2026 Edexcel IAS-Level Pure Mathematics (XPM01) 模擬試題連答案詳解

To find the equation of the tangent, we first need the \(y\)-coordinate of the point \(P\).

Substitute \(x = 4\) into the equation of the curve \(C\):

\[y = 2(4)^2 - \frac{8}{\sqrt{4}} + 3 = 32 - 4 + 3 = 31\]

So the coordinates of \(P\) are \((4, 31)\).

Next, we write the equation of \(C\) in a form ready for differentiation:

\[y = 2x^2 - 8x^{-\frac{1}{2}} + 3\]

Differentiating with respect to \(x\):

\[\frac{\text{d}y}{\text{d}x} = 4x - 8\left(-\frac{1}{2}\right)x^{-\frac{3}{2}} = 4x + 4x^{-\frac{3}{2}}\]

To find the gradient of the tangent at \(P\), substitute \(x = 4\) into \(\frac{\text{d}y}{\text{d}x}\):

\[\text{Gradient} = 4(4) + 4(4)^{-\frac{3}{2}} = 16 + 4\left(\frac{1}{8}\right) = 16 + \frac{1}{2} = \frac{33}{2}\]

The equation of the tangent at \(P(4, 31)\) is:

\[y - 31 = \frac{33}{2}(x - 4)\]

Multiply the entire equation by 2 to clear the fraction:

\[2y - 62 = 33(x - 4)\]

\[2y - 62 = 33x - 132\]

Rearranging into the form \(ax + by + c = 0\):

\[33x - 2y - 70 = 0\]

**B1**: For finding \(y = 31\) when \(x = 4\). This may be seen in the equation of the tangent or coordinate pair.

**M1**: For rewriting \(\frac{8}{\sqrt{x}}\) as \(8x^{-1/2}\) and attempting to differentiate. Award for decreasing any power by 1 (e.g. \(x^2 \to x\) or \(x^{-1/2} \to x^{-3/2}\)).

**A1**: Correct derivative: \(\frac{\text{d}y}{\text{d}x} = 4x + 4x^{-3/2}\) or equivalent (e.g., \(4x + \frac{4}{x\sqrt{x}}\)).

**M1**: Substitutes \(x = 4\) into their \(\frac{\text{d}y}{\text{d}x}\) to find a numerical gradient.

**M1**: Attempts to find the equation of the tangent using their \(y\) value and their numerical gradient at \(x = 4\). If using \(y = mx + c\), they must attempt to find a value for \(c\).

**A1**: \(33x - 2y - 70 = 0\) or any non-zero integer multiple thereof (such as \(-33x + 2y + 70 = 0\)).

To solve the simultaneous equations:

1) \( 2x + y = 5 \)

2) \( x^2 - xy + 2y^2 = 14 \)

First, rearrange the linear equation (1) to make \( y \) the subject:
\( y = 5 - 2x \)

Substitute this expression for \( y \) into the quadratic equation (2):
\( x^2 - x(5 - 2x) + 2(5 - 2x)^2 = 14 \)

Expand the terms:
\( x^2 - 5x + 2x^2 + 2(25 - 20x + 4x^2) = 14 \)
\( 3x^2 - 5x + 50 - 40x + 8x^2 = 14 \)

Combine like terms:
\( 11x^2 - 45x + 50 = 14 \)

Subtract 14 from both sides to form a quadratic equation equal to zero:
\( 11x^2 - 45x + 36 = 0 \)

Factorise the quadratic equation:
\( (11x - 12)(x - 3) = 0 \)

This gives the solutions for \( x \):
\( x = 3 \) or \( x = \frac{12}{11} \)

Now find the corresponding \( y \) values by substituting the \( x \) values back into the linear equation \( y = 5 - 2x \):

For \( x = 3 \):
\( y = 5 - 2(3) = -1 \)

For \( x = \frac{12}{11} \):
\( y = 5 - 2\left(\frac{12}{11}\right) = \frac{55}{11} - \frac{24}{11} = \frac{31}{11} \)

Thus, the solutions are:
\( x = 3, y = -1 \) and \( x = \frac{12}{11}, y = \frac{31}{11} \)

M1: Rearranges the linear equation to make \( y \) (or \( x \)) the subject and substitutes it into the quadratic equation.
A1: Obtains a correct simplified 3-term quadratic equation in one variable, e.g., \( 11x^2 - 45x + 36 = 0 \) (or \( 11y^2 - 19y - 62 = 0 \) if substituting for \( x \)).
M1: Solves their 3-term quadratic equation by any valid method, such as factorisation, completing the square, or applying the quadratic formula.
A1: Obtains correct values for one variable: \( x = 3 \) and \( x = \frac{12}{11} \) (or equivalent).
M1: Substitutes at least one of their found values back into the linear equation to find the corresponding value of the other variable.
A1: Correct, fully simplified pairs of solutions: \( x = 3, y = -1 \) and \( x = \frac{12}{11}, y = \frac{31}{11} \). Candidates must match the correct \( x \) with the correct \( y \).

(a) Expanding the inequality: \( 2x(x + 3) > x + 12 \implies 2x^2 + 6x > x + 12 \implies 2x^2 + 5x - 12 > 0 \). Factorising the quadratic expression gives \( (2x - 3)(x + 4) > 0 \). The critical values are \( x = -4 \) and \( x = \frac{3}{2} \). Since we require the expression to be greater than 0, the solution is \( x < -4 \) or \( x > \frac{3}{2} \). (b) Solving the linear inequality: \( 3(x + 1) \ge 5 - x \implies 3x + 3 \ge 5 - x \implies 4x \ge 2 \implies x \ge \frac{1}{2} \). To satisfy both inequalities, we find the intersection of the two solution sets. Since \( x < -4 \) does not overlap with \( x \ge \frac{1}{2} \), and the region \( x > \frac{3}{2} \) is entirely contained within \( x \ge \frac{1}{2} \), the combined set of values is \( x > \frac{3}{2} \) (or in set notation, \( \{x : x > \frac{3}{2}\} \)).

(a) M1: Attempts to expand and rearrange the quadratic inequality into a three-term quadratic form, e.g., \( 2x^2 + 5x - 12 > 0 \). A1: Correctly identifies the critical values \( x = -4 \) and \( x = \frac{3}{2} \). A1: Correct range of \( x < -4 \) or \( x > \frac{3}{2} \) (or equivalent set notation). (b) M1: Solves the linear inequality to find \( x \ge \frac{1}{2} \) and attempts to find the intersection with the range from part (a). A1: Correct final range \( x > \frac{3}{2} \) (or equivalent set notation).

\( \text{(a)} \) First, find the midpoint \( M \) of the line segment \( AB \):
\( M = \left(\frac{2+6}{2}, \frac{-1+7}{2}\right) = (4, 3) \)

Next, find the gradient of the line segment \( AB \):
\( m_{AB} = \frac{7 - (-1)}{6 - 2} = \frac{8}{4} = 2 \)

Since \( l_1 \) is perpendicular to \( AB \), its gradient \( m_1 \) is:
\( m_1 = -\frac{1}{m_{AB}} = -\frac{1}{2} \)

Using the point-gradient form of a straight line with point \( M(4, 3) \) and gradient \( -\frac{1}{2} \):
\( y - 3 = -\frac{1}{2}(x - 4) \)
\( 2(y - 3) = -(x - 4) \)
\( 2y - 6 = -x + 4 \)
\( x + 2y - 10 = 0 \)

\( \text{(b)} \) To find the coordinates of \( P \), solve the equations of \( l_1 \) and \( l_2 \) simultaneously:
\( l_1: x + 2y - 10 = 0 \)
\( l_2: y = 3x - 2 \)

Substitute \( y = 3x - 2 \) into the equation for \( l_1 \):
\( x + 2(3x - 2) - 10 = 0 \)
\( x + 6x - 4 - 10 = 0 \)
\( 7x - 14 = 0 \implies x = 2 \)

Substitute \( x = 2 \) back into the equation for \( l_2 \):
\( y = 3(2) - 2 = 4 \)

So, the coordinates of \( P \) are \( (2, 4) \).

\( \text{(c)} \) Let the coordinates of \( Q \) be \( (k, 3k - 2) \) since \( Q \) lies on \( l_2 \).
The distance from \( P(2, 4) \) to \( Q(k, 3k-2) \) is \( 3\sqrt{10} \). Using the distance formula:
\( \sqrt{(k - 2)^2 + ((3k - 2) - 4)^2} = 3\sqrt{10} \)

Square both sides:
\( (k - 2)^2 + (3k - 6)^2 = (3\sqrt{10})^2 \)
\( (k - 2)^2 + [3(k - 2)]^2 = 90 \)
\( (k - 2)^2 + 9(k - 2)^2 = 90 \)
\( 10(k - 2)^2 = 90 \)
\( (k - 2)^2 = 9 \)

Taking the square root of both sides:
\( k - 2 = \pm 3 \)

This gives two cases:
1) \( k - 2 = 3 \implies k = 5 \)
\( y = 3(5) - 2 = 13 \)
So, one possible position for \( Q \) is \( (5, 13) \).

