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(a) Let \(r\) be the radius of the water surface. By similar triangles, \(\frac{r}{h} = \frac{5}{15} = \frac{1}{3}\), so \(r = \frac{h}{3}\). The volume of water is \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h = \frac{1}{27}\pi h^3\). (b) We are given \(\frac{dV}{dt} = 12\pi\). Differentiating \(V\) with respect to \(h\) gives \(\frac{dV}{dh} = \frac{1}{9}\pi h^2\). When \(h = 6\), \(\frac{dV}{dh} = \frac{1}{9}\pi (6^2) = 4\pi\). Using the chain rule, \(\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh} = \frac{12\pi}{4\pi} = 3\text{ cm/s}\). (c) The area of the surface of the water is \(A = \pi r^2 = \pi \left(\frac{h}{3}\right)^2 = \frac{1}{9}\pi h^2\). Differentiating \(A\) with respect to \(h\) gives \(\frac{dA}{dh} = \frac{2}{9}\pi h\). When \(h = 6\), \(\frac{dA}{dh} = \frac{2}{9}\pi (6) = \frac{4}{3}\pi\). By the chain rule, \(\frac{dA}{dt} = \frac{dA}{dh} \times \frac{dh}{dt} = \frac{4}{3}\pi \times 3 = 4\pi\text{ cm}^2\text{s}^{-1}\).

(a) M1 for using similar triangles to express \(r\) in terms of \(h\). A1 for establishing \(V = \frac{1}{27}\pi h^3\). (b) M1 for differentiating \(V\) to find \(\frac{dV}{dh}\). M1 for using the chain rule relating \(\frac{dh}{dt}\), \(\frac{dV}{dt}\), and \(\frac{dV}{dh}\). A1 for \(3\text{ cm/s}\). (c) M1 for expressing \(A\) in terms of \(h\) and finding \(\frac{dA}{dh}\). M1 for applying chain rule \(\frac{dA}{dt} = \frac{dA}{dh} \times \frac{dh}{dt}\). A1 for \(4\pi\) (or \(12.6\)).

(a) Divide the entire equation by \(\cos^2 x\) (since \(\cos x \ne 0\)): \(3 - 4 \frac{\sin x}{\cos x} - \frac{\sin^2 x}{\cos^2 x} = 0 \implies 3 - 4 \tan x - \tan^2 x = 0\). Rearranging terms, we get \(\tan^2 x + 4 \tan x - 3 = 0\). (b) Let \(u = \tan x\). The equation is \(u^2 + 4u - 3 = 0\). Using the quadratic formula, \(u = \frac{-4 \pm \sqrt{4^2 - 4(1)(-3)}}{2} = \frac{-4 \pm \sqrt{28}}{2} = -2 \pm \sqrt{7}\). For \(\tan x = -2 + \sqrt{7} \approx 0.64575\), since \(0^\circ \le x \le 180^\circ\), \(x = \tan^{-1}(0.64575) \approx 32.9^\circ\). For \(\tan x = -2 - \sqrt{7} \approx -4.64575\), since \(0^\circ \le x \le 180^\circ\), \(x = 180^\circ - \tan^{-1}(4.64575) \approx 180^\circ - 77.9^\circ = 102.1^\circ\).

(a) M1 for dividing by \(\cos^2 x\) and using \(\tan x = \frac{\sin x}{\cos x}\). A1 for completing the derivation to show \(\tan^2 x + 4 \tan x - 3 = 0\). (b) M1 for using the quadratic formula (or completing the square) to solve for \(\tan x\). A1 for obtaining \(\tan x = -2 \pm \sqrt{7}\). M1 for finding one correct acute angle/reference angle. A1 for \(32.9^\circ\). M1 for finding the second quadrant angle for negative tangent value. A1 for \(102.1^\circ\).

(a) The expansion of \((1+ax)^n\) is \(1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \dots\). Comparing terms, we get: (1) \(an = -30\) and (2) \(\frac{n(n-1)}{2} a^2 = 375 \implies n(n-1)a^2 = 750\). From (1), \(a = -\frac{30}{n}\). Substitute this into (2): \(n(n-1)\left(-\frac{30}{n}\right)^2 = 750 \implies n(n-1)\frac{900}{n^2} = 750 \implies 900(n-1) = 750n \implies 900n - 900 = 750n \implies 150n = 900 \implies n = 6\). Substituting \(n = 6\) back into (1) gives \(a(6) = -30 \implies a = -5\). (b) The fourth term in the expansion is \(\binom{n}{3}(ax)^3 = \binom{6}{3}(-5x)^3 = 20 \times (-125)x^3 = -2500x^3\). Therefore, the coefficient of \(x^3\) is \(-2500\).

(a) M1 for writing down \(an = -30\). M1 for writing down \(\frac{n(n-1)}{2}a^2 = 375\). M1 for eliminating one variable (either \(a\) or \(n\)) to form an equation in one variable. A1 for \(n = 6\). A1 for \(a = -5\). (b) M1 for identifying the general term of \(x^3\) as \(\binom{n}{3}a^3\). M1 for substituting their values of \(n\) and \(a\). A1 for \(-2500\).

