2026 Cambridge IAL Mathematics (9709) Practice Paper with Answers

Equating the equations of the line and the curve:
\[x^2 - 4x + 7 = 2x + k\]
Rearranging into standard quadratic form:
\[x^2 - 6x + (7 - k) = 0\]
For the line and curve not to intersect, the discriminant must be strictly negative (\(b^2 - 4ac < 0\)):
\[(-6)^2 - 4(1)(7 - k) < 0\]
\[36 - 28 + 4k < 0\]
\[8 + 4k < 0\]
\[4k < -8 \implies k < -2\]

M1: Equate line and curve equations to form a 3-term quadratic in \(x\) and apply discriminant \(b^2 - 4ac < 0\).
A1: Obtain correct simplified inequality in terms of \(k\), such as \(8 + 4k < 0\).
A1: Correct final inequality \(k < -2\) (strict inequality required).

Rewrite the function in index form:
\[y = 6(2x + 1)^{-\frac{1}{2}}\]
Differentiate using the chain rule:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 6 \left(-\frac{1}{2}\right)(2)(2x + 1)^{-\frac{3}{2}} = -6(2x + 1)^{-\frac{3}{2}}\]
Substitute \(x = 4\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -6\left(2(4) + 1\right)^{-\frac{3}{2}} = -6(9)^{-\frac{3}{2}} = -6 \left(\frac{1}{27}\right) = -\frac{6}{27} = -\frac{2}{9}\]

M1: Attempt differentiation using chain rule to achieve \(k(2x + 1)^{-\frac{3}{2}}\) for non-zero constant \(k\).
A1: Correct value \(-\frac{2}{9}\) (or exact equivalent, or \(-0.222\) to 3 s.f.).

Integrate the function using the reverse chain rule:
\[\int (3x - 2)^{-\frac{1}{2}} \, \mathrm{d}x = \frac{(3x - 2)^{\frac{1}{2}}}{\frac{1}{2} \times 3} = \frac{2}{3}(3x - 2)^{\frac{1}{2}}\]
Apply the limits from 1 to 5:
\[\left[ \frac{2}{3}\sqrt{3x - 2} \right]_{1}^{5} = \frac{2}{3}\sqrt{3(5) - 2} - \frac{2}{3}\sqrt{3(1) - 2}\]
\[= \frac{2}{3}\sqrt{13} - \frac{2}{3}\sqrt{1} = \frac{2}{3}\sqrt{13} - \frac{2}{3}\]

M1: Integrate to obtain form \(k(3x - 2)^{\frac{1}{2}}\) where \(k\) is a non-zero constant.
A1: Correct integrated expression \(\frac{2}{3}(3x - 2)^{\frac{1}{2}}\).
A1: Correct substitution of limits leading to \(\frac{2}{3}\sqrt{13} - \frac{2}{3}\) (or \(a = \frac{2}{3}, b = -\frac{2}{3}\)).

The general term in the expansion of \((a + b)^6\) is given by \(\binom{6}{r} a^{6-r} b^r\).
For the term containing \(x^2\), we choose \(r = 2\):
\[\text{Term} = \binom{6}{2} (2)^{6-2} \left(-\frac{1}{2}x\right)^2\]
\[= 15 \times 2^4 \times \frac{1}{4}x^2 = 15 \times 16 \times \frac{1}{4}x^2 = 60x^2\]
Therefore, the coefficient of \(x^2\) is 60.

M1: Identify the correct term structure \(\binom{6}{2} (2)^4 \left(-\frac{1}{2}\right)^2\) with correct powers.
A1: Correct answer 60.

Set \(y = \mathrm{f}(x)\):
\[y = 2(x - 3)^2 - 5\]
Rearrange to make \(x\) the subject:
\[y + 5 = 2(x - 3)^2\]
\[(x - 3)^2 = \frac{y + 5}{2}\]
Since the domain of \(\mathrm{f}\) is \(x \leqslant 1\), we have \(x - 3 \leqslant -2 < 0\), so we take the negative square root:
\[x - 3 = -\sqrt{\frac{y + 5}{2}}\]
\[x = 3 - \sqrt{\frac{y + 5}{2}}\]
Hence,
\[\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}}\]

M1: Correct algebraic process to isolate \((x - 3)^2\) in terms of \(y\).
M1: Take square root and choose the negative branch based on the domain \(x \leqslant 1\).
A1: Fully correct inverse function \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}}\) written in terms of \(x\).

Rearrange the equation:
\[\tan^2 \theta = \frac{1}{3} \implies \tan \theta = \pm \frac{1}{\sqrt{3}}\]
For \(\tan \theta = \frac{1}{\sqrt{3}}\) in the range \(0^\circ \leqslant \theta \leqslant 180^\circ\):
\[\theta = 30^\circ\]
For \(\tan \theta = -\frac{1}{\sqrt{3}}\) in the range \(0^\circ \leqslant \theta \leqslant 180^\circ\):
\[\theta = 180^\circ - 30^\circ = 150^\circ\]
Thus, \(\theta = 30^\circ, 150^\circ\).

M1: Take square roots to obtain \(\tan \theta = \pm \frac{1}{\sqrt{3}}\) and find at least one correct principal angle.
A1: Both \(\theta = 30^\circ\) and \(\theta = 150^\circ\) with no other angles within \(0^\circ \leqslant \theta \leqslant 180^\circ\).

(a) Completing the square:
\[ 2x^2 - 4kx + 3k + 5 = 2(x^2 - 2kx) + 3k + 5 \]
\[ = 2\left[(x - k)^2 - k^2\right] + 3k + 5 \]
\[ = 2(x - k)^2 - 2k^2 + 3k + 5 \]
So \( p = k \) and \( q = -2k^2 + 3k + 5 \) (or \( 3k + 5 - 2k^2 \)).

(b) For two distinct real roots, the discriminant \( b^2 - 4ac > 0 \):
\[ (-4k)^2 - 4(2)(3k + 5) > 0 \]
\[ 16k^2 - 8(3k + 5) > 0 \]
\[ 16k^2 - 24k - 40 > 0 \]
Divide through by 8:
\[ 2k^2 - 3k - 5 > 0 \]
Factorise the quadratic equation \( 2k^2 - 3k - 5 = 0 \):
\[ (2k - 5)(k + 1) = 0 \implies k = \frac{5}{2}, k = -1 \]
For \( 2k^2 - 3k - 5 > 0 \), the values must lie outside the roots:
\[ k < -1 \quad \text{or} \quad k > \frac{5}{2} \]

(a)
- M1: Attempts to complete the square, obtaining \( 2(x - k)^2 \pm \dots \)
- A1: Fully correct expression: \( 2(x - k)^2 - 2k^2 + 3k + 5 \)

(b)
- M1: Uses discriminant \( b^2 - 4ac > 0 \) (or sets \( -q > 0 \))
- M1: Finds critical values \( k = -1 \) and \( k = \frac{5}{2} \) by factorising or formula
- A1: Correct inequality statement: \( k < -1 \text{ or } k > \frac{5}{2} \) (accept set notation; reject \( -1 > k > 2.5 \) or "and")

(a) Let \( y = 2(x - 2)^2 + 3 \).
Rearrange for \( x \):
\[ y - 3 = 2(x - 2)^2 \]
\[ (x - 2)^2 = \frac{y - 3}{2} \]
Since the domain of \( \mathrm{f} \) is \( x \ge 2 \), we take the positive square root:
\[ x - 2 = \sqrt{\frac{y - 3}{2}} \implies x = 2 + \sqrt{\frac{y - 3}{2}} \]
Thus, \( \mathrm{f}^{-1}(x) = 2 + \sqrt{\frac{x - 3}{2}} \).
The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \). Since \( x \ge 2 \), \( \mathrm{f}(x) \ge 3 \), so the domain of \( \mathrm{f}^{-1} \) is \( x \ge 3 \).

(b) \( \mathrm{gf}(x) = \mathrm{g}(2(x-2)^2 + 3) = \frac{3}{2(x-2)^2 + 3 - 1} = \frac{3}{2(x-2)^2 + 2} \).
Set \( \mathrm{gf}(x) = \frac{1}{3} \):
\[ \frac{3}{2(x-2)^2 + 2} = \frac{1}{3} \]
\[ 2(x-2)^2 + 2 = 9 \]
\[ 2(x-2)^2 = 7 \]
\[ (x-2)^2 = \frac{7}{2} \]
Since \( x \ge 2 \):
\[ x - 2 = \sqrt{\frac{7}{2}} = \frac{\sqrt{14}}{2} \implies x = 2 + \frac{\sqrt{14}}{2} \]

(a)
- M1: Sets \( y = \mathrm{f}(x) \) and makes correct algebraic steps to isolate \( (x - 2)^2 \)
- A1: Obtains \( \mathrm{f}^{-1}(x) = 2 + \sqrt{\frac{x - 3}{2}} \) (must be in terms of \( x \) with positive root chosen)
- B1: States domain \( x \ge 3 \) (or \( [3, \infty) \))

(b)
- M1: Forms correct composite function \( \frac{3}{2(x-2)^2 + 2} = \frac{1}{3} \) and solves for \( (x-2)^2 \)
- A1: Obtains exact value \( x = 2 + \sqrt{\frac{7}{2}} \) or \( 2 + \frac{\sqrt{14}}{2} \) (rejects negative square root due to domain \( x \ge 2 \))

(a) Completing the square for the circle equation:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0 \]
\[ (x - 3)^2 + (y + 2)^2 = 25 \]
Therefore, the centre is \( C(3, -2) \) and the radius is \( r = \sqrt{25} = 5 \).

(b) The gradient of the radius \( CA \) joining \( C(3, -2) \) to \( A(6, 2) \) is:
\[ m_{\text{radius}} = \frac{2 - (-2)}{6 - 3} = \frac{4}{3} \]
The tangent is perpendicular to the radius, so the gradient of the tangent is:
\[ m_{\text{tangent}} = -\frac{3}{4} \]
The equation of the tangent at \( A(6, 2) \) is:
\[ y - 2 = -\frac{3}{4}(x - 6) \]
\[ y = -\frac{3}{4}x + \frac{9}{2} + 2 = -\frac{3}{4}x + \frac{13}{2} \]
To find the \( y \)-intercept \( B \), set \( x = 0 \):
\[ y = \frac{13}{2} \]
Thus, the coordinates of \( B \) are \( \left(0, \frac{13}{2}\right) \) or \( (0, 6.5) \).

(a)
- B1: Centre \( (3, -2) \)
- B1: Radius \( r = 5 \)

(b)
- M1: Calculates gradient of radius \( CA \) and uses \( m_1 m_2 = -1 \) to find gradient of tangent (\( -\frac{3}{4} \))
- M1: Forms equation of tangent passing through \( (6, 2) \) and substitutes \( x = 0 \)
- A1: Correct coordinates \( B\left(0, \frac{13}{2}\right) \) or \( (0, 6.5) \)

(a) The perimeter of the sector is given by:
\[ P = 2r + r\theta = 30 \]
Rearranging for \( r\theta \):
\[ r\theta = 30 - 2r \implies \theta = \frac{30 - 2r}{r} \]
The area of the sector is:
\[ A = \frac{1}{2}r^2\theta = \frac{1}{2}r^2\left(\frac{30 - 2r}{r}\right) = \frac{1}{2}r(30 - 2r) = 15r - r^2 \]

(b) Given \( A = 54 \):
\[ 15r - r^2 = 54 \]
\[ r^2 - 15r + 54 = 0 \]
Factorising:
\[ (r - 6)(r - 9) = 0 \]
So the possible values of \( r \) are \( r = 6 \) and \( r = 9 \).

For \( r = 6 \):
\[ \theta = \frac{30 - 2(6)}{6} = \frac{18}{6} = 3\text{ radians} \]
(Note that \( 3 < \pi \approx 3.142 \), so this is valid.)

For \( r = 9 \):
\[ \theta = \frac{30 - 2(9)}{9} = \frac{12}{9} = \frac{4}{3}\text{ radians} \]
(Note that \( \frac{4}{3} < \pi \), so this is valid.)

(a)
- M1: Uses arc length / perimeter formula \( 2r + r\theta = 30 \) to express \( \theta \) or \( r\theta \) in terms of \( r \)
- A1: Substitutes into \( A = \frac{1}{2}r^2\theta \) and shows clear algebra to arrive at \( A = 15r - r^2 \)

(b)
- M1: Sets \( 15r - r^2 = 54 \) and attempts to solve 3-term quadratic in \( r \)
- A1: Obtains \( r = 6 \) and \( r = 9 \)
- A1: Obtains \( \theta = 3 \) (or 3.00) and \( \theta = \frac{4}{3} \) (or 1.33)

(a) Starting with the LHS:
\[ \text{LHS} = \frac{\sin\theta}{1 - \cos\theta} - \frac{1}{\sin\theta} = \frac{\sin^2\theta - (1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} \]
Using \( \sin^2\theta = 1 - \cos^2\theta \):
\[ = \frac{1 - \cos^2\theta - 1 + \cos\theta}{\sin\theta(1 - \cos\theta)} = \frac{\cos\theta - \cos^2\theta}{\sin\theta(1 - \cos\theta)} \]
Factorise the numerator:
\[ = \frac{\cos\theta(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} = \frac{\cos\theta}{\sin\theta} = \frac{1}{\tan\theta} = \text{RHS} \]

(b) Using the identity with \( \theta = 2x \):
\[ \frac{1}{\tan 2x} = 3\tan 2x \]
\[ 3\tan^2 2x = 1 \implies \tan^2 2x = \frac{1}{3} \implies \tan 2x = \pm \frac{1}{\sqrt{3}} \]
For \( 0^\circ < x < 180^\circ \), the interval for \( 2x \) is \( 0^\circ < 2x < 360^\circ \).

If \( \tan 2x = \frac{1}{\sqrt{3}} \):
\[ 2x = 30^\circ, 210^\circ \implies x = 15^\circ, 105^\circ \]
If \( \tan 2x = -\frac{1}{\sqrt{3}} \):
\[ 2x = 150^\circ, 330^\circ \implies x = 75^\circ, 165^\circ \]
Thus, the solutions are \( x = 15^\circ, 75^\circ, 105^\circ, 165^\circ \).

(a)
- M1: Puts LHS over a single common denominator \( \sin\theta(1 - \cos\theta) \)
- M1: Uses \( \sin^2\theta = 1 - \cos^2\theta \) and factorises numerator
- A1: Cancels \( (1 - \cos\theta) \) correctly to achieve \( \frac{\cos\theta}{\sin\theta} = \frac{1}{\tan\theta} \) with full justification

(b)
- M1: Replaces LHS with \( \frac{1}{\tan 2x} \) and obtains \( \tan^2 2x = \frac{1}{3} \implies \tan 2x = \pm\frac{1}{\sqrt{3}} \)
- A1: Obtains all four correct angles: \( 15^\circ, 75^\circ, 105^\circ, 165^\circ \) (no extra solutions in range)

(a) Using the binomial theorem:
\[ (2 - 3x)^5 = 2^5 + \binom{5}{1} 2^4 (-3x) + \binom{5}{2} 2^3 (-3x)^2 + \dots \]
\[ = 32 + 5(16)(-3x) + 10(8)(9x^2) + \dots \]
\[ = 32 - 240x + 720x^2 \]

(b) We consider the product:
\[ (a + x)(32 - 240x + 720x^2 + \dots) \]
The terms that contribute to the \( x^2 \) term are:
\[ a(720x^2) + x(-240x) = (720a - 240)x^2 \]
We are given that the coefficient of \( x^2 \) is \( -120 \):
\[ 720a - 240 = -120 \]
\[ 720a = 120 \]
\[ a = \frac{120}{720} = \frac{1}{6} \]

(a)
- M1: Applies binomial expansion with correct binomial coefficients for at least two terms
- A1: Fully correct simplified expansion \( 32 - 240x + 720x^2 \)

(b)
- M1: Identifies both products contributing to the \( x^2 \) coefficient: \( a(720) + 1(-240) \)
- M1: Sets their expression equal to \( -120 \) and solves linear equation for \( a \)
- A1: Obtains \( a = \frac{1}{6} \) (or exact equivalent)

(a) Rewrite \( y \) in index form:
\[ y = 8(3x + 1)^{-\frac{1}{2}} \]
Using the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 8\left(-\frac{1}{2}\right)(3x + 1)^{-\frac{3}{2}} \times 3 = -12(3x + 1)^{-\frac{3}{2}} = -\frac{12}{(3x + 1)^{\frac{3}{2}}} \]

(b) We are given \( \frac{\mathrm{d}x}{\mathrm{d}t} = 0.6 \).
By the chain rule for rates of change:
\[ \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t} \]
When \( x = 1 \):
\[ 3x + 1 = 3(1) + 1 = 4 \]
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{12}{4^{\frac{3}{2}}} = -\frac{12}{8} = -1.5 \]
Therefore:
\[ \frac{\mathrm{d}y}{\mathrm{d}t} = -1.5 \times 0.6 = -0.9 \]
The rate of change of the \( y \)-coordinate is \( -0.9\text{ units per second} \) (or decreasing at \( 0.9\text{ units per second} \)).

