2025 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

To find the inverse function, let \(y = e^{3x} - 2\).
Rearrange to make \(x\) the subject:
\(y + 2 = e^{3x}\)
Taking the natural logarithm of both sides:
\(\ln(y+2) = 3x\)
\(x = \frac{1}{3}\ln(y+2)\)

Replacing \(y\) with \(x\) gives:
\(f^{-1}(x) = \frac{1}{3}\ln(x+2)\) for \(x > -2\).

M1: Set \(y = e^{3x} - 2\) and attempt to make \(x\) the subject using logarithms.
A1: Correct expression \(f^{-1}(x) = \frac{1}{3}\ln(x+2)\) or equivalent (e.g. \(\ln(x+2)^{\frac{1}{3}}\)).

Equate the equations of the line and the curve:
\(kx - 5 = x^2 - 2x - 1\)
Rearrange into a standard quadratic form:
\(x^2 - (2+k)x + 4 = 0\)

For the line to be a tangent, the discriminant must equal zero (\(b^2 - 4ac = 0\)):
\((-(2+k))^2 - 4(1)(4) = 0\)
\((2+k)^2 - 16 = 0\)
\((2+k)^2 = 16\)
\(2+k = \pm 4\)

Solving for \(k\):
If \(2+k = 4 \implies k = 2\)
If \(2+k = -4 \implies k = -6\)

M1: Equate equations, set up a quadratic in \(x\), and apply the condition \(b^2 - 4ac = 0\).
A1: Correct values \(k = 2\) and \(k = -6\).

Using the laws of logarithms, combine the terms on the left side:
\(\log_2((x-1)(x+2)) = 2\)

Convert the logarithmic equation to its exponential form:
\((x-1)(x+2) = 2^2\)
\(x^2 + x - 2 = 4\)
\(x^2 + x - 6 = 0\)

Factorise the quadratic equation:
\((x+3)(x-2) = 0\)
This gives \(x = -3\) or \(x = 2\).

Since the term \(\log_2(x-1)\) is only defined for \(x > 1\), we must reject \(x = -3\). Therefore, the only valid solution is \(x = 2\).

M1: Combine logarithmic terms correctly into \(\log_2((x-1)(x+2))\).
M1: Convert to exponential form and solve the resulting quadratic equation.
A1: Identify \(x = 2\) as the only valid solution, explicitly rejecting \(x = -3\).

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

Factorise the quadratic expression:
\((2\sin \theta + 1)(\sin \theta - 1) = 0\)
This gives \(\sin \theta = -\frac{1}{2}\) or \(\sin \theta = 1\).

For \(0 \le \theta \le 2\pi\):
If \(\sin \theta = 1 \implies \theta = \frac{\pi}{2}\).
If \(\sin \theta = -\frac{1}{2} \implies \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\) and \(\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\).

M1: Substitute \(\cos^2\theta = 1 - \sin^2\theta\) to obtain a quadratic in \(\sin\theta\).
M1: Factorise and solve for \(\sin\theta\) to obtain \(-\frac{1}{2}\) and \(1\).
A1: Obtain all three correct solutions: \(\frac{\pi}{2}\), \(\frac{7\pi}{6}\), and \(\frac{11\pi}{6}\). Deduct marks for extra out-of-range answers.

Find the \(y\)-coordinate at \(x = 1\):
\(y = 3\ln(2(1)-1) = 3\ln(1) = 0\). The point is \((1, 0)\).

Differentiate \(y\) with respect to \(x\) using the chain rule:
\(\frac{dy}{dx} = 3 \cdot \frac{2}{2x-1} = \frac{6}{2x-1}\)

Find the gradient of the tangent at \(x = 1\):
\(m_{\text{tangent}} = \frac{6}{2(1)-1} = 6\)

Therefore, the gradient of the normal is:
\(m_{\text{normal}} = -\frac{1}{6}\)

Use the point-gradient formula with the point \((1, 0)\):
\(y - 0 = -\frac{1}{6}(x - 1)\)
\(6y = -x + 1\)
\(x + 6y - 1 = 0\)

M1: Differentiate to find \(\frac{dy}{dx} = \frac{6}{2x-1}\).
M1: Find the gradient of the normal (negative reciprocal) and set up the equation of the line using \((1,0)\).
A1: Obtain the correct equation in the form \(x + 6y - 1 = 0\) (or any integer multiple).

Let \(a\) be the first term and \(r\) be the common ratio.
The sum to infinity is given by:
\(S_{\infty} = \frac{a}{1-r} = 8 \implies a = 8(1-r)\)

The second term is given by:
\(u_2 = ar = 1.5\)

Substitute the expression for \(a\) into the second equation:
\(8(1-r)r = 1.5\)
\(8r - 8r^2 = 1.5\)
Multiply by 2 to clear the decimal:
\(16r^2 - 16r + 3 = 0\)

Factorise the quadratic equation:
\((4r - 1)(4r - 3) = 0\)

This yields the possible values for \(r\):
\(r = \frac{1}{4}\) or \(r = \frac{3}{4}\)

Both values satisfy the convergence condition \(|r| < 1\).

M1: Set up two simultaneous equations: \(a = 8(1-r)\) and \(ar = 1.5\).
M1: Substitute and obtain a quadratic equation in terms of \(r\), then attempt to solve it.
A1: Correct values: \(r = \frac{1}{4}\) and \(r = \frac{3}{4}\).

Since \(P(x)\) is exactly divisible by \(x-2\), the Factor Theorem states \(P(2) = 0\):
\(2(2)^3 + a(2)^2 + b(2) - 6 = 0\)
\(16 + 4a + 2b - 6 = 0\)
\(4a + 2b = -10\)
\(2a + b = -5\) (Equation 1)

Since dividing \(P(x)\) by \(x+1\) gives a remainder of \(-9\), the Remainder Theorem states \(P(-1) = -9\):
\(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -9\)
\(-2 + a - b - 6 = -9\)
\(a - b = -1\) (Equation 2)

Add Equation 1 and Equation 2:
\((2a + b) + (a - b) = -5 + (-1)\)
\(3a = -6 \implies a = -2\)

Substitute \(a = -2\) into Equation 2:
\(-2 - b = -1 \implies b = -1\)

M1: Correctly apply the Factor Theorem to obtain a linear equation in \(a\) and \(b\) (e.g. \(2a+b=-5\)).
M1: Correctly apply the Remainder Theorem to obtain a second equation and solve the system of simultaneous equations.
A1: Correct values: \(a = -2\) and \(b = -1\).

Find the vector \(\vec{AB}\):
\(\vec{AB} = \vec{OB} - \vec{OA}\)
\(\vec{AB} = (9\mathbf{i} + y\mathbf{j}) - (3\mathbf{i} - 2\mathbf{j}) = 6\mathbf{i} + (y+2)\mathbf{j}\)

Use the formula for the magnitude of a vector:
\(|\vec{AB}| = \sqrt{6^2 + (y+2)^2} = 10\)

Square both sides:
\(36 + (y+2)^2 = 100\)
\((y+2)^2 = 64\)

Take the square root of both sides:
\(y + 2 = \pm 8\)

Since we are given \(y > 0\):
\(y + 2 = 8 \implies y = 6\)

M1: Express \(\vec{AB}\) in the form \(6\mathbf{i} + (y+2)\mathbf{j}\).
M1: Write a correct magnitude equation \(6^2 + (y+2)^2 = 10^2\) and solve for \(y\).
A1: Deduce the correct value \(y = 6\) (rejecting \(y = -10\) due to \(y > 0\)).

