2024 Cambridge IGCSE Mathematics - Additional (0606) 模擬試題連答案詳解

Since \(x-2\) is a factor of \(p(x)\), we have \(p(2) = 0\):
\(2(2)^3 + a(2)^2 + b(2) - 12 = 0\)
\(16 + 4a + 2b - 12 = 0 \Rightarrow 4a + 2b = -4 \Rightarrow 2a + b = -2\) [Equation 1]

Since dividing \(p(x)\) by \(x+1\) gives a remainder of \(-15\), we have \(p(-1) = -15\):
\(2(-1)^3 + a(-1)^2 + b(-1) - 12 = -15\)
\(-2 + a - b - 12 = -15 \Rightarrow a - b = -1\) [Equation 2]

Solving Equations 1 and 2 simultaneously:
From Equation 2, \(b = a + 1\).
Substitute into Equation 1:
\(2a + (a + 1) = -2 \Rightarrow 3a = -3 \Rightarrow a = -1\).
Then \(b = -1 + 1 = 0\).

Thus, the polynomial is \(p(x) = 2x^3 - x^2 - 12\).

When divided by \(2x - 1\), the remainder is \(p\left(\frac{1}{2}\right)\):
\(p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 - 12 = 2\left(\frac{1}{8}\right) - \frac{1}{4} - 12 = \frac{1}{4} - \frac{1}{4} - 12 = -12\).

M1: Use the factor theorem to substitute \(x = 2\) and set equal to 0.
A1: Obtain the correct equation \(2a + b = -2\) or equivalent.
M1: Use the remainder theorem to substitute \(x = -1\) and set equal to \(-15\).
A1: Obtain the correct equation \(a - b = -1\) or equivalent.
M1: Solve their simultaneous equations to find the values of \(a\) and \(b\).
A1: Obtain \(a = -1\) and \(b = 0\).
B1: Substitute \(x = \frac{1}{2}\) into \(2x^3 - x^2 - 12\) to find the remainder \(-12\).

Let \(a\) be the first term and \(r\) be the common ratio, where \(a > 0\) and \(0 < r < 1\).

The sum of the first two terms is given by:
\(a + ar = 24 \Rightarrow a(1 + r) = 24\) [Equation 1]

The sum to infinity is given by:
\(\frac{a}{1 - r} = 32 \Rightarrow a = 32(1 - r)\) [Equation 2]

Substitute Equation 2 into Equation 1:
\(32(1 - r)(1 + r) = 24\)
\(32(1 - r^2) = 24\)
\(1 - r^2 = \frac{24}{32} = \frac{3}{4}\)
\(r^2 = 1 - \frac{3}{4} = \frac{1}{4}\)

Since the terms of the progression are positive, the common ratio must be positive:
\(r = \frac{1}{2}\)

Substitute \(r = \frac{1}{2}\) back into Equation 2 to find \(a\):
\(a = 32\left(1 - \frac{1}{2}\right) = 16\).

Thus, the common ratio is \(\frac{1}{2}\) and the first term is \(16\).

B1: Write down a correct equation for the sum of the first two terms: \(a(1 + r) = 24\) or equivalent.
B1: Write down a correct equation for the sum to infinity: \(\frac{a}{1-r} = 32\) or equivalent.
M1: Substitute the expression for \(a\) to obtain a single equation in terms of \(r\).
A1: Obtain the quadratic/simplified form \(r^2 = \frac{1}{4}\).
A1: Deduce \(r = \frac{1}{2}\) (rejecting \(r = -\frac{1}{2}\) since terms are positive).
A1: Obtain \(a = 16\).

(a) Using the quotient rule with \(u = e^{2x}\) and \(v = x - 1\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 2e^{2x}\)
\(\frac{\mathrm{d}v}{\mathrm{d}x} = 1\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2e^{2x}(x - 1) - e^{2x}(1)}{(x-1)^2}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{2x}[2(x - 1) - 1]}{(x-1)^2}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{2x}(2x - 3)}{(x-1)^2}\)

Comparing with the given form, \(A = 2\) and \(B = 3\).

(b) At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
Since \(e^{2x} \neq 0\) and \((x - 1)^2 > 0\) for \(x > 1\), we have:
\(2x - 3 = 0 \Rightarrow x = 1.5\)

Substitute \(x = 1.5\) back into the curve's equation to find the \(y\)-coordinate:
\(y = \frac{e^{2(1.5)}}{1.5 - 1} = \frac{e^3}{0.5} = 2e^3\)

Thus, the exact coordinates of the stationary point are \((1.5, 2e^3)\).

(a)
M1: Apply the quotient rule (or product rule) correctly to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
A1: Obtain correct unsimplified numerator: \(2e^{2x}(x - 1) - e^{2x}\).
A1: Simplify the expression to the form \(\frac{e^{2x}(2x-3)}{(x-1)^2}\) and state \(A = 2\), \(B = 3\).