2) \( k - 2 = -3 \implies k = -1 \)
\( y = 3(-1) - 2 = -5 \)
So, the other possible position for \( Q \) is \( (-1, -5) \).

\( \textbf{Part (a)} \)
\( \bullet \) **M1**: Attempts to find the midpoint of \( AB \) or the gradient of \( AB \). Look for correct structure using the coordinates of \( A \) and \( B \).
\( \bullet \) **A1**: Both correct midpoint \( (4, 3) \) and correct gradient of \( AB \) which is \( 2 \) (can be implied by a perpendicular gradient of \( -\frac{1}{2} \)).
\( \bullet \) **M1**: Uses the negative reciprocal of their gradient of \( AB \) to find the perpendicular gradient, and attempts to write the equation of the line passing through their midpoint with this gradient.
\( \bullet \) **A1**: Fully correct equation of \( l_1 \) in the form \( ax + by + c = 0 \) with integer coefficients (e.g. \( x + 2y - 10 = 0 \), \( 2x + 4y - 20 = 0 \), etc.).

\( \textbf{Part (b)} \)
\( \bullet \) **M1**: Attempts to solve the simultaneous equations of \( l_1 \) and \( l_2 \) by substituting \( y = 3x - 2 \) into their equation for \( l_1 \) (or equivalent method to eliminate one variable).
\( \bullet \) **A1**: Correct coordinates \( (2, 4) \) or \( x = 2, y = 4 \).

\( \textbf{Part (c)} \)
\( \bullet \) **M1**: Expresses the coordinates of \( Q \) in terms of a single variable (e.g., \( (k, 3k-2) \)) and sets up an equation for the distance between \( P \) and \( Q \) equal to \( 3\sqrt{10} \).
\( \bullet \) **M1**: Correctly expands and simplifies to a quadratic equation, solving it to find two values for the variable (e.g., \( k = 5, -1 \)).
\( \bullet \) **A1**: Both correct sets of coordinates: \( (5, 13) \) and \( (-1, -5) \).

### Part (a)
To find the perimeter \(P\) of the shape \(OACB\), we sum the lengths of its boundary lines:
\[ P = OA + \text{arc } AB + BC + CO \]

1. The length of the straight line segment \(OA\) is the radius of the sector:
\[ OA = r \]
2. The arc length of the sector \(OAB\) is given by:
\[ \text{arc } AB = r\theta \]
3. To find the length \(BC\), we apply the cosine rule to triangle \(OBC\):
\[ BC^2 = OB^2 + OC^2 - 2(OB)(OC)\cos\left(\frac{\pi}{3}\right) \]
Given that \(OB = r\) and \(OC = 2r\):
\[ BC^2 = r^2 + (2r)^2 - 2(r)(2r)(0.5) \]
\[ BC^2 = r^2 + 4r^2 - 2r^2 = 3r^2 \]
Since \(r > 0\), we have:
\[ BC = r\sqrt{3} \]
4. The length of the straight line segment \(CO\) is given as:
\[ CO = 2r \]

Summing these boundary lengths:
\[ P = r + r\theta + r\sqrt{3} + 2r = r\theta + 3r + r\sqrt{3} = r(\theta + 3 + \sqrt{3}) \]
This completes the proof.

### Part (b)
The area \(A\) of the shape is the sum of the area of the sector \(OAB\) and the area of the triangle \(OBC\):
\[ A = \text{Area}(OAB) + \text{Area}(OBC) \]

1. The area of the sector is:
\[ \text{Area}(OAB) = \frac{1}{2}r^2\theta \]
2. The area of the triangle is:
\[ \text{Area}(OBC) = \frac{1}{2}(OB)(OC)\sin\left(\frac{\pi}{3}\right) = \frac{1}{2}(r)(2r)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}r^2 \]

So the total area is:
\[ A = \frac{1}{2}r^2\theta + \frac{\sqrt{3}}{2}r^2 \]
Given that \(A = 36\):
\[ \frac{1}{2}r^2\theta + \frac{\sqrt{3}}{2}r^2 = 36 \implies r^2\theta + \sqrt{3}r^2 = 72 \]
This can be written as:
\[ r(r\theta) + \sqrt{3}r^2 = 72 \quad \text{--- (Equation 1)} \]

We are also given that the perimeter is \(30\text{ cm}\):
\[ r(\theta + 3 + \sqrt{3}) = 30 \implies r\theta + 3r + r\sqrt{3} = 30 \]
Rearranging to express \(r\theta\) in terms of \(r\):
\[ r\theta = 30 - r(3 + \sqrt{3}) \quad \text{--- (Equation 2)} \]

Substituting \text{Equation 2} into \text{Equation 1}:
\[ r[30 - r(3 + \sqrt{3})] + \sqrt{3}r^2 = 72 \]
\[ 30r - 3r^2 - r^2\sqrt{3} + \sqrt{3}r^2 = 72 \]
Notice that the terms involving \(\sqrt{3}r^2\) cancel out:
\[ 30r - 3r^2 = 72 \]
Rearranging all terms to one side:
\[ 3r^2 - 30r + 72 = 0 \]
as required.

### Part (c)
We solve the quadratic equation:
\[ 3r^2 - 30r + 72 = 0 \]
Dividing the entire equation by 3:
\[ r^2 - 10r + 24 = 0 \]
Factorizing:
\[ (r - 4)(r - 6) = 0 \]
This gives two possible values of \(r\):
\[ r = 4 \quad \text{or} \quad r = 6 \]

Now we find the corresponding values of \(\theta\) using \(\theta = \frac{30}{r} - (3 + \sqrt{3})\):

- When \(r = 4\):
\[ \theta = \frac{30}{4} - (3 + \sqrt{3}) = 7.5 - 3 - \sqrt{3} = 4.5 - \sqrt{3} \approx 2.77 \text{ radians} \]
- When \(r = 6\):
\[ \theta = \frac{30}{6} - (3 + \sqrt{3}) = 5 - 3 - \sqrt{3} = 2 - \sqrt{3} \approx 0.268 \text{ radians} \]

### Part (a)
- **M1**: Applies the cosine rule on triangle \(OBC\) to find the length of \(BC\). Accept \(BC^2 = r^2 + (2r)^2 - 2(r)(2r)\cos\left(\frac{\pi}{3}\right)\).
- **A1**: Simplifies to find \(BC = r\sqrt{3}\).
- **A1**: Writes the perimeter expression \(P = OA + \text{arc } AB + BC + OC = r + r\theta + r\sqrt{3} + 2r\) and factors out \(r\) to obtain \(r(\theta + 3 + \sqrt{3})\) with no errors seen.

### Part (b)
- **M1**: Writes down an expression for the total area \(A\) of the shape in terms of \(r\) and \(\theta\) and equates it to 36. Accept \(\frac{1}{2}r^2\theta + \frac{\sqrt{3}}{2}r^2 = 36\) or equivalent.
- **M1**: Uses the perimeter equation to write an expression for \(r\theta\) in terms of \(r\). Accept \(r\theta = 30 - r(3 + \sqrt{3})\).
- **M1**: Substitutes their expression for \(r\theta\) into their area equation.
- **A1\***: Fully correct algebraic simplification leading to the given quadratic equation \(3r^2 - 30r + 72 = 0\) with no errors or omissions in the working.

### Part (c)
- **M1**: Solves the quadratic equation to find two values of \(r\). Accept \(r = 4\) and \(r = 6\).
- **A1**: Finds the two corresponding values of \(\theta\): \(\theta \approx 2.77\) (accept answers in range \([2.76, 2.77]\)) and \(\theta \approx 0.268\) (accept answers in range \([0.267, 0.268]\)).

(a) To sketch \( y = (x+1)(x-2)^2 \):
- When \( y = 0 \), \( x = -1 \) or \( x = 2 \). Since the factor \( (x-2) \) is squared, the curve touches the \( x \)-axis at \( (2, 0) \) and crosses the \( x \)-axis at \( (-1, 0) \).
- When \( x = 0 \), \( y = (1)(-2)^2 = 4 \). So the \( y \)-intercept is at \( (0, 4) \).
- The curve is a positive cubic, so it starts in the third quadrant and goes to the first quadrant.
- A stationary point analysis (optional for the sketch but helpful) shows the local maximum is exactly at \( (0, 4) \) and the local minimum is at \( (2, 0) \).

