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We equate the two equations to find points of intersection: \( x^2 - 4x - 1 = kx - 5 \). Rearranging gives the quadratic equation \( x^2 - (4+k)x + 4 = 0 \). For the line and curve not to intersect, this quadratic equation must have no real roots, so its discriminant must be less than zero: \( B^2 - 4AC < 0 \). This gives \( (-(4+k))^2 - 4(1)(4) < 0 \), which simplifies to \( (4+k)^2 - 16 < 0 \). Solving the inequality, we get \( (4+k)^2 < 16 \), so \( -4 < 4+k < 4 \). Subtracting 4 from all parts gives the final range: \( -8 < k < 0 \).

M1 for equating equations and rearranging into a standard quadratic form. M1 for setting the discriminant less than zero and attempting to solve the quadratic inequality. A1 for the correct final range -8 < k < 0.

To find the inverse function, we first complete the square for \( \text{f}(x) \): \( \text{f}(x) = 2(x^2 - 6x) + 7 = 2[(x-3)^2 - 9] + 7 = 2(x-3)^2 - 11 \). We then set \( y = 2(x-3)^2 - 11 \) and rearrange to make \( x \) the subject: \( y + 11 = 2(x-3)^2 \) which gives \( (x-3)^2 = \frac{y+11}{2} \). Taking the square root, we get \( x - 3 = \pm\sqrt{\frac{y+11}{2}} \). Since the domain of \( \text{f} \) is \( x \le 3 \), we must choose the negative square root: \( x = 3 - \sqrt{\frac{y+11}{2}} \). Replacing \( y \) with \( x \), we obtain \( \text{f}^{-1}(x) = 3 - \sqrt{\frac{x+11}{2}} \).

M1 for expressing the quadratic function in completed square form. M1 for rearranging to express x in terms of y and correctly choosing the negative square root due to the domain restriction. A1 for the correct final expression for the inverse function.

First, find the midpoint \( M \) of the segment \( AB \): \( M = \left(\frac{2+8}{2}, \frac{-3+5}{2}\right) = (5, 1) \). Next, find the gradient of the line \( AB \): \( m_{AB} = \frac{5 - (-3)}{8 - 2} = \frac{8}{6} = \frac{4}{3} \). The gradient of the perpendicular bisector is the negative reciprocal: \( m = -\frac{3}{4} \). Using the point-slope form with point \( M(5, 1) \): \( y - 1 = -\frac{3}{4}(x - 5) \). Multiplying by 4 gives \( 4y - 4 = -3x + 15 \), which rearranges to \( 3x + 4y = 19 \).

M1 for finding the correct midpoint and gradient of the line segment AB. M1 for using the negative reciprocal gradient to set up the linear equation. A1 for the correct equation in the specified integer form.

The perimeter of the sector is given by \( P = 2r + r\theta = 24 \), so \( r\theta = 24 - 2r \). The area of the sector is given by \( A = \frac{1}{2}r^2\theta = 32 \). Substituting the expression for \( r\theta \) into the area equation: \( \frac{1}{2}r(r\theta) = 32 \implies \frac{1}{2}r(24 - 2r) = 32 \). This simplifies to \( 12r - r^2 = 32 \), or \( r^2 - 12r + 32 = 0 \). Factoring the quadratic equation gives \( (r-4)(r-8) = 0 \). Thus, the possible values for the radius are \( r = 4 \) and \( r = 8 \).

M1 for writing down equations for perimeter and area in terms of r and theta. M1 for eliminating theta to obtain a quadratic equation in r. A1 for solving the quadratic correctly to find both values of r.

Let the first term be \( a \) and the common ratio be \( r \). The second term is \( ar = 6 \), which gives \( r = \frac{6}{a} \). The sum to infinity is \( S_{\infty} = \frac{a}{1-r} = 25 \). Substituting \( r = \frac{6}{a} \) into the sum formula: \( \frac{a}{1 - 6/a} = 25 \). Multiplying the numerator and denominator of the left side by \( a \) gives \( \frac{a^2}{a-6} = 25 \). Rearranging this equation gives \( a^2 = 25a - 150 \), or \( a^2 - 25a + 150 = 0 \). Factoring the quadratic gives \( (a-10)(a-15) = 0 \). Thus, the possible values of the first term are \( a = 10 \) and \( a = 15 \).

M1 for using the formula for the nth term of a GP to write r in terms of a, and using the sum to infinity formula. M1 for substituting and setting up a quadratic equation in a. A1 for finding both correct values of a.

To find the stationary point, we first differentiate the equation of the curve with respect to \( x \): \( y = 16x^{-1} + x^2 \implies \frac{\text{d}y}{\text{d}x} = -16x^{-2} + 2x = -\frac{16}{x^2} + 2x \). At a stationary point, \( \frac{\text{d}y}{\text{d}x} = 0 \), so \( -\frac{16}{x^2} + 2x = 0 \). This gives \( 2x = \frac{16}{x^2} \), which simplifies to \( x^3 = 8 \), so \( x = 2 \). Substituting \( x = 2 \) back into the equation of the curve to find the y-coordinate: \( y = \frac{16}{2} + 2^2 = 8 + 4 = 12 \). Therefore, the coordinates of the stationary point are \( (2, 12) \).

M1 for differentiating the curve's equation correctly. M1 for setting dy/dx = 0 and solving for x. A1 for obtaining the correct coordinates (2, 12).

Let the first term be \(a\) and the common difference be \(d\).

1. Using the sum formula \(S_n = \frac{n}{2}(2a + (n-1)d)\):
\(S_{10} = \frac{10}{2}(2a + 9d) = 175\)
\(5(2a + 9d) = 175\)
\(2a + 9d = 35\) --- (Equation 1)

2. Using the general term formula \(u_n = a + (n-1)d\):
The 9th term is \(u_9 = a + 8d\).
The 3rd term is \(u_3 = a + 2d\).
Since the 9th term is 8 more than twice the 3rd term:
\(a + 8d = 2(a + 2d) + 8\)
\(a + 8d = 2a + 4d + 8\)
\(a - 4d = -8\) --- (Equation 2)

3. Solving Equations 1 and 2 simultaneously:
From Equation 2, we have \(a = 4d - 8\).
Substitute this into Equation 1:
\(2(4d - 8) + 9d = 35\)
\(8d - 16 + 9d = 35\)
\(17d = 51\)
\(d = 3\)

Now find \(a\):
\(a = 4(3) - 8 = 4\).

