2025 Cambridge IGCSE Mathematics - Additional (0606) 模拟试题及答案详解

To solve \(|3x - 5| < x + 3\), we can square both sides since both sides are non-negative for the critical values:
\[(3x - 5)^2 < (x + 3)^2\]

Expanding both sides:
\[9x^2 - 30x + 25 < x^2 + 6x + 9\]

Rearranging the terms into a single quadratic inequality:
\[8x^2 - 36x + 16 < 0\]

Divide the entire inequality by 4:
\[2x^2 - 9x + 4 < 0\]

Factorize the quadratic expression:
\[(2x - 1)(x - 4) < 0\]

Identify the critical values:
\[x = \frac{1}{2} \quad \text{and} \quad x = 4\]

Since we want the expression to be less than zero, the solution lies between the critical values:
\[\frac{1}{2} < x < 4\]

M1: For squaring both sides or setting up two linear inequalities: \(3x - 5 < x + 3\) and \(-(3x - 5) < x + 3\).
A1: For a correct quadratic equation/expression: \(2x^2 - 9x + 4 = 0\) (or equivalent).
M1: For attempting to factorize or solve their 3-term quadratic.
A1: For identifying critical values \(x = 0.5\) and \(x = 4\).
A3: For the correct range \(0.5 < x < 4\) (or \(\frac{1}{2} < x < 4\)).

Let \(y = (2x - 1)(4x + 1)^{\frac{1}{2}}\).

Applying the product rule, \(\frac{\mathrm{d}y}{\mathrm{d}x} = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}\), where:
\[u = 2x - 1 \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 2\]
\[v = (4x + 1)^{\frac{1}{2}} \implies \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \cdot 4 = 2(4x + 1)^{-\frac{1}{2}}\]

Substitute these back into the product rule formula:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = (2x - 1) \cdot 2(4x + 1)^{-\frac{1}{2}} + (4x + 1)^{\frac{1}{2}} \cdot 2\]
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(2x - 1)}{\sqrt{4x + 1}} + 2\sqrt{4x + 1}\]

Express both terms over a common denominator of \(\sqrt{4x + 1}\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(2x - 1) + 2(4x + 1)}{\sqrt{4x + 1}}\]

Expand the numerator:
\[2(2x - 1) + 2(4x + 1) = 4x - 2 + 8x + 2 = 12x\]

Therefore:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{12x}{\sqrt{4x + 1}}\]

M1: For applying product rule or chain rule.
A1: For finding \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2\) or \(\frac{\mathrm{d}v}{\mathrm{d}x} = 2(4x + 1)^{-\frac{1}{2}}\).
A1: For the unsimplified expression \(2(2x - 1)(4x + 1)^{-\frac{1}{2}} + 2(4x + 1)^{\frac{1}{2}}\).
M1: For putting the terms over a common denominator of \(\sqrt{4x + 1}\).
M1: For expanding the numerator correctly.
A2: For \(\frac{12x}{\sqrt{4x + 1}}\) (A1 for numerator \(12x\), A1 for correct final form).

Using the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\), we substitute into the equation:
\[2(1 - \sin^2 \theta) + 3\sin \theta = 3\]
\[2 - 2\sin^2 \theta + 3\sin \theta = 3\]

Rearrange into standard quadratic form:
\[2\sin^2 \theta - 3\sin \theta + 1 = 0\]

Factorize the quadratic expression:
\[(2\sin \theta - 1)(\sin \theta - 1) = 0\]

This gives two possible equations to solve:
1) \(2\sin \theta - 1 = 0 \implies \sin \theta = \frac{1}{2}\)
2) \(\sin \theta - 1 = 0 \implies \sin \theta = 1\)

For \(\sin \theta = \frac{1}{2}\) in the range \(0^\circ \le \theta \le 360^\circ\):
\[\theta = 30^\circ \quad \text{and} \quad \theta = 150^\circ\]

For \(\sin \theta = 1\) in the range \(0^\circ \le \theta \le 360^\circ\):
\[\theta = 90^\circ\]

Combining all the solutions:
\[\theta = 30^\circ, 90^\circ, 150^\circ\]

M1: For using \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a quadratic in \(\sin \theta\).
A1: For the correct quadratic equation: \(2\sin^2 \theta - 3\sin \theta + 1 = 0\) (or equivalent).
M1: For factorizing or solving their 3-term quadratic.
A1: For \(\sin \theta = 0.5\) and \(\sin \theta = 1\).
A1: For \(\theta = 30^\circ\).
A1: For \(\theta = 150^\circ\).
A1: For \(\theta = 90^\circ\).

