2025 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

The general term in the expansion of \(\left(kx + \frac{3}{x^2}\right)^6\) is given by \(\binom{6}{r} (kx)^{6-r} \left(\frac{3}{x^2}\right)^r = \binom{6}{r} k^{6-r} 3^r x^{6-3r}\). To find the term independent of \(x\), we set the power of \(x\) to 0: \(6 - 3r = 0 \implies r = 2\). Substituting \(r = 2\) into the term gives \(\binom{6}{2} k^4 3^2 = 15 \times 9 \times k^4 = 135k^4\). We are given that this term is \(2160\), so \(135k^4 = 2160 \implies k^4 = 16\). Since \(k\) is a positive constant, we have \(k = 2\). To find the coefficient of \(x^3\), we set the power of \(x\) to 3: \(6 - 3r = 3 \implies r = 1\). Substituting \(r = 1\) and \(k = 2\) into the term gives \(\binom{6}{1} (2)^5 3^1 = 6 \times 32 \times 3 = 576\). Thus, the coefficient of \(x^3\) is \(576\).

* **M1**: Attempts to find the general term and sets the power of \(x\) to 0 to obtain \(r = 2\). * **A1**: Correctly forms the equation \(135k^4 = 2160\) (or equivalent). * **A1**: Solves to find \(k = 2\) (must reject \(k = -2\) or specify \(k > 0\)). * **M1**: Identifies \(r = 1\) for the term in \(x^3\) and substitutes their positive value of \(k\). * **A1**: Obtains \(576\).

First, rewrite the equation by substituting \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\(2 \sin \theta \left(\frac{\sin \theta}{\cos \theta}\right) - 1 = 4 \cos \theta\)

Multiply the entire equation by \(\cos \theta\) (noting that \(\cos \theta \neq 0\) since \(\tan \theta\) is defined):
\(2 \sin^2 \theta - \cos \theta = 4 \cos^2 \theta\)

Now, use the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to express the equation solely in terms of \(\cos \theta\):
\(2(1 - \cos^2 \theta) - \cos \theta = 4 \cos^2 \theta\)
\(2 - 2 \cos^2 \theta - \cos \theta = 4 \cos^2 \theta\)

Rearrange the terms to form a quadratic equation in terms of \(\cos \theta\):
\(6 \cos^2 \theta + \cos \theta - 2 = 0\)

Factorise the quadratic equation:
\((3 \cos \theta + 2)(2 \cos \theta - 1) = 0\)

This yields two possible values for \(\cos \theta\):
\(\cos \theta = \frac{1}{2}\) or \(\cos \theta = -\frac{2}{3}\)

Now, find the values of \(\theta\) in the interval \(0 \leq \theta \leq 2\pi\):

1) For \(\cos \theta = \frac{1}{2}\):
\(\theta = \frac{\pi}{3}\) and \(\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\)

2) For \(\cos \theta = -\frac{2}{3}\):
The basic reference angle is \(\alpha = \arccos\left(\frac{2}{3}\right) \approx 0.8411\) radians.
Since cosine is negative in the second and third quadrants:
\(\theta =
\pi - 0.8411 \approx 2.30\) radians (to 3 s.f.)
\(\theta = \pi + 0.8411 \approx 3.98\) radians (to 3 s.f.)

Thus, the solutions are \(\theta = \frac{\pi}{3}, \frac{5\pi}{3}, 2.30, 3.98\).

M1: For substituting \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and multiplying by \(\cos \theta\) to get an equation in \(\sin^2 \theta\) and \(\cos \theta\).
M1: For substituting \(\sin^2 \theta = 1 - \cos^2 \theta\) to form a 3-term quadratic in \(\cos \theta\).
A1: For obtaining the correct quadratic equation \(6 \cos^2 \theta + \cos \theta - 2 = 0\) (or equivalent).
A1: For finding the correct roots \(\cos \theta = \frac{1}{2}\) and \(\cos \theta = -\frac{2}{3}\).
A1: For obtaining both exact solutions \(\theta = \frac{\pi}{3}\) and \(\theta = \frac{5\pi}{3}\) (or 1.05 and 5.24).
A1: For obtaining both decimal solutions \(\theta \approx 2.30\) and \(\theta \approx 3.98\) (must be rounded to 3 s.f.).

(a) Let the terms of the arithmetic progression be \(u_n = a + (n-1)d\).
The 1st, 3rd and 11th terms of the AP are:
\(u_1 = a\)
\(u_3 = a + 2d\)
\(u_{11} = a + 10d\)

Since these are the first three terms of a geometric progression, we have:
\(v_1 = a\)
\(v_2 = a + 2d\)
\(v_3 = a + 10d\)

The common ratio \(r\) is constant, so:
\(\frac{v_2}{v_1} = \frac{v_3}{v_2} \implies \frac{a+2d}{a} = \frac{a+10d}{a+2d}

Multiplying both sides by \)a(a+2d)\) gives:
\((a+2d)^2 = a(a+10d)\)
\(a^2 + 4ad + 4d^2 = a^2 + 10ad\)

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

Since we are given that \(d \neq 0\), we can divide both sides by \(2d\):
\(2d = 3a \implies d = 1.5a\) (as required).

(b) The sum of the first 20 terms of the arithmetic progression is given by:
\(S_{20} = \frac{20}{2}[2a + (20-1)d] = 610\)
\(10(2a + 19d) = 610\)
\(2a + 19d = 61\)

Substitute \(d = 1.5a\) into this equation:
\(2a + 19(1.5a) = 61\)
\(2a + 28.5a = 61\)
\(30.5a = 61 \implies a = 2\)

Next, find the common difference \(d\):
\(d = 1.5(2) = 3\)

Now, find the common ratio \(r\) of the geometric progression:
\(r = \frac{a+2d}{a} = \frac{2+6}{2} = 4\)

The 5th term of the geometric progression is:
\(v_5 = a r^4 = 2 \times 4^4 = 2 \times 256 = 512\)

(a)
M1: For writing down the terms of the AP in terms of \(a\) and \(d\) and setting up a correct ratio equation for the GP, e.g., \(\frac{a+2d}{a} = \frac{a+10d}{a+2d}\) or \((a+2d)^2 = a(a+10d)\).
A1: For obtaining a correct simplified relation, such as \(4d^2 - 6ad = 0\) (or equivalent).
A1: For concluding \(d = 1.5a\) clearly, indicating the step of division by \(d\) is valid because \(d \neq 0\).

