2024 Cambridge IAL Mathematics (9709) Practice Paper with Answers

To find the points of intersection, we set the equations of the line and curve equal to each other:
\( 2kx - 3 = x^2 + (k-2)x + 1 \)

Rearranging this into a standard quadratic form \( ax^2 + bx + c = 0 \):
\( x^2 + (k-2)x - 2kx + 1 + 3 = 0 \)
\( x^2 - (k+2)x + 4 = 0 \)

For the line and the curve to intersect at two distinct points, the discriminant of this quadratic equation must be strictly greater than zero (\( b^2 - 4ac > 0 \)):
\( (-(k+2))^2 - 4(1)(4) > 0 \)
\( (k+2)^2 - 16 > 0 \)

Expanding and simplifying:
\( k^2 + 4k + 4 - 16 > 0 \)
\( k^2 + 4k - 12 > 0 \)

Factorising the quadratic inequality:
\( (k+6)(k-2) > 0 \)

Thus, the critical values are \( k = -6 \) and \( k = 2 \).

Since we require the expression to be greater than zero, the set of values for \( k \) is:
\( k < -6 \) or \( k > 2 \).

M1: Set the line and curve equations equal and attempt to collect terms to form a 3-term quadratic equation in \( x \).
A1: Obtain the correct quadratic equation \( x^2 - (k+2)x + 4 = 0 \) or equivalent.
M1: Attempt to apply the discriminant condition \( b^2 - 4ac > 0 \).
A1: Correctly solve the inequality to find the critical values \( -6 \) and \( 2 \).
A1: State the final correct range: \( k < -6 \) or \( k > 2 \) (or equivalent interval notation).

Let the first term of the arithmetic progression (AP) be \( a \) and the common difference be \( d \).

The first, third, and eleventh terms of the AP are:
- \( T_1 = a \)
- \( T_3 = a + 2d \)
- \( T_{11} = a + 10d \)

Since these terms form the first three terms of a geometric progression, we can write:
\( \frac{a+2d}{a} = \frac{a+10d}{a+2d} \)

Cross-multiplying gives:
\( (a+2d)^2 = a(a+10d) \)

Expanding both sides:
\( a^2 + 4ad + 4d^2 = a^2 + 10ad \)

Subtracting \( a^2 \) from both sides:
\( 4ad + 4d^2 = 10ad \)
\( 4d^2 = 6ad \)

Since the common difference \( d \) is non-zero (\( d \neq 0 \)), we can divide both sides by \( 2d \):
\( 2d = 3a \implies d = 1.5a \) (or \( a = \frac{2}{3}d \))

Now, we find the common ratio \( r \) of the geometric progression:
\( r = \frac{a+2d}{a} \)

Substituting \( 2d = 3a \):
\( r = \frac{a+3a}{a} = \frac{4a}{a} = 4 \)

We can verify this with the third term:
\( r = \frac{a+10d}{a+2d} = \frac{a+5(2d)}{a+2d} = \frac{a+5(3a)}{a+3a} = \frac{16a}{4a} = 4 \).

Thus, the common ratio of the geometric progression is \( 4 \).

M1: Express the three terms of the AP correctly in terms of \( a \) and \( d \) (i.e., \( a, a+2d, a+10d \)).
M1: Set up a correct equation using the geometric progression property, e.g., \( (a+2d)^2 = a(a+10d) \).
A1: Simplify the equation to find a correct relation between \( a \) and \( d \) (such as \( 2d = 3a \) or equivalent).
M1: Substitute the relation back to find the ratio \( r = \frac{a+2d}{a} \) or \( r = \frac{a+10d}{a+2d} \).
A1: Obtain the correct common ratio \( r = 4 \).

We start with the equation:
\( 3\sin\theta\tan\theta = 8 \)

Using the trigonometric identity \( \tan\theta = \frac{\sin\theta}{\cos\theta} \):
\( 3\sin\theta \left(\frac{\sin\theta}{\cos\theta}\right) = 8 \)
\( \frac{3\sin^2\theta}{\cos\theta} = 8 \)
\( 3\sin^2\theta = 8\cos\theta \)

Using the identity \( \sin^2\theta = 1 - \cos^2\theta \):
\( 3(1 - \cos^2\theta) = 8\cos\theta \)
\( 3 - 3\cos^2\theta = 8\cos\theta \)

Rearranging to form a quadratic equation in terms of \( \cos\theta \):
\( 3\cos^2\theta + 8\cos\theta - 3 = 0 \)

Let \( y = \cos\theta \). Then the quadratic equation is:
\( 3y^2 + 8y - 3 = 0 \)

Factorising the quadratic equation:
\( (3y - 1)(y + 3) = 0 \)

This gives two possible solutions for \( y \):
\( y = \frac{1}{3} \) or \( y = -3 \)

Substituting back \( y = \cos\theta \):
1) \( \cos\theta = \frac{1}{3} \)
2) \( \cos\theta = -3 \) (which has no real solutions because the range of the cosine function is \( -1 \leq \cos\theta \leq 1 \))

Now we solve \( \cos\theta = \frac{1}{3} \) for \( 0^\circ \leq \theta \leq 360^\circ \):
- The principal value is \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.5^\circ \) (to 1 decimal place).
- The second solution in the given range is in the fourth quadrant:
\( \theta = 360^\circ - 70.53^\circ \approx 289.5^\circ \) (to 1 decimal place).

So, the solutions are \( \theta = 70.5^\circ \) and \( \theta = 289.5^\circ \).

M1: Use the identity \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) and express \( \sin^2\theta \) as \( 1-\cos^2\theta \) to obtain an equation in \( \cos\theta \) only.
A1: Form the correct 3-term quadratic equation \( 3\cos^2\theta + 8\cos\theta - 3 = 0 \).
M1: Solve the quadratic equation to get \( \cos\theta = \frac{1}{3} \) and state that \( \cos\theta = -3 \) has no solution.
A1: Find one correct solution \( \theta = 70.5^\circ \) (or \( 70.5^\circ \pm 0.1^\circ \)).
A1: Find the second correct solution \( \theta = 289.5^\circ \) (or \( 289.5^\circ \pm 0.1^\circ \)) and no other solutions in the range.

(a) Equating the line and the curve: \(x^2 - 4x + 1 = kx - 3\) gives \(x^2 - (4+k)x + 4 = 0\). For no intersection, the discriminant \(b^2 - 4ac < 0\): \\
\((-(4+k))^2 - 4(1)(4) < 0 \implies k^2 + 8k + 16 - 16 < 0 \implies k^2 + 8k < 0 \implies k(k+8) < 0\). \\
Hence, \(-8 < k < 0\).\\
\\
(b) Completing the square: \(x^2 - 4x + 1 = (x-2)^2 - 4 + 1 = (x-2)^2 - 3\). \\
The coordinates of the minimum point are \((2, -3)\).

(a) M1: Equate line and curve and collect terms.\\
A1: Correct quadratic equation \(x^2 - (4+k)x + 4 = 0\).\\
M1: Use of discriminant \(< 0\).\\
A1: Correct range \(-8 < k < 0\).\\
(b) M1: Attempt to complete the square.\\
A1: Correct expression \((x-2)^2 - 3\).\\
B1: Correct \(x\)-coordinate.\\
B1: Correct \(y\)-coordinate.

(a) Completing the square: \(2x^2 - 12x + 13 = 2(x^2 - 6x) + 13 = 2[(x-3)^2 - 9] + 13 = 2(x-3)^2 - 5\). \\
The vertex of the parabola is at \(x = 3\). For \(f\) to be a one-to-one function, we must restrict the domain to one side of the vertex. Since the domain is \(x \ge k\), the smallest possible value is \(k = 3\).\\
\\
(b) Let \(y = 2(x-3)^2 - 5\). Rearranging to make \(x\) the subject: \\
\(y + 5 = 2(x-3)^2 \implies \frac{y+5}{2} = (x-3)^2 \implies x - 3 = \pm\sqrt{\frac{y+5}{2}}\). \\
Since the domain of \(f\) is \(x \ge 3\), we choose the positive square root: \\
\(x = 3 + \sqrt{\frac{y+5}{2}}\). \\
Hence, \(f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\). \\
The domain of \(f^{-1}\) is the range of \(f\). Since \(x \ge 3\), the minimum value of \(f(x)\) is \(-5\), so the domain of \(f^{-1}\) is \(x \ge -5\).

(a) M1: Complete the square to obtain \(2(x-3)^2 + c\).\\
A1: Correct form \(2(x-3)^2 - 5\).\\
A1: Correctly state \(k=3\).\\
(b) M1: Set \(y = f(x)\) and rearrange to find \((x-3)^2\).\\
A1: Correct rearrangement.\\
M1: Take square root of both sides.\\
A1: Select the positive square root and write final expression in terms of \(x\).\\
M1: Identify that domain of \(f^{-1}\) is the range of \(f\).\\
A1: Correct domain \(x \ge -5\).

(a) Using the formula for the \(n\)th term of an arithmetic progression, \(u_n = a + (n-1)d\): \\
\(u_3 = a + 2d = 14\) and \(u_{11} = a + 10d = 46\). \\
Subtracting the first equation from the second: \(8d = 32 \implies d = 4\). \\
Substituting \(d=4\) into the first equation: \(a + 2(4) = 14 \implies a = 6\).\\
\\
(b) The sum of the first \(n\) terms is \(S_n = \frac{n}{2}[2a + (n-1)d]\). \\
Substituting \(a=6\) and \(d=4\): \\
\(\frac{n}{2}[12 + 4(n-1)] = 1560 \implies \frac{n}{2}[12 + 4n - 4] = 1560 \implies n(4n + 8) = 3120 \implies 4n^2 + 8n - 3120 = 0 \implies n^2 + 2n - 780 = 0\). \\
Factoring the quadratic equation: \((n-26)(n+30) = 0\). \\
Since \(n\) must be a positive integer, \(n = 26\).

(a) M1: Write down two simultaneous equations in \(a\) and \(d\).\\
A1: Correct value of \(d = 4\).\\
A1: Correct value of \(a = 6\).\\
(b) M1: Use of correct sum formula with their \(a\) and \(d\).\\
A1: Correct substitution leading to a quadratic equation.\\
A1: Form the simplified quadratic equation \(n^2 + 2n - 780 = 0\).\\
M1: Solve the quadratic equation.\\
A1: Correctly state \(n = 26\) and reject the negative solution.

(a) The center \(M\) of the circle is the midpoint of \(AB\): \\
\(M = \left(\frac{-1+7}{2}, \frac{5+1}{2}\right) = (3, 3)\). \\
The radius \(r\) is the distance from the center \(M(3, 3)\) to the point \(B(7, 1)\): \\
\(r^2 = (7-3)^2 + (1-3)^2 = 4^2 + (-2)^2 = 16 + 4 = 20\). \\
Thus, the equation of the circle is \((x-3)^2 + (y-3)^2 = 20\).\\
\\
(b) The gradient of the radius \(MB\) is \(m_{MB} = \frac{1-3}{7-3} = \frac{-2}{4} = -\frac{1}{2}\). \\
Since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent is \(m_t = -\frac{1}{m_{MB}} = 2\). \\
Using the point \(B(7, 1)\), the equation of the tangent is: \\
\(y - 1 = 2(x - 7) \implies y = 2x - 14 + 1 \implies y = 2x - 13\).

