- A1: Correctly factorises or solves to find \(x = -2\) and \(x = 3\).

(b)
- M1: Sets up the definite integral \(\int (y_{\text{curve}} - y_{\text{line}}) \, dx\) with their limits from part (a).
- A1: Obtains the correct simplified integrand \(6 + x - x^2\).
- M1: Integrates the terms correctly, increasing powers by 1.
- A1: Gets the correct integrated expression \(6x + \frac{1}{2}x^2 - \frac{1}{3}x^3\).
- M1: Substitutes their limits \(3\) and \(-2\) into the integrated expression.
- A1: Shows correct intermediate evaluations (\(\frac{27}{2}\) and \(-\frac{22}{3}\)).
- A1: Calculates the final exact area of \(\frac{125}{6}\) (or equivalent mixed number \(20\frac{5}{6}\)).

(a) Calculate missing \(y\) values to 4 decimal places:
- For \(x = 1.5\):
\[ y = 3\sqrt{1.5} + \frac{4}{1.5^2} \approx 3.67423 + 1.77778 = 5.4520 \]
- For \(x = 2.5\):
\[ y = 3\sqrt{2.5} + \frac{4}{2.5^2} \approx 4.74342 + 0.64000 = 5.3834 \]

Completed Table:
\begin{tabular}{|c|c|c|c|c|c|}
\hline
x & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 \\
\hline
y & 7.0000 & 5.4520 & 5.2426 & 5.3834 & 5.6406 \\
\hline
\end{tabular}

(b) The strip width is \(h = \frac{3.0 - 1.0}{4} = 0.5\).
Using the Trapezium Rule:
\[ I \approx \frac{h}{2} \left[ y_0 + y_4 + 2(y_1 + y_2 + y_3) \right] \]
\[ I \approx \frac{0.5}{2} \left[ 7.0000 + 5.6406 + 2(5.4520 + 5.2426 + 5.3834) \right] \]
\[ I \approx 0.25 \left[ 12.6406 + 2(16.0780) \right] \]
\[ I \approx 0.25 \left[ 12.6406 + 32.1560 \right] = 0.25 \times 44.7966 = 11.19915 \approx 11.2 \quad \text{(to 3 sig. figs.)} \]

(c) We find the exact integral:
\[ \int_{1}^{3} \left(3x^{1/2} + 4x^{-2}\right) \, dx = \left[ 2x^{3/2} - 4x^{-1} \right]_{1}^{3} = \left[ 2x\sqrt{x} - \frac{4}{x} \right]_{1}^{3} \]
Substitute the limits:
- At \(x = 3\):
\[ 2(3)\sqrt{3} - \frac{4}{3} = 6\sqrt{3} - \frac{4}{3} \]
- At \(x = 1\):
\[ 2(1)\sqrt{1} - \frac{4}{1} = 2 - 4 = -2 \]

Subtract the lower limit from the upper limit:
\[ \left(6\sqrt{3} - \frac{4}{3}\right) - (-2) = 6\sqrt{3} - \frac{4}{3} + 2 = 6\sqrt{3} + \frac{2}{3} \]
Thus \(a = 6\) and \(b = \frac{2}{3}\).

(a)
- B1: Calculates \(y(1.5) = 5.4520\) correctly.
- B1: Calculates \(y(2.5) = 5.3834\) correctly.

(b)
- B1: Identifies the correct strip width \(h = 0.5\).
- M1: Uses the correct trapezium rule formula structure \(\frac{h}{2}[\text{first} + \text{last} + 2(\text{middle sums})]\) with their values.
- A1: Obtains the correct unsimplified bracket sum calculation.
- A1: Finds \(11.2\) rounded to 3 significant figures.

(c)
- M1: Integrates the terms correctly, getting \(2x^{3/2}\) and \(-4x^{-1}\) (allow power errors of 1 for M1).
- A1: Obtains the correct integrated expression \(2x\sqrt{x} - \frac{4}{x}\).
- M1: Substitutes the limits \(3\) and \(1\) into their integrated expression.
- A1: Yields the correct exact simplified form \(6\sqrt{3} + \frac{2}{3}\).

(a) Differentiating the polynomial equation term by term with respect to \(x\):
\[ \frac{dy}{dx} = 3x^2 - 12x + 9 \]

(b) To find stationary points, set \(\frac{dy}{dx} = 0\):
\[ 3x^2 - 12x + 9 = 0 \]
Divide the entire equation by 3:
\[ x^2 - 4x + 3 = 0 \]
Factorise the quadratic expression:
\[ (x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3 \]

Now substitute the x-coordinates back into the original curve equation to find their corresponding y-coordinates:
- For \(x = 1\):
\[ y = 1^3 - 6(1)^2 + 9(1) + 5 = 1 - 6 + 9 + 5 = 9 \]
So, one stationary point is at \((1, 9)\).

- For \(x = 3\):
\[ y = 3^3 - 6(3)^2 + 9(3) + 5 = 27 - 54 + 27 + 5 = 5 \]
So, the second stationary point is at \((3, 5)\).

(c) Differentiating \(\frac{dy}{dx}\) with respect to \(x\) to find the second derivative:
\[ \frac{d^2y}{dx^2} = 6x - 12 \]

Now we evaluate the second derivative at each of the x-coordinates of the stationary points:
- At \(x = 1\):
\[ \frac{d^2y}{dx^2} = 6(1) - 12 = -6 \]
Since \(-6 < 0\), the stationary point \((1, 9)\) is a local maximum.

- At \(x = 3\):
\[ \frac{d^2y}{dx^2} = 6(3) - 12 = 6 \]
Since \(6 > 0\), the stationary point \((3, 5)\) is a local minimum.

(a)
- M1: Attempts to differentiate the equation term by term.
- A1: Obtains the correct derivative \(3x^2 - 12x + 9\).

(b)
- M1: Sets their derivative equal to zero and attempts to solve the quadratic equation.
- A1: Finds correct roots \(x = 1\) and \(x = 3\).
- M1: Substitutes at least one of their x-values into the original cubic equation to find the corresponding y-value.
- A1: Correctly identifies both coordinate pairs \((1, 9)\) and \((3, 5)\).

(c)
- M1: Differentiates \(\frac{dy}{dx}\) correctly to find \(\frac{d^2y}{dx^2} = 6x - 12\).
- M1: Substitutes at least one of their x-values into the second derivative to determine its sign.
- A1: Concludes that \((1, 9)\) is a local maximum because the second derivative is negative.
- A1: Concludes that \((3, 5)\) is a local minimum because the second derivative is positive.

