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(a) For the curve \(y = 2x^2 - kx + (k + 6)\) to lie entirely above the \(x\)-axis, the discriminant of the quadratic must be strictly negative (since the coefficient of \(x^2\) is \(2 > 0\)).

The discriminant \(\Delta\) is given by:
\(\Delta = b^2 - 4ac = (-k)^2 - 4(2)(k + 6)\)
\(\Delta = k^2 - 8k - 48\)

We require \(\Delta < 0\):
\(k^2 - 8k - 48 < 0\)

Factorise the quadratic inequality:
\((k - 12)(k + 4) < 0\)

Therefore, the set of values of \(k\) is \(-4 < k < 12\).

(b) When \(k = 4\), the equation of the curve becomes:
\(y = 2x^2 - 4x + 10\)

To find the vertex, we express this in completed square form:
\(y = 2(x^2 - 2x) + 10\)
\(y = 2[(x - 1)^2 - 1] + 10\)
\(y = 2(x - 1)^2 - 2 + 10\)
\(y = 2(x - 1)^2 + 8\)

Thus, the coordinates of the vertex of the curve are \((1, 8)\).

(a)
* **M1**: Sets \(\Delta = b^2 - 4ac < 0\).
* **A1**: Obtains correct quadratic expression in \(k\): \(k^2 - 8k - 48 < 0\).
* **M1**: Solves the quadratic equation to find the critical values \(k = -4\) and \(k = 12\).
* **A1**: Correct inequality range \(-4 < k < 12\).

(b)
* **M1**: Substitutes \(k = 4\) into the curve equation and attempts to complete the square or uses \(x = -b/(2a)\).
* **A1**: Correctly identifies the \(x\)-coordinate of the vertex as \(1\).
* **A1** (0.8 marks): Correctly identifies the \(y\)-coordinate of the vertex as \(8\).

(a) A function has an inverse if and only if it is a one-to-one function.
First, complete the square for \(\mathrm{f}(x)\):
\(\mathrm{f}(x) = 3(x^2 - 4x) + 7 = 3[(x - 2)^2 - 4] + 7 = 3(x - 2)^2 - 5\)

The vertex of this parabola is at \((2, -5)\). Since the domain is defined for \(x \le a\), the function will be one-to-one as long as the domain lies entirely on one side of the vertex.
Therefore, the largest value of \(a\) is the \(x\)-coordinate of the vertex, which is \(a = 2\).

(b) Using \(a = 2\), the domain of \(\mathrm{f}\) is \(x \le 2\), and the range of \(\mathrm{f}\) is \(\mathrm{f}(x) \ge -5\).

To find the inverse function, let \(y = \mathrm{f}(x)\):
\(y = 3(x - 2)^2 - 5\)
\(y + 5 = 3(x - 2)^2\)
\(\frac{y + 5}{3} = (x - 2)^2\)

Taking the square root, since the domain of \(\mathrm{f}\) is \(x \le 2\), the term \(x - 2\) must be non-positive, so we choose the negative square root:
\(x - 2 = -\sqrt{\frac{y + 5}{3}}\)
\(x = 2 - \sqrt{\frac{y + 5}{3}}\)

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

The domain of the inverse function is the range of the original function, which is \(x \ge -5\).

(a)
* **M1**: Attempts to complete the square or find the vertex of the quadratic.
* **A1** (0.8 marks): Correctly states \(a = 2\).

(b)
* **M1**: Sets \(y = 3(x-2)^2 - 5\) (or equivalent) and attempts to make \((x-2)^2\) or \(x\) the subject.
* **A1**: Obtains \((x-2)^2 = \frac{y+5}{3}\) in terms of \(y\).
* **M1**: Explains or demonstrates selection of the negative square root due to the domain \(x \le 2\).
* **A1**: Obtains correct expression \(\mathrm{f}^{-1}(x) = 2 - \sqrt{\frac{x + 5}{3}}\) (must be in terms of \(x\)).
* **B1**: Correctly states the domain as \(x \ge -5\) (accept interval notation \([-5, \infty)\)).

(a) To find the perpendicular bisector of \(AB\):
1. Find the midpoint \(M\) of \(AB\):
\(M = \left(\frac{2 + 6}{2}, \frac{8 + 0}{2}\right) = (4, 4)\)

2. Find the gradient \(m\) of \(AB\):
\(m = \frac{0 - 8}{6 - 2} = \frac{-8}{4} = -2\)

3. The gradient of the perpendicular line is \(m_{\perp} = -\frac{1}{m} = \frac{1}{2}\).

4. Find the equation of the line passing through \(M(4, 4)\) with gradient \(\frac{1}{2}\):
\(y - 4 = \frac{1}{2}(x - 4)\)
\(2(y - 4) = x - 4\)
\(2y - 8 = x - 4\)
\(2y = x + 4\)

(b) Since \(L\) is parallel to \(AB\), its gradient is also \(-2\). It passes through \((1, 5)\), so its equation is:
\(y - 5 = -2(x - 1)\)
\(y = -2x + 7\)

To find the intersection of the perpendicular bisector \(2y = x + 4\) and \(L\), substitute \(y = -2x + 7\) into the equation of the perpendicular bisector:
\(2(-2x + 7) = x + 4\)
\(-4x + 14 = x + 4\)
\(10 = 5x \implies x = 2\)

Substitute \(x = 2\) back into the equation for \(L\):
\(y = -2(2) + 7 = 3\)

So, the coordinates of the point of intersection are \((2, 3)\).

