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To find the condition for real roots, we use the discriminant \( b^2 - 4ac \ge 0 \).

(a) Here \( a = k \), \( b = 6 \), and \( c = k - 8 \).
Substituting these into the discriminant:
\( 6^2 - 4(k)(k - 8) \ge 0 \)
\( 36 - 4k^2 + 32k \ge 0 \)
Divide the entire inequality by \(-4\) and reverse the inequality sign:
\( k^2 - 8k - 9 \le 0 \) (as required).

(b) To solve \( k^2 - 8k - 9 \le 0 \), first find the critical values by setting \( k^2 - 8k - 9 = 0 \):
\( (k - 9)(k + 1) = 0 \implies k = 9 \text{ or } k = -1 \).
Since the inequality is \( \le 0 \), the range of values is \( -1 \le k \le 9 \).
Since \( k \) is a non-zero constant (as it is the coefficient of \( x^2 \)), we must exclude \( k = 0 \).
Thus, the set of possible values for \( k \) is \( -1 \le k \le 9, k \neq 0 \) (or in interval notation, \( [-1, 0) \cup (0, 9] \)).

(a) M1: For identifying coefficients \( a = k \), \( b = 6 \), \( c = k - 8 \) and using \( b^2 - 4ac \ge 0 \).
M1: For expanding \( 4k(k-8) \) correctly to get \( 36 - 4k^2 + 32k \ge 0 \).
A1: For completing the proof to show \( k^2 - 8k - 9 \le 0 \) with no errors in algebra or inequality signs.

(b) M1: For attempting to factorise or solve the quadratic \( k^2 - 8k - 9 = 0 \) to find critical values.
A1: For finding the correct critical values \( k = 9 \) and \( k = -1 \).
M1: For choosing the inside region \( -1 \le k \le 9 \).
A1.5: For identifying that \( k \neq 0 \) because the original equation is quadratic, leading to the final range \( -1 \le k \le 9, k \neq 0 \).

From the linear equation, we can express \( y \) in terms of \( x \):
\( y = 2x + 1 \).
Substitute this into the quadratic equation:
\( x^2 + (2x + 1)^2 = 7 \)
\( x^2 + 4x^2 + 4x + 1 = 7 \)
\( 5x^2 + 4x - 6 = 0 \).
Now we use the quadratic formula to solve for \( x \):
\( x = \frac{-4 \pm \sqrt{4^2 - 4(5)(-6)}}{2(5)} \)
\( x = \frac{-4 \pm \sqrt{16 + 120}}{10} = \frac{-4 \pm \sqrt{136}}{10} \)
Since \( \sqrt{136} = \sqrt{4 \times 34} = 2\sqrt{34} \):
\( x = \frac{-4 \pm 2\sqrt{34}}{10} = \frac{-2 \pm \sqrt{34}}{5} \).
Now substitute \( x \) back to find \( y \):
\( y = 2x + 1 = 2\left(\frac{-2 \pm \sqrt{34}}{5}\right) + 1 \)
\( y = \frac{-4 \pm 2\sqrt{34} + 5}{5} = \frac{1 \pm 2\sqrt{34}}{5} \).
Therefore, \( x = \frac{-2 \pm \sqrt{34}}{5} \) paired with \( y = \frac{1 \pm 2\sqrt{34}}{5} \).

M1: Rearranging the linear equation to make \( y \) or \( x \) the subject, e.g., \( y = 2x + 1 \).
M1: Substituting the expression into the quadratic equation to form an equation in one variable.
A1: Obtaining the correct simplified quadratic equation, e.g., \( 5x^2 + 4x - 6 = 0 \).
M1: Applying the quadratic formula correctly to their quadratic equation.
A1: Finding the correct simplified values for \( x = \frac{-2 \pm \sqrt{34}}{5} \).
M1: Substituting their \( x \) values back into \( y = 2x + 1 \) to find \( y \).
A1.5: Finding the correct paired values of \( y = \frac{1 \pm 2\sqrt{34}}{5} \) and presenting paired coordinates or clearly corresponding solutions.

(a) Rearranging \( l_1 \): \( y = 2x - 1 \). The gradient of \( l_1 \) is \( m_1 = 2 \).
Since \( l_2 \) is perpendicular to \( l_1 \), its gradient is \( m_2 = -\frac{1}{2} \).
The equation of \( l_2 \) passing through \( A(6, 1) \) is:
\( y - 1 = -\frac{1}{2}(x - 6) \)
Multiply by 2:
\( 2y - 2 = -x + 6 \implies x + 2y - 8 = 0 \).

(b) To find the intersection point \( B \), we solve the simultaneous equations:
1) \( y = 2x - 1 \)
2) \( x + 2y - 8 = 0 \)
Substituting (1) into (2):
\( x + 2(2x - 1) - 8 = 0 \)
\( x + 4x - 2 - 8 = 0 \implies 5x = 10 \implies x = 2 \).
Substitute \( x = 2 \) back into (1):
\( y = 2(2) - 1 = 3 \).
So, the coordinates of \( B \) are \( (2, 3) \).

(c) The distance \( AB \) between \( A(6, 1) \) and \( B(2, 3) \) is:
\( AB = \sqrt{(6 - 2)^2 + (1 - 3)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \).

(a) M1: Finding the gradient of \( l_1 \) is 2, hence gradient of perpendicular line is \( -\frac{1}{2} \).
M1: Writing down the equation of the line through \( (6, 1) \) with gradient \( -\frac{1}{2} \).
A1: Rearranging to the correct integer form \( x + 2y - 8 = 0 \) (or any integer multiple thereof).

(b) M1: Setting up and attempting to solve the simultaneous equations by substitution or elimination.
A1: Finding the correct x-coordinate \( x = 2 \).
A0.5: Finding the correct y-coordinate \( y = 3 \).

(c) M1: Using the distance formula with coordinates of \( A \) and \( B \).
A1: Simplifying to the correct exact form \( 2\sqrt{5} \) (accept \( \sqrt{20} \)).

(a) Using the identity \( \cos^2 x = 1 - \sin^2 x \):
\( 4(1 - \sin^2 x) + 9\sin x - 6 = 0 \)
\( 4 - 4\sin^2 x + 9\sin x - 6 = 0 \)
\( -4\sin^2 x + 9\sin x - 2 = 0 \)
Multiplying by \(-1\):
\( 4\sin^2 x - 9\sin x + 2 = 0 \) (as required).

(b) Let \( X = 2\theta - 30^\circ \). The equation becomes:
\( 4\sin^2 X - 9\sin X + 2 = 0 \)
Factorising the quadratic equation:
\( (4\sin X - 1)(\text{sin } X - 2) = 0 \)
This gives \( \sin X = \frac{1}{4} \) or \( \sin X = 2 \) (which has no real solutions as \( -1 \le \sin X \le 1 \)).

Given \( 0 \le \theta < 360^\circ \), the interval for \( X \) is:
\( 0 \le 2\theta < 720^\circ \implies -30^\circ \le X < 690^\circ \).

Solve \( \sin X = 0.25 \):
Principal value: \( X = \sin^{-1}(0.25) \approx 14.48^\circ \)
Other solutions in \( -30^\circ \le X < 690^\circ \):
\( X = 180^\circ - 14.48^\circ = 165.52^\circ \)
\( X = 360^\circ + 14.48^\circ = 374.48^\circ \)
\( X = 360^\circ + 165.52^\circ = 525.52^\circ \)

Now calculate the corresponding values of \( \theta \) using \( \theta = \frac{X + 30^\circ}{2} \):
\( \theta_1 = \frac{14.48^\circ + 30^\circ}{2} = 22.24^\circ \approx 22.2^\circ \)
\( \theta_2 = \frac{165.52^\circ + 30^\circ}{2} = 97.76^\circ \approx 97.8^\circ \)
\( \theta_3 = \frac{374.48^\circ + 30^\circ}{2} = 202.24^\circ \approx 202.2^\circ \)
\( \theta_4 = \frac{525.52^\circ + 30^\circ}{2} = 277.76^\circ \approx 277.8^\circ \)

So the solutions are \( \theta = 22.2^\circ, 97.8^\circ, 202.2^\circ, 277.8^\circ \).

(a) M1: Using the identity \( \cos^2 x = 1 - \sin^2 x \) to express the equation in terms of sine only.
A1.5: Expanding and simplifying correctly to show \( 4\sin^2 x - 9\sin x + 2 = 0 \) with no errors shown.

(b) M1: Factorising or using the quadratic formula to find values for \( \sin(2\theta - 30^\circ) \).
A1: Identifying that \( \sin(2\theta - 30^\circ) = 0.25 \) is the only valid branch.
M1: Finding the principal value of \( X \approx 14.5^\circ \) (or at least one other value in the transformed range).
M1: Finding at least three values of \( X \) in the range \( -30^\circ \le X < 690^\circ \).
A1: Obtaining at least two correct values of \( \theta \) (to 1 d.p.).
A1: Finding all four correct values of \( \theta \): \( 22.2^\circ, 97.8^\circ, 202.2^\circ, 277.8^\circ \) (to 1 d.p.) and no extra values in the range.

(a) By the Cosine Rule applied to triangle \( ABC \) with angle \( \angle BAC \):
\( BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC) \)
Substitute the given values:
\( (x + 2)^2 = 3^2 + (2x - 3)^2 - 2(3)(2x - 3)\cos(60^\circ) \)
Since \( \cos(60^\circ) = 0.5 \):
\( x^2 + 4x + 4 = 9 + (4x^2 - 12x + 9) - 3(2x - 3) \)
\( x^2 + 4x + 4 = 18 + 4x^2 - 12x - 6x + 9 \)
\( x^2 + 4x + 4 = 4x^2 - 18x + 27 \)
Rearranging to make one side zero:
\( 3x^2 - 22x + 23 = 0 \) (as required).

(b) Solving \( 3x^2 - 22x + 23 = 0 \) using the quadratic formula:
\( x = \frac{-(-22) \pm \sqrt{(-22)^2 - 4(3)(23)}}{2(3)} \)
\( x = \frac{22 \pm \sqrt{484 - 276}}{6} = \frac{22 \pm \sqrt{208}}{6} \)
\( x_1 = \frac{22 + \sqrt{208}}{6} \approx 6.0703... \)
\( x_2 = \frac{22 - \sqrt{208}}{6} \approx 1.2629... \)
We must check if both solutions produce valid side lengths for the triangle.
The side length \( AC \) is given by \( 2x - 3 \).
If \( x \approx 1.26 \), then \( AC = 2(1.26) - 3 = -0.48 \), which is impossible because side lengths must be positive (i.e., \( 2x - 3 > 0 \implies x > 1.5 \)).
Therefore, the only valid solution is \( x = 6.07 \) (to 3 s.f.).

(a) M1: Stating or using the Cosine Rule correctly for the given triangle.
M1: Substituting the expressions into the formula, including the correct value \( \cos(60^\circ) = 0.5 \).
A1: Expanding both \( (x+2)^2 \) and \( (2x-3)^2 \) correctly.
A1.5: Showing clear algebraic steps to simplify and obtain the correct quadratic equation \( 3x^2 - 22x + 23 = 0 \).

(b) M1: Attempting to solve the quadratic equation using the quadratic formula.
A1: Finding both numerical roots: \( x \approx 6.07 \) and \( x \approx 1.26 \).
A1: Explicitly rejecting \( x \approx 1.26 \) by stating that the side length \( AC = 2x - 3 \) must be positive (or \( x > 1.5 \)), leaving \( x = 6.07 \) as the only valid answer.