(a) M1: Attempt to find the midpoint of \(AB\).\\
A1: Correct center \((3, 3)\).\\
M1: Find the distance squared from center to point on circle.\\
A1: Correct equation \((x-3)^2 + (y-3)^2 = 20\).\\
(b) M1: Find the gradient of the radius/diameter \(MB\).\\
A1: Correct gradient of \(-\frac{1}{2}\).\\
M1: Use of perpendicular gradient rule \(m_1 m_2 = -1\).\\
A1: Gradient of tangent is \(2\).\\
A1: Correct tangent equation \(y = 2x - 13\).

(a) The perimeter of a sector is given by \(P = 2r + r\theta\). We are given \(P = 20\\, so \)2r + r\\theta = 20 \\implies r\\theta = 20 - 2r \\implies \\theta = \\frac{20 - 2r}{r}\). \\
The area of the sector is \(A = \frac{1}{2}r^2\theta\). Substituting the expression for \(\theta\): \\
\(A = \frac{1}{2}r^2\left(\frac{20-2r}{r}\right) = \frac{1}{2}r(20-2r) = 10r - r^2\).\\
\\
(b) To find the maximum area, we can differentiate \(A\) with respect to \(r\): \\
\(\frac{dA}{dr} = 10 - 2r\). \\
Setting this to zero for stationary points: \(10 - 2r = 0 \implies r = 5\). \\
(We know this is a maximum because \(\frac{d^2A}{dr^2} = -2 < 0\)). \\
The maximum area is \(A = 10(5) - 5^2 = 25\) cm\(^2\). \\
The corresponding value of \(\theta\) is \(\theta = \frac{20 - 2(5)}{5} = \frac{10}{5} = 2\) radians.

(a) B1: Correct perimeter equation \(2r + r\theta = 20\).\\
M1: Express \(\theta\) in terms of \(r\) (or vice versa).\\
M1: Substitute into area formula \(A = \frac{1}{2}r^2\theta\).\\
A1: Correctly obtain \(A = 10r - r^2\) with no errors shown.\\
(b) M1: Differentiate \(A\) and set to zero (or complete the square).\\
A1: Find \(r = 5\).\\
A1: Correct maximum area \(25\).\\
A1: Correct angle \(\theta = 2\).

(a) Combining the fractions on the left-hand side: \\
\(\frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)}\). \\
Expanding the numerator: \\
\(\frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)}\). \\
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\): \\
\(\frac{1 + 1 + 2\cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta}\).\\
\\
(b) Using the identity proven in part (a), the equation is equivalent to: \\
\(\frac{2}{\sin 2x} = 3 \implies \sin 2x = \frac{2}{3}\). \\
Since \(0^\circ \le x \le 180^\circ\), we have \(0^\circ \le 2x \le 360^\circ\). \\
Thus, \(2x = \sin^{-1}\left(\frac{2}{3}\right)\). \\
The principal value is \(2x \approx 41.810^\circ\), and the second value is \(2x \approx 180^\circ - 41.810^\circ = 138.190^\circ\). \\
Dividing by 2: \(x \approx 20.9^\circ\) or \(x \approx 69.1^\circ\) (both to 1 decimal place).

(a) M1: Express with a common denominator.\\
A1: Expand the numerator correctly.\\
M1: Use of identity \(\sin^2\theta + \cos^2\theta = 1\).\\
A1: Factorise and cancel term to complete proof.\\
(b) M1: Use identity to write \(\sin 2x = \frac{2}{3}\).\\
A1: Find one correct value of \(2x\) (e.g., \(41.8^\circ\) or \(138.2^\circ\)).\\
M1: Identify the second correct angle within range.\\
A1: Correctly find \(x = 20.9^\circ\).\\
A1: Correctly find \(x = 69.1^\circ\), and no other values in range.

(a) To find \(f(x)\), integrate \(f'(x)\): \\
\(f(x) = \int (6x^2 - 4x + 3)\\, dx = 2x^3 - 2x^2 + 3x + C\). \\
Using the point \((1, 5)\): \\
\(5 = 2(1)^3 - 2(1)^2 + 3(1) + C \implies 5 = 2 - 2 + 3 + C \implies 5 = 3 + C \implies C = 2\). \\
Therefore, the equation of the curve is \(f(x) = 2x^3 - 2x^2 + 3x + 2\).\\
\\
(b) The area is given by the definite integral: \\
\(\int_{0}^{1} (2x^3 - 2x^2 + 3x + 2)\\, dx = \left[ \frac{1}{2}x^4 - \frac{2}{3}x^3 + \frac{3}{2}x^2 + 2x \right]_{0}^{1}\). \\
Substituting the upper limit: \\
\(\left(\frac{1}{2}(1)^4 - \frac{2}{3}(1)^3 + \frac{3}{2}(1)^2 + 2(1)\right) = \frac{1}{2} - \frac{2}{3} + \frac{3}{2} + 2 = 2 - \frac{2}{3} + 2 = 4 - \frac{2}{3} = \frac{10}{3}\). \\
Substituting the lower limit: \(0\). \\
Therefore, the area of the region is \(\frac{10}{3}\) (or approximately \(3.33\)).

(a) M1: Integrate \(f'(x)\) term-by-term.\\
A1: Correct integration with constant \(C\).\\
M1: Substitute \((1, 5)\) to find \(C\).\\
A1: Correct equation \(f(x) = 2x^3 - 2x^2 + 3x + 2\).\\
(b) M1: Integrate the function \(f(x)\) from part (a).\\
A1: Correct integration of the terms.\\
M1: Substitute limits 0 and 1.\\
A1: Correct area of \(\frac{10}{3}\) or \(3\frac{1}{3}\).

To solve \(|3x - 5| = |2x + 1|\), we can square both sides or solve the equivalent linear equations.

**Method 1: Squaring both sides**
\[ (3x - 5)^2 = (2x + 1)^2 \]
\[ 9x^2 - 30x + 25 = 4x^2 + 4x + 1 \]
Rearranging into standard quadratic form:
\[ 5x^2 - 34x + 24 = 0 \]
Factoring the quadratic equation:
\[ (5x - 4)(x - 6) = 0 \]
This gives:
\[ x = 0.8 \quad \text{or} \quad x = 6 \]

**Method 2: Solving linear equations**
*Case 1:*
\[ 3x - 5 = 2x + 1 \implies x = 6 \]
*Case 2:*
\[ 3x - 5 = -(2x + 1) \implies 3x - 5 = -2x - 1 \implies 5x = 4 \implies x = 0.8 \]

M1: For an attempt to square both sides to obtain a 3-term quadratic equation, or for setting up and attempting to solve two linear equations.
A1: For obtaining one correct solution (either \(x = 6\) or \(x = 0.8\)).
A1: For obtaining the second correct solution and no others.

To find the gradient, we differentiate \(y\) with respect to \(x\) using the quotient rule:
\[ y = \frac{u}{v} \implies \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \]

Let \(u = \ln(2x+1)\) and \(v = x\).
Then \(u' = \frac{2}{2x+1}\) and \(v' = 1\).

Applying the quotient rule:
\[ \frac{dy}{dx} = \frac{\left(\frac{2}{2x+1}\right)(x) - \ln(2x+1)(1)}{x^2} \]
\[ \frac{dy}{dx} = \frac{\frac{2x}{2x+1} - \ln(2x+1)}{x^2} \]

Substitute \(x = 1\) to find the gradient at that point:
\[ \frac{dy}{dx} = \frac{\frac{2(1)}{2(1)+1} - \ln(2(1)+1)}{1^2} \]
\[ \frac{dy}{dx} = \frac{\frac{2}{3} - \ln 3}{1} = \frac{2}{3} - \ln 3 \]

M1: For applying the quotient rule (or product rule) to find \(\frac{dy}{dx}\), including a clear attempt at applying the chain rule to differentiate \(\ln(2x+1)\).
A1: For a correct derivative expression, e.g., \(\frac{dy}{dx} = \frac{\frac{2x}{2x+1} - \ln(2x+1)}{x^2}\).
A1: For substituting \(x = 1\) to obtain the correct exact value of \(\frac{2}{3} - \ln 3\) (or any equivalent exact form, such as \(\frac{2 - 3\ln 3}{3}\)).

**(i)**
To solve \(|2x - 5| < |x + 1|\), we can square both sides:
\[(2x - 5)^2 < (x + 1)^2\]
\[4x^2 - 20x + 25 < x^2 + 2x + 1\]
\[3x^2 - 22x + 24 < 0\]

Factorising the quadratic equation:
\[(3x - 4)(x - 6) < 0\]

The critical values are \(x = \frac{4}{3}\) and \(x = 6\).
Since we want the expression to be less than 0, the solution set is:
\[\frac{4}{3} < x < 6\]

**(ii)**
Using the Factor Theorem, since \((x - 6)\) is a factor of \(p(x)\), we have \(p(6) = 0\).
\[3(6)^3 - 22(6)^2 + k(6) + 12 = 0\]
\[3(216) - 22(36) + 6k + 12 = 0\]
\[648 - 792 + 6k + 12 = 0\]
\[-132 + 6k + 12 = 0\]
\[6k = 132 \implies k = 22\]

Thus, \(p(x) = 3x^3 - 22x^2 + 22x + 12\).
We can perform polynomial division of \(p(x)\) by \((x - 6)\):
\[3x^3 - 22x^2 + 22x + 12 = (x - 6)(3x^2 - 4x - 2)\]

To find the remaining roots, we solve \(3x^2 - 4x - 2 = 0\):
\[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-2)}}{2(3)}\]
\[x = \frac{4 \pm \sqrt{16 + 24}}{6} = \frac{4 \pm \sqrt{40}}{6} = \frac{4 \pm 2\sqrt{10}}{6} = \frac{2 \pm \sqrt{10}}{3}\]

Therefore, the exact roots of \(p(x) = 0\) are \(x = 6\), \(x = \frac{2 + \sqrt{10}}{3}\), and \(x = \frac{2 - \sqrt{10}}{3}\).

**(i)**
* **M1**: For squaring both sides and attempting to expand, or setting up two linear equations/inequalities.
* **A1**: Correct quadratic inequality \(3x^2 - 22x + 24 < 0\) or correct critical values \(x = \frac{4}{3}, 6\).
* **A1**: Correct final range \(\frac{4}{3} < x < 6\).
[Total: 4 marks]

**(ii)**
* **M1**: Applying the factor theorem \(p(6) = 0\) to find \(k\).
* **A1**: Finding \(k = 22\) correctly.
* **M1**: Attempting division by \(x-6\) to find the quadratic quotient \(3x^2 + ax + b\).
* **A1**: Correct quotient \(3x^2 - 4x - 2\).
* **A1**: Applying quadratic formula and finding the remaining two exact roots \(x = \frac{2 \pm \sqrt{10}}{3}\).
[Total: 4.8 marks]

**(i)**
Let \(u = 3^x\). Then \(3^{2x+1} = 3 \cdot (3^x)^2 = 3u^2\).
The equation becomes:
\[3u^2 - 10u + 3 = 0\]

Factorising the quadratic equation:
\[(3u - 1)(u - 3) = 0\]

This gives \(u = \frac{1}{3}\) or \(u = 3\).