(a) Factor out 2 from the quadratic terms: \(\mathrm{f}(x) = 2(x^2 - 6x) + 13\). Complete the square inside the bracket: \(x^2 - 6x = (x-3)^2 - 9\). Thus, \(\mathrm{f}(x) = 2[(x-3)^2 - 9] + 13 = 2(x-3)^2 - 18 + 13 = 2(x-3)^2 - 5\). So \(a = 2\), \(b = 3\), \(c = -5\). (b) Since the vertex of the parabola is at \((3, -5)\) and the coefficient of \(x^2\) is positive, the minimum value is \(-5\). For \(x \ge 3\), the range of \(\mathrm{f}\) is \(\mathrm{f}(x) \ge -5\). (c) Let \(y = 2(x-3)^2 - 5\). Make \(x\) the subject: \(y + 5 = 2(x-3)^2 \implies \frac{y+5}{2} = (x-3)^2\). Taking the square root, since \(x \ge 3\), we select the positive root: \(x - 3 = \sqrt{\frac{y+5}{2}} \implies x = 3 + \sqrt{\frac{y+5}{2}}\). Hence, \(\mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\). The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\), which is \(x \ge -5\).

(a) M1 for attempting to complete the square (e.g. factoring 2 or identifying \(b=3\)). A1 for \(2(x-3)^2\). A1 for \(c = -5\). (b) B1 for \(\mathrm{f}(x) \ge -5\) (accept \(y \ge -5\)). (c) M1 for setting \(y = 2(x-3)^2 - 5\) and attempting to make \((x-3)^2\) or \(x\) the subject. M1 for correctly taking the positive square root because of the domain \(x \ge 3\). A1 for \(\mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\). B1 for domain \(x \ge -5\) (must be in terms of \(x\)).

(a) Using the change of base formula, we have \(\log_x 3 = \frac{\log_3 3}{\log_3 x} = \frac{1}{\log_3 x}\). Substituting \(u = \log_3 x\), the equation \(\log_3 x - 4 \log_x 3 + 3 = 0\) becomes \(u - \frac{4}{u} + 3 = 0\). Multiplying the entire equation by \(u\) (where \(u \ne 0\)) yields \(u^2 + 3u - 4 = 0\). (b) Solving the quadratic equation \(u^2 + 3u - 4 = 0\) by factoring: \((u + 4)(u - 1) = 0\), which gives \(u = -4\) or \(u = 1\). Case 1: \(\log_3 x = -4 \implies x = 3^{-4} = \frac{1}{81}\). Case 2: \(\log_3 x = 1 \implies x = 3^1 = 3\). Both solutions are positive and not equal to 1, so they are both valid.

(a) M1 for using the change of base identity \(\log_x 3 = \frac{1}{\log_3 x}\). M1 for substituting \(u\) into the equation. A1 for correctly showing \(u^2 + 3u - 4 = 0\). (b) M1 for solving the quadratic equation to find \(u\). A1 for \(u = -4\) and \(u = 1\). M1 for setting up \(\log_3 x = -4\) and \(\log_3 x = 1\). A1 for \(x = 3\). A1 for \(x = \frac{1}{81}\) (or \(3^{-4}\)).

(a) Equating the line and curve: \(mx + 2 = x^2 - 4x + 6 \implies x^2 - (4+m)x + 4 = 0\). For two distinct points of intersection, the discriminant of this quadratic equation must be strictly greater than 0: \(\Delta = b^2 - 4ac > 0 \implies [-(4+m)]^2 - 4(1)(4) > 0 \implies m^2 + 8m + 16 - 16 > 0 \implies m^2 + 8m > 0\). Factorising: \(m(m + 8) > 0\). Thus, the set of values is \(m < -8\) or \(m > 0\). (b) When \(m = 1\), the quadratic equation becomes \(x^2 - 5x + 4 = 0\). Factorising: \((x - 1)(x - 4) = 0\), which gives \(x = 1\) or \(x = 4\). For \(x = 1\), \(y = 1(1) + 2 = 3\). For \(x = 4\), \(y = 1(4) + 2 = 6\). The coordinates of the intersection points are \((1, 3)\) and \((4, 6)\).

(a) M1 for equating the line and the curve. M1 for forming a three-term quadratic equation in \(x\). M1 for applying the discriminant condition \(b^2 - 4ac > 0\). A1 for finding the critical values \(m = -8, 0\). A1 for the correct ranges: \(m < -8\) or \(m > 0\). (b) M1 for substituting \(m = 1\) and attempting to solve the quadratic equation. A1 for \(x = 1\) and \(x = 4\). A1 for \((1, 3)\) and \((4, 6)\).