(a) Let \( f(x) = 4x^2 \).
By definition of the derivative from first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
Substitute \( f(x) \):
\( f(x+h) = 4(x+h)^2 = 4(x^2 + 2xh + h^2) = 4x^2 + 8xh + 4h^2 \)
Then,
\( f(x+h) - f(x) = (4x^2 + 8xh + 4h^2) - 4x^2 = 8xh + 4h^2 \)
Now, divide by \( h \):
\( \frac{f(x+h) - f(x)}{h} = \frac{8xh + 4h^2}{h} = 8x + 4h \)
Taking the limit as \( h \to 0 \):
\( f'(x) = \lim_{h \to 0} (8x + 4h) = 8x \) (as required).

(b) The curve is \( y = 4x^2 - 3x + 1 \).
Using differentiation rules (which is justified since \( \frac{d}{dx}(4x^2) = 8x \) is proven):
\( \frac{dy}{dx} = 8x - 3 \)
At the point where \( x = 1 \):
Gradient of the tangent \( m = 8(1) - 3 = 5 \).
The y-coordinate at \( x = 1 \) is:
\( y = 4(1)^2 - 3(1) + 1 = 2 \).
Using the equation of a straight line, \( y - y_1 = m(x - x_1) \):
\( y - 2 = 5(x - 1) \)
\( y - 2 = 5x - 5 \implies y = 5x - 3 \).

(a) M1: Stating the formula for differentiation from first principles.
M1: Expanding \( f(x+h) = 4(x+h)^2 \) correctly to get \( 4x^2 + 8xh + 4h^2 \).
M1: Forming the quotient \( \frac{8xh + 4h^2}{h} \) and simplifying to \( 8x + 4h \).
A1.5: Applying the limit \( h \to 0 \) correctly to obtain \( 8x \) (must have limit notation or explicit limit explanation).

(b) M1: Differentiating the polynomial to find \( \frac{dy}{dx} = 8x - 3 \) and substituting \( x = 1 \) to get the gradient.
A1: Finding both the correct gradient \( m = 5 \) and correct point \( (1, 2) \).
A1: Formulating the correct equation of the tangent as \( y = 5x - 3 \).

(a) The volume \( V \) of the rectangular box is given by:
\( V = \text{length} \times \text{width} \times \text{height} = (2x)(x)(h) = 2x^2 h \)
Given that \( V = 288 \):
\( 2x^2 h = 288 \implies h = \frac{144}{x^2} \)

The box is open-topped, so its surface area \( A \) consists of the base and four side faces:
- Base area: \( 2x \times x = 2x^2 \)
- Two side faces of area: \( x \times h \) each
- Two side faces of area: \( 2x \times h \) each

So total surface area is:
\( A = 2x^2 + 2(xh) + 2(2xh) = 2x^2 + 6xh \)
Substituting \( h = \frac{144}{x^2} \) into this expression:
\( A = 2x^2 + 6x\left(\frac{144}{x^2}\right) = 2x^2 + \frac{864}{x} \) (as required).

(b) To find the minimum surface area, we differentiate \( A \) with respect to \( x \):
\( A = 2x^2 + 864x^{-1} \)
\( \frac{dA}{dx} = 4x - 864x^{-2} = 4x - \frac{864}{x^2} \)
At the stationary point, \( \frac{dA}{dx} = 0 \):
\( 4x - \frac{864}{x^2} = 0 \implies 4x^3 = 864 \implies x^3 = 216 \implies x = 6 \).

The minimum value of \( A \) is:
\( A = 2(6)^2 + \frac{864}{6} = 72 + 144 = 216\text{ cm}^2 \).

To justify that this is a minimum, find the second derivative:
\( \frac{d^2A}{dx^2} = 4 + 1728x^{-3} = 4 + \frac{1728}{x^3} \)
At \( x = 6 \):
\( \frac{d^2A}{dx^2} = 4 + \frac{1728}{216} = 4 + 8 = 12 \)
Since \( \frac{d^2A}{dx^2} > 0 \), the value is a minimum.

(a) M1: Writing an expression for the volume, \( V = 2x^2 h = 288 \), and making \( h \) the subject.
M1: Writing an expression for the total surface area of an open-topped box, \( A = 2x^2 + 6xh \).
M1: Substituting the expression for \( h \) into the surface area equation.
A1: Obtaining the given formula \( A = 2x^2 + \frac{864}{x} \) correctly.

(b) M1: Differentiating \( A \) with respect to \( x \) to find \( \frac{dA}{dx} = 4x - \frac{864}{x^2} \).
A1: Setting \( \frac{dA}{dx} = 0 \) and solving to find \( x = 6 \).
A1: Calculating \( A = 216 \).
M0.5: Finding the second derivative \( \frac{d^2A}{dx^2} = 4 + \frac{1728}{x^3} \).
A1: Evaluating \( \frac{d^2A}{dx^2} \) at \( x = 6 \) to get 12, and concluding it is a minimum since the value is positive.

(a) To find the points of intersection, we set \( y_C = y_l \):
\( 8 + 2x - x^2 = x + 2 \)
Rearranging into a quadratic equation:
\( x^2 - x - 6 = 0 \)
Factorising the quadratic:
\( (x - 3)(x + 2) = 0 \)
So the x-coordinates of the intersection points are \( x = 3 \) and \( x = -2 \).

Substituting these x-coordinates back into the equation of \( l \):
- For \( x = -2 \), \( y = -2 + 2 = 0 \), giving the point \( (-2, 0) \).
- For \( x = 3 \), \( y = 3 + 2 = 5 \), giving the point \( (3, 5) \) (as required).

(b) The finite region is bounded between \( x = -2 \) and \( x = 3 \). The curve \( C \) lies above the line \( l \) on this interval.
The area of the region is given by:
\( \int_{-2}^{3} (y_C - y_l) \, dx = \int_{-2}^{3} (8 + 2x - x^2 - (x + 2)) \, dx \)
\( = \int_{-2}^{3} (6 + x - x^2) \, dx \)
Integrating term by term:
\( = \left[ 6x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{3} \)

Evaluating at the upper limit \( x = 3 \):
\( \left( 6(3) + \frac{3^2}{2} - \frac{3^3}{3} \right) = 18 + 4.5 - 9 = 13.5 \)

Evaluating at the lower limit \( x = -2 \):
\( \left( 6(-2) + \frac{(-2)^2}{2} - \frac{(-2)^3}{3} \right) = -12 + 2 + \frac{8}{3} = -10 + \frac{8}{3} = -\frac{22}{3} \)

Subtracting the lower limit from the upper limit:
\( \text{Area} = 13.5 - \left(-\frac{22}{3}\right) = \frac{27}{2} + \frac{22}{3} = \frac{81 + 44}{6} = \frac{125}{6} \).

(a) M1: Setting the equations equal to each other, \( 8 + 2x - x^2 = x + 2 \).
A1: Solving the resulting quadratic to find \( x = 3 \) and \( x = -2 \).
A0.5: Verifying the coordinates by substituting into either equation to get \( (-2, 0) \) and \( (3, 5) \).

(b) M1: Setting up the definite integral \( \int_{-2}^{3} (y_C - y_l) \, dx \).
A1.5: Integrating correctly to find \( 6x + \frac{x^2}{2} - \frac{x^3}{3} \).
M1: Substituting the limits \( 3 \) and \( -2 \) into their integrated expression.
A1: Finding correct values \( 13.5 \) and \( -\frac{22}{3} \).
A0.5: Subtracting to find the correct exact area of \( \frac{125}{6} \) (or equivalent mixed fraction \( 20\frac{5}{6} \)).

(a) First, find the gradient of the line \( l_1 \):
\( m_1 = \frac{2 - 5}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2} \)

Since \( l_2 \) is perpendicular to \( l_1 \), the gradient \( m_2 \) of \( l_2 \) satisfies:
\( m_2 \times \left(-\frac{1}{2}\right) = -1 \implies m_2 = 2 \)

Using the point-gradient form with \( C(1, -1) \):
\( y - (-1) = 2(x - 1) \)
\( y + 1 = 2x - 2 \)
\( 2x - y - 3 = 0 \) (or any integer multiple, e.g., \( -2x + y + 3 = 0 \))

(b) The line \( l_2 \) crosses the \( y \)-axis at \( D \). Setting \( x = 0 \) in the equation for \( l_2 \):
\( 2(0) - y - 3 = 0 \implies y = -3 \)
So, \( D(0, -3) \).

To find the area of triangle \( ACD \), we can use the vertices \( A(-2, 5) \), \( C(1, -1) \), and \( D(0, -3) \).
Using the coordinate geometry formula for the area of a triangle:
Area = \( \frac{1}{2} |x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C)| \)
Area = \( \frac{1}{2} |-2(-1 - (-3)) + 1(-3 - 5) + 0| \)
Area = \( \frac{1}{2} |-2(2) + 1(-8)| = \frac{1}{2} |-4 - 8| = \frac{1}{2} |-12| = 6 \).

(a)
- M1: Attempts to find the gradient of \( l_1 \) using \( \frac{y_2 - y_1}{x_2 - x_1} \).
- A1: Correct gradient of \( l_1 \), which is \( -\frac{1}{2} \).
- M1: Uses the perpendicular gradient rule \( m_1 m_2 = -1 \) to find the gradient of \( l_2 \) and attempts to find the equation of the line using \( C(1, -1) \).
- A1: Correct equation in the required form, e.g., \( 2x - y - 3 = 0 \) (accept any integer multiple).

(b)
- B1: Correct coordinates for \( D(0, -3) \).
- M1: A valid method to find the area of triangle \( ACD \), e.g., using the shoelace formula, a bounding box, or calculating lengths of sides and using trigonometry.
- A1: Correct area of \( 6 \).

(a) Differentiating \( y = \frac{1}{3}x^3 - 3x^2 + 8x + 5 \) with respect to \( x \):
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 3 \times \frac{1}{3}x^2 - 2 \times 3x + 8 \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = x^2 - 6x + 8 \)

(b) At stationary points, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\( x^2 - 6x + 8 = 0 \)
\( (x - 2)(x - 4) = 0 \)
So, \( x = 2 \) or \( x = 4 \).

When \( x = 2 \):
\( y = \frac{1}{3}(2)^3 - 3(2)^2 + 8(2) + 5 = \frac{8}{3} - 12 + 16 + 5 = \frac{35}{3} \)

When \( x = 4 \):
\( y = \frac{1}{3}(4)^3 - 3(4)^2 + 8(4) + 5 = \frac{64}{3} - 48 + 32 + 5 = \frac{31}{3} \)

Thus, the stationary points are \( \left(2, \frac{35}{3}\right) \) and \( \left(4, \frac{31}{3}\right) \).

(c) To determine the nature of the stationary points, find the second derivative:
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2x - 6 \)

At \( x = 2 \):
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2(2) - 6 = -2 \)
Since \( -2 < 0 \), the stationary point \( \left(2, \frac{35}{3}\right) \) is a local maximum.

At \( x = 4 \):
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2(4) - 6 = 2 \)
Since \( 2 > 0 \), the stationary point \( \left(4, \frac{31}{3}\right) \) is a local minimum.

(a)
- M1: Attempts to differentiate the given expression (at least one term differentiated correctly with power decreasing by 1).
- A1: Correct derivative \( x^2 - 6x + 8 \).

(b)
- M1: Sets their \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) and attempts to solve the quadratic equation to find \( x \).
- A1: Correct \( x \)-coordinates: \( x = 2 \) and \( x = 4 \).
- M1: Substitutes at least one of their \( x \)-coordinates back into the original equation for \( y \) to find the \( y \)-coordinate.
- A1: Correct coordinates for both stationary points: \( \left(2, \frac{35}{3}\right) \) and \( \left(4, \frac{31}{3}\right) \) (or equivalent mixed numbers or decimals).