Thus, the first term is 4 and the common difference is 3.

M1: Forms a correct linear equation in \(a\) and \(d\) using the sum formula.
M1: Forms a second correct linear equation in \(a\) and \(d\) using the term formula.
M1: Solves the two simultaneous equations to find at least one variable.
A1: Obtains \(a = 4\) and \(d = 3\) correctly.

First, write the equation of the curve as:
\(y = k(2x - 3)^{-1} + 4x\)

Differentiate with respect to \(x\) using the chain rule:
\(\frac{dy}{dx} = -k(2x - 3)^{-2} \cdot 2 + 4 = -\frac{2k}{(2x - 3)^2} + 4\)

At a stationary point, \(\frac{dy}{dx} = 0\). Given this occurs at \(x = 2\):
\(-\frac{2k}{(2(2) - 3)^2} + 4 = 0\)
\(-\frac{2k}{1^2} + 4 = 0\)
\(2k = 4 \implies k = 2\)

To determine the nature of the stationary point, find the second derivative:
\(\frac{d^2y}{dx^2} = \frac{d}{dx}\left( -2k(2x - 3)^{-2} + 4 \right) = 4k(2x - 3)^{-3} \cdot 2 = \frac{8k}{(2x - 3)^3}\)

Substitute \(k = 2\) and \(x = 2\):
\(\frac{d^2y}{dx^2} = \frac{8(2)}{(2(2) - 3)^3} = 16\)

Since \(\frac{d^2y}{dx^2} > 0\) at \(x = 2\), the stationary point is a minimum.

M1: Differentiates \(y\) correctly, showing the chain rule applied to the first term.
A1: Equates their derivative to 0 at \(x = 2\) and finds \(k = 2\).
M1: Finds the second derivative \(\frac{d^2y}{dx^2}\) and evaluates it at \(x = 2\).
A1: Correctly identifies the stationary point as a minimum with a valid reason (e.g., \(\frac{d^2y}{dx^2} = 16 > 0\)).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to rewrite the equation in terms of \(\cos \theta\):
\(2(1 - \cos^2 \theta) = 7 \cos \theta + 5\)
\(2 - 2 \cos^2 \theta = 7 \cos \theta + 5\)
\(2 \cos^2 \theta + 7 \cos \theta + 3 = 0\)

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

This gives two possible equations:
1. \(\cos \theta = -\frac{1}{2}\)
2. \(\cos \theta = -3\) (which has no real solutions since \(-1 \le \cos \theta \le 1\))

For \(\cos \theta = -\frac{1}{2}\):
The basic angle is \(\cos^{-1}(0.5) = 60^\circ\).
Since cosine is negative, the solutions lie in the second and third quadrants:
\(\theta = 180^\circ - 60^\circ = 120^\circ\)
\(\theta = 180^\circ + 60^\circ = 240^\circ\)

Both values lie within the range \(0^\circ \le \theta \le 360^\circ\).

M1: Substitutes \(\sin^2 \theta = 1 - \cos^2 \theta\) to form a quadratic equation in \(\cos \theta\).
A1: Factorises or solves the quadratic equation to obtain \(\cos \theta = -0.5\) (and optionally rejects the other root).
M1: Finds one correct angle in the interval, or shows the correct quadrant method for a negative cosine value.
A1: Obtains both \(\theta = 120^\circ\) and \(\theta = 240^\circ\) and no other values in the range.

(a) Express \(\mathrm{f}(x)\) in completed square form:
\(\mathrm{f}(x) = 2(x^2 - 6x) + 11 = 2[(x - 3)^2 - 9] + 11 = 2(x - 3)^2 - 7\)

The vertex of the parabola is at \((3, -7)\). For the quadratic function to be one-to-one on the domain \(x \le p\), we must restrict the domain to only the left half of the symmetry line \(x = 3\). Therefore, the greatest value of \(p\) is \(3\).

(b) Let \(y = 2(x - 3)^2 - 7\) with domain \(x \le 3\).
Rearrange to make \(x\) the subject:
\(y + 7 = 2(x - 3)^2\)
\(\frac{y + 7}{2} = (x - 3)^2\)

Since \(x \le 3\), the quantity \(x - 3\) is non-positive, so we take the negative square root:
\(x - 3 = -\sqrt{\frac{y + 7}{2}}\)
\(x = 3 - \sqrt{\frac{y + 7}{2}}\)

Replacing \(y\) with \(x\), we get:
\(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 7}{2}}\)

B1: States \(p = 3\) (can be earned by showing completed square form or \(x = -b/(2a) = 3\)).
M1: Sets \(y = \mathrm{f}(x)\) using their completed square form and attempts to make \((x - 3)^2\) or \(x\) the subject.
M1: Selects the negative square root correctly due to the domain condition \(x \le 3\).
A1: Obtains the correct inverse function \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 7}{2}}\) written in terms of \(x\).

First, find the gradient of the radius line \(CA\):
\(m_{CA} = \frac{1 - (-2)}{7 - 3} = \frac{3}{4}\)

Since the tangent at \(A\) is perpendicular to the radius \(CA\), the gradient of the tangent is:
\(m_{\text{tangent}} = -\frac{1}{m_{CA}} = -\frac{4}{3}\)

Now, use the point-slope formula for the equation of the tangent passing through \(A(7, 1)\):
\(y - 1 = -\frac{4}{3}(x - 7)\)

Multiply through by 3 to eliminate the fraction:
\(3(y - 1) = -4(x - 7)\)
\(3y - 3 = -4x + 28\)
\(3y + 4x = 31\)

This is in the required form \(ay + bx = c\) where \(a = 3\), \(b = 4\), and \(c = 31\).

M1: Correctly calculates the gradient of the radius \(CA\).
M1: Finds the negative reciprocal of their gradient to get the gradient of the tangent.
M1: Sets up the equation of a straight line through \((7, 1)\) using their tangent gradient.
A1: Simplifies the equation to the correct integer form \(3y + 4x = 31\) (or any integer multiple).

