2025 Edexcel IAS-Level Pure Mathematics (XPM01) 模擬試題連答案詳解

(a) By symmetry, the lowest point of the cable must occur midway between the two supporting towers, so \(x = \frac{160}{2} = 80\). The minimum height of the cable is given as \(8\text{ m}\). Therefore, the coordinates of the lowest point are \((80, 8)\).

(b) Using the vertex form of a quadratic equation, we can write:
\(y = p(x - 80)^2 + 8\)
Since the left tower is located at the origin \(x = 0\) with a height of \(24\text{ m}\), we can substitute the point \((0, 24)\) into our equation:
\(24 = p(0 - 80)^2 + 8\)
\(16 = 6400p\)
\(p = \frac{16}{6400} = 0.0025\) (or \(\frac{1}{400}\))

Now, expand the quadratic equation to express it in the form \(y = px^2 + qx + r\):
\(y = 0.0025(x^2 - 160x + 6400) + 8\)
\(y = 0.0025x^2 - 0.4x + 16 + 8\)
\(y = 0.0025x^2 - 0.4x + 24\)

Thus, \(p = 0.0025\), \(q = -0.4\), and \(r = 24\).

(c) To find the height of a vertical support wire located \(40\text{ m}\) from the left tower, we substitute \(x = 40\) into our equation:
\(y = 0.0025(40)^2 - 0.4(40) + 24\)
\(y = 0.0025(1600) - 16 + 24\)
\(y = 4 - 16 + 24 = 12\text{ m}\)

(d) To find the minimum height of the second cable, we complete the square for \(y = 0.002x^2 - 0.28x + 18\):
\(y = 0.002(x^2 - 140x) + 18\)
\(y = 0.002\left((x - 70)^2 - 4900\right) + 18\)
\(y = 0.002(x - 70)^2 - 9.8 + 18\)
\(y = 0.002(x - 70)^2 + 8.2\)

Therefore, the minimum height of this second cable is \(8.2\text{ m}\).

(a)
B1: For correctly identifying the coordinates \((80, 8)\).

(b)
M1: Writes the quadratic equation in vertex form, \(y = p(x - 80)^2 + 8\), or sets up a system of simultaneous equations using known points \((0, 24)\), \((80, 8)\), and \((160, 24)\).
M1: Substitutes \(x = 0, y = 24\) (or \(x = 160, y = 24\)) to find the value of \(p\).
A1: Correct value of \(p = 0.0025\) (or \(\frac{1}{400}\)).
A1: Fully expands to find \(q = -0.4\) (or \(-\frac{2}{5}\)) and \(r = 24\).

(c)
M1: Substitutes \(x = 40\) into their quadratic model.
A1: Correctly identifies the height as \(12\text{ m}\).

(d)
M1: Attempts to find the minimum of the second quadratic curve, either by completing the square or by finding the vertex at \(x = -\frac{b}{2a}\).
A1: Correct minimum height of \(8.2\text{ m}\).

(a) For the company to make a profit, the daily total revenue must exceed the daily total cost, so \(\mathcal{R} > \mathcal{C}\).
\(125x - x^2 > 35x + 800\)
Rearranging this inequality into standard form:
\(-x^2 + 90x - 800 > 0\)
Multiply by \(-1\) and reverse the inequality sign:
\(x^2 - 90x + 800 < 0\)
Factorise the quadratic expression:
\((x - 10)(x - 80) < 0\)
The critical values are \(x = 10\) and \(x = 80\).
Since the quadratic expression must be less than zero, the solution range is:
\(10 < x < 80\) (or, because \(x\) represents an integer number of bags, \(11 \le x \le 79\)).

(b) The profit function is:
\(\mathcal{P} = -x^2 + 90x - 800\)
To complete the square:
\(\mathcal{P} = -(x^2 - 90x) - 800\)
\(\mathcal{P} = -\left((x - 45)^2 - 2025\right) - 800\)
\(\mathcal{P} = -(x - 45)^2 + 2025 - 800\)
\(\mathcal{P} = 1225 - (x - 45)^2\)

The maximum value of \(\mathcal{P}\) is achieved when the squared term is zero, which occurs when \(x = 45\).
Thus, the company should produce \(45\) bags per day to achieve a maximum profit of \(£1225\).

(c) Two possible limitations of the model for large values of \(x\) are:
- For \(x > 125\), the revenue \(\mathcal{R} = 125x - x^2\) becomes negative, which is impossible in real life.
- For \(x > 80\), the model predicts the profit becomes negative and decreases indefinitely, which is unrealistic because in practice, production would be restricted to avoid losses, and overheads might scale differently.

(a)
M1: Sets up the inequality \(\mathcal{R} - \mathcal{C} > 0\) or equation \(\mathcal{R} - \mathcal{C} = 0\).
M1: Forms a 3-term quadratic inequality or equation, e.g., \(x^2 - 90x + 800 < 0\).
A1: Factorises or uses the quadratic formula to obtain the critical values \(x = 10\) and \(x = 80\).
A1: Identifies the correct range, accepting either \(10 < x < 80\) or \(11 \le x \le 79\).

(b)
M1: Attempts to complete the square on \(-x^2 + 90x - 800\), obtaining \(-(x - a)^2 + b\).
A1: Obtains \(-(x - 45)^2 + 1225\) or equivalent.
A1: Correctly identifies \(45\) bags and maximum profit of \(£1225\).

(c)
B1: Identifies that for \(x > 125\) the revenue becomes negative, which is impossible in a business context.
B1: Explains that for very large \(x\) the profit decreases indefinitely to negative infinity, which is unrealistic as production would stop.

(a) At the moment of launch, \(t = 0\), and the height of the rocket is the height of the platform, which is \(35\text{ m}\).
Substituting \(t = 0\) and \(h = 35\) into the equation:
\(35 = -5(0)^2 + u(0) + k \implies k = 35\).

(b) Using \(k = 35\), the model becomes \(h = -5t^2 + ut + 35\).
We can complete the square to express this quadratic in terms of its vertex:
\(h = -5\left(t^2 - \frac{u}{5}t\right) + 35\)
\(h = -5\left[\left(t - \frac{u}{10}\right)^2 - \frac{u^2}{100}\right] + 35\)
\(h = -5\left(t - \frac{u}{10}\right)^2 + \frac{u^2}{20} + 35\)

The maximum height is achieved when the squared term is zero, so the maximum height is \(\frac{u^2}{20} + 35\).
We are given that this maximum height is \(80\text{ m}\):
\(\frac{u^2}{20} + 35 = 80\)
\(\frac{u^2}{20} = 45\)
\(u^2 = 900\)

Since the rocket is launched upwards, the initial velocity constant \(u\) must be positive, so:
\(u = \sqrt{900} = 30\) (as required).