(c)
- M1: Differentiates their \( \frac{\mathrm{d}y}{\mathrm{d}x} \) to find \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \) and substitutes at least one of their \( x \) values (or uses another valid method).
- A1: Correctly identifies the nature of both points with clear mathematical justification showing \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0 \) for the maximum and \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \) for the minimum.

(a) Let the first term of the geometric series be \(a\) and the common ratio be \(r\).

The second term is \(ar = 4\).

The third term is \(ar^2 = k\).

Dividing the third term by the second term gives:
\[r = \frac{k}{4}\]

Since \(ar = 4\), we have \(a = \frac{4}{r} = \frac{16}{k}\).

The sum to infinity is given by:
\[S_{\infty} = \frac{a}{1 - r} = 18\]

Substitute \(a = \frac{16}{k}\) and \(r = \frac{k}{4}\):
\[\frac{16/k}{1 - k/4} = 18\]

Multiply the numerator and denominator of the left-hand side by \(4k\):
\[\frac{64}{4k - k^2} = 18\]
\[64 = 18(4k - k^2)\]
\[64 = 72k - 18k^2\]

Divide the entire equation by 2:
\[32 = 36k - 9k^2\]

Rearranging gives:
\[9k^2 - 36k + 32 = 0\] (as required)

(b) Solve the quadratic equation \(9k^2 - 36k + 32 = 0\):
\[(3k - 4)(3k - 8) = 0\]

Thus, \(k = \frac{4}{3}\) or \(k = \frac{8}{3}\).

We must check if the sum to infinity exists for these values:
- If \(k = \frac{4}{3}\), \(r = \frac{k}{4} = \frac{1}{3}\). Since \(|r| < 1\), the sum to infinity exists.
- If \(k = \frac{8}{3}\), \(r = \frac{k}{4} = \frac{2}{3}\). Since \(|r| < 1\), the sum to infinity exists.

So the possible values of \(k\) are \rac{4}{3}\ and \rac{8}{3}\

(a)
- M1: Attempts to find the common ratio \(r\) in terms of \(k\), or sets up equations for the terms, e.g., \(ar = 4\) and \(ar^2 = k\) and attempts to eliminate \(a\) or \(r\).
- A1: Obtains \(r = \frac{k}{4}\) and \(a = \frac{16}{k}\) (or equivalent).
- M1: Correctly substitutes their \(a\) and \(r\) into the sum to infinity formula \(S_{\infty} = \frac{a}{1 - r}\) and sets it equal to \(18\).
- A1: Fully correct algebraic manipulation leading to the given quadratic equation \(9k^2 - 36k + 32 = 0\) with no errors seen.

(b)
- M1: Attempts to solve the quadratic equation \(9k^2 - 36k + 32 = 0\) by factorisation, quadratic formula, or completing the square.
- A1: Obtains at least one correct value of \(k\) (either \(k = \frac{4}{3}\) or \(k = \frac{8}{3}\)).
- A1: Obtains both \(k = \frac{4}{3}\) and \(k = \frac{8}{3}\).
- B1: Explains that both values of \(k\) are valid because both result in a common ratio with absolute value less than 1.

Using the power law of logarithms:
\[2 \log_3(x - 2) = \log_3(x - 2)^2\]

The equation becomes:
\[\log_3(x - 2)^2 - \log_3(x + 4) = 1\]

Using the subtraction law of logarithms:
\[\log_3 \left(\frac{(x - 2)^2}{x + 4}\right) = 1\]

Rewrite in exponential form:
\[\frac{(x - 2)^2}{x + 4} = 3^1\]
\[\frac{x^2 - 4x + 4}{x + 4} = 3\]

Multiply both sides by \(x + 4\):
\[x^2 - 4x + 4 = 3(x + 4)\]
\[x^2 - 4x + 4 = 3x + 12\]

Rearrange into a quadratic equation:
\[x^2 - 7x - 8 = 0\]

Factorise the quadratic equation:
\[(x - 8)(x + 1) = 0\]

So \(x = 8\) or \(x = -1\).

We must check the validity of these solutions in the original equation:
- For \(\log_3(x - 2)\) to be defined, we require \(x - 2 > 0 \implies x > 2\).
- For \(\log_3(x + 4)\) to be defined, we require \(x + 4 > 0 \implies x > -4\).

Since \(x = -1\) does not satisfy \(x > 2\), we reject \(x = -1\).

The only valid solution is \(x = 8\).

- M1: Applies the power law of logarithms to write \(2\log_3(x - 2)\) as \( \log_3(x - 2)^2 \).
- M1: Applies the subtraction law of logarithms to combine the terms into a single logarithm: \(\log_3 \left(\frac{(x - 2)^2}{x + 4}\right)\).
- M1: Correctly removes logarithms by raising 3 to the power of 1: \(\frac{(x - 2)^2}{x + 4} = 3\).
- A1: Formulates a correct quadratic equation \(x^2 - 7x - 8 = 0\) (or equivalent).
- M1: Attempts to solve the quadratic equation, leading to \(x = 8\) and \(x = -1\).
- A1: Identifies \(x = 8\) as a solution.
- A1: Clearly rejects \(x = -1\) with a valid reason (e.g., \(\log(x-2)\) is undefined for negative arguments) and gives \(x = 8\) as the only solution.

(a) Complete the square for both \(x\) and \(y\):
\[x^2 - 6x = (x - 3)^2 - 9\]
\[y^2 + 8y = (y + 4)^2 - 16\]

Substitute these back into the circle equation:
\[(x - 3)^2 - 9 + (y + 4)^2 - 16 - 11 = 0\]
\[(x - 3)^2 + (y + 4)^2 = 36\]

Thus, the centre of \(C\) is \((3, -4)\) and the radius of \(C\) is \(\sqrt{36} = 6\).

(b) Method: Using the perpendicular distance from the centre to the tangent.
The tangent has equation \(2x - y + k = 0\).

The perpendicular distance \(d\) from the centre \((3, -4)\) to this line must be equal to the radius, which is \(6\).

Using the distance formula \(d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\):
\[6 = \frac{|2(3) - (-4) + k|}{\sqrt{2^2 + (-1)^2}}\]
\[6 = \frac{|6 + 4 + k|}{\sqrt{5}}\]
\[6\sqrt{5} = |k + 10|\]

Thus, \(k + 10 = 6\sqrt{5}\) or \(k + 10 = -6\sqrt{5}\).

So the possible values of \(k\) are:
\[k = -10 + 6\sqrt{5} \quad \text{and} \quad k = -10 - 6\sqrt{5}\]

(a)
- M1: Attempts to complete the square for both \(x\) and \(y\), looking for expressions of the form \((x \pm a)^2\) and \((y \pm b)^2\).
- A1: Correctly identifies the centre as \((3, -4)\).
- A1: Correctly identifies the radius as \(6\).

(b)
- M1: Attempts to set up a method to find \(k\) (e.g., substituting \(y = 2x + k\) into the circle equation, or using the perpendicular distance formula).
- M1: Applies the method correctly (e.g., formulating the quadratic equation in \(x\), or substituting the centre coordinates into the distance formula).
- M1: Sets the discriminant to zero for substitution, or sets up the absolute value equation \(|k + 10| = 6\sqrt{5}\) for the distance method.
- A1: Obtains a correct simplified equation in \(k\) (e.g. \(k^2 + 20k - 80 = 0\) or \(|k + 10| = 6\sqrt{5}\)).
- A1: Solves to find the correct exact values of \(k = -10 \pm 6\sqrt{5}\).

Use the identity \(\sin^2(2\theta) = 1 - \cos^2(2\theta)\):
\[6(1 - \cos^2(2\theta)) - \cos(2\theta) - 4 = 0\]
\[6 - 6\cos^2(2\theta) - \cos(2\theta) - 4 = 0\]
\[-6\cos^2(2\theta) - \cos(2\theta) + 2 = 0\]

Multiply by \(-1\):
\[6\cos^2(2\theta) + \cos(2\theta) - 2 = 0\]

Let \(y = \cos(2\theta)\):
\[6y^2 + y - 2 = 0\]

Factorise this quadratic equation:
\[(3y + 2)(2y - 1) = 0\]

This gives:
\[\cos(2\theta) = -\frac{2}{3} \quad \text{or} \quad \cos(2\theta) = \frac{1}{2}\]

Since \(0 \le \theta < 180^\circ\), we have \(0 \le 2\theta < 360^\circ\).

Case 1: \(\cos(2\theta) = \frac{1}{2}\)
\[2\theta = 60^\circ \quad \text{or} \quad 2\theta = 360^\circ - 60^\circ = 300^\circ\]
\[\theta = 30^\circ \quad \text{or} \quad \theta = 150^\circ\]

Case 2: \(\cos(2\theta) = -\frac{2}{3}\)
The principal value is:
\[2\theta = \arccos\left(-\frac{2}{3}\right) \approx 131.81^\circ\]
The other value in the range \([0, 360^\circ)\) is:
\[2\theta = 360^\circ - 131.81^\circ = 228.19^\circ\]

Dividing by 2 gives:
\[\theta \approx 65.9^\circ \quad \text{or} \quad \theta \approx 114.1^\circ\]

Combining all solutions, we get:
\[\theta = 30^\circ, 65.9^\circ, 114.1^\circ, 150^\circ\] (to 1 d.p. where appropriate).

- M1: Uses the identity \(\sin^2(2\theta) = 1 - \cos^2(2\theta)\) to form a quadratic equation in \(\cos(2\theta)\).
- A1: Obtains the correct quadratic equation \(6\cos^2(2\theta) + \cos(2\theta) - 2 = 0\) (or equivalent).
- M1: Solves the quadratic equation to find two values for \(\cos(2\theta)\).
- A1: Obtains \(\cos(2\theta) = -\frac{2}{3}\) and \(\cos(2\theta) = \frac{1}{2}\).
- M1: Finds at least one correct value of \(2\theta\) for either of their equations.
- A1: Obtains \(\theta = 30^\circ\) and \(\theta = 150^\circ\).
- A1: Obtains \(\theta = 65.9^\circ\) and \(\theta = 114.1^\circ\) (rounded correctly to 1 decimal place). Deduct 0.5 marks if extra solutions within the range are given.

(a) The formula for the total surface area of a solid cylinder is:
\[A = 2\pi r^2 + 2\pi r h\]

Given that \(A = 150\pi\):
\[2\pi r^2 + 2\pi r h = 150\pi\]

Divide the entire equation by \(2\pi\):
\[r^2 + r h = 75\]

Rearrange to express \(h\) in terms of \(r\):
\[h = \frac{75 - r^2}{r}\]

The formula for the volume of a cylinder is:
\[V = \pi r^2 h\]

Substitute \(h = \frac{75 - r^2}{r}\) into this formula:
\[V = \pi r^2 \left(\frac{75 - r^2}{r}\right)\]
\[V = \pi r (75 - r^2)\]
\[V = 75\pi r - \pi r^3\] (as required)

(b) To find the maximum value of \(V\), we differentiate \(V\) with respect to \(r\):
\[\frac{\mathrm{d}V}{\mathrm{d}r} = 75\pi - 3\pi r^2\]

Set \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\) to find the stationary points:
\[75\pi - 3\pi r^2 = 0\]
\[3\pi r^2 = 75\pi\]
\[r^2 = 25\]

Since \(r > 0\) represents a physical radius, we have:
\[r = 5\]

To confirm that this gives a maximum volume, we find the second derivative:
\[\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi r\]

When \(r = 5\):
\[\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -30\pi < 0\]

Since the second derivative is negative, \(r = 5\) gives a maximum volume.