(b)
M1: Set their numerator equal to zero to find the \(x\)-coordinate.
A1: Identify the exact value \(x = 1.5\) (or \(\frac{3}{2}\)).
A1: Correctly evaluate the exact \(y\)-coordinate as \(2e^3\).

To solve the equation:
\(3\cos^2 \theta - 5\sin \theta \cos \theta - 2\sin^2 \theta = 0\)

Divide the entire equation by \(\cos^2 \theta\) (noting \(\cos\theta \neq 0\)):
\(3 - 5\tan \theta - 2\tan^2 \theta = 0\)

Rearrange into standard quadratic form:
\(2\tan^2 \theta + 5\tan \theta - 3 = 0\)

Factor the quadratic in terms of \(\tan \theta\):
\((2\tan \theta - 1)(\tan \theta + 3) = 0\)

This gives two cases:
1) \(\tan \theta = \frac{1}{2}\)
Since \(0^\circ \le \theta \le 180^\circ\), the reference angle is:
\(\theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.6^\circ\)

2) \(\tan \theta = -3\)
Since \(0^\circ \le \theta \le 180^\circ\), the angle in the second quadrant is:
\(\theta = 180^\circ - \tan^{-1}(3) \approx 180^\circ - 71.6^\circ = 108.4^\circ\)

Therefore, the solutions in the given range are \(\theta = 26.6^\circ\) and \(\theta = 108.4^\circ\) (to 1 decimal place).

M1: Divide by \(\cos^2 \theta\) to obtain a quadratic in terms of \(\tan \theta\).
A1: Obtain the correct quadratic equation \(2\tan^2 \theta + 5\tan \theta - 3 = 0\) or equivalent.
M1: Attempt to solve their quadratic equation in \(\tan \theta\) by factorisation or formula.
A1: Identify the two values \(\tan \theta = 0.5\) and \(\tan \theta = -3\).
A1: Solve \(\tan \theta = 0.5\) to find \(\theta = 26.6^\circ\) (accept 26.57...).
A1: Solve \(\tan \theta = -3\) to find \(\theta = 108.4^\circ\) (accept 108.43...).

(a) To find \(\mathrm{f}^{-1}(x)\), let \(y = \ln(2x - 3)\):
\(e^y = 2x - 3\)
\(2x = e^y + 3\)
\(x = \frac{e^y + 3}{2}\)

Therefore, \(\mathrm{f}^{-1}(x) = \frac{e^x + 3}{2}\).

Since the range of \(\mathrm{f}(x)\) is all real numbers (\(\mathbb{R}\)), the domain of \(\mathrm{f}^{-1}(x)\) is \(x \in \mathbb{R}\).

(b) First, compute \(\mathrm{f}^{-1}(0)\):
\(\mathrm{f}^{-1}(0) = \frac{e^0 + 3}{2} = \frac{1 + 3}{2} = 2\).

We want to find \(x\) such that \(\mathrm{f}(x) = 2\):
\[ \ln(2x - 3) = 2 \]
\[ 2x - 3 = e^2 \]
\[ 2x = e^2 + 3 \]
\[ x = \frac{e^2 + 3}{2} \approx 5.19 \]

(a)
M1: Set \(y = \ln(2x-3)\) and attempt to express \(x\) in terms of \(y\) using exponentiation.
A1: Correctly obtain \(2x = e^y + 3\).
A1: Write down the correct inverse function \(\mathrm{f}^{-1}(x) = \frac{e^x + 3}{2}\) (must use variable \(x\)).
B1: State the correct domain: \(x \in \mathbb{R}\) (or 'all real numbers').

(b)
M1: Calculate \(\mathrm{f}^{-1}(0) = 2\).
M1: Set \(\ln(2x-3) = 2\) and solve for \(x\).
A1: Obtain \(x = \frac{e^2+3}{2}\) (accept approx 5.19).

(a) The perimeter of the sector is given by:
\(P = 2r + r\theta = 32\)
From this, we can express \(\theta\) in terms of \(r\):
\(r\theta = 32 - 2r \Rightarrow \theta = \frac{32 - 2r}{r}\)

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

Substitute the expression for \(\theta\) into the area formula:
\(A = \frac{1}{2}r^2 \left(\frac{32 - 2r}{r}\right)\)
\(A = \frac{1}{2}r(32 - 2r)\)
\(A = 16r - r^2\)

(b) Given that \(A = 48\):
\(16r - r^2 = 48 \Rightarrow r^2 - 16r + 48 = 0\)

Factor the quadratic equation:
\((r - 4)(r - 12) = 0\)

This gives two possible values of \(r\):
\(r = 4\) or \(r = 12\)