To find the equation of the curve, integrate \(\frac{dy}{dx}\):
\(y = \int 6(3x + 1)^{-1/2} \, dx\)

Using the reverse chain rule for integration:
\(y = 6 \cdot \frac{(3x + 1)^{1/2}}{\frac{1}{2} \cdot 3} + C\)
\(y = 6 \cdot \frac{\sqrt{3x + 1}}{\frac{3}{2}} + C\)
\(y = 4\sqrt{3x + 1} + C\)

Now, find the constant of integration \(C\) by substituting the coordinates of the point \((5, 12)\):
\(12 = 4\sqrt{3(5) + 1} + C\)
\(12 = 4\sqrt{16} + C\)
\(12 = 16 + C \implies C = -4\)

Thus, the equation of the curve is:
\(y = 4\sqrt{3x + 1} - 4\)

M1: Integrates to find \(y\), showing correct raising of power by 1 to \(\frac{1}{2}\).
A1: Obtains the correct integrated expression \(4\sqrt{3x + 1}\) (must correctly divide by the inner derivative 3).
M1: Substitutes the point \((5, 12)\) into their integrated expression containing \(C\).
A1: Obtains the correct final equation \(y = 4\sqrt{3x + 1} - 4\).

Let \(r\) be the radius and \(\theta\) be the sector angle in radians.

1. The perimeter of the sector is given by:
\(P = 2r + r\theta = 30 \implies r\theta = 30 - 2r \implies \theta = \frac{30 - 2r}{r}\)

2. The area of the sector is given by:
\(A = \frac{1}{2}r^2\theta = 50\)

3. Substitute the expression for \(\theta\) into the area equation:
\(\frac{1}{2}r^2 \left( \frac{30 - 2r}{r} \right) = 50\)
\(\frac{1}{2}r(30 - 2r) = 50\)
\(15r - r^2 = 50\)
\(r^2 - 15r + 50 = 0\)

4. Factorise the quadratic equation:
\((r - 5)(r - 10) = 0\)

This gives two possible values of \(r\):
\(r = 5\) or \(r = 10\)

B1: Writes down the correct equations for the perimeter \(2r + r\theta = 30\) and area \(\frac{1}{2}r^2\theta = 50\).
M1: Eliminates \(\theta\) to obtain a quadratic equation in terms of \(r\).
M1: Solves the resulting quadratic equation by factorisation or formula.
A1: Obtains both correct values \(r = 5\) and \(r = 10\).

For the line and curve not to intersect, the equation formed by equating them must have no real roots:
\(x^2 - 4x + 6 = kx - 3\)
\(x^2 - (4 + k)x + 9 = 0\)

This is a quadratic equation in \(x\) with coefficients:
\(a = 1\)
\(b = -(4 + k)\)
\(c = 9\)

For no intersection, the discriminant must be strictly less than zero (\(b^2 - 4ac < 0\)):
\((-(4 + k))^2 - 4(1)(9) < 0\)
\((k + 4)^2 - 36 < 0\)
\(k^2 + 8k + 16 - 36 < 0\)
\(k^2 + 8k - 20 < 0\)

Factorise the critical quadratic equation:
\((k + 10)(k - 2) < 0\)

The critical values are \(k = -10\) and \(k = 2\).
For the product to be negative, \(k\) must lie between these two values:
\(-10 < k < 2\)

M1: Equates the equations and rearranges into standard form \(ax^2 + bx + c = 0\).
M1: Uses the discriminant condition \(b^2 - 4ac < 0\) to set up an inequality in terms of \(k\).
M1: Solves the critical equation \(k^2 + 8k - 20 = 0\) to find the key boundary values.
A1: States the correct range \(-10 < k < 2\) (using strict inequalities).

To find the points of intersection, set the equations equal to each other: \(x^2 - 4x + 7 = 2x + k\). Rearranging gives the quadratic equation \(x^2 - 6x + (7 - k) = 0\). For the line and curve not to intersect, the discriminant must be strictly less than zero: \(b^2 - 4ac < 0\). Here, \(a = 1\), \(b = -6\), and \(c = 7 - k\). Substituting these values into the discriminant inequality gives \((-6)^2 - 4(1)(7 - k) < 0\), which simplifies to \(36 - 28 + 4k < 0\). This simplifies further to \(8 + 4k < 0\), leading to \(4k < -8\), and thus \(k < -2\).

M1: Set the equations equal to each other and attempt to form a quadratic equation in \(x\) of the form \(ax^2 + bx + c = 0\). M1: Use the discriminant condition \(b^2 - 4ac < 0\) for no intersection. A1: Obtain a correct simplified inequality in terms of \(k\), such as \(8 + 4k < 0\). A1: Correct final answer \(k < -2\).

First, use the trigonometric identity \(\cos^2\theta = 1 - \sin^2\theta\) to rewrite the equation in terms of \(\sin\theta\): \(3(1 - \sin^2\theta) + 5\sin\theta - 1 = 0\). This simplifies to \(3 - 3\sin^2\theta + 5\sin\theta - 1 = 0\), which rearranges to \(3\sin^2\theta - 5\sin\theta - 2 = 0\). Factoring this quadratic equation gives \((3\sin\theta + 1)(\sin\theta - 2) = 0\). This yields two possible cases: \(\sin\theta = 2\) (which has no real solutions since \(-1 \le \sin\theta \le 1\)) or \(\sin\theta = -\frac{1}{3}\). For \(\sin\theta = -\frac{1}{3}\), the basic angle is \(\alpha = \sin^{-1}(\frac{1}{3}) \approx 19.47^\circ\). Since sine is negative, \(\theta\) lies in the third and fourth quadrants: \(\theta = 180^\circ + 19.47^\circ = 199.5^\circ\) (to 1 d.p.) and \(\theta = 360^\circ - 19.47^\circ = 340.5^\circ\) (to 1 d.p.).

M1: Apply the identity \(\cos^2\theta = 1 - \sin^2\theta\) to obtain a quadratic equation in terms of \(\sin\theta\). M1: Solve the quadratic equation to find \(\sin\theta = -\frac{1}{3}\) (and reject \(\sin\theta = 2\)). A1: Identify the basic angle \(\approx 19.5^\circ\) and find one correct value of \(\theta\) (either \(199.5^\circ\) or \(340.5^\circ\)). A1: Find the second correct value of \(\theta\) and no other values in the range.

