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

First, we solve the quadratic inequality \(2x^2 - 7x - 15 > 0\). Find the critical values by solving \(2x^2 - 7x - 15 = 0\). Factoring the quadratic gives \((2x + 3)(x - 5) = 0\), which gives critical values of \(x = -\frac{3}{2}\) and \(x = 5\). Since we want \(2x^2 - 7x - 15 > 0\), we choose the region outside the critical values: \(x < -\frac{3}{2}\) or \(x > 5\). Next, we solve the linear inequality \(4(x - 1) \le 3x + 8\). Expanding the brackets gives \(4x - 4 \le 3x + 8\). Subtracting \(3x\) and adding \(4\) to both sides gives \(x \le 12\). Finally, we find the set of values of \(x\) for which both inequalities are satisfied by finding the intersection of these two regions: we need \(x \le 12\) and either \(x < -\frac{3}{2}\) or \(x > 5\). This gives the final solution: \(x < -\frac{3}{2}\) or \(5 < x \le 12\).

M1: Attempts to find the critical values of the quadratic by factorisation, completing the square, or using the quadratic formula. A1: Correct critical values of \(-\frac{3}{2}\) (or \(-1.5\)) and \(5\). M1: Chooses the outside region for their critical values, expressing it as \(x < -\frac{3}{2}\) or \(x > 5\) (or equivalent for their critical values). B1: Solves the linear inequality to obtain \(x \le 12\). M1: Attempts to find the intersection of their quadratic region and linear region. A1: Correct final answer: \(x < -\frac{3}{2}\) or \(5 < x \le 12\) (or equivalent interval/set notation).

**(a)**

To rationalize the denominator of \(\frac{2\sqrt{2} + 3}{3\sqrt{2} - 4}\), we multiply the numerator and the denominator by the conjugate of the denominator, which is \(3\sqrt{2} + 4\):

\[\frac{2\sqrt{2} + 3}{3\sqrt{2} - 4} = \frac{(2\sqrt{2} + 3)(3\sqrt{2} + 4)}{(3\sqrt{2} - 4)(3\sqrt{2} + 4)}\]

First, we expand the denominator:

\[(3\sqrt{2} - 4)(3\sqrt{2} + 4) = (3\sqrt{2})^2 - 4^2 = 9(2) - 16 = 18 - 16 = 2\]

Next, we expand the numerator:

\[(2\sqrt{2} + 3)(3\sqrt{2} + 4) = 2\sqrt{2}(3\sqrt{2}) + 2\sqrt{2}(4) + 3(3\sqrt{2}) + 3(4)\]

\[= 6(2) + 8\sqrt{2} + 9\sqrt{2} + 12 = 12 + 17\sqrt{2} + 12 = 24 + 17\sqrt{2}\]

Substitute these back into the fraction:

\[\frac{24 + 17\sqrt{2}}{2} = 12 + \frac{17}{2}\sqrt{2}\]

So, \(p = 12\) and \(q = \frac{17}{2}\) (or \(8.5\)).

**(b)**

**Method 1:**
Multiply the entire equation by \(\sqrt{3}\) to eliminate the fraction:

\[x\sqrt{12}\sqrt{3} - 5\sqrt{3} = 3x\]

Since \(\sqrt{12}\sqrt{3} = \sqrt{36} = 6\), the equation simplifies to:

\[6x - 5\sqrt{3} = 3x\]

Rearranging to make \(x\) the subject:

\[6x - 3x = 5\sqrt{3}\]

\[3x = 5\sqrt{3}\]

\[x = \frac{5}{3}\sqrt{3}\]

**Method 2:**
Simplify the terms involving surds first:

\[\sqrt{12} = 2\sqrt{3}\]

and

\[\frac{3x}{\sqrt{3}} = \frac{3x\sqrt{3}}{3} = x\sqrt{3}\]

Substituting these back into the original equation gives:

\[2\sqrt{3}x - 5 = x\sqrt{3}\]

Rearranging to group the terms in \(x\):

\[2\sqrt{3}x - x\sqrt{3} = 5\]

\[x\sqrt{3} = 5\]

\[x = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} = \frac{5}{3}\sqrt{3}\]

Hence, \(k = \frac{5}{3}\).

**(a)**
* **M1**: Attempts to multiply the numerator and the denominator by \(3\sqrt{2} + 4\). Allow one sign or arithmetic error in expansion.
* **A1**: Correctly expands the denominator to obtain \(2\) and correctly expands the numerator to obtain \(24 + 17\sqrt{2}\).
* **A1**: Correct final answer in the required form \(p + q\sqrt{2}\), giving \(12 + \frac{17}{2}\sqrt{2}\) (or \(12 + 8.5\sqrt{2}\)).

**(b)**
* **M1**: Attempts to clear fractions by multiplying all terms by \(\sqrt{3}\) (or simplifies \(\frac{3x}{\sqrt{3}}\) to \(x\sqrt{3}\) and \(\sqrt{12}\) to \(2\sqrt{3}\)).
* **A1**: Obtains a correct simplified equation in \(x\) with integer or single-surd coefficients, e.g., \(6x - 5\sqrt{3} = 3x\) or \(2\sqrt{3}x - x\sqrt{3} = 5\).
* **dM1**: Dependent on the previous M1. Collects terms in \(x\) to solve for \(x\), leading to \(3x = 5\sqrt{3}\) or \(x\sqrt{3} = 5\).
* **A1**: Fully correct final answer in the form \(k\sqrt{3}\), which is \(x = \frac{5}{3}\sqrt{3}\) or clearly states \(k = \frac{5}{3}\).

Part (a): First, rewrite the terms of \(y\) as powers of \(x\): \(y = 2x^{3/2} - 16x^{-1/2} + 3\). Now, differentiate with respect to \(x\): \(\frac{dy}{dx} = 2 \times \frac{3}{2}x^{1/2} - 16 \times \left(-\frac{1}{2}\right)x^{-3/2} = 3x^{1/2} + 8x^{-3/2}\). Part (b): Find the \(y\)-coordinate of \(P\) by substituting \(x = 4\) into the equation of \(C\): \(y = 2(4)^{3/2} - 16(4)^{-1/2} + 3 = 16 - 8 + 3 = 11\). So \(P\) has coordinates \((4, 11)\). Find the gradient of the tangent at \(P\) by substituting \(x = 4\) into \(\frac{dy}{dx}\): \(m = 3(4)^{1/2} + 8(4)^{-3/2} = 3(2) + 8\left(\frac{1}{8}\right) = 7\). Using the equation of a straight line with gradient \(m = 7\) passing through \((4, 11)\): \(y - 11 = 7(x - 4) \implies y - 11 = 7x - 28 \implies 7x - y - 17 = 0\). Part (c): To find where the tangent crosses the axes: On the \(x\)-axis, \(y = 0 \implies 7x - 17 = 0 \implies x = \frac{17}{7}\), so \(Q\) is \(\left(\frac{17}{7}, 0\right)\). On the \(y\)-axis, \(x = 0 \implies -y - 17 = 0 \implies y = -17\), so \(R\) is \((0, -17)\). The area of the triangle \(OQR\) is: \(\text{Area} = \frac{1}{2} \times OQ \times OR = \frac{1}{2} \times \frac{17}{7} \times 17 = \frac{289}{14}\).

Part (a): M1: Attempts to differentiate at least one term of the form \(Ax^n \to A n x^{n-1}\). A1: One term of the derivative correct (either \(3x^{1/2}\) or \(8x^{-3/2}\)). A1: Fully correct derivative: \(\frac{dy}{dx} = 3x^{1/2} + 8x^{-3/2}\) or equivalent simplified form. Part (b): M1: Substitution of \(x = 4\) into the equation of \(C\) to find the \(y\)-coordinate AND substitution of \(x = 4\) into their \(\frac{dy}{dx}\) to find the gradient. M1: Uses their \(y\)-coordinate and their gradient to form an equation of the tangent: \(y - y_P = m(x - 4)\). A1: Correct equation in the required form: \(7x - y - 17 = 0\) (or any non-zero integer multiple, e.g., \(y - 7x + 17 = 0\)). Part (c): M1: Correct method to find the coordinates of \(Q\) and \(R\) using their tangent equation from (b), and attempts to calculate the area of the triangle using \(\frac{1}{2} \times |x_Q| \times |y_R|\). A1: Correct area of \(\frac{289}{14}\) (or exact equivalent).