(a)
* **B1**: Finds the correct midpoint \(M(4, 4)\).
* **M1**: Calculates the gradient of \(AB\) and uses the perpendicular gradient rule \(m_1 m_2 = -1\) to find the gradient of the perpendicular bisector.
* **A1** (1.8 marks): Correctly formulates the equation of the line in the specified form \(2y = x + 4\) (or any integer multiple thereof, e.g., \(4y = 2x + 8\)).

(b)
* **B1**: Formulates the correct equation of the parallel line \(L\): \(y = -2x + 7\).
* **M1**: Sets up and attempts to solve the simultaneous equations for the intersection point.
* **A1**: Correctly identifies the coordinates of the intersection point as \((2, 3)\).

(a) The perimeter of the sector \(OPQ\) is given by:
Perimeter \(= r + r + \text{arc length } PQ = 2r + r\theta = r(2 + \theta)\)

Given that the perimeter is \(28\text{ cm}\) and \(\theta = 0.8\):
\(r(2 + 0.8) = 28\)
\(2.8r = 28 \implies r = 10\text{ cm}\).

(b) The region bounded by the arc \(PQ\), and the straight lines \(PR\) and \(RQ\) is the area of sector \(OPQ\) minus the area of the right-angled triangle \(OPR\).

1. Area of sector \(OPQ\):
\(\text{Area}_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} (10)^2 (0.8) = 40\text{ cm}^2\)

2. In the right-angled triangle \(OPR\) (where \(\angle PRO = 90^\circ\)):
\(OR = r \cos \theta = 10 \cos(0.8)\)
\(PR = r \sin \theta = 10 \sin(0.8)\)

Therefore, the area of triangle \(OPR\) is:
\(\text{Area}_{\text{triangle}} = \frac{1}{2} \times OR \times PR = \frac{1}{2} \times 10\cos(0.8) \times 10\sin(0.8) = 50 \sin(0.8) \cos(0.8)\)
Using \(\sin(0.8) \approx 0.717356\) and \(\cos(0.8) \approx 0.696707\):
\(\text{Area}_{\text{triangle}} \approx 50 \times 0.717356 \times 0.696707 \approx 24.991\text{ cm}^2\)

3. Area of the shaded region:
\(\text{Area} = 40 - 24.991 = 15.009\text{ cm}^2\)

To 3 significant figures, the area of the region is \(15.0\text{ cm}^2\).

(a)
* **M1**: Formulates the perimeter equation \(r(2 + \theta) = 28\) and substitutes \(\theta = 0.8\).
* **A1** (1.8 marks): Correctly evaluates \(r = 10\).

(b)
* **B1**: Correctly calculates the area of the sector as \(40\text{ cm}^2\).
* **M1**: Uses trigonometry to express \(OR\) and \(PR\) in terms of \(\cos(0.8)\) and \(\sin(0.8)\), and attempts to find the area of triangle \(OPR\).
* **A1**: Obtains area of triangle \(\approx 24.99\) or \(25 \sin(1.6)\).
* **A1**: Subtracts the triangle area from the sector area to obtain \(15.0\text{ cm}^2\) (accept \(15\) or \(15.0\)).

(a) To prove the identity, combine the fractions on the left-hand side (LHS) over a common denominator:
\(\text{LHS} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)}\)

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

Using the fundamental identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\text{LHS} = \frac{1 + 1 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}\)
\(\text{LHS} = \frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}\)
\(\text{LHS} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)}\)

Cancel the common factor \((1 + \cos \theta)\):
\(\text{LHS} = \frac{2}{\sin \theta} = \text{RHS}\)

(b) Using the proven identity, the equation becomes:
\(\frac{2}{\sin \theta} = 3 \tan \theta\)

Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\(\frac{2}{\sin \theta} = 3 \frac{\sin \theta}{\cos \theta}\)

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

Use \(\sin^2 \theta = 1 - \cos^2 \theta\) to form a quadratic in \(\cos \theta\):
\(2 \cos \theta = 3(1 - \cos^2 \theta)
3 \cos^2 \theta + 2 \cos \theta - 3 = 0\)

Let \(u = \cos \theta\):
\(3u^2 + 2u - 3 = 0\)

Using the quadratic formula:
\(u = \frac{-2 \pm \sqrt{2^2 - 4(3)(-3)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 36}}{6} = \frac{-2 \pm \sqrt{40}}{6}\)

Evaluating the roots:
\(u_1 = \frac{-2 + \sqrt{40}}{6} \approx 0.720759\)
\(u_2 = \frac{-2 - \sqrt{40}}{6} \approx -1.3874\) (which has no real solutions since \(\cos \theta \ge -1\))

For \(\cos \theta = 0.720759\):
\(\theta = \cos^{-1}(0.720759) \approx 43.9^\circ\) (to 1 d.p.)