Case 1: \(3^x = \frac{1}{3} = 3^{-1} \implies x = -1\)
Case 2: \(3^x = 3 = 3^1 \implies x = 1\)

**(ii)**
Using laws of logarithms, rewrite the equation:
\[\ln(2y + 3) - \ln(y^2) = \ln(3)\]
\[\ln\left(\frac{2y + 3}{y^2}\right) = \ln(3)\]

Taking exponentials of both sides:
\[\frac{2y + 3}{y^2} = 3\]
\[2y + 3 = 3y^2\]
\[3y^2 - 2y - 3 = 0\]

Solve using the quadratic formula:
\[y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-3)}}{2(3)}\]
\[y = \frac{2 \pm \sqrt{4 + 36}}{6} = \frac{2 \pm \sqrt{40}}{6} = \frac{1 \pm \sqrt{10}}{3}\]

Since \(y\) must be positive for \(\ln(y)\) to be defined:
\[y = \frac{1 + \sqrt{10}}{3} \approx \frac{1 + 3.162277}{3} \approx 1.3874\]

To 3 significant figures, \(y \approx 1.39\).

**(i)**
* **M1**: Substituting \(u = 3^x\) to form a quadratic equation in \(u\).
* **A1**: Correct quadratic equation \(3u^2 - 10u + 3 = 0\) (or equivalent).
* **M1**: Solving the quadratic for \(u\) and attempting to solve \(3^x = u\).
* **A1**: Correct values \(x = -1\) and \(x = 1\).
[Total: 4 marks]

**(ii)**
* **M1**: Applying power law \(2\ln y = \ln y^2\) and subtraction law \(\ln a - \ln b = \ln(a/b)\).
* **A1**: Correct equation without logarithms, \(\frac{2y+3}{y^2} = 3\).
* **M1**: Setting up and attempting to solve the quadratic equation \(3y^2 - 2y - 3 = 0\).
* **A1**: Finding the correct positive root \(y = \frac{1+\sqrt{10}}{3}\) and rejecting the negative root.
* **A1**: Correct final value \(y \approx 1.39\) rounded to 3 s.f.
[Total: 4.8 marks]

**(i)**
We want to write \(5\cos\theta - 12\sin\theta = R\cos(\theta + \alpha)\).
Using the compound angle identity:
\[R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha\]

Comparing coefficients:
\[R\cos\alpha = 5\]
\[R\sin\alpha = 12\]

Squaring and adding:
\[R^2(\cos^2\alpha + \sin^2\alpha) = 5^2 + 12^2 = 25 + 144 = 169 \implies R = 13\]

Dividing the equations:
\[\tan\alpha = \frac{12}{5} = 2.4\]
\[\alpha = \arctan(2.4) \approx 1.176005\text{ radians}\]

So, \(5\cos\theta - 12\sin\theta = 13\cos(\theta + 1.1760)\).

**(ii)**
Using the result from part (i), we can rewrite the equation \(5\cos(2x) - 12\sin(2x) = 4\) as:
\[13\cos(2x + 1.1760) = 4\]
\[\cos(2x + 1.1760) = \frac{4}{13}\]

Let \(\phi = 2x + 1.1760\).
Since \(0 < x < \pi\), we have \(0 < 2x < 2\pi\), which means:
\[1.1760 < \phi < 2\pi + 1.1760 \approx 7.4592\]

Now find the basic angle:
\[\phi_0 = \arccos\left(\frac{4}{13}\right) \approx 1.2586\text{ radians}\]

Since \(\cos\phi\) is positive, \\phi\) can lie in the first or fourth quadrant:
* **First quadrant:** \(\phi = 1.2586\) (which is in our range \(1.1760 < \phi < 7.4592\)).
* **Fourth quadrant:** \(\phi = 2\pi - 1.2586 \approx 6.2832 - 1.2586 = 5.0246\).
* **Next cycle:** \(\phi = 2\pi + 1.2586 \approx 7.5418\) (outside the range).

Now we solve for \(x\):

**Case 1:**
\[2x + 1.1760 = 1.2586\]
\[2x = 0.0826 \implies x \approx 0.0413\text{ to 3 s.f.}\]

**Case 2:**
\[2x + 1.1760 = 5.0246\]
\[2x = 3.8486 \implies x \approx 1.92\text{ to 3 s.f.}\]

So the solutions are \(x \approx 0.0413\) and \(x \approx 1.92\).

**(i)**
* **M1**: Finding \(R = \sqrt{5^2 + 12^2}\).
* **A1**: Correctly identifying \(R = 13\).
* **M1**: Setting up \(\tan\alpha = 12/5\) (or equivalent) to find \(\alpha\).
* **A1**: Obtaining \(\alpha \approx 1.1760\) (accept 1.176).
[Total: 3.8 marks]

**(ii)**
* **M1**: Setting up the equation \(13\cos(2x + 1.1760) = 4\).
* **M1**: Finding the principle value \(\arccos(4/13) \approx 1.2586\).
* **A1**: Finding one correct value of \(x\) (either \(0.0413\) or \(1.92\)).
* **M1**: Showing a complete method to find the second value of \(x\) in the interval \(0 < x < \pi\).
* **A1**: Finding both correct values \(x \approx 0.0413\) and \(x \approx 1.92\).
[Total: 5 marks]

**(i)**
Given: \(y^2 e^{2x} + 3y = x^2 + 4\).
Differentiate implicitly with respect to \(x\):
\[\frac{d}{dx}\left(y^2 e^{2x}\right) + \frac{d}{dx}(3y) = \frac{d}{dx}\left(x^2 + 4\right)\]

Using the product rule on the first term:
\[\left(2y \frac{dy}{dx}\right) e^{2x} + y^2 \left(2e^{2x}\right) + 3\frac{dy}{dx} = 2x\]

Group terms with \(\frac{dy}{dx}\):
\[\frac{dy}{dx}\left(2y e^{2x} + 3\right) + 2y^2 e^{2x} = 2x\]
\[\frac{dy}{dx}\left(2y e^{2x} + 3\right) = 2x - 2y^2 e^{2x}\]
\[\frac{dy}{dx} = \frac{2x - 2y^2 e^{2x}}{2y e^{2x} + 3}\]

**(ii)**
First, find the \(y\)-coordinate of the point on the curve where \(x = 0\):
\[y^2 e^0 + 3y = 0^2 + 4\]
\[y^2 + 3y - 4 = 0\]
\[(y + 4)(y - 1) = 0\]

Since \(y > 0\), we select \(y = 1\).
So the point of contact is \((0, 1)\).

Now, substitute \(x = 0\) and \(y = 1\) into the expression for \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{2(0) - 2(1^2) e^0}{2(1) e^0 + 3} = \frac{0 - 2}{2 + 3} = -\frac{2}{5}\]

Thus, the gradient of the tangent is \(-\frac{2}{5}\).
The equation of the tangent is:
\[y - 1 = -\frac{2}{5}(x - 0)\]
\[y = -\frac{2}{5}x + 1 \implies 2x + 5y = 5\]

**(i)**
* **M1**: Applying the product rule correctly to differentiate \(y^2 e^{2x}\).
* **A1**: Correct term \(2y \frac{dy}{dx} e^{2x} + 2y^2 e^{2x}\).
* **B1**: Correctly differentiating the other terms to get \(3\frac{dy}{dx}\) and \(2x\).
* **M1**: Factoring out \(\frac{dy}{dx}\) and making it the subject.
* **A1**: Correct derivative expression \(\frac{dy}{dx} = \frac{2x - 2y^2 e^{2x}}{2y e^{2x} + 3}\).
[Total: 4.8 marks]

**(ii)**
* **M1**: Substituting \(x = 0\) into the original equation to find \(y\).
* **A1**: Finding \(y = 1\) (rejecting \(y = -4\) with justification or implicitly by choosing the positive root).
* **M1**: Substituting \((0, 1)\) into their \(\frac{dy}{dx}\) expression.
* **A1**: Correct equation of the tangent, e.g. \(y = -\frac{2}{5}x + 1\) or \(2x + 5y = 5\).
[Total: 4 marks]

**(i)**
Integrate each term individually:
\[\int \frac{2}{3x-2} dx = \frac{2}{3} \ln|3x-2|\]
\[\int e^{2x-2} dx = \frac{1}{2} e^{2x-2}\]

Thus, the definite integral is:
\[\left[ \frac{2}{3} \ln|3x-2| + \frac{1}{2} e^{2x-2} \right]_{1}^{4}\]

Substitute the upper limit \(x = 4\):
\[\frac{2}{3} \ln|3(4)-2| + \frac{1}{2} e^{2(4)-2} = \frac{2}{3} \ln(10) + \frac{1}{2} e^{6}\]

Substitute the lower limit \(x = 1\):
\[\frac{2}{3} \ln|3(1)-2| + \frac{1}{2} e^{2(1)-2} = \frac{2}{3} \ln(1) + \frac{1}{2} e^0 = 0 + \frac{1}{2} = \frac{1}{2}\]

Subtract the lower limit value from the upper limit value:
\[\left( \frac{2}{3}\ln(10) + \frac{1}{2}e^6 \right) - \frac{1}{2} = \frac{2}{3}\ln(10) + \frac{1}{2}e^6 - \frac{1}{2}\]

**(ii)**
For \(\int_{1}^{4} \frac{1}{x^2 + 1} dx\) with 3 intervals:
Interval width \(h = \frac{4 - 1}{3} = 1\).

The grid points are \(x_0 = 1\), \(x_1 = 2\), \(x_2 = 3\), \(x_3 = 4\).
Let \(f(x) = \frac{1}{x^2 + 1}\):
* \(y_0 = f(1) = \frac{1}{1^2 + 1} = 0.5\)
* \(y_1 = f(2) = \frac{1}{2^2 + 1} = 0.2\)
* \(y_2 = f(3) = \frac{1}{3^2 + 1} = 0.1\)
* \(y_3 = f(4) = \frac{1}{4^2 + 1} = \frac{1}{17} \approx 0.058824\)

Applying the trapezium rule formula:
\[\text{Area} \approx \frac{h}{2} [y_0 + 2(y_1 + y_2) + y_3]\]
\[\text{Area} \approx \frac{1}{2} [0.5 + 2(0.2 + 0.1) + 0.058824]\]
\[\text{Area} \approx \frac{1}{2} [0.5 + 0.6 + 0.058824]\]
\[\text{Area} \approx \frac{1}{2} [1.158824] \approx 0.579412\]

To 3 decimal places, the approximation is \(0.579\).

**(i)**
* **M1**: Integrating \(\frac{2}{3x-2}\) to get \(k\ln|3x-2|\).
* **A1**: Correct term \(\frac{2}{3}\ln|3x-2|\).
* **A1**: Correct integration of \(e^{2x-2}\) to get \(\frac{1}{2}e^{2x-2}\).
* **M1**: Applying limits 1 and 4 correctly into their integrated expression.
* **A1**: Obtaining the correct exact answer \(\frac{2}{3}\ln(10) + \frac{1}{2}e^6 - \frac{1}{2}\).
[Total: 4.8 marks]

**(ii)**
* **B1**: Finding correct interval width \(h = 1\).
* **M1**: Calculating the functional values at \(x = 1, 2, 3, 4\).
* **M1**: Correctly substituting values into the Trapezium Rule formula.
* **A1**: Correct approximation \(0.579\) (to 3 d.p.).
[Total: 4 marks]

To find the gradient of the curve, we differentiate the equation implicitly with respect to \(x\):

\(\frac{\text{d}}{\text{d}x}(x^3 + y^3 - 3xy) = \frac{\text{d}}{\text{d}x}(3)\)

Using the chain rule and product rule, we get:

\(3x^2 + 3y^2 \frac{\text{d}y}{\text{d}x} - \left(3y + 3x \frac{\text{d}y}{\text{d}x}\right) = 0\)

Simplifying the equation:

\(3x^2 + 3y^2 \frac{\text{d}y}{\text{d}x} - 3y - 3x \frac{\text{d}y}{\text{d}x} = 0\)

Substitute the coordinates of the point \((2, 1)\) into this equation:

\(3(2)^2 + 3(1)^2 \frac{\text{d}y}{\text{d}x} - 3(1) - 3(2) \frac{\text{d}y}{\text{d}x} = 0\)

\(12 + 3 \frac{\text{d}y}{\text{d}x} - 3 - 6 \frac{\text{d}y}{\text{d}x} = 0\)

\(9 - 3 \frac{\text{d}y}{\text{d}x} = 0\)

\(3 \frac{\text{d}y}{\text{d}x} = 9 \implies \frac{\text{d}y}{\text{d}x} = 3\)

Therefore, the gradient of the curve at \((2, 1)\) is \(3\).