Substitute \(r = 5\) back into the volume equation:
\[V = 75\pi(5) - \pi(5)^3\]
\[V = 375\pi - 125\pi\]
\[V = 250\pi \approx 785.398\text{ cm}^3\]

To the nearest cubic centimetre, the maximum volume is \(785\text{ cm}^3\).

(a)
- M1: Writes down the correct formula for the total surface area of a cylinder, \(A = 2\pi r^2 + 2\pi r h\), and equates it to \(150\pi\).
- M1: Rearranges their equation to make \(h\) the subject.
- M1: Substitutes their expression for \(h\) into the volume formula \(V = \pi r^2 h\).
- A1: Achieves the given result \(V = 75\pi r - \pi r^3\) with no errors in the working.

(b)
- M1: Differentiates the given volume function to find \(\frac{\mathrm{d}V}{\mathrm{d}r}\).
- A1: Obtains \(\frac{\mathrm{d}V}{\mathrm{d}r} = 75\pi - 3\pi r^2\).
- M1: Sets their \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\) and solves to find \(r = 5\).
- M1: Calculates the second derivative \(\frac{\mathrm{d}^2V}{\mathrm{d}r^2} = -6\pi r\) and checks that it is negative at \(r=5\) to justify that it is a maximum.
- A1: Obtains the maximum volume of \(250\pi\) or \(785\) (nearest integer).

(a) For \(x = 0.5\):
\[y = 3^{0.5} - 0.5 = \sqrt{3} - 0.5 \approx 1.232\] (to 3 d.p.)

For \(x = 1.5\):
\[y = 3^{1.5} - 1.5 = 3\sqrt{3} - 1.5 \approx 3.696\] (to 3 d.p.)

(b) The interval width is \(h = \frac{2 - 0}{4} = 0.5\).

The trapezium rule formula is:
\[I \approx \frac{h}{2} [ y_0 + y_4 + 2(y_1 + y_2 + y_3) ]\]

Using the values:
\[I \approx \frac{0.5}{2} [ 1 + 7 + 2(1.232 + 2 + 3.696) ]\]
\[I \approx 0.25 [ 8 + 2(6.928) ]\]
\[I \approx 0.25 [ 8 + 13.856 ]\]
\[I \approx 0.25 [ 21.856 ] = 5.464 \approx 5.46\] (to 2 d.p.)

(c) Differentiate \(y = 3^x - x\) twice to determine the curvature:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3^x \ln 3 - 1\]
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 3^x (\ln 3)^2\]

Since \(3^x > 0\) and \((\ln 3)^2 > 0\) for all real \(x\), the second derivative is always positive (\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\)), which means the curve is convex (concave up).

Therefore, the straight lines of the trapezia lie entirely above the curve, so the trapezium rule gives an overestimate.

(a)
- B1: Obtains \(1.232\) for \(x = 0.5\).
- B1: Obtains \(3.696\) for \(x = 1.5\).

(b)
- B1: State \(h = 0.5\) (or implicitly uses this value correctly).
- M1: Applies the trapezium rule structure correctly: \(\frac{h}{2} \times \{ y_0 + y_n + 2(y_1 + \dots + y_{n-1}) \}\).
- A1: Correct substitution of values, e.g., \(\frac{0.5}{2} [ 1 + 7 + 2(1.232 + 2 + 3.696) ]\).
- A1: Obtains \(5.46\).

(c)
- M1: Mentions the curvature of the graph, or calculates/states that the second derivative is positive.
- A1: Concludes it is an overestimate with a fully correct and coherent explanation (must relate the positive second derivative, convexity, or the fact that chords lie above the curve to the overestimate).

(a) To show that \((x - 3)\) is a factor of \(\mathrm{f}(x)\), we use the factor theorem. Evaluate \(\mathrm{f}(3)\):
\[\mathrm{f}(3) = 2(3)^3 - 3(3)^2 - 11(3) + 6\]
\[\mathrm{f}(3) = 2(27) - 3(9) - 33 + 6\]
\[\mathrm{f}(3) = 54 - 27 - 33 + 6 = 0\]

Since \(\mathrm{f}(3) = 0\), by the factor theorem, \((x - 3)\) is a factor of \(\mathrm{f}(x)\).

(b) We can write \(\mathrm{f}(x) = (x - 3)(2x^2 + ax + b)\).
Comparing coefficients of \(x^3\) and constant terms:
- The coefficient of \(x^3\) is 2, so the quadratic factor starts with \(2x^2\).
- The constant term is \(+6\), so \(-3 \times b = 6 \implies b = -2\).

Comparing coefficients of \(x^2\):
\[-3(2x^2) + ax(x) = -3x^2 \implies -6x^2 + ax^2 = -3x^2 \implies a = 3\]

So, \(\mathrm{f}(x) = (x - 3)(2x^2 + 3x - 2)\).

Now, factorise the quadratic term \(2x^2 + 3x - 2\):
\[2x^2 + 3x - 2 = (2x - 1)(x + 2)\]

Thus, factorised completely:
\[\mathrm{f}(x) = (x - 3)(2x - 1)(x + 2)\]

(c) By the remainder theorem, the remainder when \(\mathrm{f}(x)\) is divided by \((x - 1)\) is \(\mathrm{f}(1)\):
\[\mathrm{f}(1) = 2(1)^3 - 3(1)^2 - 11(1) + 6\]
\[\mathrm{f}(1) = 2 - 3 - 11 + 6 = -6\]

Therefore, the remainder is \(-6\).

(a)
- M1: Attempts to evaluate \(\mathrm{f}(3)\).
- A1: Obtains \(\mathrm{f}(3) = 0\) and states a conclusion that \((x - 3)\) is a factor of \(\mathrm{f}(x)\).

(b)
- M1: Attempts algebraic division, or equates coefficients to find the quadratic factor.
- A1: Obtains the correct quadratic factor \((2x^2 + 3x - 2)\).
- A1: Factorises the quadratic factor to obtain the completely factorised form \((x - 3)(2x - 1)(x + 2)\).

(c)
- M1: Attempts to evaluate \(\mathrm{f}(1)\), or performs long division by \((x - 1)\).
- A1: Obtains \(\mathrm{f}(1) = -6\).
- A1: Concludes that the remainder is \(-6\).

(a) Since \(x\) and \(y\) are positive real numbers, their square roots \(\sqrt{x}\) and \(\sqrt{y}\) are real numbers.

For any real numbers \(\sqrt{x}\) and \(\sqrt{y}\), the square of their difference must be greater than or equal to 0:
\[(\sqrt{x} - \sqrt{y})^2 \ge 0\]

Expanding the left-hand side:
\[(\sqrt{x})^2 - 2\sqrt{x}\sqrt{y} + (\sqrt{y})^2 \ge 0\]

Since \(x\) and \(y\) are positive, \((\sqrt{x})^2 = x\) and \((\sqrt{y})^2 = y\), and \\sqrt{x}\sqrt{y} = \sqrt{xy}\\:
\[x - 2\sqrt{xy} + y \ge 0\]

Adding \(2\sqrt{xy}\) to both sides:
\[x + y \ge 2\sqrt{xy}\]
This completes the proof by deduction.

(b) To disprove the statement for general real numbers \(x\) and \(y\), we need to find a counterexample where both sides of the inequality are defined but the inequality does not hold.

Let \(x = -2\) and \(y = -8\).

The left-hand side is:
\[x + y = -2 + (-8) = -10\]

The right-hand side is:
\[2\sqrt{xy} = 2\sqrt{(-2)(-8)} = 2\sqrt{16} = 2 \times 4 = 8\]

Since \(-10 \ge 8\) is false, the statement is not always true.

(a)
- M1: Starts with a valid true statement, e.g., \((\sqrt{x} - \sqrt{y})^2 \ge 0\).
- A1: Expands the squared term correctly to obtain \(x - 2\sqrt{xy} + y \ge 0\).
- A1: Rearranges the inequality correctly to achieve \(x + y \ge 2\sqrt{xy}\).
- A1: Provides a clear step-by-step logical deduction, noting why \(\sqrt{x}\) and \(\sqrt{y}\) exist (since \(x, y > 0\)).

(b)
- M1: Chooses two negative numbers for \(x\) and \(y\) (so that \(xy > 0\) and the square root is defined).
- A1: Correctly evaluates both sides of the inequality for their chosen numbers (e.g., LHS \(= -10\), RHS \(= 8\)).
- A1: Clearly states that the inequality is false for these values, thereby completing the counterexample proof.

(a) Using the change of base formula, we have:
\[ \log_4(x^2) = \frac{\log_2(x^2)}{\log_2 4} \]
Since \( 4 = 2^2 \), \( \log_2 4 = 2 \).
Using the power law of logarithms, \( \log_2(x^2) = 2\log_2 x \) for \( x > 0 \).
Substituting these in gives:
\[ \log_4(x^2) = \frac{2\log_2 x}{2} = \log_2 x \]

(b) Substitute the result from part (a) into the given equation:
\[ \log_2(3x + 2) - \log_2 x = 3 \]
Using the subtraction law of logarithms:
\[ \log_2\left(\frac{3x + 2}{x}\right) = 3 \]
Rewrite the equation in exponential form:
\[ \frac{3x + 2}{x} = 2^3 \]
\[ \frac{3x + 2}{x} = 8 \]
Multiply both sides by \( x \):
\[ 3x + 2 = 8x \]
\[ 5x = 2 \]
\[ x = \frac{2}{5} \]
Since \( x = \frac{2}{5} > 0 \), both \( 3x+2 > 0 \) and \( x^2 > 0 \), confirming that the logarithms are well-defined and the solution is valid.

(a)
M1: Applies the change of base formula correctly to change the base of the logarithm to 2.
A1: Uses \( \log_2 4 = 2 \) and the power law \( \log_2(x^2) = 2\log_2 x \) to complete the proof convincingly.

(b)
M1: Substitutes \( \log_4(x^2) = \log_2 x \) to obtain an equation containing only base 2 logarithms.
M1: Applies the subtraction law of logarithms to combine the terms.
M1: Converts the logarithmic equation into exponential form correctly.
M1: Solves the linear equation for \( x \).
A1: Obtains the correct answer \( x = \frac{2}{5} \) (or 0.4).
B0.5: Mentions or shows that \( x = \frac{2}{5} \) is a valid solution because it satisfies \( x > 0 \) and the arguments of both logarithms in the original equation remain positive.

(a) To find the points of intersection, equate the curve and line equations:
\[ 8x - 2x^2 = 2x \]
Rearrange to form a quadratic equation:
\[ 2x^2 - 6x = 0 \]
Factorise:
\[ 2x(x - 3) = 0 \]
This gives \( x = 0 \) and \( x = 3 \).
Substitute these into the equation of \( L \) to find the corresponding y-coordinates:
For \( x = 0 \), \( y = 2(0) = 0 \).
For \( x = 3 \), \( y = 2(3) = 6 \).
So, the coordinates of the intersection points are \( (0, 0) \) and \( (3, 6) \).

(b) The area of the region \( R \) is given by:
\[ \text{Area} = \int_{0}^{3} ((8x - 2x^2) - 2x) \mathrm{d}x \]
\[ = \int_{0}^{3} (6x - 2x^2) \mathrm{d}x \]
Integrate each term:
\[ = \left[ 3x^2 - \frac{2}{3}x^3 \right]_{0}^{3} \]
Substitute the limits:
At the upper limit \( x = 3 \):
\[ 3(3)^2 - \frac{2}{3}(3)^3 = 27 - 18 = 9 \]
At the lower limit \( x = 0 \):
\[ 3(0)^2 - \frac{2}{3}(0)^3 = 0 \]
Subtracting the lower limit value from the upper limit value gives:
\[ \text{Area} = 9 - 0 = 9 \]

(a)
M1: Equates the equations of the curve and the line to set up a quadratic equation.
A1: Solves the quadratic equation to find the correct x-coordinates \( x = 0 \) and \( x = 3 \).
A0.5: Finds both y-coordinates to write the final coordinates as \( (0, 0) \) and \( (3, 6) \).

