2025 Cambridge IAS-Level Mathematics (9709) Practice Paper with Answers

To find the points of intersection, we equate the equations of the curve and the line:

\(x^2 + kx + 6 = 3x - \frac{1}{4}\)

Rearranging this into the standard quadratic form \(ax^2 + bx + c = 0\):

\(x^2 + (k-3)x + 6 + \frac{1}{4} = 0\)

\(x^2 + (k-3)x + \frac{25}{4} = 0\)

For the line and the curve to intersect at two distinct points, the discriminant of this quadratic equation must be strictly greater than zero (\(b^2 - 4ac > 0\)):

\(\Delta = (k-3)^2 - 4(1)\left(\frac{25}{4}\right) > 0\)

\(k^2 - 6k + 9 - 25 > 0\)

\(k^2 - 6k - 16 > 0\)

Factorising the quadratic inequality:

\((k-8)(k+2) > 0\)

This gives the critical values \(k = 8\) and \(k = -2\).

Since the inequality is greater than zero, the solution set is:

\(k < -2\) or \(k > 8\)

**M1**: For equating the line and the curve equations to form a 3-term quadratic in \(x\).

**A1**: For obtaining the correct quadratic equation in any equivalent form, e.g., \(x^2 + (k-3)x + \frac{25}{4} = 0\) or \(4x^2 + 4(k-3)x + 25 = 0\).

**M1**: For using the discriminant condition \(b^2 - 4ac > 0\) and attempting to solve the resulting quadratic inequality in \(k\).

**A1**: For the correct final range of \(k\), written as \(k < -2\) or \(k > 8\) (or equivalent interval notation).

Let the first term of the geometric progression be \( a \) and the common ratio be \( r \). The sum to infinity is given by \( S_{\infty} = \frac{a}{1-r} = 13.5 \). The sum of the first three terms is given by \( S_3 = \frac{a(1-r^3)}{1-r} = 13 \). We can substitute the expression for \( S_{\infty} \) into the expression for \( S_3 \): \( S_3 = S_{\infty}(1-r^3) \) which gives \( 13 = 13.5(1-r^3) \). Solving for \( r^3 \): \( 1-r^3 = \frac{13}{13.5} = \frac{26}{27} \), so \( r^3 = 1 - \frac{26}{27} = \frac{1}{27} \). Taking the cube root gives the common ratio \( r = \frac{1}{3} \). Substituting \( r = \frac{1}{3} \) back into the sum to infinity equation: \( \frac{a}{1 - 1/3} = 13.5 \implies \frac{a}{2/3} = 13.5 \implies a = 13.5 \times \frac{2}{3} = 9 \). The first term is \( a = 9 \). The fifth term \( u_5 \) is given by \( u_5 = a r^4 \). Substituting the values: \( u_5 = 9 \times \left(\frac{1}{3}\right)^4 = 9 \times \frac{1}{81} = \frac{1}{9} \).

M1: For writing down the correct formula for \( S_3 \) and \( S_{\infty} \) and attempting to eliminate \( a \) (e.g., obtaining \( 13 = 13.5(1-r^3) \)). A1: For obtaining the correct common ratio \( r = \frac{1}{3} \) (or 0.333). A1: For obtaining the correct first term \( a = 9 \). M1: For applying the term formula \( u_n = a r^{n-1} \) to find the 5th term. A1: For obtaining the correct final answer \( \frac{1}{9} \) as a fraction in its simplest form.

We are given the binomial expansion of \((1 + ax)^n = 1 + \binom{n}{1}ax + \binom{n}{2}(ax)^2 + \dots\). The term in \(x\) is \(nax\), so the coefficient of \(x\) is \(na\). This gives \(na = -12\) (Equation 1). The term in \(x^2\) is \(\frac{n(n-1)}{2} a^2 x^2\), so the coefficient of \(x^2\) is \(\frac{n(n-1)a^2}{2} = 60\), which simplifies to \(n(n-1)a^2 = 120\) (Equation 2). Expanding Equation 2 gives \(n^2 a^2 - na^2 = 120\). Since \(na = -12\) from Equation 1, squaring both sides gives \(n^2 a^2 = 144\). Substituting \(n^2 a^2 = 144\) into the expanded equation gives \(144 - na^2 = 120\), which simplifies to \(na^2 = 24\). We can write \(na^2\) as \((na)a = 24\). Substituting \(na = -12\) into this yields \(-12a = 24\), which gives \(a = -2\). Finally, substituting \(a = -2\) back into \(na = -12\) yields \(-2n = -12\), which gives \(n = 6\).

B1: Identify the coefficient of \(x\) as \(na = -12\). B1: Identify the coefficient of \(x^2\) as \(\frac{n(n-1)a^2}{2} = 60\). M1: Formulate a method to eliminate one variable and solve the simultaneous equations (e.g., substituting \(a = -12/n\) or using \(na^2 = 24\)). A1: Obtain \(n = 6\). A1: Obtain \(a = -2\).

(a) We express \( f(x) = 3x^2 - 12x + 7 \) in completed square form:
\[ f(x) = 3(x^2 - 4x) + 7 \]
\[ f(x) = 3\left[(x - 2)^2 - 4\right] + 7 \]
\[ f(x) = 3(x - 2)^2 - 12 + 7 \]
\[ f(x) = 3(x - 2)^2 - 5 \]

So, \( a = 3 \), \( b = 2 \), and \( c = -5 \).

(b) From part (a), the minimum point of the original curve \( y = f(x) \) is at \( (2, -5) \).