(b) First, applying a stretch of scale factor \( \frac{1}{2} \) parallel to the \( x \)-axis replaces \( x \) with \( 2x \):
\( y = f(2x) = (2x+1)(2x-2)^2 \)
Expanding the original equation of \( C \):
\( y = (x+1)(x^2 - 4x + 4) = x^3 - 3x^2 + 4 \)
Substituting \( 2x \) into this expanded form:
\( y = (2x)^3 - 3(2x)^2 + 4 = 8x^3 - 12x^2 + 4 \)

Next, applying a translation of 3 units in the positive \( y \)-direction adds 3 to the function:
\( y = 8x^3 - 12x^2 + 4 + 3 = 8x^3 - 12x^2 + 7 \)
So, the equation of \( C_1 \) is \( y = 8x^3 - 12x^2 + 7 \) (where \( p=8, q=-12, r=0, s=7 \)).

(c) A translation of \( a \) units in the positive \( x \)-direction replaces \( x \) with \( x - a \).
So, the equation of \( C_2 \) is:
\( y = (x-a+1)(x-a-2)^2 \)
The \( y \)-intercept is found by setting \( x = 0 \):
\( y_{intercept} = (1-a)(-a-2)^2 = (1-a)(a+2)^2 \)
Setting this equal to \( -16 \):
\( (1-a)(a^2 + 4a + 4) = -16 \)
\( a^2 + 4a + 4 - a^3 - 4a^2 - 4a = -16 \)
\( -a^3 - 3a^2 + 4 = -16 \)
\( a^3 + 3a^2 - 20 = 0 \) (as required).

To solve this cubic, we test integer values by the Factor Theorem:
For \( a = 2 \):
\( 2^3 + 3(2^2) - 20 = 8 + 12 - 20 = 0 \)
So \( a = 2 \) is a root, meaning \( (a - 2) \) is a factor.
By algebraic division:
\( a^3 + 3a^2 - 20 = (a - 2)(a^2 + 5a + 10) = 0 \)
For the quadratic equation \( a^2 + 5a + 10 = 0 \), the discriminant is:
\( \Delta = b^2 - 4ac = 5^2 - 4(1)(10) = 25 - 40 = -15 \)
Since the discriminant is negative (\( \Delta < 0 \)), there are no other real roots.
Thus, \( a = 2 \) is the only real solution.

(a)
- B1: Correct shape of a positive cubic curve with one local maximum and one local minimum touching the positive \( x \)-axis.
- B1: Curve passes through \( (-1, 0) \) and touches at \( (2, 0) \). (Coordinates must be correctly labeled on the sketch or stated in the text).
- B1: Curve crosses the \( y \)-axis at \( (0, 4) \).

(b)
- M1: Attempts to apply the horizontal stretch by replacing \( x \) with \( 2x \) in either the factored or expanded expression.
- A1: Correctly expands to obtain the intermediate cubic \( y = 8x^3 - 12x^2 + 4 \).
- A1: Adds 3 to obtain the final equation \( y = 8x^3 - 12x^2 + 7 \) (or lists the values of \( p, q, r, s \) correctly).

(c)
- M1: Substitutes \( x = 0 \) into \( y = f(x-a) \) and sets equal to \( -16 \).
- A1*: Correctly expands and simplifies with no errors shown to obtain the given cubic equation \( a^3 + 3a^2 - 20 = 0 \).
- A1: Identifies \( a = 2 \) as a root, factorises the cubic into \( (a-2)(a^2+5a+10) = 0 \), and uses the discriminant (or completing the square) to show that the quadratic factor has no real roots, concluding that \( a = 2 \) is the only real solution.

(a) The volume of a cylinder is given by \(V = \pi r^2 h\). Since \(V = 128\pi\), we have \(\pi r^2 h = 128\pi\), which simplifies to \(h = \dfrac{128}{r^2}\). The total surface area of a solid cylinder is given by \(S = 2\pi r^2 + 2\pi r h\). Substituting \(h\) into the surface area formula gives: \(S = 2\pi r^2 + 2\pi r \left(\dfrac{128}{r^2}\right) = 2\pi r^2 + \dfrac{256\pi}{r}\) (as required). (b) Differentiating \(S\) with respect to \(r\) gives \(\dfrac{\mathrm{d}S}{\mathrm{d}r} = 4\pi r - \dfrac{256\pi}{r^2}\). Setting \(\dfrac{\mathrm{d}S}{\mathrm{d}r} = 0\) for a stationary point, we get \(4\pi r = \dfrac{256\pi}{r^2}\), which leads to \(r^3 = 64\), so \(r = 4\). To check if this is a minimum, we find the second derivative: \(\dfrac{\mathrm{d}^2S}{\mathrm{d}r^2} = 4\pi + \dfrac{512\pi}{r^3}\). Substituting \(r = 4\) gives \(\dfrac{\mathrm{d}^2S}{\mathrm{d}r^2} = 4\pi + \dfrac{512\pi}{64} = 12\pi\). Since \(12\pi > 0\), the value \(r = 4\) gives a minimum value for \(S\). (c) The curve is \(y = 2\pi x^2 + \dfrac{256\pi}{x}\). At \(x = 2\), the \(y\)-coordinate is \(y = 2\pi(2)^2 + \dfrac{256\pi}{2} = 8\pi + 128\pi = 136\pi\). The gradient is given by \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4\pi x - \dfrac{256\pi}{x^2}\). At \(x = 2\), the gradient is \(m = 4\pi(2) - \dfrac{256\pi}{4} = 8\pi - 64\pi = -56\pi\). Using the line equation \(y - y_1 = m(x - x_1)\): \(y - 136\pi = -56\pi(x - 2) \implies y = -56\pi x + 112\pi + 136\pi \implies y = -56\pi x + 248\pi\).

Part (a): [3 Marks] M1: Formulates equations for volume and surface area, and attempts to eliminate \(h\). M1: Correctly expresses \(h = \dfrac{128}{r^2}\) and substitutes it into \(S = 2\pi r^2 + 2\pi r h\). A1*: Completes algebraic simplification to reach the given result without errors. Part (b): [4 Marks] M1: Differentiates \(S\) with respect to \(r\) with at least one term correct. A1: Correct first derivative \(\dfrac{\mathrm{d}S}{\mathrm{d}r} = 4\pi r - \dfrac{256\pi}{r^2}\). M1: Sets the first derivative to zero and solves to find \(r = 4\). A1: Finds \(\dfrac{\mathrm{d}^2S}{\mathrm{d}r^2} = 4\pi + \dfrac{512\pi}{r^3}\), evaluates at \(r = 4\) to get \(12\pi\) and concludes it is a minimum as \(12\pi > 0\). Part (c): [3 Marks] M1: Substitutes \(x=2\) into their derivative to find the gradient \(m = -56\pi\). M1: Evaluates \(y\)-coordinate at \(x=2\) as \(136\pi\) and attempts to form the equation of the tangent line. A1: Obtains \(y = -56\pi x + 248\pi\) (or exact equivalent in required form).

(a) To find the equation of the curve \(C\), we integrate the gradient function: \(y = \int \left(3x^2 - 4x^{-2} - 2\right) \mathrm{d}x = x^3 - \frac{4x^{-1}}{-1} - 2x + k = x^3 + \frac{4}{x} - 2x + k\). Since the point \(P(2, 5)\) lies on the curve, substitute \(x = 2\) and \(y = 5\): \(5 = 2^3 + \frac{4}{2} - 2(2) + k \implies 5 = 8 + 2 - 4 + k \implies 5 = 6 + k \implies k = -1\). Therefore, the equation of the curve is \(y = x^3 + \frac{4}{x} - 2x - 1\). (b) First find the gradient of the tangent to \(C\) at \(P(2, 5)\) by substituting \(x = 2\) into the gradient function: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3(2)^2 - \frac{4}{2^2} - 2 = 12 - 1 - 2 = 9\). The gradient of the normal is the negative reciprocal of the tangent gradient: \(m = -\frac{1}{9}\). Using the equation of a straight line: \(y - 5 = -\frac{1}{9}(x - 2) \implies 9(y - 5) = -(x - 2) \implies 9y - 45 = -x + 2 \implies x + 9y - 47 = 0\). (c) The normal cuts the \(y\)-axis at \(Q\), so substitute \(x = 0\) into the normal equation: \(0 + 9y - 47 = 0 \implies y = \frac{47}{9}\). Hence \(Q\) has coordinates \(\left(0, \frac{47}{9}\right)\). The triangle \(OPQ\) has a base along the \(y\)-axis from \(O(0,0)\) to \(Q\left(0, \frac{47}{9}\right)\), which has length \(\frac{47}{9}\). The perpendicular height is the \(x\)-coordinate of \(P\), which is 2. The area of triangle \(OPQ\) is \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{47}{9} \times 2 = \frac{47}{9}\).