The first term is \(u_1 = a\) and the second term is \(u_2 = a - 4\). The common ratio is \(r = \frac{u_2}{u_1} = \frac{a - 4}{a}\). The formula for the sum to infinity is \(S_\infty = \frac{a}{1 - r}\). Substituting the given values: \(9 = \frac{a}{1 - \frac{a-4}{a}}\). Simplify the denominator: \(1 - \frac{a-4}{a} = \frac{a - (a-4)}{a} = \frac{4}{a}\). Therefore, \(9 = \frac{a}{\frac{4}{a}} = \frac{a^2}{4}\). This gives \(a^2 = 36\), so \(a = 6\) or \(a = -6\). We must check the convergence condition \(|r| < 1\). If \(a = 6\), then \(r = \frac{6-4}{6} = \frac{1}{3}\), which satisfies \(|r| < 1\). If \(a = -6\), then \(r = \frac{-6-4}{-6} = \frac{5}{3}\), which does not satisfy \(|r| < 1\). Thus, only \(a = 6\) is a valid solution.

M1: Write an expression for the common ratio \(r\) in terms of \(a\) and substitute into the sum to infinity formula. M1: Set up and solve the resulting equation to find \(a^2 = 36\). A1: Find the two possible values \(a = 6\) or \(a = -6\). A1: Correctly reject \(a = -6\) by considering the convergence condition \(|r| < 1\) and state \(a = 6\) as the final answer.

To find \(\mathrm{f}(x)\), we integrate \(\mathrm{f}'(x)\) with respect to \(x\): \(\mathrm{f}(x) = \int 4(3x + 1)^{-\frac{1}{2}} \mathrm{d}x\). Using the reverse chain rule, \(\mathrm{f}(x) = 4 \cdot \frac{(3x + 1)^{\frac{1}{2}}}{\frac{1}{2} \cdot 3} + C = \frac{8}{3}(3x + 1)^{\frac{1}{2}} + C\). Use the coordinates of the point \((1, 5)\) to find the constant of integration \(C\): \(5 = \frac{8}{3}(3(1) + 1)^{\frac{1}{2}} + C \implies 5 = \frac{8}{3}(4)^{\frac{1}{2}} + C \implies 5 = \frac{16}{3} + C\). This gives \(C = 5 - \frac{16}{3} = -\frac{1}{3}\). Thus, the equation of the curve is \(y = \frac{8}{3}\sqrt{3x + 1} - \frac{1}{3}\).

M1: Integrate \((3x+1)^{-\frac{1}{2}}\) to obtain a term of the form \(k(3x+1)^{\frac{1}{2}}\). A1: Obtain the correct integrated expression \(\frac{8}{3}(3x + 1)^{\frac{1}{2}} + C\). M1: Substitute the coordinates \((1, 5)\) into their integrated expression to find the value of \(C\). A1: State the correct equation of the curve \(y = \frac{8}{3}\sqrt{3x + 1} - \frac{1}{3}\) (or equivalent).

Let the terms of the arithmetic progression (AP) be denoted by \(u_n\) and the terms of the geometric progression (GP) by \(v_n\).

From the given information:
\(u_1 = a = v_1\)
\(u_5 = a + 4d = v_2 = ar\)
\(u_7 = a + 6d = v_3 = ar^2\)

We can write two equations for \(d\) in terms of \(a\) and \(r\):
1) \(4d = ar - a = a(r-1)\)
2) \(6d = ar^2 - a = a(r^2-1)\)

Since \(d \neq 0\), we must have \(a \neq 0\) and \(r \neq 1\). Dividing the second equation by the first equation:
\(\frac{6d}{4d} = \frac{a(r^2-1)}{a(r-1)}\)
\(\frac{3}{2} = r + 1\)
\(r = 0.5\)

Using \(r = 0.5\) in the first equation:
\(4d = a(0.5 - 1) = -0.5a \implies d = -0.125a\)

Next, we use the sum of the first 8 terms of the AP:
\(S_8 = \frac{8}{2}[2a + 7d] = 36\)
\(4(2a + 7d) = 36\)
\(2a + 7d = 9\)

Substitute \(d = -0.125a\) into this equation:
\(2a + 7(-0.125a) = 9\)
\(2a - 0.875a = 9\)
\(1.125a = 9 \implies a = 8\)

The sum to infinity of the GP is:
\(S_{\infty} = \frac{a}{1-r} = \frac{8}{1-0.5} = 16\)

M1: Express the GP and AP terms and set up simultaneous equations in terms of \(a\), \(d\), and \(r\).
M1: Attempt to eliminate \(d\) (or \(a\)) to obtain an equation in terms of \(r\) only.
A1: Obtain \(r = 0.5\) (or equivalent).
M1: Use the sum of AP formula for 8 terms and set equal to 36.
A1: Correctly solve for \(a = 8\).
M1: Apply the sum to infinity formula with their values of \(a\) and \(r\) (where \(|r| < 1\)).
A1: Obtain 16.

(a) Starting with the equation:
\(3 \sin \theta \tan \theta + \cos \theta = 5\)

Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\(3 \sin \theta \left(\frac{\sin \theta}{\cos \theta}\right) + \cos \theta = 5\)
\(\frac{3 \sin^2 \theta}{\cos \theta} + \cos \theta = 5\)

Multiply the entire equation by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)):
\(3 \sin^2 \theta + \cos^2 \theta = 5 \cos \theta\)

Substitute the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(3(1 - \cos^2 \theta) + \cos^2 \theta = 5 \cos \theta\)
\(3 - 3\cos^2 \theta + \cos^2 \theta = 5 \cos \theta\)
\(3 - 2\cos^2 \theta = 5 \cos \theta\)

Rearranging gives:
\(2\cos^2 \theta + 5\cos \theta - 3 = 0\) (as required).

(b) Solve the quadratic equation in terms of \(\cos \theta\):
\(2\cos^2 \theta + 5\cos \theta - 3 = 0\)
\((2\cos \theta - 1)(\cos \theta + 3) = 0\)

This gives:
\(2\cos \theta - 1 = 0 \implies \cos \theta = 0.5\)
or
\(\cos \theta + 3 = 0 \implies \cos \theta = -3\)

Since \(\cos \theta = -3\) has no real solutions (as \(-1 \le \cos \theta \le 1\)), we only solve:
\(\cos \theta = 0.5\)

For the interval \(0^\circ \le \theta \le 360^\circ\):
\(\theta = 60^\circ\)
\(\theta = 360^\circ - 60^\circ = 300^\circ\)

(a)
M1: Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and multiply through by \(\cos \theta\).
M1: Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\) to obtain an equation in terms of \(\cos \theta\) only.
A1: Correctly obtain the given equation with no errors seen.