(c) The rocket reaches the ground when \(h = 0\).
Using \(u = 30\) and \(k = 35\):
\(-5t^2 + 30t + 35 = 0\)

Divide the entire equation by \(-5\):
\(t^2 - 6t - 7 = 0\)

Factorising this quadratic:
\((t - 7)(t + 1) = 0\)

This yields \(t = 7\) or \(t = -1\).
Since time must be positive (\(t \ge 0\)), we reject \(t = -1\).
Thus, the rocket reaches the ground \(7\) seconds after it was launched.

(d) Possible limitations of the model include:
- The model ignores air resistance, which would reduce both the maximum height and the time of flight.
- It assumes the acceleration due to gravity is constant.
- The model is only valid for \(0 \le t \le 7\) and ceases to represent the physical situation once the rocket hits the ground.

(a)
B1: Correctly writes down \(k = 35\).

(b)
M1: Attempts to find the maximum height in terms of \(u\) either by completing the square or using \(t = \frac{u}{10}\) (from the vertex formula \(t = -\frac{b}{2a}\)).
M1: Substitutes \(t = \frac{u}{10}\) back into the formula for \(h\) and sets the expression equal to \(80\).
A1: Obtains the equation \(u^2 = 900\) or equivalent.
A1*: Completes the proof by showing \(u = 30\), justifying why the positive root is chosen.

(c)
M1: Sets \(h = 0\) with their values of \(u\) and \(k\).
M1: Solves the resulting 3-term quadratic equation by factorisation or formula.
A1: Correct time of \(7\) seconds (rejects the negative time \(t = -1\)).

(d)
B1: Any valid physical limitation, e.g., ignores air resistance, assumes a flat ground, or only valid for the range \(0 \le t \le 7\).

**Part (a)**

To find the equation of the tangent, we first find the gradient function by differentiating \(y\):
\[ y = 2x^{3/2} - 6x + 5 \]
\[ \frac{dy}{dx} = 2 \left(\frac{3}{2}\right)x^{1/2} - 6 = 3\sqrt{x} - 6 \]

At the point \(P(9, 5)\), substitute \(x = 9\) into the gradient function:
\[ \frac{dy}{dx} = 3\sqrt{9} - 6 = 3(3) - 6 = 3 \]

So the gradient of the tangent at \(P\) is \(3\).

The equation of the tangent is:
\[ y - 5 = 3(x - 9) \]
\[ y - 5 = 3x - 27 \]
\[ 3x - y - 22 = 0 \]

**Part (b)**

The given line is \(2y + 6x = 7\), which can be rewritten as:
\[ y = -3x + \frac{7}{2} \]

Since the tangent is parallel to this line, the gradient of the tangent must be \(-3\):
\[ 3\sqrt{x} - 6 = -3 \]
\[ 3\sqrt{x} = 3 \]
\[ \sqrt{x} = 1 \implies x = 1 \]

Substitute \(x = 1\) back into the equation of the curve \(C\) to find the \(y\)-coordinate:
\[ y = 2(1)^{3/2} - 6(1) + 5 = 2 - 6 + 5 = 1 \]

Thus, the coordinates of the point are \((1, 1)\).

**(a)**
* **M1**: Attempts to differentiate the given curve with at least one term differentiated correctly (power reduced by 1).
* **A1**: Correct derivative: \(\frac{dy}{dx} = 3x^{1/2} - 6\).
* **M1**: Substitutes \(x = 9\) into their derivative to find the gradient and attempts to form the equation of the line using \(y - y_1 = m(x - x_1)\).
* **A1**: Correct equation in the required form: \(3x - y - 22 = 0\) (or any integer multiple thereof, such as \(y - 3x + 22 = 0\)).

**(b)**
* **M1**: Equates their derivative to \(-3\) (the gradient of the line \(2y + 6x = 7\)) and attempts to solve for \(x\).
* **A1**: Obtains \(x = 1\).
* **A1**: Obtains \(y = 1\), leading to the final coordinates \((1, 1)\).

**Part (a)**

To find the coordinates of the intersection points, set the equation of the curve equal to the equation of the line:
\[ 8x - x^2 = 2x + 8 \]
\[ x^2 - 6x + 8 = 0 \]

Factorising the quadratic equation:
\[ (x - 2)(x - 4) = 0 \]

This gives \(x = 2\) and \(x = 4\).

Substitute these values into the equation of the line \(y = 2x + 8\) to find the \(y\)-coordinates:
* For \(x = 2\): \(y = 2(2) + 8 = 12\)
* For \(x = 4\): \(y = 2(4) + 8 = 16\)

So the intersection points are \((2, 12)\) and \((4, 16)\).

**Part (b)**

The area of the finite region bounded by \(C\) and \(L\) is given by the integral:
\[ \text{Area} = \int_{2}^{4} (y_{\text{curve}} - y_{\text{line}}) \, dx \]
\[ \text{Area} = \int_{2}^{4} ((8x - x^2) - (2x + 8)) \, dx \]
\[ \text{Area} = \int_{2}^{4} (6x - x^2 - 8) \, dx \]

Now we integrate each term with respect to \(x\):
\[ \int (6x - x^2 - 8) \, dx = \left[ 3x^2 - \frac{1}{3}x^3 - 8x \right]_{2}^{4} \]

Substitute the upper limit \(x = 4\):
\[ 3(4)^2 - \frac{1}{3}(4)^3 - 8(4) = 48 - \frac{64}{3} - 32 = 16 - \frac{64}{3} = -\frac{16}{3} \]

Substitute the lower limit \(x = 2\):
\[ 3(2)^2 - \frac{1}{3}(2)^3 - 8(2) = 12 - \frac{8}{3} - 16 = -4 - \frac{8}{3} = -\frac{20}{3} \]

Subtract the lower limit value from the upper limit value:
\[ \text{Area} = -\frac{16}{3} - \left(-\frac{20}{3}\right) = \frac{4}{3} \]

**(a)**
* **M1**: Equates the curve and line equations to form a quadratic equation: \(8x - x^2 = 2x + 8\).
* **A1**: Solves the quadratic to find \(x = 2\) and \(x = 4\).
* **A1**: Finds both correct \(y\)-coordinates to give the intersection points \((2, 12)\) and \((4, 16)\).