Since \(0^\circ \le \theta \le 360^\circ\), the second solution is in the fourth quadrant:
\(\theta = 360^\circ - 43.9^\circ = 316.1^\circ\)

Thus, the solutions are \(\theta = 43.9^\circ\) and \(\theta = 316.1^\circ\).

(a)
* **M1**: Puts LHS over a single common denominator and expands the numerator.
* **A1**: Simplifies numerator using \(\sin^2 \theta + \cos^2 \theta = 1\).
* **A1** (0.8 marks): Factorises and cancels \((1 + \cos \theta)\) to obtain RHS.

(b)
* **M1**: Replaces the LHS of the equation with \(\frac{2}{\sin \theta}\) and substitutes \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).
* **M1**: Eliminates fractions and uses \(\sin^2 \theta = 1 - \cos^2 \theta\) to form a quadratic equation in \(\cos \theta\).
* **A1**: Obtains correct quadratic equation \(3 \cos^2 \theta + 2 \cos \theta - 3 = 0\).
* **A1**: Finds \(\theta = 43.9^\circ\) and \(\theta = 316.1^\circ\) (both required, accept 1 d.p.).

(a) The \(n\)-th term of an arithmetic progression (AP) is \(T_n = a + (n-1)d\).
The \(n\)-th term of a geometric progression (GP) is \(G_n = a r^{n-1}\).

From the first piece of information:
\(T_5 = G_3 \implies a + 4d = a r^2 \implies 4d = a(r^2 - 1) \implies d = \frac{a(r^2 - 1)}{4}\)

From the second piece of information:
\(T_6 = G_5 \implies a + 5d = a r^4\)

Substitute the expression for \(d\) into this equation:
\(a + 5\left[\frac{a(r^2 - 1)}{4}\right] = a r^4\)

Assuming \(a \ne 0\), we can divide both sides by \(a\):
\(1 + \frac{5}{4}(r^2 - 1) = r^4\)

Multiply through by 4:
\(4 + 5(r^2 - 1) = 4r^4\)
\(4 + 5r^2 - 5 = 4r^4\)
\(4r^4 - 5r^2 + 1 = 0\)

This completes the proof.

(b) To solve \(4r^4 - 5r^2 + 1 = 0\), we can factorise it as a quadratic in \(r^2\):
\((4r^2 - 1)(r^2 - 1) = 0\)

This gives:
\(r^2 = \frac{1}{4} \implies r = \pm \frac{1}{2}\)
Or
\(r^2 = 1 \implies r = \pm 1\)

Since the GP has a sum to infinity, the common ratio \(r\) must satisfy \(|r| < 1\). This excludes \(r = \pm 1\). Therefore, we must have \(r^2 = \frac{1}{4}\).

Now substitute \(r^2 = \frac{1}{4}\) back into the equation for \(d\):
\(d = \frac{a(1/4 - 1)}{4} = \frac{-3a/4}{4} = -\frac{3}{16} a\)

The sum of the first 10 terms of an AP is given by:
\(S_{10} = \frac{10}{2}[2a + 9d] = 5\left[2a + 9\left(-\frac{3}{16}a\right)\right] = 5\left[2a - \frac{27}{16}a\right] = 5\left[\frac{5}{16}a\right] = \frac{25}{16} a\)

We are given \(S_{10} = 50\):
\(\frac{25}{16} a = 50\)
\(a = 50 \times \frac{16}{25} = 32\).

Thus, the value of \(a\) is 32.

(a)
* **M1**: Uses the formulae for AP and GP terms to write equations \(a + 4d = a r^2\) and \(a + 5d = a r^4\).
* **M1**: Expresses \(d\) in terms of \(a\) and \(r\), and substitutes it into the second equation.
* **A1**: Simplifies and eliminates \(a\) (assuming \(a \ne 0\)).
* **A1** (0.8 marks): Obtains the given equation \(4r^4 - 5r^2 + 1 = 0\).

(b)
* **M1**: Solves the equation to find \(r^2 = \frac{1}{4}\) (stating why \(r^2 = 1\) is rejected based on the sum to infinity condition).
* **M1**: Obtains \(d = -\frac{3}{16} a\) and substitutes this into the AP sum formula \(S_{10}\).
* **A1**: Sets \(\frac{25}{16}a = 50\) and correctly solves to find \(a = 32\).

(a) To find the stationary points, write the curve's equation as \(y = 2x + 8x^{-1}\) and find the first derivative:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2 - 8x^{-2} = 2 - \frac{8}{x^2}\)

Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) for stationary points:
\(2 - \frac{8}{x^2} = 0 \implies \frac{8}{x^2} = 2 \implies x^2 = 4 \implies x = \pm 2\)

- For \(x = 2\):
\(y = 2(2) + \frac{8}{2} = 8\). So, the stationary point is \((2, 8)\).
- For \(x = -2\):
\(y = 2(-2) + \frac{8}{-2} = -8\). So, the stationary point is \((-2, -8)\).