First, a stretch of factor \( \frac{1}{3} \) parallel to the \( x \)-axis maps the point \( (x, y) \) to \( \left(\frac{1}{3}x, y\right) \).
Applying this to the minimum point yields:
\[ (2, -5) \to \left( \frac{2}{3}, -5 \right) \]

Second, a translation of \( \begin{pmatrix} 2 \\ -5 \end{pmatrix} \) maps the point \( (x, y) \) to \( (x + 2, y - 5) \).
Applying this translation to \( \left( \frac{2}{3}, -5 \right) \) yields:
\[ \left( \frac{2}{3}, -5 \right) \to \left( \frac{2}{3} + 2, -5 - 5 \right) = \left( \frac{8}{3}, -10 \right) \]

Therefore, the coordinates of the minimum point of the curve \( y = g(x) \) are \( \left( \frac{8}{3}, -10 \right) \).

(a)
* **M1**: For attempting to complete the square by factorising 3 or setting up \( 3(x-b)^2+c \).
* **A1**: For obtaining \( b = 2 \) (i.e. \( 3(x-2)^2 \)).
* **A1**: For obtaining \( c = -5 \) (fully correct expression \( 3(x-2)^2 - 5 \)).

(b)
* **B1**: For identifying the minimum point of the original curve is \( (2, -5) \) (can be implied by subsequent work).
* **M1**: For applying both transformations correctly to the coordinates of their minimum point (stretch divides the \( x \)-coordinate by 3, translation adds 2 to the \( x \)-coordinate and subtracts 5 from the \( y \)-coordinate).
* **A1**: For obtaining the correct final coordinates \( \left( \frac{8}{3}, -10 \right) \) or equivalent exact decimal/fractional representation.

(a) Starting with the left-hand side (LHS) of the identity, we find a common denominator:
\[ \text{LHS} = \frac{\sin \theta(1 + \sin \theta) - \sin \theta(1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} \]

Expand the numerator and denominator:
\[ \text{LHS} = \frac{\sin \theta + \sin^2 \theta - (\sin \theta - \sin^2 \theta)}{1 - \sin^2 \theta} \]
\[ \text{LHS} = \frac{2\sin^2 \theta}{\cos^2 \theta} \]

Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\[ \text{LHS} = 2\tan^2 \theta = \text{RHS} \]

(b) Using the identity proven in part (a), we can replace the left-hand side of the equation:
\[ 2\tan^2 \theta = \frac{3}{\cos \theta} \]

Rewrite \(\tan^2 \theta\) as \(\frac{\sin^2 \theta}{\cos^2 \theta}\):
\[ 2\left(\frac{\sin^2 \theta}{\cos^2 \theta}\right) = \frac{3}{\cos \theta} \]

Multiply both sides by \(\cos^2 \theta\) (noting that \(\cos \theta \neq 0\)):
\[ 2\sin^2 \theta = 3\cos \theta \]

Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[ 2(1 - \cos^2 \theta) = 3\cos \theta \]
\[ 2 - 2\cos^2 \theta = 3\cos \theta \]
\[ 2\cos^2 \theta + 3\cos \theta - 2 = 0 \]

Factor the quadratic equation:
\[ (2\cos \theta - 1)(\cos \theta + 2) = 0 \]

This gives two possible solutions for \(\cos \theta\):
1) \(\cos \theta = \frac{1}{2}\)
2) \(\cos \theta = -2\) (which has no real solutions since \(-1 \le \cos \theta \le 1\))

For \(\cos \theta = \frac{1}{2}\) in the range \(0^\circ \le \theta \le 360^\circ\):
\[ \theta = 60^\circ \quad \text{or} \quad \theta = 360^\circ - 60^\circ = 300^\circ \]

(a)
* **M1**: For putting the expression over a common denominator.
* **A1**: For obtaining the correct simplified fraction \(\frac{2\sin^2 \theta}{1 - \sin^2 \theta}\) or \(\frac{2\sin^2 \theta}{\cos^2 \theta}\).
* **A1**: For completing the proof with no errors to show \(2\tan^2 \theta\).

(b)
* **M1**: For substituting the identity from part (a) and writing \(\tan^2\theta\) as \(\frac{\sin^2\theta}{\cos^2\theta}\) or equivalent to form an equation in \(\sin\theta\) and \(\cos\theta\).
* **M1**: For using \(\sin^2\theta = 1 - \cos^2\theta\) to form a three-term quadratic in \(\cos\theta\).
* **A1**: For solving the quadratic to find \(\cos\theta = \frac{1}{2}\) and stating that \(\cos\theta = -2\) has no solutions.
* **A1**: For obtaining the correct angles \(\theta = 60^\circ\) and \(\theta = 300^\circ\) (and no others in range).

(a) To find the least value of \( a \) for which \( f \) has an inverse, we locate the vertex of the quadratic function \( f(x) = x^2 - 2x + 5 \). Completing the square: \( f(x) = (x-1)^2 + 4 \). The axis of symmetry is at \( x = 1 \). For \( f \) to have an inverse, it must be a one-to-one function. Since the domain is defined for \( x \ge a \), the least value of \( a \) that restricts the domain to a one-to-one region is \( a = 1 \).

(b) For \( a = 3 \), the domain of \( f \) is \( x \ge 3 \). Since \( x \ge 3 \) lies entirely to the right of the vertex (\( x = 1 \)), the function is strictly increasing. The minimum value of \( f \) occurs at \( x = 3 \), where \( f(3) = (3)^2 - 2(3) + 5 = 8 \). Therefore, the range of \( f \) is \( f(x) \ge 8 \).

(c) To find \( gf(x) \), we substitute \( f(x) \) into \( g(x) \):
\( gf(x) = g(f(x)) = 2(x^2 - 2x + 5) - 3 = 2x^2 - 4x + 10 - 3 = 2x^2 - 4x + 7 \).
The domain of \( gf \) consists of all values in the domain of \( f \) (which is \( x \ge 3 \)) for which the output of \( f \) falls within the domain of \( g \) (which is \( x \ge 1 \)). Since the range of \( f \) is \( f(x) \ge 8 \), and all these values are greater than or equal to \( 1 \), the composition is fully defined. Thus, the domain of \( gf \) is \( x \ge 3 \).