(b)
M1: For using the sum of an AP formula to set up a linear equation in \(a\) and \(d\) and substituting \(d = 1.5a\).
A1: For finding \(a = 2\).
A1: For finding the 5th term of the geometric progression as \(512\) (must see clear working for the common ratio \(r = 4\)).

Let \(\theta\) be the angle \(AOB\) in radians. Splitting the isosceles triangle \(OAB\) into two right-angled triangles gives: \(\sin\left(\frac{\theta}{2}\right) = \frac{6}{8} = 0.75\). Solving this, we find: \(\frac{\theta}{2} \approx 0.84806\text{ radians}\), so \(\theta \approx 1.6961\text{ radians}\). The area of the major segment can be found by adding the area of the major sector to the area of the triangle \(OAB\). First, the area of the major sector is \(\frac{1}{2} r^2 (2\pi - \theta) = \frac{1}{2} \times 8^2 \times (2\pi - 1.6961) \approx 32 \times 4.5871 \approx 146.79\text{ cm}^2\). Second, the area of the triangle \(OAB\) is \(\frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 8^2 \times \sin(1.6961) \approx 32 \times 0.9922 \approx 31.75\text{ cm}^2\). Adding these together, the total area of the major segment is \(146.79 + 31.75 = 178.54\text{ cm}^2\). Rounding to 3 significant figures gives \(179\text{ cm}^2\).

M1: For a correct trigonometric method to find the angle \(\theta\), e.g., \(\sin\left(\frac{\theta}{2}\right) = \frac{6}{8}\) or \(12^2 = 8^2 + 8^2 - 2(8)(8)\cos\theta\). A1: For obtaining \(\theta \approx 1.696\) radians (or \(97.2^\circ\)). M1: For a complete and valid method to find the area of the major segment, e.g., subtracting the minor segment from the total circle area, or adding the major sector and the triangle. M1: For correct substitution of their values into both the sector area formula and the triangle area formula. A1: For obtaining \(179\) (or any value rounding to \(179\), such as \(178.5\) to \(179.0\)).

**(a)**

**Step 1:** Reflection in the \( y \)-axis is achieved by replacing \( x \) with \( -x \):
\( y = 2(-x)^2 - 8(-x) + 5 = 2x^2 + 8x + 5 \)

**Step 2:** Translation by the vector \( \begin{pmatrix} -3 \\ 4 \end{pmatrix} \) is achieved by replacing \( x \) with \( x + 3 \) and adding 4 to the function:
\( y = 2(x+3)^2 + 8(x+3) + 5 + 4 \)
\( y = 2(x^2 + 6x + 9) + 8x + 24 + 9 \)
\( y = 2x^2 + 12x + 18 + 8x + 33 \)
\( y = 2x^2 + 20x + 51 \)

**Step 3:** Stretch parallel to the \( y \)-axis with scale factor 3 is achieved by multiplying the entire function by 3:
\( y = 3(2x^2 + 20x + 51) \)
\( y = 6x^2 + 60x + 153 \)

Thus, the equation of the curve \( C' \) is \( y = 6x^2 + 60x + 153 \).

**(b)**

We track the coordinates of the point \( P(1, -1) \) through the sequence of transformations:
1. Reflection in the \( y \)-axis maps \( (x, y) \to (-x, y) \):
\( P(1, -1) \to P_1(-1, -1) \)

2. Translation by \( \begin{pmatrix} -3 \\ 4 \end{pmatrix} \) maps \( (x, y) \to (x - 3, y + 4) \):
\( P_1(-1, -1) \to P_2(-1 - 3, -1 + 4) = P_2(-4, 3) \)

3. Stretch parallel to the \( y \)-axis with scale factor 3 maps \( (x, y) \to (x, 3y) \):
\( P_2(-4, 3) \to P'(-4, 3 \times 3) = P'(-4, 9) \)

Thus, the coordinates of \( P' \) are \( (-4, 9) \).

**(a)**
* **M1:** For substituting \( -x \) for \( x \) to obtain \( 2x^2 + 8x + 5 \).
* **M1:** For translating the function, replacing \( x \) with \( x + 3 \) and adding 4.
* **A1:** For a correct simplified quadratic expression after the translation: \( y = 2x^2 + 20x + 51 \).
* **M1:** For multiplying their quadratic expression by 3.
* **A1:** For the final correct equation \( y = 6x^2 + 60x + 153 \).

**(b)**
* **B1:** For correct coordinates after reflection: \( (-1, -1) \).
* **B1:** For correct coordinates after translation: \( (-4, 3) \) (FT from their reflection step).
* **B1:** For correct final coordinates \( (-4, 9) \) (FT from their translation step).

(i) Let \(y = 4 + \frac{3}{2x - 1}\).
Subtracting \(4\) from both sides:
\(y - 4 = \frac{3}{2x - 1}\)

Rearranging to make \(2x - 1\) the subject:
\(2x - 1 = \frac{3}{y - 4}\)

Adding \(1\) to both sides:
\(2x = 1 + \frac{3}{y - 4} = \frac{y - 4 + 3}{y - 4} = \frac{y - 1}{y - 4}\)

Dividing by \(2\):
\(x = \frac{y - 1}{2(y - 4)}\)

Replacing \(y\) with \(x\) to find the inverse function:
\(\text{f}^{-1}(x) = \frac{x - 1}{2x - 8}\) (or equivalent form such as \(\frac{1}{2} + \frac{3}{2x - 8}\)).

(ii) The domain of \(\text{f}^{-1}\) is the range of \(\text{f}\).
Since \(x > \frac{1}{2}\), we have \(2x - 1 > 0\), which implies \(\frac{3}{2x - 1} > 0\).
Therefore, \(\text{f}(x) = 4 + \frac{3}{2x - 1} > 4\).
Thus, the domain of \(\text{f}^{-1}\) is \(x > 4\).

(i)
* **M1**: For a valid attempt to make \(x\) the subject of the equation \(y = 4 + \frac{3}{2x-1}\).
* **M1**: For correctly isolating the term containing \(x\), leading to \(2x = 1 + \frac{3}{y-4}\) (or equivalent).
* **A1**: For a fully correct simplified expression for \(\text{f}^{-1}(x)\) in terms of \(x\).

(ii)
* **B1**: For stating the domain as \(x > 4\) (allow any equivalent notation, but reject \(y > 4\) or the use of other variables).