From the first equation:
\[4^x \cdot 2^y = 64\]
Rewrite \(4^x\) and \(64\) using base 2:
\[(2^2)^x \cdot 2^y = 2^6\]
\[2^{2x} \cdot 2^y = 2^{2x + y} = 2^6\]
Thus, equating indices:
\[2x + y = 6 \quad \text{--- (Equation 1)}\]

From the second equation:
\[\log_2(x + 3y) = 3\]
Convert the logarithmic equation to exponential form:
\[x + 3y = 2^3\]
\[x + 3y = 8 \quad \text{--- (Equation 2)}\]

Now, solve the simultaneous equations (1) and (2).
From Equation 1, express \(y\) in terms of \(x\):
\[y = 6 - 2x\]

Substitute \(y\) into Equation 2:
\[x + 3(6 - 2x) = 8\]
\[x + 18 - 6x = 8\]
\[-5x = -10\]
\[x = 2\]

Substitute \(x = 2\) back into the expression for \(y\):
\[y = 6 - 2(2) = 2\]

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

M1: For expressing \(4^x\) as \(2^{2x}\) or writing \(2^{2x+y} = 2^6\).
A1: For the correct first equation: \(2x + y = 6\).
M1: For rewriting the logarithmic equation as \(x + 3y = 2^3\).
A1: For the correct second equation: \(x + 3y = 8\).
M1: For attempting to solve the simultaneous equations (by substitution or elimination).
A1: For finding one variable correctly (e.g., \(x = 2\) or \(y = 2\)).
A1: For finding both variables correctly: \(x = 2\) and \(y = 2\).

The formula for the \(n\)-th term of an arithmetic progression is:
\[u_n = a + (n - 1)d\]
Given the 5th term is 19:
\[u_5 = a + 4d = 19 \quad \text{--- (Equation 1)}\]

The formula for the sum of the first \(n\) terms of an arithmetic progression is:
\[S_n = \frac{n}{2}\{2a + (n - 1)d\}\]
Given the sum of the first 10 terms is 220:
\[S_{10} = \frac{10}{2}\{2a + 9d\} = 220\]
\[5(2a + 9d) = 220\]
Divide by 5:
\[2a + 9d = 44 \quad \text{--- (Equation 2)}\]

Now, solve Equation 1 and Equation 2 simultaneously.
From Equation 1, multiply by 2:
\[2a + 8d = 38 \quad \text{--- (Equation 3)}\]

Subtract Equation 3 from Equation 2:
\[(2a + 9d) - (2a + 8d) = 44 - 38\]
\[d = 6\]

Substitute \(d = 6\) back into Equation 1:
\[a + 4(6) = 19\]
\[a + 24 = 19\]
\[a = -5\]

Thus, the values are \(a = -5\) and \(d = 6\).

M1: For using the formula for the \(n\)-th term of an AP to write \(a + 4d = 19\).
A1: For the correct first equation: \(a + 4d = 19\).
M1: For using the sum formula of an AP to write \(\frac{10}{2}(2a + 9d) = 220\) or equivalent.
A1: For the correct second equation: \(2a + 9d = 44\) (or equivalent).
M1: For a valid method to solve the simultaneous equations.
A1: For \(d = 6\).
A1: For \(a = -5\).