(b)
M1: Attempt to solve the quadratic equation to find values of \(\cos \theta\).
A1: Identify \(\cos \theta = 0.5\) and state that the other branch has no solution.
B1: Find the first angle \(\theta = 60^\circ\).
B1: Find the second angle \(\theta = 300^\circ\) and no other solutions in the range.

(a) The area \(A\) of the region \(R\) is given by:
\(A = \int_{0}^{40} \frac{3}{(2x+1)^{1/4}} \, \mathrm{d}x = \int_{0}^{40} 3(2x+1)^{-1/4} \, \mathrm{d}x\)

Using the rule for integrating \((ax+b)^n\):
\(\int 3(2x+1)^{-1/4} \, \mathrm{d}x = 3 \cdot \frac{(2x+1)^{3/4}}{\frac{3}{4} \cdot 2} = 2(2x+1)^{3/4}\)

Now, evaluate this integral between the limits \(0\) and \(40\):
\(A = \left[ 2(2x+1)^{3/4} \right]_0^{40}\)
\(A = 2(2(40)+1)^{3/4} - 2(2(0)+1)^{3/4}\)
\(A = 2(81)^{3/4} - 2(1)^{3/4}\)

Since \(81^{3/4} = 27\):
\(A = 2(27) - 2(1) = 54 - 2 = 52\)

(b) The volume \(V\) of the solid of revolution is given by:
\(V = \pi \int_{0}^{40} y^2 \, \mathrm{d}x\)

First, express \(y^2\):
\(y^2 = \left( \frac{3}{(2x+1)^{1/4}} \right)^2 = \frac{9}{(2x+1)^{1/2}} = 9(2x+1)^{-1/2}\)

Now integrate \(y^2\):
\(\int 9(2x+1)^{-1/2} \, \mathrm{d}x = 9 \cdot \frac{(2x+1)^{1/2}}{\frac{1}{2} \cdot 2} = 9(2x+1)^{1/2}\)

Evaluate the volume:
\(V = \pi \left[ 9(2x+1)^{1/2} \right]_0^{40}\)
\(V = \pi \left( 9\sqrt{81} - 9\sqrt{1} \right)\)
\(V = \pi \left( 9(9) - 9(1) \right) = 72\pi\)

(a)
M1: State the integral of \(3(2x+1)^{-1/4}\) and show an attempt to integrate.
A1: Correctly obtain the integrated expression \(2(2x+1)^{3/4}\).
M1: Substitute the limits of 40 and 0 into their integrated expression.
A1: Obtain the correct area of 52.

(b)
M1: State the correct volume formula and write down the integral for \(y^2\) as \(\int 9(2x+1)^{-1/2} \, \mathrm{d}x\).
A1: Correctly integrate to obtain \(9(2x+1)^{1/2}\).
A1: Obtain the correct volume of \(72\pi\).

(a) We complete the square for \(f(x) = 2x^2 - 12x + 13\):
\(f(x) = 2(x^2 - 6x) + 13\)
\(f(x) = 2\left[(x-3)^2 - 9\right] + 13\)
\(f(x) = 2(x-3)^2 - 18 + 13\)
\(f(x) = 2(x-3)^2 - 5\)

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

(b) For \(f\) to have an inverse, it must be a one-to-one function. The vertex of the parabola is at \(x = 3\).
Therefore, the least value of \(k\) for the domain \(x \ge k\) to define a one-to-one function is \(k = 3\).

(c) Let \(y = 2(x-3)^2 - 5\).
We make \(x\) the subject of the formula:
\(y + 5 = 2(x-3)^2\)
\(\frac{y+5}{2} = (x-3)^2\)

Taking the square root on both sides:
\(x - 3 = \pm \sqrt{\frac{y+5}{2}}\)

Since the domain of \(f\) is \(x \ge 3\), we must have \(x - 3 \ge 0\), which means we choose the positive square root:
\(x = 3 + \sqrt{\frac{y+5}{2}}\)

Replacing \(y\) with \(x\), we get:
\(f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\)

The domain of \(f^{-1}\) is the range of \(f\).
For \(x \ge 3\), the minimum value of \(f(x)\) occurs at the vertex where \(f(3) = -5\). So the range of \(f\) is \(f(x) \ge -5\).
Thus, the domain of \(f^{-1}\) is \(x \ge -5\).

(a)
M1: Attempt to complete the square by factoring out 2 or correctly setting up \(2(x-b)^2 + c\).
A1: Obtain the correct expression \(2(x-3)^2 - 5\).

(b)
B1: State \(k = 3\).

(c)
M1: Set \(y = 2(x-3)^2 - 5\) and make \((x-3)^2\) the subject.
M1: Correctly choose the positive square root and make \(x\) the subject.
A1: Obtain \(f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\).
B1: State the domain of \(f^{-1}\) is \(x \ge -5\).

Using the Remainder Theorem, since the remainder when \(p(x)\) is divided by \(x - 2\) is \(4\), we have \(p(2) = 4\).

Substituting \(x = 2\) into the expression for \(p(x)\):
\(2(2)^3 - k(2)^2 + 5(2) - 6 = 4\)
\(16 - 4k + 10 - 6 = 4\)
\(20 - 4k = 4\)
\(4k = 16\)
\(k = 4\)

M1: For applying the Remainder Theorem by setting \(p(2) = 4\) (or an equivalent long division method) [1 mark]
A1: For obtaining a correct linear equation in \(k\), e.g., \(20 - 4k = 4\) [1 mark]
A0.5: For solving to find \(k = 4\) [0.5 marks]

Let \(u = e^x\). The equation becomes a quadratic:
\(u^2 - 5u + 6 = 0\)

Factorising gives:
\((u - 2)(u - 3) = 0\)

So, \(u = 2\) or \(u = 3\).

Substituting back \(u = e^x\):
\(e^x = 2 \implies x = \ln 2\)
\(e^x = 3 \implies x = \ln 3\)

M1: For treating the equation as a quadratic in \(e^x\) and factorising or using the formula [1 mark]
A1: For obtaining correct values for \(e^x\), i.e., \(e^x = 2\) and \(e^x = 3\) [1 mark]
A0.5: For both correct logarithmic solutions \(x = \ln 2\) and \(x = \ln 3\) [0.5 marks]

Using the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\), we substitute into the equation:
\(1 + \tan^2 \theta = 3\tan \theta - 1\)

Rearranging terms to form a quadratic equation:
\(\tan^2 \theta - 3\tan \theta + 2 = 0\)

Factorising the quadratic in \(\tan \theta\):
\((\tan \theta - 1)(\tan \theta - 2) = 0\)

Thus, \(\tan \theta = 1\) or \(\tan \theta = 2\).