(a) Differentiating \( x \) and \( y \) with respect to \( \theta \):
\[ \frac{\text{d}x}{\text{d}\theta} = 2 - 2\cos(2\theta) \]
\[ \frac{\text{d}y}{\text{d}\theta} = -4\sin(\theta) \]

Using the chain rule:
\[ \frac{\text{d}y}{\text{d}x} = \frac{\text{d}y/\text{d}\theta}{\text{d}x/\text{d}\theta} = \frac{-4\sin(\theta)}{2 - 2\cos(2\theta)} \]

Using the double-angle identity \( \cos(2\theta) = 1 - 2\sin^2(\theta) \):
\[ \frac{\text{d}x}{\text{d}\theta} = 2 - 2(1 - 2\sin^2(\theta)) = 4\sin^2(\theta) \]

Substituting this back into the gradient formula:
\[ \frac{\text{d}y}{\text{d}x} = \frac{-4\sin(\theta)}{4\sin^2(\theta)} = -\frac{1}{\sin(\theta)} = -\csc(\theta) \]

(b) At \( \theta = \frac{\pi}{3} \):
\[ x = 2\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \]
\[ y = 4\cos\left(\frac{\pi}{3}\right) = 4\left(\frac{1}{2}\right) = 2 \]

The gradient of the tangent at this point is:
\[ m = -\csc\left(\frac{\pi}{3}\right) = -\frac{2}{\sqrt{3}} \]

The equation of the tangent is:
\[ y - 2 = -\frac{2}{\sqrt{3}}\left(x - \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)\right) \]

Multiply by \( \sqrt{3} \) to clear the fraction:
\[ \sqrt{3}(y - 2) = -2\left(x - \frac{2\pi}{3} + \frac{\sqrt{3}}{2}\right) \]
\[ \sqrt{3}y - 2\sqrt{3} = -2x + \frac{4\pi}{3} - \sqrt{3} \]
\[ 2x + \sqrt{3}y - \sqrt{3} - \frac{4\pi}{3} = 0 \]

Multiply by 3 to ensure all coefficients are integers as required:
\[ 6x + 3\sqrt{3}y - 3\sqrt{3} - 4\pi = 0 \]

Hence, \( a = 6 \), \( b = 3 \), \( c = -3 \), and \( d = -4 \).

(a)
- M1: Attempt to differentiate \( x \) and \( y \) with respect to \( \theta \).
- A1: Correct derivatives for both: \( \frac{\text{d}x}{\text{d}\theta} = 2 - 2\cos(2\theta) \) and \( \frac{\text{d}y}{\text{d}\theta} = -4\sin(\theta) \).
- M1: Use of \( \frac{\text{d}y}{\text{d}x} = \frac{\text{d}y/\text{d}\theta}{\text{d}x/\text{d}\theta} \) and double-angle identity to simplify the expression.
- A1*: Correctly showing the given result with no errors.

(b)
- B1: Find correct \( x \) and \( y \) coordinates for \( \theta = \frac{\pi}{3} \): \( x = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \) and \( y = 2 \).
- M1: Substitute \( \theta = \frac{\pi}{3} \) into \( \frac{\text{d}y}{\text{d}x} \) to find the tangent gradient: \( m = -\frac{2}{\sqrt{3}} \).
- M1: Form the equation of the tangent using their coordinates and gradient.
- A1: Correct unsimplified equation of the line.
- M1: Complete method to rearrange the equation into the form \( ax + b\sqrt{3}y + c\sqrt{3} + d\pi = 0 \).
- A1: Correct final equation: \( 6x + 3\sqrt{3}y - 3\sqrt{3} - 4\pi = 0 \) (or any integer multiple thereof).

(a) Let \( u = 1 + e^x \). Then \( \text{d}u = e^x \text{d}x \).
Since \( e^x = u - 1 \), we have:
\[ e^{2x} \text{d}x = e^x \cdot (e^x \text{d}x) = (u - 1) \text{d}u \]

Now, change the limits of integration:
- Lower limit: when \( x = 0 \), \( u = 1 + e^0 = 2 \).
- Upper limit: when \( x = \ln 3 \), \( u = 1 + e^{\ln 3} = 1 + 3 = 4 \).

Substituting these into the integral:
\[ \int_{0}^{\ln 3} \frac{e^{2x}}{(1+e^x)^2} \text{d}x = \int_{2}^{4} \frac{u - 1}{u^2} \text{d}u = \int_{2}^{4} \left( \frac{1}{u} - \frac{1}{u^2} \right) \text{d}u \]
This matches the given form.

(b) Integrating the expression with respect to \( u \):
\[ \int_{2}^{4} \left( \frac{1}{u} - u^{-2} \right) \text{d}u = \left[ \ln|u| + \frac{1}{u} \right]_{2}^{4} \]

Now, evaluate at the upper and lower limits:
\[ = \left( \ln 4 + \frac{1}{4} \right) - \left( \ln 2 + \frac{1}{2} \right) \]
\[ = \ln 4 - \ln 2 + \frac{1}{4} - \frac{1}{2} \]
\[ = \ln\left(\frac{4}{2}\right) - \frac{1}{4} \]
\[ = -\frac{1}{4} + \ln 2 \]

Thus, \( a = -\frac{1}{4} \) and \( b = 2 \).

(a)
- M1: Differentiate the substitution to find \( \text{d}u = e^x \text{d}x \).
- M1: Express the numerator in terms of \( u \) as \( (u-1) \text{d}u \).
- B1: Show correct limits of integration: \( u = 2 \) and \( u = 4 \).
- M1: Substitute all parts into the integral and divide through by \( u^2 \).
- A1*: Obtain the exact given expression with no errors in working.

(b)
- M1: Integrate both terms correctly to obtain \( \ln|u| + \frac{1}{u} \) (condone sign errors).
- A1: Correct integrated expression: \( \ln|u| + \frac{1}{u} \).
- M1: Substitute limits of 4 and 2 into their integrated expression.
- A1: Show unsimplified evaluation: \( \left(\ln 4 + \frac{1}{4}\right) - \left(\ln 2 + \frac{1}{2}\right) \).
- A1: Fully simplify to get the exact answer: \( -\frac{1}{4} + \ln 2 \).

(a) The curve crosses the \( x \)-axis when \( y = 0 \):
\[ e^{-x} \cos x = 0 \]
Since \( e^{-x} > 0 \) for all real \( x \), we solve \( \cos x = 0 \) in the interval \( 0 \le x \le 2\pi \):
\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \]

(b) To find stationary points, we differentiate \( y \) using the product rule:
\[ \frac{\text{d}y}{\text{d}x} = -e^{-x} \cos x + e^{-x} (-\sin x) = -e^{-x} (\cos x + \sin x) \]
Set \( \frac{\text{d}y}{\text{d}x} = 0 \):
\[ -e^{-x} (\cos x + \sin x) = 0 \implies \sin x + \cos x = 0 \implies \tan x = -1 \]
In the interval \( 0 \le x \le 2\pi \), the solutions are:
\[ x = \frac{3\pi}{4}, \quad x = \frac{7\pi}{4} \]

Now find the corresponding \( y \)-coordinates:
- For \( x = \frac{3\pi}{4} \):
\[ y = e^{-3\pi/4} \cos\left(\frac{3\pi}{4}\right) = e^{-3\pi/4} \left(-\frac{\sqrt{2}}{2}\right) = -\frac{\sqrt{2}}{2} e^{-3\pi/4} \]
- For \( x = \frac{7\pi}{4} \):
\[ y = e^{-7\pi/4} \cos\left(\frac{7\pi}{4}\right) = e^{-7\pi/4} \left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2} e^{-7\pi/4} \]

So the stationary points are \( \left(\frac{3\pi}{4}, -\frac{\sqrt{2}}{2} e^{-3\pi/4}\right) \) and \( \left(\frac{7\pi}{4}, \frac{\sqrt{2}}{2} e^{-7\pi/4}\right) \).