(a) M1: Integrates to find \(y\), raising the power by 1 on at least two terms. A1: Correct integration of any two terms. A1: Fully correct integration including the term in \(x\) (constant \(k\) not needed for this mark). M1: Substitutes \(x = 2\) and \(y = 5\) into their integrated expression containing a constant of integration. A1: Fully correct equation of the curve, \(y = x^3 + \frac{4}{x} - 2x - 1\) (or exact equivalent). (b) M1: Substitutes \(x = 2\) into the given gradient function. M1: Finds the perpendicular gradient using \(m_2 = -1/m_1\). M1: Attempts to find the equation of the straight line passing through \(P(2,5)\) with their perpendicular gradient. A1: Correct equation in the specified form, e.g., \(x + 9y - 47 = 0\) (or any non-zero integer multiple). (c) M1: Substitutes \(x = 0\) into their normal equation to find the \(y\)-coordinate of \(Q\). M1: Uses their \(y\)-coordinate of \(Q\) and \(x\)-coordinate of \(P\) in a valid area formula (e.g., \(\frac{1}{2} \times \text{base} \times \text{height}\)). A1: \(\frac{47}{9}\) (or exact equivalent).

(a) Substitute \(\cos^2 x = 1 - \sin^2 x\) into the given equation: \(6(1 - \sin^2 x) - \sin x - 5 = 0\). Expanding and rearranging gives: \(6 - 6\sin^2 x - \sin x - 5 = 0 \implies 6\sin^2 x + \sin x - 1 = 0\). Factorising this quadratic equation: \((3\sin x - 1)(2\sin x + 1) = 0\). This yields two cases: \(\sin x = \frac{1}{3}\) or \(\sin x = -\frac{1}{2}\). Case 1: For \(\sin x = \frac{1}{3}\), the principal solution is \(x = \arcsin(1/3) \approx 19.47^\circ\). The second solution in the range \(0 \le x \le 360^\circ\) is \(180^\circ - 19.47^\circ \approx 160.5^\circ\). Case 2: For \(\sin x = -\frac{1}{2}\), the solutions in the range \(0 \le x \le 360^\circ\) are \(x = 180^\circ - (-30^\circ) = 210^\circ\) and \(x = 360^\circ - 30^\circ = 330^\circ\). Hence, the solutions for part (a) are \(x = 19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\). (b) Let \(x = 2\theta\). Since \(0 \le \theta \le 180^\circ\), we have \(0 \le 2\theta \le 360^\circ\), which matches the interval for \(x\) in part (a). Therefore, the values for \(2\theta\) are the exact angles found in part (a): \(2\theta \approx 19.4712...^\circ, 160.5287...^\circ, 210^\circ, 330^\circ\). Dividing each value by 2 yields: \(\theta \approx 9.7^\circ, 80.3^\circ, 105^\circ, 165^\circ\).

Part (a): M1: Applies the identity \(\cos^2 x = 1 - \sin^2 x\) to set up an equation in \(\sin x\) only. A1: Obtains a correct three-term quadratic equation, e.g., \(6\sin^2 x + \sin x - 1 = 0\). M1: Solves their quadratic equation by factorisation, formula, or completing the square. A1: Obtains the correct values \(\sin x = \frac{1}{3}\) and \(\sin x = -\frac{1}{2}\). A1: At least two correct values of \(x\) (either both \(210^\circ, 330^\circ\) or both \(19.5^\circ, 160.5^\circ\)). A1: All four correct values of \(x\): \(19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\). Deduct 1 mark overall for any extra solutions within the range. Part (b): M1: Recognises that \(2\theta = x\) and divides their answers from part (a) by 2. A1: Correct integer solutions: \(105^\circ\) and \(165^\circ\). A1: Correct decimal solutions: \(9.7^\circ\) and \(80.3^\circ\) (rounded to 1 d.p.).

(a)
Using the binomial expansion formula:
\[ (2 + kx)^6 = 2^6 + \binom{6}{1} 2^5 (kx) + \binom{6}{2} 2^4 (kx)^2 + \binom{6}{3} 2^3 (kx)^3 + \dots \]

Calculate each term:
- First term: \( 2^6 = 64 \)
- Second term: \( 6 \times 32 \times kx = 192kx \)
- Third term: \( 15 \times 16 \times k^2 x^2 = 240k^2 x^2 \)
- Fourth term: \( 20 \times 8 \times k^3 x^3 = 160k^3 x^3 \)

So, the expansion is:
\[ 64 + 192kx + 240k^2 x^2 + 160k^3 x^3 + \dots \]

(b)
Consider the expansion of \((1 - x)(64 + 192kx + 240k^2 x^2 + 160k^3 x^3 + \dots)\).

The term in \(x^3\) is obtained from:
\[ 1 \times (160k^3 x^3) + (-x) \times (240k^2 x^2) = (160k^3 - 240k^2)x^3 \]

We are given that the coefficient of \(x^3\) is zero, so:
\[ 160k^3 - 240k^2 = 0 \]

Since \(k\) is a non-zero constant, we can divide by \(80k^2\):
\[ 2k - 3 = 0 \]
\[ k = \frac{3}{2} \]

**Part (a)**
* **M1**: Attempts binomial expansion with at least two terms correct in structure, i.e., of the form \(\binom{6}{r} 2^{6-r} (kx)^r\).
* **A1**: Two correct simplified terms from \(64\), \(192kx\), \(240k^2x^2\), \(160k^3x^3\).
* **A1**: All four terms correct and fully simplified: \(64 + 192kx + 240k^2 x^2 + 160k^3 x^3\).

**Part (b)**
* **M1**: Identifies the relevant terms to form the coefficient of \(x^3\) in the expansion of \((1-x)(2+kx)^6\), leading to an expression of the form \(A k^3 - B k^2\) where \(A\) is their coefficient of \(x^3\) and \(B\) is their coefficient of \(x^2\) from part (a).
* **M1**: Sets their expression equal to 0 and attempts to solve for a non-zero value of \(k\).
* **A1**: For \(k = \frac{3}{2}\) (or \(1.5\)) with no other non-zero solutions.

(a) Substitute \(x = 2.5\) into the equation of the curve:

\(q = \frac{8}{2.5^2 + 1} = \frac{8}{6.25 + 1} = \frac{8}{7.25} = \frac{32}{29} \approx 1.103448...\)

To 4 decimal places, \(q = 1.1034\).

(b) The strip width is \(h = \frac{3 - 1}{4} = 0.5\).

Using the trapezium rule formula:

$$\int_{1}^{3} y \, dx \approx \frac{h}{2} \left[ y_0 + y_4 + 2(y_1 + y_2 + y_3) \right]$$

$$\int_{1}^{3} \frac{8}{x^2 + 1} \, dx \approx \frac{0.5}{2} \left[ 4 + 0.8 + 2(2.4615 + 1.6 + 1.1034) \right]$$

$$\approx 0.25 \left[ 4.8 + 2(5.1649) \right]$$

$$\approx 0.25 \left[ 4.8 + 10.3298 \right] = 0.25 \times 15.1298 = 3.78245$$

To 3 decimal places, the approximation is \(3.782\).

(c) The approximation is an over-estimate.

Reason: The curve \(y = \frac{8}{x^2 + 1}\) is convex (concave upwards) for the interval \(1 \le x \le 3\) (since the second derivative \(\frac{d^2y}{dx^2} = \frac{16(3x^2 - 1)}{(x^2 + 1)^3} > 0\) on this interval). This means the straight-line chords of the trapeziums lie above the curve, so the area of the trapeziums is larger than the true area under the curve.

(a)
- B1: For \(q = 1.1034\). (1 mark)

(b)
- M1: For identifying and using a correct step width of \(h = 0.5\) (implied by a pre-factor of \(0.25\) or \(\frac{1}{2} \times 0.5\)).
- M1: For the correct structure of the bracket inside the trapezium rule, using their value of \(q\): \([4 + 0.8 + 2(2.4615 + 1.6 + \text{their } q)]\).
- A1: For \(3.782\) (must be to 3 decimal places). (3 marks)

(c)
- M1: For stating that the approximation is an over-estimate.
- A1: For a fully correct explanation referencing the shape of the curve (e.g. "the curve is convex / concave upwards" or "the chords of the trapeziums lie above the curve"). (2 marks)

**(a)**
Since \( (2x+1) \) is a factor of \( f(x) \), by the Factor Theorem:
\[ f\left(-\frac{1}{2}\right) = 0 \]

Substitute \( x = -\frac{1}{2} \) into \( f(x) \):
\[ a\left(-\frac{1}{2}\right)^3 + b\left(-\frac{1}{2}\right)^2 - 5\left(-\frac{1}{2}\right) - 6 = 0 \]
\[ -\frac{a}{8} + \frac{b}{4} + \frac{5}{2} - 6 = 0 \]

Multiply the entire equation by 8 to clear the denominators:
\[ -a + 2b + 20 - 48 = 0 \]
\[ a - 2b = -28 \quad \text{--- (Equation 1)} \]

Since the remainder when \( f(x) \) is divided by \( (x-2) \) is 100, by the Remainder Theorem:
\[ f(2) = 100 \]

Substitute \( x = 2 \) into \( f(x) \):
\[ a(2)^3 + b(2)^2 - 5(2) - 6 = 100 \]
\[ 8a + 4b - 10 - 6 = 100 \]
\[ 8a + 4b = 116 \]
\[ 2a + b = 29 \quad \text{--- (Equation 2)} \]

Now solve Equations 1 and 2 simultaneously.
From Equation 1, express \( a \) in terms of \( b \):
\[ a = 2b - 28 \]

Substitute this into Equation 2:
\[ 2(2b - 28) + b = 29 \]
\[ 4b - 56 + b = 29 \]
\[ 5b = 85 \]
\[ b = 17 \]

Substitute \( b = 17 \) back to find \( a \):
\[ a = 2(17) - 28 = 6 \]

So, \( a = 6 \) and \( b = 17 \).