M1: For an attempt to differentiate implicitly, requiring the use of the product rule for the term \(-3xy\).
A1: For a correct differentiation: \(3x^2 + 3y^2 \frac{\text{d}y}{\text{d}x} - 3y - 3x \frac{\text{d}y}{\text{d}x} = 0\) (or equivalent).
M1: For substituting \(x = 2\) and \(y = 1\) into their differentiated equation.
A1: For obtaining the correct gradient of \(3\).

We use integration by parts, \(\int u \frac{\text{d}v}{\text{d}x} \, \text{d}x = uv - \int v \frac{\text{d}u}{\text{d}x} \, \text{d}x\).

Let \(u = \ln x\) and \(\frac{\text{d}v}{\text{d}x} = x^2\).
Then \(\frac{\text{d}u}{\text{d}x} = \frac{1}{x}\) and \(v = \frac{1}{3}x^3\).

Applying the integration by parts formula:

\(\int x^2 \ln x \, \text{d}x = \left[ \frac{1}{3}x^3 \ln x \right] - \int \frac{1}{3}x^3 \cdot \frac{1}{x} \, \text{d}x\)

\(= \frac{1}{3}x^3 \ln x - \frac{1}{3} \int x^2 \, \text{d}x\)

\(= \frac{1}{3}x^3 \ln x - \frac{1}{9}x^3\)

Now we evaluate this expression from \(1\) to \(\text{e}\):

\(\left[ \frac{1}{3}x^3 \ln x - \frac{1}{9}x^3 \right]_{1}^{\text{e}} = \left( \frac{1}{3}\text{e}^3 \ln \text{e} - \frac{1}{9}\text{e}^3 \right) - \left( \frac{1}{3}(1)^3 \ln 1 - \frac{1}{9}(1)^3 \right)\)

Since \(\ln \text{e} = 1\) and \(\ln 1 = 0\), this becomes:

\(\left( \frac{1}{3}\text{e}^3 - \frac{1}{9}\text{e}^3 \right) - \left( 0 - \frac{1}{9} \right)\)

\(= \frac{2}{9}\text{e}^3 + \frac{1}{9}\)

\(= \frac{2\text{e}^3 + 1}{9}\).

M1: For attempting integration by parts of the form \(u = \ln x, v' = x^2\).
A1: For correct integration to obtain \(\frac{1}{3}x^3 \ln x - \frac{1}{9}x^3\) (or equivalent).
M1: For substituting the limits \(1\) and \(\text{e}\) into their integrated term (with correct application of \(\ln \text{e} = 1\) and \(\ln 1 = 0\)).
A1: For obtaining the correct exact value \(\frac{2\text{e}^3+1}{9}\) or equivalent form.

Let \(y = 3^x\).

Then \(3^{2x+1} = 3 \cdot 3^{2x} = 3(3^x)^2 = 3y^2\).

Substitute this into the given equation:

\(3y^2 - 5y - 2 = 0\)

Factorise the quadratic equation:

\((3y + 1)(y - 2) = 0\)

This gives:

\(y = -\frac{1}{3}\) or \(y = 2\)

Since \(3^x > 0\) for all real values of \(x\), the solution \(3^x = -\frac{1}{3}\) is rejected.

We solve \(3^x = 2\):

Taking natural logarithms on both sides:

\(\ln(3^x) = \ln 2\)

\(x \ln 3 = \ln 2\)

\(x = \frac{\ln 2}{\ln 3} \approx 0.630929\)

Correct to 3 significant figures, \(x = 0.631\).

M1: For expressing the equation as a quadratic in \(3^x\) (e.g., \(3(3^x)^2 - 5(3^x) - 2 = 0\)).
A1: For solving the quadratic equation to find the positive root \(3^x = 2\) (and correctly rejecting or ignoring the negative root).
M1: For using logarithms correctly to solve an equation of the form \(3^x = k\) where \(k > 0\).
A1: For obtaining \(x = 0.631\) correct to 3 significant figures.

We use the double-angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\) to rewrite the equation in terms of \(\sin\theta\):

\((1 - 2\sin^2\theta) + 3\sin\theta = 2\)

Rearranging terms into a standard quadratic form:

\(2\sin^2\theta - 3\sin\theta + 1 = 0\)

Factorise the quadratic equation:

\((2\sin\theta - 1)(\sin\theta - 1) = 0\)

This gives:

\(\sin\theta = \frac{1}{2}\) or \(\sin\theta = 1\)

For \(0^\circ \le \theta \le 360^\circ\):

1. If \(\sin\theta = \frac{1}{2}\):
\(\theta = 30^\circ\) or \(\theta = 180^\circ - 30^\circ = 150^\circ\)

2. If \(\sin\theta = 1\):
\(\theta = 90^\circ\)

Combining the solutions, we get:

\(\theta = 30^\circ, 90^\circ, 150^\circ\).

M1: For using the double-angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\) to obtain an equation in \(\sin\theta\) only.
A1: For obtaining the correct quadratic equation \(2\sin^2\theta - 3\sin\theta + 1 = 0\) (or equivalent).
M1: For solving the quadratic equation and finding at least one correct angle for \(\theta\).
A1: For obtaining the three correct solutions \(30^\circ\), \(90^\circ\), and \(150^\circ\), and no others in the given range.

(i) Since \(\cos^2(x) = 1 - \sin^2(x) = (1 - \sin(x))(1 + \sin(x))\), dividing by \(1 + \sin(x)\) gives \(1 - \sin(x)\). (ii) Separating variables gives \(\int e^{-2y} \mathrm{d}y = \int (1 - \sin(x)) \mathrm{d}x\), which integrates to \(-\frac{1}{2}e^{-2y} = x + \cos(x) + C\). Substituting \(x = 0\) and \(y = 0\) gives \(-\frac{1}{2} = 1 + C\), so \(C = -\frac{3}{2}\). This gives \(e^{-2y} = 3 - 2x - 2\cos(x)\), and taking logarithms yields \(y = -\frac{1}{2}\ln(3 - 2x - 2\cos(x))\).

M1: For attempt to separate variables and integrate both sides. A1: For correct integration of LHS and RHS. M1: For using the initial condition to find the constant of integration. A1: For obtaining correct value of C. M1: For making y the subject. A1.4: For correct final equation.

(i) Area is given by \(\int_0^{\ln 2} \frac{e^{2x}}{e^x + 3} \mathrm{d}x\). Let \(u = e^x + 3\), so \(\mathrm{d}u = e^x \mathrm{d}x\). When \(x=0\), \(u=4\); when \(x=\ln 2\), \(u=5\). The integral becomes \(\int_4^5 \frac{u-3}{u} \mathrm{d}u = \int_4^5 (1 - \frac{3}{u}) \mathrm{d}u = [u - 3\ln u]_4^5 = (5 - 3\ln 5) - (4 - 3\ln 4) = 1 + 3\ln(4/5)\). (ii) With 2 intervals, the width \(h = \frac{1}{2}\ln 2 \approx 0.3466\). The x-values are 0, 0.3466, 0.6931. The corresponding y-values are \(y_0 = 0.25\), \(y_1 \approx 0.4531\), \(y_2 = 0.8\). The approximation is \(\frac{h}{2}(y_0 + 2y_1 + y_2) \approx \frac{0.3466}{2}(0.25 + 2(0.4531) + 0.8) \approx 0.339\).

M1: For using substitution \(u = e^x + 3\) or equivalent. A1: For obtaining \(\int (1 - 3/u) \mathrm{d}u\). A1: For correct limits of u. A1.4: For obtaining exact area \(1 + 3\ln(4/5)\). M1: For applying trapezium rule with 2 intervals. A1: For correct approximation of 0.339.

(i) \(w = \frac{(5 - 5\mathrm{i})(2 - \mathrm{i})}{(2 + \mathrm{i})(2 - \mathrm{i})} = \frac{10 - 5\mathrm{i} - 10\mathrm{i} - 5}{5} = \frac{5 - 15\mathrm{i}}{5} = 1 - 3\mathrm{i}\). (ii) The point representing \(w\) is \((1, -3)\). The locus \(|z - w| \le \sqrt{5}\) is a circle of radius \(\sqrt{5}\) centred at \((1, -3)\). The half-line \(\arg(z) = -\frac{\pi}{4}\) is the line \(y = -x\) (for \(x \ge 0\)). The half-line \(\arg(z) = 0\) is the positive real axis. Since the circle has maximum y-coordinate \(-3 + \sqrt{5} \approx -0.76\), it lies entirely below the real axis, but intersects the line \(y = -x\). The shaded region is the segment of the circle that lies above or on the line \(y = -x\).

M1: For multiplying numerator and denominator by \(2-\mathrm{i}\). A1: For correct numerator. A1: For obtaining \(1 - 3\mathrm{i}\). B1: For drawing circle centered at \(1-3\mathrm{i}\). B1: For circle having correct radius (passing below real axis but through \(y=-x\)). B1: For drawing the ray \(\arg(z) = -\frac{\pi}{4}\). B1.4: For shading the correct region.

(i) Equating components: \(2 + \lambda = 1 + 2\mu\), \(-1 + 2\lambda = 2 - \mu\), and \(3 - \lambda = -2 + 4\mu\). Solving the first two equations gives \(\mu = 1\) and \(\lambda = 1\). Substituting these values into the third equation: \(3 - 1 = 2\) and \(-2 + 4(1) = 2\), which is consistent. Thus the lines intersect. The point of intersection \(P\) is \((3, 1, 2)\). (ii) The direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\). The dot product is \(\mathbf{d}_1 \cdot \mathbf{d}_2 = 1(2) + 2(-1) + (-1)(4) = -4\). The magnitudes are \(|\mathbf{d}_1| = \sqrt{6}\) and \(|\mathbf{d}_2| = \sqrt{21}\). Thus \(\cos \theta = \frac{|-4|}{\sqrt{6}\sqrt{21}} = \frac{4}{\sqrt{126}} \approx 0.3563\). This gives \(\theta \approx 69.1^\circ\).

M1: For setting up simultaneous equations in \(\lambda\) and \(\mu\). A1: For solving for \(\lambda\) and \(\mu\). M1: For showing consistency in the third equation. A1: For correct coordinates of P. M1: For using the scalar product formula with direction vectors. A1: For calculating correct magnitudes and dot product. A1.4: For correct angle 69.1 degrees.