(a)
- M1: Differentiates \( 8(3x+1)^{-1/2} \) to obtain form \( k(3x+1)^{-3/2} \)
- A1: Correct derivative \( -12(3x + 1)^{-3/2} \) or \( -\frac{12}{(3x + 1)^{3/2}} \)

(b)
- M1: Evaluates \( \frac{\mathrm{d}y}{\mathrm{d}x} \) at \( x = 1 \)
- M1: Uses connected rates of change formula \( \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t} \)
- A1: Correct value \( -0.9 \) (or states decreasing at \( 0.9 \))

The volume of revolution about the \( x \)-axis is given by:
\[ V = \pi \int_{a}^{b} y^2 \,\mathrm{d}x \]
Here, \( y = (2x - 1)^{\frac{3}{2}} \), so:
\[ y^2 = \left((2x - 1)^{\frac{3}{2}}\right)^2 = (2x - 1)^3 \]
The limits are \( x = 1 \) to \( x = 5 \):
\[ V = \pi \int_{1}^{5} (2x - 1)^3 \,\mathrm{d}x \]
Integrate using the reverse chain rule:
\[ \int (2x - 1)^3 \,\mathrm{d}x = \frac{(2x - 1)^4}{4 \times 2} = \frac{(2x - 1)^4}{8} \]
Evaluating between the limits:
\[ V = \pi \left[ \frac{(2x - 1)^4}{8} \right]_{1}^{5} \]
For \( x = 5 \):
\[ \frac{(2(5) - 1)^4}{8} = \frac{9^4}{8} = \frac{6561}{8} \]
For \( x = 1 \):
\[ \frac{(2(1) - 1)^4}{8} = \frac{1^4}{8} = \frac{1}{8} \]
Subtracting the lower limit from the upper limit:
\[ V = \pi \left( \frac{6561}{8} - \frac{1}{8} \right) = \pi \left( \frac{6560}{8} \right) = 820\pi \]

- M1: Uses the correct volume formula \( V = \pi \int y^2 \,\mathrm{d}x \) with \( y^2 = (2x - 1)^3 \)
- M1: Integrates \( (2x - 1)^3 \) to obtain \( \frac{(2x - 1)^4}{8} \) (or expands correctly to \( 8x^3 - 12x^2 + 6x - 1 \) and integrates term by term)
- A1: Correct integrated expression with limits \( 1 \) and \( 5 \)
- DM1: Substitutes limits \( x = 5 \) and \( x = 1 \) into their integrated function and subtracts
- A1: Obtains exact volume \( 820\pi \) (must be exact in terms of \( \pi \))

(a) Completing the square for \( \mathrm{f}(x) \):
\[ \mathrm{f}(x) = 2(x^2 - 6x) + 11 = 2\left((x - 3)^2 - 9\right) + 11 = 2(x - 3)^2 - 7 \]
The vertex of the parabola is at \( x = 3 \).
For \( \mathrm{f} \) to be a one-one function on \( x \leqslant k \), the greatest value of \( k \) is \( 3 \).

(b) Let \( y = 2(x - 3)^2 - 7 \).
\[ y + 7 = 2(x - 3)^2 \]
\[ (x - 3)^2 = \frac{y + 7}{2} \]
Since the domain of \( \mathrm{f} \) is \( x \leqslant 3 \), we have \( x - 3 \leqslant 0 \), so we take the negative square root:
\[ x - 3 = -\sqrt{\frac{y + 7}{2}} \]
\[ x = 3 - \sqrt{\frac{y + 7}{2}} \]
Therefore,
\[ \mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 7}{2}} \]
For \( x \leqslant 3 \), \( (x - 3)^2 \geqslant 0 \), so \( \mathrm{f}(x) \geqslant -7 \).
Hence, the domain of \( \mathrm{f}^{-1} \) is \( x \geqslant -7 \) (or \( [-7, \infty) \)).

(a)
B1: State \( k = 3 \) (or greatest value is 3).

(b)
M1: Attempt to complete the square to the form \( 2(x - p)^2 + q \) (where \( p = 3 \), \( q = -7 \)).
M1: Correctly rearrange \( y = 2(x - 3)^2 - 7 \) to make \( (x - 3)^2 \) or \( x \) the subject.
A1: Obtain \( \mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 7}{2}} \) or exact equivalent, with the negative square root clearly identified, written in terms of \( x \).
B1: State correct domain of \( \mathrm{f}^{-1} \): \( x \geqslant -7 \) (accept \( [-7, \infty) \) or \( y \geqslant -7 \) if clearly intended as the domain).

(a) Completing the square:
\[ 2x^2 - 12x + 13 = 2(x^2 - 6x) + 13 = 2((x - 3)^2 - 9) + 13 = 2(x - 3)^2 - 5 \]
So \( p = 3 \) and \( q = -5 \).
Since \( x \leqslant 2 \), the minimum value of \( \mathrm{f}(x) \) occurs at \( x = 2 \):
\[ \mathrm{f}(2) = 2(2 - 3)^2 - 5 = 2(1) - 5 = -3 \]
As \( x \to -\infty \), \( \mathrm{f}(x) \to \infty \), hence the range of \( \mathrm{f} \) is \( \mathrm{f}(x) \geqslant -3 \) (or \( y \geqslant -3 \)).

(b) Let \( y = 2(x - 3)^2 - 5 \).
\[ y + 5 = 2(x - 3)^2 \implies (x - 3)^2 = \frac{y + 5}{2} \]
Taking the square root and noting that \( x \leqslant 2 < 3 \), so \( x - 3 < 0 \):
\[ x - 3 = -\sqrt{\frac{y + 5}{2}} \implies x = 3 - \sqrt{\frac{y + 5}{2}} \]
Hence \( \mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}} \).

(c) \( \mathrm{gf}(x) = \frac{2}{3} \implies \frac{6}{2\mathrm{f}(x) + 11} = \frac{2}{3} \).
\[ 2(2\mathrm{f}(x) + 11) = 18 \implies 2\mathrm{f}(x) + 11 = 9 \implies 2\mathrm{f}(x) = -2 \implies \mathrm{f}(x) = -1 \]
Setting \( 2(x - 3)^2 - 5 = -1 \):
\[ 2(x - 3)^2 = 4 \implies (x - 3)^2 = 2 \]
Since \( x \leqslant 2 \):
\[ x - 3 = -\sqrt{2} \implies x = 3 - \sqrt{2} \]

(a)
B1: State \( 2(x - 3)^2 - 5 \) (or \( p = 3, q = -5 \))
M1: Attempt to evaluate \( \mathrm{f}(2) \) using their completed square form or original expression
A1: Correct range \( \mathrm{f}(x) \geqslant -3 \) or \( y \geqslant -3 \) (do not accept \( x \geqslant -3 \))

(b)
M1: Correct algebraic process of rearranging \( y = 2(x - p)^2 + q \) to make \( (x - p)^2 \) the subject
M1: Taking the negative square root due to domain restriction \( x \leqslant 2 \)
A1: Correct inverse function \( \mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}} \) (must be in terms of \( x \))

(c)
M1: Set \( \mathrm{gf}(x) = \frac{2}{3} \) and solve for \( \mathrm{f}(x) \) (obtaining \( \mathrm{f}(x) = -1 \)), then equate to \( 2(x - 3)^2 - 5 \)
A1: Correct exact answer \( x = 3 - \sqrt{2} \) only (reject \( 3 + \sqrt{2} \))

Squaring both sides of the inequality gives:
\((2x - 5)^2 < 9(x + 1)^2\)

Expand both sides:
\(4x^2 - 20x + 25 < 9(x^2 + 2x + 1)\)
\(4x^2 - 20x + 25 < 9x^2 + 18x + 9\)

Rearrange into a standard quadratic inequality:
\(0 < 5x^2 + 38x - 16\)
\(5x^2 + 38x - 16 > 0\)

Factorize the quadratic expression:
\((5x - 2)(x + 8) > 0\)

The critical values are \(x = \frac{2}{5}\) and \(x = -8\).

Since the quadratic expression is greater than zero, the solution is:
\(x < -8 \text{ or } x > \frac{2}{5}\)

M1: State a correct method involving squaring both sides or setting up linear equations/inequalities to find critical values.
A1: Obtain the correct critical values \(x = -8\) and \(x = \frac{2}{5}\) (or \(0.4\)).
A1: State the correct final inequality \(x < -8 \text{ or } x > \frac{2}{5}\) (accept equivalent notation, e.g. \((-\infty, -8) \cup (\frac{2}{5}, \infty)\)).

Use the quotient rule to differentiate \(y = \frac{\ln x}{x^2}\):
Let \(u = \ln x\) and \(v = x^2\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = 2x\).

\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{x^2\left(\frac{1}{x}\right) - (\ln x)(2x)}{(x^2)^2} = \frac{x - 2x\ln x}{x^4} = \frac{1 - 2\ln x}{x^3} \]

At a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ 1 - 2\ln x = 0 \implies \ln x = \frac{1}{2} \implies x = \mathrm{e}^{\frac{1}{2}} = \sqrt{\mathrm{e}} \]

Substitute \(x = \mathrm{e}^{\frac{1}{2}}\) into the original equation to find \(y\):
\[ y = \frac{\ln\left(\mathrm{e}^{\frac{1}{2}}\right)}{\left(\mathrm{e}^{\frac{1}{2}}\right)^2} = \frac{\frac{1}{2}}{\mathrm{e}} = \frac{1}{2\mathrm{e}} \]

Thus, the exact coordinates of the stationary point are \(\left(\mathrm{e}^{\frac{1}{2}}, \frac{1}{2\mathrm{e}}\right)\) or \(\left(\sqrt{\mathrm{e}}, \frac{1}{2\mathrm{e}}\right)\).

M1: Apply the quotient rule (or product rule) correctly to obtain an expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) of the form \(\frac{a - b\ln x}{x^3}\).
DM1: Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solve for \(x\) in terms of \(\mathrm{e}\).
A1: Obtain the correct exact coordinates \(\left(\mathrm{e}^{\frac{1}{2}}, \frac{1}{2\mathrm{e}}\right)\) or \(\left(\sqrt{\mathrm{e}}, \frac{1}{2\mathrm{e}}\right)\).

(a) Squaring both sides of \(|2x - 5| > 3|x + 1|\):
\((2x - 5)^2 > 9(x + 1)^2\)
\(4x^2 - 20x + 25 > 9(x^2 + 2x + 1)\)
\(4x^2 - 20x + 25 > 9x^2 + 18x + 9\)
\(5x^2 + 38x - 16 < 0\)

Factorising to find the critical values:
\((5x - 2)(x + 8) = 0 \implies x = \frac{2}{5} = 0.4 \text{ or } x = -8\)

Since the inequality is \(< 0\), the solution is:
\(-8 < x < \frac{2}{5}\)

(b) Replacing \(x\) with \(\mathrm{e}^y\):
\(-8 < \mathrm{e}^y < \frac{2}{5}\)

Since \(\mathrm{e}^y > 0\) for all real \(y\), this simplifies to:
\(0 < \mathrm{e}^y < \frac{2}{5}\)

Taking natural logarithms of both sides:
\(y < \ln\left(\frac{2}{5}\right)\)

(a)
M1: State or imply a valid method to solve the inequality, e.g. squaring both sides to obtain a 3-term quadratic or considering linear equations with signs.
A1: Obtain critical values \(x = -8\) and \(x = \frac{2}{5}\) (or 0.4).
A1: Correct inequality \(-8 < x < \frac{2}{5}\) (or \(-8 < x < 0.4\)).

(b)
M1: Set \(-8 < \mathrm{e}^y < \frac{2}{5}\) and recognise that \(\mathrm{e}^y > 0\) for all real \(y\).
A1: Obtain \(y < \ln\left(\frac{2}{5}\right)\) or \(y < \ln(0.4)\) (do not allow decimals unless exact form equivalent is shown).

(a) Applying logarithm laws to \(2\ln(x + 3) - \ln(2x - 1) = \ln 8\):
\(\ln\left(\frac{(x + 3)^2}{2x - 1}\right) = \ln 8\)
\(\frac{(x + 3)^2}{2x - 1} = 8\)
\(x^2 + 6x + 9 = 8(2x - 1)\)
\(x^2 + 6x + 9 = 16x - 8\)
\(x^2 - 10x + 17 = 0\)

Using the quadratic formula:
\(x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(17)}}{2} = \frac{10 \pm \sqrt{100 - 68}}{2} = \frac{10 \pm \sqrt{32}}{2} = 5 \pm 2\sqrt{2}\)

Both values satisfy \(x > \frac{1}{2}\), so both roots are valid:
\(x = 5 + 2\sqrt{2}\) and \(x = 5 - 2\sqrt{2}\).

(b) Taking natural logarithms on both sides:
\(\ln(3^{2y+1}) = \ln(5^{y-1})\)
\((2y + 1)\ln 3 = (y - 1)\ln 5\)
\(2y\ln 3 + \ln 3 = y\ln 5 - \ln 5\)
\(y(2\ln 3 - \ln 5) = -\ln 5 - \ln 3\)
\(y = \frac{-(\ln 5 + \ln 3)}{2\ln 3 - \ln 5} = \frac{-\ln 15}{\ln 9 - \ln 5} = \frac{-\ln 15}{\ln(1.8)}\)
\(y \approx \frac{-2.70805}{0.587787} \approx -4.61\)

(a)
M1: Apply power and quotient laws of logarithms correctly to combine terms into a single logarithm equation.
A1: Obtain the correct 3-term quadratic \(x^2 - 10x + 17 = 0\).
A1: Obtain \(x = 5 \pm 2\sqrt{2}\) (both roots required and in exact form).

(b)
M1: Take logarithms of both sides and expand brackets to isolate terms in \(y\).
A1: Obtain \(y = -4.61\) (correct to 3 s.f.).

(a) Expanding \(R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha\).
Comparing coefficients with \(3\sin\theta - 4\cos\theta\):
\(R\cos\alpha = 3\)
\(R\sin\alpha = 4\)

\(R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\)
\(\tan\alpha = \frac{4}{3} \implies \alpha = \arctan\left(\frac{4}{3}\right) \approx 0.9273\) radians.

Thus, \(3\sin\theta - 4\cos\theta = 5\sin(\theta - 0.9273)\).

(b) \(5\sin(\theta - 0.9273) = 2.5\)
\(\sin(\theta - 0.9273) = 0.5\)

Principal value:
\(\theta - 0.9273 = \frac{\pi}{6} \approx 0.52360\) or \(\theta - 0.9273 = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \approx 2.61799\)

Solving for \(\theta\):
\(\theta_1 = 0.52360 + 0.92730 = 1.4509 \approx 1.45\)
\(\theta_2 = 2.61799 + 0.92730 = 3.5453 \approx 3.55\)

Both values lie within the interval \(0 \leqslant \theta \leqslant 2\pi\).

(a)
B1: State \(R = 5\).
B1: State \(\alpha = 0.9273\) (accept 0.927).

(b)
M1: State \(\sin(\theta - \alpha) = 0.5\) and evaluate at least one value for \(\theta - \alpha\).
M1: Obtain a second angle in the valid range by using \(\pi - \arcsin(0.5)\).
A1: Obtain both \(\theta = 1.45\) and \(\theta = 3.55\) (allow 1.451 and 3.545; reject extra solutions in the range).

(a) The interval width is \(h = \frac{4 - 1}{3} = 1\).
The \(x\)-values are \(x_0 = 1, x_1 = 2, x_2 = 3, x_3 = 4\).

Calculating the corresponding \(y\)-values:
\(y_0 = \sqrt{2 + \ln 1} = \sqrt{2} \approx 1.414214\)
\(y_1 = \sqrt{2 + \ln 2} \approx \sqrt{2.693147} \approx 1.641081\)
\(y_2 = \sqrt{2 + \ln 3} \approx \sqrt{3.098612} \approx 1.760288\)
\(y_3 = \sqrt{2 + \ln 4} \approx \sqrt{3.386294} \approx 1.840189\)

Applying the trapezium rule formula:
\(T = \frac{h}{2}\left[y_0 + y_3 + 2(y_1 + y_2)\right]\)
\(T = \frac{1}{2}\left[1.414214 + 1.840189 + 2(1.641081 + 1.760288)\right]\)
\(T = \frac{1}{2}\left[3.254403 + 2(3.401369)\right]\)
\(T = \frac{1}{2}\left[3.254403 + 6.802738\right] = \frac{1}{2}(10.057141) \approx 5.02857\)

To 3 decimal places, \(T = 5.029\).

(b) The function \(y = \sqrt{2 + \ln x}\) is concave downwards (or \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0\)) on the interval \([1, 4]\).
Therefore, the straight line chords connecting consecutive points on the curve lie entirely below the curve, meaning the area of the trapezia is less than the area under the curve.
Hence, the trapezium rule gives an underestimate.

(a)
M1: Apply the trapezium rule with \(h = 1\) and correct structure \(\frac{1}{2}[y_0 + y_3 + 2(y_1 + y_2)]\).
A1: Calculate correct values for the ordinates (at least 3 d.p.).
A1: Obtain \(5.029\) (cao).