Using the product law of logarithms, we combine the terms on the left side: \(\log_3((x - 2)(x + 4)) = 3\). Converting this logarithmic equation to its exponential form gives \((x - 2)(x + 4) = 3^3\), which simplifies to \(x^2 + 2x - 8 = 27\). Rearranging terms leads to the quadratic equation \(x^2 + 2x - 35 = 0\). Factoring this quadratic, we get \((x + 7)(x - 5) = 0\), which gives the potential solutions \(x = -7\) and \(x = 5\). Since the argument of a logarithm must be strictly positive, we require \(x - 2 > 0\) (i.e., \(x > 2\)). Therefore, we reject \(x = -7\), leaving the only valid solution as \(x = 5\).

M1: For correctly combining the logarithms using the product rule to get \((x - 2)(x + 4) = 27\).
M1: For expanding and solving the quadratic equation to find the roots \(x = 5\) and \(x = -7\).
A1: For rejecting \(x = -7\) and identifying \(x = 5\) as the unique valid solution.

Since the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are parallel, their components are in proportion. Thus, we can set up the ratio: \(\frac{a}{8} = \frac{-2}{a - 10}\). Multiplying both sides by the denominators gives \(a(a - 10) = -16\). Expanding and rearranging this equation yields the quadratic equation: \(a^2 - 10a + 16 = 0\). Factoring the quadratic, we get \((a - 2)(a - 8) = 0\). Therefore, the possible values of \(a\) are \(a = 2\) or \(a = 8\).

M1: For setting up a correct proportional relation for parallel vectors, e.g., \(\frac{a}{8} = \frac{-2}{a-10}\).
M1: For simplifying to a standard quadratic equation \(a^2 - 10a + 16 = 0\).
A1: For finding both correct values of \(a\), which are \(a = 2\) and \(a = 8\).

The perimeter of the sector is given by \(P = 2r + r\theta = 20\), which allows us to express \(\theta\) as: \(\theta = \frac{20 - 2r}{r}\). The area of the sector is given by \(A = \frac{1}{2}r^2\theta = 25\). Substituting the expression for \(\theta\) into the area formula gives: \(\frac{1}{2}r^2\left(\frac{20 - 2r}{r}\right) = 25\). This simplifies to: \(\frac{1}{2}r(20 - 2r) = 25 \implies 10r - r^2 = 25\). Rearranging this quadratic equation yields: \(r^2 - 10r + 25 = 0\). Factoring gives \((r - 5)^2 = 0\), which results in a single positive value: \(r = 5\).

M1: For formulating the equations for perimeter and area in terms of \(r\) and \(\theta\).
M1: For substituting one equation into the other to construct a quadratic equation in \(r\).
A1: For solving the quadratic equation to obtain the correct unique radius \(r = 5\).

We apply the product rule \(\frac{\text{d}y}{\text{d}x} = u \frac{\text{d}v}{\text{d}x} + v \frac{\text{d}u}{\text{d}x}\), where \(u = 3x - 1\) and \(v = \ln(2x + 1)\). Differentiating each term, we have \(\frac{\text{d}u}{\text{d}x} = 3\) and \(\frac{\text{d}v}{\text{d}x} = \frac{2}{2x + 1}\) (using the chain rule). Thus, the derivative is: \(\frac{\text{d}y}{\text{d}x} = (3x - 1) \cdot \frac{2}{2x + 1} + 3\ln(2x + 1)\). To find the value at \(x = 0\), we substitute \(x = 0\) into the derivative expression: \(\frac{\text{d}y}{\text{d}x} = (3(0) - 1) \cdot \frac{2}{2(0) + 1} + 3\ln(2(0) + 1) = (-1) \cdot (2) + 3\ln(1) = -2 + 0 = -2\).

M1: For applying the product rule correctly to differentiate \(y\).
M1: For correctly applying the chain rule to differentiate \(\ln(2x + 1)\) to get \(\frac{2}{2x + 1}\).
A1: For substituting \(x = 0\) correctly and obtaining the final exact value of \(-2\).

To find the exact value of the integral: \(\int_1^4 \left(\frac{2x+1}{\sqrt{x}}\right) \mathrm{d}x\). First, rewrite the integrand by dividing each term by \(\sqrt{x} = x^{1/2}\): \(\frac{2x+1}{x^{1/2}} = 2x^{1/2} + x^{-1/2}\). Now, integrate term by term: \(\int (2x^{1/2} + x^{-1/2}) \mathrm{d}x = \left[ 2 \left(\frac{x^{3/2}}{3/2}\right) + \frac{x^{1/2}}{1/2} \right]_1^4 = \left[ \frac{4}{3}x^{3/2} + 2x^{1/2} \right]_1^4\). Evaluate at the upper limit \(x = 4\): \(\frac{4}{3}(4^{3/2}) + 2(4^{1/2}) = \frac{4}{3}(8) + 2(2) = \frac{32}{3} + 4 = \frac{44}{3}\). Evaluate at the lower limit \(x = 1\): \(\frac{4}{3}(1^{3/2}) + 2(1^{1/2}) = \frac{4}{3}(1) + 2(1) = \frac{10}{3}\). Subtract the lower limit evaluation from the upper limit evaluation: \(\frac{44}{3} - \frac{10}{3} = \frac{34}{3}\).

M1: For rewriting the integrand as \(2x^{1/2} + x^{-1/2}\). M1: For integrating at least one term correctly to obtain \(x^{3/2}\) or \(x^{1/2}\). A1: For the correct integrated expression \(\frac{4}{3}x^{3/2} + 2x^{1/2}\). A1: For substituting limits and obtaining the final exact answer \(\frac{34}{3}\) (or \(11\frac{1}{3}\)).

Given: \(\log_3(x-2) + \log_3(x+4) = 3\). Apply the product rule for logarithms: \(\log_3((x-2)(x+4)) = 3\). Convert the logarithmic equation into exponential form: \((x-2)(x+4) = 3^3\) which simplifies to \(x^2 + 2x - 8 = 27\). Rearrange into a quadratic equation: \(x^2 + 2x - 35 = 0\). Factorise the quadratic: \((x+7)(x-5) = 0\). This gives: \(x = -7\) or \(x = 5\). Since the original terms \(\log_3(x-2)\) and \(\log_3(x+4)\) require \(x > 2\), the solution \(x = -7\) is rejected. Thus, the only valid solution is \(x = 5\).

M1: For applying the product rule of logs: \(\log_3((x-2)(x+4))\). M1: For converting to exponential form: \((x-2)(x+4) = 27\). A1: For obtaining the quadratic equation \(x^2 + 2x - 35 = 0\). A1: For solving and correctly identifying \(x = 5\) as the only valid solution (must clearly reject \(x = -7\)).

Let the first term be \(a\) and the common ratio be \(r\). The \(n\)-th term of a geometric progression is given by \(u_n = a r^{n-1}\). We are given: \(u_3 = a r^2 = 12\) (Equation 1) and \(u_6 = a r^5 = \frac{32}{9}\) (Equation 2). Divide Equation 2 by Equation 1: \(\frac{a r^5}{a r^2} = \frac{32/9}{12}\), which gives \(r^3 = \frac{32}{108} = \frac{8}{27}\). Taking the cube root of both sides gives \(r = \frac{2}{3}\). Substitute \(r = \frac{2}{3}\) back into Equation 1 to find \(a\): \(a \left(\frac{2}{3}\right)^2 = 12 \implies a \left(\frac{4}{9}\right) = 12 \implies a = 12 \times \frac{9}{4} = 27\). The sum to infinity \(S_{\infty}\) is given by: \(S_{\infty} = \frac{a}{1-r} = \frac{27}{1 - 2/3} = \frac{27}{1/3} = 81\).