(a) First, find the gradient of the diagonal \(AC\):
\[m_{AC} = \frac{7 - 1}{3 - 1} = \frac{6}{2} = 3\]
Using the point-gradient form of a straight line with point \(A(1, 1)\):
\[y - 1 = 3(x - 1)\]
\[y - 1 = 3x - 3\]
\[y = 3x - 2\]

(b) Since \(AC\) is the line of symmetry of the kite \(ABCD\), the diagonal \(BD\) is perpendicular to the diagonal \(AC\).
Therefore, the gradient of \(BD\) is:
\[m_{BD} = -\frac{1}{m_{AC}} = -\frac{1}{3}\]
Since the line \(BD\) passes through the vertex \(B(5, 3)\), its equation is:
\[y - 3 = -\frac{1}{3}(x - 5)\]
Multiply both sides by 3 to clear the fraction:
\[3(y - 3) = -(x - 5)\]
\[3y - 9 = -x + 5\]
Rearranging into the required form:
\[x + 3y - 14 = 0\]

(c) In a kite, the diagonal which is the line of symmetry (\(AC\)) is the perpendicular bisector of the other diagonal (\(BD\)). Let \(M\) be the point of intersection of the diagonals \(AC\) and \(BD\). This point \(M\) is the midpoint of \(BD\).
To find \(M\), we solve the equations of the diagonals simultaneously:
\[y = 3x - 2\]
\[x + 3y - 14 = 0\]
Substitute the expression for \(y\) into the second equation:
\[x + 3(3x - 2) - 14 = 0\]
\[x + 9x - 6 - 14 = 0\]
\[10x - 20 = 0 \implies 10x = 20 \implies x = 2\]
Substitute \(x = 2\) back into the equation for \(AC\):
\[y = 3(2) - 2 = 4\]
So, the intersection point (midpoint of \(BD\)) is \(M(2, 4)\).

Let the coordinates of \(D\) be \((x_D, y_D)\). Since \(M(2, 4)\) is the midpoint of \(BD\) where \(B(5, 3)\):
\[\frac{5 + x_D}{2} = 2 \implies 5 + x_D = 4 \implies x_D = -1\]
\[\frac{3 + y_D}{2} = 4 \implies 3 + y_D = 8 \implies y_D = 5\]
Thus, the coordinates of \(D\) are \((-1, 5)\).

(a)
- M1: Attempts to find the gradient of \(AC\) using \(\frac{y_2 - y_1}{x_2 - x_1}\) and sets up a linear equation using \(A\) or \(C\).
- A1: Correct equation in the form \(y = mx + c\), i.e., \(y = 3x - 2\).

(b)
- M1: Recognises that \(BD\) is perpendicular to \(AC\) and determines the gradient of \(BD\) as the negative reciprocal of their gradient from part (a).
- M1: Uses their perpendicular gradient with coordinates of \(B(5, 3)\) to form an equation of the line \(BD\).
- A1*: Correctly manipulates their equation to obtain the given result \(x + 3y - 14 = 0\) with no errors in algebra, showing at least one intermediate expansion/rearrangement step.

(c)
- M1: Attempts to solve the equations \(y = 3x - 2\) and \(x + 3y - 14 = 0\) simultaneously to find the point of intersection \(M\).
- A1: Obtains the correct coordinates \(M(2, 4)\).
- M1: Uses the midpoint formula with \(B(5, 3)\) and their midpoint \(M\) to set up linear equations for the coordinates of \(D\).
- A1: Correct coordinates for \(D\) are \((-1, 5)\) (accept written as \(x = -1\), \(y = 5\)).

**(a)**
To find the \(y\)-intercept, substitute \(x = 0\):
\[f(0) = \frac{6}{0+2} - 1 = 3 - 1 = 2\]
So, the curve crosses the \(y\)-axis at \((0, 2)\).

To find the \(x\)-intercept, set \(f(x) = 0\):
\[\frac{6}{x+2} - 1 = 0 \implies \frac{6}{x+2} = 1 \implies x+2 = 6 \implies x = 4\]
So, the curve crosses the \(x\)-axis at \((4, 0)\).

**(b)**
The vertical asymptote occurs where the denominator is zero:
\[x + 2 = 0 \implies x = -2\]
As \(x \to \pm\infty\), \(f(x) \to -1\), so the horizontal asymptote is:
\[y = -1\]

**(c)**
The curve \(y = f(x-3) + 2\) represents a translation of \(C\) by the vector \(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\).

- The new vertical asymptote is \(x = -2 + 3 = 1\).
- The new horizontal asymptote is \(y = -1 + 2 = 1\).

The equation of the new curve is:
\[y = \frac{6}{(x-3)+2} - 1 + 2 = \frac{6}{x-1} + 1\]

- To find the new \(y\)-intercept, substitute \(x = 0\):
\[y = \frac{6}{0-1} + 1 = -5 \implies (0, -5)\]
- To find the new \(x\)-intercept, set \(y = 0\):
\[\frac{6}{x-1} + 1 = 0 \implies \frac{6}{x-1} = -1 \implies x-1 = -6 \implies x = -5 \implies (-5, 0)\]

**Sketch details:**
- A reciprocal curve with two branches in the first and third "quadrants" relative to the intersection of the asymptotes at \((1, 1)\).
- Dashed lines showing the asymptotes with equations \(x = 1\) and \(y = 1\) clearly labeled.
- The lower-left branch passes through the axes at \((0, -5)\) and \((-5, 0)\), with these coordinates clearly labeled.

**(a)**
- **B1**: For \((0, 2)\) (allow \(y=2\) if clearly indicated as the intercept).
- **B1**: For \((4, 0)\) (allow \(x=4\) if clearly indicated as the intercept).

**(b)**
- **B1**: For both \(x = -2\) and \(y = -1\).

**(c)**
- **B1**: Correct shape for a reciprocal curve with two branches (one in the upper-right region and one in the lower-left region relative to their asymptotes).
- **B1**: Correct asymptotes drawn and labeled with equations \(x=1\) and \(y=1\).
- **B1**: Correct \(y\)-intercept labeled at \((0, -5)\) or \(-5\) on the \(y\)-axis.
- **B1**: Correct \(x\)-intercept labeled at \((-5, 0)\) or \(-5\) on the \(x\)-axis.

**(a)**
To find the \(y\)-intercept, substitute \(x = 0\):
\[f(0) = \frac{6}{0+2} - 1 = 3 - 1 = 2\]
So, the curve crosses the \(y\)-axis at \((0, 2)\).

To find the \(x\)-intercept, set \(f(x) = 0\):
\[\frac{6}{x+2} - 1 = 0 \implies \frac{6}{x+2} = 1 \implies x+2 = 6 \implies x = 4\]
So, the curve crosses the \(x\)-axis at \((4, 0)\).

**(b)**
The vertical asymptote occurs where the denominator is zero:
\[x + 2 = 0 \implies x = -2\]
As \(x \to \pm\infty\), \(f(x) \to -1\), so the horizontal asymptote is:
\[y = -1\]

**(c)**
The curve \(y = f(x-3) + 2\) represents a translation of \(C\) by the vector \(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\).

- The new vertical asymptote is \(x = -2 + 3 = 1\).
- The new horizontal asymptote is \(y = -1 + 2 = 1\).

The equation of the new curve is:
\[y = \frac{6}{(x-3)+2} - 1 + 2 = \frac{6}{x-1} + 1\]

- To find the new \(y\)-intercept, substitute \(x = 0\):
\[y = \frac{6}{0-1} + 1 = -5 \implies (0, -5)\]
- To find the new \(x\)-intercept, set \(y = 0\):
\[\frac{6}{x-1} + 1 = 0 \implies \frac{6}{x-1} = -1 \implies x-1 = -6 \implies x = -5 \implies (-5, 0)\]

**Sketch details:**
- A reciprocal curve with two branches in the first and third "quadrants" relative to the intersection of the asymptotes at \((1, 1)\).
- Dashed lines showing the asymptotes with equations \(x = 1\) and \(y = 1\) clearly labeled.
- The lower-left branch passes through the axes at \((0, -5)\) and \((-5, 0)\), with these coordinates clearly labeled.

**(a)**
- **B1**: For \((0, 2)\) (allow \(y=2\) if clearly indicated as the intercept).
- **B1**: For \((4, 0)\) (allow \(x=4\) if clearly indicated as the intercept).

**(b)**
- **B1**: For both \(x = -2\) and \(y = -1\).

**(c)**
- **B1**: Correct shape for a reciprocal curve with two branches (one in the upper-right region and one in the lower-left region relative to their asymptotes).
- **B1**: Correct asymptotes drawn and labeled with equations \(x=1\) and \(y=1\).
- **B1**: Correct \(y\)-intercept labeled at \((0, -5)\) or \(-5\) on the \(y\)-axis.
- **B1**: Correct \(x\)-intercept labeled at \((-5, 0)\) or \(-5\) on the \(x\)-axis.

**(a)**

Using the cosine rule in triangle \(OBC\):
\[BC^2 = OB^2 + OC^2 - 2(OB)(OC)\cos\theta\]

Substitute \(OB = r\), \(OC = 2r\), and \(\theta = \frac{\pi}{3}\):
\[BC^2 = r^2 + (2r)^2 - 2(r)(2r)\cos\left(\frac{\pi}{3}\right)\]
\[BC^2 = r^2 + 4r^2 - 4r^2(0.5) = 3r^2\]

Therefore, \(BC = r\sqrt{3}\).