**(b)**
* **M1**: Sets up a subtraction of the line from the curve (or vice versa) and attempts to integrate. Must show raising of power by 1 on at least two terms.
* **A1**: Correct integrated expression: \(3x^2 - \frac{1}{3}x^3 - 8x\) (ignore limits and constant of integration).
* **M1**: Substitutes their limits \(4\) and \(2\) into their integrated expression and subtracts the lower limit from the upper limit.
* **A1**: Correct exact area of \(\frac{4}{3}\) (or equivalent, e.g., \(1\frac{1}{3}\)).

We are given the equation:
\[ 6\sin^2 \theta + 5\cos \theta - 5 = 0 \]

First, use the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to write the equation in terms of \(\cos \theta\) only:
\[ 6(1 - \cos^2 \theta) + 5\cos \theta - 5 = 0 \]
\[ 6 - 6\cos^2 \theta + 5\cos \theta - 5 = 0 \]
\[ -6\cos^2 \theta + 5\cos \theta + 1 = 0 \]

Multiply the entire equation by \(-1\) to make the leading coefficient positive:
\[ 6\cos^2 \theta - 5\cos \theta - 1 = 0 \]

Factorise the quadratic expression:
\[ (6\cos \theta + 1)(\cos \theta - 1) = 0 \]

This gives two possible cases:
1. \(\cos \theta = 1\)
2. \(\cos \theta = -\frac{1}{6}\)

Now solve each case in the interval \(0 \le \theta < 360^\circ\):

**Case 1: \(\cos \theta = 1\)**
Within the interval \(0 \le \theta < 360^\circ\), the only solution is:
\[ \theta = 0^\circ \]

**Case 2: \(\cos \theta = -\frac{1}{6}\)**
The principal value is:
\[ \theta = \arccos\left(-\frac{1}{6}\right) \approx 99.59^\circ \]

Since cosine is negative in the second and third quadrants, the second solution in the interval is:
\[ \theta = 360^\circ - 99.59^\circ \approx 260.41^\circ \]

Rounding to 1 decimal place, we obtain:
\[ \theta = 99.6^\circ \quad \text{and} \quad \theta = 260.4^\circ \]

Thus, the complete set of solutions in the given interval is:
\[ \theta = 0^\circ, \; 99.6^\circ, \; 260.4^\circ \]

* **M1**: Uses the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to form an equation in terms of \(\cos \theta\) only.
* **A1**: Obtains a correct simplified quadratic equation, e.g., \(6\cos^2 \theta - 5\cos \theta - 1 = 0\).
* **M1**: Attempts to solve their quadratic equation in \(\cos \theta\) by factorisation, quadratic formula, or completing the square.
* **A1**: Obtains correct roots \(\cos \theta = 1\) and \(\cos \theta = -\frac{1}{6}\).
* **B1**: Obtains the solution \(\theta = 0^\circ\) (ignore additional inclusion of \(360^\circ\) if present).
* **M1**: For finding one correct angle for \(\cos \theta = -\frac{1}{6}\) and showing a correct method to find the second angle in the range (e.g. doing \(360^\circ - \text{angle}\)).
* **A1**: Obtains \(\theta = 99.6^\circ\) and \(\theta = 260.4^\circ\). (Accept answers rounding to \(99.6\) and \(260.4\). Deduct 1 mark overall if extra solutions are present in the range.)

(a) First, find the midpoint \(M\) of \(AB\):
\[M = \left(\frac{-1 + 3}{2}, \frac{2 + r}{2}\right) = \left(1, \frac{2+r}{2}\right)\]

Next, find the gradient of \(l_1\), denoted by \(m_1\):
\[m_1 = \frac{r - 2}{3 - (-1)} = \frac{r-2}{4}\]

Since \(l_2\) is perpendicular to \(l_1\), the gradient of \(l_2\), denoted by \(m_2\), is:
\[m_2 = -\frac{1}{m_1} = -\frac{4}{r-2} = \frac{4}{2-r}\]

Since \(l_2\) passes through the midpoint \(M\left(1, \frac{2+r}{2}\right)\) and has \(y\)-intercept \((0, -1)\), we can also express its gradient using these two points:
\[m_2 = \frac{\frac{2+r}{2} - (-1)}{1 - 0} = \frac{2+r}{2} + 1 = \frac{r+4}{2}\]

Equating the two expressions for \(m_2\):
\[\frac{4}{2-r} = \frac{r+4}{2}\]
\[8 = (2-r)(r+4)\]
\[8 = 8 - 2r - r^2\]
\[r^2 + 2r = 0\]

(b) Solving the quadratic equation:
\[r(r+2) = 0 \implies r = 0 \text{ or } r = -2\]

Case 1: When \(r = 0\)
\[m_2 = \frac{0+4}{2} = 2\]
Using the \(y\)-intercept \((0, -1)\), the equation of \(l_2\) is:
\[y = 2x - 1 \implies 2x - y - 1 = 0\]

Case 2: When \(r = -2\)
\[m_2 = \frac{-2+4}{2} = 1\]
Using the \(y\)-intercept \((0, -1)\), the equation of \(l_2\) is:
\[y = x - 1 \implies x - y - 1 = 0\]

(a)
- M1: Attempts to find the midpoint of \(AB\) in terms of \(r\).
- A1: Correct midpoint \(\left(1, \frac{2+r}{2}\right)\).
- M1: Attempts to find the gradient of \(l_1\) in terms of \(r\).
- M1: Applies the perpendicular gradient rule \(m_2 = -\frac{1}{m_1}\) and sets up an equation equating it to the gradient computed using the \(y\)-intercept \((0, -1)\).
- A1*: Correctly simplifies to show \(r^2 + 2r = 0\) with no errors seen.

(b)
- B1: Identifies both possible values of \(r\) as \(0\) and \(-2\).
- M1: Substitutes one of their values of \(r\) to find the corresponding gradient and attempts to form the equation of \(l_2\).
- A1: Gives one correct equation in the form \(ax + by + c = 0\) (e.g., \(2x - y - 1 = 0\)).
- A1: Gives the second correct equation in the form \(ax + by + c = 0\) (e.g., \(x - y - 1 = 0\)).