(b)
M1: Sets up a correct integral for the area with limits 0 and 3.
M1: Integrates the expression, raising the powers of \( x \) by 1.
A1: Obtains the correct integrated expression \( 3x^2 - \frac{2}{3}x^3 \).
M1: Substitutes the limits 3 and 0 into their integrated expression and subtracts.
A1: Obtains the correct exact area of 9.

(a) We rewrite the numerator in terms of the denominator: \
\(2x + 5 = 2(x - 3) + 11\) \
So, \
\(\mathrm{f}(x) = \frac{2(x-3) + 11}{x-3} = 2 + \frac{11}{x-3}\) (as required). \
\
(b) Let \(y = 2 + \frac{11}{x-3}\). \
Rearranging to make \(x\) the subject: \
\(y - 2 = \frac{11}{x-3}\) \
\(x - 3 = \frac{11}{y-2}\) \
\(x = 3 + \frac{11}{y-2} = \frac{3(y-2) + 11}{y-2} = \frac{3y + 5}{y-2}\) \
So, \(\mathrm{f}^{-1}(x) = \frac{3x + 5}{x-2}\) (or \(3 + \frac{11}{x-2}\) ). \
\
(c) The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\). \
For \(x > 3\), \(x - 3 > 0\), which implies \\frac{11}{x-3} > 0\\. \
Thus \(2 + \frac{11}{x-3} > 2\), so \(\mathrm{f}(x) > 2\). \
Therefore, the domain of \(\mathrm{f}^{-1}\) is \(x > 2\). \
\
(d) \(\mathrm{f}(x) = \mathrm{f}^{-1}(x) \implies \mathrm{f}(x) = x\) \
\(\frac{2x+5}{x-3} = x \implies 2x+5 = x(x-3)\) \
\(2x + 5 = x^2 - 3x \implies x^2 - 5x - 5 = 0\) \
Using the quadratic formula: \
\(x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-5)}}{2} = \frac{5 \pm \sqrt{45}}{2} = \frac{5 \pm 3\sqrt{5}}{2}\) \
Since \(x > 3\), we reject the negative root. \
Thus, the exact value of \(x\) is \(x = \frac{5 + 3\sqrt{5}}{2}\).

Part (a): \
M1: Attempts to write the fraction in the form \(A + \frac{B}{x-3}\) by division or matching coefficients. \
A1: Correctly shows \(2 + \frac{11}{x-3}\) with working shown. \
\
Part (b): \
M1: Sets \(y = \mathrm{f}(x)\) and attempts to rearrange to make \(x\) the subject. \
M1: Reaches the stage \(x - 3 = \frac{11}{y-2}\) or equivalent. \
A1.33: Correctly obtains \(\mathrm{f}^{-1}(x) = \frac{3x+5}{x-2}\) or \(3 + \frac{11}{x-2}\) with correct function notation. \
\
Part (c): \
B1: Identifies the domain as \(x > 2\). \
\
Part (d): \
M1: Equates \(\mathrm{f}(x) = x\) (or \(\mathrm{f}^{-1}(x) = x\)) and forms a quadratic equation. \
A1: Obtains \(x = \frac{5 + 3\sqrt{5}}{2}\) and rejects the other solution with justification.

(a) Using double angle identities: \
\(\sin 2\theta = 2\sin \theta \cos \theta\) \
\(\cos 2\theta = 2\cos^2 \theta - 1\) \
Substituting these into the left-hand side: \
\(LHS = \frac{2\sin \theta \cos \theta}{1 + (2\cos^2 \theta - 1)} = \frac{2\sin \theta \cos \theta}{2\cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta = RHS\) (as required). \
\
(b) Using part (a) with \(\theta = 2x\): \
\(\frac{\sin 4x}{1 + \cos 4x} = \tan 2x\) \
So the equation becomes: \
\(\tan 2x = 3 \tan x\) \
Using the double angle identity for \(\tan 2x\): \
\(\frac{2\tan x}{1 - \tan^2 x} = 3 \tan x\) \
\(2\tan x = 3\tan x (1 - \tan^2 x)\) \
\(2\tan x = 3\tan x - 3\tan^3 x\) \
\(3\tan^3 x - \tan x = 0\) \
\(\tan x (3\tan^2 x - 1) = 0\) \
This gives: \
1) \(\tan x = 0 \implies x = 0^{\circ}\) (since \(0 \le x < 180^{\circ}\)) \
2) \(\tan^2 x = \frac{1}{3} \implies \tan x = \pm\frac{1}{\sqrt{3}}\) \
If \(\tan x = \frac{1}{\sqrt{3}} \implies x = 30^{\circ}\) \
If \(\tan x = -\frac{1}{\sqrt{3}} \implies x = 150^{\circ}\) \
Checking values in the original denominator: \
For \(x = 0^{\circ}\), \(1 + \cos 0 = 2 \neq 0\) \
For \(x = 30^{\circ}\), \(1 + \cos 120^{\circ} = 0.5 \neq 0\) \
For \(x = 150^{\circ}\), \(1 + \cos 600^{\circ} = 1 + \cos 240^{\circ} = 0.5 \neq 0\). \
So the solutions are \(x = 0^{\circ}, 30^{\circ}, 150^{\circ}\).

Part (a): \
M1: Uses standard double-angle formulas for \(\sin 2\theta\) and \(\cos 2\theta\) to express the LHS in terms of single angles. \
A1.33: Correctly simplifies the fraction to \(\tan \theta\). \
\
Part (b): \
M1: Applies part (a) to rewrite LHS as \(\tan 2x\). \
M1: Uses the double angle identity for \(\tan 2x\). \
M1: Factorises the equation to obtain \(\tan x = 0\) and \(3\tan^2 x - 1 = 0\). \
A1: Identifies \(x = 0^{\circ}\). \
A1: Identifies \(x = 30^{\circ}\). \
A1: Identifies \(x = 150^{\circ}\).

(a) Using the quotient rule with \(u = e^{3x}\) and \(v = 2x - 1\): \
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 3e^{3x}\) \
\(\frac{\mathrm{d}v}{\mathrm{d}x} = 2\) \
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3e^{3x}(2x-1) - 2(e^{3x})}{(2x-1)^2} = \frac{e^{3x}[3(2x-1) - 2]}{(2x-1)^2} = \frac{e^{3x}(6x - 5)}{(2x-1)^2}\) \
Thus, \(\mathrm{g}(x) = 6x - 5\). \
\
(b) For a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\). \
Since \(e^{3x} > 0\) for all \(x\), we solve \
\(6x - 5 = 0 \implies x = \frac{5}{6}\) \
Substituting this back into the original curve equation: \
\(y = \frac{e^{3(5/6)}}{2(5/6) - 1} = \frac{e^{5/2}}{\frac{5}{3} - 1} = \frac{e^{5/2}}{\frac{2}{3}} = \frac{3}{2}e^{5/2}\) \
So the exact coordinates of the stationary point are \(\left(\frac{5}{6}, \frac{3}{2}e^{5/2}\right)\). \
\
(c) To determine the nature, we analyze the sign of \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{3x}(6x-5)}{(2x-1)^2}\) around \(x = \frac{5}{6}\). \
For \(x > \frac{1}{2}\), both \(e^{3x}\) and \((2x-1)^2\) are always positive. \
Thus, the sign of the gradient is determined entirely by \(6x - 5\). \
- For \(x < \frac{5}{6}\), \(6x - 5 < 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} < 0\). \
- For \(x > \frac{5}{6}\), \(6x - 5 > 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} > 0\). \
Since the gradient changes from negative to positive, the stationary point is a local minimum.

Part (a): \
M1: Applies the quotient rule correctly with \(u = e^{3x}\) and \(v = 2x-1\) (allowing for minor sign slips). \
A1: Obtains a correct unsimplified numerator. \
A1: Correctly simplifies to obtain \(\mathrm{g}(x) = 6x-5\). \
\
Part (b): \
M1: Equates their derivative to 0 and solves for \(x\). \
A1: Obtains \(x = \frac{5}{6}\) (or equivalent). \
M1: Substitutes their \(x\) value back into the original function to find \(y\). \
A0.33: Obtains \(y = \frac{3}{2}e^{5/2}\) in exact form. \
\
Part (c): \
M1: Considers the sign of the gradient on both sides of \(x = \frac{5}{6}\) or attempts to find the second derivative. \
A1: Concludes it is a local minimum with a clear, valid justification.

(a) Let \(u = \ln x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\) \
Let \(\frac{\mathrm{d}v}{\mathrm{d}x} = x^2 \implies v = \frac{1}{3}x^3\) \
Using integration by parts formula \(\int u \frac{\mathrm{d}v}{\mathrm{d}x} \\, \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \\, \mathrm{d}x\): \
\(\int x^2 \ln x \\, \mathrm{d}x = \frac{1}{3}x^3 \ln x - \int \frac{1}{3}x^3 \cdot \frac{1}{x} \\, \mathrm{d}x\) \
\(= \frac{1}{3}x^3 \ln x - \frac{1}{3} \int x^2 \\, \mathrm{d}x\) \
\(= \frac{1}{3}x^3 \ln x - \frac{1}{3} \left(\frac{1}{3}x^3\right) + C\) \
\(= \frac{1}{3}x^3 \ln x - \frac{1}{9}x^3 + C\) (as required). \
\
(b) Applying limits: \
\(\int_{1}^{e} x^2 \ln x \\, \mathrm{d}x = \left[ \frac{1}{3}x^3 \ln x - \frac{1}{9}x^3 \right]_{1}^{e}\) \
\(= \left( \frac{1}{3}e^3 \ln e - \frac{1}{9}e^3 \right) - \left( \frac{1}{3}(1)^3 \ln(1) - \frac{1}{9}(1)^3 \right)\) \
Since \(\ln e = 1\) and \(\ln 1 = 0\): \
\(= \left( \frac{1}{3}e^3 - \frac{1}{9}e^3 \right) - \left( 0 - \frac{1}{9} \right)\) \
\(= \frac{2}{9}e^3 + \frac{1}{9} = \frac{2e^3 + 1}{9}\).

Part (a): \
M1: Selects correct parts \(u = \ln x\) and \(\mathrm{d}v/\mathrm{d}x = x^2\). \
A1: Obtains correct derivatives and integrals: \(\mathrm{d}u/\mathrm{d}x = 1/x\) and \(v = x^3/3\). \
M1: Correctly substitutes into the integration by parts formula. \
A1.33: Completes the integration with correct algebraic steps and constant of integration \(C\). \
\
Part (b): \
M1: Substitutes limits \(e\) and \(1\) into their expression from part (a). \
A1: Evaluates using \(\ln e = 1\) and \(\ln 1 = 0\). \
M1: Rearranges and collects terms with a common denominator. \
A1: Obtains the correct exact simplified form \(\frac{2e^3 + 1}{9}\) (or equivalent).