(a)
M1: Attempt to find the vertex of the quadratic by completing the square or using \( x = -b/(2a) \).
A1: State \( a = 1 \).

(b)
B1: State \( f(x) \ge 8 \) (or \( y \ge 8 \) or \( [8, \infty) \)).

(c)
M1: Substitute \( f(x) \) into \( g(x) \).
A1: Obtain correct simplified expression \( 2x^2 - 4x + 7 \).
M1: Explain or show that the range of \( f \) is compatible with the domain of \( g \).
A1: State domain is \( x \ge 3 \) (or \( [3, \infty) \)).

(a) The perimeter \( P \) of a sector is given by:
\[ P = 2r + r\theta \]

Given that the perimeter is constant at \( 40 \) cm, we have:
\[ 2r + r\theta = 40 \implies r\theta = 40 - 2r \]

The area \( A \) of the sector is given by:
\[ A = \frac{1}{2} r^2 \theta = \frac{1}{2} r (r\theta) \]

Substituting \( r\theta = 40 - 2r \) into the area equation:
\[ A = \frac{1}{2} r (40 - 2r) = 20r - r^2 \]

(b) We are given that \( \frac{dr}{dt} = 0.5 \text{ cm s}^{-1} \).

First, we differentiate \( A \) with respect to \( r \):
\[ \frac{dA}{dr} = 20 - 2r \]

Using the chain rule to find the rate of change of the area with respect to time:
\[ \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = (20 - 2r) \times 0.5 = 10 - r \]

To find the value of \( r \) at the instant when \( \theta = 1.2 \) radians, we use the perimeter equation:
\[ 2r + r(1.2) = 40 \implies 3.2r = 40 \implies r = \frac{40}{3.2} = 12.5 \text{ cm} \]

Substituting \( r = 12.5 \) into our expression for \( \frac{dA}{dt} \):
\[ \frac{dA}{dt} = 10 - 12.5 = -2.5 \text{ cm}^2 \text{ s}^{-1} \]

(a)
* **M1**: Attempt to express the perimeter and area in terms of \( r \) and \( \theta \) (e.g., writing \( P = 2r + r\theta \) and \( A = \frac{1}{2}r^2\theta \)).
* **M1**: Eliminate \( \theta \) (or \( r\theta \)) by substituting \( r\theta = 40 - 2r \) into the area formula.
* **A1**: Fully correct derivation leading to \( A = 20r - r^2 \).

(b)
* **M1**: Differentiate \( A = 20r - r^2 \) to obtain \( \frac{dA}{dr} = 20 - 2r \).
* **M1**: Apply the chain rule \( \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} \) using the given \( \frac{dr}{dt} = 0.5 \).
* **M1**: Use \( 2r + 1.2r = 40 \) to find the value of \( r \) when \( \theta = 1.2 \).
* **A1**: Obtain \( r = 12.5 \).
* **A1**: Obtain \( -2.5 \) (or decrease of \( 2.5 \)).

To find the volume of revolution about the \( y \)-axis, we use the formula \( V = \pi \int_{a}^{b} x^2 \, dy \). First, we express \( x^2 \) in terms of \( y \). Given the curve \( y = \frac{1}{2}(x^4 - 1) \), we rearrange for \( x \): \( 2y = x^4 - 1 \implies x^4 = 2y + 1 \). Taking the square root of both sides (since \( x \ge 0 \)), we obtain \( x^2 = \sqrt{2y + 1} = (2y + 1)^{1/2} \). The volume \( V \) is given by integrating with respect to \( y \) from \( y = 0 \) to \( y = 4 \): \( V = \pi \int_{0}^{4} (2y + 1)^{1/2} \, dy \). To perform the integration, we use the reverse chain rule: \( \int (2y + 1)^{1/2} \, dy = \frac{(2y + 1)^{3/2}}{\frac{3}{2} \times 2} = \frac{1}{3}(2y + 1)^{3/2} \). We now evaluate this expression at the limits 4 and 0: at \( y = 4 \), the value is \( \frac{1}{3}(2(4) + 1)^{3/2} = \frac{1}{3}(9)^{3/2} = \frac{27}{3} = 9 \). At \( y = 0 \), the value is \( \frac{1}{3}(2(0) + 1)^{3/2} = \frac{1}{3}(1)^{3/2} = \frac{1}{3} \). Subtracting the lower limit from the upper limit gives \( 9 - \frac{1}{3} = \frac{26}{3} \). Thus, the exact volume of the solid of revolution is \( \frac{26}{3}\pi \).

M1: Attempt to make \( x^4 \) or \( x^2 \) the subject of the formula. A1: Correct expression for \( x^2 \) as \( (2y + 1)^{1/2} \). M1: State or imply the correct volume integral \( \pi \int_{0}^{4} x^2 \, dy \). M1: Attempt to integrate \( (2y + 1)^{1/2} \) to obtain \( k(2y + 1)^{3/2} \). A1: Correct integral of \( \frac{1}{3}(2y + 1)^{3/2} \). M1: Substitute limits of 0 and 4 correctly into their integrated expression. A1: Correct exact volume of \( \frac{26}{3}\pi \) or equivalent exact form (e.g., \( 8\frac{2}{3}\pi \)).