(a) Substituting \(\theta = 15^\circ\) and \(y = 0\) into the equation of \(C\):
\[0 = 3\sin(2(15^\circ) - \alpha)\]
\[\sin(30^\circ - \alpha) = 0\]
Since \(0 < \alpha < 90^\circ\), the argument \(30^\circ - \alpha\) lies in the interval:
\[-60^\circ < 30^\circ - \alpha < 30^\circ\]
Within this interval, the only solution to \(\sin(x) = 0\) is \(x = 0\).
Therefore:
\[30^\circ - \alpha = 0 \implies \alpha = 30^\circ\]

(b) The equation is now \(y = 3\sin(2\theta - 30^\circ)\).
For the maximum point, \(\sin(2\theta - 30^\circ) = 1\):
\[2\theta - 30^\circ = 90^\circ\]
\[2\theta = 120^\circ \implies \theta = 60^\circ\]
Since \(y = 3(1) = 3\), the coordinates of the maximum point are \((60^\circ, 3)\).

For the minimum point, \(\sin(2\theta - 30^\circ) = -1\):
\[2\theta - 30^\circ = 270^\circ\]
\[2\theta = 300^\circ \implies \theta = 150^\circ\]
Since \(y = 3(-1) = -3\), the coordinates of the minimum point are \((150^\circ, -3)\).

(c) Solve \(3\sin(2\theta - 30^\circ) = -1.5\):
\[\sin(2\theta - 30^\circ) = -0.5\]
Since \(0 \le \theta \le 180^\circ\), we have \(-30^\circ \le 2\theta - 30^\circ \le 330^\circ\).
The solutions in this range for the argument are:
\[2\theta - 30^\circ = -30^\circ, 210^\circ, 330^\circ\]
Solving for \(\theta\):
\[2\theta = 0^\circ \implies \theta = 0^\circ\]
\[2\theta = 240^\circ \implies \theta = 120^\circ\]
\[2\theta = 360^\circ \implies \theta = 180^\circ\]
Thus, the solutions are \(\theta = 0^\circ, 120^\circ, 180^\circ\).

(a)
- M1: Substitutes \(\theta = 15^\circ\) and \(y = 0\) to obtain a trigonometric relation.
- M1: Evaluates the restricted domain for \(30^\circ - \alpha\) to establish the unique solution.
- A1*: Shows that \(\alpha = 30^\circ\) with correct logical steps.

(b)
- M1: Sets \(2\theta - 30^\circ = 90^\circ\) or \(270^\circ\).
- A1: Gives the correct maximum point \((60^\circ, 3)\).
- A1: Gives the correct minimum point \((150^\circ, -3)\).

(c)
- B1: Obtains \(\sin(2\theta - 30^\circ) = -0.5\).
- M1: Finds at least two valid values for \(2\theta - 30^\circ\) within the interval \([-30^\circ, 330^\circ]\).
- A1: Gives all three correct solutions: \(\theta = 0^\circ, 120^\circ, 180^\circ\).

(a) First, find the gradient of the line segment \(AB\):
\[m_{AB} = \frac{4 - 2}{5 - 1} = \frac{2}{4} = \frac{1}{2}\]

Since angle \(ABC = 90^\circ\), the line segments \(AB\) and \(BC\) are perpendicular.
Therefore, the gradient of \(BC\), denoted by \(m_{BC}\), is:
\[m_{BC} = -\frac{1}{m_{AB}} = -2\]

Using the coordinates of \(B(5, 4)\) and \(C(k, 5-k)\), we express the gradient \(m_{BC}\):
\[m_{BC} = \frac{(5-k) - 4}{k - 5} = \frac{1-k}{k-5}\]

Set this equal to \(-2\) and solve for \(k\):
\[\frac{1-k}{k-5} = -2\]
\[1-k = -2(k-5)\]
\[1-k = -2k + 10\]
\[k = 9\]

(b) With \(k = 9\), the coordinates of \(C\) are \((9, 5-9) = (9, -4)\).

Calculate the length of the base \(AB\):
\[AB = \sqrt{(5-1)^2 + (4-2)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\]

Calculate the length of the height \(BC\):
\[BC = \sqrt{(9-5)^2 + (-4-4)^2} = \sqrt{4^2 + (-8)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}\]

Since angle \(ABC = 90^\circ\), the area of triangle \(ABC\) is:
\[\text{Area} = \frac{1}{2} \times AB \times BC\]
\[\text{Area} = \frac{1}{2} \times \sqrt{20} \times \sqrt{80} = \frac{1}{2} \times \sqrt{1600} = \frac{1}{2} \times 40 = 20\]

(a)
- M1: Attempts to find the gradient of \(AB\).
- A1: Correct gradient of \(AB\) is \(\frac{1}{2}\).
- M1: Uses the perpendicular gradient rule to state that gradient of \(BC\) is \(-2\) and sets up an equation in terms of \(k\).
- A1: Correctly solves to find \(k = 9\).

(b)
- B1: Identifies the coordinates of \(C\) as \((9, -4)\).
- M1: Applies the distance formula to find the length of \(AB\) or \(BC\) (or their squares).
- M1: Applies the area formula for a right-angled triangle using their values of \(AB\) and \(BC\).
- A1*: Correctly completes the proof to show that the area is 20 with no errors seen.

**(a)**
Using the formula for the sum to infinity of a geometric series:
\[ S_{\infty} = \frac{a}{1-r} = 25 \implies a = 25(1-r) \]
Using the formula for the \(n\)-th term of a geometric series:
\[ u_3 = a r^2 = 2.4 \]
Substitute \(a = 25(1-r)\) into the second equation:
\[ 25(1-r)r^2 = 2.4 \]
\[ 25r^2 - 25r^3 = 2.4 \]
Multiply the entire equation by \(5\) to clear the decimal:
\[ 125r^2 - 125r^3 = 12 \]
Rearrange to form the required cubic equation:
\[ 125r^3 - 125r^2 + 12 = 0 \]

**(b)**
Since \(r\) is a rational number, we can test possible rational roots of the equation \(125r^3 - 125r^2 + 12 = 0\).
Using the rational root theorem, candidates for \(r\) include values such as \(\frac{2}{5} = 0.4\).
Substitute \(r = 0.4\) into the cubic equation:
\[ 125(0.4)^3 - 125(0.4)^2 + 12 = 125(0.064) - 125(0.16) + 12 = 8 - 20 + 12 = 0 \]
So \(r = 0.4\) is a root, which means \((5r-2)\) is a factor.
Factorising the cubic:
\[ 125r^3 - 125r^2 + 12 = (5r - 2)(25r^2 - 15r - 6) = 0 \]
For \(25r^2 - 15r - 6 = 0\):
\[ r = \frac{15 \pm \sqrt{(-15)^2 - 4(25)(-6)}}{50} = \frac{15 \pm \sqrt{825}}{50} = \frac{15 \pm 5\sqrt{33}}{50} \]
These roots are irrational, so the only rational root is \(r = \frac{2}{5} = 0.4\).
Thus, \(r = 0.4\).