(a) At the start of the program, \(t = 0\): \
\(P = \frac{800}{1 + 3e^0} = \frac{800}{1 + 3} = \frac{800}{4} = 200\). \
\
(b) When \(P = 600\): \
\(600 = \frac{800}{1 + 3e^{-0.2t}}\) \
\(1 + 3e^{-0.2t} = \frac{800}{600} = \frac{4}{3}\) \
\(3e^{-0.2t} = \frac{4}{3} - 1 = \frac{1}{3}\) \
\(e^{-0.2t} = \frac{1}{9}\) \
\(-0.2t = \ln\left(\frac{1}{9}\right) = -\ln 9\) \
\(t = \frac{-\ln 9}{-0.2} = 5\ln 9 \approx 10.986\) \
To 3 significant figures, \(t = 11.0\) years. \
\
(c) The rate of increase is given by \(\frac{\mathrm{d}P}{\mathrm{d}t}\). \
Using the chain rule on \(P = 800(1 + 3e^{-0.2t})^{-1}\): \
\(\frac{\mathrm{d}P}{\mathrm{d}t} = -800(1 + 3e^{-0.2t})^{-2} \cdot (3 \cdot (-0.2) e^{-0.2t})\) \
\(\frac{\mathrm{d}P}{\mathrm{d}t} = \frac{480e^{-0.2t}}{(1 + 3e^{-0.2t})^2}\) \
When \(t = 5\): \
\(e^{-0.2(5)} = e^{-1}\) \
\(\frac{\mathrm{d}P}{\mathrm{d}t} = \frac{480e^{-1}}{(1 + 3e^{-1})^2}\) \
Since \(e^{-1} \approx 0.367879\): \
Numerator \( = 480 \times 0.367879 \approx 176.582\) \
Denominator \( = (1 + 3(0.367879))^2 = (2.103638)^2 \approx 4.42529\) \
\(\frac{\mathrm{d}P}{\mathrm{d}t} \approx \frac{176.582}{4.42529} \approx 39.9\) birds per year (to 1 decimal place).

Part (a): \
M1: Substitutes \(t = 0\) into the given formula. \
A1: Obtains 200. \
\
Part (b): \
M1: Sets \(P = 600\) and attempts to solve for \(e^{-0.2t}\). \
M1: Obtains \(e^{-0.2t} = 1/9\) (or equivalent). \
M1: Uses logarithms correctly to solve for \(t\). \
A0.33: Obtains \(t \approx 11.0\) (accept 11 or 11.0). \
\
Part (c): \
M1: Differentiates the expression for \(P\) using the chain rule or quotient rule. \
A1: Obtains correct derivative \(\frac{\mathrm{d}P}{\mathrm{d}t} = \frac{480e^{-0.2t}}{(1 + 3e^{-0.2t})^2}\). \
M1: Substitutes \(t = 5\) and evaluates to 1 decimal place. \
A1: Obtains 39.9.

(a) Evaluating \(\mathrm{f}(x)\) at the endpoints: \
\(\mathrm{f}(1.5) = 3(1.5)^2 - e^{1.5-1} - 8 = 6.75 - e^{0.5} - 8 \approx 6.75 - 1.6487 - 8 = -2.8987\) \
\(\mathrm{f}(2.0) = 3(2.0)^2 - e^{2.0-1} - 8 = 12 - e^{1.0} - 8 \approx 12 - 2.7183 - 8 = 1.2817\) \
Since \(\mathrm{f}(x)\) is continuous on \([1.5, 2.0]\) and there is a change of sign between \(\mathrm{f}(1.5)\) and \(\mathrm{f}(2.0)\), there is at least one root \(\alpha \in [1.5, 2.0]\). \
\
(b) Set \(\mathrm{f}(x) = 0\): \
\(3x^2 - e^{x-1} - 8 = 0\) \
\(3x^2 = e^{x-1} + 8\) \
\(x^2 = \frac{e^{x-1} + 8}{3}\) \
Since \(x > 0\) on the given interval, we take the positive square root: \
\(x = \sqrt{\frac{e^{x-1} + 8}{3}}\) (as required). \
\
(c) Applying the iterative formula with \(x_0 = 1.5\): \
\(x_1 = \sqrt{\frac{e^{1.5-1} + 8}{3}} = \sqrt{\frac{e^{0.5} + 8}{3}} \approx 1.793388 \approx 1.7934\) \
\(x_2 = \sqrt{\frac{e^{1.793388-1} + 8}{3}} = \sqrt{\frac{e^{0.793388} + 8}{3}} \approx 1.844892 \approx 1.8449\) \
\(x_3 = \sqrt{\frac{e^{1.844892-1} + 8}{3}} = \sqrt{\frac{e^{0.844892} + 8}{3}} \approx 1.855422 \approx 1.8554\).

Part (a): \
M1: Evaluates both \(\mathrm{f}(1.5)\) and \(\mathrm{f}(2.0)\) with at least one correct calculation. \
A1: Both values correct (to 1 s.f. or more) with a valid conclusion noting sign change and continuity. \
\
Part (b): \
M1: Rearranges the equation to isolate the \(x^2\) term on one side. \
A1: Completes steps to show the required form with positive root chosen because \(x > 0\). \
\
Part (c): \
M1: Substitutes \(x_0 = 1.5\) into the formula. \
A1: Obtains \(x_1 \approx 1.7934\). \
A1: Obtains \(x_2 \approx 1.8449\). \
A1.33: Obtains \(x_3 \approx 1.8554\).

(a) We expand \(R\cos(\theta - \alpha) = R\cos \theta \cos \alpha + R\sin \theta \sin \alpha\). \
Comparing coefficients with \(5\cos \theta + 12\sin \theta\): \
\(R\cos \alpha = 5\) \
\(R\sin \alpha = 12\) \
\(R = \sqrt{5^2 + 12^2} = 13\) \
\(\tan \alpha = \frac{12}{5} = 2.4 \implies \alpha = \arctan(2.4) \approx 1.176005 \approx 1.1760\) radians. \
So, \(5\cos \theta + 12\sin \theta = 13\cos(\theta - 1.1760)\). \
\
(b) Using part (a): \
\(13\cos(\theta - 1.1760) = 6.5 \implies \cos(\theta - 1.1760) = 0.5\). \
For \(0 \le \theta < 2\pi\), we have \(-1.1760 \le \theta - 1.1760 < 2\pi - 1.1760 \approx 5.1072\). \
The solutions to \(\cos \phi = 0.5\) in this interval are: \
\(\phi = -\frac{\pi}{3} \approx -1.0472\) \
\(\phi = \frac{\pi}{3} \approx 1.0472\) \
Thus, \
\(\theta - 1.1760 = -1.0472 \implies \theta = 1.1760 - 1.0472 = 0.1288 \approx 0.129\) radians. \
\(\theta - 1.1760 = 1.0472 \implies \theta = 1.1760 + 1.0472 = 2.2232 \approx 2.223\) radians. \
So, \(\theta \approx 0.129, 2.223\). \
\
(c) The expression is \(13\cos(\theta - 1.1760)\). \
Since the maximum value of the cosine function is 1, the maximum value of the expression is 13. \
This maximum occurs when \(\cos(\theta - 1.1760) = 1\) \
\(\theta - 1.1760 = 0 \implies \theta = 1.1760 \approx 1.176\) radians.

Part (a): \
M1: Correctly uses Pythagoras to find \(R = \sqrt{5^2 + 12^2}\). \
M1: Uses \(\tan \alpha = 12/5\) to set up equation for \(\alpha\). \
A1: Obtains \(R = 13\) and \(\alpha \approx 1.1760\). \
\
Part (b): \
M1: Uses their \(R\) and \(\alpha\) to set up equation \(\cos(\theta - \alpha) = 6.5/R\). \
M1: Finds at least one valid value of \(\theta - \alpha\) (e.g. \(\pi/3\) or \(-\pi/3\)). \
A1: Obtains \(\theta \approx 0.129\). \
A0.33: Obtains \(\theta \approx 2.223\). \
\
Part (c): \
B1: States the maximum value is 13. \
B1: States \(\theta \approx 1.176\) (accept 1.18 or 1.1760).

(a) Using log laws, since \(0 < x < \pi\), \(\sin x > 0\). \
We can rewrite: \
\(y = \ln((\sin x)^2) = 2\ln(\sin x)\) \
Using the chain rule: \
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2 \cdot \frac{1}{\sin x} \cdot (\cos x) = 2\cot x\) (as required). \
\
(b) At \(x = \frac{\pi}{4}\): \
\(y = \ln\left(\sin^2\left(\frac{\pi}{4}\right)\right) = \ln\left(\left(\frac{1}{\sqrt{2}}\right)^2\right) = \ln\left(\frac{1}{2}\right) = -\ln 2\). \
The gradient of the tangent is: \
\(m = 2\cot\left(\frac{\pi}{4}\right) = 2(1) = 2\). \
Using the equation of a straight line: \
\(y - y_1 = m(x - x_1)\) \
\(y - (-\ln 2) = 2\left(x - \frac{\pi}{4}\right)\) \
\(y + \ln 2 = 2x - \frac{\pi}{2}\) \
Rearranging into the form \(ax + by + c = 0\): \
\(2x - y - \frac{\pi}{2} - \ln 2 = 0\) \
Or: \
\(2x - y - \left(\frac{\pi}{2} + \ln 2\right) = 0\). \
Here, \(a = 2\), \(b = -1\), and \(c = -\left(\frac{\pi}{2} + \ln 2\right)\).

Part (a): \
M1: Rewrites the function using logarithm laws as \(2\ln(\sin x)\) or prepares to use the chain rule on \(\ln(\sin^2 x)\). \
M1: Applies the chain rule correctly to obtain a form \(\frac{1}{\sin^2 x} \cdot A\sin x \cos x\) or \(\frac{2}{\sin x} \cdot \cos x\). \
A1: Simplifies to \(2\cot x\) with complete and clear working shown. \
\
Part (b): \
M1: Substitutes \(x = \frac{\pi}{4}\) into the original equation to find the \(y\)-coordinate. \
A1: Obtains \(y = -\ln 2\) (or \(\ln(1/2)\)). \
M1: Substitutes \(x = \frac{\pi}{4}\) into their derivative to find the gradient \(m\). \
A1: Obtains gradient \(m = 2\). \
M1: Writes down the equation of a line using their point and gradient. \
A0.33: Correctly rearranges the line equation into the form \(ax+by+c=0\), identifying constants in terms of \(\pi\) and \(\ln 2\).

(a) Let \(u = \ln(3x - 2)\) and \(v = x^2\).
Differentiating with respect to \(x\):
\[\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{3}{3x - 2}\]
\[\frac{\mathrm{d}v}{\mathrm{d}x} = 2x\]
Using the quotient rule:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}\]
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 \left(\frac{3}{3x - 2}\right) - 2x \ln(3x - 2)}{x^4}\]
Simplifying this expression by factoring out \(x\) from the numerator and denominator:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x \left[ \frac{3x}{3x - 2} - 2 \ln(3x - 2) \right]}{x^4}\]
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x - 2(3x - 2)\ln(3x - 2)}{x^3(3x - 2)}\]

(b) At a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\).
Since \(x > \frac{2}{3}\), the denominator \(x^3(3x - 2) \ne 0\).
Therefore, the numerator must be zero:
\[3x - 2(3x - 2)\ln(3x - 2) = 0\]
\[2(3x - 2)\ln(3x - 2) = 3x\]
\[\ln(3x - 2) = \frac{3x}{2(3x - 2)}\]
\[\ln(3x - 2) = \frac{3x}{6x - 4}\]
Taking exponentials of both sides:
\[3x - 2 = \exp\left(\frac{3x}{6x - 4}\right)\]
\[3x = 2 + \exp\left(\frac{3x}{6x - 4}\right)\]
Dividing both sides by 3 gives:
\[x = \frac{2}{3} + \frac{1}{3}\exp\left(\frac{3x}{6x - 4}\right)\]
which is the required equation.