(a) Since \(OP : PA = 2 : 1\), we have \(\overrightarrow{OP} = \frac{2}{3}\overrightarrow{OA} = \frac{2}{3}\mathbf{a}\). Since \(OQ : QB = 3 : 1\), we have \(\overrightarrow{OQ} = \frac{3}{4}\overrightarrow{OB} = \frac{3}{4}\mathbf{b}\). (b) Expressing \(\overrightarrow{OX}\) using \(\mu\): \(\overrightarrow{OX} = \mu \left(\frac{2}{3}\mathbf{a}\right) + (1-\mu)\mathbf{b} = \frac{2}{3}\mu \mathbf{a} + (1-\mu)\mathbf{b}\). Expressing \(\overrightarrow{OX}\) using \(\lambda\): \(\overrightarrow{OX} = \lambda \left(\frac{3}{4}\mathbf{b}\right) + (1-\lambda)\mathbf{a} = (1-\lambda)\mathbf{a} + \frac{3}{4}\lambda \mathbf{b}\). Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): For \(\mathbf{a}\): \(\frac{2}{3}\mu = 1 - \lambda \implies \lambda = 1 - \frac{2}{3}\mu\). For \(\mathbf{b}\): \(1 - \mu = \frac{3}{4}\lambda\). Substitute the expression for \(\lambda\) into the second equation: \(1 - \mu = \frac{3}{4}\left(1 - \frac{2}{3}\mu\right) \implies 1 - \mu = \frac{3}{4} - \frac{1}{2}\mu \implies 1 - \frac{3}{4} = \mu - \frac{1}{2}\mu \implies \frac{1}{4} = \frac{1}{2}\mu \implies \mu = \frac{1}{2}\). Substitute \(\mu = \frac{1}{2}\) back to find \(\lambda\): \(\lambda = 1 - \frac{2}{3}\left(\frac{1}{2}\right) = 1 - \frac{1}{3} = \frac{2}{3}\).

(a) B1 for \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\). B1 for \(\overrightarrow{OQ} = \frac{3}{4}\mathbf{b}\). (b) M1 for writing \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) using \(\mu\). M1 for writing \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) using \(\lambda\). M1 for equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to form a system of simultaneous equations. M1 for solving the simultaneous equations for one variable. A1 for \(\mu = \frac{1}{2}\). A1 for \(\lambda = \frac{2}{3}\).

(a) In the right-angled triangle \(ORQ\) (with angle \(ORQ = 90^\circ\)), the area is given by \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OR \times RQ = 24\text{ cm}^2\). Substituting \(OR = 6\): \(\frac{1}{2} \times 6 \times RQ = 24 \implies 3 RQ = 24 \implies RQ = 8\text{ cm}\). In the right-angled triangle \(ORQ\), \(\tan \theta = \frac{RQ}{OR} = \frac{8}{6} = \frac{4}{3}\). Thus, \(\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 0.927295 \approx 0.927\) radians. (b) To find the perimeter of the shaded region, we need the lengths of arc \(PQ\), segment \(RP\), and segment \(RQ\). The radius \(r = OQ\). By Pythagoras' Theorem on triangle \(ORQ\): \(r^2 = OR^2 + RQ^2 = 6^2 + 8^2 = 100 \implies r = 10\text{ cm}\). The length of arc \(PQ = r\theta = 10 \times 0.927295 \approx 9.273\text{ cm}\). The length of segment \(RP = OP - OR = r - 6 = 10 - 6 = 4\text{ cm}\). The length of segment \(RQ = 8\text{ cm}\). Therefore, the perimeter of the shaded region is \(\text{Arc } PQ + RP + RQ = 9.273 + 4 + 8 = 21.273 \approx 21.3\text{ cm}\).

(a) M1 for using the area of triangle \(ORQ\) to find \(RQ = 8\). M1 for using trigonometry, \(\tan \theta = \frac{RQ}{OR}\). A1 for establishing \(\theta \approx 0.927\) (must show at least 4 significant figures before rounding, e.g., \(0.9273\)). (b) M1 for calculating the radius \(r = 10\) using Pythagoras' Theorem or trigonometric functions. M1 for calculating arc length \(s = r\theta = 10 \times 0.927...\). M1 for calculating \(RP = r - OR = 4\). M1 for summing the three lengths (arc \(PQ\) + \(RP\) + \(RQ\)). A1 for \(21.3\text{ cm}\) (accept \(21.2 - 21.4\)).

(a) First find the gradient of the curve by differentiation:
\( \frac{dy}{dx} = \frac{d}{dx}\left[ k(2x-1)^{-2} + 3x \right] = -2k(2x-1)^{-3} \cdot 2 + 3 = \frac{-4k}{(2x-1)^3} + 3 \)

At \( x = 1 \), the gradient of the tangent is:
\( m_T = \frac{-4k}{(2(1)-1)^3} + 3 = -4k + 3 \)

The equation of the normal is \( x + 5y = c \), which can be rearranged to \( y = -\frac{1}{5}x + \frac{c}{5} \). Thus, the gradient of the normal is \( m_N = -\frac{1}{5} \).

Since the tangent and normal are perpendicular:
\( m_T \cdot m_N = -1 \implies -4k + 3 = -\frac{1}{-1/5} = 5 \)
\( -4k = 2 \implies k = -0.5 \)

Now, substitute \( k = -0.5 \) and \( x = 1 \) into the equation of the curve to find the \( y \)-coordinate of the point of contact:
\( y = \frac{-0.5}{(2(1)-1)^2} + 3(1) = -0.5 + 3 = 2.5 \)

Since this point lies on the normal line:
\( 1 + 5(2.5) = c \implies c = 13.5 \)

(b) With \( k = -0.5 \), the equation of the curve is \( y = -\frac{1}{2}(2x-1)^{-2} + 3x \).