Substitute \(r = 0.4\) back to find \(a\):
\[ a = 25(1 - 0.4) = 25(0.6) = 15 \]

**(a)**
* **M1**: Attempts to use \(S_{\infty} = \frac{a}{1-r} = 25\) to express \(a\) in terms of \(r\), or vice versa, and substitutes into \(ar^2 = 2.4\).
* **A1***: Correctly simplifies to the given cubic equation \(125r^3 - 125r^2 + 12 = 0\) with no errors in algebra. Must show intermediate steps (e.g., multiplying by 5 or showing \(25r^2 - 25r^3 = 2.4\)).

**(b)**
* **M1**: Attempts to find a rational root of the cubic equation. This could be by testing values (e.g., \(r=2/5\)) or factorising the cubic into a linear and a quadratic factor.
* **A1**: Identifies \(r = 0.4\) (or \(\frac{2}{5}\)) as the correct rational root.
* **A1**: Finds \(a = 15\) corresponding to \(r = 0.4\).

**(a)**
Using the formula for the sum of the first \(n\) terms of an arithmetic series:
\[ S_n = \frac{n}{2}[2a + (n-1)d] \]
For \(n = 8\) and \(S_8 = 80\):
\[ \frac{8}{2}(2a + 7d) = 80 \implies 4(2a + 7d) = 80 \implies 2a + 7d = 20 \quad \text{--- (Equation 1)} \]
Using the formula for the \(n\)-th term of an arithmetic series:
\[ u_n = a + (n-1)d \]
For the \(12\)th term:
\[ u_{12} = a + 11d = 25 \quad \text{--- (Equation 2)} \]

From Equation 2, we have:
\[ a = 25 - 11d \]
Substitute this into Equation 1:
\[ 2(25 - 11d) + 7d = 20 \]
\[ 50 - 22d + 7d = 20 \]
\[ 50 - 15d = 20 \]
\[ 15d = 30 \implies d = 2 \]

Substitute \(d = 2\) back into Equation 2:
\[ a = 25 - 11(2) = 3 \]
So \(d = 2\) and \(a = 3\).

**(b)**
The \(30\)th term of the series is:
\[ u_{30} = a + 29d \]
\[ u_{30} = 3 + 29(2) = 3 + 58 = 61 \]

**(a)**
* **M1**: Uses the correct sum formula for \(S_8\) to write an equation in \(a\) and \(d\), e.g., \(4(2a+7d)=80\).
* **M1**: Uses the correct term formula for \(u_{12}\) to write an equation in \(a\) and \(d\), e.g., \(a+11d=25\).
* **M1**: Solves the two equations simultaneously to find a value for \(d\) or \(a\).
* **A1**: Both \(d = 2\) and \(a = 3\) correct.

**(b)**
* **B1**: Correctly identifies the \(30\)th term as \(61\).

We use the binomial expansion formula:
\[ (a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \binom{n}{3}a^{n-3}b^3 + \dots \]
Here, \(a = 2\), \(b = -\frac{x}{4}\), and \(n = 6\).

Let's calculate each of the first four terms:
1st term:
\[ 2^6 = 64 \]

2nd term:
\[ \binom{6}{1} (2)^5 \left(-\frac{x}{4}\right) = 6 \times 32 \times \left(-\frac{x}{4}\right) = -48x \]

3rd term:
\[ \binom{6}{2} (2)^4 \left(-\frac{x}{4}\right)^2 = 15 \times 16 \times \left(\frac{x^2}{16}\right) = 15x^2 \]

4th term:
\[ \binom{6}{3} (2)^3 \left(-\frac{x}{4}\right)^3 = 20 \times 8 \times \left(-\frac{x^3}{64}\right) = -\frac{160}{64}x^3 = -\frac{5}{2}x^3 \]

Combining these terms gives the expansion:
\[ 64 - 48x + 15x^2 - \frac{5}{2}x^3 \]

* **M1**: Identifies the first term as \(2^6\) or \(64\).
* **M1**: Attempts to use the binomial expansion with correct binomial coefficients \(\binom{6}{1}\), \(\binom{6}{2}\), and \(\binom{6}{3}\) (which are \(6, 15, 20\)) and correct powers of \(2\) and \(\left(-\frac{x}{4}\right)\). At least two variable terms must be structured correctly (including the negative sign inside the bracket).
* **A1**: Any two of the four terms in the simplified form correct (can include \(64\), \(-48x\), \(15x^2\), or \(-\frac{5}{2}x^3\)).
* **A1**: Any three of the four terms in the simplified form correct.
* **A1**: Fully correct simplified expression: \(64 - 48x + 15x^2 - \frac{5}{2}x^3\) (accept \(-2.5x^3\)).

(a) Let the height of the tank be \(h\) metres. The volume of the tank is given by \(V = 3x \times x \times h = 3x^2 h\). Since the volume is \(16\text{ m}^3\), we have: \(3x^2 h = 16 \implies h = \frac{16}{3x^2}\). The total surface area \(A\) is the area of the base plus the area of five vertical walls (two of size \(3x \times h\) and three of size \(x \times h\)): \(A = 3x^2 + 2(3xh) + 3(xh) = 3x^2 + 9xh\). Substituting \(h\) into this expression: \(A = 3x^2 + 9x\left(\frac{16}{3x^2}\right) = 3x^2 + \frac{144x}{3x^2} = 3x^2 + \frac{48}{x}\). (b) Differentiating \(A\) with respect to \(x\): \(\frac{dA}{dx} = 6x - \frac{48}{x^2}\). For a stationary value, set \(\frac{dA}{dx} = 0\): \(6x - \frac{48}{x^2} = 0 \implies 6x^3 = 48 \implies x^3 = 8 \implies x = 2\). Substituting \(x = 2\) back into the formula for \(A\) to find the minimum value: \(A = 3(2)^2 + \frac{48}{2} = 12 + 24 = 36\). (c) Differentiating again to find the second derivative: \(\frac{d^2A}{dx^2} = 6 + \frac{96}{x^3}\). When \(x = 2\): \(\frac{d^2A}{dx^2} = 6 + \frac{96}{8} = 6 + 12 = 18\). Since \(\frac{d^2A}{dx^2} = 18 > 0\) at \(x = 2\), the value is a minimum.