(a) The radius squared \(r^2\) of the circle is the distance between center \(C(3, -2)\) and point \(P(7, 1)\):
\[r^2 = (7 - 3)^2 + (1 - (-2))^2\]
\[r^2 = 4^2 + 3^2 = 16 + 9 = 25\]

The standard equation of the circle with center \((h, k) = (3, -2)\) and radius squared \(r^2 = 25\) is:
\[(x - 3)^2 + (y + 2)^2 = 25\]

Expanding this equation:
\[x^2 - 6x + 9 + y^2 + 4y + 4 = 25\]
\[x^2 + y^2 - 6x + 4y + 13 - 25 = 0\]
\[x^2 + y^2 - 6x + 4y - 12 = 0\]

(b) First, find the gradient of the radius \(CP\):
\[m_{CP} = \frac{1 - (-2)}{7 - 3} = \frac{3}{4}\]

Since the tangent is perpendicular to the radius, the gradient of the tangent \(m_t\) is:
\[m_t = -\frac{1}{m_{CP}} = -\frac{4}{3}\]

The equation of the tangent line passing through \(P(7, 1)\) is:
\[y - 1 = -\frac{4}{3}(x - 7)\]

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

Part (a):
M1: For finding the radius squared using the distance formula: \((7-3)^2 + (1 - (-2))^2\).
A1: For \(r^2 = 25\).
M1: For writing down the standard equation \((x - 3)^2 + (y + 2)^2 = 25\) and expanding.
A1: For the correct expanded equation: \(x^2 + y^2 - 6x + 4y - 12 = 0\).

Part (b):
M1: For finding the gradient of the radius \(CP\): \(m = \frac{3}{4}\).
M1: For using the perpendicular gradient relationship to find the tangent's gradient \(m = -\frac{4}{3}\) and setting up the line equation.
A1: For the correct line equation in the specified format: \(3y + 4x = 31\) (or any integer multiple thereof).

(a) First, find the vector \(\overrightarrow{PQ}\):
\[\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}\]
\[\overrightarrow{PQ} = (4\mathbf{i} + 3\mathbf{j}) - (\mathbf{i} + 6\mathbf{j}) = 3\mathbf{i} - 3\mathbf{j}\]

Next, calculate the vector \(\overrightarrow{PR}\):
\[\overrightarrow{PR} = \frac{2}{3}\overrightarrow{PQ} = \frac{2}{3}(3\mathbf{i} - 3\mathbf{j}) = 2\mathbf{i} - 2\mathbf{j}\]

Now, find the position vector of \(R\), \(\overrightarrow{OR}\):
\[\overrightarrow{OR} = \overrightarrow{OP} + \overrightarrow{PR}\]
\[\overrightarrow{OR} = (\mathbf{i} + 6\mathbf{j}) + (2\mathbf{i} - 2\mathbf{j}) = 3\mathbf{i} + 4\mathbf{j}\]

(b) To find the unit vector, calculate the magnitude of \(\overrightarrow{OR}\):
\[|\overrightarrow{OR}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]

The unit vector in the direction of \(\overrightarrow{OR}\) is:
\[\frac{\overrightarrow{OR}}{|\overrightarrow{OR}|} = \frac{3\mathbf{i} + 4\mathbf{j}}{5} = \frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j} = 0.6\mathbf{i} + 0.8\mathbf{j}\]

Part (a):
M1: For finding \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}\).
A1: For \(\overrightarrow{PQ} = 3\mathbf{i} - 3\mathbf{j}\).
M1: For calculating \(\overrightarrow{OR} = \overrightarrow{OP} + \frac{2}{3}\overrightarrow{PQ}\).
A1: For \(\overrightarrow{OR} = 3\mathbf{i} + 4\mathbf{j}\) (or equivalent column vector).

Part (b):
M1: For calculating the magnitude of \(\overrightarrow{OR}\): \(\sqrt{3^2 + 4^2}\).
A1: For magnitude = 5.
A1: For the correct unit vector: \(\frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j}\) (or \(0.6\mathbf{i} + 0.8\mathbf{j}\)).

(a) The perimeter \(P\) of a sector is given by:
\[P = 2r + r\theta\]
We are given \(P = 20\), so:
\[2r + r\theta = 20\]
\[r\theta = 20 - 2r \quad \text{--- (Equation 1)}\]

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

Substitute \(r\theta\) from Equation 1 into the area formula:
\[A = \frac{1}{2}r(r\theta) = \frac{1}{2}r(20 - 2r)\]
\[A = 10r - r^2 \quad \text{(as required)}\]

(b) To find the maximum area, differentiate \(A\) with respect to \(r\):
\[\frac{\mathrm{d}A}{\mathrm{d}r} = 10 - 2r\]

For a maximum or stationary point, set \(\frac{\mathrm{d}A}{\mathrm{d}r} = 0\):
\[10 - 2r = 0 \implies 2r = 10 \implies r = 5\text{ cm}\]

To verify this is a maximum, check the second derivative:
\[\frac{\mathrm{d}^2A}{\mathrm{d}r^2} = -2 < 0\]
Since the second derivative is negative, \(r = 5\) indeed yields a maximum area.