**Part (a)**

The 5th, 8th, and 14th terms of the arithmetic progression can be written as:
\(u_5 = a + 4d\)
\(u_8 = a + 7d\)
\(u_{14} = a + 13d\)

Since these terms form consecutive terms of a geometric progression, we have:
\((a + 7d)^2 = (a + 4d)(a + 13d)\)

Expanding both sides:
\(a^2 + 14ad + 49d^2 = a^2 + 17ad + 52d^2\)

Subtracting \(a^2 + 14ad + 49d^2\) from both sides gives:
\(3ad + 3d^2 = 0\)
\(3d(a + d) = 0\)

Since \(d \neq 0\), we can divide by \(3d\) to obtain:
\(a + d = 0 \implies d = -a\)

We are given that the sum of the first 10 terms is 175:
\(S_{10} = \frac{10}{2}[2a + 9d] = 175\)
\(5[2a + 9d] = 175 \implies 2a + 9d = 35\)

Substituting \(a = -d\) into the equation:
\(2(-d) + 9d = 35\)
\(7d = 35 \implies d = 5\)

Since \(a = -d\), we find:
\(a = -5\)

**Part (b)**

To find the sum of the first 20 terms of this arithmetic progression, we use the formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) with \(n = 20\), \(a = -5\), and \(d = 5\):
\(S_{20} = \frac{20}{2}[2(-5) + (20-1)(5)]\)
\(S_{20} = 10[-10 + 19(5)]\)
\(S_{20} = 10[-10 + 95]\)
\(S_{20} = 10[85] = 850\)

**Part (a)**
* **M1**: For expressing the geometric property, e.g., using \((a + 7d)^2 = (a + 4d)(a + 13d)\) or \(\frac{a + 7d}{a + 4d} = \frac{a + 13d}{a + 7d}\).
* **A1**: For obtaining the correct expanded equation \(a^2 + 14ad + 49d^2 = a^2 + 17ad + 52d^2\) (or equivalent) and simplifying to show \(d = -a\) clearly.
* **M1**: For using \(S_{10} = 175\) to set up the equation \(5(2a + 9d) = 175\) or \(2a + 9d = 35\).
* **A1**: For correct substitution of \(a = -d\) (or \(d = -a\)) to find either \(a\) or \(d\).
* **A1**: For obtaining both \(a = -5\) and \(d = 5\).

**Part (b)**
* **M1**: For using the correct sum formula \(S_{n} = \frac{n}{2}[2a + (n-1)d]\) with \(n = 20\).
* **M1**: For substituting their values of \(a\) and \(d\) into the sum formula.
* **A1**: For obtaining \(850\).

(a) Substitute \(y = 2x - 2\) into the equation of the circle:
\(x^2 + (2x - 2)^2 - 4x + 6(2x - 2) - 12 = 0\)

Expand and simplify the equation:
\(x^2 + (4x^2 - 8x + 4) - 4x + 12x - 12 - 12 = 0\)
\(5x^2 - 20 = 0\)
\(5x^2 = 20 \implies x^2 = 4 \implies x = \pm 2\)

Substitute the \(x\)-values back into the linear equation to find the corresponding \(y\)-values:
For \(x = 2\):
\(y = 2(2) - 2 = 2 \implies A(2, 2)\)

For \(x = -2\):
\(y = 2(-2) - 2 = -6 \implies B(-2, -6)\)

Thus, the coordinates of the points of intersection are \(A(2, 2)\) and \(B(-2, -6)\).

(b) First, find the midpoint \(M\) of the chord \(AB\):
\(M = \left(\frac{2 + (-2)}{2}, \frac{2 + (-6)}{2}\right) = (0, -2)\)

The gradient of the line \(L\) is \(2\), so the gradient of the perpendicular bisector is the negative reciprocal:
\(m = -\frac{1}{2}\)

The equation of the perpendicular bisector is:
\(y - (-2) = -\frac{1}{2}(x - 0) \implies y = -\frac{1}{2}x - 2\)

Now, find the center of the circle \(C\) by completing the square on the original equation:
\(x^2 - 4x + y^2 + 6y - 12 = 0\)
\((x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0\)
\((x - 2)^2 + (y + 3)^2 = 25\)

The center of the circle is \((2, -3)\).

Substitute the center's coordinates into the equation of the perpendicular bisector to verify:
\(y = -\frac{1}{2}(2) - 2 = -1 - 2 = -3\)

Since the equation holds true, the perpendicular bisector passes through the center of the circle.

(a)
**M1**: For substituting \(y = 2x - 2\) into the circle equation.
**A1**: For simplifying correctly to a 3-term quadratic or simpler (e.g., \(5x^2 = 20\)).
**A1**: For obtaining \(x = 2\) and \(x = -2\).
**A1**: For obtaining both correct coordinates \(A(2, 2)\) and \(B(-2, -6)\).

(b)
**M1**: For finding the midpoint \((0, -2)\) and the perpendicular gradient \(-\frac{1}{2}\), and using them to formulate the equation of the line.
**A1**: For the correct equation of the perpendicular bisector: \(y = -\frac{1}{2}x - 2\) (or any equivalent form like \(x + 2y + 4 = 0\)).
**B1**: For correctly finding the circle's center as \((2, -3)\) (by completing the square) and showing that it lies on the perpendicular bisector.