**(a)**
The radius \(r\) of the circle is the distance between the center \((2, 1)\) and the point \(P(6, 4)\):
\(r = \sqrt{(6-2)^2 + (4-1)^2} = \sqrt{4^2 + 3^2} = 5\)

So, the equation of the circle is:
\((x-2)^2 + (y-1)^2 = 25\)

**(b)**
The gradient of the radius \(CP\) is:
\(m_{CP} = \frac{4-1}{6-2} = \frac{3}{4}\)

Since the tangent is perpendicular to the radius, the gradient of the tangent at \(P\) is:
\(m = -\frac{4}{3}\)

Using the point-slope form with \(P(6, 4)\):
\(y - 4 = -\frac{4}{3}(x - 6)\)
\(3(y - 4) = -4(x - 6)\)
\(3y - 12 = -4x + 24\)
\(4x + 3y - 36 = 0\)

**(c)**
To find the coordinates of \(Q\), set \(x = 0\) in the tangent equation:
\(4(0) + 3y - 36 = 0 \implies 3y = 36 \implies y = 12\)
So, \(Q\) is \((0, 12)\).

**Method 1: Using simultaneous equations**
Since \(QR\) is tangent to the circle, the length of \(QR\) is equal to the length of \(QP\):
\(QP^2 = (6-0)^2 + (4-12)^2 = 6^2 + (-8)^2 = 100\)
So, \(QR^2 = 100 \implies x^2 + (y-12)^2 = 100\)

The point \(R(x, y)\) lies on both this circle and the original circle \(C\):
1) \(x^2 + y^2 - 4x - 2y - 20 = 0\)
2) \(x^2 + y^2 - 24y + 44 = 0\)

Subtracting the second equation from the first:
\(-4x + 22y - 64 = 0 \implies 2x - 11y + 32 = 0 \implies x = \frac{11y - 32}{2}\)

Substitute this expression for \(x\) into \(x^2 + (y-12)^2 = 100\):
\(\left(\frac{11y-32}{2}\right)^2 + y^2 - 24y + 144 = 100\)
\(\frac{121y^2 - 704y + 1024}{4} + y^2 - 24y + 44 = 0\)

Multiply the entire equation by 4:
\(121y^2 - 704y + 1024 + 4y^2 - 96y + 176 = 0\)
\(125y^2 - 800y + 1200 = 0\)

Divide by 25:
\(5y^2 - 32y + 48 = 0\)
\((y - 4)(5y - 12) = 0\)

Since \(y = 4\) corresponds to the point \(P\), the \(y\)-coordinate of \(R\) is:
\(y = \frac{12}{5} = 2.4\)

The \(x\)-coordinate of \(R\) is:
\(x = \frac{11(2.4) - 32}{2} = -2.8\)

Thus, the coordinates of \(R\) are \((-2.8, 2.4)\).

**Method 2: Using vectors and symmetry**
Let \(M\) be the midpoint of \(PR\). The line of symmetry between the two tangents \(QP\) and \(QR\) is the line \(QC\).
\(\vec{QC} = \begin{pmatrix} 2 \\ -11 \end{pmatrix}\) and \(\vec{QP} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}\).

Projecting \(\vec{QP}\) onto \(\vec{QC}\) to find the projection vector \(\vec{QM}\):
\(\vec{QM} = \left( \frac{\vec{QP} \cdot \vec{QC}}{|\vec{QC}|^2} \right) \vec{QC} = \left( \frac{12 + 88}{125} \right) \begin{pmatrix} 2 \\ -11 \end{pmatrix} = 0.8 \begin{pmatrix} 2 \\ -11 \end{pmatrix} = \begin{pmatrix} 1.6 \\ -8.8 \end{pmatrix}\)

So, \(M = (1.6, 3.2)\).
Since \(M\) is the midpoint of \(PR\):
\(\mathbf{r} = 2\mathbf{m} - \mathbf{p} = 2\begin{pmatrix} 1.6 \\ 3.2 \end{pmatrix} - \begin{pmatrix} 6 \\ 4 \end{pmatrix} = \begin{pmatrix} -2.8 \\ 2.4 \end{pmatrix}\)

So \(R\) is \((-2.8, 2.4)\).

**(a)**
M1: For calculating the radius or radius squared using the distance formula.
A1: For the correct equation of the circle \((x-2)^2 + (y-1)^2 = 25\) (or equivalent expanded form).

**(b)**
M1: For finding the gradient of the radius \(CP\) and attempting to find the perpendicular gradient.
M1: For a correct method to find the equation of the line through \(P\) using their perpendicular gradient.
A1: For the correct equation in the form \(4x + 3y - 36 = 0\) (or any integer multiple thereof).

**(c)**
B1: For finding the coordinates of \(Q(0, 12)\).
M1: For a valid method to find the coordinates of \(R\) (such as setting up simultaneous circle/tangent-length equations to eliminate quadratic terms, or using vector projection to find the midpoint of \(PR\)).
A1: For obtaining the correct coordinates of \(R(-2.8, 2.4)\) or \(\left(-\frac{14}{5}, \frac{12}{5}\right)\).

(a) The volume \(V\) of the solid of revolution is given by:
\(V = \pi \int_{0}^{4} y^2 \, dx\)

First, find the expression for \(y^2\):
\(y^2 = \left( \frac{k}{(2x+1)^{1/4}} \right)^2 = \frac{k^2}{(2x+1)^{1/2}} = k^2(2x+1)^{-1/2}\)

Next, integrate this expression with respect to \(x\):
\(V = \pi \int_{0}^{4} k^2(2x+1)^{-1/2} \, dx\)
\(V = \pi k^2 \left[ \frac{(2x+1)^{1/2}}{\frac{1}{2} \times 2} \right]_{0}^{4}\)
\(V = \pi k^2 \left[ (2x+1)^{1/2} \right]_{0}^{4}\)

Now substitute the limits \(x = 4\) and \(x = 0\):
\(V = \pi k^2 \left( \sqrt{2(4)+1} - \sqrt{2(0)+1} \right)\)
\(V = \pi k^2 ( \sqrt{9} - \sqrt{1} ) = \pi k^2 (3 - 1) = 2\pi k^2\)

This shows that the volume of the solid of revolution is \(2\pi k^2\).