Now calculate the maximum area:
\[A_{\text{max}} = 10(5) - 5^2 = 50 - 25 = 25\text{ cm}^2\]

Part (a):
M1: For writing the correct perimeter formula: \(2r + r\theta = 20\).
A1: For expressing \(r\theta\) (or \(\theta\)) in terms of \(r\): \(r\theta = 20 - 2r\).
M1: For substituting their expression into the area formula \(A = \frac{1}{2}r^2\theta\).
A1: For successfully showing \(A = 10r - r^2\) with no errors.

Part (b):
M1: For differentiating \(A\) with respect to \(r\) and setting to 0: \(\frac{\mathrm{d}A}{\mathrm{d}r} = 10 - 2r = 0\).
A1: For finding \(r = 5\).
A1: For calculating the maximum area = 25.

(a) Using the product rule for differentiation with \( u = 2x - 3 \) and \( v = e^{2x} \):
\( \frac{du}{dx} = 2 \) and \( \frac{dv}{dx} = 2e^{2x} \).

\( \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \)
\( \frac{dy}{dx} = (2x - 3)(2e^{2x}) + e^{2x}(2) \)
\( \frac{dy}{dx} = (4x - 6)e^{2x} + 2e^{2x} = (4x - 4)e^{2x} \).

(b) At the stationary point, \( \frac{dy}{dx} = 0 \).
\( (4x - 4)e^{2x} = 0 \)
Since \( e^{2x} > 0 \) for all real \( x \):
\( 4x - 4 = 0 \implies x = 1 \).

Substitute \( x = 1 \) back into the curve's equation:
\( y = (2(1) - 3)e^{2(1)} = -e^2 \).

The exact coordinates of the stationary point are \( (1, -e^2) \).

(c) To determine the nature of the stationary point, differentiate \( \frac{dy}{dx} = (4x - 4)e^{2x} \) with respect to \( x \) to find the second derivative:
\( \frac{d^2y}{dx^2} = 4e^{2x} + (4x - 4)(2e^{2x}) = (8x - 4)e^{2x} \).

At \( x = 1 \):
\( \frac{d^2y}{dx^2} = (8(1) - 4)e^{2} = 4e^2 \).

Since \( 4e^2 > 0 \), the stationary point is a local minimum.

(a)
M1: Attempt to use the product rule to differentiate \( (2x - 3)e^{2x} \).
A1: One term differentiated correctly, e.g., \( 2e^{2x} \) or \( 2(2x - 3)e^{2x} \).
A1: Fully correct derivative: \( (4x - 4)e^{2x} \).

(b)
M1: Set their \( \frac{dy}{dx} = 0 \) and solve for \( x \).
A1: Correct coordinates \( (1, -e^2) \).

(c)
M1: Attempt to find the second derivative \( \frac{d^2y}{dx^2} \) (or use first derivative sign test).
A1: Conclude minimum with valid reasoning from a positive second derivative value \( 4e^2 \).

(a) Applying the change of base formula to \( \log_4(x - 1) \):
\( \log_4(x - 1) = \frac{\log_2(x - 1)}{\log_2 4} = \frac{1}{2} \log_2(x - 1) \).

Substitute this back into the equation:
\( \log_2(x + 3) - \frac{1}{2} \log_2(x - 1) = 2 \)

Multiply both sides by 2:
\( 2 \log_2(x + 3) - \log_2(x - 1) = 4 \)

Use logarithm laws to combine terms:
\( \log_2(x + 3)^2 - \log_2(x - 1) = 4 \)
\( \log_2\left( \frac{(x + 3)^2}{x - 1} \right) = 4 \)

Convert from logarithmic to exponential form:
\( \frac{(x + 3)^2}{x - 1} = 2^4 = 16 \)
\( x^2 + 6x + 9 = 16(x - 1) \)
\( x^2 + 6x + 9 = 16x - 16 \)
\( x^2 - 10x + 25 = 0 \).