(a) M1: Attempts to write an expression for the volume, \(V = 3x^2 h = 16\). A1: Correctly expresses \(h\) in terms of \(x\), e.g., \(h = \frac{16}{3x^2}\). M1: Writes a correct total area expression in terms of \(x\) and \(h\), i.e., \(A = 3x^2 + 9xh\). A1*: Substitutes \(h\) and simplifies to show the given result \(A = 3x^2 + \frac{48}{x}\) with no errors seen. (b) M1: Differentiates \(A\) to obtain at least one term correct. A1: Correct derivative: \(\frac{dA}{dx} = 6x - \frac{48}{x^2}\). M1: Sets their derivative equal to 0 and attempts to solve for \(x\). A1: Obtains \(x = 2\). A1: Obtains the minimum surface area of \(36\). (c) M1: Attempts to find the second derivative \(\frac{d^2A}{dx^2}\). A1: Finds \(\frac{d^2A}{dx^2} = 6 + \frac{96}{x^3}\), evaluates at \(x = 2\) to get \(18\), and concludes it is a minimum because \(18 > 0\).

(a) First, find \(\frac{dy}{dx}\): \(\frac{dy}{dx} = 16 \left(\frac{1}{2}x^{-1/2}\right) - \frac{1}{2}(2x) = 8x^{-1/2} - x\). At the stationary point, \(\frac{dy}{dx} = 0\): \(8x^{-1/2} - x = 0 \implies \frac{8}{\sqrt{x}} = x \implies x^{3/2} = 8 \implies x = 8^{2/3} = 4\). Substitute \(x = 4\) back into the equation of \(C\) to find the y-coordinate: \(y = 16(4)^{1/2} - \frac{1}{2}(4)^2 = 16(2) - \frac{1}{2}(16) = 32 - 8 = 24\). Thus, the coordinates of the stationary point are \((4, 24)\). (b) The area of \(R\) is given by: \(\int_{1}^{4} \left(16x^{1/2} - \frac{1}{2}x^2\right) dx = \left[ \frac{16x^{3/2}}{3/2} - \frac{x^3}{6} \right]_{1}^{4} = \left[ \frac{32}{3}x^{3/2} - \frac{1}{6}x^3 \right]_{1}^{4}\). Evaluating at the upper limit \(x = 4\): \(\frac{32}{3}(4)^{3/2} - \frac{1}{6}(4)^3 = \frac{32}{3}(8) - \frac{64}{6} = \frac{256}{3} - \frac{32}{3} = \frac{224}{3}\). Evaluating at the lower limit \(x = 1\): \(\frac{32}{3}(1)^{3/2} - \frac{1}{6}(1)^3 = \frac{32}{3} - \frac{1}{6} = \frac{64 - 1}{6} = \frac{63}{6} = \frac{21}{2}\). The exact area of \(R\) is: \(\frac{224}{3} - \frac{21}{2} = \frac{448 - 63}{6} = \frac{385}{6}\). (c) At \(x = 1\), the y-coordinate is: \(y = 16(1)^{1/2} - \frac{1}{2}(1)^2 = 16 - 0.5 = 15.5 = \frac{31}{2}\). The gradient of the tangent at \(x = 1\) is: \(\frac{dy}{dx} = 8(1)^{-1/2} - 1 = 7\). The equation of the tangent line is: \(y - \frac{31}{2} = 7(x - 1) \implies 2y - 31 = 14(x - 1) \implies 2y - 31 = 14x - 14 \implies 14x - 2y + 17 = 0\).

(a) M1: Differentiates \(y\) with respect to \(x\) to obtain at least one term correct. A1: Correct derivative: \(\frac{dy}{dx} = 8x^{-1/2} - x\). M1: Sets their derivative equal to 0 and solves for \(x\). A1: Correctly identifies \(x = 4\) and finds the corresponding \(y = 24\) to give coordinates \((4, 24)\). (b) M1: Integrates the expression with at least one term correct. A1: Correct integration: \(\frac{32}{3}x^{3/2} - \frac{1}{6}x^3\). M1: Substitutes the limits 4 and 1 into their integrated expression. A1: Correct evaluation of both limits (e.g., \(\frac{224}{3}\) and \(\frac{21}{2}\)). A1: Yields the exact value of \(\frac{385}{6}\). (c) M1: Evaluates both the gradient and \(y\)-value at \(x = 1\) and attempts to write down a tangent equation. A1: Simplifies to the form \(14x - 2y + 17 = 0\) (or any integer multiple thereof).

(a) Since \( (x-2) \) is a factor of \( f(x) \), by the Factor Theorem we have:
\( f(2) = 0 \)
\( 3(2)^3 + a(2)^2 + b(2) - 6 = 0 \)
\( 24 + 4a + 2b - 6 = 0 \)
\( 4a + 2b = -18 \implies 2a + b = -9 \) [Equation 1]

Since dividing \( f(x) \) by \( (x+1) \) gives a remainder of \( -12 \), by the Remainder Theorem we have:
\( f(-1) = -12 \)
\( 3(-1)^3 + a(-1)^2 + b(-1) - 6 = -12 \)
\( -3 + a - b - 6 = -12 \)
\( a - b = -3 \) [Equation 2]

From Equation 2, \( a = b - 3 \). Substituting this into Equation 1:
\( 2(b - 3) + b = -9 \)
\( 2b - 6 + b = -9 \)
\( 3b = -3 \implies b = -1 \)

Substitute \( b = -1 \) back into Equation 2:
\( a = -1 - 3 = -4 \)

So \( a = -4 \) and \( b = -1 \).