(b) We are given that the volume is \(72\pi\):
\(2\pi k^2 = 72\pi \implies k^2 = 36\)
Since \(k\) is a positive constant, we have:
\(k = 6\)

The equation of the curve with this value of \(k\) is:
\(y = 6(2x+1)^{-1/4}\)

Now, differentiate \(y\) with respect to \(x\) to find the gradient function:
\(\frac{dy}{dx} = 6 \times \left(-\frac{1}{4}\right)(2x+1)^{-5/4} \times 2\)
\(\frac{dy}{dx} = -3(2x+1)^{-5/4}\)

Set the gradient equal to \(-\frac{3}{32}\) to find the x-coordinate of the point:
\(-3(2x+1)^{-5/4} = -\frac{3}{32}\)
\((2x+1)^{-5/4} = \frac{1}{32}\)
\((2x+1)^{5/4} = 32\)
\(2x+1 = 32^{4/5} = (2^5)^{4/5} = 2^4 = 16\)
\(2x = 15 \implies x = 7.5\)

Substitute \(x = 7.5\) back into the equation of the curve to find the y-coordinate:
\(y = 6(2(7.5)+1)^{-1/4} = 6(16)^{-1/4} = 6 \times \frac{1}{2} = 3\)

Thus, the coordinates of the point on the curve are \((7.5, 3)\) or \(\left(\frac{15}{2}, 3\right)\).

(a)
M1: Attempt to use \(V = \pi \int y^2 \, dx\) with limits \(0\) and \(4\).
A1: Obtain the correct squared integrand \(y^2 = k^2(2x+1)^{-1/2}\).
M1: Integrate \(k^2(2x+1)^{-1/2}\) correctly to obtain \(k^2(2x+1)^{1/2}\) (must show division by both the power \(\frac{1}{2}\) and the coefficient of \(x\), which is \(2\)).
M1: Substitute the limits \(4\) and \(0\) correctly into their integrated expression.
A1: Completely correct algebraic working leading convincingly to the given answer \(2\pi k^2\).

(b)
B1: Set \(2\pi k^2 = 72\pi\) and solve to find \(k = 6\) (rejecting \(k = -6\)).
M1: Differentiate \(y = 6(2x+1)^{-1/4}\) using the chain rule, set the derivative equal to \(-\frac{3}{32}\), and solve the index equation to find a value for \(x\).
A1: Obtain the correct coordinates of the point, \((7.5, 3)\) or \(\left(\frac{15}{2}, 3\right)\).

(a) To find the intersection of the curve and the line, we equate their expressions:

\( (k-1)x^2 + 2kx + (k+3) = 2x + k \)

Subtracting \( 2x + k \) from both sides to form a quadratic equation:

\( (k-1)x^2 + (2k-2)x + 3 = 0 \)

This is a quadratic equation of the form \( ax^2 + bx + c = 0 \) where:
\( a = k-1 \)
\( b = 2(k-1) \)
\( c = 3 \)

For the line and curve to intersect at two distinct points, the discriminant must be strictly positive (\( b^2 - 4ac > 0 \)):

\( [2(k-1)]^2 - 4(k-1)(3) > 0 \)

\( 4(k-1)^2 - 12(k-1) > 0 \)

Factorising the inequality:

\( 4(k-1)[(k-1) - 3] > 0 \)

\( 4(k-1)(k-4) > 0 \)

The critical values are \( k = 1 \) and \( k = 4 \).
Since the inequality is \( > 0 \), the solution set is:

\( k < 1 \) or \( k > 4 \)

(b) When \( k = 5 \), the equation of intersection becomes:

\( (5-1)x^2 + 2(5-1)x + 3 = 0 \)

\( 4x^2 + 8x + 3 = 0 \)

Factorising the quadratic expression:

\( (2x + 1)(2x + 3) = 0 \)

This gives the \( x \)-coordinates:

\( x = -0.5 \) and \( x = -1.5 \)

Substitute these values into the line equation \( y = 2x + 5 \) to find the corresponding \( y \)-coordinates:

For \( x = -0.5 \):
\( y = 2(-0.5) + 5 = 4 \)

For \( x = -1.5 \):
\( y = 2(-1.5) + 5 = 2 \)

Thus, the points of intersection are \( (-0.5, 4) \) and \( (-1.5, 2) \).

(c) For \( L \) to be a tangent to the curve, the discriminant must be equal to zero:

\( 4(k-1)(k-4) = 0 \implies k = 1 \text{ or } k = 4 \)

Since we are given that \( k > 1 \), we have:

\( k = 4 \)

Substituting \( k = 4 \) into the intersection equation:

\( (4-1)x^2 + 2(4-1)x + 3 = 0 \)

\( 3x^2 + 6x + 3 = 0 \implies x^2 + 2x + 1 = 0 \implies (x+1)^2 = 0 \)

This gives \( x = -1 \).

Substituting \( x = -1 \) and \( k = 4 \) into the equation of the line \( y = 2x + 4 \):

\( y = 2(-1) + 4 = 2 \)

The coordinates of the point of contact are \( (-1, 2) \).

(a)
M1: Equates curve and line to form a three-term quadratic in \( x \).
M1: Attempts to find the discriminant \( b^2 - 4ac \) in terms of \( k \).
A1: Obtains a correct simplified quadratic in \( k \) or factored form, e.g., \( 4(k-1)(k-4) > 0 \).
A1: Identifies correct range: \( k < 1 \) or \( k > 4 \) (do not accept integrated inequalities like \( 1 > k > 4 \)).

(b)
M1: Substitutes \( k = 5 \) into the quadratic equation and attempts to solve for \( x \).
A1: Obtains both correct \( x \)-coordinates: \( x = -0.5 \) and \( x = -1.5 \) (or equivalent fractions).
A1: Obtains both correct coordinates: \( (-0.5, 4) \) and \( (-1.5, 2) \).

(c)
B1: States \( k = 4 \) (rejects \( k = 1 \) based on the condition \( k > 1 \)).
B1: Calculates and states the correct coordinates of the point of contact: \( (-1, 2) \).

(a) We differentiate the equation of the curve with respect to \( x \):
\( y = 8x + 4x^{-2} \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 8 - 8x^{-3} = 8 - \frac{8}{x^3} \)

To find the stationary point, we set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\( 8 - \frac{8}{x^3} = 0 \implies \frac{8}{x^3} = 8 \implies x^3 = 1 \implies x = 1 \) (since \( x > 0 \)).