(i) We find \(\frac{\mathrm{d}x}{\mathrm{d}t} = 2t + 2 = 2(t + 1)\) and \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{1}{t+1} - 1 = \frac{-t}{t+1}\). Thus \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{-t}{2(t+1)^2}\). (ii) For stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies -t = 0 \implies t = 0\). The coordinates are \(x = 0^2 + 2(0) = 0\) and \(y = \ln(1) - 0 = 0\). Thus the stationary point is \((0, 0)\). To determine its nature: since \(2(t+1)^2 > 0\) for \(t > -1\), the sign of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) is determined by \(-t\). For \(-1 < t < 0\), \(\frac{\mathrm{d}y}{\mathrm{d}x} > 0\). For \(t > 0\), \(\frac{\mathrm{d}y}{\mathrm{d}x} < 0\). Since \(x\) is strictly increasing with \(t\), the gradient changes from positive to negative, indicating a local maximum.

M1: For finding \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\). A1: For both correct. M1: For using chain rule. A1: For correct simplified expression. M1: For setting derivative to 0 to find t. A1: For correct coordinates (0,0). M1: For checking the sign of the gradient or using second derivative. A1.4: For concluding it is a local maximum with sound reasoning.

(i) Write \(5x^2 - 3x + 4 = A(x-1)(2x+1) + B(2x+1) + C(x-1)^2\). Substituting \(x = 1\) gives \(6 = 3B \implies B = 2\). Substituting \(x = -1/2\) gives \(27/4 = (9/4)C \implies C = 3\). Equating coefficients of \(x^2\) gives \(5 = 2A + C \implies 5 = 2A + 3 \implies A = 1\). Thus \(f(x) = \frac{1}{x-1} + \frac{2}{(x-1)^2} + \frac{3}{2x+1}\). (ii) Rewrite the terms: \(\frac{1}{x-1} = -(1-x)^{-1} = -(1 + x + x^2 + \dots)\), \(\frac{2}{(x-1)^2} = 2(1-x)^{-2} = 2(1 + 2x + 3x^2 + \dots)\), and \(\frac{3}{2x+1} = 3(1+2x)^{-1} = 3(1 - 2x + 4x^2 - \dots)\). Summing these gives \((-1 - x - x^2) + (2 + 4x + 6x^2) + (3 - 6x + 12x^2) = 4 - 3x + 17x^2\).

M1: For correct form of partial fractions. A1: For finding one of A, B, or C correctly. A1: For finding a second coefficient correctly. A1: For finding all three coefficients correctly. M1: For binomial expansions of the terms. A1.4: For correct final expansion 4 - 3x + 17x^2.

(i) Let \(3\cos\theta - 2\sin\theta = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha\). Thus \(R\cos\alpha = 3\) and \(R\sin\alpha = 2\). This gives \(R = \sqrt{3^2 + 2^2} = \sqrt{13}\) and \(\tan\alpha = \frac{2}{3} \implies \alpha \approx 33.69^\circ\). (ii) The equation is \(\sqrt{13}\cos(2x + 33.69^\circ) = 1.5 \implies \cos(2x + 33.69^\circ) = \frac{1.5}{\sqrt{13}} \approx 0.41599\). Thus \(2x + 33.69^\circ = 65.42^\circ\) or \(2x + 33.69^\circ = 360^\circ - 65.42^\circ = 294.58^\circ\). This gives \(2x = 31.73^\circ \implies x \approx 15.9^\circ\) or \(2x = 260.89^\circ \implies x \approx 130.4^\circ\).

M1: For using \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = b/a\). A1: For \(R = \sqrt{13}\). A1: For \(\alpha = 33.69^\circ\). M1: For relating the equation to the R-formula expression. M1: For finding one correct value of \(2x + \alpha\). A1: For \(x = 15.9^\circ\). A1.4: For \(x = 130.4^\circ\).

Let \(R\) be the normal reaction force between the particle and the plane. Resolving perpendicular to the incline: \(R = mg \cos 30^\circ = 4 \times 10 \cos 30^\circ = 40 \cos 30^\circ = 20\sqrt{3}\text{ N} \approx 34.64\text{ N}\). The maximum frictional force is \(F = \mu R = 0.2 \times 20\sqrt{3} = 4\sqrt{3}\text{ N} \approx 6.928\text{ N}\). For the maximum value of \(P\), the particle is on the point of slipping up the plane, so the frictional force acts down the plane. Resolving parallel to the incline: \(P - mg \sin 30^\circ - F = 0\) which gives \(P = 40 \sin 30^\circ + 4\sqrt{3} = 20 + 4\sqrt{3} \approx 26.928\text{ N}\). To 3 significant figures, the maximum force is \(26.9\text{ N}\).

M1: For resolving forces perpendicular to the incline to find the normal reaction. A1: For obtaining \(R = 20\sqrt{3}\) or \(34.6\text{ N}\). M1: For using \(F = \mu R\) and setting up the equilibrium equation along the incline with the frictional force acting down the plane. A1.5: For obtaining the correct value of \(P = 26.9\).

Using the relation between power, driving force, and speed: \(P = DF \times v\). Given \(P = 18000\text{ W}\) and \(v = 15\text{ m s}^{-1}\), the driving force is \(DF = 18000 / 15 = 1200\text{ N}\). Applying Newton's second law in the direction of motion: \(DF - R = ma\), where \(R = 400\text{ N}\) is the resistance and \(m = 1000\text{ kg}\) is the mass. Substitute the known values: \(1200 - 400 = 1000a\), which gives \(800 = 1000a\), and solving for acceleration gives \(a = 0.8\text{ m s}^{-2}\).

M1: For using \(P = DF \times v\) to find the driving force. A1: For obtaining \(DF = 1200\text{ N}\). M1: For applying Newton's second law \(DF - R = ma\). A1.5: For obtaining \(a = 0.8\text{ m s}^{-2}\).

(i) Since \(\tan \alpha = 0.75\), we have \(\sin \alpha = 0.6\) and \(\cos \alpha = 0.8\).
Let \(R\) be the normal reaction force. Resolving perpendicular to the plane:
\(R = 40 \cos \alpha + P \sin \alpha = 40(0.8) + 0.6P = 32 + 0.6P\).
The maximum frictional force is \(F_{\text{max}} = \mu R = 0.25(32 + 0.6P) = 8 + 0.15P\).
For the least value of \(P\), the block is on the point of slipping down the plane, so friction acts up the inclined plane.
Resolving parallel to the plane:
\(P \cos \alpha + F_{\text{max}} = 40 \sin \alpha\)
\(0.8P + 8 + 0.15P = 24\)
\(0.95P = 16\)
\(P = \frac{16}{0.95} = \frac{320}{19} \approx 16.8\text{ N}\).

(ii) For the greatest value of \(P\), the block is on the point of slipping up the plane, so friction acts down the inclined plane.
Resolving parallel to the plane:
\(P \cos \alpha = 40 \sin \alpha + F_{\text{max}}\)
\(0.8P = 24 + 8 + 0.15P\)
\(0.65P = 32\)
\(P = \frac{32}{0.65} = \frac{640}{13} \approx 49.2\text{ N}\).

Part (i):
- M1: For resolving forces perpendicular to the plane to find \(R\) in terms of \(P\).
- A1: Correct expression \(R = 32 + 0.6P\).
- M1: For setting up the equilibrium equation parallel to the plane with friction acting upwards.
- A1: Correct calculation giving \(P = 16.8\) (or \(\frac{320}{19}\)).

Part (ii):
- M1: For setting up the equilibrium equation parallel to the plane with friction acting downwards.
- A1: Correct equation \(0.8P = 24 + 8 + 0.15P\).
- A1: Correct calculation giving \(P = 49.2\) (or \(\frac{640}{13}\)).

(i) To find the velocity \(v(t)\), we integrate the acceleration \(a(t)\):
\(v(t) = \int (6 - 2t) \, dt = 6t - t^2 + C\).
Since \(v(0) = 7\), we have \(C = 7\).
Thus, \(v(t) = 7 + 6t - t^2\).

The maximum velocity occurs when \(a(t) = 0\):
\(6 - 2t = 0 \implies t = 3\text{ s}\).

Substituting \(t = 3\) into the velocity equation:
\(v(3) = 7 + 6(3) - (3)^2 = 7 + 18 - 9 = 16 \text{ m s}^{-1}\).

(ii) To find the displacement \(s(t)\), we integrate the velocity \(v(t)\):
\(s(t) = \int (7 + 6t - t^2) \, dt = 7t + 3t^2 - \frac{1}{3}t^3 + D\).
Since \(s(0) = 0\), we have \(D = 0\).

Substituting \(t = 6\) into the displacement equation:
\(s(6) = 7(6) + 3(6)^2 - \frac{1}{3}(6)^3\)
\(s(6) = 42 + 3(36) - \frac{1}{3}(216) = 42 + 108 - 72 = 78\text{ m}\).

Part (i):
- M1: For integrating the acceleration function to find the velocity expression.
- A1: Correct velocity equation \(v(t) = 7 + 6t - t^2\) including the constant of integration.
- M1: For identifying that maximum velocity occurs when \(a = 0\) to find \(t = 3\).
- A1: Correct maximum velocity of \(16 \text{ m s}^{-1}\).

Part (ii):
- M1: For integrating the velocity function to obtain the displacement expression.
- A1: Correct displacement function \(s(t) = 7t + 3t^2 - \frac{1}{3}t^3\).
- A1: Correct substitution of \(t = 6\) to find the displacement as \(78\text{ m}\).

(i) Let the acceleration of the system be \(a\) and the tension in the string be \(T\).
For particle \(A\), the normal reaction is \(R = 0.6g = 6\text{ N}\).
The frictional force on \(A\) is \(F = \mu R = 0.25 \times 6 = 1.5\text{ N}\).
Applying Newton's second law to \(A\):
\(T - 1.5 = 0.6a\) --- (Equation 1)

Applying Newton's second law to \(B\):
\(0.4g - T = 0.4a \implies 4 - T = 0.4a\) --- (Equation 2)

Adding Equations 1 and 2:
\(4 - 1.5 = 1.0a \implies 2.5 = 1.0a \implies a = 2.5 \text{ m s}^{-2}\) (shown).

Substitute \(a\) back into Equation 2:
\(T = 4 - 0.4(2.5) = 3\text{ N}\).

(ii) Before the string breaks, the distance \(s_1\) travelled in \(t = 1.2\text{ s}\) is:
\(s_1 = \frac{1}{2} a t^2 = 0.5(2.5)(1.2)^2 = 1.8\text{ m}\).

The velocity of \(A\) at the instant the string breaks is:
\(v = u + at = 0 + 2.5(1.2) = 3 \text{ m s}^{-1}\).

After the string breaks, the tension in the string becomes zero. The only horizontal force acting on \(A\) is friction.
Applying Newton's second law to \(A\) post-break:
\(-F = m_A a' \implies -1.5 = 0.6a' \implies a' = -2.5 \text{ m s}^{-2}\).

Let \(s_2\) be the distance travelled by \(A\) after the string breaks until it comes to rest:
\(v_{\text{final}}^2 = v^2 + 2a's_2 \implies 0 = 3^2 + 2(-2.5)s_2\)
\(5s_2 = 9 \implies s_2 = 1.8\text{ m}\).

The total distance travelled by \(A\) is:
\(s_{\text{total}} = s_1 + s_2 = 1.8 + 1.8 = 3.6\text{ m}\).

Part (i):
- M1: For applying Newton's second law to both particles, including a friction force of \(1.5\text{ N}\).
- A1: For deriving the acceleration \(a = 2.5 \text{ m s}^{-2}\).
- A1: For calculating the tension \(T = 3\text{ N}\).