To determine the nature of the stationary points, find the second derivative:
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}(2 - 8x^{-2}) = 16x^{-3} = \frac{16}{x^3}\)

- At \(x = 2\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{16}{2^3} = 2 > 0\). Since the second derivative is positive, \((2, 8)\) is a **minimum point**.
- At \(x = -2\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{16}{(-2)^3} = -2 < 0\). Since the second derivative is negative, \((-2, -8)\) is a **maximum point**.

(b) We are given that \(\frac{\mathrm{d}x}{\mathrm{d}t} = 0.5\) units/sec.
We need to find \(\frac{\mathrm{d}y}{\mathrm{d}t}\) when \(x = 4\).

By the chain rule:
\(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\)

First find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) at \(x = 4\):
\(\frac{\mathrm{d}y}{\mathrm{d}x}\Big|_{x=4} = 2 - \frac{8}{4^2} = 2 - \frac{8}{16} = 1.5\)

Then calculate \(\frac{\mathrm{d}y}{\mathrm{d}t}\):
\(\frac{\mathrm{d}y}{\mathrm{d}t} = 1.5 \times 0.5 = 0.75\) units per second.

(a)
* **M1**: Differentiates \(y\) with respect to \(x\).
* **A1**: Finds correct stationary values \(x = 2\) and \(x = -2\).
* **A1**: Finds correct coordinates of both stationary points: \((2, 8)\) and \((-2, -8)\).
* **M1**: Finds the second derivative \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{16}{x^3}\) and substitutes the \(x\)-values.
* **A1** (0.8 marks): Correctly concludes \((2, 8)\) is a minimum point and \((-2, -8)\) is a maximum point.

(b)
* **M1**: Applies the chain rule \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\) and calculates \(\frac{\mathrm{d}y}{\mathrm{d}x}\) at \(x=4\).
* **A1**: Correctly evaluates \(\frac{\mathrm{d}y}{\mathrm{d}t} = 0.75\).

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

Using the reverse chain rule for integration:
\(\int 3(2x + 1)^{-1/4} \mathrm{d}x = \frac{3(2x + 1)^{3/4}}{\frac{3}{4} \times 2} = \frac{3(2x + 1)^{3/4}}{1.5} = 2(2x + 1)^{3/4}\)

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

Since \(81 = 3^4\), we have:
\(81^{3/4} = (3^4)^{3/4} = 3^3 = 27\)

Therefore:
\(A = 2(27) - 2(1) = 54 - 2 = 52\).

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

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

Now integrate \(y^2\) with respect to \(x\) over the same interval:
\(V = \pi \int_{0}^{40} 9(2x + 1)^{-1/2} \mathrm{d}x\)

Using the reverse chain rule:
\(\int 9(2x + 1)^{-1/2} \mathrm{d}x = \frac{9(2x + 1)^{1/2}}{\frac{1}{2} \times 2} = 9\sqrt{2x + 1}\)

Evaluate at the limits:
\(V = \pi \left[ 9\sqrt{2x + 1} \right]_{0}^{40}
= \pi \left( 9\sqrt{2(40) + 1} - 9\sqrt{2(0) + 1} \right)
= \pi \left( 9\sqrt{81} - 9\sqrt{1} \right)
= \pi (9(9) - 9(1))
= \pi (81 - 9) = 72\pi\).

So the exact volume is \(72\pi\).

(a)
* **M1**: Expresses the area as \(\int 3(2x + 1)^{-1/4} \mathrm{d}x\) and attempts integration.
* **A1**: Obtains correct integrated term \(2(2x+1)^{3/4}\).
* **M1**: Correctly applies limits of \(0\) and \(40\) to their integrated expression.
* **A1** (0.8 marks): Obtains final area value \(52\).

(b)
* **M1**: Uses the volume formula \(V = \pi \int y^2 \mathrm{d}x\), correctly squaring the function to get \(y^2 = 9(2x+1)^{-1/2}\).
* **A1**: Performs integration to get \(9\sqrt{2x+1}\) (condone missing \(\pi\) during integration but must be present in final answer).
* **A1**: Substitutes limits and finds exact volume as \(72\pi\) (or equivalent exact expression like \(72 \pi\)).