Substituting \( x = 1 \) back into the curve equation:
\( y = 8(1) + \frac{4}{1^2} = 12 \).

Thus, the coordinates of the stationary point are \( (1, 12) \).

(b) We find the second derivative by differentiating \( \frac{\mathrm{d}y}{\mathrm{d}x} = 8 - 8x^{-3} \):
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 24x^{-4} = \frac{24}{x^4} \).

At the stationary point where \( x = 1 \):
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{24}{1^4} = 24 \).

Since \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0 \), the stationary point \( (1, 12) \) is a minimum.

(c) First, find the gradient of the tangent at \( P(2, 17) \):
Substitute \( x = 2 \) into \( \frac{\mathrm{d}y}{\mathrm{d}x} \):
\( m_{\text{tangent}} = 8 - \frac{8}{2^3} = 8 - 1 = 7 \).

The gradient of the normal, \( m_{\text{normal}} \), is given by:
\( m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{7} \).

The equation of the normal at \( P(2, 17) \) is:
\( y - 17 = -\frac{1}{7}(x - 2) \implies y = -\frac{1}{7}x + \frac{121}{7} \).

The normal meets the \( y \)-axis when \( x = 0 \). Substituting \( x = 0 \) into the equation:
\( y = \frac{121}{7} \).

So, the coordinates of \( Q \) are \( \left(0, \frac{121}{7}\right) \).

(d) The area under the curve is given by the definite integral:
\( \text{Area} = \int_{1}^{2} \left(8x + 4x^{-2}\right) \mathrm{d}x \)
\( = \left[ 4x^2 - 4x^{-1} \right]_{1}^{2} = \left[ 4x^2 - \frac{4}{x} \right]_{1}^{2} \)
\( = \left(4(2)^2 - \frac{4}{2}\right) - \left(4(1)^2 - \frac{4}{1}\right) \)
\( = (16 - 2) - (4 - 4) = 14 - 0 = 14 \).

(a)
* M1: For attempting to differentiate the equation of the curve, reducing at least one power of \( x \) by 1.
* A1: For obtaining the correct derivative \( \frac{\mathrm{d}y}{\mathrm{d}x} = 8 - \frac{8}{x^3} \).
* M1: For setting their derivative to 0 and solving for \( x \).
* A1: For obtaining \( x = 1 \) and \( y = 12 \) (or the coordinates \( (1, 12) \)).

(b)
* B1: For obtaining the correct second derivative \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{24}{x^4} \).
* B1: For substituting \( x = 1 \) into their second derivative, showing it is positive (e.g., \( 24 > 0 \)), and concluding that it is a minimum point.

(c)
* M1: For substituting \( x = 2 \) into their derivative to find the tangent gradient.
* M1: For finding the normal gradient using \( m_1 m_2 = -1 \) and attempting the equation of the normal.
* A1: For a correct equation of the normal, e.g., \( y - 17 = -\frac{1}{7}(x - 2) \) or equivalent.
* A1: For finding \( x = 0 \implies y = \frac{121}{7} \) and stating the coordinates of \( Q \) as \( \left(0, \frac{121}{7}\right) \) (accept equivalent fraction or decimal \( 17.3 \) to 3 s.f.).

(d)
* M1: For attempting to integrate \( 8x + 4x^{-2} \), increasing at least one power of \( x \) by 1.
* A1: For correct integration yielding \( 4x^2 - \frac{4}{x} \) (condone lack of constant of integration).
* A1: For substituting the limits 1 and 2 correctly and obtaining the final exact area of \( 14 \).

We use the double angle identity for cosine, \(\cos(4x) = 1 - 2\sin^2(2x)\), to rewrite the integrand: \(4\sin^2(2x) = 2(1 - \cos(4x)) = 2 - 2\cos(4x)\). Now, we integrate this expression with respect to \(x\): \(\int (2 - 2\cos(4x)) \, \mathrm{d}x = 2x - \frac{2}{4}\sin(4x) = 2x - \frac{1}{2}\sin(4x)\). Next, we apply the limits of integration from \(0\) to \(\frac{\pi}{6}\): \(\left[ 2x - \frac{1}{2}\sin(4x) \right]_{0}^{\frac{\pi}{6}} = \left( 2\left(\frac{\pi}{6}\right) - \frac{1}{2}\sin\left(\frac{4\pi}{6}\right) \right) - \left( 2(0) - \frac{1}{2}\sin(0) \right) = \frac{\pi}{3} - \frac{1}{2}\sin\left(\frac{2\pi}{3}\right)\). Since \(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\), this simplifies to: \(\frac{\pi}{3} - \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} - \frac{\sqrt{3}}{4}\).

**M1**: For using the double angle identity to express \(4\sin^2(2x)\) in the form \(a + b\cos(4x)\) (where \(a, b \neq 0\)). **A1**: For obtaining the correct integrated expression \(2x - \frac{1}{2}\sin(4x)\) (or equivalent). **A1**: For substituting limits correctly and obtaining the exact answer \(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\) (or equivalent single fraction such as \(\frac{4\pi - 3\sqrt{3}}{12}\)).

We start with the given equation: \(e^x + 12e^{-x} = 7\). Multiplying both sides by \(e^x\), we get: \(e^{2x} + 12 = 7e^x\). Rearranging this into a quadratic equation form gives: \(e^{2x} - 7e^x + 12 = 0\). Let \(u = e^x\). The equation becomes: \(u^2 - 7u + 12 = 0\). Factoring the quadratic expression, we have: \((u - 3)(u - 4) = 0\). This yields the solutions: \(u = 3\) or \(u = 4\). Substituting back \(u = e^x\), we get: \(e^x = 3 \implies x = \ln 3\) and \(e^x = 4 \implies x = \ln 4\).

M1: For multiplying by \(e^x\) to form a three-term quadratic equation in \(e^x\).
A1: For solving the quadratic equation to find the correct values of \(e^x\), which are \(e^x = 3\) and \(e^x = 4\).
A1: For obtaining both correct exact solutions \(x = \ln 3\) and \(x = \ln 4\) (or \(2\ln 2\)).