(b)
B1: State that the approximation is an underestimate.
B1: Provide a correct reason referring to the curve being concave downwards / chords lying below the curve.

(a) Let \(f(x) = x^3 - 5x + 1\).
\(f(2.1) = (2.1)^3 - 5(2.1) + 1 = 9.261 - 10.5 + 1 = -0.239 < 0\)
\(f(2.2) = (2.2)^3 - 5(2.2) + 1 = 10.648 - 11.0 + 1 = 0.648 > 0\)

Since \(f(x)\) is a continuous polynomial and there is a change of sign between \(x = 2.1\) and \(x = 2.2\), the root \(\alpha\) must lie in the interval \(2.1 < \alpha < 2.2\).

(b) Applying the iteration \(x_{n+1} = (5x_n - 1)^{1/3}\) with \(x_1 = 2.15\):
\(x_2 = (5(2.15) - 1)^{1/3} = (9.75)^{1/3} \approx 2.13632\)
\(x_3 = (5(2.13632) - 1)^{1/3} = (9.68160)^{1/3} \approx 2.13129\)
\(x_4 = (5(2.13129) - 1)^{1/3} = (9.65645)^{1/3} \approx 2.12945\)
\(x_5 = (5(2.12945) - 1)^{1/3} = (9.64725)^{1/3} \approx 2.12877\)
\(x_6 = (5(2.12877) - 1)^{1/3} = (9.64385)^{1/3} \approx 2.12852\)
\(x_7 = (5(2.12852) - 1)^{1/3} = (9.64260)^{1/3} \approx 2.12843\)

Rounding to 3 decimal places, \(\alpha = 2.128\).

(a)
M1: Evaluate \(f(2.1)\) and \(f(2.2)\) for \(f(x) = x^3 - 5x + 1\).
A1: Obtain \(-0.239\) and \(0.648\) and state a valid conclusion referring to the change of sign.

(b)
M1: Use the iterative formula correctly at least once.
A1: Obtain iterations \(x_2 = 2.13632\), \(x_3 = 2.13129\), \(x_4 = 2.12945\) (to 5 d.p.).
A1: Obtain final answer \(\alpha = 2.128\) after sufficient iterations demonstrating convergence.

(a) Using the trigonometric double-angle identity \(\cos 2x = 2\cos^2 x - 1\), we have \(\cos^2 x = \frac{1 + \cos 2x}{2}\).
Therefore, \(6\cos^2 x = 3(1 + \cos 2x) = 3 + 3\cos 2x\).

Rewriting the integral:
\[\int_0^{\frac{1}{4}\pi} \left(3 + 3\cos 2x + \frac{2}{2x + 1}\right) \mathrm{d}x\]

Integrating term by term:
\[\left[ 3x + \frac{3}{2}\sin 2x + \ln|2x + 1| \right]_0^{\frac{1}{4}\pi}\]

Substituting the upper limit \(x = \frac{1}{4}\pi\):
\[3\left(\frac{1}{4}\pi\right) + \frac{3}{2}\sin\left(\frac{1}{2}\pi\right) + \ln\left(2\left(\frac{1}{4}\pi\right) + 1\right) = \frac{3}{4}\pi + \frac{3}{2}(1) + \ln\left(\frac{1}{2}\pi + 1\right) = \frac{3}{4}\pi + \frac{3}{2} + \ln\left(\frac{\pi + 2}{2}\right)\]

Substituting the lower limit \(x = 0\):
\[3(0) + \frac{3}{2}\sin(0) + \ln(1) = 0 + 0 + 0 = 0\]

Subtracting the lower limit value gives:
\[\frac{3}{4}\pi + \frac{3}{2} + \ln\left(\frac{\pi + 2}{2}\right) \quad \left(\text{or } \frac{3}{4}\pi + \frac{3}{2} + \ln(\pi + 2) - \ln 2\right)\]

(b) The integrand \(\frac{1}{2x + 1}\) has a vertical asymptote (discontinuity) at \(x = -\frac{1}{2}\).
Since \(x = -\frac{1}{2}\) lies inside the interval of integration \([-1, 1]\), the integral is improper and undefined.

(a)
M1: Use the identity \(\cos^2 x = \frac{1}{2}(1 + \cos 2x)\) to express \(6\cos^2 x\) as \(3 + 3\cos 2x\).
M1: Integrate correctly to obtain \(3x + \frac{3}{2}\sin 2x + \ln(2x + 1)\) (or with \(k\ln(2x+1)\) where \(k=1\)).
M1: Substitute limits \(0\) and \(\frac{1}{4}\pi\) into their integrated expression.
A1: Obtain the correct exact value \(\frac{3}{4}\pi + \frac{3}{2} + \ln\left(\frac{\pi + 2}{2}\right)\) or any exact equivalent.

(b)
B1: State that the integrand is discontinuous / undefined at \(x = -\frac{1}{2}\) (or \(x = -0.5\)) which lies within the interval of integration.

(a) Squaring both sides:
\[ (2x - 5)^2 = (3x + 1)^2 \]
\[ 4x^2 - 20x + 25 = 9x^2 + 6x + 1 \]
\[ 5x^2 + 26x - 24 = 0 \]
\[ (5x - 4)(x + 6) = 0 \]
Thus, \(x = \frac{4}{5}\) or \(x = -6\).

(b) The inequality \(|2x - 5| > |3x + 1|\) corresponds to:
\[ (2x - 5)^2 > (3x + 1)^2 \]
\[ 5x^2 + 26x - 24 < 0 \]
\[ (5x - 4)(x + 6) < 0 \]
Considering the region between the critical values, we obtain:
\[ -6 < x < \frac{4}{5} \]

(a)
M1: State a correct method to solve the equation (e.g., squaring both sides or setting up two linear equations \(2x - 5 = \pm(3x + 1)\)).
A1: Obtain both correct values \(x = -6\) and \(x = \frac{4}{5}\) (or \(0.8\)).

(b)
M1: Relate the inequality to the critical values from part (a) or establish \(5x^2 + 26x - 24 < 0\).
M1: Identify the interior region between the two critical values.
A1: Obtain correct solution \(-6 < x < \frac{4}{5}\) (or \(-6 < x < 0.8\)).

(a)
Differentiating \(x\) with respect to \(\theta\):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2 - 2\cos 2\theta = 2(1 - \cos 2\theta) \]
Using the double-angle identity \(\cos 2\theta = 1 - 2\sin^2 \theta\), we have \(1 - \cos 2\theta = 2\sin^2 \theta\):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2(2\sin^2 \theta) = 4\sin^2 \theta \]

Differentiating \(y\) with respect to \(\theta\) using the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 4 \times 2\sin \theta \cos \theta = 8\sin \theta \cos \theta \]

Using the chain rule for parametric differentiation:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{8\sin \theta \cos \theta}{4\sin^2 \theta} = \frac{2\cos \theta}{\sin \theta} = 2\cot \theta \]

---

(b)
At \(\theta = \frac{1}{3}\pi\):
\[ x_P = 2\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \]
\[ y_P = 4\sin^2\left(\frac{\pi}{3}\right) = 4\left(\frac{\sqrt{3}}{2}\right)^2 = 4 \times \frac{3}{4} = 3 \]

The gradient of the tangent at \(P\) is:
\[ m = 2\cot\left(\frac{\pi}{3}\right) = \frac{2}{\tan(\frac{\pi}{3})} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \]

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

Since \(Q\) lies on the \(y\)-axis, substitute \(x = 0\):
\[ y - 3 = \frac{2}{\sqrt{3}}\left(-\frac{2\pi}{3} + \frac{\sqrt{3}}{2}\right) \]
\[ y - 3 = -\frac{4\pi}{3\sqrt{3}} + 1 \]
\[ y = 4 - \frac{4\pi}{3\sqrt{3}} = 4 - \frac{4\sqrt{3}\pi}{9} \]

Thus, the exact coordinates of \(Q\) are \(\left(0, 4 - \frac{4\sqrt{3}\pi}{9}\right)\).

---

(c)
The given line is \(x + 2\sqrt{3}y = 5 \implies y = -\frac{1}{2\sqrt{3}}x + \frac{5}{2\sqrt{3}}\).
So the gradient of the line is \(m_{\text{line}} = -\frac{1}{2\sqrt{3}}\).

The gradient of the normal to the curve at parameter \(\theta\) is:
\[ m_{\text{normal}} = -\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}} = -\frac{1}{2\cot \theta} = -\frac{1}{2}\tan \theta \]

Since the normal is parallel to the line:
\[ -\frac{1}{2}\tan \theta = -\frac{1}{2\sqrt{3}} \implies \tan \theta = \frac{1}{\sqrt{3}} \]

For \(0 < \theta < \pi\):
\[ \theta = \frac{1}{6}\pi \]

(a)
- M1: Differentiate \(x\) to obtain form \(a + b\cos 2\theta\) and \(y\) to obtain form \(c\sin\theta\cos\theta\) (or \(k\sin 2\theta\)).
- A1: Correct derivatives: \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 2 - 2\cos 2\theta\) and \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 8\sin\theta\cos\theta\).
- A1: Fully correct use of \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}\) and the identity \(1 - \cos 2\theta = 2\sin^2 \theta\) leading convincingly to \(2\cot\theta\) with no algebraic errors.

(b)
- M1: Evaluate \(x\), \(y\), and \(\frac{\mathrm{d}y}{\mathrm{d}x}\) at \(\theta = \frac{1}{3}\pi\).
- M1: Form the equation of the tangent line and set \(x = 0\) (or use \(y = y_1 - mx_1\)).
- A1: Obtain correct exact coordinates \(\left(0, 4 - \frac{4\sqrt{3}\pi}{9}\right)\) or \(\left(0, 4 - \frac{4\pi}{3\sqrt{3}}\right)\).

(c)
- M1: Find gradient of the line \(-\frac{1}{2\sqrt{3}}\) and equate to gradient of normal \(-\frac{1}{2\cot\theta}\) or \(-\frac{1}{2}\tan\theta\).
- A1: Obtain \(\theta = \frac{1}{6}\pi\) (or exact equivalent in radians; do not accept degrees unless converted).

Squaring both sides of the inequality gives:
\[ (3x - 2)^2 > 4(x + 1)^2 \]
\[ 9x^2 - 12x + 4 > 4(x^2 + 2x + 1) \]
\[ 9x^2 - 12x + 4 > 4x^2 + 8x + 4 \]
\[ 5x^2 - 20x > 0 \]
\[ 5x(x - 4) > 0 \]

The critical values are \(x = 0\) and \(x = 4\).

Since the quadratic expression \(5x(x - 4)\) must be strictly positive, the solution is:
\[ x < 0 \quad\text{or}\quad x > 4 \]

M1: State or imply non-negative squaring of both sides or set up two linear equations/inequalities to find critical values.
A1: Obtain the correct critical values \(x = 0\) and \(x = 4\).
A1: State the correct final range \(x < 0\) or \(x > 4\) (accept separate inequalities with 'or', reject 'and').

Apply laws of logarithms to the left-hand side:
\[ \ln(2x + 5) - \ln(x^2) = \ln 3 \]
\[ \ln\left(\frac{2x + 5}{x^2}\right) = \ln 3 \]

Equating arguments:
\[ \frac{2x + 5}{x^2} = 3 \]
\[ 3x^2 - 2x - 5 = 0 \]

Factorising the quadratic equation:
\[ (3x - 5)(x + 1) = 0 \]
\[ x = \frac{5}{3} \quad\text{or}\quad x = -1 \]

Since \(\ln x\) is only defined for \(x > 0\), we reject \(x = -1\).

Therefore, \(x = \frac{5}{3} \approx 1.67\).

M1: Use power and quotient laws of logarithms correctly to combine terms into a single logarithmic expression or remove logarithms.
A1: Obtain correct 3-term quadratic \(3x^2 - 2x - 5 = 0\) and solve to find roots.
A1: State \(x = 1.67\) (or exact \(\frac{5}{3}\)) and explicitly reject the negative root \(x = -1\).

Expand each compound angle expression using the addition formulae:
\[ \cos(\theta + 60^\circ) = \cos\theta\cos 60^\circ - \sin\theta\sin 60^\circ \]
\[ \sin(\theta + 30^\circ) = \sin\theta\cos 30^\circ + \cos\theta\sin 30^\circ \]

Substitute the exact values of the trigonometric ratios:
\[ \left(\frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta\right) + 2\left(\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta\right) = 0 \]
\[ \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta + \sqrt{3}\sin\theta + \cos\theta = 0 \]

Collect like terms:
\[ \frac{3}{2}\cos\theta + \frac{\sqrt{3}}{2}\sin\theta = 0 \]
\[ \frac{\sqrt{3}}{2}\sin\theta = -\frac{3}{2}\cos\theta \]
\[ \tan\theta = -\frac{3}{\sqrt{3}} = -\sqrt{3} \]

M1: Expand both compound angle terms using standard addition formulae.
M1: Substitute exact values for \(\sin 30^\circ, \cos 30^\circ, \sin 60^\circ, \cos 60^\circ\) and group terms in \(\sin\theta\) and \(\cos\theta\).
A1: Obtain \(\tan\theta = -\sqrt{3}\) with complete and convincing working.

Differentiating each parametric equation with respect to \(t\):
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = 2\mathrm{e}^{2t} \]
\[ \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{3}{3t + 1} \]

Using the chain rule for parametric differentiation:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{\frac{3}{3t + 1}}{2\mathrm{e}^{2t}} = \frac{3}{2(3t + 1)\mathrm{e}^{2t}} \]

At \(t = 1\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{2(4)\mathrm{e}^2} = \frac{3}{8\mathrm{e}^2} \]

The gradient of the normal is given by \(-\frac{1}{\mathrm{d}y/\mathrm{d}x}\):
\[ m_{\text{normal}} = -\frac{8\mathrm{e}^2}{3} = -\frac{8}{3}\mathrm{e}^2 \]

M1: Correctly differentiate \(x\) and \(y\) with respect to \(t\) and use \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\).
M1: Substitute \(t = 1\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and use \(m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\).
A1: Obtain exact value \(-\frac{8}{3}\mathrm{e}^2\) (or \(-\frac{8\mathrm{e}^2}{3}\)).

Use integration by parts with \(u = x\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \cos(3x)\):
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = 1, \quad v = \frac{1}{3}\sin(3x) \]

Applying the integration by parts formula:
\[ \int x \cos(3x) \, \mathrm{d}x = \left[ \frac{1}{3}x\sin(3x) \right]_0^{\frac{\pi}{6}} - \int_0^{\frac{\pi}{6}} \frac{1}{3}\sin(3x) \, \mathrm{d}x \]
\[ = \left( \frac{1}{3}\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{2}\right) - 0 \right) - \left[ -\frac{1}{9}\cos(3x) \right]_0^{\frac{\pi}{6}} \]
\[ = \frac{\pi}{18}(1) + \frac{1}{9}\left(\cos\left(\frac{\pi}{2}\right) - \cos(0)\right) \]
\[ = \frac{\pi}{18} + \frac{1}{9}(0 - 1) = \frac{\pi}{18} - \frac{1}{9} = \frac{\pi - 2}{18} \]

M1: Apply integration by parts to obtain an expression of the form \(kx\sin(3x) \pm \int m\sin(3x)\,\mathrm{d}x\).
A1: Correct indefinite integral \(\frac{1}{3}x\sin(3x) + \frac{1}{9}\cos(3x)\).
A1: Substitute limits correctly and obtain exact value \(\frac{\pi - 2}{18}\) (or \(\frac{\pi}{18} - \frac{1}{9}\)).

Multiply the numerator and denominator by the complex conjugate of the denominator:
\[ w = \frac{(1 + 7\mathrm{i})(4 - 3\mathrm{i})}{(4 + 3\mathrm{i})(4 - 3\mathrm{i})} = \frac{4 - 3\mathrm{i} + 28\mathrm{i} - 21\mathrm{i}^2}{4^2 + 3^2} \]
\[ w = \frac{4 + 25\mathrm{i} + 21}{16 + 9} = \frac{25 + 25\mathrm{i}}{25} = 1 + \mathrm{i} \]

Now determine the modulus \(r\) and argument \(\theta\):
\[ r = |w| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
\[ \theta = \arg(w) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]

Therefore, in exponential form:
\[ w = \sqrt{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{4}} \]

M1: Multiply numerator and denominator by \(4 - 3\mathrm{i}\) and simplify to Cartesian form \(x + \mathrm{i}y\).
A1: Obtain \(w = 1 + \mathrm{i}\).
A1: State the correct modulus \(r = \sqrt{2}\) and argument \(\theta = \frac{\pi}{4}\), writing \(w = \sqrt{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\) (or \(\sqrt{2}\mathrm{e}^{\frac{1}{4}\pi\mathrm{i}}\)).

First find the displacement vector \(\overrightarrow{AB}\):
\[ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ 1 \\ -3 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ -6 \end{pmatrix} \]

The direction vector of line \(l\) is \(\mathbf{d} = \begin{pmatrix} 4 \\ -2 \\ p \end{pmatrix}\).