Part (a): At a stationary point, \(\frac{dy}{dx} = 0\). Substituting \(x = 1\) into \(\frac{dy}{dx}\) gives \(6(1)^{1/2} - \frac{k}{1^2} = 0\), which simplifies to \(6 - k = 0\), so \(k = 6\). Part (b): Using \(k = 6\), we have \(\frac{dy}{dx} = 6x^{1/2} - 6x^{-2}\). Integrating this with respect to \(x\) gives \(y = \int (6x^{1/2} - 6x^{-2}) dx = 4x^{3/2} + \frac{6}{x} + C\). Substituting the coordinates of the point \((4, 35)\) into the equation gives \(35 = 4(4)^{3/2} + \frac{6}{4} + C\), which simplifies to \(35 = 32 + 1.5 + C\). Thus, \(C = 1.5\). The equation of the curve is \(y = 4x^{3/2} + \frac{6}{x} + 1.5\). Part (c): Let \(m = \frac{dy}{dx} = 6x^{1/2} - 6x^{-2}\). Differentiating \(m\) with respect to \(x\) gives the rate of change of the gradient with respect to \(x\): \(\frac{dm}{dx} = 3x^{-1/2} + 12x^{-3}\). At \(x = 4\), \(\frac{dm}{dx} = 3(4)^{-1/2} + 12(4)^{-3} = \frac{3}{2} + \frac{12}{64} = 1.5 + 0.1875 = 1.6875\). Using the chain rule, \(\frac{dm}{dt} = \frac{dm}{dx} \times \frac{dx}{dt}\). Given that \(\frac{dx}{dt} = 0.4\), we find \(\frac{dm}{dt} = 1.6875 \times 0.4 = 0.675\) units per second.

Part (a): M1 for setting \(\frac{dy}{dx} = 0\) when \(x = 1\). A1 for obtaining the equation \(6 - k = 0\). A1 for \(k = 6\). Part (b): M1 for integration of \(6x^{1/2} - kx^{-2}\) (at least one term integrated correctly). A1 for \(4x^{3/2}\). A1 for \(\frac{6}{x}\) (or \(6x^{-1}\)). M1 for substituting \(x = 4\) and \(y = 35\) to find the constant of integration \(C\). A1 for the correct final equation \(y = 4x^{3/2} + \frac{6}{x} + 1.5\). Part (c): M1 for differentiating \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\). A1 for finding \(\frac{dm}{dx} = 1.6875\) at \(x = 4\). A1 for correctly applying the chain rule to obtain the final answer \(0.675\).

We use the double-angle identity for cosine: \(\cos(4x) = 1 - 2\sin^2(2x)\), which gives \(\sin^2(2x) = \frac{1}{2}(1 - \cos(4x))\). Integrating this term-by-term: \(\int_{0}^{\frac{\pi}{6}} \sin^2(2x) \, dx = \int_{0}^{\frac{\pi}{6}} \left(\frac{1}{2} - \frac{1}{2}\cos(4x)\right) \, dx = \left[ \frac{1}{2}x - \frac{1}{8}\sin(4x) \right]_{0}^{\frac{\pi}{6}}\). Substituting the upper and lower limits: \(\left( \frac{1}{2}\left(\frac{\pi}{6}\right) - \frac{1}{8}\sin\left(\frac{2\pi}{3}\right) \right) - 0 = \frac{\pi}{12} - \frac{1}{8}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{12} - \frac{\sqrt{3}}{16}\).

M1: Use the correct double-angle identity to express \(\sin^2(2x)\) in the form \(a + b\cos(4x)\). A1: Obtain the correct integration \(\frac{1}{2}x - \frac{1}{8}\sin(4x)\). A1: Obtain the exact value \(\frac{\pi}{12} - \frac{\sqrt{3}}{16}\) (or equivalent exact simplified expression, e.g., \(\frac{4\pi - 3\sqrt{3}}{48}\)).

Let \( y = 2^x \). Note that \( 2^{2x+1} = 2^1 \cdot (2^x)^2 = 2y^2 \). Rewrite the equation in terms of \( y \): \( 2y^2 - 9y + 4 = 0 \). Factoring the quadratic equation gives \( (2y - 1)(y - 4) = 0 \), which has solutions \( y = \frac{1}{2} \) and \( y = 4 \). Substituting back \( y = 2^x \):
1) \( 2^x = \frac{1}{2} \implies 2^x = 2^{-1} \implies x = -1 \)
2) \( 2^x = 4 \implies 2^x = 2^2 \implies x = 2 \)
Thus, the solutions are \( x = -1 \) and \( x = 2 \).

**M1**: Substitute \( y = 2^x \) (or equivalent) to form a quadratic equation of the form \( 2y^2 - 9y + 4 = 0 \).
**A1**: Solve the quadratic equation to find the correct values of \( 2^x \), namely \( 2^x = \frac{1}{2} \) and \( 2^x = 4 \).
**A1**: Obtain both correct final answers \( x = -1 \) and \( x = 2 \).

We can solve the equation \(|2\sin x - 1| = |\sin x|\) by squaring both sides:

\[ (2\sin x - 1)^2 = (\sin x)^2 \]

\[ 4\sin^2 x - 4\sin x + 1 = \sin^2 x \]

\[ 3\sin^2 x - 4\sin x + 1 = 0 \]

Alternatively, we can solve this by considering the two possible linear cases directly:

Case 1:
\[ 2\sin x - 1 = \sin x \implies \sin x = 1 \]

Case 2:
\[ 2\sin x - 1 = -\sin x \implies 3\sin x = 1 \implies \sin x = \frac{1}{3} \]

Now we find the values of \(x\) within the interval \(0^\circ \leq x \leq 180^\circ\):

For \(\sin x = 1\):
\[ x = 90^\circ \]

For \(\sin x = \frac{1}{3}\):
\[ x = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.5^\circ \]
\[ x = 180^\circ - 19.471^\circ \approx 160.5^\circ \]

Thus, the solutions in the interval are \(x = 19.5^\circ\), \(90^\circ\), and \(160.5^\circ\).

M1: For attempting to solve the equation by squaring both sides or by setting up two linear equations.
A1: For obtaining the correct quadratic equation \(3\sin^2 x - 4\sin x + 1 = 0\) or the two correct linear equations \(2\sin x - 1 = \sin x\) and \(2\sin x - 1 = -\sin x\).
A1: For obtaining the correct values \(\sin x = 1\) and \(\sin x = \frac{1}{3}\).
B1: For the solution \(x = 90^\circ\).
B1: For the solution \(x = 19.5^\circ\) (accept 19.47).
B1ft: For the solution \(x = 160.5^\circ\) (accept 160.53, follow through from \(180^\circ - \text{their acute angle}\)).