Now, calculate the corresponding values of \(\theta\) using \(\theta = \frac{32 - 2r}{r}\):

- If \(r = 4\):
\(\theta = \frac{32 - 2(4)}{4} = \frac{24}{4} = 6\text{ radians}\)

- If \(r = 12\):
\(\theta = \frac{32 - 2(12)}{12} = \frac{8}{12} = \frac{2}{3}\text{ radians}\)

(a)
M1: Use the perimeter formula \(2r + r\theta = 32\) to express \(\theta\) or \(r\theta\) in terms of \(r\).
A1: Obtain a correct expression, e.g., \(\theta = \frac{32 - 2r}{r}\).
M1: Substitute their expression into the area formula \(A = \frac{1}{2}r^2\theta\).
A1: Simplify to obtain \(A = 16r - r^2\).

(b)
M1: Set their area expression equal to 48 and attempt to solve the quadratic equation.
A1: Find the two values of radius, \(r = 4\) and \(r = 12\).
A1: Obtain corresponding correct values of \(\theta = 6\) and \(\theta = \frac{2}{3}\) (both pairs must be correct and matched).

For the line and the curve not to intersect, the equation:
\(kx - 5 = 2x^2 + 3x - 3\)
must have no real solutions.

Rearrange this into standard quadratic form:
\(2x^2 + (3 - k)x + 2 = 0\)

For no real solutions, the discriminant \(\Delta = b^2 - 4ac\) must be less than zero:
\((3 - k)^2 - 4(2)(2) < 0\)
\((3 - k)^2 - 16 < 0\)
\((3 - k)^2 < 16\)

Taking square roots on both sides:
\(-4 < 3 - k < 4\)

Subtract 3 from all parts:
\(-7 < -k < 1\)

Multiply by \(-1\) (reversing the inequalities):
\(-1 < k < 7\).

Thus, the set of values of \(k\) is \(-1 < k < 7\).

M1: Equate the line and the curve to form a single equation.
A1: Write the equation in standard quadratic form \(2x^2 + (3 - k)x + 2 = 0\).
M1: Apply the condition for no real roots, \(\Delta < 0\), on their quadratic equation.
A1: Obtain correct inequality in terms of \(k\), e.g., \((3 - k)^2 - 16 < 0\) or \(k^2 - 6k - 7 < 0\).
M1: Find the critical values \(k = -1\) and \(k = 7\).
A1: Express final answer as the single range \(-1 < k < 7\).

Given:
\(\log_5(2x + 1) + \log_5(x - 2) - \log_5(x) = 1\)

Use the addition and subtraction laws of logarithms:
\(\log_5\left(\frac{(2x + 1)(x - 2)}{x}\right) = 1\)

Convert the logarithmic equation to its exponential form:
\(\frac{(2x + 1)(x - 2)}{x} = 5^1 = 5\)

Multiply both sides by \(x\):
\((2x + 1)(x - 2) = 5x\)

Expand the left-hand side:
\(2x^2 - 4x + x - 2 = 5x\)
\(2x^2 - 3x - 2 = 5x\)

Subtract \(5x\) to form a standard quadratic equation:
\(2x^2 - 8x - 2 = 0\)

Divide by 2:
\(x^2 - 4x - 1 = 0\)

Using the quadratic formula:
\(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2(1)}\)
\(x = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2}\)
\(x = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}\)

Since the original log expression requires \(x > 2\) for \(\log_5(x - 2)\) to be defined, the negative root \(2 - \sqrt{5} \approx -0.24\) must be rejected.

Thus, the only valid solution is:
\(x = 2 + \sqrt{5}\) (or approx \(4.24\)).

M1: Apply the addition law of logs correctly to combine terms: \(\log_5[(2x + 1)(x - 2)]\).
M1: Apply the subtraction law of logs correctly: \(\log_5\left(\frac{(2x + 1)(x - 2)}{x}\right)\).
M1: Convert the log equation to index form: \(\frac{(2x + 1)(x - 2)}{x} = 5\).
A1: Form the correct quadratic equation, e.g., \(2x^2 - 8x - 2 = 0\) or \(x^2 - 4x - 1 = 0\).
M1: Solve their quadratic equation using formula/completing the square.
A1: Identify both roots: \(x = 2 \pm \sqrt{5}\).
A1: Deduce the unique solution \(x = 2 + \sqrt{5}\) by explicitly rejecting the negative root.

First, we find the derivative of \(\mathrm{p}(x)\):
\(\mathrm{p}'(x) = 6x^2 + 2ax + b\).

Using the given condition \(\mathrm{p}'(0) = -4\):
\(\mathrm{p}'(0) = 6(0)^2 + 2a(0) + b = -4 \implies b = -4\).