**(b)**
With \( a = 6 \) and \( b = 17 \), the function is:
\[ f(x) = 6x^3 + 17x^2 - 5x - 6 \]

Since \( (2x+1) \) is a factor, we can express \( f(x) \) as:
\[ f(x) = (2x+1)(3x^2 + px + q) \]
By inspection, comparing the constant terms: \( 1 \times q = -6 \implies q = -6 \).
Comparing the \( x^2 \) coefficients:
\[ 2p + 3 = 17 \implies 2p = 14 \implies p = 7 \]

So the quadratic factor is \( (3x^2 + 7x - 6) \).

Factorising this quadratic:
\[ 3x^2 + 7x - 6 = (3x - 2)(x + 3) \]

Therefore, completely factorised, \( f(x) = (2x + 1)(3x - 2)(x + 3) \).

**(a)**
* **M1**: Attempts \( f\left(-\frac{1}{2}\right) = 0 \) or uses long division by \( (2x+1) \) and sets the remainder to 0. Must show substitution of \( x = -\frac{1}{2} \).
* **A1**: Correct simplified linear equation in \( a \) and \( b \), e.g., \( a - 2b = -28 \) or any equivalent form.
* **M1**: Attempts \( f(2) = 100 \) or uses long division by \( (x-2) \) and sets the remainder to 100. Must show substitution of \( x = 2 \).
* **A1**: Correct simplified linear equation in \( a \) and \( b \), e.g., \( 2a + b = 29 \) or any equivalent form.
* **A1**: Both \( a = 6 \) and \( b = 17 \) correct. (Solving of simultaneous equations must be shown; do not award if values are written down without working).

**(b)**
* **M1**: Attempts to divide \( 6x^3 + 17x^2 - 5x - 6 \) by \( (2x+1) \) using long division, synthetic division, or equating coefficients to find a quadratic quotient. Must achieve a quadratic of the form \( 3x^2 + kx
\pm 6 \).
* **A1**: Completely factorised form \( (2x + 1)(3x - 2)(x + 3) \). (Accept equivalent forms such as \( 3(2x+1)\left(x - \frac{2}{3}\right)(x+3) \)).

We begin by rewriting the terms of the integrand in index form:
\(\frac{3}{\sqrt{x}} - 2x + \frac{x^2}{2} = 3x^{-\frac{1}{2}} - 2x + \frac{1}{2}x^2\)

Now we integrate term by term, increasing the power of \(x\) by 1 and dividing by the new power:
\(\int \left( 3x^{-\frac{1}{2}} - 2x + \frac{1}{2}x^2 \right) \text{d}x = \left[ \frac{3x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{2x^2}{2} + \frac{\frac{1}{2}x^3}{3} \right] = \left[ 6x^{\frac{1}{2}} - x^2 + \frac{1}{6}x^3 \right]\)

Applying the limits of integration \(1\) and \(4\):

At the upper limit \(x = 4\):
\(6(4)^{\frac{1}{2}} - (4)^2 + \frac{1}{6}(4)^3 = 6(2) - 16 + \frac{64}{6} = 12 - 16 + \frac{32}{3} = -4 + \frac{32}{3} = \frac{20}{3}\)

At the lower limit \(x = 1\):
\(6(1)^{\frac{1}{2}} - (1)^2 + \frac{1}{6}(1)^3 = 6 - 1 + \frac{1}{6} = 5 + \frac{1}{6} = \frac{31}{6}\)

Subtracting the lower limit value from the upper limit value:
\(\int_{1}^{4} \left( \frac{3}{\sqrt{x}} - 2x + \frac{x^2}{2} \right) \text{d}x = \frac{20}{3} - \frac{31}{6} = \frac{40}{6} - \frac{31}{6} = \frac{9}{6} = \frac{3}{2}\)

M1: Attempts to integrate at least two terms of the expression correctly (increases the power by 1 for those terms).
A1: Any two terms correct from the integrated expression: \(6x^{\frac{1}{2}} - x^2 + \frac{1}{6}x^3\).
A1: Fully correct integrated expression: \(6x^{\frac{1}{2}} - x^2 + \frac{1}{6}x^3\) (ignore constant of integration).
M1: Substitutes both limits \(4\) and \(1\) into their integrated expression and subtracts the correct way round: \(F(4) - F(1)\).
A1: Correct exact value of \(\frac{3}{2}\) or \(1.5\) (or exact equivalent fraction).

(a) By completing the square for both \(x\) and \(y\):

\((x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0\)

\((x - 3)^2 + (y + 2)^2 = 25\)

Thus, the centre of \(C\) is \((3, -2)\) and the radius is \(\sqrt{25} = 5\).

(b) Method 1: Using the perpendicular distance formula

The perpendicular distance from the centre of the circle \((3, -2)\) to the line \(kx - y - 12 = 0\) must equal the radius \(5\).

\(\frac{|k(3) - (-2) - 12|}{\sqrt{k^2 + (-1)^2}} = 5\)

\(\frac{|3k - 10|}{\sqrt{k^2 + 1}} = 5\)

\(|3k - 10| = 5\sqrt{k^2 + 1}\)

Squaring both sides:

\((3k - 10)^2 = 25(k^2 + 1)\)

\(9k^2 - 60k + 100 = 25k^2 + 25\)

\(16k^2 + 60k - 75 = 0\)

Using the quadratic formula:

\(k = \frac{-60 \pm \sqrt{60^2 - 4(16)(-75)}}{2(16)}\)

\(k = \frac{-60 \pm \sqrt{3600 + 4800}}{32} = \frac{-60 \pm \sqrt{8400}}{32}\)

\(k = \frac{-60 \pm 20\sqrt{21}}{32} = \frac{-15 \pm 5\sqrt{21}}{8}\)

Since \(k\) is a positive constant, we select the positive root:

\(k = \frac{-15 + 5\sqrt{21}}{8}\)

Method 2: Using the discriminant

Substitute \(y = kx - 12\) into the circle equation:

\(x^2 + (kx - 12)^2 - 6x + 4(kx - 12) - 12 = 0\)

\(x^2 + k^2x^2 - 24kx + 144 - 6x + 4kx - 48 - 12 = 0\)

\((1 + k^2)x^2 - (20k + 6)x + 84 = 0\)

Since the line is a tangent to the circle, the discriminant \(\Delta = b^2 - 4ac\) must be zero:

\((20k + 6)^2 - 4(1 + k^2)(84) = 0\)

\(4(10k + 3)^2 - 336(1 + k^2) = 0\)

\(4(100k^2 + 60k + 9) - 336 - 336k^2 = 0\)

\(400k^2 + 240k + 36 - 336 - 336k^2 = 0\)

\(64k^2 + 240k - 300 = 0\)

Dividing by 4 gives:

\(16k^2 + 60k - 75 = 0\)

Solving for \(k\) as above yields the positive root:

\(k = \frac{-15 + 5\sqrt{21}}{8}\)

(a)
M1: Attempt to complete the square for both \(x\) and \(y\). Look for expressions of the form \((x \pm 3)^2\) and \((y \pm 2)^2\).
A1: Correct coordinates for the centre \((3, -2)\) and correct radius \(r = 5\) (or \(r^2 = 25\)).

(b)
M1: Formulates a condition for tangency. This can be setting the perpendicular distance from the centre \((3, -2)\) to the line \(kx - y - 12 = 0\) equal to their radius, OR substituting \(y = kx - 12\) into the circle equation to obtain a quadratic equation in \(x\).
M1: Sets up the algebraic equation in \(k\). Either by squaring the distance equation to obtain \((3k-10)^2 = 25(k^2+1)\), OR by setting the discriminant of their quadratic in \(x\) to 0.
A1: Obtains a correct simplified quadratic equation in \(k\), such as \(16k^2 + 60k - 75 = 0\) (or any non-zero constant multiple of this equation).
A1: Correct exact value of \(k = \frac{-15 + 5\sqrt{21}}{8}\). Must be fully simplified and the negative root rejected.