We start with the equation: \(e^x + 12e^{-x} = 7\). Multiply the entire equation by \(e^x\): \(e^{2x} + 12 = 7e^x\). Rearranging into a quadratic form: \(e^{2x} - 7e^x + 12 = 0\). Let \(u = e^x\). The equation becomes: \(u^2 - 7u + 12 = 0\). Factorising the quadratic equation: \((u - 3)(u - 4) = 0\). This gives: \(u = 3\) or \(u = 4\). Substitute back \(u = e^x\): \(e^x = 3 \implies x = \ln 3\), \(e^x = 4 \implies x = \ln 4\).

M1: For multiplying by \(e^x\) and forming a three-term quadratic equation in \(e^x\).
A1: For obtaining the correct values \(e^x = 3\) and \(e^x = 4\).
A1: For obtaining both correct exact solutions \(x = \ln 3\) and \(x = \ln 4\) (or \(2\ln 2\)).

We are given the equation: \(|5\cos \theta - 1| = 2\cos \theta + 1\) for \(0 \le \theta \le 2\pi\). Step 1: Solve the modulus equation. We can solve this by squaring both sides or by splitting the equation into two cases. Method 1: Squaring both sides: \((5\cos \theta - 1)^2 = (2\cos \theta + 1)^2\) which expands to \(25\cos^2 \theta - 10\cos \theta + 1 = 4\cos^2 \theta + 4\cos \theta + 1\). Rearranging and simplifying gives \(21\cos^2 \theta - 14\cos \theta = 0\), which factorises to \(7\cos \theta(3\cos \theta - 2) = 0\). Method 2: Splitting into two cases: Case A: \(5\cos \theta - 1 = 2\cos \theta + 1 \implies 3\cos \theta = 2 \implies \cos \theta = \frac{2}{3}\). Case B: \(5\cos \theta - 1 = -(2\cos \theta + 1) \implies 7\cos \theta = 0 \implies \cos \theta = 0\). In both methods, we obtain: \(\cos \theta = 0\) or \(\cos \theta = \frac{2}{3}\). Step 2: Solve for \(\theta\) in the range \(0 \le \theta \le 2\pi\). For \(\cos \theta = 0\): \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\) (or \(1.57\) and \(4.71\) to 3 s.f.). For \(\cos \theta = \frac{2}{3}\): The principal value is \(\theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 0.841\) (to 3 s.f.). The second solution in the range is \(\theta = 2\pi - 0.84107... \approx 5.44\) (to 3 s.f.). Therefore, the complete set of solutions in the given range is: \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}, 0.841, 5.44\).

M1: For a valid method to solve the modulus equation (e.g., squaring both sides to obtain a quadratic in \(\cos \theta\), or setting up and solving the two linear cases: \(5\cos \theta - 1 = 2\cos \theta + 1\) and \(5\cos \theta - 1 = -2\cos \theta - 1\)). A1: For obtaining both correct values: \(\cos\theta = 0\) and \(\cos\theta = \frac{2}{3}\) (or equivalent equations). B1: For \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\) (accept \(1.57\) and \(4.71\)). M1: For a correct method to find at least one value of \(\theta\) from \(\cos\theta = \frac{2}{3}\) in the given interval (e.g., \(\cos^{-1}(2/3)\)). A1: For \(\theta = 0.841\). A1: For \(\theta = 5.44\) and no other solutions in the range.

**Part (i)**
To find the \(x\)-coordinates where the curve intersects the \(x\)-axis, we set \( y = 0 \):
\[ e^{2x} - 5e^x + 4 = 0 \]

Let \( u = e^x \). The equation becomes a quadratic in \( u \):
\[ u^2 - 5u + 4 = 0 \]
\[ (u - 1)(u - 4) = 0 \]

This gives:
\[ u = 1 \implies e^x = 1 \implies x = 0 \]
\[ u = 4 \implies e^x = 4 \implies x = \ln 4 \text{ (or } 2\ln 2\text{)} \]

So the \(x\)-coordinates of \(A\) and \(B\) are \( x = 0 \) and \( x = \ln 4 \).

**Part (ii)**
For the interval \( 0 < x < \ln 4 \), the curve lies below the \(x\)-axis (since, for example, when \( x = \ln 2 \), \( y = 4 - 10 + 4 = -2 < 0 \)).

Therefore, the area of the enclosed region is given by:
\[ \text{Area} = \int_{0}^{\ln 4} (0 - y) \, dx = \int_{0}^{\ln 4} (5e^x - e^{2x} - 4) \, dx \]

Integrating term by term:
\[ \int (5e^x - e^{2x} - 4) \, dx = \left[ 5e^x - \frac{1}{2}e^{2x} - 4x \right]_{0}^{\ln 4} \]

Substitute the upper limit \( x = \ln 4 \):
\[ 5e^{\ln 4} - \frac{1}{2}e^{2\ln 4} - 4\ln 4 = 5(4) - \frac{1}{2}(16) - 4(2\ln 2) \]
\[ = 20 - 8 - 8\ln 2 = 12 - 8\ln 2 \]

Substitute the lower limit \( x = 0 \):
\[ 5e^0 - \frac{1}{2}e^0 - 4(0) = 5 - \frac{1}{2} = \frac{9}{2} \]

Subtract the lower limit evaluation from the upper limit evaluation:
\[ \text{Area} = (12 - 8\ln 2) - \frac{9}{2} \]
\[ \text{Area} = \frac{24}{2} - \frac{9}{2} - 8\ln 2 = \frac{15}{2} - 8\ln 2 \]

Comparing this to the form \( a - b\ln 2 \), we find:
\[ a = \frac{15}{2} \text{ (or } 7.5\text{)} \quad \text{and} \quad b = 8 \]

**Part (i)**
* **M1**: Set \( e^{2x} - 5e^x + 4 = 0 \) and attempt to solve as a quadratic in \( e^x \).
* **A1**: Obtain \( e^x = 1 \) and \( e^x = 4 \) (or equivalent).
* **A1**: Show correct \(x\)-coordinates: \( x = 0 \) and \( x = \ln 4 \) (or \( x = 2\ln 2 \)).

**Part (ii)**
* **M1**: For integrating \( e^{2x} - 5e^x + 4 \) (or its negative), obtaining at least one integrated term of the form \( k e^{2x} \) where \( k \ne 1 \).
* **A1**: Obtain correct integral: \( \frac{1}{2}e^{2x} - 5e^x + 4x \) (or its negative).
* **M1**: Apply the limits \( 0 \) and their upper limit \( \ln 4 \) (or \( 2\ln 2 \)) to their integrated expression, and use the rule \( e^{2\ln 4} = 16 \) (or \( e^{4\ln 2} = 16 \)) and \( \ln 4 = 2\ln 2 \) correctly.
* **A1**: Obtain the correct exact area value of \( \frac{15}{2} - 8\ln 2 \), identifying \( a = \frac{15}{2} \) (or \( 7.5 \)) and \( b = 8 \).