(b) Substituting \( a = -4 \) and \( b = -1 \) into \( f(x) \):
\( f(x) = 3x^3 - 4x^2 - x - 6 \)

Since \( (x-2) \) is a factor, we can express \( f(x) \) as:
\( f(x) = (x-2)(3x^2 + kx + 3) \) for some constant \( k \).

Expanding this:
\( (x-2)(3x^2 + kx + 3) = 3x^3 + (k-6)x^2 + (3-2k)x - 6 \)

Comparing coefficients of \( x^2 \):
\( k - 6 = -4 \implies k = 2 \)

So \( f(x) = (x-2)(3x^2 + 2x + 3) \).

For the quadratic factor \( 3x^2 + 2x + 3 = 0 \), we calculate the discriminant:
\( \Delta = b^2 - 4ac = 2^2 - 4(3)(3) = 4 - 36 = -32 \)

Since \( \Delta < 0 \), the quadratic equation has no real roots.
Therefore, the only real root of \( f(x) = 0 \) is \( x = 2 \).

Part (a):
- M1: Attempts to use the factor theorem with \( f(2) = 0 \) to obtain a linear equation in \( a \) and \( b \).
- A1: Obtains the correct equation \( 2a + b = -9 \) (or equivalent).
- M1: Attempts to use the remainder theorem with \( f(-1) = -12 \) to obtain a second linear equation in \( a \) and \( b \).
- A1: Obtains the correct equation \( a - b = -3 \) (or equivalent).
- A1: Solves the simultaneous equations to find \( a = -4 \) and \( b = -1 \).

Part (b):
- M1: Attempts algebraic division or coefficient matching to find the quadratic factor of \( f(x) \).
- A1: Correctly identifies the quadratic factor as \( 3x^2 + 2x + 3 \).
- A1: Calculates the discriminant of the quadratic factor to be \( -32 \) (or uses completion of square) and correctly concludes that since \( -32 < 0 \), there are no real roots from this factor, hence \( f(x) = 0 \) has only one real root.

(a) Using the laws of logarithms:
\( 2\log_3(y - 2) = \log_3(y - 2)^2 \)

So the equation becomes:
\( \log_3(y - 2)^2 - \log_3(y + 4) = 1 \)

Using the subtraction law:
\( \log_3\left( \frac{(y-2)^2}{y+4} \right) = 1 \)

Converting from logarithmic form to exponential form:
\( \frac{(y-2)^2}{y+4} = 3^1 \)
\( (y-2)^2 = 3(y+4) \)
\( y^2 - 4y + 4 = 3y + 12 \)
\( y^2 - 7y - 8 = 0 \)

Factorising the quadratic:
\( (y-8)(y+1) = 0 \)

This gives \( y = 8 \) or \( y = -1 \).
Since the original equation requires \( y > 2 \) for the logarithm to be defined, we reject \( y = -1 \).
Therefore, the only solution is \( y = 8 \).

(b) Rewrite the term \( 3^{2w+1} \) as \( 3 \cdot (3^w)^2 \).
Let \( u = 3^w \). The equation becomes:
\( 3u^2 - 10u + 3 = 0 \)

Factorising this quadratic:
\( (3u-1)(u-3) = 0 \)

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

Substitute back \( u = 3^w \):
- For \( u = \frac{1}{3} \):
\( 3^w = 3^{-1} \implies w = -1 \)

- For \( u = 3 \):
\( 3^w = 3^1 \implies w = 1 \)

So the solutions are \( w = -1 \) and \( w = 1 \).

Part (a):
- M1: Uses the power law of logarithms to write \( 2\log_3(y-2) \) as \( \log_3(y-2)^2 \).
- M1: Uses the subtraction law of logarithms to combine the terms into a single logarithm.
- M1: Removes logarithms correctly to obtain a quadratic equation, e.g., \( (y-2)^2 = 3(y+4) \).
- A1: Obtains the correct simplified quadratic equation \( y^2 - 7y - 8 = 0 \).
- A1: Solves the quadratic and correctly rejects \( y = -1 \) to give \( y = 8 \) as the only solution.

Part (b):
- M1: Recognises the quadratic structure in \( 3^w \) and substitutes \( u = 3^w \) to obtain \( 3u^2 - 10u + 3 = 0 \) (or equivalent).
- A1: Solves the quadratic to find \( u = \frac{1}{3} \) and \( u = 3 \) (or equivalent in terms of \( 3^w \)).
- A1: Finds both correct integer values \( w = -1 \) and \( w = 1 \).

(a) Factorise each of the quadratic and cubic expressions:
- \( x^3 - 3x^2 - 10x = x(x^2 - 3x - 10) = x(x-5)(x+2) \)
- \( 2x^2 - 7x - 15 = (2x+3)(x-5) \)
- \( x^2 - 4 = (x-2)(x+2) \)
- \( 2x^2 + x - 3 = (2x+3)(x-1) \)

Now, substitute these factorised forms into the expression and change the division to multiplication by the reciprocal:
\( \frac{x(x-5)(x+2)}{(2x+3)(x-5)} \div \frac{(x-2)(x+2)}{(2x+3)(x-1)} = \frac{x(x-5)(x+2)}{(2x+3)(x-5)} \times \frac{(2x+3)(x-1)}{(x-2)(x+2)} \)

Cancel out common terms (for \( x \ne 5, -1.5, -2 \)):
- Cancel \( (x-5) \)
- Cancel \( (2x+3) \)
- Cancel \( (x+2) \)

This leaves:
\( \frac{x(x-1)}{x-2} \)

Comparing this to \( \frac{x(x+a)}{x+b} \), we get:
\( a = -1 \) and \( b = -2 \).

(b) Using the result from part (a), the equation is:
\( \frac{x(x-1)}{x-2} = 2x - 3 \)

Multiply both sides by \( (x-2) \):
\( x(x-1) = (2x-3)(x-2) \)
\( x^2 - x = 2x^2 - 7x + 6 \)

Rearrange into a standard quadratic equation:
\( x^2 - 6x + 6 = 0 \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(6)}}{2} \)
\( x = \frac{6 \pm \sqrt{36 - 24}}{2} \)
\( x = \frac{6 \pm \sqrt{12}}{2} \)
\( x = \frac{6 \pm 2\sqrt{3}}{2} = 3 \pm \sqrt{3} \)

We are given that \( x > 2 \).
- \( 3 + \sqrt{3} \approx 4.732 \) (which is \( > 2 \))
- \( 3 - \sqrt{3} \approx 1.268 \) (which is \( < 2 \))

Therefore, we reject \( 3 - \sqrt{3} \).
The final solution is \( x = 3 + \sqrt{3} \).