(c) Using the iterative formula with \(x_1 = 1.2\):
\[x_2 = \frac{2}{3} + \frac{1}{3}\exp\left(\frac{3(1.2)}{6(1.2)-4}\right)\]
\[x_2 = \frac{2}{3} + \frac{1}{3}\exp(1.125) \approx 1.6934056...\]
So \(x_2 \approx 1.6934\) (to 4 d.p.)

Now substituting \(x_2 \approx 1.6934056...\):
\[x_3 = \frac{2}{3} + \frac{1}{3}\exp\left(\frac{3(1.6934056)}{6(1.6934056)-4}\right) \approx 1.427028...\]
So \(x_3 \approx 1.4270\) (to 4 d.p.)

(a)
- M1: Applies the quotient rule (or product rule) to obtain an expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) of the form \(\frac{x^2 \cdot u' - 2x \cdot \ln(3x - 2)}{x^4}\) with some attempt at differentiating the numerator.
- A1: Correctly differentiates \(u = \ln(3x - 2)\) to obtain \(u' = \frac{3}{3x - 2}\) and \(v = x^2\) to obtain \(2x\).
- A1: Fully correct simplified derivative equivalent to \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x - 2(3x - 2)\ln(3x - 2)}{x^3(3x - 2)}\).

(b)
- M1: Sets their derivative equal to 0, isolates the logarithmic term, and attempts to write \(\ln(3x - 2)\) as a single algebraic fraction.
- M1: Eliminates the logarithm by correctly applying exponentials to both sides, leading to an expression for \(3x - 2\).
- A1*: Fully correct proof with no errors, leading to the given result.

(c)
- M1: Attempts to substitute \(x_1 = 1.2\) into the iterative formula to find \(x_2\).
- A1: Both \(x_2 \approx 1.6934\) and \(x_3 \approx 1.4270\) (accept 1.427) correct to 4 decimal places.

Assume the contradiction: Suppose that there is an integer \(n\) such that \(n^3 + 5\) is odd, and \(n\) is odd.

Since \(n\) is odd, we can write \(n = 2k + 1\) for some integer \(k\).

Substituting this expression for \(n\) into \(n^3 + 5\):
\[n^3 + 5 = (2k + 1)^3 + 5\]
\[= (8k^3 + 12k^2 + 6k + 1) + 5\]
\[= 8k^3 + 12k^2 + 6k + 6\]
\[= 2(4k^3 + 6k^2 + 3k + 3)\]

Since \(k\) is an integer, \(4k^3 + 6k^2 + 3k + 3\) is also an integer.
Therefore, \(n^3 + 5\) is a multiple of 2, which means \(n^3 + 5\) is even.

This contradicts our initial assumption that \(n^3 + 5\) is odd.

Hence, the assumption that \(n\) is odd must be false, which proves that if \(n^3 + 5\) is odd, then \(n\) must be even.

M1: Set up the contradiction by assuming that \(n\) is odd and \(n^3 + 5\) is odd.
M1: Express \(n\) as \(2k + 1\) (or any equivalent odd integer representation) where \(k \in \mathbb{Z}\).
M1: Expand \((2k + 1)^3 + 5\) algebraically.
A1: Correctly factorise out 2 to show that \(n^3 + 5 = 2(4k^3 + 6k^2 + 3k + 3)\), and state that this is even.
A1: Conclude the proof with a clear statement pointing out the contradiction.

(a) First, write \((4 - 9x)^{-\frac{1}{2}}\) in the standard form \((1 + X)^n\):
\[(4 - 9x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \left(1 - \frac{9}{4}x\right)^{-\frac{1}{2}} = \frac{1}{2} \left(1 - \frac{9}{4}x\right)^{-\frac{1}{2}}\]

Using the binomial expansion formula:
\[(1 + X)^n = 1 + nX + \frac{n(n-1)}{2!}X^2 + \frac{n(n-1)(n-2)}{3!}X^3 + \dots\]
with \(n = -\frac{1}{2}\) and \(X = -\frac{9}{4}x\):
\[\left(1 - \frac{9}{4}x\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{9}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\frac{9}{4}x\right)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}\left(-\frac{9}{4}x\right)^3 + \dots\]
\[= 1 + \frac{9}{8}x + \frac{3}{8}\left(\frac{81}{16}x^2\right) + \frac{5}{16}\left(\frac{729}{64}x^3\right) + \dots\]
\[= 1 + \frac{9}{8}x + \frac{243}{128}x^2 + \frac{3645}{1024}x^3 + \dots\]

Multiplying through by the scale factor of \(\frac{1}{2}\):
\[(4 - 9x)^{-\frac{1}{2}} = \frac{1}{2} + \frac{9}{16}x + \frac{243}{256}x^2 + \frac{3645}{2048}x^3\]

(b) The binomial expansion of \((1 + X)^n\) is valid for \(|X| < 1\).
Here, \(X = -\frac{9}{4}x\), so:
\[\left|-\frac{9}{4}x\right| < 1 \implies \frac{9}{4}|x| < 1 \implies |x| < \frac{4}{9}\]

M1: Factor out 4 correctly to get \(4^{-\frac{1}{2}}\) or \(\frac{1}{2}\) as a multiplier.
M1: Use the binomial expansion formula with \(n = -\frac{1}{2}\) and \(X = -\frac{9}{4}x\) with at least two terms correct.
A1: Correctly expand the bracketed term to get \(1 + \frac{9}{8}x + \frac{243}{128}x^2 + \frac{3645}{1024}x^3\).
A1: Multiply by \(\frac{1}{2}\) to get the final simplified coefficients: \(\frac{1}{2} + \frac{9}{16}x + \frac{243}{256}x^2 + \frac{3645}{2048}x^3\).
B1: State the correct validity condition: \(|x| < \frac{4}{9}\) or \(-\frac{4}{9} < x < \frac{4}{9}\).

(a) Differentiating the parametric equations with respect to \(t\):
\[\frac{dx}{dt} = 6t\]
\[\frac{dy}{dt} = 3t^2 - 4\]

Using the chain rule to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 - 4}{6t}\]

(b) At the point where \(t = 2\):
\[x = 3(2)^2 - 1 = 11\]
\[y = (2)^3 - 4(2) = 0\]

So the tangent passes through the point \((11, 0)\).

The gradient of the tangent at this point is found by substituting \(t = 2\) into the derivative:
\[m = \frac{3(2)^2 - 4}{6(2)} = \frac{12 - 4}{12} = \frac{8}{12} = \frac{2}{3}\]

Using the equation of a straight line:
\[y - y_1 = m(x - x_1)\]
\[y - 0 = \frac{2}{3}(x - 11)\]
\[3y = 2x - 22\]
\[2x - 3y - 22 = 0\]

M1: Differentiate both \(x\) and \(y\) with respect to \(t\).
A1: Obtain correct derivatives: \(\frac{dx}{dt} = 6t\) and \(\frac{dy}{dt} = 3t^2 - 4\).
M1: Apply the parametric differentiation formula \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).
A1: Correctly write \(\frac{dy}{dx} = \frac{3t^2 - 4}{6t}\).
B1: Find the coordinates of the point at \(t = 2\) to be \((11, 0)\).
M1: Substitute \(t = 2\) into their expression for \(\frac{dy}{dx}\) to find the gradient \(m = \frac{2}{3}\).
M1: Form the equation of a straight line with their gradient and point.
A1: Rearrange to the required integer form, e.g., \(2x - 3y - 22 = 0\) (or equivalent such as \(3y - 2x + 22 = 0\)).

(a) Differentiate both sides of the equation with respect to \(x\), using the product rule for \(-2xy\):
\[\frac{d}{dx}(2x^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(10)\]
\[4x - \left(2y + 2x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0\]
\[4x - 2y - 2x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\]

Rearranging to make \(\frac{dy}{dx}\) the subject:
\[(2y - 2x)\frac{dy}{dx} = 2y - 4x\]
\[\frac{dy}{dx} = \frac{2y - 4x}{2y - 2x} = \frac{y - 2x}{y - x}\]

(b) For the tangent to be parallel to the \(x\)-axis, the gradient must be zero:
\[\frac{dy}{dx} = 0 \implies y - 2x = 0 \implies y = 2x\]

Substitute \(y = 2x\) back into the original equation of the curve:
\[2x^2 - 2x(2x) + (2x)^2 = 10\]
\[2x^2 - 4x^2 + 4x^2 = 10\]
\[2x^2 = 10 \implies x^2 = 5 \implies x = \pm\sqrt{5}\]

Using \(y = 2x\) to find the corresponding \(y\)-coordinates:
- When \(x = \sqrt{5}\), \(y = 2\sqrt{5}\).
- When \(x = -\sqrt{5}\), \(y = -2\sqrt{5}\).

Therefore, the exact coordinates of the points are \((\sqrt{5}, 2\sqrt{5})\) and \((-\sqrt{5}, -2\sqrt{5})\).

M1: Differentiate \(2x^2\) to get \(4x\) and implicitly differentiate \(y^2\) to get \(2y\frac{dy}{dx}\).
M1: Apply the product rule to differentiate \(-2xy\) to obtain \(-2y - 2x\frac{dy}{dx}\).
A1: Correct differentiated equation: \(4x - 2y - 2x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\).
A1: Rearrange and simplify to find \(\frac{dy}{dx} = \frac{y - 2x}{y - x}\).
M1: Set \(\frac{dy}{dx} = 0\) to obtain \(y = 2x\).
M1: Substitute \(y = 2x\) into the original equation.
A1: Solve to obtain \(x = \pm\sqrt{5}\).
A1: State both correct exact coordinate points: \((\sqrt{5}, 2\sqrt{5})\) and \((-\sqrt{5}, -2\sqrt{5})\).

Let \(u = 2x + 1\). Then \(du = 2 \, dx\), which means \(dx = \frac{1}{2} \, du\).

Rearranging the substitution for \(x\):
\[2x = u - 1 \implies x = \frac{u - 1}{2}\]

Next, change the limits of integration from \(x\) to \(u\):
- When \(x = 0\), \(u = 2(0) + 1 = 1\).
- When \(x = 4\), \(u = 2(4) + 1 = 9\).

Substitute all components into the integral:
\[\int_0^4 \frac{x}{\sqrt{2x + 1}} \, dx = \int_1^9 \frac{\frac{u - 1}{2}}{\sqrt{u}} \cdot \left(\frac{1}{2} \, du\right)\]
\[= \frac{1}{4} \int_1^9 \frac{u - 1}{\sqrt{u}} \, du\]
\[= \frac{1}{4} \int_1^9 \left(u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right) \, du\]

Now perform the integration:
\[= \frac{1}{4} \left[ \frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}} \right]_1^9\]

Evaluate at the upper and lower limits:
\[= \frac{1}{4} \left[ \left(\frac{2}{3}(9)^{\frac{3}{2}} - 2(9)^{\frac{1}{2}}\right) - \left(\frac{2}{3}(1)^{\frac{3}{2}} - 2(1)^{\frac{1}{2}}\right) \right]\]
\[= \frac{1}{4} \left[ \left(\frac{2}{3}(27) - 2(3)\right) - \left(\frac{2}{3} - 2\right) \right]\]
\[= \frac{1}{4} \left[ (18 - 6) - \left(-\frac{4}{3}\right) \right]\]
\[= \frac{1}{4} \left[ 12 + \frac{4}{3} \right] = \frac{1}{4} \left[ \frac{40}{3} \right] = \frac{10}{3}\]

Thus, the value of the integral is \(\frac{10}{3}\), so \(a = 10\).