Using the factor theorem with factor \(x-1\), we have \(\mathrm{p}(1) = 0\):
\(\mathrm{p}(1) = 2(1)^3 + a(1)^2 + b(1) + c = 0\)
\(2 + a - 4 + c = 0 \implies a + c = 2\) (Equation 1).

Using the remainder theorem with divisor \(x+2\), we have \(\mathrm{p}(-2) = -15\):
\(\mathrm{p}(-2) = 2(-2)^3 + a(-2)^2 + b(-2) + c = -15\)
\(-16 + 4a - 2(-4) + c = -15\)
\(-16 + 4a + 8 + c = -15\)
\(4a + c - 8 = -15 \implies 4a + c = -7\) (Equation 2).

Subtract Equation 1 from Equation 2:
\((4a + c) - (a + c) = -7 - 2\)
\(3a = -9 \implies a = -3\).

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

Thus, the values are \(a = -3\), \(b = -4\), and \(c = 5\).

B1: Differentiates to find \(\mathrm{p}'(x) = 6x^2 + 2ax + b\) and uses \(\mathrm{p}'(0) = -4\) to show \(b = -4\).
M1: Applies the factor theorem by substituting \(x = 1\) and equating to \(0\).
A1: Obtains the simplified correct equation \(a + c = 2\) (or equivalent with their \(b\)).
M1: Applies the remainder theorem by substituting \(x = -2\) and equating to \(-15\).
A1: Obtains the simplified correct equation \(4a + c = -7\) (or equivalent with their \(b\)).
A1: Solves the system of equations simultaneously to find both \(a = -3\) and \(c = 5\).

(a) Let \(Y = \ln y\) and \(X = \ln x\).
The gradient of the straight line is:
\(m = \frac{9 - 3}{5 - 2} = \frac{6}{3} = 2\).

Using the point-slope form with the point \((2, 3)\):
\(Y - 3 = 2(X - 2)\)
\(Y = 2X - 1\).

Substituting back \(Y = \ln y\) and \(X = \ln x\):
\(\ln y = 2\ln x - 1\)
\(\ln y = \ln(x^2) - 1\).

Taking exponentials on both sides:
\(y = \mathrm{e}^{\ln(x^2) - 1}\)
\(y = \mathrm{e}^{-1} x^2\).

(b) When \(y = 4\mathrm{e}^3\):
\(4\mathrm{e}^3 = \mathrm{e}^{-1} x^2\)
\(x^2 = 4\mathrm{e}^4\).

Since \(\ln x\) is defined, \(x > 0\), so taking the positive square root:
\(x = 2\mathrm{e}^2\).

(a)
M1: For finding the gradient of the straight line, \(m = 2\).
M1: For using a correct point to set up the equation of the line, e.g. \(\ln y - 3 = 2(\ln x - 2)\).
M1: For correct use of the laws of logarithms or indices to make \(y\) the subject, e.g. \(y = \mathrm{e}^{2\ln x - 1}\).
A1: For the correct expression \(y = \mathrm{e}^{-1}x^2\) (or \(y = \frac{x^2}{\mathrm{e}}\)).

(b)
M1: For setting up the equation \(4\mathrm{e}^3 = \mathrm{e}^{-1}x^2\) and solving for \(x^2\).
A1: For \(x = 2\mathrm{e}^2\) only.

We use the identity \(\sec^2\theta = 1 + \tan^2\theta\) to rewrite the equation in terms of \(\tan\theta\):
\(3(1 + \tan^2\theta) - 2\tan\theta = 4\)
\(3 + 3\tan^2\theta - 2\tan\theta - 4 = 0\)
\(3\tan^2\theta - 2\tan\theta - 1 = 0\).

Letting \(u = \tan\theta\), we obtain a quadratic equation:
\(3u^2 - 2u - 1 = 0\)
\((3u + 1)(u - 1) = 0\).

So, \(\tan\theta = 1\) or \(\tan\theta = -\frac{1}{3}\).

Case 1: \(\tan\theta = 1\) in the range \(-180^\circ \leqslant \theta \leqslant 180^\circ\).
The basic angle is \(45^\circ\).
\(\theta = 45^\circ\) or \(\theta = 45^\circ - 180^\circ = -135^\circ\).

Case 2: \(\tan\theta = -\frac{1}{3}\) in the range \(-180^\circ \leqslant \theta \leqslant 180^\circ\).
The basic angle is \(\tan^{-1}(1/3) \approx 18.43^\circ\).
Since tangent is negative in quadrants 2 and 4:
In quadrant 4: \(\theta = -18.43^\circ \approx -18.4^\circ\).
In quadrant 2: \(\theta = 180^\circ - 18.43^\circ = 161.57^\circ \approx 161.6^\circ\).