Part (ii):
- M1: For calculating the distance \(s_1 = 1.8\text{ m}\) or the velocity \(v = 3 \text{ m s}^{-1}\) before the string breaks.
- M1: For calculating the deceleration \(a' = -2.5 \text{ m s}^{-2}\) after the string breaks.
- M1: For using kinematic equations to find the stopping distance \(s_2 = 1.8\text{ m}\).
- A1: For finding the total distance as \(3.6\text{ m}\).

(i) Since the car is moving up the hill at a constant speed, the acceleration is zero.
Let the driving force of the engine be \(DF\).
Resolving parallel to the incline:
\(DF = R + mg \sin \theta\)
\(DF = 400 + 1200(10)(0.05) = 400 + 600 = 1000\text{ N}\).

Power \(P = DF \times v = 1000 \times 15 = 15000\text{ W} = 15\text{ kW}\).

(ii) When the power is increased to \(20 \text{ kW} = 20000\text{ W}\) and the speed is \(16 \text{ m s}^{-1}\):
New driving force \(DF' = \frac{P_{\text{new}}}{v} = \frac{20000}{16} = 1250\text{ N}\).

Using Newton's second law along the incline:
\(DF' - R - mg \sin \theta = ma\)
\(1250 - 400 - 600 = 1200a\)
\(250 = 1200a\)
\(a = \frac{250}{1200} = \frac{5}{24} \approx 0.208 \text{ m s}^{-2}\).

Part (i):
- M1: For resolving forces along the plane to find the driving force.
- A1: For obtaining \(DF = 1000\text{ N}\).
- A1: For calculating the power as \(15\text{ kW}\) (or \(15000\text{ W}\)).

Part (ii):
- M1: For using \(P = DF \times v\) to find the new driving force at the given speed.
- A1: For obtaining \(DF' = 1250\text{ N}\).
- M1: For setting up the Newton's second law equation \(DF' - R - mg\sin\theta = ma\).
- A1: For obtaining \(a = 0.208 \text{ m s}^{-2}\) (or \(\frac{5}{24}\)).

(i) Let the direction of the initial velocity of \(A\) be positive.
Initial velocities: \(u_A = 4 \text{ m s}^{-1}\), \(u_B = -2 \text{ m s}^{-1}\).
Since both directions are reversed post-collision:
Final velocities: \(v_A = -1 \text{ m s}^{-1}\), \(v_B = 1.5 \text{ m s}^{-1}\).

By Conservation of Linear Momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\(m(4) + 3(-2) = m(-1) + 3(1.5)\)
\(4m - 6 = -m + 4.5\)
\(5m = 10.5\)
\(m = 2.1\text{ kg}\).

(ii) Kinetic energy before collision:
\(E_{\text{initial}} = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2\)
\(E_{\text{initial}} = \frac{1}{2}(2.1)(4)^2 + \frac{1}{2}(3)(-2)^2 = 16.8 + 6 = 22.8\text{ J}\).

Kinetic energy after collision:
\(E_{\text{final}} = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2\)
\(E_{\text{final}} = \frac{1}{2}(2.1)(-1)^2 + \frac{1}{2}(3)(1.5)^2 = 1.05 + 3.375 = 4.425\text{ J}\).

Loss of kinetic energy:
\(E_{\text{loss}} = 22.8 - 4.425 = 18.375\text{ J}\) (or \(18.4\text{ J}\)).

Part (i):
- M1: For using the principle of conservation of linear momentum with correct signs for velocities.
- A1: For writing a correct equation like \(4m - 6 = -m + 4.5\).
- A1: For finding \(m = 2.1\).

Part (ii):
- M1: For calculating the total initial kinetic energy.
- A1: For obtaining \(22.8\text{ J}\).
- M1: For calculating the total final kinetic energy.
- A1: For obtaining \(18.4\text{ J}\) (or \(18.375\text{ J}\)) as the loss in kinetic energy.

The team of 5 players must satisfy two conditions:
1. At least 3 field hockey players (out of 6 available).
2. At least 1 ice hockey player (out of 5 available).

The possible compositions of (field hockey, ice hockey) players are:
- **Case 1**: 3 field hockey players and 2 ice hockey players.
Number of ways: \(\binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200\)
- **Case 2**: 4 field hockey players and 1 ice hockey player.
Number of ways: \(\binom{6}{4} \times \binom{5}{1} = 15 \times 5 = 75\)
- **Case 3**: 5 field hockey players and 0 ice hockey players.
This is not allowed, since we must have at least 1 ice hockey player.

Total number of ways = \(200 + 75 = 275\).

M1 for identifying the two valid cases: (3 FH, 2 IH) and (4 FH, 1 IH).
M1 for calculating the combinations for both cases correctly.
A1 for the correct final answer of 275.

We are given:
\(\text{P}(A) = 0.6\)
\(\text{P}(B) = 0.45\)
\(\text{P}(A \cup B) = 0.8\)

Using the addition rule of probability:
\(\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B)\)
\(0.8 = 0.6 + 0.45 - \text{P}(A \cap B)\)
\(\text{P}(A \cap B) = 1.05 - 0.8 = 0.25\)

We need to find \(\text{P}(A' | B)\):
\(\text{P}(A' | B) = \frac{\text{P}(A' \cap B)}{\text{P}(B)}\)

Since \(\text{P}(A' \cap B) = \text{P}(B) - \text{P}(A \cap B)\):
\(\text{P}(A' \cap B) = 0.45 - 0.25 = 0.2\)

Therefore,
\(\text{P}(A' | B) = \frac{0.2}{0.45} = \frac{4}{9}\) (or \(0.444\) to 3 significant figures).

M1 for finding \(\text{P}(A \cap B) = 0.25\) using the addition rule.
M1 for applying the conditional probability formula for \(\text{P}(A' | B)\) correctly.
A1 for obtaining \(\frac{4}{9}\) or \(0.444\).

First, we use the expectation \(\text{E}(X)\) to find \(k\):
\(\text{E}(X) = \sum x \cdot \text{P}(X=x)\)
\(1.75 = -2(0.15) + 1(0.4) + 3(0.3) + k(0.15)\)
\(1.75 = -0.3 + 0.4 + 0.9 + 0.15k\)
\(1.75 = 1.0 + 0.15k\)
\(0.75 = 0.15k \implies k = 5\)

Next, we calculate \(\text{E}(X^2)\):
\(\text{E}(X^2) = \sum x^2 \cdot \text{P}(X=x)\)
\(\text{E}(X^2) = (-2)^2(0.15) + 1^2(0.4) + 3^2(0.3) + 5^2(0.15)\)
\(\text{E}(X^2) = 4(0.15) + 1(0.4) + 9(0.3) + 25(0.15)\)
\(\text{E}(X^2) = 0.6 + 0.4 + 2.7 + 3.75 = 7.45\)

Now, we calculate the variance \(\text{Var}(X)\):
\(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\)
\(\text{Var}(X) = 7.45 - 1.75^2 = 7.45 - 3.0625 = 4.3875\) (or \(\frac{351}{80}\)).

M1 for setting up the expectation equation and solving for \(k = 5\).
M1 for calculating \(\text{E}(X^2) = 7.45\).
A1 for finding the correct variance of \(4.3875\) (or \(\frac{351}{80}\)).

(i) The total number of ways to choose 5 people from the 12 available is:
\(\binom{12}{5} = 792\)

We must exclude selections that do not meet the criteria:
- Selections with no doctors (only chosen from the 4 nurses and 3 administrators): \(\binom{7}{5} = 21\) ways.
- Selections with no nurses (only chosen from the 5 doctors and 3 administrators): \(\binom{8}{5} = 56\) ways.
- Selections with neither doctors nor nurses is impossible since there are only 3 administrators in total.

Thus, the number of successful committee selections is:
\(792 - 21 - 56 = 715\)

(ii) First, treat the 3 administrators as a single block \(A_{\text{block}}\). Within this block, the administrators can be arranged in:
\(3! = 6\) ways.

This block and the 5 doctors form 6 units to be arranged in a line:
\(6! = 720\) ways.

These 6 units create 7 available spaces (before, between, and after) to insert the 4 nurses:
_ U1 _ U2 _ U3 _ U4 _ U5 _ U6 _

We choose 4 spaces out of the 7 for the nurses:
\(\binom{7}{4} = 35\) ways.

The 4 nurses can be arranged among themselves in these chosen spaces in:
\(4! = 24\) ways.

Multiplying these together gives the total number of arrangements:
\(3! \times 6! \times \binom{7}{4} \times 4! = 6 \times 720 \times 35 \times 24 = 3,628,800\)

(iii) To arrange the 5 doctors such that Alice (A), Bob (B), and Carol (C) are not adjacent, we first arrange the other 2 doctors (D4, D5) in:
\(2! = 2\) ways.

They create 3 spaces:
_ D4 _ D5 _

We must place Alice, Bob, and Carol in these 3 spaces. The number of ways to arrange them is:
\(3! = 6\) ways.

Total number of ways is:
\(2! \times 3! = 2 \times 6 = 12\)

(i)
- M1: Attempting to calculate total combinations and subtracting combinations with no doctors and no nurses.
- A1: Correctly calculating \(\binom{12}{5} = 792\), \(\binom{7}{5} = 21\), and \(\binom{8}{5} = 56\).
- A1: Correct subtraction of 21 and 56 from 792.
- A1: Final answer 715.

(ii)
- M1: Treating the 3 administrators as a block and arranging \(3!\).
- M1: Multiplying by \(6!\) for the arrangement of the block and 5 doctors.
- M1: Using the insertion method with 7 spaces, multiplying by \(\binom{7}{4} \times 4!\).
- A1: Final answer 3,628,800.

(iii)
- M1: Arranging the 2 non-specific doctors in \(2!\) ways.
- A1: Arranging the 3 specific doctors in the 3 gaps in \(3!\) ways.
- A0.2: Correct final answer 12.

(i) We are given:
\(P(X < 93.5) = 0.0968\)
Standardising:
\(P\left(Z < \frac{93.5 - \mu}{\sigma}\right) = 0.0968\)
Using the normal distribution table, \(\Phi(1.300) = 1 - 0.0968 = 0.9032\). Since \(0.0968 < 0.5\), the z-score is negative:
\(\frac{93.5 - \mu}{\sigma} = -1.300 \implies \mu - 1.3\sigma = 93.5\) — (Equation 1)

We are also given:
\(P(X > 104.2) = 0.2005 \implies P(X < 104.2) = 0.7995\)
Standardising:
\(P\left(Z < \frac{104.2 - \mu}{\sigma}\right) = 0.7995\)
Using the normal table, \(\Phi(0.840) = 0.7995\):
\(\frac{104.2 - \mu}{\sigma} = 0.840 \implies \mu + 0.84\sigma = 104.2\) — (Equation 2)

Subtracting Equation 1 from Equation 2:
\((\mu + 0.84\sigma) - (\mu - 1.3\sigma) = 104.2 - 93.5\)
\(2.14\sigma = 10.7 \implies \sigma = 5.00\)

Substituting \(\sigma = 5.00\) back into Equation 2:
\[\mu + 0.84(5.00) = 104.2 \implies \mu = 100.0\]

(ii) We want to find \(P(98 < X < 105)\) given \(\mu = 100\) and \(\sigma = 5\):
\(P(98 < X < 105) = P\left(\frac{98 - 100}{5} < Z < \frac{105 - 100}{5}\right) = P(-0.4 < Z < 1.0)\)
\(= \Phi(1.0) - \Phi(-0.4)\)
\(= \Phi(1.0) - (1 - \Phi(0.4))\)
\(= 0.8413 - (1 - 0.6554)\)
\(= 0.8413 - 0.3446 = 0.4967\) (or 0.497 to 3 s.f.).