The area of the region is given by the definite integral:
\( \text{Area} = \int_{1}^{2} \left( -\frac{1}{2}(2x-1)^{-2} + 3x \right) dx \)
\( \text{Area} = \left[ \frac{-1/2 \cdot (2x-1)^{-1}}{-1 \cdot 2} + \frac{3x^2}{2} \right]_{1}^{2} \)
\( \text{Area} = \left[ \frac{1}{4(2x-1)} + \frac{3x^2}{2} \right]_{1}^{2} \)

Evaluating at the limits:
At \( x = 2 \):
\( \frac{1}{4(3)} + \frac{3(4)}{2} = \frac{1}{12} + 6 = \frac{73}{12} \)

At \( x = 1 \):
\( \frac{1}{4(1)} + \frac{3(1)}{2} = \frac{1}{4} + \frac{3}{2} = \frac{7}{4} = \frac{21}{12} \)

\( \text{Area} = \frac{73}{12} - \frac{21}{12} = \frac{52}{12} = \frac{13}{3} \)

(a)
M1: Differentiates the curve's equation to find \( \frac{dy}{dx} \) in terms of \( k \).
A1: Obtains \( \frac{dy}{dx} = \frac{-4k}{(2x-1)^3} + 3 \) or equivalent.
M1: Uses the perpendicular gradient rule \( m_T = 5 \) to form an equation in \( k \) and solve for \( k \).
A1: Obtains \( k = -0.5 \) (or \( -\frac{1}{2} \)).
A1: Finds the point \( (1, 2.5) \) and substitutes it into the normal equation to find \( c = 13.5 \) (or \( \frac{27}{2} \)).

(b)
M1: Integrates \( -\frac{1}{2}(2x-1)^{-2} + 3x \) to obtain \( \frac{1}{4(2x-1)} + \frac{3x^2}{2} \) (allow sign/coefficient errors for M1).
A1: Correct integration.
A1: Applies the limits of 1 and 2 correctly to obtain \( \frac{13}{3} \) (or \( 4\frac{1}{3} \) or \( 4.33 \)).

(a) Given the equation:
\( 3 \sin^2 \theta + 5 \cos \theta \sin \theta - 2 \cos^2 \theta = 0 \)

Divide the entire equation by \( \cos^2 \theta \) (assuming \( \cos \theta \neq 0 \)):
\( \frac{3 \sin^2 \theta}{\cos^2 \theta} + \frac{5 \cos \theta \sin \theta}{\cos^2 \theta} - \frac{2 \cos^2 \theta}{\cos^2 \theta} = 0 \)

Since \( \frac{\sin \theta}{\cos \theta} = \tan \theta \), this simplifies to:
\( 3 \left(\frac{\sin \theta}{\cos \theta}\right)^2 + 5 \left(\frac{\sin \theta}{\cos \theta}\right) - 2 = 0 \)
\( 3 \tan^2 \theta + 5 \tan \theta - 2 = 0 \) (as required).

(b) Solve the quadratic in \( \tan \theta \):
\( 3 \tan^2 \theta + 5 \tan \theta - 2 = 0 \)
\( (3 \tan \theta - 1)(\tan \theta + 2) = 0 \)

This gives two cases:
1) \( 3 \tan \theta - 1 = 0 \implies \tan \theta = \frac{1}{3} \)
2) \( \tan \theta + 2 = 0 \implies \tan \theta = -2 \)

Case 1: \( \tan \theta = \frac{1}{3} \)
The basic angle is \( \alpha = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.43^\circ \).
Since \( \tan \theta \) is positive, \( \theta \) lies in Quadrants 1 and 3:
\( \theta = 18.43^\circ \approx 18.4^\circ \)
\( \theta = 180^\circ + 18.43^\circ = 198.43^\circ \approx 198.4^\circ \)

Case 2: \( \tan \theta = -2 \)
The basic angle is \( \beta = \tan^{-1}(2) \approx 63.43^\circ \).
Since \( \tan \theta \) is negative, \( \theta \) lies in Quadrants 2 and 4:
\( \theta = 180^\circ - 63.43^\circ = 116.57^\circ \approx 116.6^\circ \)
\( \theta = 360^\circ - 63.43^\circ = 296.57^\circ \approx 296.6^\circ \)

Thus, the solutions in the interval \( 0^\circ \le \theta \le 360^\circ \) are:
\( \theta = 18.4^\circ, 116.6^\circ, 198.4^\circ, 296.6^\circ \)

(a)
M1: Divides the equation through by \( \cos^2 \theta \) (or equivalent method like dividing by \( \sin^2 \theta \) and then taking reciprocal).
A1: Shows clearly that \( \frac{\sin \theta}{\cos \theta} = \tan \theta \) is used to arrive at the given equation.

(b)
M1: Factorises or uses the quadratic formula on \( 3 \tan^2 \theta + 5 \tan \theta - 2 = 0 \).
A1: Obtains \( \tan \theta = \frac{1}{3} \) and \( \tan \theta = -2 \).
M1: Finds one correct angle for \( \tan \theta = \frac{1}{3} \) (usually \( 18.4^\circ \)).
A1: Obtains both \( 18.4^\circ \) and \( 198.4^\circ \) (accept answers that round to these values; deduct 1 mark overall for extra answers inside the range).
M1: Finds one correct angle for \( \tan \theta = -2 \) (usually \( 116.6^\circ \) or \( 296.6^\circ \)).
A1: Obtains both \( 116.6^\circ \) and \( 296.6^\circ \) (accept answers that round to these values).