(a) Let the terms of the arithmetic progression be \(u_n = a + (n-1)d\). The 9th term is \(u_9 = a + 8d\), the 3rd term is \(u_3 = a + 2d\), and the 1st term is \(u_1 = a\). Since these represent the first three consecutive terms of a geometric progression, the ratio between consecutive terms is constant: \(\frac{a+2d}{a+8d} = \frac{a}{a+2d}\). Cross-multiplying gives \((a+2d)^2 = a(a+8d)\). Expanding both sides: \(a^2 + 4ad + 4d^2 = a^2 + 8ad\). Subtracting \(a^2\) and \(4ad\) from both sides gives \(4d^2 = 4ad\). Since \(d \neq 0\), we divide by \(4d\) to obtain \(d = a\). (b) The sum of the first 8 terms of an arithmetic progression is given by \(S_8 = \frac{8}{2}(2a + 7d) = 4(2a + 7d)\). We are given \(S_8 = 360\), so \(4(2a + 7d) = 360 \implies 2a + 7d = 90\). Since \(d = a\), substitute \(d\) with \(a\): \(2a + 7a = 90 \implies 9a = 90 \implies a = 10\). Thus, \(a = 10\) and \(d = 10\). (c) The first term of the geometric progression is \(v_1 = u_9 = a + 8d = 10 + 8(10) = 90\). The second term is \(v_2 = u_3 = a + 2d = 10 + 2(10) = 30\). Thus, the common ratio is \(r = \frac{v_2}{v_1} = \frac{30}{90} = \frac{1}{3}\). The sum to infinity is given by \(S_{\infty} = \frac{v_1}{1-r} = \frac{90}{1 - 1/3} = \frac{90}{2/3} = 135\).

(a) M1: For setting up the ratio equation based on geometric progression properties. A1: For expanding and simplifying to a quadratic relation in \(a\) and \(d\). A1: For correctly concluding \(d = a\). (b) M1: For using the sum of arithmetic progression formula and substituting \(d = a\). A1: For finding \(a = 10\) and \(d = 10\). (c) M1: For finding the first term and the common ratio of the geometric progression, and using the sum to infinity formula. A1: For the final answer of 135.

(a) To find the equation of the curve, we integrate the derivative: \(f(x) = \int \left(6x^{-1/2} - 3\right) dx = \frac{6x^{1/2}}{1/2} - 3x + C = 12x^{1/2} - 3x + C\). We are given that the curve passes through \((9, 12)\). Substituting these coordinates: \(12 = 12\sqrt{9} - 3(9) + C \implies 12 = 12(3) - 27 + C \implies 12 = 36 - 27 + C \implies 12 = 9 + C \implies C = 3\). Thus, the equation of the curve is \(y = 12\sqrt{x} - 3x + 3\). (b) At the stationary point, \(f'(x) = 0\): \(\frac{6}{\sqrt{x}} - 3 = 0 \implies \frac{6}{\sqrt{x}} = 3 \implies \sqrt{x} = 2 \implies x = 4\). To find the y-coordinate, substitute \(x = 4\) into the equation of the curve: \(y = 12\sqrt{4} - 3(4) + 3 = 12(2) - 12 + 3 = 15\). Thus, the stationary point is at \((4, 15)\). To determine its nature, find the second derivative: \(f''(x) = \frac{d}{dx}\left(6x^{-1/2} - 3\right) = -3x^{-3/2}\). Evaluating the second derivative at \(x = 4\): \(f''(4) = -3(4)^{-3/2} = -3\left(\frac{1}{8}\right) = -\frac{3}{8}\). Since \(f''(4) < 0\), the stationary point is a maximum.

(a) M1: For integrating the derivative with at least one power increased correctly. A1: For the correct integration terms \(12x^{1/2} - 3x\). M1: For substituting \((9, 12)\) into their integrated expression including the constant \(C\). A1: For the correct curve equation with \(C = 3\). (b) M1: For setting \(f'(x) = 0\) and solving for \(x\). A1: For finding the coordinates \((4, 15)\). A1: For evaluating the second derivative (or alternative method) and concluding that the stationary point is a maximum.

(a) Combine the fractions on the LHS over a common denominator: \(\frac{\sin\theta(1+\cos\theta) + \sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}\). Expand the numerator and the denominator: \(\frac{\sin\theta + \sin\theta\cos\theta + \sin\theta - \sin\theta\cos\theta}{1 - \cos^2\theta}\). Simplify the expression: \(\frac{2\sin\theta}{\sin^2\theta}\) (since \(1 - \cos^2\theta = \sin^2\theta\)). Dividing both the numerator and denominator by \\sin\theta\\, we get: \(\frac{2}{\sin\theta}\). Thus the identity is proven. (b) Using the identity, rewrite the equation as: \(\frac{2}{\sin\theta} = \frac{4}{3}\tan\theta\). Since \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), we have: \(\frac{2}{\sin\theta} = \frac{4\sin\theta}{3\cos\theta}\). Cross-multiplying yields: \(6\cos\theta = 4\sin^2\theta \implies 3\cos\theta = 2\sin^2\theta\). Use \(\sin^2\theta = 1 - \cos^2\theta\) to form a quadratic equation in \(\cos\theta\): \(3\cos\theta = 2(1 - \cos^2\theta) \implies 2\cos^2\theta + 3\cos\theta - 2 = 0\). Factoring the quadratic: \((2\cos\theta - 1)(\cos\theta + 2) = 0\). This gives two possible cases: 1) \(\cos\theta = \frac{1}{2}\), which within the range \(0^{\circ} \le \theta \le 360^{\circ}\) gives \(\theta = 60^{\circ}\) and \(\theta = 300^{\circ}\). 2) \(\cos\theta = -2\), which has no solutions since \(-1 \le \cos\theta \le 1\). Therefore, the solutions are \(\theta = 60^{\circ}\) and \(\theta = 300^{\circ}\).