M1: For writing down two equations: \(a r^2 = 12\) and \(a r^5 = \frac{32}{9}\), and attempting to solve for \(r\). A1: For finding \(r = \frac{2}{3}\). A1: For finding \(a = 27\). A1: For calculating the sum to infinity \(S_{\infty} = 81\).

Given: \(2 \cos^2 \theta + 3 \sin \theta - 3 = 0\). Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\): \(2(1 - \sin^2 \theta) + 3 \sin \theta - 3 = 0 \implies 2 - 2 \sin^2 \theta + 3 \sin \theta - 3 = 0 \implies -2 \sin^2 \theta + 3 \sin \theta - 1 = 0\). Multiply by \(-1\): \(2 \sin^2 \theta - 3 \sin \theta + 1 = 0\). Factorise the quadratic in \(\sin \theta\): \((2 \sin \theta - 1)(\sin \theta - 1) = 0\). This gives two possible equations: (1) \(2 \sin \theta - 1 = 0 \implies \sin \theta = \frac{1}{2}\). For \(0^\circ \le \theta \le 360^\circ\): \(\theta = 30^\circ\) or \(\theta = 180^\circ - 30^\circ = 150^\circ\). (2) \(\sin \theta - 1 = 0 \implies \sin \theta = 1\). For \(0^\circ \le \theta \le 360^\circ\): \(\theta = 90^\circ\). Combining these, the solutions are: \(\theta = 30^\circ, 90^\circ, 150^\circ\).

M1: For substituting \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a quadratic in \(\sin \theta\). M1: For factorising or solving the quadratic equation to get \(\sin \theta = \frac{1}{2}\) and \(\sin \theta = 1\). A1: For finding \(\theta = 30^\circ, 150^\circ\). A1: For finding \(\theta = 90^\circ\). Deduct 1 mark if there are any extra incorrect angles within the range.

According to the Factor Theorem, since \(p(x)\) is exactly divisible by \(x-2\), we have \(p(2) = 0 \implies 2(2)^3 + a(2)^2 + b(2) - 6 = 0 \implies 16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10\). Dividing by 2 gives \(2a + b = -5\) (Equation 1). According to the Remainder Theorem, since \(p(x)\) leaves a remainder of \(-15\) when divided by \(x+1\), we have \(p(-1) = -15 \implies 2(-1)^3 + a(-1)^2 + b(-1) - 6 = -15 \implies -2 + a - b - 6 = -15 \implies a - b = -7\) (Equation 2). Solve the simultaneous equations: from Equation 2, \(b = a + 7\). Substitute into Equation 1: \(2a + (a + 7) = -5 \implies 3a = -12 \implies a = -4\). Substitute \(a = -4\) back to find \(b\): \(b = -4 + 7 = 3\). Therefore, \(a = -4\) and \(b = 3\).

M1: For using the factor theorem to write \(p(2) = 0 \implies 2a + b = -5\) (or equivalent). M1: For using the remainder theorem to write \(p(-1) = -15 \implies a - b = -7\) (or equivalent). M1: For attempting to solve the simultaneous equations to find one of the variables. A1: For both correct values: \(a = -4\) and \(b = 3\).

(i) To find the unit vector in the direction of \(\mathbf{u}\), first calculate the magnitude of \(\mathbf{u}\): \(|\mathbf{u}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\). The unit vector is \(\hat{\mathbf{u}} = \frac{\mathbf{u}}{|\mathbf{u}|} = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}\). (ii) For the two vectors to be perpendicular, their dot product must be equal to 0: \(\mathbf{u} \cdot \mathbf{v} = 0 \implies (3\mathbf{i} - 4\mathbf{j}) \cdot (k\mathbf{i} + 6\mathbf{j}) = 0 \implies 3(k) + (-4)(6) = 0 \implies 3k - 24 = 0 \implies 3k = 24 \implies k = 8\).

M1: For finding the magnitude of \(\mathbf{u}\) to be 5. A1: For the correct unit vector \(\frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}\) (or equivalent column vector notation). M1: For setting up the dot product equation for perpendicular vectors: \(3k - 24 = 0\). A1: For finding \(k = 8\).

(i) If there are no restrictions, we need to choose any 5 students out of the total 11 students (6 boys + 5 girls). Number of ways = \(\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 11 \times 3 \times 2 \times 7 = 462\). (ii) If the committee must contain exactly 3 boys and 2 girls, we choose 3 boys from 6, which can be done in \(\binom{6}{3}\) ways, and 2 girls from 5, which can be done in \(\binom{5}{2}\) ways. Total number of ways = \(\binom{6}{3} \times \binom{5}{2} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1} = 20 \times 10 = 200\).

M1: For identifying the combination formula \(\binom{11}{5}\) for part (i). A1: For the correct value 462. M1: For using the product of two combinations \(\binom{6}{3} \times \binom{5}{2}\) for part (ii). A1: For the correct value 200.

(i) Let \(\theta\) be the angle of the sector in radians. The perimeter of the sector is given by \(P = 2r + r\theta = 24 \implies r\theta = 24 - 2r \implies \theta = \frac{24 - 2r}{r}\). The area of the sector is \(A = \frac{1}{2}r^2\theta\). Substituting \(\theta\) gives \(A = \frac{1}{2}r^2\left(\frac{24 - 2r}{r}\right) = \frac{1}{2}r(24 - 2r) = 12r - r^2\) (as required). (ii) To find the maximum area, differentiate \(A\) with respect to \(r\): \(\frac{\mathrm{d}A}{\mathrm{d}r} = 12 - 2r\). Setting \(\frac{\mathrm{d}A}{\mathrm{d}r} = 0\) for stationary points gives \(12 - 2r = 0 \implies r = 6\) cm. Substituting \(r = 6\) back into the area formula gives the maximum area \(A = 12(6) - 6^2 = 72 - 36 = 36\) cm\(^2\).

M1: For writing the perimeter equation \(2r + r\theta = 24\) and expressing \(\theta\) or \(r\theta\) in terms of \(r\). A1: For substituting into the area formula \(A = \frac{1}{2}r^2\theta\) and correctly simplifying to show \(12r - r^2\). M1: For differentiating \(A\) to find \(\frac{\mathrm{d}A}{\mathrm{d}r}\) and setting to 0 to find \(r = 6\). A1: For finding the maximum area of 36.

To find the stationary point, we first differentiate \(y\) with respect to \(x\) using the product rule:
Let \(u = 2x - 3 \implies \frac{du}{dx} = 2\).
Let \(v = e^{-2x} \implies \frac{dv}{dx} = -2e^{-2x}\).

Applying the product rule:
\(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)
\(\frac{dy}{dx} = (2x - 3)(-2e^{-2x}) + 2e^{-2x}\)
\(\frac{dy}{dx} = e^{-2x}(-4x + 6 + 2) = e^{-2x}(8 - 4x)\)

At the stationary point, \(\frac{dy}{dx} = 0\).
Since \(e^{-2x} \neq 0\) for all real \(x\):
\(8 - 4x = 0 \implies x = 2\)

Substitute \(x = 2\) back into the original curve equation to find the \(y\)-coordinate:
\(y = (2(2) - 3)e^{-2(2)} = (1)e^{-4} = e^{-4}\)
So, the stationary point is \((2, e^{-4})\).

To determine the nature of the stationary point, find the second derivative \(\frac{d^2y}{dx^2}\):
\(\frac{d^2y}{dx^2} = \frac{d}{dx}[e^{-2x}(8 - 4x)]\)
\(\frac{d^2y}{dx^2} = -2e^{-2x}(8 - 4x) + e^{-2x}(-4)\)
\(\frac{d^2y}{dx^2} = e^{-2x}(-16 + 8x - 4) = e^{-2x}(8x - 20)\)

Substitute \(x = 2\) into the second derivative:
\(\frac{d^2y}{dx^2} = e^{-4}(8(2) - 20) = -4e^{-4}\)
Since \(-4e^{-4} < 0\), the stationary point is a maximum.