For \(\tan \theta = 1\):
\(\theta = \tan^{-1}(1) = 45^\circ\)

For \(\tan \theta = 2\):
\(\theta = \tan^{-1}(2) \approx 63.43^\circ \approx 63.4^\circ\)

Both values lie within the interval \(0^\circ \le \theta \le 180^\circ\).

M1: For using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to form a quadratic equation in \(\tan \theta\) [1 mark]
A1: For obtaining \(\tan \theta = 1\) and \(\tan \theta = 2\) [1 mark]
A0.5: For both correct angles \(\theta = 45^\circ\) and \(\theta = 63.4^\circ\) (deduct 0.5 marks if extra incorrect angles within the range are given) [0.5 marks]

To find the gradient of the curve, we differentiate \(y\) with respect to \(x\) using the quotient rule:

Let \(u = \ln(3x - 1)\) and \(v = x\). The derivatives are:
\(\frac{du}{dx} = \frac{3}{3x - 1}\)
\(\frac{dv}{dx} = 1\)

Applying the quotient rule:
\(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} = \frac{x \left(\frac{3}{3x - 1}\right) - \ln(3x - 1)}{x^2}\)

Now substitute \(x = 1\) into the derivative:
\(\frac{dy}{dx}\Big|_{x=1} = \frac{1 \left(\frac{3}{3(1) - 1}\right) - \ln(3(1) - 1)}{1^2} = \frac{3}{2} - \ln 2\)

M1: For applying the quotient rule (or product rule) to differentiate \(y\), with at least one term differentiated correctly [1 mark]
A1: For obtaining a correct expression for \(\frac{dy}{dx}\), e.g., \(\frac{\frac{3x}{3x-1} - \ln(3x-1)}{x^2}\) [1 mark]
A0.5: For substituting \(x=1\) and obtaining the exact value \%\frac{3}{2} - \ln 2\u0025 (or \(1.5 - \ln 2\)) [0.5 marks]

We can solve the inequality by squaring both sides:

\( (2x - 5)^2 < 9(x + 1)^2 \)

Expanding both sides:

\( 4x^2 - 20x + 25 < 9(x^2 + 2x + 1) \)

\( 4x^2 - 20x + 25 < 9x^2 + 18x + 9 \)

Rearranging to form a quadratic inequality:

\( 0 < 5x^2 + 38x - 16 \)

Factoring the quadratic:

\( (5x - 2)(x + 8) > 0 \)

The critical values are \( x = 0.4 \) and \( x = -8 \).

Since we require the expression to be greater than zero, the solution is:

\( x < -8 \) or \( x > 0.4 \)

M1: For squaring both sides and expanding, or for setting up two equations to find critical values.
A1: For finding the correct critical values \( x = -8 \) and \( x = 0.4 \).
M1: For a correct method of determining the outside intervals.
A1: For the correct final answer \( x < -8 \text{ or } x > 0.4 \) (or equivalent interval notation).

Let \( u = 2^x \). Since \( 4^x = (2^x)^2 = u^2 \), we can rewrite the equation as:

\( u^2 - 5u - 14 = 0 \)

Factorizing the quadratic equation:

\( (u - 7)(u + 2) = 0 \)

This gives \( u = 7 \) or \( u = -2 \).

Since \( u = 2^x \) and \( 2^x > 0 \) for all real values of \( x \), we reject \( u = -2 \).

Thus, we solve:

\( 2^x = 7 \)

Taking natural logarithms of both sides:

\( x \ln 2 = \ln 7 \implies x = \frac{\ln 7}{\ln 2} \approx 2.80735 \)

Correct to 3 significant figures, \( x = 2.81 \).

M1: Substitute \( u = 2^x \) to form a quadratic equation in \( u \).
A1: Obtain the correct roots \( u = 7 \) and \( u = -2 \).
M1: Recognize that \( 2^x = -2 \) has no real solution, and use logarithms to solve \( 2^x = 7 \).
A1: Obtain the correct final answer \( x = 2.81 \).

We use the double-angle identity for cosine:

\( \cos 2\theta = 1 - 2\sin^2\theta \)

Substituting this into the given equation:

\( 1 - 2\sin^2\theta + 3\sin\theta = 2 \)

Rearranging into a quadratic equation in terms of \( \sin\theta \):

\( 2\sin^2\theta - 3\sin\theta + 1 = 0 \)

Factorizing the quadratic:

\( (2\sin\theta - 1)(\sin\theta - 1) = 0 \)

This gives two cases:

Case 1: \( \sin\theta = \frac{1}{2} \)
Within the range \( 0^\circ \le \theta \le 360^\circ \), we have:
\( \theta = 30^\circ \) and \( \theta = 180^\circ - 30^\circ = 150^\circ \).

Case 2: \( \sin\theta = 1 \)
Within the range \( 0^\circ \le \theta \le 360^\circ \), we have:
\( \theta = 90^\circ \).

Combining all the solutions: \( \theta = 30^\circ, 90^\circ, 150^\circ \).

M1: Substitute the correct identity \( \cos 2\theta = 1 - 2\sin^2\theta \) into the equation.
A1: Form the correct quadratic equation \( 2\sin^2\theta - 3\sin\theta + 1 = 0 \) and solve to obtain \( \sin\theta = \frac{1}{2} \) and \( \sin\theta = 1 \).
A1: Find \( \theta = 30^\circ \) and \( \theta = 150^\circ \).
A1: Find \( \theta = 90^\circ \) and ensure no other solutions are incorrectly included.

We use the quotient rule to differentiate \( y = \frac{u}{v} \), where \( u = \ln(2x + 1) \) and \( v = x^2 \).