(iii) Let \(Y\) be the number of chocolate bars with a mass between \(98\text{ g}\) and \(105\text{ g}\).
Then \(Y \sim \text{B}(5, 0.4967)\).
We require \(P(Y = 3)\):
\(P(Y = 3) = \binom{5}{3} (0.4967)^3 (1 - 0.4967)^2\)
\(= 10 \times (0.4967)^3 \times (0.5033)^2\)
\(= 10 \times 0.12255 \times 0.25331 \approx 0.310\) (to 3 s.f.).

(i)
- M1: Setting up an equation with standardisation for \(X < 93.5\) using a z-score.
- A1: Obtaining \(\mu - 1.3\sigma = 93.5\).
- M1: Setting up an equation with standardisation for \(X > 104.2\) using a z-score.
- A1: Solving simultaneous equations to find \(\sigma = 5.00\).
- A1: Finding \(\mu = 100.0\).

(ii)
- M1: Standardising both limits with their values of \(\mu\) and \(\sigma\).
- M1: Showing correct use of normal tables for \(\Phi(1.0) - (1 - \Phi(0.4))\).
- A1: Correct probability 0.497 (or 0.4967).

(iii)
- M1: Using Binomial probability formula \(\binom{5}{3} p^3 (1-p)^2\).
- A1: Substituting their probability from part (ii).
- A0.2: Correct final answer 0.310 (or 0.311 from 0.497).

(i) Total number of balls is \(4 + 3 + 2 = 9\).
We select 3 balls without replacement. The total number of ways to choose 3 balls from 9 is:
\(\binom{9}{3} = 84\)

To choose exactly 1 red, 1 blue, and 1 green ball:
\(\binom{4}{1} \times \binom{3}{1} \times \binom{2}{1} = 4 \times 3 \times 2 = 24\) ways.

Thus, the probability is:
\(P(\text{all different}) = \frac{24}{84} = \frac{2}{7}\) (or 0.286 to 3 s.f.).

(ii) Given that the first ball was red, there are now 8 balls remaining in the bag: 3 red, 3 blue, and 2 green. We draw two more balls.
Let \(B_3\) be the event that the third ball is blue.
The second ball drawn can be Red, Blue, or Green.
- Case 1: 2nd is Red. Prob is \(\frac{3}{8}\). 3rd is Blue with prob \(\frac{3}{7}\).
- Case 2: 2nd is Blue. Prob is \(\frac{3}{8}\). 3rd is Blue with prob \(\frac{2}{7}\).
- Case 3: 2nd is Green. Prob is \(\frac{2}{8}\). 3rd is Blue with prob \(\frac{3}{7}\).

\(P(B_3 | R_1) = \left(\frac{3}{8} \times \frac{3}{7}\right) + \left(\frac{3}{8} \times \frac{2}{7}\right) + \left(\frac{2}{8} \times \frac{3}{7}\right)\)
\(= \frac{9}{56} + \frac{6}{56} + \frac{6}{56} = \frac{21}{56} = \frac{3}{8}\) (or 0.375).

(Alternatively: Since the second draw is unobserved, the probability of the third ball being blue is simply the proportion of blue balls remaining after the first draw, which is \(\frac{3}{8}\)).

(iii) The random variable \(R\) can take values 0, 1, 2, or 3.
- \(P(R = 0)\): Choosing 3 non-red balls (from the 5 blue/green balls):
\(P(R = 0) = \frac{\binom{5}{3}}{\binom{9}{3}} = \frac{10}{84} = \frac{5}{42}\)

- \(P(R = 1)\): Choosing 1 red ball and 2 non-red balls:
\(P(R = 1) = \frac{\binom{4}{1} \times \binom{5}{2}}{\binom{9}{3}} = \frac{4 \times 10}{84} = \frac{40}{84} = \frac{10}{21}\)

- \(P(R = 2)\): Choosing 2 red balls and 1 non-red ball:
\(P(R = 2) = \frac{\binom{4}{2} \times \binom{5}{1}}{\binom{9}{3}} = \frac{6 \times 5}{84} = \frac{30}{84} = \frac{5}{14}\)

- \(P(R = 3)\): Choosing 3 red balls:
\(P(R = 3) = \frac{\binom{4}{3}}{\binom{9}{3}} = \frac{4}{84} = \frac{1}{21}\)

Therefore, the probability distribution of \(R\) is:

\[
\begin{array}{c|c|c|c|c}
r & 0 & 1 & 2 & 3 \\
\hline
P(R = r) & \frac{5}{42} & \frac{10}{21} & \frac{5}{14} & \frac{1}{21}
\end{array}
\]

To find \(\text{E}(R)\):
\(\text{E}(R) = \sum r P(R=r) = 0\left(\frac{5}{42}\right) + 1\left(\frac{10}{21}\right) + 2\left(\frac{5}{14}\right) + 3\left(\frac{1}{21}\right)\)
\(= \frac{10}{21} + \frac{10}{14} + \frac{3}{21}\)
\(= \frac{13}{21} + \frac{15}{21} = \frac{28}{21} = \frac{4}{3}\) (or 1.33).

(i)
- M1: Finding total combinations \(\binom{9}{3} = 84\) (or attempting probabilities for individual routes).
- M1: Finding required combinations \(\binom{4}{1}\binom{3}{1}\binom{2}{1} = 24\).
- A1: Correct probability \(\frac{2}{7}\).

(ii)
- M1: Identifying that 8 balls remain with 3 blue balls.
- M1: Summing the probabilities of the three paths or using symmetry.
- A1: Correct probability \(\frac{3}{8}\).

(iii)
- M1: Calculating probabilities for at least three of the values \(r = 0, 1, 2, 3\).
- A1: Correct probabilities for all four outcomes.
- M1: Correct formula for \(\text{E}(R) = \sum r P(R = r)\).
- A1.2: Correct expectation \(\frac{4}{3}\).

(i) Group A data values in ascending order are:
22, 25, 28, 31, 33, 36, 39, 40, 44, 47, 53
Number of values, \(n = 11\).

- Median is the 6th value:
\(\text{Median} = 36\)

- Lower quartile (\(Q_1\)) is the median of the lower half (22, 25, 28, 31, 33):
\(Q_1 = 28\)

- Upper quartile (\(Q_3\)) is the median of the upper half (39, 40, 44, 47, 53):
\(Q_3 = 44\)

- Interquartile Range (\(IQR\)) of Group A:
\(IQR = Q_3 - Q_1 = 44 - 28 = 16\)

(ii) Group B data values in ascending order are:
12, 15, 24, 27, 29, 30, 32, 35, 38, 41, 43, 52
Number of values, \(n = 12\).

- Median is the average of the 6th and 7th values:
\(\text{Median} = \frac{30 + 32}{2} = 31\)

- Lower quartile (\(Q_1\)) is the median of the lower half (12, 15, 24, 27, 29, 30):
\(Q_1 = \frac{24 + 27}{2} = 25.5\)

- Upper quartile (\(Q_3\)) is the median of the upper half (32, 35, 38, 41, 43, 52):
\(Q_3 = \frac{38 + 41}{2} = 39.5\)

- Interquartile Range (\(IQR\)) of Group B:
\(IQR = Q_3 - Q_1 = 39.5 - 25.5 = 14\)

(iii) The mean time of Group B is:
\(\text{Mean} = \frac{12 + 15 + 24 + 27 + 29 + 30 + 32 + 35 + 38 + 41 + 43 + 52}{12} = \frac{378}{12} = 31.5\text{ minutes}\)

Comparing consistency:
- Using IQR: Group B is more consistent because its IQR (14 minutes) is smaller than Group A's IQR (16 minutes).
- Using Standard Deviation: The standard deviation of Group B is 10.9 minutes (variance is 119.6). Since Group A's standard deviation (9.11 minutes) is smaller than Group B's, Group A is more consistent.

(Both comparisons are acceptable if supported by correct values).

(i)
- M1: Listing Group A values in order.
- A1: Finding correct median = 36.
- M1: Finding correct quartiles (\(Q_1 = 28\), \(Q_3 = 44\)).
- A1: Finding IQR = 16.

(ii)
- M1: Finding the median as the average of 30 and 32 to get 31.
- M1: Finding the lower quartile \(Q_1 = 25.5\) and upper quartile \(Q_3 = 39.5\).
- A1: Finding IQR = 14.
- A1: Correct values for all of them.

(iii)
- B1: Finding mean of Group B = 31.5.
- B1.2: Correct comparison of consistency based on either IQR (Group B is more consistent because 14 < 16) or standard deviation (Group A is more consistent because 9.11 < 10.9).

We need to find \( P(X > 1.5) = \int_{1.5}^{3} \frac{1}{9}(4x - x^2) \\, dx \). First, find the indefinite integral: \( \int \frac{1}{9}(4x - x^2) \\, dx = \frac{1}{9} \left( 2x^2 - \frac{1}{3}x^3 \right) + c \). Evaluating from 1.5 to 3: At \( x = 3 \): \( \frac{1}{9} \left( 2(3)^2 - \frac{1}{3}(3)^3 \right) = \frac{1}{9}(18 - 9) = 1 \). At \( x = 1.5 = \frac{3}{2} \): \( \frac{1}{9} \left( 2\left(\frac{3}{2}\right)^2 - \frac{1}{3}\left(\frac{3}{2}\right)^3 \right) = \frac{1}{9} \left( \frac{9}{2} - \frac{9}{8} \right) = \frac{1}{9} \left( \frac{27}{8} \right) = \frac{3}{8} = 0.375 \). Therefore, \( P(X > 1.5) = 1 - 0.375 = 0.625 \) (or \( \frac{5}{8} \)).

M1: For attempting to integrate \( f(x) \) with correct limits. A1: For obtaining the correct integrated expression \( \frac{1}{9}(2x^2 - \frac{1}{3}x^3) \). M1: For substituting the limits 1.5 and 3 correctly into their integrated function. A1: For obtaining the correct final probability of 0.625 (or \( \frac{5}{8} \)).

Let \( Y \) be the number of particles emitted in a 2-minute interval. Since the rate is 2.4 per minute, for 2 minutes we have \( \lambda = 2.4 \times 2 = 4.8 \). Thus, \( Y \sim \text{Po}(4.8) \). We want to find the conditional probability \( P(Y \ge 2 \mid Y \le 3) = \frac{P(2 \le Y \le 3)}{P(Y \le 3)} \). First, calculate the individual probabilities: \( P(Y = 0) = e^{-4.8} \), \( P(Y = 1) = 4.8e^{-4.8} \), \( P(Y = 2) = \frac{4.8^2}{2} e^{-4.8} = 11.52e^{-4.8} \), \( P(Y = 3) = \frac{4.8^3}{6} e^{-4.8} = 18.432e^{-4.8} \). The numerator is \( P(2 \le Y \le 3) = P(Y=2) + P(Y=3) = e^{-4.8}(11.52 + 18.432) = 29.952 e^{-4.8} \). The denominator is \( P(Y \le 3) = P(Y=0) + P(Y=1) + P(Y=2) + P(Y=3) = e^{-4.8}(1 + 4.8 + 11.52 + 18.432) = 35.752 e^{-4.8} \). The conditional probability is \( \frac{29.952 e^{-4.8}}{35.752 e^{-4.8}} = \frac{29.952}{35.752} \approx 0.83777 \). To 3 significant figures, the probability is 0.838.