Since \(l\) is perpendicular to \(\overrightarrow{AB}\), their scalar (dot) product is zero:
\[ \mathbf{d} \cdot \overrightarrow{AB} = 0 \]
\[ (4)(3) + (-2)(2) + (p)(-6) = 0 \]
\[ 12 - 4 - 6p = 0 \]
\[ 8 - 6p = 0 \implies p = \frac{8}{6} = \frac{4}{3} \]

B1: Obtain correct vector \(\overrightarrow{AB} = 3\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}\) (or \(\begin{pmatrix} 3 \\ 2 \\ -6 \end{pmatrix}\)).
M1: Set the scalar product of their \(\overrightarrow{AB}\) and \(\begin{pmatrix} 4 \\ -2 \\ p \end{pmatrix}\) equal to zero.
A1: Obtain \(p = \frac{4}{3}\) (or exact equivalent).

Using the iteration formula \(x_{n+1} = \frac{2x_n^3 + 1}{4}\) with \(x_1 = 0.3\):
\[ x_2 = \frac{2(0.3)^3 + 1}{4} = \frac{2(0.027) + 1}{4} = \frac{1.054}{4} = 0.2635 \]
\[ x_3 = \frac{2(0.2635)^3 + 1}{4} = \frac{2(0.018294) + 1}{4} \approx 0.2591 \]
\[ x_4 = \frac{2(0.259147)^3 + 1}{4} = \frac{2(0.017403) + 1}{4} \approx 0.2587 \]

Continuing to observe convergence:
\[ x_5 = \frac{2(0.258702)^3 + 1}{4} \approx 0.2587 \]

Rounding to 3 decimal places gives:
\[ \alpha = 0.259 \]

M1: Calculate at least two consecutive iterates correctly using the given formula.
A1: Obtain \(x_2 = 0.2635\), \(x_3 = 0.2591\), \(x_4 = 0.2587\) correct to 4 decimal places.
A1: State final value \(\alpha = 0.259\) correct to 3 decimal places with evidence of convergence.

(a) Let \( \frac{7x + 4}{(1 - 2x)(2 + x)} = \frac{A}{1 - 2x} + \frac{B}{2 + x} \).
Multiplying across gives \( 7x + 4 = A(2 + x) + B(1 - 2x) \).
Setting \( x = \frac{1}{2} \): \( 7\left(\frac{1}{2}\right) + 4 = A\left(2 + \frac{1}{2}\right) \implies \frac{15}{2} = \frac{5}{2}A \implies A = 3 \).
Setting \( x = -2 \): \( 7(-2) + 4 = B(1 - 2(-2)) \implies -10 = 5B \implies B = -2 \).
Thus, \( f(x) = \frac{3}{1 - 2x} - \frac{2}{2 + x} \).

(b) Rewrite as powers:
\( f(x) = 3(1 - 2x)^{-1} - 2(2 + x)^{-1} = 3(1 - 2x)^{-1} - \left(1 + \frac{1}{2}x\right)^{-1} \).
Expanding each part:
\( 3(1 - 2x)^{-1} = 3\left[ 1 + (-1)(-2x) + \frac{(-1)(-2)}{2!}(-2x)^2 + \dots \right] = 3(1 + 2x + 4x^2) = 3 + 6x + 12x^2 \).
\( \left(1 + \frac{1}{2}x\right)^{-1} = 1 + (-1)\left(\frac{1}{2}x\right) + \frac{(-1)(-2)}{2!}\left(\frac{1}{2}x\right)^2 + \dots = 1 - \frac{1}{2}x + \frac{1}{4}x^2 \).
Subtracting the two series:
\( f(x) = (3 + 6x + 12x^2) - \left(1 - \frac{1}{2}x + \frac{1}{4}x^2\right) = 2 + \left(6 + \frac{1}{2}\right)x + \left(12 - \frac{1}{4}\right)x^2 = 2 + \frac{13}{2}x + \frac{47}{4}x^2 \).

(a)
M1: Correct method to find at least one constant using substitution or equating coefficients.
A1: Obtains \( \frac{3}{1 - 2x} - \frac{2}{2 + x} \) (or equivalent).

(b)
M1: Correct method to expand \( (1 - 2x)^{-1} \) or \( \left(1 + \frac{1}{2}x\right)^{-1} \) up to the term in \( x^2 \).
M1: Combines the expansions correctly according to their partial fractions.
A1: Obtains correct simplified expansion \( 2 + \frac{13}{2}x + \frac{47}{4}x^2 \) (or \( 2 + 6.5x + 11.75x^2 \)).

The equation can be written as:
\( 3 \cdot (3^x)^2 - 5(3^x) - 2 = 0 \).
Let \( u = 3^x \):
\( 3u^2 - 5u - 2 = 0 \)
\( (3u + 1)(u - 2) = 0 \).
This gives \( u = -\frac{1}{3} \) or \( u = 2 \).
Since \( 3^x > 0 \) for all real \( x \), \( u = -\frac{1}{3} \) has no real solution.
For \( 3^x = 2 \):
\( x \ln 3 = \ln 2 \implies x = \frac{\ln 2}{\ln 3} \approx 0.6309297... \)
Rounding to 3 significant figures gives \( x = 0.631 \).

M1: Recognises the quadratic in \( 3^x \) and attempts to factorise or use quadratic formula.
A1: Obtains \( 3^x = 2 \) and rejects \( 3^x = -\frac{1}{3} \).
M1: Applies logarithms correctly to solve \( 3^x = 2 \).
A1: Obtains \( 0.631 \) (must be to 3 s.f.).

(a) \( R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha \).
Equating coefficients:
\( R\cos\alpha = 4 \)
\( R\sin\alpha = 3 \)
\( R = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \).
\( \tan\alpha = \frac{3}{4} \implies \alpha = 36.8698...^\circ \approx 36.87^\circ \).
So, \( 4\sin\theta - 3\cos\theta = 5\sin(\theta - 36.87^\circ) \).

(b) Using the result from part (a) with angle \( 2\theta \):
\( 5\sin(2\theta - 36.87^\circ) = 2.5 \implies \sin(2\theta - 36.87^\circ) = 0.5 \).
For \( 0^\circ \leqslant \theta \leqslant 180^\circ \), the range of \( 2\theta - 36.87^\circ \) is \( -36.87^\circ \leqslant 2\theta - 36.87^\circ \leqslant 323.13^\circ \).
Possible angles:
\( 2\theta - 36.87^\circ = 30^\circ \implies 2\theta = 66.87^\circ \implies \theta \approx 33.4^\circ \).
\( 2\theta - 36.87^\circ = 180^\circ - 30^\circ = 150^\circ \implies 2\theta = 186.87^\circ \implies \theta \approx 93.4^\circ \).

(a)
B1: State \( R = 5 \).
B1: Calculate \( \alpha = 36.87^\circ \).

(b)
M1: State \( \sin(2\theta - 36.87^\circ) = 0.5 \) and find at least one valid principal value.
A1: Obtains \( \theta = 33.4^\circ \) (to 1 d.p.).
A1: Obtains second solution \( \theta = 93.4^\circ \) and no others in the given range.

Use integration by parts:
Let \( u = 2x + 1 \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 2 \).
Let \( \frac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^{-2x} \implies v = -\frac{1}{2}\mathrm{e}^{-2x} \).

Applying the formula \( \int u \frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x \):
\( \int (2x + 1)\mathrm{e}^{-2x}\,\mathrm{d}x = \left[ -\frac{1}{2}(2x + 1)\mathrm{e}^{-2x} \right] - \int \left(-\frac{1}{2}\mathrm{e}^{-2x}\right)(2)\,\mathrm{d}x \)
\( = -\frac{1}{2}(2x + 1)\mathrm{e}^{-2x} + \int \mathrm{e}^{-2x}\,\mathrm{d}x \)
\( = -\frac{1}{2}(2x + 1)\mathrm{e}^{-2x} - \frac{1}{2}\mathrm{e}^{-2x} \)
\( = -(x + 1)\mathrm{e}^{-2x} \).

Evaluate between limits \( 0 \) and \( 1 \):
\( \left[ -(x + 1)\mathrm{e}^{-2x} \right]_0^1 = \left( -(1 + 1)\mathrm{e}^{-2} \right) - \left( -(0 + 1)\mathrm{e}^{0} \right) \)
\( = -2\mathrm{e}^{-2} - (-1) = 1 - 2\mathrm{e}^{-2} \) (or \( \frac{\mathrm{e}^2 - 2}{\mathrm{e}^2} \)).

M1: Applies integration by parts in the correct form \( u = 2x + 1 \) and \( v' = \mathrm{e}^{-2x} \).
A1: Obtains \( -\frac{1}{2}(2x + 1)\mathrm{e}^{-2x} - \int -\frac{1}{2}\mathrm{e}^{-2x}(2)\,\mathrm{d}x \).
A1: Obtains correct integrated expression \( -(x + 1)\mathrm{e}^{-2x} \) (or equivalent unsimplified form).
DM1: Correctly substitutes limits \( 0 \) and \( 1 \) into their integrated expression.
A1: Obtains exact answer \( 1 - 2\mathrm{e}^{-2} \) (or \( \frac{\mathrm{e}^2 - 2}{\mathrm{e}^2} \)).

(a) Let \( f(x) = x^3 - 3x - 7 \).
\( f(2) = 2^3 - 3(2) - 7 = 8 - 6 - 7 = -5 < 0 \).
\( f(3) = 3^3 - 3(3) - 7 = 27 - 9 - 7 = 11 > 0 \).
Since there is a sign change and \( f(x) \) is continuous, \( \alpha \) lies between 2 and 3.

(b) If the sequence converges to a limit \( L \), then \( L = \sqrt[3]{3L + 7} \).
Cubing both sides gives \( L^3 = 3L + 7 \), which is the original equation defining \( \alpha \). Thus, \( L = \alpha \).

(c) Choosing \( x_1 = 2.5 \):
\( x_2 = \sqrt[3]{3(2.5) + 7} = \sqrt[3]{14.5} \approx 2.4384 \)
\( x_3 = \sqrt[3]{3(2.4384) + 7} \approx 2.4280 \)
\( x_4 = \sqrt[3]{3(2.4280) + 7} \approx 2.4263 \)
\( x_5 = \sqrt[3]{3(2.4263) + 7} \approx 2.4260 \)
Thus, \( \alpha = 2.43 \) correct to 2 decimal places.

(a)
M1: Evaluates values of \( f(x) = x^3 - 3x - 7 \) (or equivalent) at \( x = 2 \) and \( x = 3 \).
A1: Shows correct values \( -5 \) and \( 11 \) and concludes with reference to sign change.

(b)
B1: Replaces \( x_{n+1} \) and \( x_n \) with \( L \) (or \( \alpha \)) and derives \( L^3 = 3L + 7 \).

(c)
M1: Uses the iterative formula with starting value in \( [2, 3] \) and calculates at least two iterations to 4 d.p.
A1: Obtains \( 2.43 \) with iterations clearly displayed.

First, find the direction vector of the line passing through \( A \) and \( B \):
\( \vec{AB} = \mathbf{b} - \mathbf{a} = (3 - 1)\mathbf{i} + (-1 - 2)\mathbf{j} + (2 - (-1))\mathbf{k} = 2\mathbf{i} - 3\mathbf{j} + 3\mathbf{k} \).

The direction vector of line \( l \) is \( \mathbf{d} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \).

Calculate the scalar product \( \vec{AB} \cdot \mathbf{d} \):
\( \vec{AB} \cdot \mathbf{d} = (2)(1) + (-3)(-2) + (3)(2) = 2 + 6 + 6 = 14 \).

Calculate the magnitudes of the two direction vectors:
\( |\vec{AB}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{4 + 9 + 9} = \sqrt{22} \).
\( |\mathbf{d}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).

Let \( \theta \) be the angle between the lines:
\( \cos\theta = \frac{|\vec{AB} \cdot \mathbf{d}|}{|\vec{AB}||\mathbf{d}|} = \frac{14}{3\sqrt{22}} \approx \frac{14}{14.071247} \approx 0.994936 \).
\( \theta = \arccos(0.994936) \approx 5.768^\circ \approx 5.8^\circ \).

B1: Obtains direction vector \( \vec{AB} = 2\mathbf{i} - 3\mathbf{j} + 3\mathbf{k} \) (or opposite).
M1: Calculates scalar product of \( \vec{AB} \) and \( \mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \).
M1: Calculates magnitudes of both direction vectors and uses \( \cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|} \).
A1: Obtains \( \cos\theta = \frac{14}{3\sqrt{22}} \) (or decimal equivalent \( 0.9949... \)).
A1: Obtains acute angle \( 5.8^\circ \) (or \( 0.101 \) rad).

(a) Multiply numerator and denominator by the complex conjugate of the denominator, \( 3 + 4\mathrm{i} \):
\( u = \frac{(1 + 7\mathrm{i})(3 + 4\mathrm{i})}{(3 - 4\mathrm{i})(3 + 4\mathrm{i})} = \frac{3 + 4\mathrm{i} + 21\mathrm{i} + 28\mathrm{i}^2}{3^2 + 4^2} \)
\( = \frac{3 + 25\mathrm{i} - 28}{9 + 16} = \frac{-25 + 25\mathrm{i}}{25} = -1 + \mathrm{i} \).

(b) Modulus:
\( |u| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \).

Argument:
Since \( u \) lies in the second quadrant (real part negative, imaginary part positive):
\( \arg(u) = \pi - \arctan(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).

(c) The locus \( |z - u| = 2 \) represents a circle with centre at \( u = -1 + \mathrm{i} \) (coordinates \( (-1, 1) \)) and radius 2.

(a)
M1: Multiplies numerator and denominator by \( 3 + 4\mathrm{i} \) and simplifies denominator.
A1: Obtains \( -1 + \mathrm{i} \).

(b)
B1: Obtains \( |u| = \sqrt{2} \).
B1: Obtains \( \arg(u) = \frac{3}{4}\pi \) (accept \( 2.36 \) or \( 135^\circ \)).

(c)
B1: States circle with centre \( (-1, 1) \) (or \( -1 + \mathrm{i} \)) and radius 2.

(a) Let \(\frac{9 - 7x}{(1 + x)(3 - x)} = \frac{A}{1 + x} + \frac{B}{3 - x}\).

Then \(9 - 7x = A(3 - x) + B(1 + x)\).

Setting \(x = -1\):
\(9 - 7(-1) = A(3 - (-1)) \implies 16 = 4A \implies A = 4\).

Setting \(x = 3\):
\(9 - 7(3) = B(1 + 3) \implies -12 = 4B \implies B = -3\).

Thus, \(f(x) = \frac{4}{1 + x} - \frac{3}{3 - x} = \frac{4}{1 + x} - \frac{1}{1 - \frac{1}{3}x}\).

(b) Expanding each term using the binomial series:
\(4(1 + x)^{-1} = 4\left(1 - x + x^2 - \dots\right) = 4 - 4x + 4x^2\)

\(-\left(1 - \frac{1}{3}x\right)^{-1} = -\left(1 + \frac{1}{3}x + \left(-\frac{1}{3}x\right)^2\frac{(-1)(-2)}{2} + \dots\right) = -\left(1 + \frac{1}{3}x + \frac{1}{9}x^2\right) = -1 - \frac{1}{3}x - \frac{1}{9}x^2\)

Combining the expansions:
\(f(x) = (4 - 1) + \left(-4 - \frac{1}{3}\right)x + \left(4 - \frac{1}{9}\right)x^2 = 3 - \frac{13}{3}x + \frac{35}{9}x^2\).

(a)
- M1: State correct form \(\frac{A}{1+x} + \frac{B}{3-x}\) and use a valid method to find at least one constant.
- A1: Obtain correct partial fractions \(\frac{4}{1+x} - \frac{3}{3-x}\).

(b)
- M1: Use binomial series to expand \((1+x)^{-1}\) or \((1 - \frac{1}{3}x)^{-1}\) up to the term in \(x^2\).
- M1: Combine the two expansions correctly using their partial fractions.
- A1: Obtain fully correct simplified expansion \(3 - \frac{13}{3}x + \frac{35}{9}x^2\).

To evaluate the definite integral \(\int_0^{\frac{1}{2}\pi} x^2 \sin(2x) \, \mathrm{d}x\), we use integration by parts:

Let \(u = x^2\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \sin(2x)\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\) and \(v = -\frac{1}{2}\cos(2x)\).