M1: Differentiate the substitution to find \(du = 2 \, dx\) or equivalent.
B1: Correctly change the integration limits from \(x \in [0, 4]\) to \(u \in [1, 9]\).
M1: Substitute \(x = \frac{u-1}{2}\) and \(dx\) to express the integral completely in terms of \(u\).
A1: Obtain the simplified integral form \(\frac{1}{4} \int_1^9 (u^{1/2} - u^{-1/2}) \, du\).
M1: Integrate to obtain \(\frac{1}{4} \left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right]\).
M1: Substitute the limits 9 and 1 and show calculations.
A1: Reach the final exact value \(\frac{10}{3}\) and state that \(a = 10\).

(a) Set up the partial fractions identity:
\[\frac{2x^2 + 7x}{(x+2)^2(x-1)} = \frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{x-1}\]
Multiplying both sides by \((x+2)^2(x-1)\):
\[2x^2 + 7x = A(x+2)(x-1) + B(x-1) + C(x+2)^2\]

To find \(C\), substitute \(x = 1\):
\[2(1)^2 + 7(1) = C(1+2)^2 \implies 9 = 9C \implies C = 1\]

To find \(B\), substitute \(x = -2\):
\[2(-2)^2 + 7(-2) = B(-2-1) \implies 8 - 14 = -3B \implies -6 = -3B \implies B = 2\]

To find \(A\), substitute \(x = 0\) (and use \(B=2\), \(C=1\)):
\[0 = A(2)(-1) + 2(-1) + 1(2)^2\]
\[0 = -2A - 2 + 4 \implies 2A = 2 \implies A = 1\]

Therefore, the partial fraction decomposition is:
\[f(x) = \frac{1}{x+2} + \frac{2}{(x+2)^2} + \frac{1}{x-1}\]

(b) Integrate \(f(x)\) using the partial fractions:
\[\int_2^3 f(x) \, dx = \int_2^3 \left( \frac{1}{x+2} + \frac{2}{(x+2)^2} + \frac{1}{x-1} \right) \, dx\]
\[= \left[ \ln|x+2| - \frac{2}{x+2} + \ln|x-1| \right]_2^3\]

Substitute the upper limit \(x = 3\):
\[\left( \ln(5) - \frac{2}{5} + \ln(2) \right) = \ln(10) - \frac{2}{5}\]

Substitute the lower limit \(x = 2\):
\[\left( \ln(4) - \frac{2}{4} + \ln(1) \right) = \ln(4) - \frac{1}{2}\]

Subtract the lower limit evaluation from the upper limit evaluation:
\[\left( \ln(10) - \frac{2}{5} \right) - \left( \ln(4) - \frac{1}{2} \right)\]
\[= \ln(10) - \ln(4) - \frac{2}{5} + \frac{1}{2}\]
\[= \ln\left(\frac{10}{4}\right) + \frac{1}{10} = \ln\left(\frac{5}{2}\right) + \frac{1}{10}\]

So \(p = \frac{5}{2}\) and \(q = \frac{1}{10}\).

M1: Set up the correct algebraic identity for partial fractions with a repeated linear factor.
M1: Attempt to solve for the constants by substituting values or equating coefficients.
A1: Correct values for two of the constants (e.g., \(B=2, C=1\)).
A1: All constants correct: \(A=1, B=2, C=1\).
M1: Integrate their partial fractions to obtain log and reciprocal terms.
A1: Correct integration: \(\ln(x+2) - \frac{2}{x+2} + \ln(x-1)\).
M1: Substitute the limits 3 and 2 and simplify terms using logarithm properties.
A1: State the final exact answer: \(\ln\left(\frac{5}{2}\right) + \frac{1}{10}\) (or equivalent with \(p = 2.5\), \(q = 0.1\)).

First, separate the variables:
\[\int \frac{1}{y^2} \, dy = \int \frac{1}{2x + 3} \, dx\]

Integrate both sides:
\[-\frac{1}{y} = \frac{1}{2} \ln|2x + 3| + C\]

Since \(x > -1.5\), \(2x + 3 > 0\), so we can drop the absolute value bars:
\[-\frac{1}{y} = \frac{1}{2} \ln(2x + 3) + C\]

Apply the boundary conditions \(y = 1\) when \(x = 3\) to find \(C\):
\[-\frac{1}{1} = \frac{1}{2} \ln(2(3) + 3) + C\]
\[-1 = \frac{1}{2} \ln(9) + C\]

Using log properties, \(\frac{1}{2} \ln(9) = \ln(9^{\frac{1}{2}}) = \ln(3)\):
\[-1 = \ln(3) + C \implies C = -1 - \ln(3)\]

Substitute the expression for \(C\) back into the equation:
\[-\frac{1}{y} = \frac{1}{2} \ln(2x + 3) - \ln(3) - 1\]
\[-\frac{1}{y} = \ln\left(\sqrt{2x+3}\right) - \ln(3) - 1\]
\[-\frac{1}{y} = \ln\left(\frac{\sqrt{2x + 3}}{3}\right) - 1\]

Multiply by \(-1\):
\[\frac{1}{y} = 1 - \ln\left(\frac{\sqrt{2x + 3}}{3}\right)\]

Take the reciprocal to get \(y\):
\[y = \frac{1}{1 - \ln\left(\frac{\sqrt{2x + 3}}{3}\right)}\]

M1: Separate variables correctly: \(\int \frac{1}{y^2} \, dy = \int \frac{1}{2x + 3} \, dx\).
A1: Perform integration on both sides to get \(-\frac{1}{y}\) and \(\frac{1}{2} \ln(2x + 3)\) (+ \(C\)).
M1: Substitute the given condition \(x=3, y=1\) into an equation containing a constant of integration.
A1: Determine \(C = -1 - \ln 3\) (or any equivalent form like \(C = -1 - \frac{1}{2}\ln 9\)).
M1: Rearrange the integrated expression to make \(y\) the subject.
A1: Obtain the final simplified solution: \(y = \frac{1}{1 - \ln\left(\frac{\sqrt{2x + 3}}{3}\right)}\) (or any equivalent correct algebraic form, such as \(y = \frac{2}{2 - \ln(2x+3) + \ln 9}\)).

(a) To show they do not intersect, we assume they do and set up simultaneous equations by equating the components of \(l_1\) and \(l_2\):
\[2 + \lambda = 5 + 2\mu \implies \lambda - 2\mu = 3 \quad \text{(Equation 1)}\]
\[-1 + 3\lambda = 1 - \mu \implies 3\lambda + \mu = 2 \quad \text{(Equation 2)}\]
\[4 - 2\lambda = -3 + \mu \implies 2\lambda + \mu = 7 \quad \text{(Equation 3)}\]

From Equation 2, we can express \(\mu\) as:
\[\mu = 2 - 3\lambda\]

Substitute this into Equation 1:
\[\lambda - 2(2 - 3\lambda) = 3\]
\[\lambda - 4 + 6\lambda = 3\]
\[7\lambda = 7 \implies \lambda = 1\]

Now find \(\mu\):
\[\mu = 2 - 3(1) = -1\]

Test these values of \(\lambda\) and \(\mu\) in Equation 3:
\[\text{LHS of Equation 3} = 2\lambda + \mu = 2(1) + (-1) = 1\]
\[\text{RHS of Equation 3} = 7\]

Since \(1 \neq 7\), the values \(\lambda = 1\) and \(\mu = -1\) do not satisfy Equation 3.
Therefore, the system of equations has no solution, which means the lines \(l_1\) and \(l_2\) do not intersect.

(b) The direction vectors of the lines are:
\[\mathbf{d_1} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} \quad \text{and} \quad \mathbf{d_2} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\]

Calculate the dot product of the direction vectors:
\[\mathbf{d_1} \cdot \mathbf{d_2} = (1)(2) + (3)(-1) + (-2)(1) = 2 - 3 - 2 = -3\]

Calculate the magnitudes of each direction vector:
\[|\mathbf{d_1}| = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{14}\]
\[|\mathbf{d_2}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}\]

Let \(\theta\) be the acute angle between the lines:
\[\cos\theta = \frac{|\mathbf{d_1} \cdot \mathbf{d_2}|}{|\mathbf{d_1}| |\mathbf{d_2}|} = \frac{|-3|}{\sqrt{14}\sqrt{6}} = \frac{3}{\sqrt{84}} = \frac{3}{2\sqrt{21}} = \frac{\sqrt{21}}{14}\]

M1: Write three simultaneous equations by equating components of the two vector line equations.
M1: Solve any two of the equations to find values for \(\lambda\) and \(\mu\).
A1: Correctly find \(\lambda = 1\) and \(\mu = -1\).
A1: Substitute into the remaining equation, show that it is not satisfied, and conclude that the lines do not intersect.
M1: Use the direction vectors \(\begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\) in the scalar product formula.
A1: Correct magnitudes of \(\sqrt{14}\) and \(\sqrt{6}\).
A1: Correct dot product of \(-3\) (or absolute value \(3\)).
A1: Correct exact value for the cosine of the acute angle: \(\frac{\sqrt{21}}{14}\) (or equivalent like \(\frac{3}{\sqrt{84}}\)).

(a) Let \(\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}\). Then \(1 = A(x+1) + Bx\). Setting \(x = 0\) gives \(A = 1\). Setting \(x = -1\) gives \(B = -1\). Thus, \(\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}\). (b) Separating the variables in the differential equation gives \(\int \frac{1}{y \ln y} \text{d}y = \int \frac{1}{x(x+1)} \text{d}x\). Integrating the left-hand side using the substitution \(u = \ln y\), we have \(\text{d}u = \frac{1}{y} \text{d}y\), so \(\int \frac{1}{u} \text{d}u = \ln |u| = \ln(\ln y)\) since \(y > 1\). Integrating the right-hand side using the result from part (a) gives \(\int \left(\frac{1}{x} - \frac{1}{x+1}\right) \text{d}x = \ln x - \ln(x+1) + c = \ln\left(\frac{x}{x+1}\right) + c\) since \(x > 0\). Equating the two sides: \(\ln(\ln y) = \ln\left(\frac{x}{x+1}\right) + c\). Since \(y = e^2\) when \(x = 1\), we substitute these values: \(\ln(\ln e^2) = \ln\left(\frac{1}{1+1}\right) + c \implies \ln 2 = \ln\left(\frac{1}{2}\right) + c \implies \ln 2 = -\ln 2 + c \implies c = 2\ln 2 = \ln 4\). Thus, \(\ln(\ln y) = \ln\left(\frac{x}{x+1}\right) + \ln 4 = \ln\left(\frac{4x}{x+1}\right)\). Taking exponentials of both sides: \(\ln y = \frac{4x}{x+1}\). Taking exponentials again: \(y = e^{\frac{4x}{x+1}}\).

(a) M1: Writes \(\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}\) and attempts to find \(A\) and \(B\) by clearing denominators and substituting or equating coefficients. A1: Correct partial fractions \(\frac{1}{x} - \frac{1}{x+1}\). (b) M1: Separates variables to get \(\int \frac{1}{y \ln y} \text{d}y = \int \frac{1}{x(x+1)} \text{d}x\). M1: Integrates LHS to obtain \(\ln(\ln y)\) or equivalent. A1ft: Integrates RHS to obtain \(\ln x - \ln(x+1)\) or equivalent based on their part (a). M1: Substitutes the boundary condition \(x=1\), \(y=e^2\) into their integrated equation containing a constant of integration. A1: Obtains a correct constant of integration, e.g., \(c = \ln 4\) when the equation is written as \(\ln(\ln y) = \ln\left(\frac{x}{x+1}\right) + c\). A1: Correct final equation in the form \(y = f(x)\), which is \(y = e^{\frac{4x}{x+1}}\).