To find the gradient function, we use the quotient rule on \(y = \frac{ax-3}{2x+1}\):
\(\frac{dy}{dx} = \frac{a(2x+1) - 2(ax-3)}{(2x+1)^2} = \frac{2ax + a - 2ax + 6}{(2x+1)^2} = \frac{a+6}{(2x+1)^2}\).

We are given that at \(x = 1\), the gradient is \(\frac{7}{9}\):
\(\frac{a+6}{(2(1)+1)^2} = \frac{7}{9}\)
\(\frac{a+6}{9} = \frac{7}{9}\)
\(a + 6 = 7 \implies a = 1\).

Since the point \(P(1, b)\) lies on the curve:
\(b = \frac{1(1) - 3}{2(1) + 1} = \frac{-2}{3}\).

Thus, \(a = 1\) and \(b = -\frac{2}{3}\).

M1: Attempt to differentiate \(y\) using the quotient rule or product rule.
A1: Obtain the correct derivative \(\frac{dy}{dx} = \frac{a+6}{(2x+1)^2}\).
M1: Set \(\frac{dy}{dx} = \frac{7}{9}\) at \(x=1\).
A1: Solve to find \(a = 1\).
M1: Substitute \(x=1\) and \(a=1\) into the original equation for \(y\).
A1: Obtain \(b = -\frac{2}{3}\).

Substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) into the equation:
\(6(1 - \sin^2 \theta) - \sin \theta - 5 = 0\)
\(6 - 6\sin^2 \theta - \sin \theta - 5 = 0\)
\(6\sin^2 \theta + \sin \theta - 1 = 0\)

Factorising the quadratic:
\((2\sin \theta + 1)(3\sin \theta - 1) = 0\)

This gives two cases:

Case 1: \(\sin \theta = -\frac{1}{2}\)
The basic angle is \(30^\circ\). Since sine is negative in the 3rd and 4th quadrants:
\(\theta = 180^\circ + 30^\circ = 210^\circ\)
\(\theta = 360^\circ - 30^\circ = 330^\circ\)

Case 2: \(\sin \theta = \frac{1}{3}\)
The basic angle is \(\sin^{-1}(1/3) \approx 19.47^\circ\). Since sine is positive in the 1st and 2nd quadrants:
\(\theta \approx 19.5^\circ\)
\(\theta = 180^\circ - 19.47^\circ \approx 160.5^\circ\)

All solutions in the range \(0^\circ \le \theta \le 360^\circ\) are \(\theta = 19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\).

M1: Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a quadratic equation in \(\sin \theta\).
A1: Obtain the correct quadratic \(6\sin^2 \theta + \sin \theta - 1 = 0\).
M1: Attempt to solve or factorise the quadratic equation.
A1: Obtain \(\sin \theta = \frac{1}{3}\) and \(\sin \theta = -\frac{1}{2}\).
B1: Find at least two correct angles.
A1: Obtain all four correct angles: \(19.5^\circ, 160.5^\circ, 210^\circ, 330^\circ\) (rounded to 1 decimal place).

The formula for the sum to infinity is \(S_\infty = \frac{a}{1-r} = 4.5\).
The formula for the sum of the first three terms is \(S_3 = \frac{a(1-r^3)}{1-r} = 3.5\).

(a) We can write \(S_3\) as \(S_\infty (1-r^3)\):
\(4.5(1-r^3) = 3.5\)
\(1-r^3 = \frac{3.5}{4.5} = \frac{7}{9}\)
\(r^3 = 1 - \frac{7}{9} = \frac{2}{9}\).

(b) Solving for \(r\):
\(r = \left(\frac{2}{9}\right)^{1/3} \approx 0.6057\)
To 3 decimal places, \(r = 0.606\).

(c) Solving for \(a\):
Using \(\frac{a}{1-r} = 4.5\):
\(a = 4.5(1 - r) \approx 4.5(1 - 0.6057) = 4.5(0.3943) = 1.774\).
To 2 decimal places, \(a = 1.77\).

M1: Write the correct formulas for \(S_\infty\) and \(S_3\).
M1: Substitute \(S_\infty\) into the formula for \(S_3\).
A1: Convincingly show that \(r^3 = \frac{2}{9}\).
M1: Calculate \(r\) by taking the cube root.
A1: Obtain \(r = 0.606\).
M1: Substitute \(r\) into the sum to infinity equation to solve for \(a\).
A1: Obtain \(a = 1.77\).

(a) Completing the square for \(\mathrm{f}(x)\):
\(\mathrm{f}(x) = 2(x^2 - 2x) + 5 = 2(x-1)^2 + 3\).
Since the domain is restricted to \(x \ge 1\), the function is strictly increasing, which means it is a one-to-one function. Therefore, its inverse exists.