M1: For applying the product rule to find the derivative.
A1: For obtaining a correct simplified derivative \(\frac{dy}{dx} = e^{-2x}(8 - 4x)\).
M1: For setting their derivative to 0, finding \(x = 2\), and finding the corresponding y-coordinate \(y = e^{-4}\).
A1: For confirming the coordinates \((2, e^{-4})\) and showing the nature is a maximum (using \(\frac{d^2y}{dx^2} < 0\) or a valid sign-table method).

Let the first term of the progression be \(a\) and the common ratio be \(r\).
The \(n\)-th term of a geometric progression is given by \(u_n = ar^{n-1}\).

From the given information:
\(u_3 = ar^2 = 12\) (Equation 1)
\(u_6 = ar^5 = -96\) (Equation 2)

Divide Equation 2 by Equation 1:
\(\frac{ar^5}{ar^2} = \frac{-96}{12}\)
\(r^3 = -8\)
\(r = -2\)

Substitute \(r = -2\) back into Equation 1:
\(a(-2)^2 = 12\)
\(4a = 12 \implies a = 3\)

The sum of the first \(n\) terms of a geometric progression is:
\(S_n = \frac{a(1 - r^n)}{1 - r}\)

For \(n = 8\):
\(S_8 = \frac{3(1 - (-2)^8)}{1 - (-2)}\)
\(S_8 = \frac{3(1 - 256)}{3}\)
\(S_8 = 1 - 256 = -255\)

M1: For setting up the two term equations: \(ar^2 = 12\) and \(ar^5 = -96\).
A1: For solving to find \(r = -2\).
A1: For finding the first term \(a = 3\).
M1: For using the sum of a geometric progression formula with their values of \(a\) and \(r\) for \(n = 8\).
A1: For obtaining \(S_8 = -255\).

**(a)**
To find the stationary point, we first differentiate the curve equation with respect to \( x \):
\[ y = 4e^{2x} + 9e^{-2x} \]
\[ \frac{dy}{dx} = 8e^{2x} - 18e^{-2x} \]

At the stationary point, \( \frac{dy}{dx} = 0 \):
\[ 8e^{2x} - 18e^{-2x} = 0 \]
\[ 8e^{2x} = \frac{18}{e^{2x}} \]
\[ e^{4x} = \frac{18}{8} = \frac{9}{4} \]

Since \( e^{2x} > 0 \) for all real \( x \), we take the positive square root:
\[ e^{2x} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]
Taking the natural logarithm on both sides:
\[ 2x = \ln\left(\frac{3}{2}\right) \]
\[ x = \frac{1}{2}\ln\left(\frac{3}{2}\right) \]

Now, we find the corresponding \( y \)-coordinate by substituting \( e^{2x} = \frac{3}{2} \) and \( e^{-2x} = \frac{2}{3} \) back into the curve equation:
\[ y = 4\left(\frac{3}{2}\right) + 9\left(\frac{2}{3}\right) \]
\[ y = 6 + 6 = 12 \]

So, the exact coordinates of the stationary point are \( \left(\frac{1}{2}\ln\left(\frac{3}{2}\right), 12\right) \).

**(b)**
To determine the nature of the stationary point, we find the second derivative:
\[ \frac{d^2y}{dx^2} = 16e^{2x} + 36e^{-2x} \]

At the stationary point, since \( e^{2x} = \frac{3}{2} > 0 \) and \( e^{-2x} = \frac{2}{3} > 0 \):
\[ \frac{d^2y}{dx^2} = 16\left(\frac{3}{2}\right) + 36\left(\frac{2}{3}\right) = 24 + 24 = 48 \]

Since \( \frac{d^2y}{dx^2} = 48 > 0 \), the stationary point is a **minimum**.

**(c)**
First, find the coordinates of the point on the curve where \( x = 0 \):
\[ y = 4e^0 + 9e^0 = 4 + 9 = 13 \]
So the point of contact is \( (0, 13) \).

Next, find the gradient of the tangent at \( x = 0 \):
\[ \left. \frac{dy}{dx} \right|_{x=0} = 8e^0 - 18e^0 = 8 - 18 = -10 \]

The gradient of the normal, \( m_n \), is perpendicular to the tangent gradient:
\[ m_n = -\frac{1}{-10} = \frac{1}{10} \]

Now, find the equation of the normal passing through \( (0, 13) \) with gradient \( \frac{1}{10} \):
\[ y - 13 = \frac{1}{10}(x - 0) \]
\[ 10(y - 13) = x \]
\[ 10y - 130 = x \]
\[ x - 10y + 130 = 0 \]

**(a)**
* **M1**: Correct differentiation of at least one term of \( y \) to obtain \( 8e^{2x} \) or \( -18e^{-2x} \).
* **M1**: Equating their derivative to \( 0 \) and correctly rearranging to the form \( e^{4x} = k \) or \( e^{2x} = \sqrt{k} \).
* **A1**: Obtaining the correct exact value of \( x = \frac{1}{2}\ln\left(\frac{3}{2}\right) \) or equivalent exact form (e.g. \( \frac{1}{4}\ln\left(\frac{9}{4}\right) \)).
* **A1**: Obtaining the correct exact value of \( y = 12 \).

**(b)**
* **M1**: Differentiating their \( \frac{dy}{dx} \) to get \( a e^{2x} + b e^{-2x} \) where \( a > 0, b > 0 \).
* **A1**: Showing that \( \frac{d^2y}{dx^2} > 0 \) (or evaluating it to \( 48 \)) and concluding it is a **minimum**.

**(c)**
* **B1**: Correctly finding \( y = 13 \) when \( x = 0 \).
* **M1**: Substituting \( x = 0 \) into their \( \frac{dy}{dx} \) to find the tangent gradient (gets \( -10 \)).
* **M1**: Finding the normal gradient as the negative reciprocal of their tangent gradient (gets \( \frac{1}{10} \)).
* **M1**: Applying the straight-line formula with their point \( (0, y_0) \) and normal gradient.
* **A1**: Correct final equation in the form \( ax + by + c = 0 \) with integer coefficients, e.g., \( x - 10y + 130 = 0 \) (or any non-zero integer multiple).

At \(x = 1\), \(y = 3\ln(1) + 4(1) = 4\). The point of contact is \((1, 4)\). Differentiating \(y\) with respect to \(x\) gives \(\frac{dy}{dx} = \frac{6}{2x-1} + 4\). At \(x = 1\), \(\frac{dy}{dx} = \frac{6}{1} + 4 = 10\). The gradient of the normal is the negative reciprocal of the gradient of the tangent, which is \(-\frac{1}{10}\). The equation of the normal is \(y - 4 = -\frac{1}{10}(x - 1)\). Multiplying by 10 and rearranging gives \(x + 10y = 41\).

M1: For finding \(y = 4\) and correctly differentiating the curve to find \(\frac{dy}{dx}\).
M1: For substituting \(x = 1\) and finding the gradient of the normal as the negative reciprocal.
A1: For the correct equation of the normal, e.g., \(x + 10y = 41\) or equivalent.