The perimeter of the shaded region is given by:
\[P = AC + BC + \text{arc } AB\]

Since \(OC = 2r\) and \(OA = r\), we have \(AC = OC - OA = 2r - r = r\).

The arc length \(AB = r\theta = r\left(\frac{\pi}{3}\right)\).

So,
\[P = r + r\sqrt{3} + r\left(\frac{\pi}{3}\right) = r\left(1 + \frac{\pi}{3} + \sqrt{3}\right)\text{ cm}\]
(as required).


**(b)**

The area of the shaded region is the area of triangle \(OBC\) minus the area of the sector \(OAB\):
\[\text{Area of } \triangle OBC = \frac{1}{2}(OB)(OC)\sin\theta = \frac{1}{2}(r)(2r)\sin\theta = r^2\sin\theta\]
\[\text{Area of sector } OAB = \frac{1}{2}r^2\theta\]

So the shaded area \(S = r^2\sin\theta - \frac{1}{2}r^2\theta = \frac{1}{2}r^2(2\sin\theta - \theta)\).

Substituting \(r = 6\) and \(S = 9\):
\[9 = \frac{1}{2}(6^2)(2\sin\theta - \theta)\]
\[9 = 18(2\sin\theta - \theta)\]
\[2\sin\theta - \theta = \frac{9}{18} = \frac{1}{2}\]
(as required).


**(c)**

Given \(\theta = \frac{\pi}{6}\) and the area is \(12\text{ cm}^2\):
\[12 = \frac{1}{2}r^2\left(2\sin\left(\frac{\pi}{6}\right) - \frac{\pi}{6}\right)\]

Since \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\):
\[12 = \frac{1}{2}r^2\left(2\left(\frac{1}{2}\right) - \frac{\pi}{6}\right)\]
\[12 = \frac{1}{2}r^2\left(1 - \frac{\pi}{6}\right)\]
\[24 = r^2\left(\frac{6 - \pi}{6}\right)\]
\[144 = r^2(6 - \pi)\]
\[r^2 = \frac{144}{6 - \pi}\]

Since \(r > 0\), we have:
\[r = \frac{12}{\sqrt{6 - \pi}}\]

Thus \(k = 12\) and \(a = 6\).

**(a)**

M1: Attempts to use the cosine rule in triangle \(OBC\) with \(OB = r\), \(OC = 2r\), and \(\theta = \frac{\pi}{3}\). Must be a correct formula structure.

A1: Correctly simplifies to find \(BC = r\sqrt{3}\).

M1: Identifies the components of the perimeter: \(AC = r\), \(\text{arc } AB = r\left(\frac{\pi}{3}\right)\), and adds them to their \(BC\).

A1*: Fully correct proof with no errors seen, leading to the given answer.

**(b)**

M1: Writes down a correct expression for the shaded area in terms of \(r\) and \(\theta\), namely \(\text{Area} = r^2\sin\theta - \frac{1}{2}r^2\theta\) or equivalent.

M1: Substitutes \(r = 6\) and \(\text{Area} = 9\) into their expression.

A1*: Obtains the given answer of \(2\sin\theta - \theta = \frac{1}{2}\) with clear intermediate steps and no errors.

**(c)**

M1: Substitutes \(\theta = \frac{\pi}{6}\) and \(\text{Area} = 12\) into the area formula \(\frac{1}{2}r^2(2\sin\theta - \theta)\).

A1: Correctly substitutes \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) to get \(12 = \frac{1}{2}r^2\left(1 - \frac{\pi}{6}\right)\) or equivalent.

M1: Rearranges the equation to solve for \(r^2\), leading to \(r^2 = \frac{144}{6 - \pi}\).

A1: Finds the exact value \(r = \frac{12}{\sqrt{6 - \pi}}\) (accept \(k = 12, a = 6\)).

(a) To find \(f(x)\), we integrate \(f'(x)\):

\[f(x) = \int \left(3x^2 - 4x^{-2} + 2\right) \text{d}x\]

Integrating term by term:

\[f(x) = \frac{3x^3}{3} - \frac{4x^{-1}}{-1} + 2x + C\]

\[f(x) = x^3 + \frac{4}{x} + 2x + C\]

Since \(P(2, 9)\) lies on the curve \(C\), we can substitute \(x = 2\) and \(f(x) = 9\) to find the constant of integration \(C\):

\[9 = 2^3 + \frac{4}{2} + 2(2) + C\]

\[9 = 8 + 2 + 4 + C\]

\[9 = 14 + C \implies C = -5\]

So,

\[f(x) = x^3 + \frac{4}{x} + 2x - 5\]

(b) To find the equation of the normal at \(P(2, 9)\), we first find the gradient of the tangent at this point by substituting \(x = 2\) into \(f'(x)\):

\[f'(2) = 3(2)^2 - \frac{4}{2^2} + 2 = 12 - 1 + 2 = 13\]

Since the normal is perpendicular to the tangent, the gradient of the normal, \(m_N\), is:

\[m_N = -\frac{1}{13}\]

Using the equation of a straight line with point \((2, 9)\) and gradient \(-\frac{1}{13}\):

\[y - 9 = -\frac{1}{13}(x - 2)\]

Multiplying through by 13 to eliminate the fraction:

\[13(y - 9) = -(x - 2)\]

\[13y - 117 = -x + 2\]

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

\[x + 13y - 119 = 0\]

(a)
- **M1**: Attempts to integrate \(f'(x)\) by raising the power of at least one term by 1 (e.g. \(x^2 \to x^3\) or \(x^{-2} \to x^{-1}\)).
- **A1**: Correct integration of any two terms (constant of integration not required at this stage).
- **A1**: Fully correct integrated expression including the constant of integration: \(x^3 + \frac{4}{x} + 2x + C\).
- **dM1**: Dependent on the first M1. Substitutes \(x = 2\) and \(y = 9\) into their integrated expression to find a value for \(C\).
- **A1**: \(f(x) = x^3 + \frac{4}{x} + 2x - 5\) (or equivalent).

(b)
- **M1**: Substitutes \(x = 2\) into the given expression for \(f'(x)\) to find the gradient of the tangent.
- **M1**: Uses \(m_1 m_2 = -1\) to find the gradient of the normal from their tangent gradient.
- **M1**: Formulates the equation of the line using their normal gradient and the point \((2, 9)\).
- **A1**: \(x + 13y - 119 = 0\) or any integer multiple of this equation (e.g. \(-x - 13y + 119 = 0\)).

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

Rearranging the terms to form a quadratic equation in the form \(Ax^2 + Bx + C = 0\):
\[ x^2 + (k-3)x - x + 2k - 1 - k = 0 \]
\[ x^2 + (k-4)x + (k-1) = 0 \]

Since \(C\) and \(L\) do not intersect, this quadratic equation has no real roots. Therefore, the discriminant of this quadratic equation must be strictly negative (\(b^2 - 4ac < 0\)).

Identifying the coefficients:
\[ a = 1, \quad b = k-4, \quad c = k-1 \]

We require:
\[ (k-4)^2 - 4(1)(k-1) < 0 \]

Expanding and simplifying:
\[ (k^2 - 8k + 16) - (4k - 4) < 0 \]
\[ k^2 - 12k + 20 < 0 \]

To solve this inequality, we first find the critical values by solving the equation:
\[ k^2 - 12k + 20 = 0 \]
\[ (k-2)(k-10) = 0 \]
So, the critical values are \(k = 2\) and \(k = 10\).

Since the inequality is \(k^2 - 12k + 20 < 0\), we require the region between the two critical values.

Thus, the set of possible values for \(k\) is:
\[ 2 < k < 10 \]

* **M1**: Sets the equation of the curve \(C\) equal to the equation of the line \(L\) and attempts to collect terms to form a three-term quadratic equation in \(x\).
* **A1**: Obtains a correct simplified quadratic equation, which is \(x^2 + (k-4)x + (k-1) = 0\) or equivalent.
* **M1**: Uses the discriminant condition \(b^2 - 4ac < 0\) with their coefficients \(a = 1\), \(b = k-4\), and \(c = k-1\).
* **A1**: Obtains the correct quadratic inequality in \(k\): \(k^2 - 12k + 20 < 0\) (or equivalent).
* **M1**: Attempts to solve their three-term quadratic in \(k\) to find the critical values and chooses the inside region (e.g., \(c_1 < k < c_2\) for their critical values \(c_1\) and \(c_2\)).
* **A1**: Fully correct final answer: \(2 < k < 10\) or equivalent set notation, such as \(\{k : 2 < k < 10\}\) or \((2, 10)\).