Apply the compound angle expansion formulae:
\[\sin(A + B) =
\sin A \cos B + \cos A \sin B\]
\[\cos(A + B) = \cos A \cos B - \sin A \sin B\]

Expanding both sides of the given equation:
\[\sin(\theta + 60^\circ) = \sin\theta \cos 60^\circ + \cos\theta \sin 60^\circ = \frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta\]
\[2 \cos(\theta + 30^\circ) = 2(\cos\theta \cos 30^\circ - \sin\theta \sin 30^\circ) = 2\left(\frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta\right) = \sqrt{3}\cos\theta - \sin\theta\]

Now, equate the expanded forms:
\[\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta = \sqrt{3}\cos\theta - \sin\theta\]

Rearrange to group the \(\sin\theta\) terms on one side and the \(\cos\theta\) terms on the other:
\[\frac{1}{2}\sin\theta + \sin\theta = \sqrt{3}\cos\theta - \frac{\sqrt{3}}{2}\cos\theta\]
\[\frac{3}{2}\sin\theta = \frac{\sqrt{3}}{2}\cos\theta\]

Multiply by 2 and divide by \(\cos\theta\) (where \(\cos\theta \ne 0\)):
\[3\tan\theta = \sqrt{3}\]
\[\tan\theta = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}\]

For the interval \(0^\circ \le \theta \le 360^\circ\), the solutions are:
\[\theta = 30^\circ\]
\[\theta = 180^\circ + 30^\circ = 210^\circ\]

M1: State or imply compound angle formula expansions for either \(\sin(\theta + 60^\circ)\) or \(\cos(\theta + 30^\circ)\).
A1: Obtain correct expanded equation: \(\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta = \sqrt{3}\cos\theta - \sin\theta\) (or equivalent).
M1: Group terms and use \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) to obtain an equation in \(\tan\theta\).
A1: Obtain \(\tan\theta = \frac{1}{\sqrt{3}}\) (or equivalent exact form).
A1: Obtain both \(\theta = 30^\circ\) and \(\theta = 210^\circ\), and no other solutions in the range.

(i) At the point \(P\), \(y = 0\).

\[e^{2x} - 3e^x = 0\]

Since \(e^x \neq 0\) for all real \(x\), we can divide both sides by \(e^x\):

\[e^x - 3 = 0 \implies e^x = 3\]

Taking the natural logarithm of both sides gives the exact x-coordinate of \(P\):

\[x = \ln 3\]

(ii) The region is bounded by the y-axis (where \(x = 0\)), the curve, and the x-axis up to \(P\) (where \(x = \ln 3\)).

For \(0 \le x < \ln 3\), the curve lies below the x-axis because \(e^x < 3\), which implies \(e^{2x} - 3e^x < 0\). Therefore, the area \(A\) of this region is given by:

\[A = \int_{0}^{\ln 3} -y \, dx = \int_{0}^{\ln 3} (3e^x - e^{2x}) \, dx\]

Integrating term by term:

\[\int (3e^x - e^{2x}) \, dx = 3e^x - \frac{1}{2}e^{2x}\]

Now, substitute the upper and lower limits:

\[A = \left[ 3e^x - \frac{1}{2}e^{2x} \right]_{0}^{\ln 3}\]

Substituting the upper limit \(x = \ln 3\):

\[3e^{\ln 3} - \frac{1}{2}e^{2\ln 3} = 3(3) - \frac{1}{2}(3^2) = 9 - \frac{9}{2} = \frac{9}{2}\]

Substituting the lower limit \(x = 0\):

\[3e^0 - \frac{1}{2}e^0 = 3 - \frac{1}{2} = \frac{5}{2}\]

Subtracting the lower limit value from the upper limit value:

\[A = \frac{9}{2} - \frac{5}{2} = 2\]

Thus, the exact area of the region is \(2\).

Part (i):
M1: Set \(y = 0\) and attempt to solve the equation for \(e^x\).
A1: Obtain the correct exact value \(x = \ln 3\).

Part (ii):
M1: State a correct integral for the area, with or without limits, using the correct sign (either \(\int (3e^x - e^{2x}) \, dx\) or absolute value of \(\int (e^{2x} - 3e^x) \, dx\)).
A1: Integrate at least one term correctly (obtain \(3e^x\) or \(-\frac{1}{2}e^{2x}\)).
A1: Fully correct integrated expression \(3e^x - \frac{1}{2}e^{2x}\) (or signs reversed if working with a negative area initially).
M1: Substitute limits \(0\) and their \(\ln 3\) from part (i) into their integrated expression.
A1: Obtain the final exact area of \(2\) with clear, correct working showing both limit evaluations.

(i) Differentiating both \(x\) and \(y\) with respect to \(t\):
\[ \frac{\text{d}x}{\text{d}t} = e^t - 2 \]
\[ \frac{\text{d}y}{\text{d}t} = 2e^{2t} - 5e^t \]
Using the chain rule, we obtain:
\[ \frac{\text{d}y}{\text{d}x} = \frac{\frac{\text{d}y}{\text{d}t}}{\frac{\text{d}x}{\text{d}t}} = \frac{2e^{2t} - 5e^t}{e^t - 2} \]

(ii) When \(t = 0\):
\[ x = e^0 - 2(0) = 1 \]
\[ y = e^0 - 5e^0 = -4 \]
The coordinates of the point are \((1, -4)\).