**(i)**

First, differentiate \(x\) with respect to \(t\):
\[\frac{\text{d}x}{\text{d}t} = \frac{1}{2t+1} \cdot 2 + 6t = \frac{2}{2t+1} + 6t\]

Combine this into a single fraction:
\[\frac{\text{d}x}{\text{d}t} = \frac{2 + 6t(2t+1)}{2t+1}\]

Next, differentiate \(y\) with respect to \(t\):
\[\frac{\text{d}y}{\text{d}t} = -e^{-t} + 4t\]

Using the chain rule for parametric differentiation:
\[\frac{\text{d}y}{\text{d}x} = \frac{\frac{\text{d}y}{\text{d}t}}{\frac{\text{d}x}{\text{d}t}} = \frac{4t - e^{-t}}{\frac{2 + 6t(2t+1)}{2t+1}}\]

Multiply the numerator by the denominator's reciprocal:
\[\frac{\text{d}y}{\text{d}x} = \frac{(4t - e^{-t})(2t+1)}{2 + 6t(2t+1)}\]

This completes the proof.

**(ii)**

To find the equation of the normal at \(t = 0\):

1. Find the coordinates of the point of contact:
\[x = \ln(2(0)+1) + 3(0)^2 = \ln(1) + 0 = 0\]
\[y = e^{-(0)} + 2(0)^2 = 1 + 0 = 1\]
So the point is \((0, 1)\).

2. Calculate the gradient of the tangent at \(t = 0\):
\[\frac{\text{d}y}{\text{d}x} = \frac{(4(0) - e^{0})(2(0)+1)}{2 + 6(0)(2(0)+1)} = \frac{(-1)(1)}{2 + 0} = -\frac{1}{2}\]

3. Calculate the gradient of the normal:
\[m_{\text{normal}} = -\frac{1}{-\frac{1}{2}} = 2\]

4. Use the point-slope formula for the normal:
\[y - 1 = 2(x - 0)\]
\[y - 1 = 2x\]
\[y - 2x = 1\]

Alternatively, written as \(2x - y = -1\).

**Part (i):**
* **M1**: Attempt to differentiate \(x\) with respect to \(t\) to obtain a form \(\frac{k}{2t+1} + ct\).
* **A1**: Obtain correct \(\frac{\text{d}x}{\text{d}t} = \frac{2}{2t+1} + 6t\) (any equivalent form).
* **B1**: Obtain correct \(\frac{\text{d}y}{\text{d}t} = -e^{-t} + 4t\).
* **A1**: Show complete and convincing algebraic steps to reach the given expression for \(\frac{\text{d}y}{\text{d}x}\).

**Part (ii):**
* **B1**: Obtain both coordinates \(x = 0\) and \(y = 1\) correctly.
* **M1**: Substitute \(t = 0\) into the expression for \(\frac{\text{d}y}{\text{d}x}\) to find the numerical gradient of the tangent.
* **M1**: Use the negative reciprocal of their tangent gradient to find the gradient of the normal.
* **M1**: Attempt to form the equation of a straight line through their point \((0, 1)\) using their numerical normal gradient.
* **A1**: Obtain correct final equation in the required form, e.g., \(y - 2x = 1\) or \(2x - y = -1\).

**(a)**
Using the Factor Theorem, since \( (2x+1) \) is a factor of \( p(x) \), we must have \( p\left(-\frac{1}{2}\right) = 0 \):
\[ p\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right)^2 + a\left(-\frac{1}{2}\right) + b = 0 \]
\[ 2\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) - \frac{1}{2}a + b = 0 \]
\[ -\frac{1}{4} - \frac{3}{4} - \frac{1}{2}a + b = 0 \]
\[ -1 - \frac{1}{2}a + b = 0 \implies 2b - a = 2 \quad \text{(Equation 1)} \]

Using the Remainder Theorem, since the remainder when \( p(x) \) is divided by \( (x-2) \) is 15, we must have \( p(2) = 15 \):
\[ p(2) = 2(2)^3 - 3(2)^2 + a(2) + b = 15 \]
\[ 16 - 12 + 2a + b = 15 \]
\[ 4 + 2a + b = 15 \implies 2a + b = 11 \quad \text{(Equation 2)} \]

Solving Equations 1 and 2 simultaneously:
From Equation 1, \( a = 2b - 2 \).
Substitute this into Equation 2:
\[ 2(2b - 2) + b = 11 \implies 4b - 4 + b = 11 \implies 5b = 15 \implies b = 3 \]
Then substitute \( b = 3 \) back into the expression for \( a \):
\[ a = 2(3) - 2 = 4 \]
Thus, \( a = 4 \) and \( b = 3 \).

**(b)**
With \( a = 4 \) and \( b = 3 \), the polynomial is \( p(x) = 2x^3 - 3x^2 + 4x + 3 \).
The equation to be solved is:
\[ p(x) = 15 - 5\ln(x+2) \implies 2x^3 - 3x^2 + 4x + 3 = 15 - 5\ln(x+2) \]
\[ 2x^3 - 3x^2 + 4x - 12 + 5\ln(x+2) = 0 \]

Let \( f(x) = 2x^3 - 3x^2 + 4x - 12 + 5\ln(x+2) \). We evaluate \( f(x) \) at the boundaries of the interval \( [1, 2] \):
- For \( x = 1 \):
\[ f(1) = 2(1)^3 - 3(1)^2 + 4(1) - 12 + 5\ln(3) = -9 + 5\ln(3) \approx -9 + 5.4931 = -3.5069 \]
Since \( f(1) < 0 \).
- For \( x = 2 \):
\[ f(2) = 2(2)^3 - 3(2)^2 + 4(2) - 12 + 5\ln(4) = 16 - 12 + 8 - 12 + 5\ln(4) = 5\ln(4) \approx 6.9315 \]
Since \( f(2) > 0 \).

There is a change of sign between \( f(1) \) and \( f(2) \). Since \( f(x) \) is continuous on the interval \( [1, 2] \), there must be a root \( \alpha \) such that \( 1 < \alpha < 2 \).