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

Rearranging the terms to form a quadratic equation in the form \(Ax^2 + Bx + C = 0\):
\[ x^2 + (k-3)x - x + 2k - 1 - k = 0 \]
\[ x^2 + (k-4)x + (k-1) = 0 \]

Since \(C\) and \(L\) do not intersect, this quadratic equation has no real roots. Therefore, the discriminant of this quadratic equation must be strictly negative (\(b^2 - 4ac < 0\)).

Identifying the coefficients:
\[ a = 1, \quad b = k-4, \quad c = k-1 \]

We require:
\[ (k-4)^2 - 4(1)(k-1) < 0 \]

Expanding and simplifying:
\[ (k^2 - 8k + 16) - (4k - 4) < 0 \]
\[ k^2 - 12k + 20 < 0 \]

To solve this inequality, we first find the critical values by solving the equation:
\[ k^2 - 12k + 20 = 0 \]
\[ (k-2)(k-10) = 0 \]
So, the critical values are \(k = 2\) and \(k = 10\).

Since the inequality is \(k^2 - 12k + 20 < 0\), we require the region between the two critical values.

Thus, the set of possible values for \(k\) is:
\[ 2 < k < 10 \]

* **M1**: Sets the equation of the curve \(C\) equal to the equation of the line \(L\) and attempts to collect terms to form a three-term quadratic equation in \(x\).
* **A1**: Obtains a correct simplified quadratic equation, which is \(x^2 + (k-4)x + (k-1) = 0\) or equivalent.
* **M1**: Uses the discriminant condition \(b^2 - 4ac < 0\) with their coefficients \(a = 1\), \(b = k-4\), and \(c = k-1\).
* **A1**: Obtains the correct quadratic inequality in \(k\): \(k^2 - 12k + 20 < 0\) (or equivalent).
* **M1**: Attempts to solve their three-term quadratic in \(k\) to find the critical values and chooses the inside region (e.g., \(c_1 < k < c_2\) for their critical values \(c_1\) and \(c_2\)).
* **A1**: Fully correct final answer: \(2 < k < 10\) or equivalent set notation, such as \(\{k : 2 < k < 10\}\) or \((2, 10)\).

Using the identity \( \sin^2(2\theta) + \cos^2(2\theta) = 1 \), we can rewrite the equation in terms of \( \cos(2\theta) \): \( 3(1 - \cos^2(2\theta)) + 7\cos(2\theta) - 5 = 0 \). Simplifying this gives: \( 3 - 3\cos^2(2\theta) + 7\cos(2\theta) - 5 = 0 \), which simplifies to \( 3\cos^2(2\theta) - 7\cos(2\theta) + 2 = 0 \). Factoring the quadratic expression, we get: \( (3\cos(2\theta) - 1)(\cos(2\theta) - 2) = 0 \). This yields two possible cases: \( \cos(2\theta) = \frac{1}{3} \) or \( \cos(2\theta) = 2 \). Since the range of the cosine function is \( -1 \le \cos(x) \le 1 \), the equation \( \cos(2\theta) = 2 \) has no real solutions. We now find the number of solutions for \( \cos(2\theta) = \frac{1}{3} \) in the interval \( 0 \le \theta \le 2\pi \). Let \( x = 2\theta \). Since \( 0 \le \theta \le 2\pi \), the interval for \( x \) is \( 0 \le x \le 4\pi \). In the interval \( [0, 2\pi] \), the equation \( \cos(x) = \frac{1}{3} \) has exactly 2 solutions. Since the cosine function is periodic with a period of \( 2\pi \), there are also exactly 2 solutions in the interval \( [2\pi, 4\pi] \). Thus, there are a total of \( 2 + 2 = 4 \) solutions in the interval \( 0 \le \theta \le 2\pi \).

M1: Attempts to use the trigonometric identity \( \sin^2(2\theta) + \cos^2(2\theta) = 1 \) to form a quadratic equation in terms of \( \cos(2\theta) \). A1: Correctly simplifies the equation to \( 3\cos^2(2\theta) - 7\cos(2\theta) + 2 = 0 \) and solves it to obtain \( \cos(2\theta) = \frac{1}{3} \), clearly stating or demonstrating that \( \cos(2\theta) = 2 \) has no solutions. M1: Considers the interval for the transformed variable, e.g. stating \( 0 \le 2\theta \le 4\pi \), or uses a graphical approach to show the number of intersections of \( y = \cos(2\theta) \) and \( y = \frac{1}{3} \) in the interval \( [0, 2\pi] \). A1: Concludes that there are exactly 4 solutions with a clear, mathematically sound justification.

Using the identity \( \sin^2(2\theta) + \cos^2(2\theta) = 1 \), we can rewrite the equation in terms of \( \cos(2\theta) \): \( 3(1 - \cos^2(2\theta)) + 7\cos(2\theta) - 5 = 0 \). Simplifying this gives: \( 3 - 3\cos^2(2\theta) + 7\cos(2\theta) - 5 = 0 \), which simplifies to \( 3\cos^2(2\theta) - 7\cos(2\theta) + 2 = 0 \). Factoring the quadratic expression, we get: \( (3\cos(2\theta) - 1)(\cos(2\theta) - 2) = 0 \). This yields two possible cases: \( \cos(2\theta) = \frac{1}{3} \) or \( \cos(2\theta) = 2 \). Since the range of the cosine function is \( -1 \le \cos(x) \le 1 \), the equation \( \cos(2\theta) = 2 \) has no real solutions. We now find the number of solutions for \( \cos(2\theta) = \frac{1}{3} \) in the interval \( 0 \le \theta \le 2\pi \). Let \( x = 2\theta \). Since \( 0 \le \theta \le 2\pi \), the interval for \( x \) is \( 0 \le x \le 4\pi \). In the interval \( [0, 2\pi] \), the equation \( \cos(x) = \frac{1}{3} \) has exactly 2 solutions. Since the cosine function is periodic with a period of \( 2\pi \), there are also exactly 2 solutions in the interval \( [2\pi, 4\pi] \). Thus, there are a total of \( 2 + 2 = 4 \) solutions in the interval \( 0 \le \theta \le 2\pi \).

M1: Attempts to use the trigonometric identity \( \sin^2(2\theta) + \cos^2(2\theta) = 1 \) to form a quadratic equation in terms of \( \cos(2\theta) \). A1: Correctly simplifies the equation to \( 3\cos^2(2\theta) - 7\cos(2\theta) + 2 = 0 \) and solves it to obtain \( \cos(2\theta) = \frac{1}{3} \), clearly stating or demonstrating that \( \cos(2\theta) = 2 \) has no solutions. M1: Considers the interval for the transformed variable, e.g. stating \( 0 \le 2\theta \le 4\pi \), or uses a graphical approach to show the number of intersections of \( y = \cos(2\theta) \) and \( y = \frac{1}{3} \) in the interval \( [0, 2\pi] \). A1: Concludes that there are exactly 4 solutions with a clear, mathematically sound justification.

Part (a)
The equation of the curve is \( y = x(x-2)(x+3) \).
Since the coefficient of \( x^3 \) is positive, the curve starts in the third quadrant and ends in the first quadrant.
The \( x \)-intercepts occur where \( y = 0 \), which gives \( x = -3 \), \( x = 0 \), and \( x = 2 \). Thus, the intercepts are \( (-3, 0) \), \( (0, 0) \), and \( (2, 0) \).
The \( y \)-intercept is also at \( (0, 0) \).

Part (b)
To find the intersection points, equate the two equations:
\( x(x-2)(x+3) = mx \)
\( x(x^2 + x - 6) = mx \)
\( x^3 + x^2 - 6x - mx = 0 \)
\( x(x^2 + x - (6 + m)) = 0 \)

One intersection is always at \( x = 0 \). For there to be exactly three distinct points of intersection, the quadratic equation \( x^2 + x - (6 + m) = 0 \) must have two distinct real roots, neither of which is \( x = 0 \).

1. For two distinct real roots, the discriminant must be strictly positive:
\( b^2 - 4ac > 0 \)
\( 1^2 - 4(1)(-(6+m)) > 0 \)
\( 1 + 24 + 4m > 0 \)
\( 4m > -25 \implies m > -\frac{25}{4} \)

2. To ensure neither root is \( 0 \), substitute \( x = 0 \) into the quadratic:
\( 0^2 + 0 - (6+m) = 0 \implies m = -6 \).
Thus, we must have \( m \neq -6 \) to ensure that \( x = 0 \) is not a repeated root.

Combining these results, the set of values of \( m \) is:
\( m > -\frac{25}{4} \) and \( m \neq -6 \).

Part (a)
M1: Sketch of a positive cubic curve (starts bottom left, ends top right).
A1: Curve passes through the origin \( (0, 0) \).
A1: Correct \( x \)-intercepts at \( (-3, 0) \) and \( (2, 0) \) clearly labeled on the axes.