(a) Differentiating the equation of the curve with respect to \(x\):

\[\frac{dy}{dx} = x^2 - 4kx + 4k^2 - k\]

(b) At the stationary points, \(\frac{dy}{dx} = 0\):

\[x^2 - 4kx + 4k^2 - k = 0\]

Using the quadratic formula or completing the square:

Completing the square gives:
\[(x - 2k)^2 - 4k^2 + 4k^2 - k = 0\]
\[(x - 2k)^2 = k\]
\[x - 2k = \pm \sqrt{k}\]
\[x = 2k \pm \sqrt{k}\]

(c) Let the \(x\)-coordinates of the stationary points be \(x_1 = 2k - \sqrt{k}\) and \(x_2 = 2k + \sqrt{k}\).

Method 1: Algebraic Simplification (Polynomial division/identity)
Since \(x^2 - 4kx + 4k^2 - k = 0\) at the stationary points, we can write:
\[y = \frac{1}{3}x(x^2 - 4kx + 4k^2 - k) - \frac{2}{3}kx^2 + \left(\frac{8}{3}k^2 - \frac{2}{3}k\right)x + 7\]
Substituting \(x^2 = 4kx - 4k^2 + k\) into the remaining quadratic terms:
\[y = -\frac{2}{3}k(4kx - 4k^2 + k) + \left(\frac{8}{3}k^2 - \frac{2}{3}k\right)x + 7\]
\[y = -\frac{8}{3}k^2x + \frac{8}{3}k^3 - \frac{2}{3}k^2 + \frac{8}{3}k^2x - \frac{2}{3}kx + 7\]
\[y = \frac{8}{3}k^3 - \frac{2}{3}k^2 + 7 - \frac{2}{3}kx\]
Now substitute \(x_1 = 2k - \sqrt{k}\) and \(x_2 = 2k + \sqrt{k}\) into this simplified expression:
\[y_1 = \frac{8}{3}k^3 - \frac{2}{3}k^2 + 7 - \frac{2}{3}k(2k - \sqrt{k}) = \frac{8}{3}k^3 - 2k^2 + 7 + \frac{2}{3}k^{3/2}\]
\[y_2 = \frac{8}{3}k^3 - \frac{2}{3}k^2 + 7 - \frac{2}{3}k(2k + \sqrt{k}) = \frac{8}{3}k^3 - 2k^2 + 7 - \frac{2}{3}k^{3/2}\]
Subtracting the two \(y\)-coordinates:
\[y_1 - y_2 = \left(\frac{8}{3}k^3 - 2k^2 + 7 + \frac{2}{3}k^{3/2}\right) - \left(\frac{8}{3}k^3 - 2k^2 + 7 - \frac{2}{3}k^{3/2}\right) = \frac{4}{3}k^{3/2}\]

Method 2: Direct Expansion
Substituting \(x = 2k \pm \sqrt{k}\) directly into \(y\):
\[y = \frac{1}{3}(2k \pm \sqrt{k})^3 - 2k(2k \pm \sqrt{k})^2 + (4k^2 - k)(2k \pm \sqrt{k}) + 7\]
Using \((2k \pm \sqrt{k})^3 = 8k^3 \pm 12k^2\sqrt{k} + 6k^2 \pm k\sqrt{k}\) and \((2k \pm \sqrt{k})^2 = 4k^2 \pm 4k\sqrt{k} + k\):
\[y = \frac{1}{3}\left(8k^3 \pm 12k^2\sqrt{k} + 6k^2 \pm k\sqrt{k}\right) - 2k\left(4k^2 \pm 4k\sqrt{k} + k\right) + (4k^2 - k)(2k \pm \sqrt{k}) + 7\]
\[y = \frac{8}{3}k^3 \pm 4k^2\sqrt{k} + 2k^2 \pm \frac{1}{3}k\sqrt{k} - 8k^3 \mp 8k^2\sqrt{k} - 2k^2 + 8k^3 \pm 4k^2\sqrt{k} - 2k^2 \mp k\sqrt{k} + 7\]
Collecting terms yields:
\[y = \left(\frac{8}{3}k^3 - 8k^3 + 8k^3\right) + \left(4k^2 - 8k^2 + 4k^2\right)\sqrt{k} + \left(2k^2 - 2k^2 - 2k^2\right) + \left(\frac{1}{3}k - k\right)\sqrt{k} + 7\]
\[y = \frac{8}{3}k^3 - 2k^2 + 7 \mp \frac{2}{3}k^{3/2}\]
Hence, the two \(y\)-coordinates are:
\[y_1 = \frac{8}{3}k^3 - 2k^2 + 7 + \frac{2}{3}k^{3/2} \quad \text{and} \quad y_2 = \frac{8}{3}k^3 - 2k^2 + 7 - \frac{2}{3}k^{3/2}\]
Subtracting the two values gives:
\[y_1 - y_2 = \frac{4}{3}k^{3/2}\]

(a)
B1: For \(x^2 - 4kx + 4k^2 - k\) or equivalent.

(b)
M1: Sets their derivative to 0 and attempts to solve the quadratic equation.
M1: Completes the square or applies the quadratic formula correctly.
A1*: Reaches the given answer \(2k \pm \sqrt{k}\) with no errors in working.

(c)
M1: Identifies \(x_1 = 2k - \sqrt{k}\) and \(x_2 = 2k + \sqrt{k}\) and attempts to substitute them into the original equation, or uses polynomial division to simplify \(y\).
M1: Correctly expands the algebraic terms or simplifies the curve equation to \(y = \frac{8}{3}k^3 - \frac{2}{3}k^2 + 7 - \frac{2}{3}kx\).
A1: Obtains the two simplified \(y\)-values: \(\frac{8}{3}k^3 - 2k^2 + 7 \pm \frac{2}{3}k^{3/2}\) (or equivalent).
A1*: Reaches \(\frac{4}{3}k^{3/2}\) through a correct subtraction with clear working.

(a)
Using the power law of logarithms, we rewrite the first term:
$$2\log_a(x-3) = \log_a((x-3)^2)$$

Using the subtraction law of logarithms on the left-hand side:
$$\log_a\left(\frac{(x-3)^2}{x+42}\right) = -\log_a 2$$

Using the power law on the right-hand side:
$$-\log_a 2 = \log_a(2^{-1}) = \log_a\left(\frac{1}{2}\right)$$

Equating the arguments of the logarithms:
$$\frac{(x-3)^2}{x+42} = \frac{1}{2}$$

Multiply both sides by \(2(x+42)\):
$$2(x-3)^2 = x+42$$
$$2(x^2 - 6x + 9) = x + 42$$
$$2x^2 - 12x + 18 = x + 42$$
$$2x^2 - 13x - 24 = 0$$

Factorising the quadratic equation:
$$(2x+3)(x-8) = 0$$

This gives potential solutions:
$$x = -\frac{3}{2} \quad \text{or} \quad x = 8$$

Since the term \(\log_a(x-3)\) requires \(x - 3 > 0\) (i.e., \(x > 3\)), the solution \(x = -\frac{3}{2}\) is invalid and must be rejected.

Therefore, the only valid solution is:
$$x = 8$$

(b)
From the first equation:
$$\log_2 y - \log_2 x = 3$$
$$\log_2\left(\frac{y}{x}\right) = 3$$
$$\frac{y}{x} = 2^3 = 8$$
$$y = 8x \quad \text{--- [Equation 1]}$$

From the second equation:
$$2^y = 4^{3x+1}$$
$$2^y = (2^2)^{3x+1}$$
$$2^y = 2^{6x+2}$$

Equating the exponents:
$$y = 6x + 2 \quad \text{--- [Equation 2]}$$

Substitute Equation 1 into Equation 2:
$$8x = 6x + 2$$
$$2x = 2$$
$$x = 1$$

Substitute \(x = 1\) back into Equation 1:
$$y = 8(1) = 8$$

Thus, the solutions are:
$$x = 1, \quad y = 8$$

(a)
* M1: Applies the power law of logarithms to write \(2\log_a(x-3)\) as \(\log_a(x-3)^2\).
* M1: Combines logarithms correctly to obtain \(\frac{(x-3)^2}{x+42} = \frac{1}{2}\) or equivalent form without logarithms.
* A1: Correctly expands and simplifies to a three-term quadratic equation, e.g., \(2x^2 - 13x - 24 = 0\).
* M1: Solves their 3-term quadratic equation by factorisation, quadratic formula, or completing the square, yielding two solutions.
* A1: Identifies \(x = 8\) as the only valid solution and rejects \(x = -1.5\) with a valid reason (e.g., stating that \(x > 3\) is required for \(\log_a(x-3)\) to be defined).