(b) Let \(y = 2(x-1)^2 + 3\).
Rearranging to express \(x\) in terms of \(y\):
\(y - 3 = 2(x-1)^2 \implies \frac{y-3}{2} = (x-1)^2\).
Since \(x \ge 1\), \(x - 1 \ge 0\), we take the positive square root:
\(x - 1 = \sqrt{\frac{y-3}{2}} \implies x = 1 + \sqrt{\frac{y-3}{2}}\).
Thus, \(\mathrm{f}^{-1}(x) = 1 + \sqrt{\frac{x-3}{2}}\).
The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\). For \(x \ge 1\), \(\mathrm{f}(x) \ge 3\), so the domain is \(x \ge 3\).

(c) To solve \(\mathrm{gf}(x) = \frac{1}{3}\):
\(\mathrm{g}(\mathrm{f}(x)) = \frac{3}{\mathrm{f}(x) + 2} = \frac{1}{3}\)
\(9 = \mathrm{f}(x) + 2 \implies \mathrm{f}(x) = 7\).

Setting \(2x^2 - 4x + 5 = 7\):
\(2x^2 - 4x - 2 = 0 \implies x^2 - 2x - 1 = 0\).
Using the quadratic formula:
\(x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}\).
Since the domain of \(\mathrm{f}\) is \(x \ge 1\), we choose \(x = 1 + \sqrt{2}\).

B1: Explain that \(\mathrm{f}\) is one-to-one on the given domain (by completing the square or discussing the vertex).
M1: Complete the square or rearrange the equation to make \(x\) the subject.
A1: Obtain the correct expression \(\mathrm{f}^{-1}(x) = 1 + \sqrt{\frac{x-3}{2}}\).
B1: State the correct domain \(x \ge 3\).
M1: Solve \(\mathrm{g}(\mathrm{f}(x)) = \frac{1}{3}\) to find \(\mathrm{f}(x) = 7\).
M1: Solve \(2x^2 - 4x + 5 = 7\) using the quadratic formula.
A1: State the final correct answer \(x = 1 + \sqrt{2}\) and reject \(1 - \sqrt{2}\).

From the first equation:
\(\log_2(x - 2y) = 3 \implies x - 2y = 2^3 = 8 \implies x = 2y + 8\). (Equation 1)

From the second equation, write all terms with a common base of 3:
\(3^{x+1} \cdot (3^2)^{y-2} = (3^3)^y\)
\(3^{x+1} \cdot 3^{2y-4} = 3^{3y}\)
\(3^{x + 2y - 3} = 3^{3y}\)

Equating exponents:
\(x + 2y - 3 = 3y \implies x - y = 3 \implies x = y + 3\). (Equation 2)

Substitute Equation 2 into Equation 1:
\(y + 3 = 2y + 8 \implies y = -5\).

Now substitute \(y = -5\) back into Equation 2:
\(x = -5 + 3 = -2\).

Check validation: \(x - 2y = -2 - 2(-5) = 8 > 0\), which is valid for the logarithm.

Thus, the solution is \(x = -2, y = -5\).

M1: Convert the log equation to the exponential form \(x - 2y = 8\).
M1: Rewrite the second equation using base 3.
A1: Obtain the correct linear equation \(x - y = 3\) from the index properties.
M1: Attempt to solve the two linear simultaneous equations.
A1: Solve for \(y = -5\).
A1: Solve for \(x = -2\).
B1: State that \(x - 2y = 8 > 0\), showing the solution is valid for the log term.

(a) First find the vector \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (5\mathbf{i} - 2\mathbf{j}) - (\mathbf{i} + 4\mathbf{j}) = 4\mathbf{i} - 6\mathbf{j}\).

Since \(\overrightarrow{AC} = \lambda \overrightarrow{AB}\):
\(\overrightarrow{AC} = 4\lambda\mathbf{i} - 6\lambda\mathbf{j}\).

The position vector of \(C\) is:
\(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = (1 + 4\lambda)\mathbf{i} + (4 - 6\lambda)\mathbf{j}\).

We are given that the \(\mathbf{j}\)-component of \(\overrightarrow{OC}\) is \(-11\):
\(4 - 6\lambda = -11 \implies 6\lambda = 15 \implies \lambda = 2.5\).

(b) Substituting \(\lambda = 2.5\) into the expression for \(\overrightarrow{OC}\):
\(\overrightarrow{OC} = (1 + 4(2.5))\mathbf{i} - 11\mathbf{j} = 11\mathbf{i} - 11\mathbf{j}\).

(c) To find the unit vector in the direction of \(\overrightarrow{OC}\):
\(|\overrightarrow{OC}| = \sqrt{11^2 + (-11)^2} = \sqrt{242} = 11\sqrt{2}\).

The unit vector is:
\(\frac{\overrightarrow{OC}}{|\overrightarrow{OC}|} = \frac{11\mathbf{i} - 11\mathbf{j}}{11\sqrt{2}} = \frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}\).

M1: Obtain the vector \(\overrightarrow{AB} = 4\mathbf{i} - 6\mathbf{j}\).
M1: State an expression for \(\overrightarrow{OC}\) in terms of \(\lambda\).
A1: Equate the \(\mathbf{j}\)-component to \(-11\) and find \(\lambda = 2.5\).
A1: Substitute \(\lambda\) to find \(\overrightarrow{OC} = 11\mathbf{i} - 11\mathbf{j}\).
M1: Find the magnitude \(|\overrightarrow{OC}| = 11\sqrt{2}\) (or \(\approx 15.56\)).
A1: Divide the vector \(\overrightarrow{OC}\) by its magnitude to obtain \(\frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}\) (or equivalent decimal form).