(a) M1: For combining fractions with a common denominator. A1: For expanding the numerator and denominator correctly and using \(1 - \cos^2\theta = \sin^2\theta\). A1: For obtaining the final expression \(\frac{2}{\sin\theta}\) with all steps clearly shown. (b) M1: For substituting the identity to simplify the left-hand side of the equation. M1: For expressing \(\tan\theta\) as \(\frac{\sin\theta}{\cos\theta}\) and obtaining a relation between \(\sin^2\theta\) and \(\cos\theta\). M1: For substituting \(\sin^2\theta = 1 - \cos^2\theta\) and forming a three-term quadratic in \(\cos\theta\). A1: For solving the quadratic to get \(\cos\theta = 0.5\) and finding both solutions \(\theta = 60^{\circ}, 300^{\circ}\) and no others in the range.

(i) To solve \( |2x - 5| < 3x + 1 \), we first require that the right-hand side is positive, so \( 3x + 1 > 0 \implies x > -\frac{1}{3} \).

Now, we solve the inequality by considering two cases:

Case 1: \( 2x - 5 \ge 0 \implies x \ge 2.5 \)
\( 2x - 5 < 3x + 1 \implies -6 < x \).
Intersecting with \( x \ge 2.5 \), we get \( x \ge 2.5 \).

Case 2: \( 2x - 5 < 0 \implies x < 2.5 \)
\( -(2x - 5) < 3x + 1 \implies -2x + 5 < 3x + 1 \implies 4 < 5x \implies x > 0.8 \).
Intersecting with \( x < 2.5 \), we get \( 0.8 < x < 2.5 \).

Combining both cases, the complete solution is \( x > 0.8 \).

(ii) Substituting \( x = e^{-y} \) into the inequality from part (i), we have \( e^{-y} > 0.8 \).
Taking the natural logarithm of both sides:
\( \ln(e^{-y}) > \ln(0.8) \)
\( -y > \ln\left(\frac{4}{5}\right) \)
\( y < -\ln\left(\frac{4}{5}\right) = \ln\left(\frac{5}{4}\right) \)
\( y < \ln(1.25) \).

(i)
M1: For attempting to solve the critical boundary equation, e.g., \( -(2x - 5) = 3x + 1 \) or squaring both sides to obtain a quadratic equation.
A1: For finding the correct critical boundary value of \( x = 0.8 \).
M1: For identifying that the inequality requires \( 3x + 1 > 0 \).
A1: For the correct solution interval \( x > 0.8 \).

(ii)
M1: For substituting \( e^{-y} > 0.8 \) using the result from part (i).
M1: For taking the natural logarithm of both sides correctly.
A1.1: For obtaining \( y < \ln(1.25) \) or any equivalent exact log expression.

(i) Taking the natural logarithm of both sides of \( y = A b^x \):
\( \ln y = \ln(A b^x) \implies \ln y = \ln A + x \ln b \).

This is in the linear form \( \ln y = mx + c \), where gradient \( m = \ln b \) and vertical intercept \( c = \ln A \).

Using the points \( (1, 2.45) \) and \( (4, 4.25) \) to find the gradient \( m \):
\( m = \frac{4.25 - 2.45}{4 - 1} = \frac{1.8}{3} = 0.6 \).

To find \( c \), substitute \( (1, 2.45) \) into the equation \( \ln y = 0.6x + c \):
\( 2.45 = 0.6(1) + c \implies c = 1.85 \).

(ii) From part (i), we have:
\( \ln b = m = 0.6 \implies b = e^{0.6} \approx 1.8221 \approx 1.82 \) (to 2 d.p.).

\( \ln A = c = 1.85 \implies A = e^{1.85} \approx 6.3598 \approx 6.36 \) (to 2 d.p.).

(i)
M1: For attempting to find the gradient of the line using the two given coordinates.
A1: For obtaining \( m = 0.6 \).
A1: For substituting a point into the equation of the line and obtaining \( c = 1.85 \).

(ii)
M1: For equating \( \ln b = 0.6 \) and attempting to solve for \( b \).
A1: For obtaining \( b \approx 1.82 \).
M1: For equating \( \ln A = 1.85 \) and attempting to solve for \( A \).
A1.1: For obtaining \( A \approx 6.36 \).

(i) Using the double angle identity \( \cos 2\theta = 1 - 2\sin^2\theta \):
\( f(\theta) = 3(1 - 2\sin^2\theta) + 8\sin\theta - 5 \)
\( f(\theta) = 3 - 6\sin^2\theta + 8\sin\theta - 5 \)
\( f(\theta) = -6\sin^2\theta + 8\sin\theta - 2 \).