First, compute the individual derivatives:

\( \frac{du}{dx} = \frac{2}{2x + 1} \)

\( \frac{dv}{dx} = 2x \)

Applying the quotient rule:

\( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)

\( \frac{dy}{dx} = \frac{x^2 \left( \frac{2}{2x + 1} \right) - 2x\ln(2x + 1)}{x^4} \)

Substitute \( x = 1 \) into the derivative to find the gradient:

\( \frac{dy}{dx}\Big|_{x=1} = \frac{(1)^2 \left( \frac{2}{3} \right) - 2(1)\ln(3)}{1^4} \)

\( \frac{dy}{dx}\Big|_{x=1} = \frac{2}{3} - 2\ln 3 \)

Comparing with \( a\ln 3 + b \), we have \( a = -2 \) and \( b = \frac{2}{3} \), which are both rational.

M1: Apply the quotient rule (or product rule) to differentiate \( y \).
A1: Correctly differentiate \( \ln(2x + 1) \) to get \( \frac{2}{2x + 1} \).
A1: Obtain the correct full expression for \( \frac{dy}{dx} \).
A1: Substitute \( x = 1 \) and write the final exact answer in the form \( -2\ln 3 + \frac{2}{3} \) (or equivalent).

Integrate each term individually:

1) \( \int e^{2x-2} dx = \frac{1}{2}e^{2x-2} \)

2) \( \int \frac{6}{3x-1} dx = 6 \times \frac{1}{3}\ln|3x-1| = 2\ln|3x-1| \)

Now apply the limits of integration from 1 to 2:

\( \left[ \frac{1}{2}e^{2x-2} + 2\ln|3x-1| \right]_{1}^{2} \)

Substitute the upper limit \( x = 2 \):

\( \left( \frac{1}{2}e^{2(2)-2} + 2\ln|3(2)-1| \right) = \frac{1}{2}e^2 + 2\ln 5 \)

Substitute the lower limit \( x = 1 \):

\( \left( \frac{1}{2}e^{2(1)-2} + 2\ln|3(1)-1| \right) = \frac{1}{2}e^0 + 2\ln 2 = \frac{1}{2} + 2\ln 2 \)

Subtract the lower limit evaluation from the upper limit evaluation:

\( \left( \frac{1}{2}e^2 + 2\ln 5 \right) - \left( \frac{1}{2} + 2\ln 2 \right) = \frac{1}{2}e^2 - \frac{1}{2} + 2\ln 5 - 2\ln 2 \)

Using log properties to simplify:

\( = \frac{1}{2}e^2 - \frac{1}{2} + 2\ln\left(\frac{5}{2}\right) \)

M1: Integrate \( e^{2x-2} \) correctly to obtain \( \frac{1}{2}e^{2x-2} \).
M1: Integrate \( \frac{6}{3x-1} \) correctly to obtain \( 2\ln(3x-1) \).
M1: Substitute the limits of 2 and 1 into their integrated expression.
A1: Obtain the exact value \( \frac{1}{2}e^2 - \frac{1}{2} + 2\ln\left(\frac{5}{2}\right) \) or any equivalent simplified exact form.

(a) Rearranging the equation:

\( x^3 = 5x + 3 \)

Dividing both sides by \( x \) (since \( x \neq 0 \)):

\( x^2 = 5 + \frac{3}{x} \)

Taking the positive square root of both sides:

\( x = \sqrt{5 + \frac{3}{x}} \)

(b) We perform the iterations with \( x_1 = 2.5 \):

\( x_2 = \sqrt{5 + \frac{3}{2.5}} = \sqrt{5 + 1.2} = \sqrt{6.2} \approx 2.48998 \)

\( x_3 = \sqrt{5 + \frac{3}{2.48998}} \approx 2.49095 \)

\( x_4 = \sqrt{5 + \frac{3}{2.49095}} \approx 2.49086 \)

\( x_5 = \sqrt{5 + \frac{3}{2.49086}} \approx 2.49086 \)

Since the successive iterations converge to \( 2.49086... \), the root correct to 3 significant figures is \( 2.49 \).

B1: (For part a) Show the rearrangement steps clearly to derive \( x = ∕(5 + 3/x) \).
M1: (For part b) Apply the iterative formula correctly at least twice.
A1: Obtain the values \( x_2 ≈ 2.48998 \) and \( x_3 ≈ 2.49095 \).
A1: Conclude with the correct root of \( 2.49 \) to 3 significant figures.

By the Factor Theorem, since \( p(x) \) is divisible by \( x - 2 \), we have \( p(2) = 0 \):

\( 2(2)^3 + a(2)^2 + b(2) - 6 = 0 \)

\( 16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5 \) (Equation 1)

By the Remainder Theorem, since dividing by \( x + 1 \) leaves a remainder of \( -9 \), we have \( p(-1) = -9 \):

\( 2(-1)^3 + a(-1)^2 + b(-1) - 6 = -9 \)

\( -2 + a - b - 6 = -9 \implies a - b = -1 \) (Equation 2)

We now solve the simultaneous equations:

From Equation 2, we have \( a = b - 1 \).

Substitute this into Equation 1:

\( 2(b - 1) + b = -5 \)

\( 2b - 2 + b = -5 \)

\( 3b = -3 \implies b = -1 \)

Substitute \( b = -1 \) back to find \( a \):

\( a = -1 - 1 = -2 \)

So, \( a = -2 \) and \( b = -1 \).

M1: Apply the Factor Theorem by setting \( p(2) = 0 \) to obtain a linear equation in \( a \) and \( b \).
M1: Apply the Remainder Theorem by setting \( p(-1) = -9 \) to obtain a second linear equation in \( a \) and \( b \).
M1: Correctly solve the simultaneous equations for \( a \) and \( b \).
A1: Obtain the correct values \( a = -2 \) and \( b = -1 \).