The gradient of the tangent at \(t = 0\) is:
\[ m_{\text{tangent}} = \frac{2(1) - 5(1)}{1 - 2} = \frac{-3}{-1} = 3 \]

The gradient of the normal is the negative reciprocal of the gradient of the tangent:
\[ m_{\text{normal}} = -\frac{1}{3} \]

Using the point-gradient form, the equation of the normal is:
\[ y - (-4) = -\frac{1}{3}(x - 1) \]
\[ 3(y + 4) = -(x - 1) \]
\[ 3y + 12 = -x + 1 \]
\[ x + 3y + 11 = 0 \]

(iii) Since the tangent is parallel to \(y = 3x - 5\), the gradient of the tangent is equal to 3:
\[ \frac{2e^{2t} - 5e^t}{e^t - 2} = 3 \]
\[ 2e^{2t} - 5e^t = 3(e^t - 2) \]
\[ 2e^{2t} - 5e^t = 3e^t - 6 \]
\[ 2e^{2t} - 8e^t + 6 = 0 \]
\[ e^{2t} - 4e^t + 3 = 0 \]

Factorising this quadratic equation in terms of \(e^t\):
\[ (e^t - 3)(e^t - 1) = 0 \]

Since \(t > 0\), we must have \(e^t = 3\) (as \(e^t = 1 \implies t = 0\)).
Taking the natural logarithm of both sides gives:
\[ t = \ln 3 \]

Substituting \(e^t = 3\) and \(t = \ln 3\) into the expression for \(x\):
\[ x = e^t - 2t = 3 - 2\ln 3 \]

(i)
M1: Attempt to differentiate \(x\) and \(y\) with respect to \(t\) (at least one derivative correct).
M1: Apply the parametric differentiation formula \(\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y/\text{d}t}{\text{d}x/\text{d}t}\).
A1: Obtain correct derivative \(\frac{2e^{2t} - 5e^t}{e^t - 2}\) or equivalent.

(ii)
B1: Determine the point \((1, -4)\) and show that the gradient of the tangent is 3.
M1: Calculate the gradient of the normal as \(-\frac{1}{3}\) and attempt to form the equation of the line passing through \((1, -4)\).
A1: Obtain \(x + 3y + 11 = 0\) (or any integer multiple thereof).

(iii)
M1: Equate their expression for \(\frac{\text{d}y}{\text{d}x}\) to 3 and reduce to a three-term quadratic equation in \(e^t\).
A1: Solve the quadratic to find \(e^t = 3\) (showing rejection of \(e^t = 1\) due to the condition \(t > 0\)).
A1: Substitute \(e^t = 3\) and \(t = \ln 3\) to obtain the exact coordinate \(x = 3 - 2\ln 3\) (or \(3 - \ln 9\)).

(i) Since \( (x - 2) \) is a factor, we have \( \mathrm{p}(2) = 0 \):
\( a(2)^3 + b(2)^2 - 14(2) + 8 = 0 \)
\( 8a + 4b - 28 + 8 = 0 \implies 8a + 4b = 20 \implies 2a + b = 5 \) (Equation 1)

Since dividing by \( (x + 1) \) gives a remainder of 24, we have \( \mathrm{p}(-1) = 24 \):
\( a(-1)^3 + b(-1)^2 - 14(-1) + 8 = 24 \)
\( -a + b + 14 + 8 = 24 \implies -a + b = 2 \) (Equation 2)

Subtracting Equation 2 from Equation 1:
\( (2a + b) - (-a + b) = 5 - 2 \implies 3a = 3 \implies a = 1 \)

Substituting \( a = 1 \) into Equation 2:
\( -1 + b = 2 \implies b = 3 \).

(ii) Using the values \( a = 1 \) and \( b = 3 \), the polynomial is \( \mathrm{p}(x) = x^3 + 3x^2 - 14x + 8 \).
Evaluate \( \mathrm{p}(x) \) at the boundaries:
\( \mathrm{p}(0.5) = (0.5)^3 + 3(0.5)^2 - 14(0.5) + 8 = 0.125 + 0.75 - 7 + 8 = 1.875 \)
\( \mathrm{p}(1.0) = (1.0)^3 + 3(1.0)^2 - 14(1.0) + 8 = 1 + 3 - 14 + 8 = -2 \)

Since there is a change of sign between \( \mathrm{p}(0.5) > 0 \) and \( \mathrm{p}(1.0) < 0 \), and \( \mathrm{p}(x) \) is a continuous function, there must be a root of \( \mathrm{p}(x) = 0 \) in the interval \( [0.5, 1.0] \).

(iii) Set \( \mathrm{p}(x) = 0 \):
\( x^3 + 3x^2 - 14x + 8 = 0 \)
\( 14x = x^3 + 3x^2 + 8 \)
\( x = \frac{x^3 + 3x^2 + 8}{14} \).

(iv) The iterative formula is:
\( x_{n+1} = \frac{x_n^3 + 3x_n^2 + 8}{14} \)

Using \( x_1 = 0.7 \):
\( x_2 = \frac{0.7^3 + 3(0.7)^2 + 8}{14} = 0.70093 \)
\( x_3 = \frac{(0.70093)^3 + 3(0.70093)^2 + 8}{14} = 0.70131 \)
\( x_4 = \frac{(0.70131)^3 + 3(0.70131)^2 + 8}{14} = 0.70146 \)
\( x_5 = \frac{(0.70146)^3 + 3(0.70146)^2 + 8}{14} = 0.70152 \)
\( x_6 = \frac{(0.70152)^3 + 3(0.70152)^2 + 8}{14} = 0.70155 \)

Both \( x_5 \) and \( x_6 \) round to \( 0.702 \) to 3 decimal places. Thus, the root is \( 0.702 \) correct to 3 decimal places.