**(c)**
Starting with the equation:
\[ 2x^3 - 3x^2 + 4x - 12 + 5\ln(x+2) = 0 \]
Rearrange to isolate the term with \( x^3 \):
\[ 2x^3 = 12 + 3x^2 - 4x - 5\ln(x+2) \]
Divide the entire equation by 2:
\[ x^3 = 6 + 1.5x^2 - 2x - 2.5\ln(x+2) \]
Take the cube root of both sides:
\[ x = \left( 6 + 1.5x^2 - 2x - 2.5\ln(x+2) \right)^{\frac{1}{3}} \]
This is the required iterative formula.

**(d)**
Applying the iterative formula \( x_{n+1} = \left( 6 + 1.5x_n^2 - 2x_n - 2.5\ln(x_n+2) \right)^{\frac{1}{3}} \) with \( x_1 = 1.5 \):
- \( x_1 = 1.5 \)
- \( x_2 = \left( 6 + 1.5(1.5)^2 - 2(1.5) - 2.5\ln(3.5) \right)^{\frac{1}{3}} \approx 1.4803 \)
- \( x_3 = \left( 6 + 1.5(1.480313)^2 - 2(1.480313) - 2.5\ln(3.480313) \right)^{\frac{1}{3}} \approx 1.4750 \)
- \( x_4 = \left( 6 + 1.5(1.474965)^2 - 2(1.474965) - 2.5\ln(3.474965) \right)^{\frac{1}{3}} \approx 1.4735 \)
- \( x_5 = \left( 6 + 1.5(1.473549)^2 - 2(1.473549) - 2.5\ln(3.473549) \right)^{\frac{1}{3}} \approx 1.4732 \)
- \( x_6 = \left( 6 + 1.5(1.473167)^2 - 2(1.473167) - 2.5\ln(3.473167) \right)^{\frac{1}{3}} \approx 1.4731 \)

Since the values converge to \( 1.47 \) when rounded to 2 decimal places, we have \( \alpha = 1.47 \).

**(a)**
* **M1**: For substituting \( x = -0.5 \) into \( p(x) \) and setting to 0, or performing equivalent division and setting remainder to 0.
* **A1**: For obtaining a correct simplified equation, e.g., \( 2b - a = 2 \).
* **M1**: For substituting \( x = 2 \) into \( p(x) \) and setting to 15, and attempting to solve the resulting simultaneous equations.
* **A1**: For obtaining both correct values: \( a = 4 \) and \( b = 3 \).

**(b)**
* **M1**: For defining a function \( f(x) = p(x) - 15 + 5\ln(x+2) \) and attempting to evaluate \( f(1) \) and \( f(2) \) (must involve substituting values and showing steps).
* **A1**: For obtaining correct values \( f(1) \approx -3.51 \) (or \( -9+5\ln 3 \)) and \( f(2) \approx 6.93 \) (or \( 5\ln 4 \)).
* **A1**: For a complete explanation mentioning both the sign change and the continuity of the function over the interval.

**(c)**
* **M1**: For algebraic steps showing isolation of the \( x^3 \) term (or equivalent rearrange showing clear steps).
* **A1**: For completing the steps to arrive at the given iterative formula with no errors.

**(d)**
* **M1**: For calculating at least the first two iterations correctly to 4 decimal places (i.e., \( x_2 = 1.4803 \) and \( x_3 = 1.4750 \)).
* **A1**: For obtaining the final value of \( 1.47 \) to 2 decimal places, supported by convergent iterations shown up to at least \( x_4 \) or \( x_5 \).

To find the stationary points, we first differentiate the given equation with respect to \(x\) using the product rule.

Let \(u = x^2 - x - 1\) and \(v = \mathrm{e}^{-3x}\).

Then, we differentiate both terms:
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 2x - 1\)
\(\frac{\mathrm{d}v}{\mathrm{d}x} = -3\mathrm{e}^{-3x}\)

Applying the product rule, \(\frac{\mathrm{d}y}{\mathrm{d}x} = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = (x^2 - x - 1)(-3\mathrm{e}^{-3x}) + (2x - 1)\mathrm{e}^{-3x}\)

Factor out \(\mathrm{e}^{-3x}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-3x} \left[ -3(x^2 - x - 1) + (2x - 1) \right]\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-3x} \left[ -3x^2 + 3x + 3 + 2x - 1 \right]\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-3x} \left[ -3x^2 + 5x + 2 \right]\)

At stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\).
Since \(\mathrm{e}^{-3x} \neq 0\) for all real \(x\), we solve:
\(-3x^2 + 5x + 2 = 0\)
\(3x^2 - 5x - 2 = 0\)
\((3x + 1)(x - 2) = 0\)

This yields the \(x\)-coordinates:
\(x = 2\) and \(x = -\frac{1}{3}\).

Next, we find the corresponding \(y\)-coordinates by substituting these \(x\)-values back into the original curve equation:

For \(x = 2\):
\(y = (2^2 - 2 - 1)\mathrm{e}^{-3(2)} = (4 - 3)\mathrm{e}^{-6} = \mathrm{e}^{-6}\)

For \(x = -\frac{1}{3}\):
\(y = \left[\left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right) - 1\right]\mathrm{e}^{-3\left(-\frac{1}{3}\right)} = \left(\frac{1}{9} + \frac{1}{3} - 1\right)\mathrm{e}^1 = \left(\frac{4}{9} - 1\right)\mathrm{e} = -\frac{5}{9}\mathrm{e}\)

Thus, the exact coordinates of the stationary points are \(\left(2, \mathrm{e}^{-6}\right)\) and \(\left(-\frac{1}{3}, -\frac{5}{9}\mathrm{e}\right)\).

M1: Attempt to apply the product rule to differentiate \(y = (x^2 - x - 1)\mathrm{e}^{-3x}\).
A1: Obtain correct derivative of the first term or second term as part of the product rule.
A1: Obtain fully correct derivative, equivalent to \(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-3x}(-3x^2 + 5x + 2)\).
M1: Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and attempt to solve the quadratic equation \(3x^2 - 5x - 2 = 0\).
A1: Obtain both correct \(x\)-coordinates: \(x = 2\) and \(x = -\frac{1}{3}\).
A1: Obtain both correct exact \(y\)-coordinates: \(y = \mathrm{e}^{-6}\) and \(y = -\frac{5}{9}\mathrm{e}\). (Accept equivalents, but exact form is required.)