Applying integration by parts:
\[ \int x^2 \sin(2x) \, \mathrm{d}x = -\frac{1}{2}x^2 \cos(2x) - \int \left(-\frac{1}{2}\cos(2x)\right)(2x) \, \mathrm{d}x \]
\[ = -\frac{1}{2}x^2 \cos(2x) + \int x \cos(2x) \, \mathrm{d}x \]

Apply integration by parts a second time to \(\int x \cos(2x) \, \mathrm{d}x\):
Let \(u = x\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \cos(2x)\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = 1\) and \(v = \frac{1}{2}\sin(2x)\).

\[ \int x \cos(2x) \, \mathrm{d}x = \frac{1}{2}x \sin(2x) - \int \frac{1}{2}\sin(2x) \, \mathrm{d}x \]
\[ = \frac{1}{2}x \sin(2x) + \frac{1}{4}\cos(2x) \]

Combining the terms gives the indefinite integral:
\[ \int x^2 \sin(2x) \, \mathrm{d}x = -\frac{1}{2}x^2 \cos(2x) + \frac{1}{2}x \sin(2x) + \frac{1}{4}\cos(2x) \]

Now substitute the limits \(0\) and \(\frac{1}{2}\pi\):
At the upper limit \(x = \frac{1}{2}\pi\):
\[ -\frac{1}{2}\left(\frac{\pi}{2}\right)^2 \cos(\pi) + \frac{1}{2}\left(\frac{\pi}{2}\right) \sin(\pi) + \frac{1}{4}\cos(\pi) = -\frac{1}{2}\left(\frac{\pi^2}{4}\right)(-1) + 0 + \frac{1}{4}(-1) = \frac{\pi^2}{8} - \frac{1}{4} \]

At the lower limit \(x = 0\):
\[ -\frac{1}{2}(0)^2 \cos(0) + \frac{1}{2}(0)\sin(0) + \frac{1}{4}\cos(0) = 0 + 0 + \frac{1}{4}(1) = \frac{1}{4} \]

Subtracting the lower limit value from the upper limit value:
\[ \left(\frac{\pi^2}{8} - \frac{1}{4}\right) - \frac{1}{4} = \frac{\pi^2}{8} - \frac{1}{2} = \frac{\pi^2 - 4}{8} \]

M1: Attempt integration by parts to obtain an expression of the form \(a x^2 \cos(2x) + b \int x \cos(2x) \, \mathrm{d}x\), where \(a \neq 0\) and \(b \neq 0\).
A1: Obtain correct first-stage indefinite integral: \(-\frac{1}{2}x^2 \cos(2x) + \int x \cos(2x) \, \mathrm{d}x\) (or equivalent).
M1: Attempt integration by parts on \(\int x \cos(2x) \, \mathrm{d}x\) to obtain \(c x \sin(2x) \pm d \int \sin(2x) \, \mathrm{d}x\).
A1: Obtain completely correct integral \(\left[ -\frac{1}{2}x^2 \cos(2x) + \frac{1}{2}x \sin(2x) + \frac{1}{4}\cos(2x) \right]_0^{\frac{1}{2}\pi}\).
A1: Substitute limits correctly and obtain exact value \(\frac{1}{8}\pi^2 - \frac{1}{2}\) or \(\frac{\pi^2 - 4}{8}\) (isw if decimal value is also given, but decimal alone scores A0).

(a) Let \(\frac{1}{(t+1)(2t+1)} = \frac{A}{t+1} + \frac{B}{2t+1}\).
\(1 = A(2t+1) + B(t+1)\).
Setting \(t = -1 \implies 1 = A(-1) \implies A = -1\).
Setting \(t = -\frac{1}{2} \implies 1 = B\left(\frac{1}{2}\right) \implies B = 2\).
Hence, \(\frac{1}{(t+1)(2t+1)} = \frac{2}{2t+1} - \frac{1}{t+1}\).

(b) Separating variables:
\[\int \frac{1}{5-m}\,\mathrm{d}m = \int \frac{2}{(t+1)(2t+1)}\,\mathrm{d}t\]
\[\int \frac{1}{5-m}\,\mathrm{d}m = 2\int \left(\frac{2}{2t+1} - \frac{1}{t+1}\right)\,\mathrm{d}t\]
\[-\ln(5-m) = 2\left(\ln(2t+1) - \ln(t+1)\right) + c\]
\[-\ln(5-m) = 2\ln\left(\frac{2t+1}{t+1}\right) + c\]
Using the initial condition \(m = 0\) when \(t = 0\):
\[-\ln(5) = 2\ln(1) + c \implies c = -\ln 5\]
Substituting \(c\):
\[-\ln(5-m) = 2\ln\left(\frac{2t+1}{t+1}\right) - \ln 5\]
\[\ln(5-m) = \ln 5 - \ln\left(\frac{2t+1}{t+1}\right)^{\!2} = \ln\left(\frac{5(t+1)^2}{(2t+1)^2}\right)\]
\[5 - m = \frac{5(t+1)^2}{(2t+1)^2}\]
\[m = 5 - \frac{5(t+1)^2}{(2t+1)^2} = 5\left[1 - \left(\frac{t+1}{2t+1}\right)^{\!2}\right] = \frac{5(3t^2 + 2t)}{(2t+1)^2}\]

(c) As \(t \to \infty\), \(\frac{t+1}{2t+1} = \frac{1 + 1/t}{2 + 1/t} \to \frac{1}{2}\).
Therefore,
\[m \to 5\left(1 - \left(\frac{1}{2}\right)^{\!2}\right) = 5\left(1 - \frac{1}{4}\right) = \frac{15}{4} = 3.75\text{ g}.\]

(a)
M1: Correct method to find partial fractions (equating coefficients or substitution)
A1: Correct partial fractions \(\frac{2}{2t+1} - \frac{1}{t+1}\)

(b)
M1: Separating variables correctly
M1: Correct integration of LHS to \(-\ln(5-m)\) (ignore sign error for M mark)
M1: Correct integration of RHS to \(2\ln(2t+1) - 2\ln(t+1)\) using part (a)
M1: Using \(t=0, m=0\) in an equation containing an arbitrary constant to find \(c\)
A1: Correct expression for \(m\) in terms of \(t\), e.g. \(m = 5\left[1 - \left(\frac{t+1}{2t+1}\right)^2\right]\) or \(m = \frac{15t^2 + 10t}{(2t+1)^2}\)

(c)
M1: Evaluating \(\lim_{t\to\infty} m(t)\) by considering the limiting value of the fraction
A1: Correct limiting value \(\frac{15}{4}\) or \(3.75\)

(a) Equating the components of \(l_1\) and \(l_2\):
\(1 + 2\lambda = 2 + \mu \implies 2\lambda - \mu = 1\) --- (1)
\(3 - \lambda = -5 + 2\mu \implies \lambda + 2\mu = 8\) --- (2)
\(2\lambda = 10 - 2\mu \implies 2\lambda + 2\mu = 10\) --- (3)

From (1), \(\mu = 2\lambda - 1\).
Substituting into (2): \(\lambda + 2(2\lambda - 1) = 8 \implies 5\lambda - 2 = 8 \implies 5\lambda = 10 \implies \lambda = 2\).
Then \(\mu = 2(2) - 1 = 3\).

Checking in the third component:
For \(l_1\): \(z = 0 + 2(2) = 4\).
For \(l_2\): \(z = 10 - 2(3) = 4\).
Since both values agree, the lines intersect.

The position vector of the point of intersection \(P\) is:
\[\vec{OP} = \begin{pmatrix} 1 + 2(2) \\ 3 - 2 \\ 0 + 2(2) \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \\ 4 \end{pmatrix} = 5\mathbf{i} + \mathbf{j} + 4\mathbf{k}.\]

(b) Let \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\).
\[\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-1)(2) + (2)(-2) = 2 - 2 - 4 = -4\]
\[|\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = 3\]
\[|\mathbf{d}_2| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = 3\]
Let \(\theta\) be the acute angle between the lines:
\[\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|} = \frac{|-4|}{3 \times 3} = \frac{4}{9}\]
\[\theta = \cos^{-1}\left(\frac{4}{9}\right) \approx 63.612^\circ \approx 63.6^\circ.\]

(c) The unit direction vector along \(l_1\) is \(\frac{1}{3}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\).
Since \(Q\) lies on \(l_1\) and \(PQ = 9\):
\[\vec{OQ} = \vec{OP} \pm 9\left(\frac{1}{3}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\right) = \begin{pmatrix} 5 \\ 1 \\ 4 \end{pmatrix} \pm 3\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\]
\[\vec{OQ}_1 = \begin{pmatrix} 5 \\ 1 \\ 4 \end{pmatrix} + \begin{pmatrix} 6 \\ -3 \\ 6 \end{pmatrix} = \begin{pmatrix} 11 \\ -2 \\ 10 \end{pmatrix}\]
\[\vec{OQ}_2 = \begin{pmatrix} 5 \\ 1 \\ 4 \end{pmatrix} - \begin{pmatrix} 6 \\ -3 \\ 6 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ -2 \end{pmatrix}\]

(a)
M1: Equate components of both lines to form a pair of simultaneous equations in \(\lambda\) and \(\mu\)
M1: Solve simultaneous equations to find values for \(\lambda\) and \(\mu\) (\(\lambda = 2, \mu = 3\))
B1: Check consistency in the third equation to confirm intersection
A1: Correct position vector \(\begin{pmatrix} 5 \\ 1 \\ 4 \end{pmatrix}\) or \(5\mathbf{i} + \mathbf{j} + 4\mathbf{k}\)

(b)
M1: Correct use of the scalar product formula \(\mathbf{d}_1 \cdot \mathbf{d}_2 = |\mathbf{d}_1||\mathbf{d}_2|\cos\theta\) with the direction vectors
A1: Obtain \(\cos\theta = \pm\frac{4}{9}\)
A1: Correct acute angle \(63.6^\circ\) (or \(1.11\) radians)

(c)
M1: Realise that \(\vec{OQ} = \vec{OP} \pm k\mathbf{d}_1\) where \(k = \frac{9}{|\mathbf{d}_1|} = 3\)
A1: One correct position vector (\(11\mathbf{i} - 2\mathbf{j} + 10\mathbf{k}\) or \(-\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\))
A1: Second correct position vector

The motion consists of two stages:
1. Acceleration stage (from \(t = 0\) to \(t = 6\)):
\(\text{Distance } s_1 = \frac{1}{2} \times 6 \times 18 = 54\text{ m}\)

2. Constant speed stage (from \(t = 6\) to \(t = T\)):
\(\text{Distance } s_2 = 18 \times (T - 6)\)

Total distance travelled:
\(s_1 + s_2 = 243\)
\(54 + 18(T - 6) = 243\)
\(18(T - 6) = 189\)
\(T - 6 = 10.5\)
\(T = 16.5\)

M1: For using area under the \(v\)-\(t\) graph or constant acceleration formula to form an equation in \(T\) for total distance, e.g. \(\frac{1}{2}(6)(18) + 18(T - 6) = 243\) or \(\frac{1}{2}(T + T - 6)(18) = 243\).
A1: For obtaining \(T = 16.5\) (or \(\frac{33}{2}\)).

First, find the time \(t\) when acceleration is \(14\text{ m s}^{-2}\):
\(6t - 4 = 14 \implies 6t = 18 \implies t = 3\text{ s}\).

Next, integrate acceleration to find the expression for velocity \(v\):
\(v = \int (6t - 4)\,\mathrm{d}t = 3t^2 - 4t + c\).

Using the initial condition \(v = 5\) when \(t = 0\):
\(5 = 3(0)^2 - 4(0) + c \implies c = 5\).

Thus, \(v(t) = 3t^2 - 4t + 5\).

Substituting \(t = 3\):
\(v(3) = 3(3)^2 - 4(3) + 5 = 27 - 12 + 5 = 20\text{ m s}^{-1}\).

M1: For setting \(6t - 4 = 14\) to find \(t = 3\).
M1: For integrating \(a\) with respect to \(t\) to find \(v(t)\) and using \(v(0) = 5\) to find the constant of integration.
A1: For the correct final velocity \(20\text{ m s}^{-1}\).

Using the constant acceleration equation \(s = ut - \frac{1}{2}gt^2\) with \(u = 24\), \(g = 10\), and \(s = 20\):
\(20 = 24t - 5t^2\)
\(5t^2 - 24t + 20 = 0\)

Solving for \(t\) using the quadratic formula:
\(t = \frac{24 \pm \sqrt{(-24)^2 - 4(5)(20)}}{2(5)} = \frac{24 \pm \sqrt{576 - 400}}{10} = \frac{24 \pm \sqrt{176}}{10}\)
\(t_1 = \frac{24 - \sqrt{176}}{10} \approx 1.0733\text{ s}\)
\(t_2 = \frac{24 + \sqrt{176}}{10} \approx 3.7267\text{ s}\)

The time duration during which the height is at least \(20\text{ m}\) is:
\(\Delta t = t_2 - t_1 = \frac{2\sqrt{176}}{10} = \frac{4\sqrt{11}}{5} \approx 2.65\text{ s}\) (to 3 s.f.).

M1: For setting up a 3-term quadratic in \(t\) using \(s = ut - \frac{1}{2}gt^2\) with \(s = 20\), \(u = 24\), and \(g = 10\), and finding the difference between the two positive roots.
A1: For obtaining \(2.65\text{ s}\) (accept exact surd form \(\frac{4\sqrt{11}}{5}\) or \(0.8\sqrt{11}\)).

Let \(u = 28\text{ m s}^{-1}\), \(v = 12\text{ m s}^{-1}\), and \(s = 80\text{ m}\).

Using the constant acceleration equation \(v^2 = u^2 + 2as\):
\(12^2 = 28^2 + 2a(80)\)
\(144 = 784 + 160a\)
\(160a = 144 - 784 = -640\)
\(a = -4\text{ m s}^{-2}\)

Therefore, the magnitude of the deceleration is \(4\text{ m s}^{-2}\).

M1: For applying \(v^2 = u^2 + 2as\) with \(u = 28\), \(v = 12\), and \(s = 80\) to solve for acceleration/deceleration.
A1: For stating the magnitude of deceleration as \(4\text{ m s}^{-2}\).

(a)
Change in kinetic energy:
\(\Delta \text{KE} = \frac{1}{2} m (v^2 - u^2) = \frac{1}{2} \times 1200 \times (18^2 - 12^2) = 600 \times (324 - 144) = 600 \times 180 = 108\,000\text{ J}\).

Change in gravitational potential energy:
Vertical height \(h = 200 \sin\alpha = 200 \times \frac{1}{15} = \frac{40}{3}\text{ m}\).
\(\Delta \text{PE} = mgh = 1200 \times 10 \times \frac{40}{3} = 160\,000\text{ J}\).

Work done against resistance:
\(W_{\text{res}} = F_{\text{res}} \times d = 450 \times 200 = 90\,000\text{ J}\).

By the work-energy principle:
\(W_{\text{engine}} = \Delta \text{KE} + \Delta \text{PE} + W_{\text{res}}\)
\(W_{\text{engine}} = 108\,000 + 160\,000 + 90\,000 = 358\,000\text{ J}\).

(b)
Average power \(P = \frac{\text{Work done}}{\text{time}}\):
\(P = \frac{358\,000}{13.5} \approx 26\,518.5\text{ W} = 26.5\text{ kW}\) (to 3 s.f.).

(a)
- M1: For calculating \(\Delta \text{KE}\) and \(\Delta \text{PE}\) using \(g = 10\).
- M1: For applying the work-energy principle: \(W = \Delta \text{KE} + \Delta \text{PE} + W_{\text{res}}\).
- A1: For obtaining \(358\,000\text{ J}\) (or \(358\text{ kJ}\)).

(b)
- M1: For using \(P = \frac{W}{t}\) with their work done from (a).
- A1: For obtaining \(26.5\text{ kW}\) or \(26\,500\text{ W}\) (accept \(26\,520\text{ W}\)).

(a)
On the horizontal road, at steady speed \(v_1\), the driving force \(F_1\) equals the resistance \(R\):
\(F_1 = 24\text{ N}\).

Using \(P = F_1 v_1\):
\(360 = 24 v_1 \implies v_1 = \frac{360}{24} = 15\text{ m s}^{-1}\).

(b)
When moving up the hill at steady speed \(v_2\), acceleration is zero, so the driving force \(F_2\) balances the resistance and the component of weight down the slope:
\(F_2 = R + mg \sin\theta\)
\(F_2 = 24 + 80 \times 10 \times 0.03 = 24 + 24 = 48\text{ N}\).

Using \(P = F_2 v_2\):
\(360 = 48 v_2 \implies v_2 = \frac{360}{48} = 7.5\text{ m s}^{-1}\).

(a)
- M1: For using \(P = Fv\) with driving force \(F = 24\text{ N}\).
- A1: For \(v = 15\text{ m s}^{-1}\).

(b)
- M1: For setting up the equilibrium equation along the slope: \(F = 24 + 80(10)(0.03)\).
- A1: For correctly finding the driving force \(F = 48\text{ N}\).
- A1: For \(v = 7.5\text{ m s}^{-1}\).

(a)
Vertical drop \(h = 4 \sin 30^\circ = 4 \times 0.5 = 2\text{ m}\).
Loss in \(\text{PE} = mgh = 0.6 \times 10 \times 2 = 12\text{ J}\).

(b)
Gain in \(\text{KE} = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.6 \times 5^2 = 0.3 \times 25 = 7.5\text{ J}\).
Work done against friction \(W_f = \text{Loss in PE} - \text{Gain in KE} = 12 - 7.5 = 4.5\text{ J}\).