Therefore, the solutions are \(\theta = -135^\circ\), \(-18.4^\circ\), \(45^\circ\), and \(161.6^\circ\).

B1: For using \(\sec^2\theta = 1 + \tan^2\theta\) to form a three-term quadratic equation in \(\tan\theta\).
M1: For attempting to factorise or solve the quadratic equation \(3\tan^2\theta - 2\tan\theta - 1 = 0\).
A1: For obtaining both correct roots \(\tan\theta = 1\) and \(\tan\theta = -\frac{1}{3}\).
M1: For finding one correct angle from \(\tan\theta = 1\) (usually \(45^\circ\)).
A1: For finding both correct angles: \(45^\circ\) and \(-135^\circ\) (and no extra values from this branch).
M1: For finding one correct angle from \(\tan\theta = -\frac{1}{3}\) (either \(-18.4^\circ\) or \(161.6^\circ\)).
A1: For both correct angles: \(-18.4^\circ\) and \(161.6^\circ\) (rounded to 1 decimal place, and no extra values in the range).

(a) To find the equation of the tangent at \(x = 0\):
First, find the y-coordinate:
\(y = 8\mathrm{e}^{-2(0)} = 8\).

Now, differentiate to find the gradient function:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 8(-2)\mathrm{e}^{-2x} = -16\mathrm{e}^{-2x}\).

At \(x = 0\), the gradient is:
\(m = -16\mathrm{e}^0 = -16\).

The equation of the tangent is:
\(y - 8 = -16(x - 0)\)
\(y = -16x + 8\).

(b) The area of the region is given by the integral:
\(A = \int_{0}^{\ln 3} 8\mathrm{e}^{-2x} \mathrm{d}x\).

Integrating \(8\mathrm{e}^{-2x}\):
\(\int 8\mathrm{e}^{-2x} \mathrm{d}x = \frac{8}{-2}\mathrm{e}^{-2x} = -4\mathrm{e}^{-2x}\).

Applying the limits:
\(A = \left[ -4\mathrm{e}^{-2x} \right]_{0}^{\ln 3}\)
\(A = -4\mathrm{e}^{-2\ln 3} - (-4\mathrm{e}^{0})\)
\(A = -4\mathrm{e}^{\ln(3^{-2})} + 4\)
\(A = -4\left(\frac{1}{9}\right) + 4\)
\(A = -\frac{4}{9} + 4 = \frac{32}{9}\).

(a)
B1: Differentiates correctly to obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = -16\mathrm{e}^{-2x}\).
M1: Evaluates both the y-coordinate (\(y = 8\)) and the gradient (\(m = -16\)) at \(x = 0\), and uses them to write down the equation of the line.
A1: Obtains \(y = -16x + 8\) (or any equivalent form).

(b)
M1: Writes down the correct definite integral \(\int_{0}^{\ln 3} 8\mathrm{e}^{-2x} \mathrm{d}x\).
A1: For the correct integration \(\left[ -4\mathrm{e}^{-2x} \right]\).
M1: Applies limits correctly, using laws of logarithms to simplify \(\mathrm{e}^{-2\ln 3}\) to \(\frac{1}{9}\).
A1: Obtains the correct exact value \(\frac{32}{9}\) (or \(3\frac{5}{9}\)).

First, find the \(y\)-coordinate at \(x = \frac{\pi}{4}\):
\(y = 3\left(\frac{\pi}{4}\right) \sin\left(2 \cdot \frac{\pi}{4}\right) = \frac{3\pi}{4} \sin\left(\frac{\pi}{2}\right) = \frac{3\pi}{4}\).
So the point is \(\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)\).

Next, differentiate \(y = 3x \sin 2x\) using the product rule:
\(\frac{dy}{dx} = 3\sin 2x + 6x\cos 2x\).

Evaluate the derivative at \(x = \frac{\pi}{4}\):
\(\frac{dy}{dx} = 3\sin\left(\frac{\pi}{2}\right) + 6\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{2}\right) = 3(1) + 0 = 3\).

The gradient of the tangent is 3, so the gradient of the normal is \(-\frac{1}{3}\).

The equation of the normal is:
\(y - \frac{3\pi}{4} = -\frac{1}{3}\left(x - \frac{\pi}{4}\right)\)
\(y = -\frac{1}{3}x + \frac{\pi}{12} + \frac{3\pi}{4}\)
\(y = -\frac{1}{3}x + \frac{5\pi}{6}\).