(b) Factorise the quadratic equation:
\( (x - 5)^2 = 0 \implies x = 5 \).

Check validity: For \( x = 5 \), \( x + 3 = 8 > 0 \) and \( x - 1 = 4 > 0 \). Thus, the solution is valid.

(a)
M1: Apply change of base formula to get \( \log_4(x-1) = \frac{1}{2}\log_2(x-1) \).
M1: Apply logarithm power law, e.g., \( 2\log_2(x+3) = \log_2(x+3)^2 \).
M1: Combine logarithms to get \( \log_2\left(\frac{(x+3)^2}{x-1}\right) = 4 \).
A1: Eliminate logarithm and expand correctly to obtain \( x^2 - 10x + 25 = 0 \).

(b)
M1: Factorise or solve their quadratic equation.
A1: Correct answer \( x = 5 \) nfww.

(a) Using the trigonometric identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\( 2(1 - \sin^2 \theta) + 3\sin \theta = 3 \)
\( 2 - 2\sin^2 \theta + 3\sin \theta = 3 \)

Rearranging all terms to one side:
\( 2\sin^2 \theta - 3\sin \theta + 1 = 0 \).

(b) Factorising the quadratic equation in \( \sin \theta \):
\( (2\sin \theta - 1)(\sin \theta - 1) = 0 \)

This gives:
\( \sin \theta = \frac{1}{2} \) or \( \sin \theta = 1 \).

For \( 0 \le \theta \le 2\pi \):
- If \( \sin \theta = \frac{1}{2} \), then \( \theta = \frac{\pi}{6} \) or \( \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \).
- If \( \sin \theta = 1 \), then \( \theta = \frac{\pi}{2} \).

Therefore, the solutions are \( \theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \).

(a)
M1: Substitute \( \cos^2 \theta = 1 - \sin^2 \theta \) into the equation.
A1: Simplify correctly to show \( 2\sin^2 \theta - 3\sin \theta + 1 = 0 \).

(b)
M1: Solve the quadratic equation to obtain values for \( \sin\theta \).
A1: Identify both correct values \( \sin \theta = \frac{1}{2} \) and \( \sin \theta = 1 \).
A1: Find at least two correct angles in radians.
A1: Fully correct set of values: \( \theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \) and no extras.

(a) The sum of the first \( n \) terms of an arithmetic progression is given by:
\( S_n = \frac{n}{2}[2a + (n - 1)d] \).

For \( S_4 = 28 \):
\( \frac{4}{2}[2a + 3d] = 28 \implies 2(2a + 3d) = 28 \implies 2a + 3d = 14 \) (Equation 1)

For \( S_8 = 88 \):
\( \frac{8}{2}[2a + 7d] = 88 \implies 4(2a + 7d) = 88 \implies 2a + 7d = 22 \) (Equation 2)

Subtract Equation 1 from Equation 2:
\( (2a + 7d) - (2a + 3d) = 22 - 14 \)
\( 4d = 8 \implies d = 2 \).

Substitute \( d = 2 \) back into Equation 1:
\( 2a + 3(2) = 14 \implies 2a + 6 = 14 \implies 2a = 8 \implies a = 4 \).

Thus, \( a = 4 \) and \( d = 2 \).

(b) (i) Since the terms are in geometric progression:
\( \frac{y}{9} = \frac{4}{y} \implies y^2 = 36 \).
Since \( y > 0 \), \( y = 6 \).

(ii) The common ratio is \( r = \frac{y}{9} = \frac{6}{9} = \frac{2}{3} \).
Since \( |r| < 1 \), the sum to infinity is:
\( S_{\infty} = \frac{a}{1 - r} = \frac{9}{1 - 2/3} = \frac{9}{1/3} = 27 \).

(a)
M1: Use the sum formula \( S_n \) to set up at least one correct equation.
A1: Establish both equations correctly: \( 2a + 3d = 14 \) and \( 2a + 7d = 22 \) (or equivalent).
M1: Solve the simultaneous equations to find one variable.
A1: Correct values: \( a = 4 \) and \( d = 2 \).