(b)
* M1: Uses subtraction law and base definitions to correctly rewrite the first equation as \(y = 8x\) or equivalent.
* M1: Expresses \(4^{3x+1}\) as \(2^{6x+2}\) (or writes both sides in terms of base 4 or taking logs of both sides) to obtain a correct linear relation, e.g., \(y = 6x + 2\).
* M1: Solves the resulting simultaneous equations to find a value for \(x\) or \(y\).
* A1: Obtains \(x = 1\) and \(y = 8\).

**(a)**

Starting with the given equation:
\[4 \sin x \tan x = 15 - 11\cos x\]

Using the trigonometric identity \(\tan x = \frac{\sin x}{\cos x}\):
\[4 \sin x \left(\frac{\sin x}{\cos x}\right) = 15 - 11\cos x\]

Multiplying both sides by \(\cos x\) (noting that \(\cos x \neq 0\)) gives:
\[4 \sin^2 x = 15\cos x - 11\cos^2 x\]

Using the identity \(\sin^2 x = 1 - \cos^2 x\):
\[4(1 - \cos^2 x) = 15\cos x - 11\cos^2 x\]
\[4 - 4\cos^2 x = 15\cos x - 11\cos^2 x\]

Rearranging all terms to one side of the equation:
\[11\cos^2 x - 4\cos^2 x - 15\cos x + 4 = 0\]
\[7\cos^2 x - 15\cos x + 4 = 0\]
(as required)

---

**(b)**

Let \(x = 2\theta - 30^\circ\).

Since \(0 \le \theta < 180^\circ\), we have:
\[0 \le 2\theta < 360^\circ\]
\[-30^\circ \le 2\theta - 30^\circ < 330^\circ\]
So, \(-30^\circ \le x < 330^\circ\).

From part (a), the equation can be written as:
\[7\cos^2 x - 15\cos x + 4 = 0\]

Using the quadratic formula to solve for \(\cos x\):
\[\cos x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(7)(4)}}{2(7)}\]
\[\cos x = \frac{15 \pm \sqrt{225 - 112}}{14}\]
\[\cos x = \frac{15 \pm \sqrt{113}}{14}\]

Evaluating these two possible values:
1. \(\cos x = \frac{15 + \sqrt{113}}{14} \approx 1.8307\)
Since \(|\cos x| \le 1\), this equation has no real solutions.

2. \(\cos x = \frac{15 - \sqrt{113}}{14} \approx 0.31213\)
Finding the principal value of \(x\):
\[x = \arccos(0.31213) \approx 71.8105^\circ\]

Using the symmetry of the cosine graph, other potential values for \(x\) in the general interval are:
\[x = -71.8105^\circ \quad \text{and} \quad x = 360^\circ - 71.8105^\circ = 288.1895^\circ\]

Checking which values lie in the range \([-30^\circ, 330^\circ)\):
- \(x = 71.8105^\circ\) (inside range)
- \(x = -71.8105^\circ\) (outside range)
- \(x = 288.1895^\circ\) (inside range)

Now, convert these back to \(\theta\) using \(2\theta - 30^\circ = x \implies \theta = \frac{x + 30^\circ}{2}\):

For \(x = 71.8105^\circ\):
\[\theta = \frac{71.8105^\circ + 30^\circ}{2} = 50.9052^\circ \approx 50.9^\circ\]

For \(x = 288.1895^\circ\):
\[\theta = \frac{288.1895^\circ + 30^\circ}{2} = 159.0948^\circ \approx 159.1^\circ\]

Thus, the solutions in the interval \(0 \le \theta < 180^\circ\) to 1 decimal place are:
\[\theta = 50.9^\circ, 159.1^\circ\]

**Part (a)**
- **M1**: Replaces \(\tan x\) with \(\frac{\sin x}{\cos x}\) and multiplies the equation by \(\cos x\) to eliminate the fraction.
- **M1**: Applies the identity \(\sin^2 x = 1 - \cos^2 x\) to obtain an equation in terms of \(\cos x\) only.
- **A1\***: Rearranges to the given quadratic equation, showing all necessary intermediate algebraic steps correctly with no errors in notation.

**Part (b)**
- **M1**: Realises the connection to part (a) with \(x = 2\theta - 30^\circ\) and applies the quadratic formula (or completes the square) to solve \(7\cos^2 x - 15\cos x + 4 = 0\) for \(\cos x\).
- **A1**: Obtains \(\cos(2\theta - 30^\circ) = \frac{15 - \sqrt{113}}{14}\) (or \(\approx 0.312\)) and notes that \(\cos(2\theta - 30^\circ) = \frac{15 + \sqrt{113}}{14}\) has no real solutions.
- **B1**: Identifies the correct range of the substituted variable, \(-30^\circ \le 2\theta - 30^\circ < 330^\circ\) (or equivalent interval processing for \(\theta\)).
- **M1**: Finds at least one correct angle for \(2\theta - 30^\circ\), e.g., \(71.8^\circ\) (or \(71.81^\circ\)).
- **M1**: Uses a correct method to find a second angle in the required range for \(2\theta - 30^\circ\), e.g., \(360^\circ - 71.8^\circ = 288.2^\circ\).
- **M1**: Undoes the substitution to find at least one value of \(\theta\) by adding \(30^\circ\) and dividing by \(2\).
- **A1**: Both answers \(\theta \approx 50.9^\circ\) and \(\theta \approx 159.1^\circ\) (or to 1 decimal place), and no other solutions in the range.

**(a)**

Starting with the given equation:
\[4 \sin x \tan x = 15 - 11\cos x\]

Using the trigonometric identity \(\tan x = \frac{\sin x}{\cos x}\):
\[4 \sin x \left(\frac{\sin x}{\cos x}\right) = 15 - 11\cos x\]

Multiplying both sides by \(\cos x\) (noting that \(\cos x \neq 0\)) gives:
\[4 \sin^2 x = 15\cos x - 11\cos^2 x\]

Using the identity \(\sin^2 x = 1 - \cos^2 x\):
\[4(1 - \cos^2 x) = 15\cos x - 11\cos^2 x\]
\[4 - 4\cos^2 x = 15\cos x - 11\cos^2 x\]

Rearranging all terms to one side of the equation:
\[11\cos^2 x - 4\cos^2 x - 15\cos x + 4 = 0\]
\[7\cos^2 x - 15\cos x + 4 = 0\]
(as required)

---

**(b)**

Let \(x = 2\theta - 30^\circ\).

Since \(0 \le \theta < 180^\circ\), we have:
\[0 \le 2\theta < 360^\circ\]
\[-30^\circ \le 2\theta - 30^\circ < 330^\circ\]
So, \(-30^\circ \le x < 330^\circ\).

From part (a), the equation can be written as:
\[7\cos^2 x - 15\cos x + 4 = 0\]

Using the quadratic formula to solve for \(\cos x\):
\[\cos x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(7)(4)}}{2(7)}\]
\[\cos x = \frac{15 \pm \sqrt{225 - 112}}{14}\]
\[\cos x = \frac{15 \pm \sqrt{113}}{14}\]

Evaluating these two possible values:
1. \(\cos x = \frac{15 + \sqrt{113}}{14} \approx 1.8307\)
Since \(|\cos x| \le 1\), this equation has no real solutions.

2. \(\cos x = \frac{15 - \sqrt{113}}{14} \approx 0.31213\)
Finding the principal value of \(x\):
\[x = \arccos(0.31213) \approx 71.8105^\circ\]

Using the symmetry of the cosine graph, other potential values for \(x\) in the general interval are:
\[x = -71.8105^\circ \quad \text{and} \quad x = 360^\circ - 71.8105^\circ = 288.1895^\circ\]

Checking which values lie in the range \([-30^\circ, 330^\circ)\):
- \(x = 71.8105^\circ\) (inside range)
- \(x = -71.8105^\circ\) (outside range)
- \(x = 288.1895^\circ\) (inside range)

Now, convert these back to \(\theta\) using \(2\theta - 30^\circ = x \implies \theta = \frac{x + 30^\circ}{2}\):

For \(x = 71.8105^\circ\):
\[\theta = \frac{71.8105^\circ + 30^\circ}{2} = 50.9052^\circ \approx 50.9^\circ\]

For \(x = 288.1895^\circ\):
\[\theta = \frac{288.1895^\circ + 30^\circ}{2} = 159.0948^\circ \approx 159.1^\circ\]

Thus, the solutions in the interval \(0 \le \theta < 180^\circ\) to 1 decimal place are:
\[\theta = 50.9^\circ, 159.1^\circ\]

**Part (a)**
- **M1**: Replaces \(\tan x\) with \(\frac{\sin x}{\cos x}\) and multiplies the equation by \(\cos x\) to eliminate the fraction.
- **M1**: Applies the identity \(\sin^2 x = 1 - \cos^2 x\) to obtain an equation in terms of \(\cos x\) only.
- **A1\***: Rearranges to the given quadratic equation, showing all necessary intermediate algebraic steps correctly with no errors in notation.