(ii) Setting \( f(\theta) = 0 \):
\( -6\sin^2\theta + 8\sin\theta - 2 = 0 \)
Dividing by \( -2 \):
\( 3\sin^2\theta - 4\sin\theta + 1 = 0 \)

This factors as:
\( (3\sin\theta - 1)(\sin\theta - 1) = 0 \)

Thus, we have two cases:
1) \( \sin\theta = 1 \implies \theta = 90^\circ \).
2) \( \sin\theta = \frac{1}{3} \implies \theta = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ \approx 19.5^\circ \).
Another solution in the range is:
\( \theta = 180^\circ - 19.47^\circ \approx 160.5^\circ \).

Therefore, the solutions are \( \theta = 19.5^\circ, 90^\circ, 160.5^\circ \).

(i)
M1: For applying the correct double angle identity \( \cos 2\theta = 1 - 2\sin^2\theta \).
A1: For obtaining \( -6\sin^2\theta + 8\sin\theta - 2 \).

(ii)
M1: For setting their expression to 0 and attempting to factorise or solve the quadratic in \( \sin\theta \).
A1: For obtaining the correct roots \( \sin\theta = 1 \) and \( \sin\theta = \frac{1}{3} \).
B1: For the solution \( \theta = 90^\circ \).
M1: For finding one correct angle for \( \sin\theta = \frac{1}{3} \) (approx. \( 19.5^\circ \)).
A1.1: For finding both \( 19.5^\circ \) and \( 160.5^\circ \) correctly.

(i) Differentiating both \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = 2t + 2 = 2(t + 1) \)
\( \frac{dy}{dt} = \frac{2}{2t + 1} \)

Now, use the chain rule to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{2}{2t+1}}{2(t+1)} = \frac{1}{(2t+1)(t+1)} = \frac{1}{2t^2 + 3t + 1} \).

(ii) Set \( \frac{dy}{dx} = \frac{1}{3} \):
\( \frac{1}{2t^2 + 3t + 1} = \frac{1}{3} \implies 2t^2 + 3t + 1 = 3 \implies 2t^2 + 3t - 2 = 0 \).

Factorising this quadratic equation:
\( (2t - 1)(t + 2) = 0 \).

Since \( t > -\frac{1}{2} \), we reject \( t = -2 \), which leaves \( t = \frac{1}{2} \).

Substituting \( t = \frac{1}{2} \) back into the parametric equations:
\( x = \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) = 1.25 \) or \( \frac{5}{4} \).
\( y = \ln\left(2\left(\frac{1}{2}\right) + 1\right) = \ln(2) \).

So, the coordinates of the point are \( \left(\frac{5}{4}, \ln 2\right) \).

(i)
M1: For differentiating \( x \) correctly to obtain \( 2t + 2 \).
M1: For differentiating \( y \) correctly using chain rule to obtain \( \frac{2}{2t+1} \).
A1: For combining correctly to get \( \frac{dy}{dx} = \frac{1}{(2t+1)(t+1)} \).

(ii)
M1: For setting their \( \frac{dy}{dx} \) equal to \( \frac{1}{3} \) and attempting to solve the resulting quadratic equation for \( t \).
A1: For finding the correct value \( t = 0.5 \) (and rejecting \( t = -2 \)).
M1: For substituting their positive \( t \) back into both parametric equations.
A1.1: For obtaining the exact coordinates \( \left(\frac{5}{4}, \ln 2\right) \).

(i) To find \( \int_{0}^{2} e^{3x-1} \, dx \):
Using the standard result \( \int e^{ax+b} \, dx = \frac{1}{a} e^{ax+b} \):
\( \int_{0}^{2} e^{3x-1} \, dx = \left[ \frac{1}{3} e^{3x-1} \right]_{0}^{2} = \frac{1}{3} e^5 - \frac{1}{3} e^{-1} = \frac{1}{3}(e^5 - e^{-1}) \).

(ii) To find \( \int_{0}^{\pi/4} \cos^2(2x) \, dx \):
Using the identity \( \cos^2(\theta) = \frac{1}{2}(1 + \cos 2\theta) \) where \( \theta = 2x \):
\( \cos^2(2x) = \frac{1}{2}(1 + \cos 4x) \).

Integrating this expression:
\( \int_{0}^{\pi/4} \cos^2(2x) \, dx = \int_{0}^{\pi/4} \frac{1}{2}(1 + \cos 4x) \, dx = \frac{1}{2} \left[ x + \frac{1}{4} \sin 4x \right]_{0}^{\pi/4} \)
\( = \frac{1}{2} \left( \left(\frac{\pi}{4} + \frac{1}{4} \sin\pi \right) - (0 + 0) \right) = \frac{1}{2} \left(\frac{\pi}{4}\right) = \frac{\pi}{8} \).

(i)
M1: For attempting to integrate \( e^{3x-1} \) correctly to obtain \( k e^{3x-1} \) where \( k = \frac{1}{3} \).
A1: For the correct integrated expression \( \frac{1}{3} e^{3x-1} \).
A1: For substituting the limits correctly and obtaining the exact value \( \frac{1}{3}(e^5 - e^{-1}) \).