Part (a):
- M1: Attempts to factorise \( x^3 - 3x^2 - 10x \) to \( x(x-5)(x+2) \) or \( 2x^2 - 7x - 15 \) to \( (2x+3)(x-5) \).
- A1: Correctly factorises both terms of the first fraction.
- M1: Attempts to factorise \( x^2 - 4 \) to \( (x-2)(x+2) \) and \( 2x^2 + x - 3 \) to \( (2x+3)(x-1) \).
- M1: Inverts the second fraction and shows the intention to multiply, cancelling at least one common term.
- A1: Obtains the fully simplified fraction \( \frac{x(x-1)}{x-2} \) (with \( a = -1 \) and \( b = -2 \)).

Part (b):
- M1: Sets the simplified fraction equal to \( 2x-3 \) and expands to obtain a 3-term quadratic equation.
- A1: Obtains the correct quadratic equation \( x^2 - 6x + 6 = 0 \).
- A1: Uses the quadratic formula or completing the square to find the roots, and correctly rejects the root less than 2, leaving the final answer as \( x = 3 + \sqrt{3} \).

We use the trigonometric identity \(\cos^2(2\theta) = 1 - \sin^2(2\theta)\) to write the entire equation in terms of sine: \( 6(1 - \sin^2(2\theta)) - \sin(2\theta) - 5 = 0 \). Expanding this gives \( 6 - 6\sin^2(2\theta) - \sin(2\theta) - 5 = 0 \), which simplifies to the quadratic equation: \( 6\sin^2(2\theta) + \sin(2\theta) - 1 = 0 \). Factoring the quadratic, we obtain: \( (3\sin(2\theta) - 1)(2\sin(2\theta) + 1) = 0 \). This yields two possible cases: \( \sin(2\theta) = \frac{1}{3} \) or \( \sin(2\theta) = -\frac{1}{2} \). Since \(0 \le \theta < \pi\), the range of values for \(2\theta\) is \(0 \le 2\theta < 2\pi\). **Case 1:** \( \sin(2\theta) = -\frac{1}{2} \). In the interval \(0 \le 2\theta < 2\pi\), the solutions are: \( 2\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \implies \theta = \frac{7\pi}{12} \), and \( 2\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \implies \theta = \frac{11\pi}{12} \). **Case 2:** \( \sin(2\theta) = \frac{1}{3} \). The reference angle is \( \arcsin\left(\frac{1}{3}\right) \approx 0.339837 \) radians. The solutions for \(2\theta\) are: \( 2\theta \approx 0.339837 \implies \theta \approx 0.17 \) (to 2 decimal places), and \( 2\theta \approx \pi - 0.339837 \approx 2.801756 \implies \theta \approx 1.40 \) (to 2 decimal places). Thus, the complete set of solutions in the interval is \( \theta = 0.17, \; 1.40, \; \frac{7\pi}{12}, \; \frac{11\pi}{12} \).

- **M1**: Uses the identity \(\cos^2(2\theta) = 1 - \sin^2(2\theta)\) to express the equation in terms of \(\sin(2\theta)\) only. - **A1**: Obtains a correct three-term quadratic equation, e.g., \(6\sin^2(2\theta) + \sin(2\theta) - 1 = 0\). - **M1**: Solves their quadratic equation to find two values for \(\sin(2\theta)\). - **A1**: Correctly identifies \(\sin(2\theta) = \frac{1}{3}\) and \(\sin(2\theta) = -\frac{1}{2}\). - **A1**: Obtains the exact solutions \(\theta = \frac{7\pi}{12}\) and \(\theta = \frac{11\pi}{12}\). - **A1**: Obtains the decimal solutions \(\theta \approx 0.17\) and \(\theta \approx 1.40\) rounded to 2 decimal places.

(a) To show that the statement is false, we need to find a single positive integer \(n\) for which \(2^n + 3^n\) is composite. Let \(n = 3\). Then: \( 2^3 + 3^3 = 8 + 27 = 35 \). Since 35 is divisible by 5 and 7, it is not a prime number. Therefore, the statement is false. (b) We begin by rearranging the terms of the inequality we wish to prove: \( x^2 + y^2 \ge 2(x + y - 1) \) becomes \( x^2 + y^2 \ge 2x + 2y - 2 \). Grouping all terms on one side of the inequality yields: \( x^2 - 2x + y^2 - 2y + 2 \ge 0 \). We can rewrite the constant 2 as \(1 + 1\) and complete the square for \(x\) and \(y\) separately: \( (x^2 - 2x + 1) + (y^2 - 2y + 1) \ge 0 \), which simplifies to: \( (x - 1)^2 + (y - 1)^2 \ge 0 \). Since \(x\) and \(y\) are real numbers, \( (x - 1)^2 \ge 0 \) and \( (y - 1)^2 \ge 0 \) because the square of any real number is always non-negative. Consequently, their sum must also be non-negative: \( (x - 1)^2 + (y - 1)^2 \ge 0 \). Since this final inequality is always true, and each step is reversible, the original inequality \( x^2 + y^2 \ge 2(x + y - 1) \) must hold for all real numbers \(x\) and \(y\).

Part (a):
- **M1**: Chooses a valid positive integer candidate (e.g., \(n = 3\) or \(n = 5\)) and attempts to evaluate \(2^n + 3^n\).
- **A1**: Correctly evaluates \(2^3 + 3^3 = 35\) (or \(2^5 + 3^5 = 275\)) and concludes with a valid reason that 35 is not prime, disproving the statement.

Part (b):
- **M1**: Rearranges the inequality to group terms on one side: \(x^2 - 2x + y^2 - 2y + 2 \ge 0\).
- **M1**: Attempts to complete the square for both the \(x\) and \(y\) terms.
- **A1**: Obtains the correct completed square expression: \( (x - 1)^2 + (y - 1)^2 \ge 0 \).
- **M1**: Explains that \( (x - 1)^2 \ge 0 \) and \( (y - 1)^2 \ge 0 \) for all real numbers \(x\) and \(y\).
- **A1**: Concludes the proof by linking the non-negativity of the sum of squares back to the truth of the original inequality.