(c)
Work done against friction is \(W_f = F_r \times d = 4.5\text{ J}\),
so the frictional force \(F_r = \frac{4.5}{4} = 1.125\text{ N}\).

Resolving perpendicular to the plane:
\(R = mg \cos 30^\circ = 0.6 \times 10 \times \cos 30^\circ = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \approx 5.196\text{ N}\).

Using \(F_r = \mu R\):
\(\mu = \frac{1.125}{3\sqrt{3}} = \frac{1.125}{5.196} \approx 0.2165 = 0.217\) (to 3 s.f.).

(a)
- B1: For \(12\text{ J}\).

(b)
- M1: For calculating \(\text{KE} = \frac{1}{2}(0.6)(5^2)\) and forming the energy balance \(W_f = \text{PE loss} - \text{KE gain}\).
- A1: For \(4.5\text{ J}\).

(c)
- M1: For finding normal reaction \(R = 0.6 \times 10 \cos 30^\circ\) and using \(F_r = \mu R\) with \(F_r = \frac{W_f}{4}\).
- A1: For \(\mu = 0.217\) (or \(\frac{\sqrt{3}}{8}\)).

(a)
Resolving vertically for the sledge in equilibrium:
\(R + P \sin 20^\circ = mg = 30 \times 10 = 300\)
\(R = 300 - P \sin 20^\circ\).

(b)
The frictional force is \(F = \mu R = 0.35 (300 - P \sin 20^\circ) = 105 - 0.35 P \sin 20^\circ\).

Applying Newton's second law horizontally:
\(P \cos 20^\circ - F = m a\)
\(P \cos 20^\circ - (105 - 0.35 P \sin 20^\circ) = 30 \times 0.8\)
\(P (\cos 20^\circ + 0.35 \sin 20^\circ) - 105 = 24\)
\(P (\cos 20^\circ + 0.35 \sin 20^\circ) = 129\).

Evaluating the coefficient of \(P\):
\(\cos 20^\circ + 0.35 \sin 20^\circ \approx 0.93969 + 0.35 \times 0.34202 = 0.93969 + 0.11971 = 1.0594\).

Therefore,
\(P = \frac{129}{1.0594} \approx 121.766 \approx 122\text{ N}\) (to 3 s.f.).

(a)
- B1: For \(R = 300 - P \sin 20^\circ\).

(b)
- M1: For using \(F = \mu R\) with their expression for \(R\).
- M1: For applying Newton's second law horizontally: \(P \cos 20^\circ - F = 30 \times 0.8\).
- DM1: For isolating \(P\) in the linear equation.
- A1: For \(P = 122\text{ N}\) (accept \(121.8\)).

(a)
At constant speed on a horizontal road, the driving force \(F\) equals the resistance \(R\).
Using \(P = Fv\):
\(30\,000 = F \times 25 \implies F = \frac{30\,000}{25} = 1200\text{ N}\).
Hence, \(R = 1200\text{ N}\).

(b)
When the power increases to \(42\text{ kW} = 42\,000\text{ W}\) at \(v = 25\text{ m s}^{-1}\), the new driving force \(F_{\text{new}}\) is:
\(F_{\text{new}} = \frac{42\,000}{25} = 1680\text{ N}\).

Applying Newton's second law:
\(F_{\text{new}} - R = ma\)
\(1680 - 1200 = 1400 a\)
\(480 = 1400 a \implies a = \frac{480}{1400} = \frac{12}{35} \approx 0.343\text{ m s}^{-2}\) (to 3 s.f.).

(a)
- B1: For \(F = \frac{30\,000}{25} = 1200\text{ N}\) and concluding \(R = 1200\text{ N}\).

(b)
- M1: For calculating the new driving force \(F_{\text{new}} = \frac{42\,000}{25} = 1680\text{ N}\).
- M1: For applying Newton's second law: \(F_{\text{new}} - 1200 = 1400 a\).
- A1: For \(a = 0.343\text{ m s}^{-2}\) (or \(\frac{12}{35}\)).

(a)
Resolving perpendicular to the plane:
\(R = mg \cos 25^\circ = 8 \times 10 \times \cos 25^\circ = 80 \cos 25^\circ \approx 72.505\text{ N}\).

Frictional force acting down the plane:
\(F_r = \mu R = 0.3 \times 80 \cos 25^\circ = 24 \cos 25^\circ \approx 21.751\text{ N}\).

Component of weight down the plane:
\(W_{\parallel} = mg \sin 25^\circ = 80 \sin 25^\circ \approx 33.809\text{ N}\).

Total retarding force up the plane:
\(F_{\text{net}} = W_{\parallel} + F_r = 80 \sin 25^\circ + 24 \cos 25^\circ \approx 33.809 + 21.751 = 55.560\text{ N}\).

Applying Newton's second law:
\(m a = -F_{\text{net}} \implies 8 a = -55.560 \implies a = -6.945\text{ m s}^{-2}\).
So the deceleration is \(6.94\text{ m s}^{-2}\) (to 3 s.f.) (or \(6.95\text{ m s}^{-2}\)).

(b)
Using \(v^2 = u^2 + 2as\) with \(v = 0\), \(u = 6\text{ m s}^{-1}\), and \(a = -6.945\text{ m s}^{-2}\):
\(0 = 6^2 - 2(6.945)s\)
\(36 = 13.89 s \implies s = \frac{36}{13.89} \approx 2.59\text{ m}\).

(a)
- M1: For resolving forces along the plane: \(8a = -(80 \sin 25^\circ + F_r)\).
- M1: For using \(F_r = 0.3 \times 80 \cos 25^\circ\).
- A1: For deceleration \(= 6.94\text{ m s}^{-2}\) or \(6.95\text{ m s}^{-2}\).

(b)
- M1: For using \(v^2 = u^2 + 2as\) with \(v = 0\), \(u = 6\) and their \(a\).
- A1: For \(s = 2.59\text{ m}\).

(a) At \(v = 18\text{ m s}^{-1}\), the driving force \(F_D\) provided by the engine is given by:
\[ F_D = \frac{P}{v} = \frac{36\,000}{18} = 2000\text{ N} \]
The component of the car's weight acting down the slope is:
\[ W_{\parallel} = mg\sin\alpha = 1200 \times 10 \times \frac{1}{15} = 800\text{ N} \]
Applying Newton's second law parallel to the slope in the direction of motion:
\[ F_D - mg\sin\alpha - R = ma \]
\[ 2000 - 800 - 450 = 1200a \]
\[ 750 = 1200a \implies a = \frac{750}{1200} = 0.625\text{ m s}^{-2} \]

(b) When travelling at a steady speed, the acceleration is zero (\(a = 0\)), so the driving force equals the sum of the opposing forces:
\[ F_D = mg\sin\alpha + R = 800 + 450 = 1250\text{ N} \]
Using \(P = F_D v\):
\[ v = \frac{P}{F_D} = \frac{36\,000}{1250} = 28.8\text{ m s}^{-1} \]

(a)
M1: Use \(P = Fv\) to find driving force \(F_D = \frac{36\,000}{18} = 2000\text{ N}\)
M1: Form equation of motion \(F_D - 1200g\sin\alpha - 450 = 1200a\) with correct resolution of weight
A1: Correct acceleration \(a = 0.625\text{ m s}^{-2}\) (or \(\frac{5}{8}\text{ m s}^{-2}\))

(b)
M1: Set net force to zero to obtain \(F_D = 1250\text{ N}\) and use \(v = \frac{P}{F_D}\)
A1: Correct steady speed \(v = 28.8\text{ m s}^{-1}\)

(a)
Resolving vertically for particle \(A\):
\[ R_A = 0.4g = 0.4 \times 10 = 4\text{ N} \]

The frictional force acting on \(A\) is:
\[ F_A = \mu R_A = 0.25 \times 4 = 1.0\text{ N} \]

Applying Newton's second law horizontally in the direction of motion:
\[ -F_A = m_A a_A \implies -1.0 = 0.4 a_A \implies a_A = -2.5\text{ m s}^{-2} \]

Using the constant acceleration formula \(v^2 = u^2 + 2as\):
\[ v^2 = 5^2 + 2(-2.5)(1.8) = 25 - 9 = 16 \]
\[ v = \sqrt{16} = 4\text{ m s}^{-1} \]

(b)
Applying conservation of linear momentum for the collision:
\[ m_A u_A + m_B u_B = (m_A + m_B) v_C \]
\[ (0.4)(4) + (0.6)(0) = (0.4 + 0.6) v_C \]
\[ 1.6 = 1.0 v_C \implies v_C = 1.6\text{ m s}^{-1} \]

(c)
For particle \(C\) of mass \(1.0\text{ kg}\):
\[ R_C = 1.0g = 10\text{ N} \]
\[ F_C = \mu_C R_C = 0.2 \times 10 = 2.0\text{ N} \]

Deceleration of \(C\):
\[ a_C = -\frac{F_C}{m_C} = -\frac{2.0}{1.0} = -2.0\text{ m s}^{-2} \]

Let \(s_C\) be the distance travelled by \(C\) before coming to rest (\(v = 0\)):
\[ 0^2 = 1.6^2 + 2(-2.0)s_C \]
\[ 4.0 s_C = 2.56 \implies s_C = 0.64\text{ m} \]

Total distance travelled:
\[ s_{\text{total}} = 1.8 + 0.64 = 2.44\text{ m} \]

(a)
- M1: For resolving vertically and calculating the frictional force \(F = \mu R\) and using \(F = ma\) to find the deceleration of \(A\).
- M1: For using \(v^2 = u^2 + 2as\) with their deceleration and \(s = 1.8\) (or using the work-energy principle: \(\frac{1}{2}m u^2 - \text{WD against friction} = \frac{1}{2}m v^2\)).
- A1: For fully correct working leading to the given answer \(v = 4\text{ m s}^{-1}\).

(b)
- M1: For applying conservation of momentum to the collision: \(0.4 \times 4 = (0.4 + 0.6) v\).
- A1: For \(v = 1.6\text{ m s}^{-1}\) (or \(\frac{8}{5}\text{ m s}^{-1}\)).

(c)
- M1: For finding the deceleration of \(C\) (\(a = 2\text{ m s}^{-2}\)) and using \(v^2 = u^2 + 2as\) to find the distance moved by \(C\) (or using \(\text{KE} = \text{work done}\)).
- A1: For total distance \(= 2.44\text{ m}\) (or \(\frac{61}{25}\text{ m}\)).

The letters in \(\text{PARALLELS}\) are \(\text{P}, \text{A}, \text{R}, \text{A}, \text{L}, \text{L}, \text{E}, \text{L}, \text{S}\), which comprise \(1\text{ P}, 2\text{ As}, 1\text{ R}, 3\text{ Ls}, 1\text{ E}, 1\text{ S}\).

Treating the three \(\text{L}\)s as a single entity \([\text{LLL}]\), we have \(7\) items to arrange: \(\text{P}, \text{A}, \text{A}, \text{R}, [\text{LLL}], \text{E}, \text{S}\).

Among these \(7\) items, the letter \(\text{A}\) is repeated \(2\) times.

Therefore, the number of distinct arrangements is:
\[\frac{7!}{2!} = \frac{5040}{2} = 2520\]

M1: For considering \(\text{LLL}\) as a single block and dividing \(7!\) by \(2!\)
A1: Correct answer 2520

Let \(y = x - 15\).

Mean of \(y\):
\[\bar{y} = \frac{\sum y}{n} = \frac{48}{20} = 2.4\]

Mean of \(x\):
\[\bar{x} = 15 + \bar{y} = 15 + 2.4 = 17.4\]

Variance of \(x\) (which equals variance of \(y\)):
\[\text{Var}(x) = \frac{\sum y^2}{n} - (\bar{y})^2 = \frac{360}{20} - (2.4)^2 = 18 - 5.76 = 12.24\]

Standard deviation of \(x\):
\[\sigma = \sqrt{12.24} \approx 3.50\text{ (to 3 s.f.)}\]

B1: Correct mean of 17.4
M1: Correct method for variance or standard deviation using \(\frac{\sum y^2}{20} - (\bar{y})^2\)
A1: Correct standard deviation of 3.50 (or \(\sqrt{12.24}\) or \(\frac{3\sqrt{34}}{5}\))

The random variable \(X\) follows a geometric distribution with parameter \(p = \frac{2}{6} = \frac{1}{3}\).

\(\text{P}(X \geqslant 4) = (1 - p)^{4-1} = (1 - p)^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \approx 0.296\)

\(\text{E}(X) = \frac{1}{p} = \frac{1}{\frac{1}{3}} = 3\)

M1: Identifying geometric distribution with \(p = \frac{1}{3}\) and using \(\text{P}(X \geqslant 4) = (1 - p)^3\)
A1: Correct values \(\text{P}(X \geqslant 4) = \frac{8}{27}\) (or \(0.296\)) and \(\text{E}(X) = 3\)

(i) We have:
\[\text{P}(A \cap B) = \text{P}(A) - \text{P}(A \cap B') = 0.65 - 0.45 = 0.20\]

(ii) For \(A\) and \(B\) to be independent, we must have \(\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B)\).
\[\text{P}(A) \times \text{P}(B) = 0.65 \times 0.40 = 0.26\]
Since \(\text{P}(A \cap B) = 0.20 \neq 0.26\), events \(A\) and \(B\) are not independent.

B1: Correct value \(\text{P}(A \cap B) = 0.20\)
M1: Evaluating \(\text{P}(A) \times \text{P}(B)\) and comparing with \(\text{P}(A \cap B)\)
A1: Correct conclusion with valid numerical justification (not independent since \(0.20 \neq 0.26\))

Standardising \(W\):
\[\text{P}(W > 75) = \text{P}\left(Z > \frac{75 - \mu}{12}\right) = 0.0885\]

This gives:
\[\Phi\left(\frac{75 - \mu}{12}\right) = 1 - 0.0885 = 0.9115\]

From the standard normal distribution table:
\[\Phi(1.35) = 0.9115\]

Equating the values:
\[\frac{75 - \mu}{12} = 1.35\]
\[75 - \mu = 1.35 \times 12 = 16.2\]
\[\mu = 75 - 16.2 = 58.8\]

M1: For standardising and equating to \(\pm z\) where \(\Phi(z) = 0.9115\) (i.e. \(\frac{75 - \mu}{12} = z\))
B1: For identifying critical \(z = 1.35\)
A1: For obtaining \(\mu = 58.8\)

Let \(X\) be the number of heads obtained, so \(X \sim \text{B}(8, 0.35)\).

\[\text{P}(X \geqslant 2) = 1 - \text{P}(X = 0) - \text{P}(X = 1)\]

Calculating the individual probabilities:
\[\text{P}(X = 0) = (0.65)^8 \approx 0.031864\]
\[\text{P}(X = 1) = \binom{8}{1} (0.35)^1 (0.65)^7 = 8 \times 0.35 \times (0.65)^7 \approx 0.137262\]

Therefore:
\[\text{P}(X \geqslant 2) = 1 - (0.031864 + 0.137262) = 1 - 0.169126 = 0.830874 \approx 0.831\text{ (to 3 s.f.)}\]

M1: Using \(1 - [\text{P}(X = 0) + \text{P}(X = 1)]\) with correct binomial probability terms
A1: Correct answer 0.831

We consider the cases for the number of teachers chosen:

- \(2\) teachers and \(2\) students:
\[\binom{5}{2} \times \binom{7}{2} = 10 \times 21 = 210\]

- \(3\) teachers and \(1\) student:
\[\binom{5}{3} \times \binom{7}{1} = 10 \times 7 = 70\]

- \(4\) teachers and \(0\) students:
\[\binom{5}{4} \times \binom{7}{0} = 5 \times 1 = 5\]

Total number of ways:
\[210 + 70 + 5 = 285\]

M1: Summing the number of combinations for \(2\), \(3\), and \(4\) teachers (or subtracting \(0\) and \(1\) teacher cases from \(\binom{12}{4}\))
A1: Correct answer 285

(a) Let \(X\) be the weight of a loaf in grams, so \(X \sim \text{N}(\mu, \sigma^2)\).

Given \(\text{P}(X < 140) = 0.0668\):
\[\text{P}\left(Z < \frac{140 - \mu}{\sigma}\right) = 0.0668\]
Using standard normal tables, \(\Phi(1.50) = 0.9332\), so:
\[\frac{140 - \mu}{\sigma} = -1.50 \implies \mu - 1.50\sigma = 140 \quad \text{--- (1)}\]

Given \(\text{P}(X > 215) = 0.1587\):
\[\text{P}\left(Z > \frac{215 - \mu}{\sigma}\right) = 0.1587 \implies \Phi\left(\frac{215 - \mu}{\sigma}\right) = 1 - 0.1587 = 0.8413\]
From tables, \(\Phi(1.00) = 0.8413\), so:
\[\frac{215 - \mu}{\sigma} = 1.00 \implies \mu + 1.00\sigma = 215 \quad \text{--- (2)}\]

Subtracting equation (1) from equation (2):
\[2.50\sigma = 75 \implies \sigma = 30\]
Substituting \(\sigma = 30\) into (2):
\[\mu + 30 = 215 \implies \mu = 185\]

(b) \(\text{P}(170 < X < 200) = \text{P}\left(\frac{170 - 185}{30} < Z < \frac{200 - 185}{30}\right) = \text{P}(-0.5 < Z < 0.5)\)
\[= \Phi(0.5) - \Phi(-0.5) = 2\Phi(0.5) - 1 = 2(0.6915) - 1 = 0.383\]

(a)
- M1: Standardising with \(\mu\) and \(\sigma\) and equating to a \(z\)-value (at least one correct from tables: \(\pm 1.50\) or \(\pm 1.00\)).
- A1: Both standard normal equations correct with correct signs: \(\frac{140 - \mu}{\sigma} = -1.50\) and \(\frac{215 - \mu}{\sigma} = 1.00\).
- M1: Method to solve simultaneous linear equations for \(\mu\) and \(\sigma\).
- A1: Correct values: \(\mu = 185\) and \(\sigma = 30\).