M1: For finding the correct \(y\)-coordinate \(\frac{3\pi}{4}\)
M1: For differentiating using the product rule to get \(k\sin 2x + k'x\cos 2x\)
A1: For the correct derivative \(\frac{dy}{dx} = 3\sin 2x + 6x\cos 2x\)
M1: For substituting \(x = \frac{\pi}{4}\) into their derivative to find the gradient of the tangent
M1: For finding the gradient of the normal as the negative reciprocal of their tangent gradient
M1: For formulating the equation of the line using their point and normal gradient
A1: For the correct exact equation \(y = -\frac{1}{3}x + \frac{5\pi}{6}\) oe

Use the identity \(\cot^2 \theta = \csc^2 \theta - 1\):
\(2(\csc^2 \theta - 1) - \csc \theta - 1 = 0\)
\(2\csc^2 \theta - \csc \theta - 3 = 0\)

Factorise the quadratic equation:
\((2\csc \theta - 3)(\csc \theta + 1) = 0\)

This gives:
\(\csc \theta = 1.5\) or \(\csc \theta = -1\)

Convert to sine:
\(\sin \theta = \frac{2}{3}\) or \(\sin \theta = -1\)

For \(\sin \theta = \frac{2}{3}\):
\(\theta \approx 41.8^\circ\) or \(\theta \approx 180^\circ - 41.8^\circ = 138.2^\circ\)

For \(\sin \theta = -1\):
\(\theta = 270^\circ\)

Therefore, \(\theta = 41.8^\circ, 138.2^\circ, 270^\circ\).

B1: For using the correct identity to obtain a quadratic in terms of \(\csc \theta\)
M1: For obtaining the 3-term quadratic \(2\csc^2 \theta - \csc \theta - 3 = 0\)
M1: For attempt to solve/factorise their 3-term quadratic to obtain values for \(\csc \theta\) or \(\sin \theta\)
A1: For \(\sin \theta = \frac{2}{3}\) and \(\sin \theta = -1\) oe
B1: For \(\theta = 270^\circ\)
B1: For \(\theta = 41.8^\circ\) (accept 41.8)
B1: For \(\theta = 138.2^\circ\) (accept 138.2)

(a) Let the first term of the arithmetic progression (AP) be \(a\) and the common difference be \(d\).
The terms are:
First term \(T_1 = a\)
Third term \(T_3 = a + 2d\)
Ninth term \(T_9 = a + 8d\)

Since these are in geometric progression (GP):
\((a + 2d)^2 = a(a + 8d)\)
\(a^2 + 4ad + 4d^2 = a^2 + 8ad\)
\(4d^2 - 4ad = 0\)
\(4d(d - a) = 0\)

Since the common ratio is non-zero (and progression is not constant), \(d \ne 0\), which gives \(d = a\).

The terms of the GP are \(a\), \(3a\), \(9a\).
The common ratio is \(r = \frac{3a}{a} = 3\).

(b) The sum of the first 10 terms of the AP is:
\(S_{10} = \frac{10}{2}[2a + (10 - 1)d] = 165\)
\(5[2a + 9d] = 165\)
\(2a + 9d = 33\)

Substitute \(d = a\):
\(2a + 9a = 33\)
\(11a = 33\)
\(a = 3\).

Thus, the first term of the progression is 3.

(a)
M1: For writing expressions for the three terms of the AP: \(a\), \(a + 2d\), \(a + 8d\)
M1: For setting up the GP relation: \((a + 2d)^2 = a(a + 8d)\) oe
A1: For solving to show \(d = a\) (or equivalent relationship)
A1: For finding the correct common ratio \(r = 3\)

(b)
M1: For using the sum of AP formula \(S_{10} = 5(2a + 9d) = 165\)
M1: For substituting \(d = a\) into their sum equation and attempting to solve for \(a\)
A1: For the correct first term \(a = 3\)

From the first equation:
\(\log_3(x - 2y) = 2 \implies x - 2y = 3^2 = 9\)
\(x = 2y + 9\) [Equation 1]

From the second equation:
\(\log_3\left(\frac{x}{y}\right) = \log_3 3 + \log_3 2\)
\(\log_3\left(\frac{x}{y}\right) = \log_3(3 \times 2) = \log_3 6\)
\(\frac{x}{y} = 6 \implies x = 6y\) [Equation 2]

Substitute Equation 2 into Equation 1:
\(6y = 2y + 9\)
\(4y = 9 \implies y = 2.25\)

Find \(x\):
\(x = 6(2.25) = 13.5\).

Check validity: both \(x > 0\), \(y > 0\), and \(x - 2y = 9 > 0\). Thus, the solutions are valid.

B1: For converting the first logarithmic equation to linear form: \(x - 2y = 9\)
M1: For applying subtraction/addition laws of logarithms to the second equation
A1: For converting the second equation to linear form: \(\frac{x}{y} = 6\) or \(x = 6y\) oe
M1: For setting up a method to solve the two linear equations simultaneously
A1: For finding \(y = 2.25\) oe
A1: For finding \(x = 13.5\) oe
B1: For confirming validity of both solutions (implicitly or explicitly)

(a) Total number of ways to choose 5 people from 10 without restrictions is:
\(\binom{10}{5} = \frac{10!}{5!5!} = 252\).