First, calculate the width of each interval:

\( h = \frac{\frac{3\pi}{4} - 0}{3} = \frac{\pi}{4} \)

The boundary values of \( x \) for the 3 intervals are:

\( x_0 = 0 \)

\( x_1 = \frac{\pi}{4} \)

\( x_2 = \frac{\pi}{2} \)

\( x_3 = \frac{3\pi}{4} \)

Now, calculate the corresponding ordinate values of \( y = \sqrt{1 + \sin^2 x} \):

\( y_0 = \sqrt{1 + \sin^2 0} = \sqrt{1 + 0} = 1.00000 \)

\( y_1 = \sqrt{1 + \sin^2\left(\frac{\pi}{4}\right)} = \sqrt{1 + 0.5} = \sqrt{1.5} \approx 1.22474 \)

\( y_2 = \sqrt{1 + \sin^2\left(\frac{\pi}{2}\right)} = \sqrt{1 + 1} = \sqrt{2} \approx 1.41421 \)

\( y_3 = \sqrt{1 + \sin^2\left(\frac{3\frac{\pi}{4}}\right)} = \sqrt{1 + 0.5} = \sqrt{1.5} \approx 1.22474 \)

Applying the Trapezium Rule formula:

\( I \approx \frac{h}{2} \left[ y_0 + y_3 + 2(y_1 + y_2) \right] \)

\( I \approx \frac{\pi}{8} \left[ 1.00000 + 1.22474 + 2(1.22474 + 1.41421) \right] \)

\( I \approx \frac{\pi}{8} \left[ 2.22474 + 2(2.63895) \right] \)

\( I \approx \frac{\pi}{8} \left[ 2.22474 + 5.27790 \right] \)

\( I \approx \frac{\pi}{8} \times 7.50264 \approx 2.94628 \)

Correct to 3 decimal places, the estimate is \( 2.946 \).

B1: Identify the correct interval width \( h = \frac{\pi}{4} \).
M1: Find the 4 correct \( y \)-values corresponding to the chosen \( x \)-values (rounding must be minimal, in radians).
M1: Apply the correct Trapezium Rule formula \( \frac{h}{2}[y_0 + y_3 + 2(y_1 + y_2)] \).
A1: Obtain the correct estimated value of \( 2.946 \).

We use the identity \( \sec^2 2\theta = 1 + \tan^2 2\theta \) to rewrite the equation in terms of \( \tan 2\theta \): \( 3(1 + \tan^2 2\theta) + 2 \tan 2\theta = 8 \). Simplifying this gives: \( 3 + 3 \tan^2 2\theta + 2 \tan 2\theta = 8 \implies 3\tan^2 2\theta + 2\tan 2\theta - 5 = 0 \). Let \( y = \tan 2\theta \). Then we have the quadratic equation \( 3y^2 + 2y - 5 = 0 \), which factorises to: \( (3y + 5)(y - 1) = 0 \). This yields two possible values: \( \tan 2\theta = 1 \) or \( \tan 2\theta = -\frac{5}{3} \). Since \( 0^\circ \le \theta \le 180^\circ \), the interval for \( 2\theta \) is \( 0^\circ \le 2\theta \le 360^\circ \). For \( \tan 2\theta = 1 \): \( 2\theta = 45^\circ \) or \( 2\theta = 225^\circ \), which gives \( \theta = 22.5^\circ \) or \( \theta = 112.5^\circ \). For \( \tan 2\theta = -\frac{5}{3} \): the acute basic angle is \( \tan^{-1}\left(\frac{5}{3}\right) \approx 59.04^\circ \). Since tangent is negative, \( 2\theta \) lies in the second or fourth quadrant: \( 2\theta = 180^\circ - 59.04^\circ = 120.96^\circ \) or \( 2\theta = 360^\circ - 59.04^\circ = 300.96^\circ \). Dividing by 2 gives \( \theta \approx 60.5^\circ \) or \( \theta \approx 150.5^\circ \) (to 1 decimal place).

M1: Use the trigonometric identity \( \sec^2 2\theta = 1 + \tan^2 2\theta \) to obtain an equation in terms of \( \tan 2\theta \) only. A1: Correctly write the quadratic equation as \( 3\tan^2 2\theta + 2\tan 2\theta - 5 = 0 \). M1: Solve the quadratic equation to find two values for \( \tan 2\theta \) (specifically, 1 and \( -\frac{5}{3} \)). A1: Find any two correct solutions for \( \theta \) (e.g., \( 22.5^\circ \) and \( 112.5^\circ \)). A1: Find a third correct solution (e.g., \( 60.5^\circ \) or \( 150.5^\circ \)). A1: Find the fourth correct solution and no extra values in the range.

To integrate the function, we first rewrite the equation using the double-angle identity \( \cos^2 A = \frac{1 + \cos 2A}{2} \). Substituting \( A = x - \frac{\pi}{6} \) gives: \( \cos^2\left(x - \frac{\pi}{6}\right) = \frac{1 + \cos\left(2x - \frac{\pi}{3}\right)}{2} \). Thus, \( y = 4 \left( \frac{1 + \cos\left(2x - \frac{\pi}{3}\right)}{2} \right) = 2 + 2\cos\left(2x - \frac{\pi}{3}\right) \). The area of the region is given by the definite integral: \( \int_{0}^{\frac{\pi}{3}} \left( 2 + 2\cos\left(2x - \frac{\pi}{3}\right) \right) dx \). Integrating term by term: \( \int 2 \, dx = 2x \) and \( \int 2\cos\left(2x - \frac{\pi}{3}\right) dx = \sin\left(2x - \frac{\pi}{3}\right) \). So the antiderivative is \( F(x) = 2x + \sin\left(2x - \frac{\pi}{3}\right) \). We now evaluate this antiderivative at the limits of integration. At the upper limit \( x = \frac{\pi}{3} \): \( F\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right) + \sin\left(2\left(\frac{\pi}{3}\right) - \frac{\pi}{3}\right) = \frac{2\pi}{3} + \sin\left(\frac{\pi}{3}\right) = \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \). At the lower limit \( x = 0 \): \( F(0) = 2(0) + \sin\left(2(0) - \frac{\pi}{3}\right) = 0 + \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \). Subtracting the lower limit value from the upper limit value: \( \text{Area} = \left(\frac{2\pi}{3} + \frac{\sqrt{3}}{2}\right) - \left(-\frac{\sqrt{3}}{2}\right) = \frac{2\pi}{3} + \sqrt{3} \).

M1: Attempt to use the double-angle formula to express \( \cos^2\left(x - \frac{\pi}{6}\right) \) in terms of \( \cos\left(2x - \frac{\pi}{3}\right) \). A1: Obtain the correct simplified expression for the integrand: \( 2 + 2\cos\left(2x - \frac{\pi}{3}\right) \). M1: Attempt to integrate the terms, resulting in an expression of the form \( ax + b\sin\left(2x - \frac{\pi}{3}\right) \). A1: Correct integration to get \( 2x + \sin\left(2x - \frac{\pi}{3}\right) \). M1: Correctly substitute limits \( 0 \) and \( \frac{\pi}{3} \) into their integrated expression. A1: Obtain the correct exact area \( \frac{2}{3}\pi + \sqrt{3} \) (or equivalent exact form).