Part (b)
M1: Equates the curve and line, and factorizes out \( x \) to obtain a quadratic equation.
A1: Correct quadratic equation \( x^2 + x - (6+m) = 0 \).
M1: Uses the discriminant \( b^2 - 4ac > 0 \) on their quadratic.
A1: Solves to find \( m > -\frac{25}{4} \) (or \( m > -6.25 \)).
A1: Deduces that \( m \neq -6 \) is required to avoid a repeated root, and states the final correct set of values: \( m > -\frac{25}{4} \) and \( m \neq -6 \) (or equivalent correct inequality notation like \( -\frac{25}{4} < m < -6 \) and \( m > -6 \)).

(a) Using the formula for the \(n\)-th term of an arithmetic series, \(u_n = a + (n-1)d\):
\(u_{11} = a + 10d = 35\) (Equation 1)

Using the formula for the sum of the first \(n\) terms, \(S_n = \frac{n}{2}[2a + (n-1)d]\):
\(S_4 = \frac{4}{2}[2a + 3d] = 38 \implies 2(2a + 3d) = 38 \implies 2a + 3d = 19\) (Equation 2)

To solve simultaneously, multiply Equation 1 by 2:
\(2a + 20d = 70\) (Equation 3)

Subtract Equation 2 from Equation 3:
\(17d = 51 \implies d = 3\)

Substitute \(d = 3\) back into Equation 1:
\(a + 10(3) = 35 \implies a = 5\)

So, \(a = 5\) and \(d = 3\).

(b) Using the sum formula for \(S_{50}\):
\(S_{50} = \frac{50}{2}[2a + 49d]\)
\(S_{50} = 25[2(5) + 49(3)]\)
\(S_{50} = 25[10 + 147]\)
\(S_{50} = 25(157) = 3925\)

(a)
M1: Sets up a correct equation for the 11th term of the series, e.g., \(a + 10d = 35\).
M1: Sets up a correct equation for the sum of the first 4 terms, e.g., \(\frac{4}{2}(2a+3d) = 38\) or \(2a+3d = 19\).
dM1: Solves the two equations simultaneously to find a value for \(a\) or \(d\) (dependent on at least one of the previous M marks).
A1: For both \(a = 5\) and \(d = 3\).

(b)
M1: Attempts to use the sum formula for 50 terms with their values of \(a\) and \(d\), e.g., \(\frac{50}{2}(2a + 49d)\).
A1: Correct sum of \(3925\).

(a) Using the recurrence relation:
\(a_1 = k\)
\(a_2 = 3a_1 - 5 = 3k - 5\)
\(a_3 = 3a_2 - 5 = 3(3k - 5) - 5\)
\(a_3 = 9k - 15 - 5 = 9k - 20\)

(b) First, find \(a_4\):
\(a_4 = 3a_3 - 5 = 3(9k - 20) - 5 = 27k - 60 - 5 = 27k - 65\)

Now set up the summation:
\(\sum_{r=1}^{4} a_r = a_1 + a_2 + a_3 + a_4\)
\(\sum_{r=1}^{4} a_r = k + (3k - 5) + (9k - 20) + (27k - 65)\)
\(\sum_{r=1}^{4} a_r = 40k - 90\)

Set this sum equal to 110:
\(40k - 90 = 110\)
\(40k = 200\)
\(k = 5\)

(a)
M1: Attempts to find \(a_2\) in terms of \(k\), e.g. \(a_2 = 3k - 5\), and substitutes this into the recurrence relation to find an expression for \(a_3\).
A1: Correct simplified expression: \(9k - 20\) (or equivalent).

(b)
M1: Attempts to find \(a_4\) in terms of \(k\) using their expression for \(a_3\), e.g., \(a_4 = 3(\text{their } a_3) - 5\).
M1: Attempts to find the sum of their four terms \(a_1 + a_2 + a_3 + a_4\) and sets the sum equal to 110.
A1: Obtains a correct simplified equation in \(k\), e.g., \(40k - 90 = 110\) (or equivalent).
A1: \(k = 5\) only.

(a) Substitute \(x = 2\) into the equation of the curve:
\[y = 2^{1.5} - 3(2^{0.5}) + 4 = 2\sqrt{2} - 3\sqrt{2} + 4 = 4 - \sqrt{2}\]
Using a calculator to find the decimal value:
\[y \approx 2.585786...\]
To 3 decimal places, \(y = 2.586\).

(b) The step size is \(h = \frac{4 - 1}{3} = 1\).
Using the trapezium rule:
\[\text{Area} \approx \frac{h}{2} \left[ y_0 + y_3 + 2(y_1 + y_2) \right]\]
\[\text{Area} \approx \frac{1}{2} \left[ 2 + 6 + 2(2.586 + 4) \right]\]
\[\text{Area} \approx \frac{1}{2} \left[ 8 + 2(6.586) \right] = \frac{1}{2} \left[ 8 + 13.172 \right] = \frac{21.172}{2} = 10.586\]
To 2 decimal places, the approximate value is \(10.59\).

(c) To find the exact value, integrate each term:
\[\int_{1}^{4} \left(x^{1.5} - 3x^{0.5} + 4\right) \, \mathrm{d}x = \left[ \frac{x^{2.5}}{2.5} - \frac{3x^{1.5}}{1.5} + 4x \right]_{1}^{4}\]
\[= \left[ \frac{2}{5}x^{2.5} - 2x^{1.5} + 4x \right]_{1}^{4}\]
Substitute the upper limit \(x = 4\):
\[\left(\frac{2}{5}(4^{2.5}) - 2(4^{1.5}) + 4(4)\right) = \left(\frac{2}{5}(32) - 2(8) + 16\right) = 12.8 - 16 + 16 = 12.8\]
Substitute the lower limit \(x = 1\):
\[\left(\frac{2}{5}(1^{2.5}) - 2(1^{1.5}) + 4(1)\right) = 0.4 - 2 + 4 = 2.4\]
Subtract the lower limit evaluation from the upper limit evaluation:
\[12.8 - 2.4 = 10.4\]

(a)
- **B1**: For \(2.586\) (accept \(4 - \sqrt{2}\)).

(b)
- **B1**: For step size \(h = 1\) (can be implied by the working).
- **M1**: For a correct application of the Trapezium Rule formula with their value from part (a). Look for \(\frac{1}{2} \times 1 \times [2 + 6 + 2(\text{their } y_1 + 4)]\).
- **A1**: For \(10.59\) (or follow through to 2 d.p. from their part (a) value, e.g., if using exact \(4-\sqrt{2}\), the value is \(12 - \sqrt{2} \approx 10.59\)).

(c)
- **M1**: For an attempt to integrate: at least one term has its power increased by 1.
- **A1**: For \(10.4\) (or exact equivalent such as \(\frac{52}{5}\)). Note: A correct answer with no integration working shown scores 0 marks.

(a) Substitute \(x = 2\) into the equation of the curve:
\[y = 2^{1.5} - 3(2^{0.5}) + 4 = 2\sqrt{2} - 3\sqrt{2} + 4 = 4 - \sqrt{2}\]
Using a calculator to find the decimal value:
\[y \approx 2.585786...\]
To 3 decimal places, \(y = 2.586\).

(b) The step size is \(h = \frac{4 - 1}{3} = 1\).
Using the trapezium rule:
\[\text{Area} \approx \frac{h}{2} \left[ y_0 + y_3 + 2(y_1 + y_2) \right]\]
\[\text{Area} \approx \frac{1}{2} \left[ 2 + 6 + 2(2.586 + 4) \right]\]
\[\text{Area} \approx \frac{1}{2} \left[ 8 + 2(6.586) \right] = \frac{1}{2} \left[ 8 + 13.172 \right] = \frac{21.172}{2} = 10.586\]
To 2 decimal places, the approximate value is \(10.59\).

(c) To find the exact value, integrate each term:
\[\int_{1}^{4} \left(x^{1.5} - 3x^{0.5} + 4\right) \, \mathrm{d}x = \left[ \frac{x^{2.5}}{2.5} - \frac{3x^{1.5}}{1.5} + 4x \right]_{1}^{4}\]
\[= \left[ \frac{2}{5}x^{2.5} - 2x^{1.5} + 4x \right]_{1}^{4}\]
Substitute the upper limit \(x = 4\):
\[\left(\frac{2}{5}(4^{2.5}) - 2(4^{1.5}) + 4(4)\right) = \left(\frac{2}{5}(32) - 2(8) + 16\right) = 12.8 - 16 + 16 = 12.8\]
Substitute the lower limit \(x = 1\):
\[\left(\frac{2}{5}(1^{2.5}) - 2(1^{1.5}) + 4(1)\right) = 0.4 - 2 + 4 = 2.4\]
Subtract the lower limit evaluation from the upper limit evaluation:
\[12.8 - 2.4 = 10.4\]

(a)
- **B1**: For \(2.586\) (accept \(4 - \sqrt{2}\)).