(b) If there must be at least 3 women, the possible scenarios are:
1) 3 women and 2 men:
\(\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120\)

2) 4 women and 1 man:
\(\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60\)

3) 5 women and 0 men:
\(\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6\)

Total number of ways = \(120 + 60 + 6 = 186\).

(a)
M1: For attempting to calculate \(\binom{10}{5}\)
A1: For 252

(b)
M1: For identifying the three valid cases: (3W, 2M), (4W, 1M), (5W, 0M)
M1: For calculating the combinations for any one case correctly (e.g. \(\binom{6}{3} \times \binom{4}{2}\))
A1: For obtaining any two cases correct (e.g., 120 and 60)
A1: For all three cases calculated correctly (120, 60, 6)
A1: For the correct sum of 186

Equate the equations of the line and the curve:
\(kx - 5 = x^2 - 4x + 4\)
\(x^2 - (k + 4)x + 9 = 0\)

For no intersection, the discriminant of this quadratic equation must be negative (\(b^2 - 4ac < 0\)):
\(a = 1\), \(b = -(k + 4)\), \(c = 9\)

\((-(k + 4))^2 - 4(1)(9) < 0\)
\((k + 4)^2 - 36 < 0\)
\((k + 4)^2 < 36\)

Taking the square root:
\(-6 < k + 4 < 6\)
\(-10 < k < 2\).

M1: For equating the line and the curve equation
A1: For obtaining a correct 3-term quadratic equation, e.g., \(x^2 - (k+4)x + 9 = 0\)
M1: For using the discriminant \(b^2 - 4ac\) with their coefficients
M1: For setting up the inequality \(b^2 - 4ac < 0\) for no intersection
A1: For obtaining the critical values \(-10\) and \(2\)
M1: For correctly solving the quadratic inequality to find the inside region
A1: For the correct range \(-10 < k < 2\) (or equivalent)

(a) Since \(OB : BC = 2 : 3\), the total length \(OC\) is \(2 + 3 = 5\) parts where \(OB\) is 2 parts.
Thus, \(\overrightarrow{OC} = 2.5 \mathbf{b}\) (or \(\frac{5}{2}\mathbf{b}\)).

Therefore, \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = 2.5\mathbf{b} - \mathbf{a}\).

(b) Find \(\overrightarrow{OD}\):
\(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AD} = \mathbf{a} + \lambda \overrightarrow{AC}\)
\(\overrightarrow{OD} = \mathbf{a} + \lambda(2.5\mathbf{b} - \mathbf{a}) = (1 - \lambda)\mathbf{a} + 2.5\lambda\mathbf{b}\).

Since \(\overrightarrow{OD}\) is parallel to \(\mathbf{a} + 2\mathbf{b}\), there exists a scalar \(k\) such that:
\(\overrightarrow{OD} = k(\mathbf{a} + 2\mathbf{b})\)

Comparing coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
1) \(1 - \lambda = k\)
2) \(2.5\lambda = 2k\)

Substitute (1) into (2):
\(2.5\lambda = 2(1 - \lambda)\)
\(2.5\lambda = 2 - 2\lambda\)
\(4.5\lambda = 2\)
\(\lambda = \frac{2}{4.5} = \frac{4}{9}\).

(a)
B1: For finding \(\overrightarrow{OC} = 2.5\mathbf{b}\) oe
B1: For the correct expression \(\overrightarrow{AC} = 2.5\mathbf{b} - \mathbf{a}\) oe

(b)
M1: For expressing \(\overrightarrow{OD}\) as \(\overrightarrow{OA} + \lambda \overrightarrow{AC}\)
A1: For the correct simplified vector \(\overrightarrow{OD} = (1 - \lambda)\mathbf{a} + 2.5\lambda\mathbf{b}\) oe
M1: For using the parallel condition to set up a ratio or system of equations, e.g., \(\frac{2.5\lambda}{1-\lambda} = 2\) oe
M1: For solving their linear equation in terms of \(\lambda\)
A1: For \(\lambda = \frac{4}{9}\) oe

(a) The area of a sector is given by \(A = \frac{1}{2}r^2\theta = 48 \implies r^2\theta = 96\).

The perimeter of a sector is given by \(P = 2r + r\theta = 28 \implies r\theta = 28 - 2r\).

Substitute \(r\theta\) from the perimeter equation into the area equation:
\(r(r\theta) = 96\)
\(r(28 - 2r) = 96\)
\(28r - 2r^2 = 96\)
\(2r^2 - 28r + 96 = 0\)
Divide by 2:
\(r^2 - 14r + 48 = 0\) (as required).