Apply the power law and subtraction law of logarithms: \(\log_3 \left(\frac{(x - 2)^2}{x + 4}\right) = 1\). Convert this to exponential form: \(\frac{(x - 2)^2}{x + 4} = 3^1\), which simplifies to \((x - 2)^2 = 3(x + 4)\). Expanding and rearranging the terms gives \(x^2 - 4x + 4 = 3x + 12 \implies x^2 - 7x - 8 = 0\). Factoring the quadratic equation gives \((x - 8)(x + 1) = 0\), which yields \(x = 8\) or \(x = -1\). Since \(x - 2 > 0\) is required for the logarithm to be defined, we must have \(x > 2\). Thus, we reject \(x = -1\) and the only valid solution is \(x = 8\).

M1: For applying laws of logarithms to obtain a single logarithmic expression.
M1: For removing logarithms correctly and forming a quadratic equation.
A1: For solving the quadratic equation to find \(x = 8\) and identifying that \(x = -1\) must be rejected.

Divide both sides of the equation by 3: \(\sin\left(2\theta - \frac{\pi}{6}\right) = 0.5\). Let \(\phi = 2\theta - \frac{\pi}{6}\). Since \(0 \le \theta \le \pi\), the range for \(\phi\) is \(-\frac{\pi}{6} \le \phi \le \frac{11\pi}{6}\). In this range, the solutions for \(\sin\phi = 0.5\) are \(\phi = \frac{\pi}{6}\) and \(\phi = \frac{5\pi}{6}\). Now, solve for \(\theta\):\
1) \(2\theta - \frac{\pi}{6} = \frac{\pi}{6} \implies 2\theta = \frac{\pi}{3} \implies \theta = \frac{\pi}{6}\n2) \)2\theta - \frac{\pi}{6} = \frac{5\pi}{6} \implies 2\theta = \pi \implies \theta = \frac{\pi}{2}\\nBoth solutions lie within the required range \([0, \pi]\).

M1: For finding at least one correct value for \(2\theta - \frac{\pi}{6}\), such as \(\frac{\pi}{6}\) or \(\frac{5\pi}{6}\).
M1: For a correct method to solve for \(\theta\) from their angle values.
A1: For both correct solutions \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{\pi}{2}\).

Let the first term be \(a\) and the common ratio be \(r\). The third term is \(ar^2 = 18\) and the sixth term is \(ar^5 = -\frac{16}{3}\). Dividing the sixth term by the third term gives \(\frac{ar^5}{ar^2} = \frac{-16/3}{18} \implies r^3 = -\frac{8}{27}\), which yields \(r = -\frac{2}{3}\). Substitute \(r = -\frac{2}{3}\) back into \(ar^2 = 18\) to find \(a\): \(a\left(-\frac{2}{3}\right)^2 = 18 \implies a\left(\frac{4}{9}\right) = 18 \implies a = \frac{18 \times 9}{4} = 40.5\). The sum to infinity is given by \(S_{\infty} = \frac{a}{1 - r} = \frac{40.5}{1 - (-2/3)} = \frac{40.5}{5/3} = 24.3\).

M1: For setting up the ratio of terms to find the value of \(r\).
M1: For finding the first term \(a\) and substituting both \(a\) and \(r\) into the sum to infinity formula.
A1: For obtaining the correct sum to infinity, \(24.3\) (or \(\frac{243}{10}\)).

Set the equations of the line and the curve equal to find their intersection: \(kx - 5 = x^2 - 4x - 1\). 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: \(D = [-(4+k)]^2 - 4(1)(4) < 0\). This simplifies to \((4+k)^2 - 16 < 0 \implies (4+k)^2 < 16\). Taking the square root gives \(-4 < 4+k < 4\). Subtracting 4 from all parts yields the interval \(-8 < k < 0\).

M1: For equating the line and curve and rearranging into a standard quadratic form.
M1: For setting the discriminant of their quadratic equation to be less than zero and attempting to solve the inequality.
A1: For the correct set of values \(-8 < k < 0\) (or equivalent interval notation).

By the Factor Theorem, since \(x - 2\) is a factor of \(p(x)\), we have \(p(2) = 0\). This gives \(2(2)^3 + a(2)^2 + b(2) - 6 = 0 \implies 16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5\). By the Remainder Theorem, since dividing \(p(x)\) by \(x + 1\) leaves a remainder of \(-9\), we have \(p(-1) = -9\). This gives \(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -9 \implies -2 + a - b - 6 = -9 \implies a - b = -1\). Now solve the simultaneous equations: \(2a + b = -5\) and \(a - b = -1\). Adding the two equations gives \(3a = -6 \implies a = -2\). Substituting \(a = -2\) into \(a - b = -1\) gives \(-2 - b = -1 \implies b = -1\).

M1: For applying the Factor Theorem and Remainder Theorem to write down two linear equations in \(a\) and \(b\).
M1: For solving the simultaneous equations to find the values of \(a\) and \(b\).
A1: For both correct values \(a = -2\) and \(b = -1\).

The committee of 5 must contain at least 3 women. This means the committee can have 3, 4, or 5 women.
Case 1: 3 women and 2 men. The number of ways is \(\binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150\).
Case 2: 4 women and 1 man. The number of ways is \(\binom{5}{4} \times \binom{6}{1} = 5 \times 6 = 30\).
Case 3: 5 women and 0 men. The number of ways is \(\binom{5}{5} \times \binom{6}{0} = 1 \times 1 = 1\).
Total number of different committees is the sum of these cases: \(150 + 30 + 1 = 181\).

M1: For identifying the three valid scenarios (3, 4, or 5 women) and using combinations.
M1: For calculating the combinations for at least two of the scenarios correctly.
A1: For the correct total of \(181\).

Since the point \(C\) divides the line segment \(AB\) in the ratio \(3:1\), we can use the ratio theorem for position vectors: \(\mathbf{c} = \frac{1\mathbf{a} + 3\mathbf{b}}{3+1}\). Substituting the given vectors: \(\mathbf{c} = \frac{(3\mathbf{i} - 2\mathbf{j}) + 3(-5\mathbf{i} + 4\mathbf{j})}{4} = \frac{3\mathbf{i} - 2\mathbf{j} - 15\mathbf{i} + 12\mathbf{j}}{4} = \frac{-12\mathbf{i} + 10\mathbf{j}}{4} = -3\mathbf{i} + 2.5\mathbf{j}\).

M1: For using a valid vector method to find the position vector of \(C\), such as \(\mathbf{c} = \mathbf{a} + \frac{3}{4}(\mathbf{b} - \mathbf{a})\) or the ratio theorem.
M1: For calculating \(\mathbf{b} - \mathbf{a} = -8\mathbf{i} + 6\mathbf{j}\) or establishing the correct weighted sum of the components.
A1: For the correct final position vector \(-3\mathbf{i} + 2.5\mathbf{j}\) (or equivalent form).

To find where the line and curve do not intersect, we set their equations equal to each other:
\(kx - 3 = x^2 + 5x + 6\)

Rearranging into a standard quadratic equation form:
\(x^2 + (5 - k)x + 9 = 0\)

For no points of intersection, the discriminant must be strictly less than zero (\(b^2 - 4ac < 0\)):
\((5 - k)^2 - 4(1)(9) < 0\)
\((5 - k)^2 - 36 < 0\)
\((5 - k)^2 < 36\)

Taking the square root on both sides:
\(-6 < 5 - k < 6\)

Subtracting 5 from all parts:
\(-11 < -k < 1\)

Multiplying by \(-1\) (and reversing the inequality signs):
\(-1 < k < 11\)

M1: Set up the quadratic equation \(x^2 + (5 - k)x + 9 = 0\) and attempt to find the discriminant.
M1: Apply the condition \(b^2 - 4ac < 0\) and solve the quadratic inequality.
A1: Obtain the correct final range \(-1 < k < 11\).