(ii)
M1: For expressing \( \cos^2(2x) \) in terms of \( \cos(4x) \) using a double angle identity.
A1: For the correct integration to obtain \( \frac{1}{2}x + \frac{1}{8}\sin(4x) \).
M1: For substituting limits \( \frac{\pi}{4} \) and \( 0 \) into their integrated expression.
A1.1: For obtaining the exact value \( \frac{\pi}{8} \).

(i) Let \( f(x) = x^3 - 5x - 3 \).
Evaluate \( f(x) \) at \( x = 2 \) and \( x = 3 \):
\( f(2) = (2)^3 - 5(2) - 3 = 8 - 10 - 3 = -5 < 0 \)
\( f(3) = (3)^3 - 5(3) - 3 = 27 - 15 - 3 = 9 > 0 \)

Since \( f(x) \) is continuous and there is a change of sign between \( x = 2 \) and \( x = 3 \), the root \( \alpha \) must lie between 2 and 3.

(ii) If the sequence converges to a limit \( L \), then:
\( L = \sqrt{\frac{3}{L} + 5} \)
Squaring both sides:
\( L^2 = \frac{3}{L} + 5 \)
Multiplying by \( L \):
\( L^3 = 3 + 5L \implies L^3 - 5L - 3 = 0 \).
This is the original equation, which is satisfied by \( L = \alpha \). Thus, if the formula converges, it converges to \( \alpha \).

(iii) Using \( x_1 = 2.5 \):
\( x_2 = \sqrt{\frac{3}{2.5} + 5} = \sqrt{6.2} \approx 2.4900 \)
\( x_3 = \sqrt{\frac{3}{2.4900} + 5} \approx 2.4909 \)
\( x_4 = \sqrt{\frac{3}{2.4909} + 5} \approx 2.4909 \)

Since successive iterations converge to \( 2.49 \) to two decimal places, we conclude that \( \alpha \approx 2.49 \).

(i)
M1: For evaluating \( f(2) \) and \( f(3) \) with at least one correct calculation.
A1: For both correct values (\( -5 \) and \( 9 \)) and a conclusion referencing the sign change.

(ii)
M1: For substituting \( x_{n+1} = x_n = L \) into the iterative formula.
A1: For algebraic manipulation showing the equation reduces to \( L^3 - 5L - 3 = 0 \).

(iii)
M1: For calculating \( x_2 \) correctly to at least 3 d.p.
A1: For showing iterations \( x_2 \approx 2.4900 \) and \( x_3 \approx 2.4909 \).
A1.1: For concluding that the root is \( 2.49 \) correct to 2 decimal places, supported by convergent iterations.

(i) Since \( (x - 2) \) is a factor of \( p(x) \), by the factor theorem, \( p(2) = 0 \):
\( 2(2)^3 + a(2)^2 - 11(2) + b = 0 \implies 16 + 4a - 22 + b = 0 \implies 4a + b = 6 \) (Equation 1)

Since the remainder when \( p(x) \) is divided by \( (x + 1) \) is 6, by the remainder theorem, \( p(-1) = 6 \):
\( 2(-1)^3 + a(-1)^2 - 11(-1) + b = 6 \implies -2 + a + 11 + b = 6 \implies a + b = -3 \) (Equation 2)

Subtracting Equation 2 from Equation 1:
\( (4a + b) - (a + b) = 6 - (-3) \implies 3a = 9 \implies a = 3 \).
Substitute \( a = 3 \) back into Equation 2:
\( 3 + b = -3 \implies b = -6 \).

Thus, \( a = 3 \) and \( b = -6 \).

(ii) Using the values of \( a \) and \( b \), the polynomial is:
\( p(x) = 2x^3 + 3x^2 - 11x - 6 \).

Since \( (x - 2) \) is a factor, we can perform polynomial division or use algebraic inspection to find the quadratic factor:
\( 2x^3 + 3x^2 - 11x - 6 = (x - 2)(2x^2 + 7x + 3) \).

Factorising this quadratic expression:
\( 2x^2 + 7x + 3 = (2x + 1)(x + 3) \).

Therefore, the completely factorised form is:
\( p(x) = (x - 2)(2x + 1)(x + 3) \).

(i)
M1: For applying the factor theorem to write the equation \( p(2) = 0 \).
A1: For obtaining \( 4a + b = 6 \) (or equivalent).
M1: For applying the remainder theorem to write the equation \( p(-1) = 6 \).
A1: For obtaining both correct values: \( a = 3 \) and \( b = -6 \).

(ii)
M1: For attempting division or inspection to find the quadratic factor of \( 2x^3 + 3x^2 - 11x - 6 \) using their values of \( a \) and \( b \).
A1: For obtaining the correct quadratic factor \( 2x^2 + 7x + 3 \).
A1.1: For factorising completely to obtain \( (x - 2)(2x + 1)(x + 3) \).