(b) Factorise the quadratic equation:
\((r - 6)(r - 8) = 0\)
So, the two possible values of \(r\) are \(r = 6\text{ cm}\) and \(r = 8\text{ cm}\).

If \(r = 6\):
\(r\theta = 28 - 2(6) = 16 \implies 6\theta = 16 \implies \theta = \frac{16}{6} = \frac{8}{3}\text{ rad}\).

If \(r = 8\):
\(r\theta = 28 - 2(8) = 12 \implies 8\theta = 12 \implies \theta = \frac{12}{8} = \frac{3}{2}\text{ rad}\).

(a)
B1: For the correct area equation \(\frac{1}{2}r^2\theta = 48\) oe
B1: For the correct perimeter equation \(2r + r\theta = 28\) oe
M1: For substituting one equation into the other to eliminate \(\theta\)
A1: For showing the correct given quadratic equation cleanly

(b)
M1: For solving the quadratic equation to obtain two values of \(r\)
A1: For \(r = 6\) and \(r = 8\)
A1: For the correct corresponding values \(\theta = \frac{8}{3}\) (or \(2.67\)) and \(\theta = 1.5\) (or \(\frac{3}{2}\))

Using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\), the equation becomes \(3(1 + \tan^2 \theta) - 5 \tan \theta - 5 = 0\). Simplifying, we get \(3\tan^2 \theta - 5 \tan \theta - 2 = 0\). Factorising the quadratic yields \((3\tan \theta + 1)(\tan \theta - 2) = 0\), which gives \(\tan \theta = 2\) or \(\tan \theta = -\frac{1}{3}\). For \(\tan \theta = 2\), \(\theta = 63.4^\circ\) or \(243.4^\circ\). For \(\tan \theta = -\frac{1}{3}\), \(\theta = 161.6^\circ\) or \(341.6^\circ\).

B1 for substituting identity to get quadratic in tan. M1 for correct simplified quadratic. M1 for solving quadratic. A1 for 63.4 and 243.4. A1 for 161.6 and 341.6. A1 for no extra solutions in range.

Using the product rule, the derivative is \(\frac{dy}{dx} = 2e^{-x} - (2x - 3)e^{-x} = e^{-x}(5 - 2x)\). Setting \(\frac{dy}{dx} = 0\) gives \(5 - 2x = 0\), so \(x = 2.5\). Substituting \(x = 2.5\) into the original equation gives the y-coordinate \(y = (2(2.5) - 3)e^{-2.5} = 2e^{-2.5}\). To determine its nature, find \(\frac{d^2y}{dx^2} = -2e^{-x} - (5 - 2x)e^{-x} = e^{-x}(2x - 7)\). At \(x = 2.5\), \(\frac{d^2y}{dx^2} = -2e^{-2.5} < 0\), hence the point is a maximum.

M1 for product rule differentiation. A1 for correct first derivative. M1 for setting to zero and solving for x. A1 for x = 2.5. A1 for y = 2e^(-2.5). M1 for finding second derivative or testing gradient. A1 for showing the second derivative is negative and concluding it is a maximum.

The sum of the first two terms is \(a + ar = 15 \implies a(1 + r) = 15\). The sum to infinity is \(\frac{a}{1 - r} = 27 \implies a = 27(1 - r)\). Substituting \(a\) into the first equation yields \(27(1 - r)(1 + r) = 15 \implies 27(1 - r^2) = 15\). This simplifies to \(1 - r^2 = \frac{5}{9} \implies r^2 = \frac{4}{9}\). Since \(r > 0\), \(r = \frac{2}{3}\). Substituting back gives \(a = 27(1 - \frac{2}{3}) = 9\).

B1 for sum of first two terms equation. B1 for sum to infinity equation. M1 for substituting to form one equation. A1 for obtaining 27(1 - r^2) = 15. M1 for solving for r. A1 for r = 2/3. A1 for a = 9.

Using log laws, we can write the equation as \(\log_2 \left(\frac{5x + 1}{(x - 1)^2}\right) = 3\). Converting to exponential form, \(\frac{5x + 1}{(x - 1)^2} = 2^3 = 8\). Thus, \(5x + 1 = 8(x^2 - 2x + 1)\), which expands to \(5x + 1 = 8x^2 - 16x + 8\). Rearranging gives the quadratic equation \(8x^2 - 21x + 7 = 0\). Solving via quadratic formula yields \(x = \frac{21 \pm \sqrt{217}}{16}\), giving \(x \approx 2.23\) or \(x \approx 0.392\). Since the domain requires \(x > 1\), the only valid solution is \(x \approx 2.23\).

M1 for power law on log. M1 for subtraction law on logs. M1 for converting to index form. A1 for 8x^2 - 21x + 7 = 0. M1 for solving the quadratic. A1 for rejecting the extraneous solution and stating the correct final answer.