(b)
- **B1**: For step size \(h = 1\) (can be implied by the working).
- **M1**: For a correct application of the Trapezium Rule formula with their value from part (a). Look for \(\frac{1}{2} \times 1 \times [2 + 6 + 2(\text{their } y_1 + 4)]\).
- **A1**: For \(10.59\) (or follow through to 2 d.p. from their part (a) value, e.g., if using exact \(4-\sqrt{2}\), the value is \(12 - \sqrt{2} \approx 10.59\)).

(c)
- **M1**: For an attempt to integrate: at least one term has its power increased by 1.
- **A1**: For \(10.4\) (or exact equivalent such as \(\frac{52}{5}\)). Note: A correct answer with no integration working shown scores 0 marks.

**(a)**

Using the remainder theorem with \(x = 2\):
\[\mathrm{f}(2) = -10\]
\[2(2)^3 + p(2)^2 - 13(2) + q = -10\]
\[16 + 4p - 26 + q = -10\]
\[4p + q - 10 = -10 \implies 4p + q = 0 \quad \text{--- (Equation 1)}\]

Using the factor theorem with \(x = -3\):
\[\mathrm{f}(-3) = 0\]
\[2(-3)^3 + p(-3)^2 - 13(-3) + q = 0\]
\[-54 + 9p + 39 + q = 0\]
\[9p + q - 15 = 0 \implies 9p + q = 15 \quad \text{--- (Equation 2)}\]

Subtracting Equation 1 from Equation 2:
\[(9p + q) - (4p + q) = 15 - 0\]
\[5p = 15 \implies p = 3\]

Substituting \(p = 3\) back into Equation 1:
\[4(3) + q = 0 \implies q = -12\]

So, \(p = 3\) and \(q = -12\).

**(b)**

Substituting \(p\) and \(q\) into the expression for \(\mathrm{f}(x)\):
\[\mathrm{f}(x) = 2x^3 + 3x^2 - 13x - 12\]

We divide \(\mathrm{f}(x)\) by \((x + 3)\) using algebraic long division or equating coefficients:
\[2x^3 + 3x^2 - 13x - 12 = (x + 3)(ax^2 + bx + c)\]

Equating \(x^3\) terms: \(a = 2\)
Equating constant terms: \(3c = -12 \implies c = -4\)
Equating \(x^2\) terms: \(3a + b = 3 \implies 3(2) + b = 3 \implies b = -3\)

Thus,
\[\mathrm{f}(x) = (x + 3)(2x^2 - 3x - 4)\]

**(c)**

To solve \(\mathrm{f}(x) = 0\):
\[(x + 3)(2x^2 - 3x - 4) = 0\]

This gives \(x = -3\) or \(2x^2 - 3x - 4 = 0\).

Applying the quadratic formula to the quadratic part:
\[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-4)}}{2(2)}\]
\[x = \frac{3 \pm \sqrt{9 + 32}}{4}\]
\[x = \frac{3 \pm \sqrt{41}}{4}\]

So the exact solutions are:
\[x = -3, \quad x = \frac{3 + \sqrt{41}}{4}, \quad x = \frac{3 - \sqrt{41}}{4}\]

**(a)**
* **M1**: Attempts to use the remainder theorem with \(x = 2\) or the factor theorem with \(x = -3\) (e.g., substitutes \(x = \pm 2\) or \(x = \pm 3\) and sets equal to \(-10\) or \(0\) respectively).
* **A1**: One correct equation, either \(4p + q = 0\) or \(9p + q = 15\) (or unsimplified equivalent).
* **M1**: Solves their two linear equations simultaneously to find a value for \(p\) or \(q\).
* **A1**: Both \(p = 3\) and \(q = -12\).

**(b)**
* **M1**: Attempts to divide \(\mathrm{f}(x)\) by \((x + 3)\) or uses equating of coefficients. Must obtain a quadratic expression with at least two of \(a\), \(b\) or \(c\) correct for their \(p\) and \(q\).
* **A1**: Correct expression \((x + 3)(2x^2 - 3x - 4)\) or states \(a = 2, b = -3, c = -4\) clearly.

**(c)**
* **B1**: Identifies \(x = -3\) as one solution (can be implied by the final list of solutions).
* **M1**: Applies the quadratic formula (or completing the square) to their three-term quadratic factor of \(\mathrm{f}(x)\).
* **A1**: Correct exact solutions: \(x = -3\) and \(x = \frac{3 \pm \sqrt{41}}{4}\) (must be exact, decimal approximations receive A0).

To find the value of \( k \), we first integrate the expression with respect to \( x \):

\[ \int (3x^2 - 4x + 1) \, \mathrm{d}x = \left[ x^3 - 2x^2 + x \right] \]

Applying the limits \( 1 \) and \( k \):

\[ \left[ x^3 - 2x^2 + x \right]_{1}^{k} = (k^3 - 2k^2 + k) - (1^3 - 2(1)^2 + 1) \]

\[ = k^3 - 2k^2 + k - 0 \]

\[ = k^3 - 2k^2 + k \]

We are given that this definite integral is equal to 12:

\[ k^3 - 2k^2 + k = 12 \]

\[ k^3 - 2k^2 + k - 12 = 0 \]

By testing small integer values, we find that \( k = 3 \) is a root of the equation, since:

\[ 3^3 - 2(3)^2 + 3 - 12 = 27 - 18 + 3 - 12 = 0 \]

Using polynomial division or factorisation:

\[ (k - 3)(k^2 + k + 4) = 0 \]

For the quadratic factor \( k^2 + k + 4 \), we check the discriminant:

\[ b^2 - 4ac = 1^2 - 4(1)(4) = -15 \]

Since the discriminant is less than 0, there are no other real roots.

Since we are given \( k > 1 \), the only real solution is:

\[ k = 3 \]

- **M1**: Attempts to integrate \( 3x^2 - 4x + 1 \), increasing at least one power of \( x \) by 1 (e.g. \( x^2 \to x^3 \) or \( x \to x^2 \)).
- **A1**: Correct integration resulting in \( x^3 - 2x^2 + x \) (the constant of integration is not required).
- **M1**: Substitutes the limits \( 1 \) and \( k \) into their integrated expression, subtracts the correct way around, and sets the result equal to 12.
- **dM1**: Formulates a cubic equation of the form \( k^3 - 2k^2 + k - 12 = 0 \) and attempts to solve it (at least showing that \( k = 3 \) is a root by substitution or division). Dependent on the first M mark.
- **A1**: Obtains \( k = 3 \) with clear working showing that there are no other real roots (e.g., by evaluating the discriminant of the quadratic factor \( k^2 + k + 4 \) to show it is negative, or completing the square).

(a) Substituting the coordinates of \(P(2, 1)\) into the circle equation: \(2^2 + 1^2 - 10(2) + 8(1) + k = 0 \implies 4 + 1 - 20 + 8 + k = 0 \implies -7 + k = 0\), which gives \(k = 7\). (b) Substituting \(k = 7\) gives the equation \(x^2 - 10x + y^2 + 8y + 7 = 0\). Completing the square: \((x - 5)^2 - 25 + (y + 4)^2 - 16 + 7 = 0 \implies (x - 5)^2 + (y + 4)^2 = 34\). Thus, the centre \(M\) is \((5, -4)\) and the exact radius is \(\sqrt{34}\). (c) The gradient of the radius \(MP\) is \(m_{MP} = \frac{1 - (-4)}{2 - 5} = \frac{5}{-3} = -\frac{5}{3}\). The tangent \(l\) is perpendicular to \(MP\), so its gradient is \(m_l = \frac{3}{5}\). The equation of \(l\) is \(y - 1 = \frac{3}{5}(x - 2) \implies 5(y - 1) = 3(x - 2) \implies 5y - 5 = 3x - 6 \implies 3x - 5y - 1 = 0\). (d) Since \(l\) is tangent to \(C\) at \(P\), angle \(MPQ\) is \(90^{\circ}\). By Pythagoras' Theorem in the right-angled triangle \(MPQ\): \(MP^2 + PQ^2 = MQ^2 \implies (\sqrt{34})^2 + PQ^2 = (\sqrt{170})^2 \implies 34 + PQ^2 = 170 \implies PQ^2 = 136 \implies PQ = \sqrt{136} = 2\sqrt{34}\).