The committee of 4 people must contain at least 3 women. There are two mutually exclusive cases to consider:

Case 1: Exactly 3 women and 1 man
Number of ways to choose 3 women from 6: \(\binom{6}{3} = 20\)
Number of ways to choose 1 man from 5: \(\binom{5}{1} = 5\)
Total ways for Case 1: \(20 \times 5 = 100\)

Case 2: Exactly 4 women and 0 men
Number of ways to choose 4 women from 6: \(\binom{6}{4} = 15\)
Number of ways to choose 0 men from 5: \(\binom{5}{0} = 1\)
Total ways for Case 2: \(15 \times 1 = 15\)

Total number of different committees: \(100 + 15 = 115\).

M1: Calculate the number of ways for 3 women and 1 man using combinations: \(\binom{6}{3} \times \binom{5}{1}\).
M1: Calculate the number of ways for 4 women using combinations: \(\binom{6}{4}\).
A1: Add the two cases together to get the correct final answer of 115.

First, find the vector \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\)
\(\overrightarrow{AB} = (x\mathbf{i} + 5\mathbf{j}) - (3\mathbf{i} - \mathbf{j})\)
\(\overrightarrow{AB} = (x - 3)\mathbf{i} + 6\mathbf{j}\)

Next, use the magnitude of \(\overrightarrow{AB}\):
\(|\overrightarrow{AB}| = \sqrt{(x - 3)^2 + 6^2} = \sqrt{61}\)

Squaring both sides:
\((x - 3)^2 + 36 = 61\)
\((x - 3)^2 = 25\)

Taking the square root:
\(x - 3 = 5\) or \(x - 3 = -5\)
\(x = 8\) or \(x = -2\)

Since we are given that \(x > 0\), we reject \(x = -2\). Thus, \(x = 8\).

M1: Express \(\overrightarrow{AB}\) in terms of \(x\) as \((x - 3)\mathbf{i} + 6\mathbf{j}\).
M1: Set up the magnitude equation \((x - 3)^2 + 36 = 61\) and solve for \(x\).
A1: Deduce that \(x = 8\) (rejecting \(-2\) due to \(x > 0\)).

Using the laws of logarithms, combine the terms on the left-hand side:
\(\log_3((y + 2)(y - 4)) = 3\)

Convert the logarithmic equation into exponential form:
\((y + 2)(y - 4) = 3^3\)
\(y^2 - 2y - 8 = 27\)

Rearrange to form a quadratic equation equal to zero:
\(y^2 - 2y - 35 = 0\)

Factorise the quadratic expression:
\((y - 7)(y + 5) = 0\)

This gives \(y = 7\) or \(y = -5\).

We must check the domain of the original logarithmic terms:
For \(\log_3(y + 2)\) to be defined, \(y > -2\).
For \(\log_3(y - 4)\) to be defined, \(y > 4\).

Since \(y = -5\) is not within the valid domain, the only solution is \(y = 7\).

M1: Use the product rule of logarithms to obtain \(\log_3((y + 2)(y - 4))\).
M1: Correctly convert the logarithm equation to quadratic form: \(y^2 - 2y - 8 = 27\).
A1: Solve the quadratic equation and identify \(y = 7\) as the only valid solution (rejecting \(y = -5\)).

To find the stationary point, we first differentiate \(y = 4x \ln(x)\) with respect to \(x\) using the product rule:
Let \(u = 4x\) and \(v = \ln(x)\).
Then \(\frac{du}{dx} = 4\) and \(\frac{dv}{dx} = \frac{1}{x}\).

\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\)
\(\frac{dy}{dx} = 4x \left(\frac{1}{x}\right) + 4 \ln(x)\)
\(\frac{dy}{dx} = 4 + 4 \ln(x)\)

Set \(\frac{dy}{dx} = 0\) to find the stationary point:
\(4 + 4 \ln(x) = 0\)
\(\ln(x) = -1\)
\(x = e^{-1} = \frac{1}{e}\)

Now, find the corresponding \(y\)-coordinate:
\(y = 4\left(\frac{1}{e}\right) \ln\left(\frac{1}{e}\right)\)
Since \(\ln\left(\frac{1}{e}\right) = -1\):
\(y = \frac{4}{e} (-1) = -\frac{4}{e}\)

Thus, the coordinates of the stationary point are \(\left(\frac{1}{e}, -\frac{4}{e}\right)\).

M1: Apply the product rule correctly to differentiate \(4x \ln(x)\) and obtain \(4 + 4 \ln(x)\).
M1: Set the derivative equal to zero and solve for \(x\) to obtain \(x = e^{-1}\).
A1: Substitute \(x\) back into the original curve equation to find the correct coordinates \(\left(\frac{1}{e}, -\frac{4}{e}\right)\).

First, find the midpoint of the line segment \(PQ\):
\(\text{Midpoint} = \left(\frac{-2 + 4}{2}, \frac{5 - 7}{2}\right) = (1, -1)\)

Next, find the gradient of the line segment \(PQ\) (let this be \(m_1\)):
\(m_1 = \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2\)

The gradient of the perpendicular bisector (let this be \(m_2\)) is the negative reciprocal of \(m_1\):
\(m_2 = -\frac{1}{-2} = \frac{1}{2}\)

Now, use the point-slope form with the midpoint \((1, -1)\) and gradient \(m_2 = \frac{1}{2}\):
\(y - (-1) = \frac{1}{2}(x - 1)\)
\(y + 1 = \frac{1}{2}x - \frac{1}{2}\)
\(y = \frac{1}{2}x - \frac{3}{2}\) (or \(y = 0.5x - 1.5\))

M1: Find the correct midpoint \((1, -1)\) of \(PQ\) and the gradient of \(PQ\) as \(-2\).
M1: Determine the perpendicular gradient as \(\frac{1}{2}\) and construct the equation of the line.
A1: Obtain the correct final equation \(y = 0.5x - 1.5\) or \(y = \frac{1}{2}x - \frac{3}{2}\).

The perimeter of a sector is given by:
\(P = 2r + r\theta = 20\) --- (Equation 1)

This gives:
\(r\theta = 20 - 2r \implies \theta = \frac{20 - 2r}{r}\)

The area of a sector is given by:
\(A = \frac{1}{2}r^2\theta = 24\) --- (Equation 2)

Substitute \(\theta = \frac{20 - 2r}{r}\) into Equation 2:
\(\frac{1}{2}r^2 \left(\frac{20 - 2r}{r}\right) = 24\)
\(\frac{1}{2}r(20 - 2r) = 24\)
\(r(10 - r) = 24\)
\(10r - r^2 = 24\)
\(r^2 - 10r + 24 = 0\)

Factorising the quadratic equation:
\((r - 4)(r - 6) = 0\)

Therefore, the two possible values of \(r\) are \(r = 4\) and \(r = 6\).

M1: Set up the simultaneous equations for perimeter \(2r + r\theta = 20\) and area \(\frac{1}{2}r^2\theta = 24\).
M1: Substitute and obtain a quadratic equation in terms of \(r\), such as \(r^2 - 10r + 24 = 0\).
A1: Factorise/solve correctly to find the two values \(r = 4\) and \(r = 6\).

Since the terms are in a geometric progression, the ratio between consecutive terms is constant:
\(\frac{k}{k + 4} = \frac{2k - 15}{k}\)

Cross-multiplying:
\(k^2 = (k + 4)(2k - 15)\)
\(k^2 = 2k^2 - 15k + 8k - 60\)
\(k^2 = 2k^2 - 7k - 60\)

Rearranging to form a quadratic equation:
\(k^2 - 7k - 60 = 0\)

Factorising the equation:
\((k - 12)(k + 5) = 0\)

This gives \(k = 12\) or \(k = -5\). Since \(k\) is a positive constant, we have \(k = 12\).