(c) At any stationary point, we have \( \tan x = -1 \). This implies:
\[ \cos^2 x = \frac{1}{1 + \tan^2 x} = \frac{1}{1 + (-1)^2} = \frac{1}{2} \]

Squaring both sides of the equation for the curve \( y = e^{-x} \cos x \):
\[ y^2 = e^{-2x} \cos^2 x \]
Substituting \( \cos^2 x = \frac{1}{2} \):
\[ y^2 = \frac{1}{2} e^{-2x} \implies 2y^2 = e^{-2x} \] (as required)

(a)
- M1: Set \( y = 0 \) and deduce \( \cos x = 0 \).
- A1: Correctly find both \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

(b)
- M1: Apply product rule to differentiate \( y = e^{-x} \cos x \).
- A1: Correct derivative \( \frac{\text{d}y}{\text{d}x} = -e^{-x}(\sin x + \cos x) \).
- M1: Set \( \frac{\text{d}y}{\text{d}x} = 0 \) and attempt to solve for \( \tan x = -1 \).
- A1: Correct \( x \)-coordinates: \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \).
- M1: Substitute \( x \)-coordinates back into the curve equation to find exact \( y \)-coordinates.
- A1: Correct coordinates for both stationary points: \( \left(\frac{3\pi}{4}, -\frac{\sqrt{2}}{2} e^{-3\pi/4}\right) \) and \( \left(\frac{7\pi}{4}, \frac{\sqrt{2}}{2} e^{-7\pi/4}\right) \).

(c)
- M1: Use identity \( \cos^2 x = \frac{1}{2} \) derived from \( \tan x = -1 \), or show that squaring their coordinates satisfies the equation.
- A1: Correctly show \( 2y^2 = e^{-2x} \) with convincing algebraic steps.

(a) First, separate the variables:
\[ \int \frac{1}{y^2} \text{d}y = \int \frac{\ln x}{x} \text{d}x \]

Integrate the left-hand side:
\[ \int y^{-2} \text{d}y = -y^{-1} = -\frac{1}{y} \]

Integrate the right-hand side using the substitution \( u = \ln x \), so \( \text{d}u = \frac{1}{x} \text{d}x \):
\[ \int \frac{\ln x}{x} \text{d}x = \int u \text{d}u = \frac{1}{2} u^2 + C = \frac{1}{2} (\ln x)^2 + C \]

Thus, the general solution is:
\[ -\frac{1}{y} = \frac{1}{2} (\ln x)^2 + C \]

Apply the boundary condition \( y = -1 \) when \( x = 1 \):
\[ -\frac{1}{-1} = \frac{1}{2} (\ln 1)^2 + C \implies 1 = 0 + C \implies C = 1 \]

Substitute \( C = 1 \) back into the equation:
\[ -\frac{1}{y} = \frac{1}{2} (\ln x)^2 + 1 \]

Rearrange to make \( y \) the subject:
\[ \frac{1}{y} = -\frac{1}{2} (\ln x)^2 - 1 = -\frac{(\ln x)^2 + 2}{2} \]
\[ y = -\frac{2}{(\ln x)^2 + 2} \]

(b) Since \( (\ln x)^2 \ge 0 \) for all \( x > 0 \):
\[ (\ln x)^2 + 2 \ge 2 \]
Taking the reciprocal and multiplying by \( -2 \):
\[ 0 < \frac{1}{(\ln x)^2 + 2} \le \frac{1}{2} \]
\[ -1 \le -\frac{2}{(\ln x)^2 + 2} < 0 \]

Thus, the range of the function is \( [-1, 0) \) or \( -1 \le y < 0 \).

(a)
- M1: Attempt to separate variables to get \( \int \frac{1}{y^2} \text{d}y = \int \frac{\ln x}{x} \text{d}x \).
- A1: Correct separated terms.
- M1: Integrate LHS to obtain \( -\frac{1}{y} \).
- M1: Integrate RHS using substitution or inspection to obtain \( \frac{1}{2}(\ln x)^2 \).
- A1: Correct integration of both sides with constant of integration: \( -\frac{1}{y} = \frac{1}{2}(\ln x)^2 + C \).
- M1: Substitute \( y = -1 \) and \( x = 1 \) into their integrated expression to find \( C \).
- A1: Correct value of \( C = 1 \).
- A1: Correct particular solution: \( y = -\frac{2}{(\ln x)^2 + 2} \).

(b)
- M1: Establish that \( (\ln x)^2 \ge 0 \) and use it to bound the denominator.
- A1: Correct range \( -1 \le y < 0 \) (accept interval notation \( [-1, 0) \)).

(a) Let \(n\) be an odd integer. Then \(n\) can be written in the form \(n = 2m + 1\) where \(m\) is an integer.

Squaring \(n\) gives:
\(n^2 = (2m + 1)^2 = 4m^2 + 4m + 1 = 4m(m + 1) + 1\)

For any integer \(m\), one of \(m\) or \(m + 1\) must be even. Therefore, their product \(m(m + 1)\) must be even, so we can write \(m(m + 1) = 2k\) for some integer \(k\).

Substituting this back gives:
\(n^2 = 4(2k) + 1 = 8k + 1\)

Since \(k\) is an integer, this is of the form \(8k + 1\), which completes the proof for part (a).

(b) We wish to prove that \(n^4 - 1\) is divisible by 16 for any odd integer \(n\).

We can factorise \(n^4 - 1\) as a difference of two squares:
\(n^4 - 1 = (n^2 - 1)(n^2 + 1)\)

From part (a), we know that for any odd integer \(n\), \(n^2 = 8k + 1\) for some integer \(k\).

Substituting \(n^2 = 8k + 1\) into the expression:
\(n^2 - 1 = 8k\)
\(n^2 + 1 = (8k + 1) + 1 = 8k + 2 = 2(4k + 1)\)

Now multiply these two factors:
\(n^4 - 1 = (8k) \times 2(4k + 1) = 16k(4k + 1)\)

Since \(k(4k + 1)\) is an integer, \(n^4 - 1\) is a multiple of 16. Thus, \(n^4 - 1\) is divisible by 16.

Part (a) [5 Marks]:
- M1: Attempts to write an odd integer in the form \(2m + 1\) (or \(2m - 1\)) and squares it.
- A1: Correct expansion to \(4m^2 + 4m + 1\) or equivalent.
- M1: Factorises to obtain \(4m(m + 1) + 1\).
- M1: Explains that \(m(m+1)\) is the product of two consecutive integers, hence must be even (equal to \(2k\)).
- A1: Completes the proof by showing \(n^2 = 8k + 1\) with a clear concluding statement.

Part (b) [3 Marks]:
- M1: Factorises \(n^4 - 1\) to \((n^2 - 1)(n^2 + 1)\).
- M1: Substitutes \(n^2 = 8k + 1\) to obtain \(8k(8k + 2)\) or \(16k(4k + 1)\).
- A1: Fully correct proof showing a factor of 16 with a clear concluding statement.

(a) Consider the difference between \((x + y)\) and \((x - y)\):
\((x + y) - (x - y) = 2y\)

Since \(y\) is an integer, \(2y\) is an even integer.

If the difference between two integers is even, they must have the same parity. (Alternatively, if \((x-y)\) is even, then \((x+y) = (x-y) + 2y\) is even. If \((x-y)\) is odd, then \((x+y) = (x-y) + 2y\) is odd. Thus, they must have the same parity.)

(b) Suppose there exist positive integers \(x\) and \(y\) such that \(x^2 - y^2 = 22\).

We can factorise the left-hand side as a difference of two squares:
\((x - y)(x + y) = 22\)

Since \(x\) and \(y\) are positive integers, \(x + y\) and \(x - y\) must be integers, and since \(y > 0\), we must have \(x + y > x - y > 0\).

We now find all pairs of positive integer factors of 22. These are:
Case 1: \(x - y = 1\) and \(x + y = 22\)
Case 2: \(x - y = 2\) and \(x + y = 11\)

From part (a), \((x - y)\) and \((x + y)\) must have the same parity.