(b)
- B1: Correct probability \(0.383\) (or exact \(0.3830\)).

(a) The sum of probabilities equals 1:
\[0.1 + p + 0.2 + q + 0.1 = 1 \implies p + q + 0.4 = 1 \implies p + q = 0.6 \quad \text{--- (1)}\]

Using \(\text{E}(X) = \sum x \text{P}(X = x) = 0.8\):
\[(-2)(0.1) + 0(p) + 1(0.2) + 2(q) + 4(0.1) = 0.8\]
\[-0.2 + 0 + 0.2 + 2q + 0.4 = 0.8\]
\[2q + 0.4 = 0.8 \implies 2q = 0.4 \implies q = 0.2\]

Substituting \(q = 0.2\) into equation (1):
\[p = 0.6 - 0.2 = 0.4\]

(b) \(\text{E}(X^2) = \sum x^2 \text{P}(X = x)\):
\[\text{E}(X^2) = (-2)^2(0.1) + 0^2(0.4) + 1^2(0.2) + 2^2(0.2) + 4^2(0.1)\]
\[= 4(0.1) + 0 + 1(0.2) + 4(0.2) + 16(0.1) = 0.4 + 0.2 + 0.8 + 1.6 = 3.0\]

Using \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\):
\[\text{Var}(X) = 3.0 - (0.8)^2 = 3.0 - 0.64 = 2.36\]

(a)
- B1: States or uses \(p + q = 0.6\).
- M1: Forms correct equation for \(\text{E}(X)\) using \(\sum x \text{P}(X = x) = 0.8\).
- A1: Correct values: \(p = 0.4\) and \(q = 0.2\).

(b)
- M1: Correct method to find \(\text{E}(X^2)\) and evaluates \(\text{E}(X^2) - [\text{E}(X)]^2\).
- A1: Correct answer \(2.36\).

Let \(X\) be the number of bottles with minor defects, then \(X \sim \text{B}(250, 0.08)\).

Check approximation conditions:
\(np = 250 \times 0.08 = 20 > 5\) and \(n(1 - p) = 250 \times 0.92 = 230 > 5\).
So \(X\) can be approximated by \(Y \sim \text{N}(\mu, \sigma^2)\), where:
\[\mu = np = 20\]
\[\sigma^2 = np(1 - p) = 250 \times 0.08 \times 0.92 = 18.4\]

We want \(\text{P}(X < 16) = \text{P}(X \le 15)\).
Applying the continuity correction:
\[\text{P}(X \le 15) \approx \text{P}(Y < 15.5)\]
Standardising:
\[\text{P}\left(Z < \frac{15.5 - 20}{\sqrt{18.4}}\right) = \text{P}\left(Z < \frac{-4.5}{4.2895}\right) = \text{P}(Z < -1.049)\]
\[= 1 - \Phi(1.049) = 1 - 0.8529 = 0.1471 \approx 0.147\]

- B1: States \(\text{N}(20, 18.4)\) or uses \(\mu = 20\) and \(\sigma = \sqrt{18.4} \approx 4.290\).
- M1: Uses continuity correction of \(15.5\) (or \(16.5\)).
- M1: Standardises using their mean, their standard deviation and their continuity corrected value: \(\frac{15.5 - 20}{\sqrt{18.4}}\).
- A1: Correct answer \(0.147\) (accept \(0.1471\)).

Let \(X\) be the number of throws until the first six, so \(X \sim \text{Geo}(0.35)\).

(a) \(\text{P}(X = 4) = (1 - 0.35)^3(0.35) = (0.65)^3(0.35) = 0.274625 \times 0.35 = 0.09611875 \approx 0.0961\)

(b) \(\text{P}(X \ge 5) = \text{P}(\text{first } 4 \text{ throws are not six}) = (1 - 0.35)^4 = (0.65)^4 = 0.17850625 \approx 0.179\)

(c) Using the conditional probability definition or memoryless property of the geometric distribution:
\[\text{P}(X < 6 \mid X > 2) = \text{P}(X \le 5 \mid X > 2) = \frac{\text{P}(2 < X \le 5)}{\text{P}(X > 2)}\]
\[\text{P}(X > 2) = (0.65)^2 = 0.4225\]
\[\text{P}(2 < X \le 5) = \text{P}(X = 3) + \text{P}(X = 4) + \text{P}(X = 5) = (0.65)^2(0.35) + (0.65)^3(0.35) + (0.65)^4(0.35) = 0.3064671875\]
\[\text{P}(X < 6 \mid X > 2) = \frac{0.3064671875}{0.4225} = 0.725375 \approx 0.725\]
Alternatively, by the memoryless property:
\[\text{P}(X \le 5 \mid X > 2) = \text{P}(X \le 3) = 1 - (0.65)^3 = 1 - 0.274625 = 0.725375 \approx 0.725\]

(a)
- B1: Correct answer \(0.0961\) (or exact \(0.09611875\)).

(b)
- B1: Correct answer \(0.179\) (or exact \(0.17850625\)).

(c)
- M1: Correct method using conditional probability \(\frac{\text{P}(2 < X \le 5)}{\text{P}(X > 2)}\) or memoryless property \(1 - (0.65)^3\).
- A1: Correct answer \(0.725\) (or \(0.7254\)).

(a) \(L \sim \text{N}(45.0, 1.8^2)\).

Probability a rod is shorter than \(42.5\text{ cm}\):
\[\text{P}(L < 42.5) = \text{P}\left(Z < \frac{42.5 - 45.0}{1.8}\right) = \text{P}\left(Z < -1.389\right) = 1 - \Phi(1.389) = 1 - 0.9176 = 0.0824\]

Probability a rod is longer than \(47.7\text{ cm}\):
\[\text{P}(L > 47.7) = \text{P}\left(Z > \frac{47.7 - 45.0}{1.8}\right) = \text{P}(Z > 1.50) = 1 - \Phi(1.50) = 1 - 0.9332 = 0.0668\]

Total probability of rejection:
\[\text{P}(\text{rejected}) = 0.0824 + 0.0668 = 0.1492 \approx 0.149\]

(b) Let \(Y\) be the number of rejected rods in the sample of \(8\), so \(Y \sim \text{B}(8, 0.1492)\).
\[\text{P}(Y \ge 2) = 1 - \text{P}(Y = 0) - \text{P}(Y = 1)\]
\[\text{P}(Y = 0) = (1 - 0.1492)^8 = (0.8508)^8 \approx 0.27407\]
\[\text{P}(Y = 1) = \binom{8}{1}(0.8508)^7(0.1492)^1 = 8(0.32213)(0.1492) \approx 0.38450\]
\[\text{P}(Y \ge 2) = 1 - (0.27407 + 0.38450) = 1 - 0.65857 = 0.34143 \approx 0.341\]

(a)
- M1: Standardising at least once with \(\mu = 45.0\) and \(\sigma = 1.8\).
- M1: Finding both tail probabilities \(1 - \Phi(1.389)\) and \(1 - \Phi(1.50)\) and adding them.
- A1: Correct probability \(0.149\) (accept \(0.1492\)).

(b)
- M1: Uses \(1 - \text{P}(Y=0) - \text{P}(Y=1)\) with binomial term structure for \(n = 8\) and their \(p\).
- A1: Correct answer \(0.341\) (accept \(0.341\) to \(0.342\)).

The letters of \(\text{MILLENNIUM}\) and their frequencies are:
\(\text{M}: 2,\; \text{I}: 2,\; \text{L}: 2,\; \text{N}: 2,\; \text{E}: 1,\; \text{U}: 1\) (Total \(= 10\) letters).

(a)
The total number of different arrangements of the 10 letters is:
\[ \frac{10!}{2! \times 2! \times 2! \times 2!} = \frac{3\,628\,800}{16} = 226\,800 \]

(b)
Place the two \(\text{M}\)s at the two ends: \(\text{M} \; \underline{\hspace{2cm}} \; \text{M}\). Since the two \(\text{M}\)s are identical, there is only \(1\) way to place them.

Between them, we arrange the remaining 8 letters with the two \(\text{N}\)s together. Treat \((\text{NN})\) as a single unit.

The items to arrange are: \((\text{NN}), \text{I}, \text{I}, \text{L}, \text{L}, \text{E}, \text{U}\), which is \(7\) items with \(2\) \(\text{I}\)s and \(2\) \(\text{L}\)s.

The number of arrangements is:
\[ \frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260 \]

(c)
We must select 4 letters containing at least one \(\text{I}\) and at least one \(\text{L}\).
The available other letters are \(\{\text{M}, \text{M}, \text{N}, \text{N}, \text{E}, \text{U}\}\).

We consider the possible cases for the numbers of \(\text{I}\)s and \(\text{L}\)s:

1. **\(1\text{I}, 1\text{L}\) and 2 other letters:**
- 2 identical other letters: \(\text{MM}\) or \(\text{NN}\) \(\Rightarrow 2\) ways.
- 2 different other letters from \(\{\text{M}, \text{N}, \text{E}, \text{U}\}\): \(\displaystyle \binom{4}{2} = 6\) ways.
- Total for this case \(= 2 + 6 = 8\) selections.

2. **\(2\text{I}, 1\text{L}\) and 1 other letter:**
- 1 other letter from \(\{\text{M}, \text{N}, \text{E}, \text{U}\}\): \(\displaystyle \binom{4}{1} = 4\) selections.

3. **\(1\text{I}, 2\text{L}\) and 1 other letter:**
- 1 other letter from \(\{\text{M}, \text{N}, \text{E}, \text{U}\}\): \(\displaystyle \binom{4}{1} = 4\) selections.

4. **\(2\text{I}, 2\text{L}\) and 0 other letters:**
- \(1\) selection (\(\text{IILL}\)).

Total number of selections \(= 8 + 4 + 4 + 1 = 17\).

(d)
Total number of ways to choose 4 letters from the 10 letters is:
\[ \binom{10}{4} = 210 \]
To get 4 different letters, we select 4 distinct letter types from the 6 available types: \(4\) types with multiplicity 2 (\(\text{M, I, L, N}\)) and \(2\) types with multiplicity 1 (\(\text{E, U}\)).

- 4 types with multiplicity 2:
\[ \binom{4}{4} \times 2^4 = 1 \times 16 = 16 \]
- 3 types with multiplicity 2 and 1 type with multiplicity 1:
\[ \binom{4}{3} \times 2^3 \times \binom{2}{1} \times 1^1 = 4 \times 8 \times 2 \times 1 = 64 \]
- 2 types with multiplicity 2 and 2 types with multiplicity 1:
\[ \binom{4}{2} \times 2^2 \times \binom{2}{2} \times 1^2 = 6 \times 4 \times 1 \times 1 = 24 \]
Total favourable outcomes \(= 16 + 64 + 24 = 104\).

\[ \text{Probability} = \frac{104}{210} = \frac{52}{105} \approx 0.495 \]

(a)
- B1: \(226\,800\) (or \(\frac{10!}{(2!)^4}\)).

(b)
- M1: For \(\frac{7!}{2! \times 2!}\) or recognizing \(7\) items with repeats of \(2\) and \(2\).
- A1: \(1260\).

(c)
- B1: Identifying the four valid combinations of \(\text{I}\)s and \(\text{L}\)s (\(1\text{I}\,1\text{L}\), \(2\text{I}\,1\text{L}\), \(1\text{I}\,2\text{L}\), \(2\text{I}\,2\text{L}\)).
- M1: For correctly finding the number of selections of 2 other letters from \(\{\text{M, M, N, N, E, U}\}\) as \(2 + \binom{4}{2} = 8\).
- M1: Summing all 4 cases: \(8 + 4 + 4 + 1\).
- A1: \(17\).

(d)
- M1: Finding the total number of outcomes \(\binom{10}{4} = 210\) used as the denominator.
- M1: Correct method to count favourable outcomes with 4 distinct letters (e.g. splitting into cases according to number of repeated-letter types chosen).
- A1: \(\frac{52}{105}\) (or exact fraction equivalent, or \(0.495\) to 3 s.f.).

A Type I error occurs when the null hypothesis \(\text{H}_0\) is rejected when it is actually true.

In this context, it means concluding that the new fertilizer increases the mean harvest yield (or that \(\mu > 45\)) when in reality the mean harvest yield is still \(45\text{ kg}\) (the fertilizer has no effect).

B1: Stating rejection of \(\text{H}_0\) when \(\text{H}_0\) is true (or concluding \(\mu > 45\) when \(\mu = 45\)).
B1: Correct context given (mentioning fertilizer increasing yield and actual yield remaining 45 kg / no increase).

For a Poisson distribution to model the number of customer inquiries:
1. Inquiries must occur independently of each other (or singly/one at a time).
2. Inquiries must occur at a constant average rate throughout the one-hour period.

B1: For stating that inquiries occur independently (or singly / not simultaneously).
B1: For stating that inquiries occur at a constant mean/average rate.

Using the properties of variance for independent random variables:
\[ \text{Var}(3X_1 - 2X_2 + 5) = 3^2 \text{Var}(X_1) + (-2)^2 \text{Var}(X_2) + 0 \]
\[ = 9 \text{Var}(X_1) + 4 \text{Var}(X_2) \]
Since \(\text{Var}(X_1) = \text{Var}(X_2) = 8\):
\[ \text{Var}(3X_1 - 2X_2 + 5) = 9(8) + 4(8) = 72 + 32 = 104 \]

M1: For using \(3^2 \text{Var}(X_1) + (-2)^2 \text{Var}(X_2)\) (or \(9\text{Var}(X) + 4\text{Var}(X)\)).
A1: For correct value \(104\).

By the Central Limit Theorem, since the sample size \(n = 120\) is large (\(n \ge 30\)), the distribution of the sample mean \(\bar{X}\) is approximately normal:
\[ \text{E}(\bar{X}) = \mu = 14.5 \]
\[ \text{Var}(\bar{X}) = \frac{\sigma^2}{n} = \frac{32.4}{120} = 0.27 \]
Thus, \(\bar{X} \sim \text{N}(14.5, 0.27)\), justified by the Central Limit Theorem.

B1: Stating \(\bar{X} \sim \text{N}(14.5, 0.27)\) (or \(\text{N}\left(14.5, \frac{32.4}{120}\right)\)).
B1: Naming the Central Limit Theorem (accept CLT).

A Type I error occurs if \(\text{H}_0\) is rejected given that \(\text{H}_0\) is true (i.e. \(X \ge 4\) when \(\lambda = 1.2\)):
\[ \text{P}(\text{Type I error}) = \text{P}(X \ge 4 \mid \lambda = 1.2) = 1 - \text{P}(X \le 3 \mid \lambda = 1.2) \]
\[ \text{P}(X \le 3) = \text{e}^{-1.2} \left( 1 + 1.2 + \frac{1.2^2}{2!} + \frac{1.2^3}{3!} \right) \]
\[ = \text{e}^{-1.2} (1 + 1.2 + 0.72 + 0.288) = \text{e}^{-1.2}(3.208) \approx 0.96623 \]
\[ \text{P}(\text{Type I error}) = 1 - 0.96623 = 0.03377 \approx 0.0338 \text{ (to 3 s.f.)} \]

M1: For evaluating \(1 - \text{e}^{-1.2}\left(1 + 1.2 + \frac{1.2^2}{2} + \frac{1.2^3}{6}\right)\) or equivalent sum for \(\text{P}(X \ge 4)\).
A1: For correct answer \(0.0338\) (accept 0.03377).

(a) Let $X$ be the number of calls received in a 1-hour period.
Under the null hypothesis, $X \sim \text{Po}(4.5)$.

State the hypotheses:
$H_0: \lambda = 4.5$
$H_1: \lambda < 4.5$

Calculate the $p$-value for the observed value $X = 1$:
$$\text{P}(X \le 1) = \text{P}(X = 0) + \text{P}(X = 1)$$
$$\text{P}(X \le 1) = \text{e}^{-4.5} + \text{e}^{-4.5}(4.5) = 5.5\text{e}^{-4.5} \approx 0.0611$$

Compare with the significance level $\alpha = 0.05$:
Since $0.0611 > 0.05$, the result is not significant.
Do not reject $H_0$.
There is insufficient evidence at the 5% significance level to conclude that the mean rate of calls has decreased.