B1: For identifying the correct parameter \( \lambda = 4.8 \). M1: For expressing the conditional probability formula \( \frac{P(2 \le Y \le 3)}{P(Y \le 3)} \) or equivalent. M1: For attempting to calculate the required Poisson terms \( P(Y=0), P(Y=1), P(Y=2), P(Y=3) \). A1: For obtaining the correct final probability of 0.838.

Let \( \mu \) be the population mean weight of the boxes. We state the hypotheses: \( H_0: \mu = 500 \), \( H_1: \mu < 500 \). Under the null hypothesis, the sample mean \( \bar{X} \) has distribution \( \bar{X} \sim N\left(500, \frac{12^2}{40}\right) \). The standard error is \( \frac{12}{\sqrt{40}} \approx 1.8974 \). The test statistic is \( z = \frac{496.2 - 500}{1.8974} \approx -2.003 \). For a one-tailed test at the 5% level of significance, the critical value is \( -1.645 \). Since \( -2.003 < -1.645 \), the test statistic falls in the critical region. Therefore, we reject \( H_0 \). There is significant evidence at the 5% level to suggest that the machine is underfilling the boxes.

B1: For correctly stating both null and alternative hypotheses. M1: For calculating the standard error and standardizing to find the z-value. M1: For comparing the z-value with the critical value -1.645 (or finding p-value \\approx 0.0226 and comparing with 0.05). A1: For a correct conclusion in context, rejecting \( H_0 \).

\(X \sim N(120, 5^2)\) and \(Y \sim N(450, 12^2)\).

(i) Let \(W = X_1 + X_2 + X_3 + X_4\) be the total weight of 4 pastries.
\(E(W) = 4 \times 120 = 480\)
\(Var(W) = 4 \times 5^2 = 100\)
So \(W \sim N(480, 100)\).
We want \(P(W > 490)\):
\(Z = \frac{490 - 480}{\sqrt{100}} = \frac{10}{10} = 1.0\)
\(P(Z > 1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587 \approx 0.159\).

(ii) We want \(P(Y > 3.6X) \Rightarrow P(Y - 3.6X > 0)\).
Let \(D = Y - 3.6X\).
\(E(D) = E(Y) - 3.6E(X) = 450 - 3.6(120) = 450 - 432 = 18\)
\(Var(D) = Var(Y) + 3.6^2 Var(X) = 12^2 + 12.96 \times 25 = 144 + 324 = 468\)
So \(D \sim N(18, 468)\).
We want \(P(D > 0)\):
\(Z = \frac{0 - 18}{\sqrt{468}} = \frac{-18}{21.633} \approx -0.832\)
\(P(Z > -0.832) = \Phi(0.832) \approx 0.797\).

(iii) Let \(T = Y + X_1 + X_2\) be the total weight of the box contents.
\(E(T) = E(Y) + 2E(X) = 450 + 240 = 690\)
\(Var(T) = Var(Y) + 2Var(X) = 144 + 2(25) = 194\)
So \(T \sim N(690, 194)\).
We want \(P(T < 680)\):
\(Z = \frac{680 - 690}{\sqrt{194}} = \frac{-10}{13.928} \approx -0.718\)
\(P(Z < -0.718) = 1 - \Phi(0.718) = 1 - 0.7636 = 0.2364 \approx 0.236\).

(i)
M1: For finding correct mean and variance of 4 pastries (480 and 100).
A1: For standardizing correctly with their mean and variance.
A1: For obtaining correct final probability 0.159 (or 0.1587).

(ii)
M1: For identifying the combination \(Y - 3.6X\) and finding its mean (18).
M1: For finding the correct variance of \(Y - 3.6X\) (i.e. \(12^2 + 3.6^2 \times 5^2 = 468\)).
M1: For standardizing and calculating the probability.
A1: For obtaining correct final probability 0.797.

(iii)
M1: For finding correct mean and variance of \(Y + X_1 + X_2\) (690 and 194).
M1: For standardizing with their values.
A1: For obtaining correct final probability 0.236.

(i) Since \(f(x)\) is a probability density function, the total area under the curve must be 1:
\(\int_0^3 k x^2 (3 - x) dx = 1\)
\(k \int_0^3 (3x^2 - x^3) dx = 1\)
\(k \left[ x^3 - \frac{x^4}{4} \right]_0^3 = 1\)
\(k \left( 27 - \frac{81}{4} \right) = 1\)
\(k \left( \frac{108 - 81}{4} \right) = 1 \Rightarrow k \left( \frac{27}{4} \right) = 1 \Rightarrow k = \frac{4}{27}\). (Shown)

(ii) \(E(X) = \int_0^3 x f(x) dx = \frac{4}{27} \int_0^3 (3x^3 - x^4) dx\)
\(= \frac{4}{27} \left[ \frac{3x^4}{4} - \frac{x^5}{5} \right]_0^3\)
\(= \frac{4}{27} \left( \frac{243}{4} - \frac{243}{5} \right)
= \frac{4}{27} \times 243 \left( \frac{1}{4} - \frac{1}{5} \right)
= 36 \times \frac{1}{20} = 1.8\).

(iii) We want to find \(P(X > 1.8)\):
\(P(X > 1.8) = \int_{1.8}^3 \frac{4}{27} (3x^2 - x^3) dx\)
\(= \frac{4}{27} \left[ x^3 - \frac{x^4}{4} \right]_{1.8}^3\)
\(= \frac{4}{27} \left( \left( 27 - \frac{81}{4} \right) - \left( 1.8^3 - \frac{1.8^4}{4} \right) \right)
= \frac{4}{27} \left( 6.75 - (5.832 - 2.6244) \right)
= \frac{4}{27} \left( 6.75 - 3.2076 \right)
= \frac{4}{27} \times 3.5424 = 0.5248 \approx 0.525\).

(i)
M1: For integrating the function \(k(3x^2 - x^3)\) with limits 0 and 3.
A1: For obtaining the correct integrated expression \(k [x^3 - x^4/4]\) and substituting the limits to get \(27/4 k\).
A1: For setting the integral equal to 1 and showing \(k = 4/27\) with clear, convincing steps.

(ii)
M1: For setting up the integral for \(E(X)\) as \(\int x f(x) dx\).
M1: For integrating correctly to obtain \(\frac{4}{27} [\frac{3x^4}{4} - \frac{x^5}{5}]\).
A1: For substituting the limits 0 and 3 correctly to obtain 1.8.

(iii)
M1: For setting up the integral \(\int_{1.8}^3 f(x) dx\).
M1: For using the correct integration from part (i) on the new limits.
M1: For evaluating the expression correctly at 1.8 (obtaining 3.2076) and at 3 (obtaining 6.75).
A1: For obtaining correct final probability 0.525 (or 0.5248).

(i) \(H_0: \mu = 15\)
\(H_1: \mu > 15\)

(ii) Sample size \(n = 40\), sample mean \(\bar{x} = 16.8\).
Since \(n\) is large, we can use the sample standard deviation \(s = 5.4\) as an estimate of the population standard deviation \(\sigma\).
Standard error of the mean = \(\frac{s}{\sqrt{n}} = \frac{5.4}{\sqrt{40}} \approx 0.8538\).
Under \(H_0\), the test statistic \(z\) is:
\(z = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{16.8 - 15}{0.8538} = \frac{1.8}{0.8538} \approx 2.108\).
At the 5% level of significance for a one-tailed test, the critical value of \(z\) is 1.645.
Since \(2.108 > 1.645\), we reject \(H_0\).
There is significant evidence at the 5% level to support the advocate group's claim that the mean waiting time is greater than 15 minutes.

(iii) Since we rejected the null hypothesis \(H_0\), we could have made a Type I error (rejecting a true null hypothesis).

(iv) The Central Limit Theorem is necessary because the distribution of the individual waiting times is unknown and not specified as normal. The CLT allows us to assume that the sample mean \(\bar{X}\) is approximately normally distributed because the sample size (\(40\)) is large (\(n \ge 30\)). Therefore, we do not need to assume that the population of individual waiting times is normally distributed.

(i)
B1: For stating both hypotheses correctly using \(\mu\) (accept \(H_0: \mu = 15\) and \(H_1: \mu > 15\)).

(ii)
M1: For calculating the standard error \(\frac{5.4}{\sqrt{40}}\) (approx 0.854).
M1: For calculating the test statistic \(z\) using their standard error.
A1: For obtaining \(z \approx 2.11\) (or 2.108).
M1: For comparing their test statistic with the critical value 1.645 (or comparing the p-value 0.0175 with 0.05).
A1: For correctly concluding to reject \(H_0\) in context (i.e. support the claim that the mean waiting time is greater than 15 minutes).

(iii)
B1: For stating "Type I error" because \(H_0\) was rejected.

(iv)
B1: For stating that the distribution of individual waiting times is unknown.
B1: For stating that the Central Limit Theorem applies because the sample size \(n = 40\) is large (or \(\ge 30\)).
B1: For stating that it was unnecessary to assume that the population was normally distributed.

(i) Let \(X\) be the number of vehicles arriving per minute. \(X \sim \text{Po}(4.5)\).
\(P(X = 6) = \frac{e^{-4.5} \times 4.5^6}{6!} \approx 0.1281 \approx 0.128\).

(ii) Let \(Y\) be the number of vehicles arriving in a 30-second interval.
\(\lambda = 4.5 \times 0.5 = 2.25\), so \(Y \sim \text{Po}(2.25)\).
We want \(P(Y \ge 3) = 1 - P(Y \le 2) = 1 - e^{-2.25} \left(1 + 2.25 + \frac{2.25^2}{2}\right)\)
\(= 1 - e^{-2.25} (1 + 2.25 + 2.53125) = 1 - e^{-2.25} (5.78125)\)
\(\approx 1 - 0.1054 \times 5.78125 = 1 - 0.6093 = 0.3907 \approx 0.391\).

(iii) Let \(W\) be the number of vehicles arriving in a 1-hour (60-minute) period.
\(\lambda = 4.5 \times 60 = 270\), so \(W \sim \text{Po}(270)\).
Since \(\lambda > 15\), we can approximate \(W\) using a normal distribution: \(W \approx N(270, 270)\).
We want \(P(W < 250) = P(W \le 249)\).
Applying a continuity correction:
\(P(W \le 249) \approx P(W_{\text{normal}} < 249.5)\).
\(Z = \frac{249.5 - 270}{\sqrt{270}} = \frac{-20.5}{16.4317} \approx -1.248\).
\(P(Z < -1.248) = 1 - \Phi(1.248) \approx 1 - 0.8940 = 0.106\).

(i)
M1: For using Poisson formula with \(\lambda = 4.5\) and \(x = 6\).
A1: For obtaining 0.128 (or 0.1281).

(ii)
M1: For identifying \(\lambda = 2.25\) as the parameter for a 30-second interval.
M1: For expressing \(P(Y \ge 3)\) as \(1 - P(Y \le 2)\) and writing out the correct terms.
A1: For obtaining 0.391 (or 0.3907).

(iii)
M1: For finding the correct mean \(\lambda = 270\) and stating the normal approximation \(N(270, 270)\).
M1: For applying a continuity correction to get 249.5.
M1: For standardizing with their mean and variance.
M1: For calculating the correct probability using standard normal tables.
A1: For obtaining 0.106 (accept 0.106 to 0.1065).