However, in Case 1, 1 is odd and 22 is even (different parities).
In Case 2, 2 is even and 11 is odd (different parities).

Therefore, there are no integer factor pairs of 22 with the same parity.

This is a contradiction, hence there are no positive integer solutions to \(x^2 - y^2 = 22\).

Part (a) [3 Marks]:
- M1: Considers the difference \((x+y) - (x-y) = 2y\) or the sum \((x+y) + (x-y) = 2x\).
- M1: States that \(2y\) (or \(2x\)) is even.
- A1: Explains clearly why this implies they must have the same parity (e.g., adding an even number to an integer does not change its parity).

Part (b) [5 Marks]:
- M1: Sets up the contradiction by assuming positive integer solutions exist, and factorises to \((x - y)(x + y) = 22\).
- M1: States that \(x-y\) and \(x+y\) must be positive integers.
- A1: Identifies the positive factor pairs of 22 as \((1, 22)\) and \((2, 11)\).
- M1: Refers to part (a) to state that the factors must have the same parity, but notes that both factor pairs consist of one odd and one even number.
- A1: Concludes the proof by contradiction with a clear final statement.

(a) We group the terms to factorise:
\(a^3 + b^3 - a^2 b - a b^2 = a^2(a - b) - b^2(a - b)\)
\(= (a^2 - b^2)(a - b)\)
\(= (a - b)(a + b)(a - b)\)
\(= (a + b)(a - b)^2\)

(b) We wish to show that \(a^3 + b^3 - a^2 b - a b^2 \ge 0\) for positive real numbers \(a\) and \(b\).

Using the factorisation from part (a):
\(a^3 + b^3 - a^2 b - a b^2 = (a + b)(a - b)^2\)

Since \(a\) and \(b\) are positive real numbers, \(a > 0\) and \(b > 0\), so \(a + b > 0\).

Since the square of any real number is non-negative, \((a - b)^2 \ge 0\).

The product of a positive number and a non-negative number is non-negative, so:
\((a + b)(a - b)^2 \ge 0\)
\(\implies a^3 + b^3 \ge a^2 b + a b^2\)

(c) Consider the difference \(\frac{x^3 + y^3}{2} - \left(\frac{x+y}{2}\right)^3\):
\(\frac{x^3 + y^3}{2} - \frac{(x+y)^3}{8} = \frac{4(x^3 + y^3) - (x^3 + 3x^2y + 3xy^2 + y^3)}{8}\)
\(= \frac{3x^3 - 3x^2y - 3xy^2 + 3y^3}{8}\)
\(= \frac{3}{8}(x^3 + y^3 - x^2y - xy^2)\)

From part (b), we know that for all positive real numbers \(x\) and \(y\), \(x^3 + y^3 - x^2y - xy^2 \ge 0\).

Since \(\frac{3}{8} > 0\), we must have:
\(\frac{3}{8}(x^3 + y^3 - x^2y - xy^2) \ge 0\)

Therefore, \(\frac{x^3 + y^3}{2} \ge \left(\frac{x+y}{2}\right)^3\), as required.

Part (a) [2 Marks]:
- M1: Groups terms to obtain \(a^2(a - b) - b^2(a - b)\) or equivalent.
- A1: Factorises completely to \((a + b)(a - b)^2\).

Part (b) [2 Marks]:
- M1: Explains that \((a - b)^2 \ge 0\) and \(a + b > 0\) since \(a, b\) are positive real numbers.
- A1: Concludes with a clear explanation that the product of a positive and a non-negative number is non-negative, proving \(a^3 + b^3 \ge a^2b + ab^2\).

Part (c) [4 Marks]:
- M1: Writes the difference \(\frac{x^3 + y^3}{2} - \left(\frac{x+y}{2}\right)^3\) and finds a common denominator of 8.
- A1: Correctly expands \((x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\).
- M1: Simplifies the numerator to \(3(x^3 + y^3 - x^2y - xy^2)\) and factors out \(\frac{3}{8}\).
- A1: Completes the proof by linking to the inequality from part (b).

(a) Let the three consecutive integers be \(n-1\), \(n\), and \(n+1\). Their product is \(P = (n-1)n(n+1)\).

First, we prove \(P\) is divisible by 2. Among any two consecutive integers, one must be even. Therefore, either \(n\) is even, or \(n-1\) (and \(n+1\)) is even. In either case, the product \(P\) contains an even factor, so \(P\) is divisible by 2.

Second, we prove \(P\) is divisible by 3. Any integer \(n\) can be written in one of three forms: \(3k\), \(3k+1\), or \(3k+2\), where \(k\) is an integer.
- Case 1: If \(n = 3k\), then \(n\) is divisible by 3, so \(P\) is divisible by 3.
- Case 2: If \(n = 3k+1\), then \(n-1 = 3k\) is divisible by 3, so \(P\) is divisible by 3.
- Case 3: If \(n = 3k+2\), then \(n+1 = 3k+3 = 3(k+1)\) is divisible by 3, so \(P\) is divisible by 3.
In all possible cases, one of the three factors is divisible by 3, so \(P\) is divisible by 3.

Since 2 and 3 are prime numbers (and coprime), any number divisible by both 2 and 3 must be divisible by \(2 \times 3 = 6\). Thus, the product of any three consecutive integers is divisible by 6.

(b) Let \(f(n) = n^3 + 11n\). We can rewrite \(f(n)\) as:
\(f(n) = n^3 - n + 12n\)
\(= n(n^2 - 1) + 12n\)
\(= (n-1)n(n+1) + 12n\)

From part (a), the term \((n-1)n(n+1)\) is the product of three consecutive integers and is therefore divisible by 6.

The term \(12n\) is clearly divisible by 6 because \(12 = 6 \times 2\), so \(12n = 6(2n)\).

Since \(f(n)\) is the sum of two terms that are both divisible by 6, \(f(n)\) must also be divisible by 6. Thus, \(n^3 + 11n\) is divisible by 6 for all integers \(n\).

Part (a) [5 Marks]:
- M1: Defines the product of three consecutive integers as \(P = (n-1)n(n+1)\) (or equivalent).
- M1: Explains clearly why the product must be even (divisible by 2) by considering the presence of an even integer.
- M1: Uses a systematic method (such as modulo 3 or cases \(3k, 3k+1, 3k+2\)) to show that one of the terms is always divisible by 3.
- A1: Shows all three cases for divisibility by 3 correctly.
- A1: Concludes that since \(P\) is divisible by both 2 and 3, it must be divisible by 6.

Part (b) [3 Marks]:
- M1: Rewrites \(n^3 + 11n\) in terms of the consecutive product, i.e., \((n-1)n(n+1) + 12n\).
- A1: States that \((n-1)n(n+1)\) is divisible by 6 (referring to part a) and \(12n\) is divisible by 6.
- A1: Completes the proof with a clear concluding statement.

(a) Since \(\tan\alpha = \frac{3}{4}\), we have \(\sin\alpha = 0.6\) and \(\cos\alpha = 0.8\).
For \(P\), resolving perpendicular to the plane:
\(R = 3mg \cos\alpha = 3mg(0.8) = 2.4mg\).
The maximum frictional force is \(F = \mu R = 0.25 \times 2.4mg = 0.6mg\).
Equation of motion for \(P\) moving down the plane:
\(3mg \sin\alpha - F - T = 3ma\)
\(3mg(0.6) - 0.6mg - T = 3ma \implies 1.2mg - T = 3ma\) (Equation 1)

For \(Q\) moving vertically upwards:
\(T - mg = ma\) (Equation 2)

Adding Equation 1 and Equation 2:
\(1.2mg - mg = 4ma \implies 0.2mg = 4ma \implies a = 0.05g\).