(a) M1: Attempts to substitute \(x = 2\) and \(y = 1\) into the circle equation and sets equal to 0. A1: \(k = 7\). (b) M1: Completes the square for both \(x\) and \(y\) terms. Allow one sign or arithmetic error. A1: Correct coordinates for the centre \(M(5, -4)\). A1: Correct exact radius \(\sqrt{34}\). (c) M1: Attempts to find the gradient of the radius \(MP\) using their centre \(M\) and \(P(2, 1)\). M1: Uses the perpendicular gradient rule \(m_1 m_2 = -1\) to find the gradient of the tangent. M1: Attempts to find the equation of the line \(l\) using their perpendicular gradient and point \(P(2, 1)\). A1: Correct equation in the form \(ax + by + c = 0\) with integer coefficients, e.g. \(3x - 5y - 1 = 0\) (or any integer multiple). (d) M1: Uses Pythagoras' Theorem in triangle \(MPQ\) to set up \(PQ^2 = MQ^2 - r^2\) with their radius and \(\sqrt{170}\). A1: Correct exact distance of \(2\sqrt{34}\) or \(\sqrt{136}\).

(a) The total area \(A\) of the flowerbed is composed of the area of the rectangle and the area of the semi-circle:
\[A = 2xy + \frac{1}{2}\pi x^2 = 30\]
Rearranging this expression to express \(2y\) in terms of \(x\):
\[2xy = 30 - \frac{1}{2}\pi x^2 \implies 2y = \frac{30}{x} - \frac{\pi}{2}x\]
The perimeter \(P\) of the flowerbed consists of the three outer straight edges of the rectangle and the semi-circular arc:
\[P = 2x + 2y + \pi x\]
Substituting the expression for \(2y\) into this perimeter formula:
\[P = 2x + \left(\frac{30}{x} - \frac{\pi}{2}x\right) + \pi x\]
\[P = 2x + \frac{\pi}{2}x + \frac{30}{x} = \left(2 + \frac{\pi}{2}\right)x + \frac{30}{x}\]
This completes the proof.

(b) To find the value of \(x\) for which \(P\) is a minimum, we differentiate \(P\) with respect to \(x\):
\[P = \left(2 + \frac{\pi}{2}\right)x + 30x^{-1}\]
\[\frac{\mathrm{d}P}{\mathrm{d}x} = \left(2 + \frac{\pi}{2}\right) - 30x^{-2}\]
At the minimum value, \[\frac{\mathrm{d}P}{\mathrm{d}x} = 0\]
\[\left(2 + \frac{\pi}{2}\right) - \frac{30}{x^2} = 0 \implies x^2 = \frac{30}{2 + \frac{\pi}{2}} = \frac{60}{4 + \pi}\]
Since \(x > 0\), we take the positive square root:
\[x = \sqrt{\frac{60}{4 + \pi}} \approx 2.8985...\]
To 3 significant figures, \(x = 2.90\).

(c) To justify that this value gives a minimum, we find the second derivative:
\[\frac{\mathrm{d}^2P}{\mathrm{d}x^2} = 60x^{-3} = \frac{60}{x^3}\]
Since \(x \approx 2.90 > 0\), \(\frac{\mathrm{d}^2P}{\mathrm{d}x^2} > 0\), which confirms that the value of \(x\) gives a minimum.

To find the minimum value of \(P\):
\[P = \left(2 + \frac{\pi}{2}\right)(2.8985) + \frac{30}{2.8985} \approx 20.6999...\]
To 1 decimal place, the minimum perimeter \(P = 20.7\text{ m}\).

Part (a):
- M1: Formulates a correct expression for the area of the flowerbed in terms of \(x\) and \(y\): \(2xy + \frac{1}{2}\pi x^2 = 30\).
- A1: Rearranges the area equation to correctly express \(y\) or \(2y\) in terms of \(x\). (e.g., \(2y = \frac{30}{x} - \frac{\pi}{2}x\)).
- M1: Writes a correct expression for the perimeter of the shape: \(P = 2x + 2y + \pi x\).
- A1*: Substitutes the expression for \(2y\) into the perimeter expression and simplifies to show the given result cleanly with no errors.

Part (b):
- M1: Attempts to differentiate \(P\) with respect to \(x\) with at least one term correct (e.g., \(\left(2+\frac{\pi}{2}\right)\) or \(-\frac{30}{x^2}\)).
- A1: Obtains the correct derivative: \(\frac{\mathrm{d}P}{\mathrm{d}x} = \left(2 + \frac{\pi}{2}\right) - \frac{30}{x^2}\).
- M1: Sets their derivative to 0 and attempts to make \(x^2\) or \(x\) the subject.
- A1: Identifies the exact expression for \(x\): \(x = \sqrt{\frac{60}{4+\pi}}\).
- A1: Calculates the decimal value as \(x \approx 2.90\) (3 sf).

Part (c):
- M1: Finds the second derivative \(\frac{\mathrm{d}^2P}{\mathrm{d}x^2} = \frac{60}{x^3}\) and evaluates its sign, or uses another valid justification method.
- A1: Correctly evaluates \(P \approx 20.7\) (1 dp) and provides a clear conclusion that \(\frac{\mathrm{d}^2P}{\mathrm{d}x^2} > 0\) means the stationary point is a minimum.

(a) For a geometric sequence, the ratio between consecutive terms is constant. Therefore:
$$\frac{2\sin\theta}{4\cos\theta} = \frac{3\cos\theta}{2\sin\theta}$$
Using cross-multiplication (or the relation \(u_2^2 = u_1 u_3\)):
$$(2\sin\theta)^2 = (4\cos\theta)(3\cos\theta)$$
$$4\sin^2\theta = 12\cos^2\theta$$
Dividing both sides by \(4\cos^2\theta\) (since \(\cos\theta \neq 0\)):
$$\frac{\sin^2\theta}{\cos^2\theta} = \frac{12}{4}$$
$$\tan^2\theta = 3$$

(b) Since \(0 < \theta < \frac{\pi}{2}\), \(\tan\theta > 0\).
Taking the square root of both sides of \(\tan^2\theta = 3\):
$$\tan\theta = \sqrt{3}$$
Thus, the exact value of \(\theta\) in radians is:
$$\theta = \frac{\pi}{3}$$

(c) The first term is:
$$a = u_1 = 4\cos\left(\frac{\pi}{3}\right) = 4 \times \frac{1}{2} = 2$$
The second term is:
$$u_2 = 2\sin\left(\frac{\pi}{3}\right) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$$
The common ratio \(r\) is:
$$r = \frac{u_2}{u_1} = \frac{\sqrt{3}}{2}$$

(d) The sum to infinity is given by:
$$S_{\infty} = \frac{a}{1 - r}$$
Substituting \(a = 2\) and \(r = \frac{\sqrt{3}}{2}\):
$$S_{\infty} = \frac{2}{1 - \frac{\sqrt{3}}{2}} = \frac{4}{2 - \sqrt{3}}$$
Rationalizing the denominator:
$$S_{\infty} = \frac{4(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{8 + 4\sqrt{3}}{4 - 3} = 8 + 4\sqrt{3}$$
Thus, the exact value of the sum to infinity is \(8 + 4\sqrt{3}\).

(a)
M1: Attempts to use the property of a geometric sequence, e.g., \(u_2^2 = u_1 u_3\) or \(\frac{u_2}{u_1} = \frac{u_3}{u_2}\).
M1: Obtains a correct equation in terms of \(\sin^2\theta\) and \(\cos^2\theta\), e.g., \(4\sin^2\theta = 12\cos^2\theta\).
A1: Reaches the given result \(\tan^2\theta = 3\) with no errors shown.

(b)
B1: \(\theta = \frac{\pi}{3}\) (or exact equivalent in radians).

(c)
M1: Substitutes their value of \(\theta\) to find the first term \(a\) or uses \(r = \frac{1}{2}\tan\theta\).
A1: \(r = \frac{\sqrt{3}}{2}\) (exact value).

(d)
M1: Uses the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\) with their \(a\) and \(r\).
M1: Multiplies numerator and denominator by the conjugate of the denominator to rationalize.
A1: \(8 + 4\sqrt{3}\) (or \(a=8, b=4\)).