(a) The area of the sector is:
\(A = \frac{1}{2}r^2\theta = 120 \implies \theta = \frac{240}{r^2}\).

The perimeter of the sector is:
\(P = 2r + r\theta = 46\).

Substitute \(\theta = \frac{240}{r^2}\) into the perimeter formula:
\(2r + r\left(\frac{240}{r^2}\right) = 46\)
\(2r + \frac{240}{r} = 46\).

Multiply the entire equation by \(r\):
\(2r^2 + 240 = 46r \implies 2r^2 - 46r + 240 = 0\). (Showed)

(b) Dividing the quadratic equation by 2:
\(r^2 - 23r + 120 = 0\)
\((r - 8)(r - 15) = 0\).

So the possible values of \(r\) are \(r = 8\) or \(r = 15\).

(c) For \(r = 8\):
\(\theta = \frac{240}{64} = 3.75\) radians.

For \(r = 15\):
\(\theta = \frac{240}{225} = 1.07\) radians (to 3 s.f.).

Since \(\pi \approx 3.14\), the value \(\theta = 3.75\) radians is a reflex angle (\(\theta > \pi\)). Thus, the corresponding pair is \(r = 8\text{ cm}\) and \(\theta = 3.75\text{ radians}\).

M1: State the formula for the area of a sector and obtain \(\theta = \frac{240}{r^2}\).
M1: State the formula for the perimeter of a sector and substitute the expression for \(\theta\).
A1: Correctly manipulate the terms to obtain the given quadratic equation \(2r^2 - 46r + 240 = 0\).
M1: Factorise or solve the quadratic equation to get two values for \(r\).
A1: Obtain the values \(r = 8\) and \(r = 15\).
M1: Substitute the \(r\) values back to calculate both values of \(\theta\).
A1: Identify that \(r = 8, \theta = 3.75\) is the correct pair because \(3.75 > \pi\).

(a) Since \(x - 2\) is a factor of \(\mathrm{p}(x)\), by the Factor Theorem:
\(\mathrm{p}(2) = 0 \implies 2(2)^3 + a(2)^2 + b(2) - 10 = 0\)
\(16 + 4a + 2b - 10 = 0 \implies 2a + b = -3\). (Equation 1)

Since the remainder when divided by \(x + 1\) is \(-9\), by the Remainder Theorem:
\(\mathrm{p}(-1) = -9 \implies 2(-1)^3 + a(-1)^2 + b(-1) - 10 = -9\)
\(-2 + a - b - 10 = -9 \implies a - b = 3\). (Equation 2)

Adding Equation 1 and Equation 2:
\(3a = 0 \implies a = 0\).

Substitute \(a = 0\) into Equation 2:
\(0 - b = 3 \implies b = -3\).

(b) Using \(a = 0\) and \(b = -3\), the polynomial is:
\(\mathrm{p}(x) = 2x^3 - 3x - 10\).

Dividing \(\mathrm{p}(x)\) by the factor \((x - 2)\):
\(2x^3 - 3x - 10 = (x-2)(2x^2 + cx + 5)\)
Comparing the \(x\) coefficients:
\(5 - 2c = -3 \implies 2c = 8 \implies c = 4\).

So the quadratic factor is \(2x^2 + 4x + 5\).

Let's check the discriminant of \(2x^2 + 4x + 5\):
\(\Delta = b^2 - 4ac = 4^2 - 4(2)(5) = 16 - 40 = -24 < 0\).
Since the discriminant is negative, the quadratic cannot be factorised further over the real numbers.
Thus, the complete factorisation over the real numbers is:
\(\mathrm{p}(x) = (x-2)(2x^2 + 4x + 5)\).

(c) To solve \(\mathrm{p}(x) = 0\):
\((x-2)(2x^2 + 4x + 5) = 0\).
Since the quadratic factor has no real roots, the only real solution is \(x = 2\).

M1: Set up the first linear equation using the Factor Theorem \(\mathrm{p}(2) = 0\).
M1: Set up the second linear equation using the Remainder Theorem \(\mathrm{p}(-1) = -9\).
A1: Solve the simultaneous equations to obtain \(a = 0\) and \(b = -3\).
M1: Perform algebraic long division or equate coefficients to divide \(\mathrm{p}(x)\) by \((x - 2)\).
A1: Obtain the correct quadratic factor \(2x^2 + 4x + 5\).
M1: Show that the quadratic factor has no real roots (e.g., using discriminant \(\Delta < 0\)).
A1: Identify that \(x = 2\) is the only real solution.