(b) Substitute \(a = 0.05g\) into Equation 2:
\(T = m(g + a) = m(g + 0.05g) = 1.05mg\).

(c) For the first stage of motion, \(Q\) moves with acceleration \(a = 0.05g\) over a distance \(s_1 = 1.6\text{ m}\).
Using \(v^2 = u^2 + 2as_1\):
\(v^2 = 0 + 2(0.05g)(1.6) = 0.16g\).
When the string breaks, \(Q\) moves freely under gravity, so its acceleration is \(-g\).
Using \(v^2 = u^2 + 2as\) for this stage with final velocity 0:
\(0 = 0.16g - 2g(s_2) \implies s_2 = \frac{0.16g}{2g} = 0.08\text{ m}\).
Total distance traveled by \(Q\) is \(s_1 + s_2 = 1.6 + 0.08 = 1.68\text{ m}\).

(d) The tension in the string on both sides of the pulley is assumed to have the same magnitude.

(a)
M1: Resolving forces perpendicular to the plane to get \(R = 3mg \cos\alpha\).
A1: Correct reaction \(R = 2.4mg\) and limiting friction \(F = 0.6mg\).
M1: Applying Newton's second law to \(P\) down the incline and \(Q\) vertically.
A1: Correct equations: \(1.2mg - T = 3ma\) and \(T - mg = ma\).
A1: Solving simultaneously to show \(a = 0.05g\).

(b)
M1: Substituting \(a = 0.05g\) into one of the equations of motion.
A1: Finding \(T = 1.05mg\) or \(\frac{21}{20}mg\).
B1: Consistent units/form in terms of \(m\) and \(g\).

(c)
M1: Setting up \(v^2 = u^2 + 2as\) to find speed squared when string breaks.
A1: Obtaining \(v^2 = 0.16g\) (or \(1.568\)).
A1: Finding the upward free-fall distance \(0.08\text{ m}\) and summing to get \(1.68\text{ m}\).

(d)
B1: Stating that tension is the same on both sides of the pulley.

(a) Since the center of mass is at a distance \(x\text{ m}\) from \(A\), and \(C\) is at \(1.5\text{ m}\) from \(A\), the distance of the center of mass from \(C\) is \(x - 1.5\text{ m}\).

(b) When the beam is on the point of tilting about \(C\), the reaction force at support \(D\) is zero.
Taking moments about \(C\):
\(12g \times 1.5 = Mg \times (x - 1.5) \implies 18 = M(x - 1.5)\) (Equation 1)

When instead the particle of mass \(52\text{ kg}\) is placed at \(B\), the beam is on the point of tilting about \(D\), so the reaction force at \(C\) is zero.
The distance of \(D\) from \(A\) is \(6 - 1 = 5\text{ m}\).
The distance of the center of mass from \(D\) is \(5 - x\text{ m}\).
Taking moments about \(D\):
\(52g \times 1 = Mg \times (5 - x) \implies 52 = M(5 - x)\) (Equation 2)

Dividing Equation 1 by Equation 2:
\(\frac{18}{52} = \frac{x - 1.5}{5 - x} \implies \frac{9}{26} = \frac{x - 1.5}{5 - x}\)
\(9(5 - x) = 26(x - 1.5)\)
\(45 - 9x = 26x - 39\)
\(35x = 84 \implies x = 2.4\).

(c) Substituting \(x = 2.4\) into Equation 1:
\(18 = M(2.4 - 1.5)\)
\(18 = 0.9M \implies M = 20\).

(d) The center of mass is not assumed to be at the midpoint of the beam (which would be at \(3\text{ m}\)).

(e) The reaction force at support \(D\) is zero.

(a)
B1: Correct expression \(x - 1.5\) (or equivalent).

(b)
M1: Correctly stating or using the condition that the reaction at \(D\) is zero.
M1: Setting up the moment equation about \(C\) with correct forces and distances.
A1: Correct Equation 1: \(18 = M(x - 1.5)\) (accept with \(g\)).
M1: Setting up the moment equation about \(D\) with correct forces and distances.
A1: Correct Equation 2: \(52 = M(5 - x)\) (accept with \(g\)).
A1: Solving the simultaneous equations by eliminating \(M\) to arrive at \(x = 2.4\).

(c)
M1: Substituting \(x = 2.4\) back into either Equation 1 or Equation 2.
A1: Correct calculation leading to \(M = 20\).
B1: Units stated correctly as kg.

(d)
B1: Explaining that a non-uniform model means the center of mass is not at the geometric midpoint.

(e)
B1: Stating that the reaction force at \(D\) is zero.

(a) Since \(v = \int a \text{ dt}\):
\(v = \int (3t^2 - 12t + 9) \text{ dt} = t^3 - 6t^2 + 9t + C\)
Given that the rocket starts from rest, \(v(0) = 0 \implies C = 0\).
So, \(v = t^3 - 6t^2 + 9t\) for \(0 \le t \le 4\).

(b) To find stationary points of velocity, set \(a = 0\):
\(3t^2 - 12t + 9 = 0 \implies 3(t-1)(t-3) = 0\)
So stationary points occur at \(t = 1\) and \(t = 3\).
Checking the nature of these points using \(\frac{d^2v}{dt^2} = \frac{da}{dt} = 6t - 12\):
At \(t = 3\), \(\frac{d^2v}{dt^2} = 6(3) - 12 = 6 > 0\), so \(t = 3\) is a local minimum.
The velocity at this minimum is:
\(v(3) = 3^3 - 6(3^2) + 9(3) = 27 - 54 + 27 = 0\text{ m/s}\).
At the boundary points:
\(v(0) = 0\text{ m/s}\)
\(v(4) = 4^3 - 6(4^2) + 9(4) = 64 - 96 + 36 = 4\text{ m/s}\).
Thus, the minimum velocity of the rocket during \(0 \le t \le 4\) is \(0\text{ m/s}\). Since \(v(t) \ge 0\) for all \(t\) in this interval, the rocket never moves downwards.

(c) Since height \(s = \int v \text{ dt}\):
\(s = \int (t^3 - 6t^2 + 9t) \text{ dt} = \frac{1}{4}t^4 - 2t^3 + 4.5t^2 + C_2\)
Since \(s(0) = 0\), \(C_2 = 0\).
At \(t = 4\):
\(s(4) = \frac{1}{4}(4^4) - 2(4^3) + 4.5(4^2) = 64 - 128 + 72 = 8\text{ m}\).

(d) One limitation is that the model ignores air resistance, which would reduce the maximum height and velocity of the rocket. Another limitation is that the model's acceleration function is only valid up to \(t = 4\), so it does not capture the behavior after the engine burns out.

(a)
M1: Attempting to integrate the acceleration function with respect to \(t\).
A1: Correctly integrating terms to get \(t^3 - 6t^2 + 9t\).
A1: Using \(v = 0\) when \(t = 0\) to show the constant of integration is zero.

(b)
M1: Setting \(a = 0\) to find stationary values of velocity.
A1: Solving to get \(t = 1\) and \(t = 3\).
M1: Substituting \(t = 3\) into \(v(t)\) to find the minimum value.
A1: Correctly finding \(v(3) = 0\text{ m/s}\) and reasoning that \(v(t) \ge 0\) throughout, hence no downward motion.

(c)
M1: Attempting to integrate \(v(t)\) with respect to \(t\).
A1: Correct integration to find \(\frac{1}{4}t^4 - 2t^3 + 4.5t^2\).
A1: Substituting \(t = 4\) to obtain \(8\text{ m}\).