(b)(i)
B1: State \( y = 6 \) (must discard the negative root).

(b)(ii)
M1: Calculate the common ratio \( r = \frac{2}{3} \) and substitute into the sum to infinity formula.
A1: Correct answer \( 27 \).

We start by finding the first derivative of the curve:
\(y = 2x^3 - 9x^2 - 24x + 5\)

\(\frac{dy}{dx} = 6x^2 - 18x - 24\)

To find the stationary points, set \(\frac{dy}{dx} = 0\):
\(6(x^2 - 3x - 4) = 0\)
\((x-4)(x+1) = 0\)

This yields the \(x\)-coordinates:
\(x = 4\) or \(x = -1\)

Substitute these values back into the original curve equation to find the corresponding \(y\)-coordinates:
For \(x = 4\):
\(y = 2(4)^3 - 9(4)^2 - 24(4) + 5 = 128 - 144 - 96 + 5 = -107\)
So, one stationary point is \((4, -107)\).

For \(x = -1\):
\(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 5 = -2 - 9 + 24 + 5 = 18\)
So, the other stationary point is \((-1, 18)\).

To determine the nature of these points, find the second derivative:
\(\frac{d^2y}{dx^2} = 12x - 18\)

Test each point:
At \(x = 4\):
\(\frac{d^2y}{dx^2} = 12(4) - 18 = 30 > 0\) \(\implies\) Local minimum.

At \(x = -1\):
\(\frac{d^2y}{dx^2} = 12(-1) - 18 = -30 < 0\) \(\implies\) Local maximum.

M1: For attempting to differentiate the curve.
A1: For correct first derivative \(\frac{dy}{dx} = 6x^2 - 18x - 24\).
M1: For setting their derivative to 0 and attempting to solve the resulting quadratic equation.
A1: For obtaining \(x = 4\) and \(x = -1\).
A1: For finding the correct y-coordinates, yielding \((4, -107)\) and \((-1, 18)\).
M1: For finding the second derivative \(\frac{d^2y}{dx^2} = 12x - 18\) (or using first derivative test).
A1: For correctly identifying the nature of both points.

First, we simplify the exponential equation by writing all terms with base 3:
\(3^{2x} \times (3^2)^y = (3^3)^{x+1}\)
\(3^{2x} \times 3^{2y} = 3^{3x + 3}\)
\(3^{2x + 2y} = 3^{3x + 3}\)

Equating the powers:
\(2x + 2y = 3x + 3\)
\(x = 2y - 3\) --- (Equation 1)

Next, simplify the logarithmic equation:
\(\log_2(y - x) + \log_2(x + 5) = 3\)
\(\log_2((y - x)(x + 5)) = 3\)

Convert this to exponential form:
\((y - x)(x + 5) = 2^3 = 8\) --- (Equation 2)

Substitute Equation 1 into Equation 2:
\((y - (2y - 3))((2y - 3) + 5) = 8\)
\((3 - y)(2y + 2) = 8\)
\(2(3 - y)(y + 1) = 8\)
\((3 - y)(y + 1) = 4\)
\(3y + 3 - y^2 - y = 4\)
\(y^2 - 2y + 1 = 0\)
\((y - 1)^2 = 0\)
\(y = 1\)

Now, find \(x\):
\(x = 2(1) - 3 = -1\)

Check validation within logarithm domains:
\(y - x = 1 - (-1) = 2 > 0\) (valid)
\(x + 5 = -1 + 5 = 4 > 0\) (valid)

Thus, the solution is \(x = -1\), \(y = 1\).

M1: For expressing the indices in terms of base 3.
A1: For obtaining the linear relation \(x = 2y - 3\) (or equivalent).
M1: For combining the log terms on one side using logarithmic laws.
A1: For converting log equation to algebraic form: \((y-x)(x+5) = 8\).
M1: For substituting their linear relation into their quadratic equation.
A1: For solving the quadratic equation to find \(y = 1\).
A1: For finding \(x = -1\) and confirming the values are valid.