**Part (b)**
- **M1**: Realises the connection to part (a) with \(x = 2\theta - 30^\circ\) and applies the quadratic formula (or completes the square) to solve \(7\cos^2 x - 15\cos x + 4 = 0\) for \(\cos x\).
- **A1**: Obtains \(\cos(2\theta - 30^\circ) = \frac{15 - \sqrt{113}}{14}\) (or \(\approx 0.312\)) and notes that \(\cos(2\theta - 30^\circ) = \frac{15 + \sqrt{113}}{14}\) has no real solutions.
- **B1**: Identifies the correct range of the substituted variable, \(-30^\circ \le 2\theta - 30^\circ < 330^\circ\) (or equivalent interval processing for \(\theta\)).
- **M1**: Finds at least one correct angle for \(2\theta - 30^\circ\), e.g., \(71.8^\circ\) (or \(71.81^\circ\)).
- **M1**: Uses a correct method to find a second angle in the required range for \(2\theta - 30^\circ\), e.g., \(360^\circ - 71.8^\circ = 288.2^\circ\).
- **M1**: Undoes the substitution to find at least one value of \(\theta\) by adding \(30^\circ\) and dividing by \(2\).
- **A1**: Both answers \(\theta \approx 50.9^\circ\) and \(\theta \approx 159.1^\circ\) (or to 1 decimal place), and no other solutions in the range.

(a) Under Plan A, the extraction volumes form an arithmetic sequence with first term \(a = 5000\) and common difference \(d = -400\).
The volume extracted in Year \(N\) is given by:
\[u_N = a + (N-1)d = 5000 - 400(N-1)\]
For Year 13:
\[u_{13} = 5000 - 400(12) = 200\text{ barrels}\]
For Year 14:
\[u_{14} = 5000 - 400(13) = -200\text{ barrels}\]
Since \(u_{13} > 0\) and \(u_{14} < 0\), the volume extracted in Year 14 would be negative, so the extraction must stop in Year 13.

(b) The total volume of oil extracted under Plan A from Year 1 to Year 13 is the sum of the first 13 terms of the arithmetic series:
\[S_n = \frac{n}{2}[2a + (n-1)d]\]
\[S_{13} = \frac{13}{2}[2(5000) + 12(-400)] = \frac{13}{2}[10000 - 4800] = \frac{13}{2}[5200] = 33800\text{ barrels}\]

(c) Under Plan B, the extraction volumes form a geometric series with first term \(a = 5000\) and common ratio \(r = 0.92\).
Since \(|r| < 1\), the sum to infinity exists and is given by:
\[S_{\infty} = \frac{a}{1-r} = \frac{5000}{1 - 0.92} = \frac{5000}{0.08} = 62500\text{ barrels}\]

(d) Under Plan B, the volume of oil extracted in Year \(N\) is:
\[v_N = a r^{N-1} = 5000 \times 0.92^{N-1}\]
We want to find the smallest integer \(N\) such that:
\[5000 \times 0.92^{N-1} < 1000\]
\[0.92^{N-1} < 0.2\]
Taking natural logarithms of both sides:
\[\ln(0.92^{N-1}) < \ln(0.2)\]
\[(N-1)\ln(0.92) < \ln(0.2)\]
Since \(\ln(0.92) < 0\), dividing by \(\ln(0.92)\) reverses the inequality:
\[N-1 > \frac{\ln(0.2)}{\ln(0.92)}\]
\[N-1 > 19.302...\]
\[N > 20.302...\]
So the first year in which the extraction falls below 1000 barrels is Year 21.

(e) We require the sum of the first \(k\) years under Plan B to exceed the total lifetime extraction under Plan A, which is 33800 barrels.
\[S_k = \frac{a(1 - r^k)}{1 - r} > 33800\]
\[\frac{5000(1 - 0.92^k)}{1 - 0.92} > 33800\]
\[62500(1 - 0.92^k) > 33800\]
\[1 - 0.92^k > \frac{33800}{62500}\]
\[1 - 0.92^k > 0.5408\]
\[0.92^k < 0.4592\]
Taking natural logarithms:
\[k \ln(0.92) < \ln(0.4592)\]
\[k > \frac{\ln(0.4592)}{\ln(0.92)}\]
\[k > 9.333...\]
Since \(k\) must be an integer, the number of years required is 10.

(a)
B1: Calculates \(u_{13} = 200\) and \(u_{14} = -200\) (or solves \(5000 - 400(N-1) < 0\) to get \(N > 13.5\)) and concludes extraction stops in Year 13. (1)

(b)
M1: Attempts to use the sum formula for an arithmetic series with \(n = 13\), \(a = 5000\), and \(d = -400\). (1)
A1: Correct total of 33800 barrels. (1)

(c)
M1: Attempts to use the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\) with \(a = 5000\) and \(r = 0.92\). (1)
A1: Correct total of 62500 barrels. (1)

(d)
M1: Sets up the inequality \(5000 \times 0.92^{N-1} < 1000\) and uses logarithms to solve for \(N\). (1)
A1: Obtains the critical value \(N - 1 \approx 19.3\) or \(N \approx 20.3\) (accept equivalent expressions). (1)
A1: Concludes Year 21. (1)

(e)
M1: Sets up the inequality \(62500(1 - 0.92^k) > 33800\) and uses logarithms to solve for \(k\). (1)
A1: Concludes 10 years. (1)

(a) Write the equation of \(C\) in index form:
\[ y = 3x^{1/2} - 6x^{-1/2} + 2 \]
Differentiating with respect to \(x\):
\[ \frac{dy}{dx} = \frac{3}{2}x^{-1/2} + 3x^{-3/2} \]
At the point \(P(4, 5)\), substituting \(x = 4\):
\[ \frac{dy}{dx} = \frac{3}{2(2)} + 3(4^{-3/2}) = \frac{3}{4} + \frac{3}{8} = \frac{9}{8} \]
The equation of the tangent at \(P\) is:
\[ y - 5 = \frac{9}{8}(x - 4) \]
\[ y = \frac{9}{8}x - \frac{9}{2} + 5 \]
\[ y = \frac{9}{8}x + \frac{1}{2} \]

(b) The region \(R\) is bounded by the tangent, the curve \(C\), and the line \(x = 1\). Since the tangent lies above the curve for the interval \(1 \le x \le 4\), the area of \(R\) is given by:
\[ \text{Area} = \int_{1}^{4} \left( \left(\frac{9}{8}x + \frac{1}{2}\right) - \left(3x^{1/2} - 6x^{-1/2} + 2\right) \right) \mathrm{d}x \]
\[ \text{Area} = \int_{1}^{4} \left( \frac{9}{8}x - \frac{3}{2} - 3x^{1/2} + 6x^{-1/2} \right) \mathrm{d}x \]
Integrating term by term:
\[ \text{Area} = \left[ \frac{9}{16}x^2 - \frac{3}{2}x - 2x^{3/2} + 12x^{1/2} \right]_{1}^{4} \]
Substitute the upper limit \(x = 4\):
\[ \left(\frac{9}{16}(16) - \frac{3}{2}(4) - 2(8) + 12(2)\right) = 9 - 6 - 16 + 24 = 11 \]
Substitute the lower limit \(x = 1\):
\[ \left(\frac{9}{16}(1) - \frac{3}{2}(1) - 2(1) + 12(1)\right) = \frac{9}{16} - \frac{3}{2} - 2 + 12 = \frac{145}{16} \]
Subtract the lower limit from the upper limit:
\[ \text{Area} = 11 - \frac{145}{16} = \frac{176 - 145}{16} = \frac{31}{16} \]
So the exact area of \(R\) is \(\frac{31}{16}\) (or \(1.9375\)).

(a)
M1: Attempts to differentiate the equation of \(C\) with at least one fractional index term differentiated correctly (\(x^{n} \to x^{n-1}\)).
A1: Correct derivative: \(\frac{dy}{dx} = \frac{3}{2}x^{-1/2} + 3x^{-3/2}\) or equivalent.
A1: Correct equation of the tangent in the required form: \(y = \frac{9}{8}x + \frac{1}{2}\) (or exact decimal equivalent \(y = 1.125x + 0.5\)).

(b)
M1: Sets up a correct integral for the area with limits 1 and 4, subtracting the curve equation from their tangent equation.
M1: Integrates at least two terms of their integrand, with at least one fractional power integrated correctly (\(x^{n} \to x^{n+1}\)).
A1: Fully correct integrated expression: \(\frac{9}{16}x^2 - \frac{3}{2}x - 2x^{3/2} + 12x^{1/2}\) (condone missing \(+ C\)).
M1: Substitutes the limits 4 and 1 into their integrated expression and subtracts the lower limit value from the upper limit value.
A1: Correct exact area of \(\frac{31}{16}\) (or \(1.9375\)).