(d)
B2: A fully explained physical limitation of the model (B1 for a basic mention of air resistance or variable mass).

(a) The force diagram should include:
1. Weight \(mg\) acting vertically downwards.
2. Normal reaction \(R\) acting perpendicular to the inclined plane.
3. Pulling force \(P = 18\text{ N}\) acting up the inclined plane.
4. Friction force \(F\) acting down the inclined plane (opposing the potential motion up the plane).

(b) Let \(\theta\) be the angle of inclination.
Resolving perpendicular to the plane:
\(R = mg \cos \theta\)

Resolving parallel to the plane (on the point of slipping up the plane):
\(P = mg \sin \theta + F\)
Since the peg is on the point of slipping, \(F = \mu R = \frac{1}{3} R\).
Thus, the equations are:
\(R = mg \cos \theta\) and \(18 = mg \sin \theta + \frac{1}{3} R\).

(c) Since \(\sin \theta = \frac{5}{13}\), we have \(\cos \theta = \frac{12}{13}\).
Substitute \(R\) into the parallel equation:
\(18 = mg \left(\frac{5}{13}\right) + \frac{1}{3} \left(mg \left(\frac{12}{13}\right)\right)\)
\(18 = mg \left(\frac{5}{13}\right) + mg \left(\frac{4}{13}\right) = mg \left(\frac{9}{13}\right)\)
\(mg = 18 \times \frac{13}{9} = 26\text{ N}\).
Using \(g = 9.8\text{ m/s}^2\):
\(m = \frac{26}{9.8} \approx 2.653\text{ kg}\).
To 3 significant figures, \(m = 2.65\text{ kg}\).

(d) When \(P\) is removed, the only forces acting along the plane are the component of weight pulling the peg down and friction opposing it.
Component of weight down the plane: \(mg \sin \theta = 26 \times \frac{5}{13} = 10\text{ N}\).
Maximum possible static frictional force: \(F_{\max} = \mu R = \frac{1}{3} mg \cos \theta = \frac{1}{3} \times 26 \times \frac{12}{13} = 8\text{ N}\).
Since \(10\text{ N} > 8\text{ N}\), the component of weight down the plane exceeds the maximum frictional force. Therefore, the peg does not remain in equilibrium and will slide down the plane.

(a)
B1: Drawing normal reaction and weight forces correctly oriented.
B1: Drawing force of 18 N up the slope and friction acting down the slope.

(b)
M1: Attempting to resolve perpendicular to the plane.
A1: Correct equation \(R = mg \cos\theta\).
M1: Attempting to resolve parallel to the plane with friction opposing the motion.
A1: Correct equation \(18 = mg \sin\theta + F\).

(c)
M1: Using \(F = \frac{1}{3}R\) and substituting trigonometric values.
A1: Simplifying to \(\frac{9}{13}mg = 18\).
M1: Substituting \(g = 9.8\) and solving for \(m\).
A1: Correct value \(m \approx 2.65\text{ kg}\) (allow 2.7 to 2 s.f.).

(d)
M1: Calculating weight component down slope (10 N) and max friction (8 N).
A1: Correctly concluding that the peg will slide down the plane since weight component exceeds max friction.

(a) Power \(P = 24000\text{ W}\).
At maximum constant speed up the road, acceleration is zero, so the driving force \(F\) is balanced by the components of weight and resistance down the slope:
\(F = mg \sin \alpha + R\)
Using \(P = F v \implies F = \frac{24000}{15} = 1600\text{ N}\).
Therefore:
\(1600 = 1200(9.8)\left(\frac{1}{20}\right) + R\)
\(1600 = 60(9.8) + R\)
\(1600 = 588 + R \implies R = 1012\).

(b) On the horizontal road, gravity has no component along the direction of motion.
At speed \(v = 20\text{ m/s}\), the driving force is:
\(F' = \frac{P}{v} = \frac{24000}{20} = 1200\text{ N}\).
Using Newton's second law along the horizontal direction:
\(F' - R = ma\)
\(1200 - 1012 = 1200 a\)
\(188 = 1200 a \implies a = \frac{188}{1200} \approx 0.157\text{ m/s}^2\).

(c) The maximum speed on the horizontal road occurs when acceleration is zero, so \(F' = R\):
\(\frac{24000}{v_{\max}} = 1012 \implies v_{\max} = \frac{24000}{1012} \approx 23.7\text{ m/s}\).

(d) The resistance force could be modeled as varying with speed, such as incorporating a term proportional to the square of the speed of the car (air resistance).

(a)
M1: Using \(P = Fv\) with \(P=24000\) and \(v=15\).
A1: Calculating driving force \(F = 1600\text{ N\).
M1: Writing the force balance equation along the incline.
A1: Correctly calculating the weight component \(mg\sin\alpha = 588\text{ N\).
A1: Obtaining the given value \(R = 1012\).

(b)
M1: Finding the driving force at \(v = 20\text{ m/s}\).
A1: \(F' = 1200\text{ N\).
M1: Setting up \(F' - R = ma\) horizontal equation.
A1: Correct acceleration \(a \approx 0.157\text{ m/s}^2\) (accept \(0.16\)).

(c)
M1: Recognizing that maximum speed occurs when the driving force equals resistance.
A1: Correct calculation giving \(v \approx 23.7\text{ m/s}\).

(d)
B1: Stating that resistance should vary with speed (e.g., \(R = kv^2\) or similar).

(a) Since \(\tan\theta = \frac{3}{4}\), we have \(\cos\theta = 0.8\) and \(\sin\theta = 0.6\).
Using horizontal motion at constant speed:
\(x = (U \cos\theta) t \implies 24 = 0.8U t\)
\(t = \frac{24}{0.8U} = \frac{30}{U}\).

(b) Using the vertical displacement formula under gravity:
\(y = (U \sin\theta) t - \frac{1}{2}gt^2\)
Substitute the wall details \(y = 8\text{ m}\), \(t = \frac{30}{U}\), and \(\sin\theta = 0.6\):
\(8 = U(0.6)\left(\frac{30}{U}\right) - 4.9 \left(\frac{30}{U}\right)^2\)
\(8 = 18 - 4.9 \left(\frac{900}{U^2}\right)\)
\(8 = 18 - \frac{4410}{U^2}\)
\(\frac{4410}{U^2} = 10 \implies U^2 = 441\)
Since \(U > 0\), \(U = 21\).

(c) The ball hits the ground when vertical displacement \(y = 0\):
\(y = (21 \times 0.6) t - 4.9 t^2 = 0\)
\(12.6 t - 4.9 t^2 = 0\)
Since \(t \ne 0\):
\(t = \frac{12.6}{4.9} = \frac{18}{7}\text{ s}\).
Total horizontal range is:
\(R = (U \cos\theta) t_{\text{total}} = 21(0.8) \left(\frac{18}{7}\right) = 3 \times 0.8 \times 18 = 43.2\text{ m}\).
The horizontal distance from the wall to the landing point is:
\(43.2 - 24 = 19.2\text{ m}\).

(d) Air resistance is negligible.

(a)
M1: Finding \(\cos\theta = 0.8\) from \(\tan\theta = \frac{3}{4}\).
M1: Using horizontal displacement formula with \(x = 24\).
A1: Correctly showing \(t = \frac{30}{U}\).