We use the trigonometric identity \(\tan^2 x = \sec^2 x - 1\):
\(3(\sec^2 x - 1) - 5\sec x + 1 = 0\)
\(3\sec^2 x - 3 - 5\sec x + 1 = 0\)
\(3\sec^2 x - 5\sec x - 2 = 0\)

Let \(u = \sec x\). The equation becomes:
\(3u^2 - 5u - 2 = 0\)
\((3u + 1)(u - 2) = 0\)

This gives two cases:
1. \(u = -\frac{1}{3} \implies \sec x = -\frac{1}{3} \implies \cos x = -3\)
Since \(-1 \le \cos x \le 1\), this case has no real solutions.

2. \(u = 2 \implies \sec x = 2 \implies \cos x = \frac{1}{2}\)

For \(0 \le x \le 2\pi\):
\(x = \frac{\pi}{3}\) (Quadrant I)
\(x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\) (Quadrant IV)

Thus, the solutions are \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\).

M1: For using the identity \(\tan^2 x = \sec^2 x - 1\).
A1: For obtaining the quadratic equation \(3\sec^2 x - 5\sec x - 2 = 0\).
M1: For factorising or solving the quadratic equation in terms of \(\sec x\).
A1: For identifying that \(\sec x = 2\) is the only valid case (and noting that \(\sec x = -1/3\) has no solutions).
M1: For translating to \(\cos x = \frac{1}{2}\).
A1: For \(x = \frac{\pi}{3}\).
A1: For \(x = \frac{5\pi}{3}\).

(a) The general term in the binomial expansion of \(\left(x^2 - \frac{2}{x}\right)^6\) is:
\(T_{r+1} = \binom{6}{r} (x^2)^{6-r} \left(-\frac{2}{x}\right)^r\)
\(T_{r+1} = \binom{6}{r} (-2)^r x^{12-2r-r}\)
\(T_{r+1} = \binom{6}{r} (-2)^r x^{12-3r}\)

For the term independent of \(x\), we set the exponent of \(x\) to 0:
\(12 - 3r = 0 \implies r = 4\)

The term independent of \(x\) is:
\(T_5 = \binom{6}{4} (-2)^4 = 15 \times 16 = 240\).

(b) We expand the product:
\(\left(1 + x^3\right)\left(x^2 - \frac{2}{x}\right)^6 = 1\left(x^2 - \frac{2}{x}\right)^6 + x^3\left(x^2 - \frac{2}{x}\right)^6\)

To find the overall coefficient of \(x^3\), we need:
1. The coefficient of the term containing \(x^3\) from the first part of the expression:
Setting \(12 - 3r = 3 \implies 3r = 9 \implies r = 3\).
This coefficient is \(\binom{6}{3} (-2)^3 = 20 \times (-8) = -160\).

2. The term containing \(x^0\) (constant term) from the second part, which becomes multiplied by \(x^3\):
This corresponds to the constant term calculated in part (a), which is \(240\).

Summing these contributions:
\(\text{Total coefficient of } x^3 = -160 + 240 = 80\).

(a)
M1: For stating the general term of the expansion.
A1: For simplifying the index of \(x\) to \(12 - 3r\).
M1: For setting their index equal to 0 to solve for \(r\).
A1: For the correct independent term of \(240\).

(b)
M1: For identifying that the coefficient of \(x^3\) consists of two parts.
A1: For correctly finding the coefficient of \(x^3\) in \((x^2 - 2/x)^6\) as \(-160\).
A1: For combining \(-160\) and \(240\) to get the correct final answer of \(80\).

(a) Complete the square for the circle equation:
\((x^2 - 6x) + (y^2 + 4y) - 12 = 0\)
\((x-3)^2 - 9 + (y+2)^2 - 4 - 12 = 0\)
\((x-3)^2 + (y+2)^2 = 25\)

Thus, the centre is \((3, -2)\) and the radius is \(\sqrt{25} = 5\).