Now, substitute \(k = 12\) to find the terms:
First term \(a = 12 + 4 = 16\)
Second term \(ar = 12\)

Thus, the common ratio \(r\) is:
\(r = \frac{12}{16} = \frac{3}{4} = 0.75\).

M1: Create the relation \(\frac{k}{k + 4} = \frac{2k - 15}{k}\) and expand to form a quadratic equation \(k^2 - 7k - 60 = 0\).
M1: Solve the quadratic equation to get \(k = 12\) (explicitly rejecting the negative solution).
A1: Find the correct value of the common ratio \(r = 0.75\) (or \(\frac{3}{4}\)).

First, find the derivative of \(y\) with respect to \(x\) using the chain rule: \(\frac{dy}{dx} = 5(2x - 3)^4 \times 2 + 4 = 10(2x - 3)^4 + 4\). Set the gradient equal to 164: \(10(2x - 3)^4 + 4 = 164\) which simplifies to \(10(2x - 3)^4 = 160\), so \((2x - 3)^4 = 16\). Taking the fourth root on both sides gives: \(2x - 3 = 2\) or \(2x - 3 = -2\). Solving these linear equations: If \(2x - 3 = 2 \implies 2x = 5 \implies x = 2.5\). If \(2x - 3 = -2 \implies 2x = 1 \implies x = 0.5\). Now, find the corresponding \(y\)-coordinates: For \(x = 2.5\): \(y = (2(2.5) - 3)^5 + 4(2.5) = 2^5 + 10 = 42\). For \(x = 0.5\): \(y = (2(0.5) - 3)^5 + 4(0.5) = (-2)^5 + 2 = -30\). Thus, the coordinates are \((2.5, 42)\) and \((0.5, -30)\).

M1: For attempt to differentiate using chain rule to get \(k(2x-3)^4 + 4\). A1: For correct derivative \(10(2x-3)^4 + 4\). M1: For setting their derivative to 164 and finding at least one correct value of \(x\) (e.g., \(x = 2.5\) or \(x = 0.5\)). M1: For finding the corresponding \(y\)-coordinates for both of their \(x\) values. A1: For both correct coordinates: \((2.5, 42)\) and \((0.5, -30)\).

Using the trigonometric identity \(\cos^2\theta = 1 - \sin^2\theta\), substitute into the equation: \(2(1 - \sin^2\theta) - \sin\theta - 1 = 0\). Expand and rearrange into a quadratic form: \(2 - 2\sin^2\theta - \sin\theta - 1 = 0 \implies 2\sin^2\theta + \sin\theta - 1 = 0\). Factorise the quadratic equation: \((2\sin\theta - 1)(\sin\theta + 1) = 0\). This gives two possible cases: 1) \(2\sin\theta - 1 = 0 \implies \sin\theta = 0.5\). In the range \(0^\circ \le \theta \le 360^\circ\), the solutions are \(\theta = 30^\circ\) and \(\theta = 150^\circ\). 2) \(\sin\theta + 1 = 0 \implies \sin\theta = -1\). In the range \(0^\circ \le \theta \le 360^\circ\), the solution is \(\theta = 270^\circ\). Combining these, the complete set of solutions is \(\theta = 30^\circ, 150^\circ, 270^\circ\).

M1: For using \(\cos^2\theta = 1 - \sin^2\theta\) to form an equation in terms of \(\sin\theta\) only. A1: For obtaining the correct quadratic equation \(2\sin^2\theta + \sin\theta - 1 = 0\). M1: For solving the quadratic equation to get \(\sin\theta = 0.5\) and \(\sin\theta = -1\). A1: For finding \(\theta = 30^\circ\) and \(\theta = 150^\circ\). A1: For finding \(\theta = 270^\circ\) and no extra solutions in range.

First, find the relevant terms of the binomial expansion of \((1 + 2x)^6\). Using the binomial theorem, \((1 + 2x)^6 = \binom{6}{0} + \binom{6}{1}(2x) + \binom{6}{2}(2x)^2 + \dots\). This simplifies to \(1 + 6(2x) + 15(4x^2) + \dots = 1 + 12x + 60x^2 + \dots\). We need the coefficient of \(x^2\) in the product \((3 - x)(1 + 12x + 60x^2 + \dots)\). The term in \(x^2\) is obtained from: \(3 \times (60x^2) + (-x) \times (12x) = 180x^2 - 12x^2 = 168x^2\). Thus, the coefficient of \(x^2\) is 168.

M1: For identifying the general term or attempting the expansion of \((1 + 2x)^6\). M1: For finding the correct term in \(x\): \(12x\). A1: For finding the correct term in \(x^2\): \(60x^2\). M1: For correctly combining terms from the product to form the overall \(x^2\) term: \(3(60) - 1(12)\). A1: For the final answer 168.

Apply the power law of logarithms to the first term: \(\log_3(x-2)^2 - \log_3(x+4) = 1\). Apply the subtraction law of logarithms: \(\log_3\left(\frac{(x-2)^2}{x+4}\right) = 1\). Convert from logarithmic form to exponential form: \(\frac{(x-2)^2}{x+4} = 3^1 = 3\). Multiply both sides by \(x+4\): \((x-2)^2 = 3(x+4)\). Expand and simplify: \(x^2 - 4x + 4 = 3x + 12 \implies x^2 - 7x - 8 = 0\). Factorise the quadratic: \((x-8)(x+1) = 0\), which gives \(x = 8\) or \(x = -1\). Since \(x-2\) must be positive for \(\log_3(x-2)\) to be defined, we must have \(x > 2\). Therefore, \(x = -1\) is rejected, and the only solution is \(x = 8\).

M1: For applying the power law to get \(\log_3(x-2)^2\). M1: For applying the division law to get \(\log_3\left(\frac{(x-2)^2}{x+4}\right)\). M1: For writing the equation in index form: \(\frac{(x-2)^2}{x+4} = 3\). A1: For solving the quadratic equation to get \(x = 8\) and \(x = -1\). A1: For rejecting \(x = -1\) and stating the final solution is \(x = 8\).

We can express \(\vec{OP}\) using the ratio theorem: \(\vec{OP} = \frac{2\vec{OA} + \vec{OB}}{1 + 2} = \frac{2(3\mathbf{i} + 5\mathbf{j}) + (12\mathbf{i} - 7\mathbf{j})}{3}\). Simplify the expression: \(\vec{OP} = \frac{6\mathbf{i} + 10\mathbf{j} + 12\mathbf{i} - 7\mathbf{j}}{3} = \frac{18\mathbf{i} + 3\mathbf{j}}{3} = 6\mathbf{i} + \mathbf{j}\). Alternatively, \(\vec{AB} = \vec{OB} - \vec{OA} = 9\mathbf{i} - 12\mathbf{j}\). Then, \(\vec{AP} = \frac{1}{3}\vec{AB} = 3\mathbf{i} - 4\mathbf{j}\), so \(\vec{OP} = \vec{OA} + \vec{AP} = (3\mathbf{i} + 5\mathbf{j}) + (3\mathbf{i} - 4\mathbf{j}) = 6\mathbf{i} + \mathbf{j}\). Next, find the magnitude of \(\vec{OP}\): \(|\vec{OP}| = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37}\). The unit vector in the direction of \(\vec{OP}\) is \(\frac{\vec{OP}}{|\vec{OP}|} = \frac{1}{\sqrt{37}}(6\mathbf{i} + \mathbf{j})\).