**(a)**
The second term is given by \(u_2 = ar = 2.4 \implies r = \frac{2.4}{a}\).
The sum to infinity is given by \(S_{\infty} = \frac{a}{1-r} = 10\).
Substituting \(r = \frac{2.4}{a}\) into the sum formula:
\(\frac{a}{1 - \frac{2.4}{a}} = 10\)
\(\frac{a^2}{a - 2.4} = 10\)
\(a^2 = 10(a - 2.4)\)
\(a^2 = 10a - 24\)
\(a^2 - 10a + 24 = 0\) (as required)

**(b)**
Solving \(a^2 - 10a + 24 = 0\) by factoring:
\((a - 6)(a - 4) = 0\)
This gives \(a = 6\) or \(a = 4\).
If \(a = 6\), then \(r = \frac{2.4}{6} = 0.4\).
If \(a = 4\), then \(r = \frac{2.4}{4} = 0.6\).

**(c)**
For the first progression (\(a = 6, r = 0.4\)):
\(u_4 = ar^3 = 6 \times (0.4)^3 = 6 \times 0.064 = 0.384\)
For the second progression (\(a = 4, r = 0.6\)):
\(u_4 = ar^3 = 4 \times (0.6)^3 = 4 \times 0.216 = 0.864\)
The positive difference is:
\(0.864 - 0.384 = 0.48\)

*M1* for using \(u_2 = ar\) and \(S_{\infty} = \frac{a}{1-r}\) to set up two equations.
*A1* for substituting \(r = \frac{2.4}{a}\) and simplifying correctly to show the quadratic equation \(a^2 - 10a + 24 = 0\).
*M1* for solving the quadratic equation to find two values of \(a\).
*A1* for obtaining both pairs: \(a=6, r=0.4\) and \(a=4, r=0.6\).
*M1* for calculating the fourth term of at least one of the progressions using \(u_4 = ar^3\).
*M1* for finding both fourth terms (\(0.384\) and \(0.864\)).
*A1* for the final correct positive difference of \(0.48\) (or \(\frac{12}{25}\)).

**(a)**
To find the intersection points, set the equations equal to each other:
\(3 + 2x - x^2 = 3 - x\)
Rearranging terms:
\(x^2 - 3x = 0\)
\(x(x - 3) = 0\)
So, the \(x\)-coordinates of \(A\) and \(B\) are \(x = 0\) and \(x = 3\).

**(b)**
The area of the enclosed region is given by the integral of the upper curve minus the lower line, from \(x = 0\) to \(x = 3\):
\(\text{Area} = \int_{0}^{3} \left[ (3 + 2x - x^2) - (3 - x) \right] \, dx\)
\(\text{Area} = \int_{0}^{3} (3x - x^2) \, dx\)
Integrating term by term:
\(\text{Area} = \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_{0}^{3}\)
Evaluating at the upper limit \(x = 3\):
\(\left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right) = \left( \frac{27}{2} - \frac{27}{3} \right) = 13.5 - 9 = 4.5\)
Evaluating at the lower limit \(x = 0\) gives \(0\).
Therefore, the area of the region is \(4.5\) (or \(\frac{9}{2}\)).

*M1* for equating the curve and the line equation.
*A1* for finding correct \(x\)-coordinates \(x=0\) and \(x=3\).
*M1* for setting up the definite integral of \(\int (y_{\text{curve}} - y_{\text{line}}) \, dx\) with their limits.
*M1* for integrating to obtain \(\frac{3x^2}{2} - \frac{x^3}{3}\) (at least one term integrated correctly).
*A1* for completely correct integration.
*M1* for substituting their limits \(0\) and \(3\) into the integrated expression.
*A1* for final area of \(4.5\) or \(\frac{9}{2}\).

**(a)**
Express the terms over a common denominator:
\(\text{LHS} = \frac{(1 + \sin x)^2 + \cos^2 x}{\cos x (1 + \sin x)}
= \frac{1 + 2\sin x + \sin^2 x + \cos^2 x}{\cos x (1 + \sin x)}\)
Using the identity \(\sin^2 x + \cos^2 x = 1\):
\(\text{LHS} = \frac{1 + 2\sin x + 1}{\cos x (1 + \sin x)} = \frac{2(1 + \sin x)}{\cos x (1 + \sin x)}\)
Cancelling the common factor \((1 + \sin x)\):
\(\text{LHS} = \frac{2}{\cos x} = 2\sec x\) (as required)

**(b)**
Using the identity from part (a) with \(x = 2\theta\):
\(2\sec 2\theta = 4 \implies \sec 2\theta = 2\)
Since \(\sec 2\theta = \frac{1}{\cos 2\theta}\):
\(\cos 2\theta = \frac{1}{2}\)
Given \(0 < \theta < \pi\), the range for \(2\theta\) is \(0 < 2\theta < 2\pi\).
Finding the solutions for \(2\theta\):
\(2\theta = \frac{\pi}{3}\) or \(2\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\)
Dividing by 2 to find \(\theta\):
\(\theta = \frac{\pi}{6}\) or \(\theta = \frac{5\pi}{6}\)

*M1* for putting the LHS of (a) over a common denominator.
*M1* for expanding and using the identity \(\sin^2 x + \cos^2 x = 1\).
*A1* for simplifying the numerator to \(2(1+\sin x)\) and completing the proof.
*M1* for using the identity in (b) to obtain \(\cos 2\theta = \frac{1}{2}\).
*M1* for finding at least one correct value for \(2\theta\) in radians.
*A1* for \(\theta = \frac{\pi}{6}\).
*A1* for \(\theta = \frac{5\pi}{6}\) and no extra solutions in the range.