(b) At the 1% significance level, $\alpha = 0.01$.
Since $0.0611 > 0.01$, $H_0$ is still not rejected; there is insufficient evidence to suggest the mean rate of calls has decreased.

(a)
B1: State correct null and alternative hypotheses $H_0: \lambda = 4.5$ and $H_1: \lambda < 4.5$ (or $\mu = 4.5$ vs $\mu < 4.5$).
M1: Calculate $\text{P}(X \le 1) = \text{e}^{-4.5}(1 + 4.5)$.
A1: Correct probability $0.0611$ (accept $0.061$ or better).
A1: Correct comparison ($0.0611 > 0.05$) and non-assertive contextual conclusion (insufficient evidence that the mean rate of calls has decreased).

(b)
B1: State that the conclusion is unchanged / do not reject $H_0$ since $0.0611 > 0.01$.

(a) Sample size $n = 80$.
Unbiased estimate of the population mean:
$$\bar{x} = \frac{\sum x}{n} = \frac{3320}{80} = 41.5$$

Unbiased estimate of the population variance:
$$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$$
$$s^2 = \frac{1}{79}\left(140\,250 - \frac{3320^2}{80}\right) = \frac{1}{79}(140\,250 - 137\,780) = \frac{2470}{79} \approx 31.266 = 31.3\text{ (to 3 s.f.)}$$

(b) For a 96% confidence interval, the critical value $z$ satisfies $\Phi(z) = 0.98 \implies z = 2.054$.

The confidence interval is:
$$\bar{x} \pm z \sqrt{\frac{s^2}{n}}$$
$$41.5 \pm 2.054 \sqrt{\frac{31.266}{80}} = 41.5 \pm 2.054 \times 0.62516 = 41.5 \pm 1.284$$

Thus, the interval is $[40.216, 42.784] \approx [40.2, 42.8]$.

(a)
B1: State correct mean $\bar{x} = 41.5$.
M1: Use correct formula for unbiased variance $s^2 = \frac{1}{79}\left(140250 - \frac{3320^2}{80}\right)$.
A1: Obtain $s^2 = 31.3$ or $\frac{2470}{79}$ (accept $31.26$ to $31.27$).

(b)
B1: Identify critical value $z = 2.054$ (or $2.053$ to $2.055$).
M1: Compute $\bar{x} \pm z \sqrt{\frac{s^2}{80}}$ using their values.
A1: Correct interval $[40.2, 42.8]$ (or $[40.22, 42.78]$).

(a) The population distribution of packet masses is not known to be normal. Because the sample size is large ($n = 100 > 30$), the Central Limit Theorem ensures that the distribution of the sample mean $\bar{X}$ is approximately normal.

(b) State the hypotheses:
$H_0: \mu = 500$
$H_1: \mu < 500$

Sample statistics: $n = 100$, $\bar{x} = 497.6$, $s^2 = 108.16 \implies s = 10.4$.

Standard error of the mean:
$$\sigma_{\bar{x}} = \sqrt{\frac{108.16}{100}} = \frac{10.4}{10} = 1.04$$

Calculate the test statistic:
$$z = \frac{497.6 - 500}{1.04} = \frac{-2.4}{1.04} \approx -2.308$$

Critical value for a 1-tailed test at the 2% significance level:
$$z_{\text{crit}} = -2.054 \quad (\text{since } \Phi(2.054) = 0.98)$$

Comparison:
Since $-2.308 < -2.054$ (or $p\text{-value} = 1 - \Phi(2.308) = 0.0105 < 0.02$), the test statistic falls in the rejection region.
Reject $H_0$.
There is sufficient evidence at the 2% significance level to support the consumer group's suspicion that the mean mass of cereal packets is less than 500 g.

(a)
B1: State that the population distribution is not known/not normal and the sample size is large ($n=100$), so $\bar{X}$ is approximately normal.

(b)
B1: State correct hypotheses $H_0: \mu = 500$ and $H_1: \mu < 500$.
M1: Calculate test statistic $z = \frac{497.6 - 500}{\sqrt{108.16/100}}$ or find critical sample mean $\bar{x}_{\text{crit}} = 500 - 2.054(1.04) = 497.86$.
A1: Correct $z$-value $-2.31$ or $-2.308$ (or critical sample mean $497.9$).
A1: Correct comparison with $-2.054$ and correct contextual conclusion supporting the consumer group's suspicion.

(a) Let $p$ be the probability that the die lands on 8.
$H_0: p = \frac{1}{8} = 0.125$
$H_1: p > 0.125$

(b) Under $H_0$, $X \sim \text{B}(25, 0.125)$.
We need to find the smallest integer $k$ such that $\text{P}(X \ge k) \le 0.05$.

Calculate upper tail probabilities:
$$\text{P}(X = r) = \binom{25}{r} (0.125)^r (0.875)^{25-r}$$
$$\text{P}(X = 0) = (0.875)^{25} \approx 0.03549$$
$$\text{P}(X = 1) = 25(0.125)(0.875)^{24} \approx 0.12675$$
$$\text{P}(X = 2) = 300(0.125)^2(0.875)^{23} \approx 0.21728$$
$$\text{P}(X = 3) = 2300(0.125)^3(0.875)^{22} \approx 0.23798$$
$$\text{P}(X = 4) = 12650(0.125)^4(0.875)^{21} \approx 0.18698$$
$$\text{P}(X = 5) = 53130(0.125)^5(0.875)^{20} \approx 0.11219$$
$$\text{P}(X = 6) = 177100(0.125)^6(0.875)^{19} \approx 0.05342$$

Sum of probabilities for $X \le 5$:
$$\text{P}(X \le 5) = 0.03549 + 0.12675 + 0.21728 + 0.23798 + 0.18698 + 0.11219 = 0.91667$$
$$\implies \text{P}(X \ge 6) = 1 - 0.91667 = 0.08333 > 0.05$$

Sum of probabilities for $X \le 6$:
$$\text{P}(X \le 6) = 0.91667 + 0.05342 = 0.97009$$
$$\implies \text{P}(X \ge 7) = 1 - 0.97009 = 0.02991 \le 0.05$$

Hence, the rejection region is $X \ge 7$ (or $\{7, 8, \dots, 25\}$).

(c) The probability of a Type I error is the probability of rejecting $H_0$ given that $H_0$ is true:
$$\text{P}(\text{Type I error}) = \text{P}(X \ge 7 \mid p = 0.125) = 0.0299\text{ (to 3 s.f.)}$$

(a)
B1: State correct hypotheses $H_0: p = 0.125$ and $H_1: p > 0.125$ (or $p = \frac{1}{8}$).

(b)
M1: Attempt binomial probability calculation for $\text{P}(X \ge 6)$ or $\text{P}(X \ge 7)$ with $n = 25, p = 0.125$.
A1: Show $\text{P}(X \ge 6) = 0.0833$ (or $0.083$) and $\text{P}(X \ge 7) = 0.0299$ (or $0.030$).
A1: Correct rejection region $X \ge 7$ (or $X > 6$).

(c)
B1: Correct Type I error probability $0.0299$ (or $0.030$).

(a) Sample size $n = 250$, number supporting $x = 165$.
Sample proportion:
$$\hat{p} = \frac{165}{250} = 0.66$$

For a 98% confidence interval, the critical value $z$ is $2.326$ (from $\Phi(2.326) = 0.99$).

Estimated standard error of the proportion:
$$\text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.66 \times 0.34}{250}} = \sqrt{\frac{0.2244}{250}} = \sqrt{0.0008976} \approx 0.02996$$

Margin of error:
$$z \times \text{SE} = 2.326 \times 0.02996 \approx 0.06969$$

Confidence interval:
$$0.66 \pm 0.0697 = [0.5903, 0.7297] \approx [0.590, 0.730]$$

(b) The value 0.70 (70%) lies inside the 98% confidence interval $[0.590, 0.730]$, so the planner's claim is plausible and consistent with the sample data at this confidence level.

(c) Any one of the following:
- The normal distribution is used as a continuous approximation to the discrete binomial distribution.
- The sample proportion $\hat{p}$ is used as an estimate for the unknown true proportion $p$ in calculating the standard error.

(a)
B1: Identify $z = 2.326$ (accept $2.33$).
M1: Use formula $\hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ with $\hat{p} = 0.66$ and $n = 250$.
A1: Correct confidence interval $[0.590, 0.730]$ (accept $[0.590, 0.730]$ or $[0.59, 0.73]$).

(b)
B1: State that 0.70 lies within the interval, so the claim is supported / plausible.

(c)
B1: Mention normal approximation to binomial OR estimation of $p$ by $\hat{p}$ in the standard error.

(a) State hypotheses:
$H_0: \mu = 750$
$H_1: \mu
e 750$

Sample statistics:
$n = 36$, $\sigma = 4.8$, $\bar{x} = \frac{26\,946}{36} = 748.5$.

Standard error:
$$\sigma_{\bar{x}} = \frac{4.8}{\sqrt{36}} = \frac{4.8}{6} = 0.8$$

Calculate the test statistic:
$$z = \frac{748.5 - 750}{0.8} = \frac{-1.5}{0.8} = -1.875$$

For a two-tailed test at the 5% significance level, the critical values are $\pm 1.960$.

Comparison:
Since $-1.960 < -1.875 < 1.960$ (or $|z| = 1.875 < 1.960$, or two-tailed $p\text{-value} = 2(1 - \Phi(1.875)) = 2(1 - 0.9696) = 0.0608 > 0.05$), we do not reject $H_0$.

Conclusion:
There is insufficient evidence at the 5% significance level to conclude that the mean volume dispensed has changed from 750 ml.

(b) A Type II error occurs when the test fails to reject $H_0$ given that $H_0$ is false.
In context: Concluding that the mean volume dispensed has not changed (remains 750 ml) when in fact the true mean volume has changed.

(a)
B1: State correct hypotheses $H_0: \mu = 750$ and $H_1: \mu
e 750$.
M1: Calculate standard error $\frac{4.8}{\sqrt{36}} = 0.8$ and test statistic $z = \frac{748.5 - 750}{0.8}$.
A1: Obtain $z = -1.875$ (or $\bar{x}_{\text{crit}} = 748.43$ and $751.57$).
A1: Correct comparison with $\pm 1.960$ and correct contextual conclusion (insufficient evidence that the mean has changed).

(b)
B1: Correct contextual definition of Type II error (accepting/failing to reject $\mu = 750$ when $\mu
e 750$).

(a) Over a 50-week period, the expected number of incidents under the null hypothesis is $\lambda = 50 \times 1.4 = 70$. Because $\lambda = 70 > 15$ (or sufficiently large), the Poisson distribution $\text{Po}(70)$ is well approximated by a normal distribution.

(b) Let $Y$ be the total number of incidents in 50 weeks.
State hypotheses:
$H_0: \lambda = 70$ (or mean weekly rate $\mu = 1.4$)
$H_1: \lambda < 70$ (or mean weekly rate $\mu < 1.4$)

Under $H_0$, $Y \sim \text{Po}(70) \approx \text{N}(70, 70)$.

Apply continuity correction for $\text{P}(Y \le 54) \approx \text{P}(W \le 54.5)$, where $W \sim \text{N}(70, 70)$:
$$z = \frac{54.5 - 70}{\sqrt{70}} = \frac{-15.5}{\sqrt{70}} \approx \frac{-15.5}{8.3666} \approx -1.853$$

For a 1-tailed test at the 5% significance level:
$$z_{\text{crit}} = -1.645$$

Comparison:
Since $-1.853 < -1.645$ (or $p\text{-value} = 1 - \Phi(1.853) = 1 - 0.9681 = 0.0319 < 0.05$), the test statistic falls in the rejection region.
Reject $H_0$.

Conclusion:
There is sufficient evidence at the 5% significance level that the mean rate of traffic incidents has decreased.

(a)
B1: State that the mean $\lambda = 70$ is large (or $\lambda > 15$).

(b)
B1: State correct hypotheses $H_0: \lambda = 70$ and $H_1: \lambda < 70$ (or weekly rate $\mu = 1.4$ vs $\mu < 1.4$).
M1: Standardise using normal approximation with mean 70, variance 70, and continuity correction $54.5$ (or $53.5$).
A1: Correct $z$-value $-1.853$ (or $-1.85$).
A1: Correct comparison with critical value $-1.645$ and correct contextual conclusion indicating evidence of a decrease.

(a)

Let $\mu$ be the population mean lifetime of the bulbs in hours.

State the null and alternative hypotheses:
$H_0: \mu = 12\,000$
$H_1: \mu < 12\,000$

Under $H_0$, by the Central Limit Theorem, the distribution of the sample mean is:
$\bar{X} \sim \text{N}\left(12\,000, \frac{450^2}{60}\right)$

Calculate the test statistic $z$:
$z = \frac{11\,880 - 12\,000}{\frac{450}{\sqrt{60}}} = \frac{-120}{58.094757\dots} \approx -2.066$

For a one-tailed test at the 2.5% significance level, the critical value is $-z_{0.025} = -1.960$.

Since $-2.066 < -1.960$ (or equivalently, $p\text{-value} = 1 - \Phi(2.066) = 0.0194 < 0.025$), we reject $H_0$.

There is sufficient evidence at the 2.5% significance level to suggest that the mean lifetime of the bulbs is less than 12\,000 hours.

(b)

The sample size ($n = 60$) is large ($n \ge 30$), so by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed regardless of the underlying population distribution.

(c)

A Type I error occurs when the test concludes that the mean lifetime is less than 12\,000 hours when in fact the true mean lifetime is 12\,000 hours (rejecting $H_0$ when $H_0$ is true).

(a)
- B1: State both hypotheses correctly in terms of $\mu$: $H_0: \mu = 12\,000$ and $H_1: \mu < 12\,000$.
- M1: Use standardisation formula $\frac{\bar{x} - 12\,000}{450/\sqrt{60}}$ or find the critical value for $\bar{x}$ using $\pm 1.960$.
- A1: Obtain $z = -2.07$ (or $-2.066$) or critical sample mean $11\,886$ hours or $p\text{-value} = 0.0194$.
- A1: Correct comparison ($|-2.066| > 1.960$ or $p < 0.025$) and correct conclusion in context (reject $H_0$, sufficient evidence that mean lifetime is less than 12\,000 hours). Must not be deterministic.

(b)
- B1: State that sample size is large ($n = 60$) so Central Limit Theorem applies.

(c)
- B1: Correct statement of Type I error in context: Concluding that the mean lifetime is less than 12\,000 hours when it is actually 12\,000 hours.

(a)

Let $\bar{X}$ be the sample mean volume of 16 cartons.

Under $H_0$, $\bar{X} \sim \text{N}\left(250, \frac{4.8^2}{16}\right) = \text{N}(250, 1.2^2)$.

For a two-tailed test at the 5% significance level, the critical $z$-values are $\pm 1.960$.

The critical values for $\bar{X}$ are:
$c_1 = 250 - 1.960 \times 1.2 = 250 - 2.352 = 247.648 \approx 247.6$
$c_2 = 250 + 1.960 \times 1.2 = 250 + 2.352 = 252.352 \approx 252.4$

So the acceptance region is $247.648 < \bar{X} < 252.352$, and the critical region is $\bar{X} < 247.6$ or $\bar{X} > 252.4$.

(b)

Since the sample mean $\bar{x} = 252.6 > 252.352$, $\bar{x}$ falls in the critical region.

Therefore, reject $H_0$. There is sufficient evidence at the 5% significance level to suggest that the mean volume of juice in a carton is not equal to 250 ml.

(c)

A Type II error occurs when $H_0$ is accepted when the true mean is $\mu = 253.0$.

When $\mu = 253.0$, $\bar{X} \sim \text{N}(253.0, 1.2^2)$.

$\text{P}(\text{Type II error}) = \text{P}(247.648 < \bar{X} < 252.352 \mid \mu = 253.0)$

Standardise the bounds:
$z_1 = \frac{247.648 - 253.0}{1.2} = \frac{-5.352}{1.2} = -4.46$
$z_2 = \frac{252.352 - 253.0}{1.2} = \frac{-0.648}{1.2} = -0.54$

$\text{P}(\text{Type II error}) = \Phi(-0.54) - \Phi(-4.46) = (1 - \Phi(0.54)) - 0 = 1 - 0.7054 = 0.2946 \approx 0.295$

(a)
- M1: State standard error $\frac{4.8}{\sqrt{16}} = 1.2$ and use $z = \pm 1.960$.
- M1: Form expressions $250 \pm 1.960 \times 1.2$.
- A1: Obtain critical values $247.6$ (or $247.65$) and $252.4$ (or $252.35$).

(b)
- B1: Correct comparison of $252.6$ with critical value and conclusion in context to reject $H_0$ (there is sufficient evidence that the population mean volume has changed from 250 ml).

(c)
- M1: Standardise upper critical value $252.352$ (or their critical value) using mean $253.0$ and standard error $1.2$: $\frac{252.352 - 253.0}{1.2}$.
- A1: Obtain $0.295$ (or $0.2946$ to 3 s.f.).