M1: For calculating the vector \(\vec{AB} = 9\mathbf{i} - 12\mathbf{j}\) or attempting to use the ratio formula. M1: For finding the vector \(\vec{AP} = 3\mathbf{i} - 4\mathbf{j}\) or equivalent. A1: For the correct vector \(\vec{OP} = 6\mathbf{i} + \mathbf{j}\). M1: For calculating the magnitude of their \(\vec{OP}\): \(|\vec{OP}| = \sqrt{6^2 + 1^2} = \sqrt{37}\). A1: For the correct unit vector \(\frac{1}{\sqrt{37}}(6\mathbf{i} + \mathbf{j})\) (or equivalent form).

(a) Since there are no restrictions, we choose 5 people from the total of 11 people (6 men + 5 women). The number of ways is \(\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462\).
(b) For a committee of 5 with at least 3 men and at least 1 woman, the possible combinations of (Men, Women) are:
Case 1: 3 Men and 2 Women
Number of ways = \(\binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200\).
Case 2: 4 Men and 1 Woman
Number of ways = \(\binom{6}{4} \times \binom{5}{1} = 15 \times 5 = 75\).
Note that 5 Men and 0 Women is not allowed as we need at least 1 woman.
Total number of different committees = \(200 + 75 = 275\).

B1: For the correct answer 462 in part (a). M1: For identifying the two valid cases for part (b): (3 Men, 2 Women) and (4 Men, 1 Woman). M1: For calculating the number of ways for Case 1: \(\binom{6}{3} \times \binom{5}{2} = 200\). M1: For calculating the number of ways for Case 2: \(\binom{6}{4} \times \binom{5}{1} = 75\). A1: For summing the cases to get the correct answer 275.

The perimeter of a sector of a circle of radius \(r\) and angle \(\theta\) is given by \(P = 2r + r\theta\). We are given \(P = 20\), so: \(2r + r\theta = 20 \implies \theta = \frac{20 - 2r}{r} = \frac{20}{r} - 2\). The area of the sector is given by \(A = \frac{1}{2}r^2\theta\). Substitute the expression for \(\theta\) into the area formula: \(A = \frac{1}{2}r^2\left(\frac{20}{r} - 2\right) = 10r - r^2\). To find the maximum area, differentiate \(A\) with respect to \(r\) and set it to 0: \(\frac{dA}{dr} = 10 - 2r = 0 \implies r = 5\). Since \(\frac{d^2A}{dr^2} = -2 < 0\), this value of \(r\) gives a maximum area. Substituting \(r = 5\) back to find \(\theta\): \(\theta = \frac{20}{5} - 2 = 2\) radians.

M1: For using the perimeter formula to express \(\theta\) in terms of \(r\): \(\theta = \frac{20-2r}{r}\) (or equivalent relation). M1: For substituting this into the area formula to obtain a quadratic expression in terms of \(r\) only: \(A = 10r - r^2\). M1: For differentiating and setting the derivative equal to 0: \(10 - 2r = 0\). A1: For finding \(r = 5\). A1: For finding \(\theta = 2\) radians.

(a) Let \(Y = \ln y\) and \(X = x^2\). The straight line passes through \((2, 5)\) and \((6, 17)\). The gradient \(m\) is: \(m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3\). Using the point-slope form with \((2, 5)\): \(Y - 5 = 3(X - 2) \implies Y = 3X - 1\). Substituting back \(Y = \ln y\) and \(X = x^2\) gives: \(\ln y = 3x^2 - 1\).
(b) To express \(y\) in terms of \(x\), rewrite the equation in exponential form: \(y = e^{3x^2 - 1}\). Using exponent laws: \(y = e^{-1} \cdot e^{3x^2} = e^{-1} \cdot (e^3)^{x^2}\). Comparing this with \(y = Ab^{x^2}\), we have: \(A = e^{-1}\) (or \(\frac{1}{e}\)) and \(b = e^3\).

M1: For finding the gradient of the line: \(m = 3\). A1: For writing the correct equation \(\ln y = 3x^2 - 1\) (or equivalent). M1: For converting the logarithmic equation to exponential form: \(y = e^{3x^2 - 1}\). M1: For applying index laws to separate the terms: \(y = e^{-1}(e^3)^{x^2}\). A1: For identifying both constants correctly: \(A = e^{-1}\) (or \(0.368\)) and \(b = e^3\) (or \(20.1\)).

(a) At \(x = 1\), the \(y\)-coordinate is:
\( y = \frac{12}{2(1)+1} = 4 \)

Using the chain rule to differentiate \(y = 12(2x+1)^{-1}\):
\( \frac{dy}{dx} = -12(2x+1)^{-2} \times 2 = -\frac{24}{(2x+1)^2} \)

At \(x = 1\), the gradient of the tangent is:
\( m = -\frac{24}{(2(1)+1)^2} = -\frac{24}{9} = -\frac{8}{3} \)

The equation of the tangent at \(P(1, 4)\) is:
\( y - 4 = -\frac{8}{3}(x - 1) \)
\( y = -\frac{8}{3}x + \frac{20}{3} \) (or \( 8x + 3y = 20 \))

(b) The tangent crosses the \(x\)-axis when \(y = 0\):
\( 0 = -\frac{8}{3}x + \frac{20}{3} \implies 8x = 20 \implies x = 2.5 \)

So the coordinates of \(Q\) are \( (2.5, 0) \) or \( \left(\frac{5}{2}, 0\right) \).

(c) The area of the region bounded by the curve, the tangent, and the \(y\)-axis (from \(x = 0\) to \(x = 1\)) is found by subtracting the area under the tangent from the area under the curve.

Area under the curve:
\( \int_{0}^{1} \frac{12}{2x+1} dx = \left[ 6\ln(2x+1) \right]_0^1 = 6\ln 3 - 6\ln 1 = 6\ln 3 \)

Area under the tangent line (a trapezium with parallel vertical sides \(y = \frac{20}{3}\) at \(x=0\) and \(y = 4\) at \(x=1\)):
\( \text{Area} = \frac{1}{2} \left( \frac{20}{3} + 4 \right) \times 1 = \frac{16}{3} \)

Alternatively, by integration:
\( \int_{0}^{1} \left( -\frac{8}{3}x + \frac{20}{3} \right) dx = \left[ -\frac{4}{3}x^2 + \frac{20}{3}x \right]_0^1 = -\frac{4}{3} + \frac{20}{3} = \frac{16}{3} \)

Subtracting the two areas gives the exact area of the region:
\( \text{Exact Area} = 6\ln 3 - \frac{16}{3} \)

(a)
- M1: Attempts to differentiate \(y = 12(2x+1)^{-1}\) obtaining a form of \(k(2x+1)^{-2}\)
- A1: Correctly obtains \( \frac{dy}{dx} = -\frac{24}{(2x+1)^2} \)
- M1: Substitutes \(x=1\) to find the gradient and uses their \(y(1) = 4\) to form the tangent equation
- A1: Obtains \( y = -\frac{8}{3}x + \frac{20}{3} \) (or any equivalent form like \( 8x + 3y = 20 \))

(b)
- B1: Correctly identifies \( Q(2.5, 0) \) or \( \left(\frac{5}{2}, 0\right) \)

(c)
- M1: Integrates \(\frac{12}{2x+1}\) to get the form \(k\ln(2x+1)\)
- A1: Correctly finds the area under the curve is \( 6\ln 3 \) (accept \(\ln 729\))
- M1: Attempts to find the area under the tangent line from \(x=0\) to \(x=1\) (either via trapezium formula or integration)
- M1: Subtracts the area under the tangent from the area under the curve using the correct limits \(x=0\) and \(x=1\)
- A1: Obtains the exact area \( 6\ln 3 - \frac{16}{3} \) (or equivalent exact expression, e.g., \(\ln 729 - \frac{16}{3}\))