(i)
M1: Attempt to use the factor theorem with \( p(2) = 0 \) to obtain a linear equation in \( a \) and \( b \).
A1: Obtain correct equation \( 2a + b = 5 \) (or equivalent).
M1: Attempt to use the remainder theorem with \( p(-1) = 24 \) to obtain a second equation in \( a \) and \( b \).
A1: Solve simultaneously to obtain \( a = 1 \) and \( b = 3 \).

(ii)
M1: Attempt to evaluate both \( \mathrm{p}(0.5) \) and \( \mathrm{p}(1.0) \) with their polynomial.
A1: Obtain \( 1.875 \) (or \( \frac{15}{8} \)) and \( -2 \), and make a conclusion mentioning the sign change.

(iii)
B1: Show clear algebraic steps to derive the rearrangement.

(iv)
M1: Substitute \( x_1 = 0.7 \) into the iterative formula to calculate \( x_2 \).
A1: Obtain \( x_2 = 0.70093 \) and \( x_3 = 0.70131 \) (or equivalent 5 d.p. values).
A1: Obtain further iterations \( x_4 = 0.70146 \), \( x_5 = 0.70152 \), etc.
A1: Conclude with the correct root of \( 0.702 \) to 3 d.p.

To find the stationary points, we first differentiate \(y = (x^2 - 2x - 1)\mathrm{e}^{2x}\) with respect to \(x\) using the product rule:

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

Then:
\(\frac{du}{dx} = 2x - 2\)
\(\frac{dv}{dx} = 2\mathrm{e}^{2x}\)

Applying the product rule:
\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\)
\(\frac{dy}{dx} = (x^2 - 2x - 1)(2\mathrm{e}^{2x}) + (2x - 2)\mathrm{e}^{2x}\)
\(\frac{dy}{dx} = \mathrm{e}^{2x} [ 2(x^2 - 2x - 1) + 2x - 2 ]\)
\(\frac{dy}{dx} = \mathrm{e}^{2x} [ 2x^2 - 4x - 2 + 2x - 2 ]\)
\(\frac{dy}{dx} = (2x^2 - 2x - 4)\mathrm{e}^{2x}\)

At stationary points, \(\frac{dy}{dx} = 0\):
Since \(\mathrm{e}^{2x} \neq 0\) for all real values of \(x\), we solve:
\(2x^2 - 2x - 4 = 0\)
\(2(x^2 - x - 2) = 0\)
\(2(x - 2)(x + 1) = 0\)

This gives \(x = 2\) and \(x = -1\).

Now find the corresponding exact \(y\)-coordinates:
- For \(x = 2\):
\(y = (2^2 - 2(2) - 1)\mathrm{e}^{2(2)} = (4 - 4 - 1)\mathrm{e}^4 = -\mathrm{e}^4\)
So, one stationary point is \((2, -\mathrm{e}^4)\).

- For \(x = -1\):
\(y = ((-1)^2 - 2(-1) - 1)\mathrm{e}^{2(-1)} = (1 + 2 - 1)\mathrm{e}^{-2} = 2\mathrm{e}^{-2}\)
So, the other stationary point is \((-1, 2\mathrm{e}^{-2})\).

To determine the nature of these stationary points, we find the second derivative \(\frac{d^2 y}{dx^2}\) using the product rule on \(\frac{dy}{dx} = (2x^2 - 2x - 4)\mathrm{e}^{2x}\):

Let \(U = 2x^2 - 2x - 4\) and \(V = \mathrm{e}^{2x}\).

Then:
\(\frac{dU}{dx} = 4x - 2\)
\(\frac{dV}{dx} = 2\mathrm{e}^{2x}\)

\(\frac{d^2 y}{dx^2} = (4x - 2)\mathrm{e}^{2x} + (2x^2 - 2x - 4)(2\mathrm{e}^{2x})\)
\(\frac{d^2 y}{dx^2} = \mathrm{e}^{2x} [ 4x - 2 + 4x^2 - 4x - 8 ]\)
\(\frac{d^2 y}{dx^2} = (4x^2 - 10)\mathrm{e}^{2x}\)

Now substitute the \(x\)-values into the second derivative expression:
- At \(x = 2\):
\(\frac{d^2 y}{dx^2} = (4(2)^2 - 10)\mathrm{e}^4 = (16 - 10)\mathrm{e}^4 = 6\mathrm{e}^4 > 0\)
Since the second derivative is positive, \((2, -\mathrm{e}^4)\) is a **local minimum**.

- At \(x = -1\):
\(\frac{d^2 y}{dx^2} = (4(-1)^2 - 10)\mathrm{e}^{-2} = (4 - 10)\mathrm{e}^{-2} = -6\mathrm{e}^{-2} < 0\)
Since the second derivative is negative, \((-1, 2\mathrm{e}^{-2})\) is a **local maximum**.

M1: Apply the product rule to differentiate \(y\). Must see correct form \(u'v + uv'\).
A1: Obtain correct derivative \(\frac{dy}{dx} = (2x^2 - 2x - 4)\mathrm{e}^{2x}\) (or equivalent form).
M1: Set \(\frac{dy}{dx} = 0\) and solve the resulting quadratic equation to find the two \(x\)-values.
A1: Obtain correct exact coordinates for both stationary points: \((2, -\mathrm{e}^4)\) and \((-1, 2\mathrm{e}^{-2})\) (accept \((-1, \frac{2}{\mathrm{e}^2})\)). Both must be in exact form.
M1: Attempt to find the second derivative \(\frac{d^2 y}{dx^2}\) (or use a first derivative sign table) and substitute at least one of their \(x\)-values.
A1: Correctly conclude that \((2, -\mathrm{e}^4)\) is a local minimum and \((-1, 2\mathrm{e}^{-2})\) is a local maximum, supported by correct signs of the second derivative (or correct signs in the table).