(a) For a geometric sequence, the ratio between consecutive terms is constant. Therefore:
$$\frac{2\sin\theta}{4\cos\theta} = \frac{3\cos\theta}{2\sin\theta}$$
Using cross-multiplication (or the relation \(u_2^2 = u_1 u_3\)):
$$(2\sin\theta)^2 = (4\cos\theta)(3\cos\theta)$$
$$4\sin^2\theta = 12\cos^2\theta$$
Dividing both sides by \(4\cos^2\theta\) (since \(\cos\theta \neq 0\)):
$$\frac{\sin^2\theta}{\cos^2\theta} = \frac{12}{4}$$
$$\tan^2\theta = 3$$

(b) Since \(0 < \theta < \frac{\pi}{2}\), \(\tan\theta > 0\).
Taking the square root of both sides of \(\tan^2\theta = 3\):
$$\tan\theta = \sqrt{3}$$
Thus, the exact value of \(\theta\) in radians is:
$$\theta = \frac{\pi}{3}$$

(c) The first term is:
$$a = u_1 = 4\cos\left(\frac{\pi}{3}\right) = 4 \times \frac{1}{2} = 2$$
The second term is:
$$u_2 = 2\sin\left(\frac{\pi}{3}\right) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$$
The common ratio \(r\) is:
$$r = \frac{u_2}{u_1} = \frac{\sqrt{3}}{2}$$

(d) The sum to infinity is given by:
$$S_{\infty} = \frac{a}{1 - r}$$
Substituting \(a = 2\) and \(r = \frac{\sqrt{3}}{2}\):
$$S_{\infty} = \frac{2}{1 - \frac{\sqrt{3}}{2}} = \frac{4}{2 - \sqrt{3}}$$
Rationalizing the denominator:
$$S_{\infty} = \frac{4(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{8 + 4\sqrt{3}}{4 - 3} = 8 + 4\sqrt{3}$$
Thus, the exact value of the sum to infinity is \(8 + 4\sqrt{3}\).

(a)
M1: Attempts to use the property of a geometric sequence, e.g., \(u_2^2 = u_1 u_3\) or \(\frac{u_2}{u_1} = \frac{u_3}{u_2}\).
M1: Obtains a correct equation in terms of \(\sin^2\theta\) and \(\cos^2\theta\), e.g., \(4\sin^2\theta = 12\cos^2\theta\).
A1: Reaches the given result \(\tan^2\theta = 3\) with no errors shown.

(b)
B1: \(\theta = \frac{\pi}{3}\) (or exact equivalent in radians).

(c)
M1: Substitutes their value of \(\theta\) to find the first term \(a\) or uses \(r = \frac{1}{2}\tan\theta\).
A1: \(r = \frac{\sqrt{3}}{2}\) (exact value).

(d)
M1: Uses the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\) with their \(a\) and \(r\).
M1: Multiplies numerator and denominator by the conjugate of the denominator to rationalize.
A1: \(8 + 4\sqrt{3}\) (or \(a=8, b=4\)).

From the first equation, we use the subtraction law of logarithms:

\(\log_3\left(\frac{x}{y}\right) = 2\)

Using the definition of logarithms, this can be written as:

\(\frac{x}{y} = 3^2\)

\(x = 9y\) [Equation 1]

From the second equation, we remove the logarithm:

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

\(x - 5y = 8\) [Equation 2]

Substitute Equation 1 into Equation 2:

\(9y - 5y = 8\)

\(4y = 8\)

\(y = 2\)

Substitute \(y = 2\) back into Equation 1:

\(x = 9(2) = 18\)

Thus, the solutions are \(x = 18\) and \(y = 2\).

M1: Applies the subtraction law of logarithms to the first equation to obtain \(\log_3\left(\frac{x}{y}\right) = 2\) or writes \(\log_3 x = \log_3 y + 2 = \log_3(9y)\) to obtain \(x = 9y\).

M1: Converts the second equation to a non-logarithmic linear form, showing \(x - 5y = 2^3\) or \(x - 5y = 8\).

M1: Substitutes \(x = 9y\) into their linear equation in \(x\) and \(y\) and attempts to solve for at least one variable.

A1: Correct values of both \(x = 18\) and \(y = 2\).

(a) Using the binomial expansion formula:

\[(2 + kx)^7 = 2^7 + \binom{7}{1}2^6(kx) + \binom{7}{2}2^5(kx)^2 + \binom{7}{3}2^4(kx)^3 + \dots\]

Evaluating each term:
- \(2^7 = 128\)
- \(\binom{7}{1}2^6(kx) = 7 \times 64 \times kx = 448kx\)
- \(\binom{7}{2}2^5(kx)^2 = 21 \times 32 \times k^2x^2 = 672k^2x^2\)
- \(\binom{7}{3}2^4(kx)^3 = 35 \times 16 \times k^3x^3 = 560k^3x^3\)

So the first 4 terms are:
\[128 + 448kx + 672k^2x^2 + 560k^3x^3\]

(b) Consider the product:
\[\left(1 - \frac{3}{4}x\right)(128 + 448kx + 672k^2x^2 + \dots)\]

The terms in \(x^2\) in this product arise from:
\[1 \times (672k^2x^2) + \left(-\frac{3}{4}x\right) \times (448kx) = \left(672k^2 - 336k\right)x^2\]

We are given that the coefficient of \(x^2\) is zero, so:
\[672k^2 - 336k = 0\]

Since \(k\) is a non-zero constant, we can factorise:
\[336k(2k - 1) = 0\]

As \(k \neq 0\):
\[2k - 1 = 0 \implies k = \frac{1}{2}\]

(a)
- **B1**: For 128.
- **M1**: For a correct binomial coefficient combined with correct powers of 2 and \(kx\) for either the \(x\) term or the \(x^2\) term, e.g., \(7 \times 2^6 \times (kx)\) or \(21 \times 2^5 \times (kx)^2\). The powers of \(k\) and \(x\) must be consistent, i.e., \(k^2x^2\) or at least \((kx)^2\).
- **A1**: For \(448kx + 672k^2x^2\).
- **A1**: For \(560k^3x^3\). (Allow if listed as a comma-separated list or as a sum).

(b)
- **M1**: Attempts to identify the terms in \(x^2\) from the expansion of \(\left(1 - \frac{3}{4}x\right)(128 + 448kx + 672k^2x^2 + \dots)\). Must see an expression of the form \(A k^2 + B k\) where \(A\) is their coefficient of \(x^2\) and \(B = -\frac{3}{4} \times\) (their coefficient of \(x\)).
- **A1ft**: For a correct equation for the coefficient of \(x^2\), e.g., \(672k^2 - 336k = 0\), follow through their coefficients of \(x\) and \(x^2\) from part (a).
- **dM1**: Dependent on the previous M mark. Solves the quadratic/linear equation in \(k\) by factorising or dividing by \(k\) (since \(k \neq 0\)).
- **A1**: For \(k = \frac{1}{2}\) (or 0.5) only.

(a) Using the binomial expansion formula:

\[(2 + kx)^7 = 2^7 + \binom{7}{1}2^6(kx) + \binom{7}{2}2^5(kx)^2 + \binom{7}{3}2^4(kx)^3 + \dots\]

Evaluating each term:
- \(2^7 = 128\)
- \(\binom{7}{1}2^6(kx) = 7 \times 64 \times kx = 448kx\)
- \(\binom{7}{2}2^5(kx)^2 = 21 \times 32 \times k^2x^2 = 672k^2x^2\)
- \(\binom{7}{3}2^4(kx)^3 = 35 \times 16 \times k^3x^3 = 560k^3x^3\)

So the first 4 terms are:
\[128 + 448kx + 672k^2x^2 + 560k^3x^3\]

(b) Consider the product:
\[\left(1 - \frac{3}{4}x\right)(128 + 448kx + 672k^2x^2 + \dots)\]

The terms in \(x^2\) in this product arise from:
\[1 \times (672k^2x^2) + \left(-\frac{3}{4}x\right) \times (448kx) = \left(672k^2 - 336k\right)x^2\]

We are given that the coefficient of \(x^2\) is zero, so:
\[672k^2 - 336k = 0\]

Since \(k\) is a non-zero constant, we can factorise:
\[336k(2k - 1) = 0\]

As \(k \neq 0\):
\[2k - 1 = 0 \implies k = \frac{1}{2}\]

(a)
- **B1**: For 128.
- **M1**: For a correct binomial coefficient combined with correct powers of 2 and \(kx\) for either the \(x\) term or the \(x^2\) term, e.g., \(7 \times 2^6 \times (kx)\) or \(21 \times 2^5 \times (kx)^2\). The powers of \(k\) and \(x\) must be consistent, i.e., \(k^2x^2\) or at least \((kx)^2\).
- **A1**: For \(448kx + 672k^2x^2\).
- **A1**: For \(560k^3x^3\). (Allow if listed as a comma-separated list or as a sum).

(b)
- **M1**: Attempts to identify the terms in \(x^2\) from the expansion of \(\left(1 - \frac{3}{4}x\right)(128 + 448kx + 672k^2x^2 + \dots)\). Must see an expression of the form \(A k^2 + B k\) where \(A\) is their coefficient of \(x^2\) and \(B = -\frac{3}{4} \times\) (their coefficient of \(x\)).
- **A1ft**: For a correct equation for the coefficient of \(x^2\), e.g., \(672k^2 - 336k = 0\), follow through their coefficients of \(x\) and \(x^2\) from part (a).
- **dM1**: Dependent on the previous M mark. Solves the quadratic/linear equation in \(k\) by factorising or dividing by \(k\) (since \(k \neq 0\)).
- **A1**: For \(k = \frac{1}{2}\) (or 0.5) only.