(b)
M1: Setting up the vertical displacement equation under gravity.
A1: Correctly substituting \(\sin\theta = 0.6\) and \(t = \frac{30}{U}\).
M1: Rearranging to get a linear equation in \(\frac{1}{U^2}\).
A1: Correctly showing \(U^2 = 441\).
A1: Finding \(U = 21\).

(c)
M1: Setting \(y = 0\) to solve for total flight time.
A1: Correct flight time \(t = \frac{18}{7}\text{ s}\) (or approx \(2.57\text{ s}\)).
A1: Finding horizontal range \(43.2\text{ m}\) and final distance \(19.2\text{ m}\).

(d)
B1: Stating that air resistance is negligible / no wind effects.

(a) The number of tourists who dined at local restaurants only is given as 25. The total number of tourists who dined at local restaurants is 57. The regions making up \(D\) are: \(D\text{ only}\), \(M \cap D \cap B'\), \(D \cap B \cap M'\), and \(M \cap D \cap B\). Substituting the given values: \(25 + (22 - x) + (18 - x) + x = 57 \implies 65 - x = 57 \implies x = 8\).

(b) The number of tourists in each region is:
- \(M \cap D \cap B = 8\)
- \(M \cap D \cap B' = 22 - 8 = 14\)
- \(D \cap B \cap M' = 18 - 8 = 10\)
- \(M \cap B \cap D' = 12\)
- \(M\text{ only} = 55 - (14 + 12 + 8) = 21\)
- \(D\text{ only} = 25\)
- \(B\text{ only} = 45 - (12 + 10 + 8) = 15\)
- Outside all three circles = 15

(c)(i) The number of tourists who participated in at least two activities is: \(14 + 10 + 12 + 8 = 44\). Probability = \(\frac{44}{120} = \frac{11}{30}\).

(c)(ii) \(P(D | M) = \frac{P(D \cap M)}{P(M)} = \frac{22}{55} = \frac{2}{5} = 0.4\).

(a)
M1: Sets up a correct equation for the elements of D in terms of x.
A1: Correct simplification: \(65 - x = 57\).
A1*: Reaches \(x = 8\) with no errors.

(b)
B1: Outside region (15) and center (8) correctly placed.
B1: Two-set intersections correctly calculated: \(M \cap D \text{ only} = 14\), \(D \cap B \text{ only} = 10\), \(M \cap B \text{ only} = 12\).
B1: Single-set regions correctly calculated: \(M\text{ only} = 21\), \(D\text{ only} = 25\), \(B\text{ only} = 15\).
B1: Fully drawn Venn diagram with correct labels for circles and universal set.

(c)(i)
M1: Attempts sum of intersection regions: \(14 + 10 + 12 + 8\).
A1: Correct probability of \(\frac{11}{30}\) or equivalent.

(c)(ii)
M1: Identifies the conditional probability formula \(P(D | M) = \frac{N(D \cap M)}{N(M)}\).
A1: Correct probability of \(\frac{2}{5}\) or equivalent.

(a) Total frequency \(n = 80\). The median is located at the \(40\)-th value. Cumulative frequencies are: 8, 22, 44, 64, 75, 80. The median class is \(1400 \le h < 1500\). By linear interpolation: \(M = 1400 + \frac{40 - 22}{22} \times 100 = 1400 + \frac{18}{22} \times 100 \approx 1481.8\) hours (or 1480 hours to 3 s.f.).

(b) Using midpoints \(x\) of classes: 1100, 1300, 1450, 1550, 1700, 2000:
\(\sum f = 80\)
\(\sum fx = 8 \times 1100 + 14 \times 1300 + 22 \times 1450 + 20 \times 1550 + 11 \times 1700 + 5 \times 2000 = 118600\)
\(\bar{x} = \frac{118600}{80} = 1482.5\) hours (or 1480 hours to 3 s.f.).

\(\sum fx^2 = 8 \times 1100^2 + 14 \times 1300^2 + 22 \times 1450^2 + 20 \times 1550^2 + 11 \times 1700^2 + 5 \times 2000^2 = 179435000\)
\(\sigma^2 = \frac{179435000}{80} - 1482.5^2 = 2242937.5 - 2197806.25 = 45131.25\)
\(\sigma = \sqrt{45131.25} \approx 212.4\) hours (or 212 hours to 3 s.f.).

(c) The middle 50% of the bulbs lie between the lower quartile (\(Q_1\)) and the upper quartile (\(Q_3\)). For all of these to have a lifetime of at least 1450 hours, the lower quartile must be at least 1450 hours.
\(Q_1\) is at the 20th value, which lies in \(1200 \le h < 1400\).
\(Q_1 = 1200 + \frac{20 - 8}{14} \times 200 \approx 1371.4\) hours.
Since \(Q_1 \approx 1371.4 < 1450\), the claim is not supported.

(a)
M1: Identifies the correct median interval \([1400, 1500)\) and sets up interpolation formula.
A1: Correct fraction: \(\frac{40-22}{22}\) or equivalent.
A1: Correct median of 1481.8 (or 1480) hours.

(b)
M1: Clear attempt to find \(\sum fx\) using midpoints.
A1: Correct mean of 1482.5 hours (or 1480).
M1: Clear attempt to find \(\sum fx^2\) and use standard deviation formula.
A1: Correct variance of 45131.25.
A1: Correct standard deviation of 212.4 (or 212) hours.

(c)
M1: Identifies that the lower quartile must be evaluated to test the claim.
A1: Correct calculation of \(Q_1 = 1371.4\) (or 1370) hours.
A1: Correct conclusion with justification comparing \(Q_1\) with 1450.

(a) \(n = 10\).
\(S_{TT} = \sum T^2 - \frac{(\sum T)^2}{n} = 3615 - \frac{185^2}{10} = 192.5\)
\(S_{GG} = \sum G^2 - \frac{(\sum G)^2}{n} = 1844 - \frac{128^2}{10} = 205.6\)
\(S_{TG} = \sum TG - \frac{\sum T \sum G}{n} = 2512 - \frac{185 \times 128}{10} = 144\)
\(r = \frac{S_{TG}}{\sqrt{S_{TT} S_{GG}}} = \frac{144}{\sqrt{192.5 \times 205.6}} \approx 0.7238 \approx 0.724\).

(b) Yes, a linear regression model is suitable because the PMCC of 0.724 indicates a moderately strong positive linear correlation between temperature and weekly growth.

(c) \(b = \frac{S_{TG}}{S_{TT}} = \frac{144}{192.5} \approx 0.74805 \approx 0.748\).
\(\bar{T} = 18.5\), \(\bar{G} = 12.8\).
\(a = \bar{G} - b\bar{T} = 12.8 - 0.74805 \times 18.5 \approx -1.039 \approx -1.04\).
Therefore, the regression line equation is \(G = -1.04 + 0.748T\).

(d) The gradient \(b = 0.748\) means that for every 1°C increase in average weekly temperature, the weekly growth of the plant increases by approximately 0.748 mm.

(e) At \(T = 24\): \(G = -1.039 + 0.74805 \times 24 \approx 16.9\) mm. Since 24°C lies outside the range of the recorded temperature data (12°C to 22°C), this estimate is an extrapolation and is therefore unreliable.

(a)
M1: Attempt to find \(S_{TT}\), \(S_{GG}\), or \(S_{TG}\).
A1: Correct values for \(S_{TT} = 192.5\), \(S_{GG} = 205.6\), and \(S_{TG} = 144\).
M1: Correct formula used for PMCC \(r\).
A1: Correct value of \(r = 0.724\) (accept 0.723 to 0.725).

(b)
B1: States 'yes' and links to positive linear correlation.

(c)
M1: Correct formula for \(b\) used with their values.
A1: Correct value of \(b = 0.748\) (or 0.7481).
M1: Correct formula for \(a\) used with their mean values.
A1: Correct value of \(a = -1.04\) (or -1.039) leading to \(G = -1.04 + 0.748T\).

(d)
B1: Correctly interprets the rate of change including units (°C and mm).

(e)
B1: Calculates 16.9 mm and clearly states that it is an extrapolation/unreliable.

(a) The sum of probabilities is 1:
\(a + b + 0.3 + b + c = 1 \implies a + 2b + c = 0.7\).

(b) \(\text{E}(X) = -2a - b + 0 + b + 2c = 2c - 2a\).
Since \(\text{E}(X) = 0.2\), we have:
\(2c - 2a = 0.2 \implies c - a = 0.1\).

(c) \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = 1.56\).
\(\text{E}(X^2) = 1.56 + 0.2^2 = 1.6\).
\(\text{E}(X^2) = 4a + b + 0 + b + 4c = 4a + 2b + 4c = 1.6\).
Using \(2b = 0.7 - a - c\) from part (a):
\(4a + (0.7 - a - c) + 4c = 1.6 \implies 3a + 3c + 0.7 = 1.6 \implies 3a + 3c = 0.9 \implies a + c = 0.3\).
We solve simultaneous equations:
1) \(a + c = 0.3\)
2) \(c - a = 0.1\)
Adding the two equations: \(2c = 0.4 \implies c = 0.2\).
Then \(a = 0.3 - 0.2 = 0.1\).
From part (a): \(2b = 0.7 - 0.1 - 0.2 = 0.4 \implies b = 0.2\).
Thus, \(a = 0.1, b = 0.2, c = 0.2\).

(d) \(\text{P}(3X - 1 > 1) = \text{P}(3X > 2) = \text{P}(X > \frac{2}{3})\).
Since \(X\) is discrete, this is equivalent to:
\(\text{P}(X = 1) + \text{P}(X = 2) = b + c = 0.2 + 0.2 = 0.4\).

(a)
B1: Correctly sums probabilities and shows \(a + 2b + c = 0.7\) clearly.

(b)
M1: Sets up the formula for \(\text{E}(X)\) in terms of variables.
A1*: Equates to 0.2 and algebraically shows \(c - a = 0.1\).

(c)
M1: Uses \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\) to find \(\text{E}(X^2) = 1.6\).
A1: Sets up correct expression \(4a + 2b + 4c = 1.6\).
M1: Substitutes \(2b\) expression to eliminate \(b\).
A1: Obtains \(a + c = 0.3\).
A1: Solves simultaneously for two variables (e.g., \(a=0.1\) and \(c=0.2\)).
A1: Obtains correct value for \(b = 0.2\).

(d)
M1: Simplifies inequality to \(X > \frac{2}{3}\).
A1: Correctly identifies \(P(X=1) + P(X=2)\) and evaluates to 0.4.

(a) Let \(M \sim \text{N}(\mu, \sigma^2)\).
For 12% less than 110 g: \(P(M < 110) = 0.12 \implies P\left(Z < \frac{110-\mu}{\sigma}\right) = 0.12\).
From the standard normal table: \(\frac{110-\mu}{\sigma} = -1.175 \implies \mu - 1.175\sigma = 110\).

For 15% greater than 165 g: \(P(M > 165) = 0.15 \implies P\left(Z < \frac{165-\mu}{\sigma}\right) = 0.85\).
From the standard normal table: \(\frac{165-\mu}{\sigma} = 1.036 \implies \mu + 1.036\sigma = 165\).

(b) Subtracting the first equation from the second:
\(2.211\sigma = 55 \implies \sigma \approx 24.876 \approx 24.9\) g.
Substituting \(\sigma\) back into the second equation:
\[\mu = 165 - 1.036 \times 24.876 \approx 139.228 \approx 139.2\) g.

(c) The probability of rejection is \(P(M < 100)\):
\(Z = \frac{100 - 139.228}{24.876} \approx -1.577\).
Using the standard normal table: \(P(Z < -1.58) = 1 - \Phi(1.58) = 1 - 0.9429 = 0.0571\) (or \(0.0574\) if using a calculator with unrounded values).

(a)
M1: Attempts to use normal distribution table for 0.12 tail (accepts \(z = -1.17\) to \(-1.18\)).
A1: Correct equation: \(\mu - 1.175\sigma = 110\).
M1: Attempts to use normal distribution table for 0.15 tail (accepts \(z = 1.03\) to \(1.04\)).
A1: Correct equation: \(\mu + 1.036\sigma = 165\).

(b)
M1: Solves simultaneous equations to find \(\sigma\).
A1: Correct value of \(\sigma = 24.9\) (accept 24.8 to 25.1).
M1: Solves for \(\mu\).
A1: Correct value of \(\mu = 139.2\) (accept 139.0 to 139.4).

(c)
M1: Standardizes with 100 using their \(\mu\) and \(\sigma\).
A1: Obtains a z-value around \(-1.58\).
A1: Correct probability in the range \(0.057 - 0.058\).

(a) The tree diagram should show:
- First branch stage: \(R_1\) with probability \(\frac{5}{12}\) and \(B_1\) with probability \(\frac{7}{12}\).
- Second branch stage from \(R_1\): \(R_2\) with probability \(\frac{4}{11}\) and \(B_2\) with probability \(\frac{7}{11}\).
- Second branch stage from \(B_1\): \(R_2\) with probability \(\frac{5}{11}\) and \(B_2\) with probability \(\frac{6}{11}\).

(b)(i) Both balls of the same colour:
\(P(\text{Same}) = P(R_1 \cap R_2) + P(B_1 \cap B_2) = \left(\frac{5}{12} \times \frac{4}{11}\right) + \left(\frac{7}{12} \times \frac{6}{11}\right) = \frac{20}{132} + \frac{42}{132} = \frac{62}{132} = \frac{31}{66}\) (or approx 0.470).

(b)(ii) At least one red ball is drawn:
\(P(\text{At least one red}) = 1 - P(B_1 \cap B_2) = 1 - \frac{42}{132} = \frac{90}{132} = \frac{15}{22}\) (or approx 0.682).

(c) Let \(A\) be the event that at least one red ball is drawn. We want \(P(B_1 | A) = \frac{P(B_1 \cap A)}{P(A)}\).
The event \(B_1 \cap A\) represents drawing a blue ball first and at least one red ball in total, which is only possible for the outcome \(B_1 \cap R_2\).
\(P(B_1 \cap R_2) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132}\).
\(P(B_1 | A) = \frac{35/132}{90/132} = \frac{35}{90} = \frac{7}{18}\) (or approx 0.389).

(a)
B1: First draw branches correct with labels and probabilities.
B1: Second draw branches from Red correct.
B1: Second draw branches from Blue correct.

(b)(i)
M1: Sums products of branches for RR and BB.
A1: Correct probability of \(\frac{31}{66}\) or equivalent.

(b)(ii)
M1: Uses complement of BB or sums RR, RB, BR branches.
A1: Correct probability of \(\frac{15}{22}\) or equivalent.

(c)
M1: Identifies conditional probability structure \(\frac{P(B_1 \cap A)}{P(A)}\).
M1: Correctly identifies \(B_1 \cap A\) as outcome \(B_1 R_2\).
A1: Correct value of \(P(B_1 \cap R_2) = \frac{35}{132}\).
A1: Correct conditional probability of \(\frac{7}{18}\) or equivalent.