(b) The point \(P(6, 2)\) is on the circle. First find the gradient of the normal (radius) from \((3, -2)\) to \((6, 2)\):
\(m_r = \frac{2 - (-2)}{6 - 3} = \frac{4}{3}\)

Since the tangent is perpendicular to the radius, the gradient of the tangent, \(m_t\), is:
\(m_t = -\frac{3}{4}\)

Using the point-slope form at \(P(6, 2)\):
\(y - 2 = -\frac{3}{4}(x - 6)\)
\(4(y - 2) = -3(x - 6)\)
\(3x + 4y - 26 = 0\)

(c) Find coordinates of \(A\) (intercept on the \(x\)-axis, where \(y = 0\)) and \(B\) (intercept on the \(y\)-axis, where \(x = 0\)):
For \(A\):
\(3x - 26 = 0 \implies x = \frac{26}{3}\). So, \(A\) is \(\left(\frac{26}{3}, 0\right)\).

For \(B\):
\(4y - 26 = 0 \implies y = \frac{13}{2}\). So, \(B\) is \(\left(0, \frac{13}{2}\right)\).

Area of the right-angled triangle \(OAB\):
\(\text{Area} = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times \frac{26}{3} \times \frac{13}{2} = \frac{169}{6}\) (or approximately \(28.2\)).

(a)
M1: For attempting to complete the square for \(x\) and \(y\).
A1: For correctly identifying the centre \((3, -2)\) and radius \(5\).

(b)
M1: For finding the gradient of the radius to the point \(P\).
M1: For applying \(m_1 m_2 = -1\) and writing the linear equation using \(P(6,2)\).
A1: For \(3x + 4y - 26 = 0\) (or any equivalent form).

(c)
M1: For finding both axis intercepts, \(A\) and \(B\).
A1: For the correct exact area \(\frac{169}{6}\) (or \(28.2\)).

(a) The given linear equation is:
\(\ln y = x\ln b + \ln a\)

This is in the linear form \(Y = mX + c\), where:
- The vertical coordinate is \(Y = \ln y\)
- The horizontal coordinate is \(X = x\)
- The gradient is \(m = \ln b\)
- The vertical intercept is \(c = \ln a\)

First, find the gradient \(m\) using the points \((2, 3.5)\) and \((5, 8.0)\):
\(m = \frac{8.0 - 3.5}{5 - 2} = \frac{4.5}{3} = 1.5\)

Therefore:
\(\ln b = 1.5\)

Next, use the point \((2, 3.5)\) to find \(\ln a\):
\(\ln y - 3.5 = 1.5(x - 2)\)
\(\ln y = 1.5x - 3 + 3.5\)
\(\ln y = 1.5x + 0.5\)

This gives the intercept:
\(\ln a = 0.5\)

(b) Solve for \(a\) and \(b\):
\(\ln a = 0.5 \implies a = e^{0.5} \approx 1.65\)
\(\ln b = 1.5 \implies b = e^{1.5} \approx 4.48\)

(a)
M1: For showing that the gradient represents \(\ln b\) and the intercept represents \(\ln a\).
M1: For calculating the gradient \(\frac{8.0 - 3.5}{5 - 2}\).
A1: For \(\ln b = 1.5\).
A1: For \(\ln a = 0.5\).

(b)
M1: For correct method to solve for \(a\) and \(b\) using the exponential function.
A1: For \(a = 1.65\).
A1: For \(b = 4.48\).

(a) The natural logarithm function \(f(x) = \ln(x - 3)\) outputs all real numbers. Thus, the range of \(f\) is \(\mathbb{R}\) (all real numbers).

(b) To find the inverse function, let \(y = \ln(x - 3)\):
\(e^y = x - 3\)
\(x = e^y + 3\)

Switching variables, we get:
\(f^{-1}(x) = e^x + 3\)

(c) The composite function \(gf(x)\) is:
\(gf(x) = g(f(x)) = e^{2\ln(x-3)} + 3\)

Using log properties:
\(e^{2\ln(x-3)} = e^{\ln(x-3)^2} = (x - 3)^2\)

So:
\(gf(x) = (x - 3)^2 + 3\)

The domain of \(gf\) is the same as the domain of the inner function \(f\), which is \(x > 3\).

(a)
B1: For stating range is \(\mathbb{R}\) (or all real values).

(b)
M1: For changing the subject of the equation \(y = \ln(x-3)\) to \(x\).
A1: For \(f^{-1}(x) = e^x + 3\).

(c)
M1: For writing down the composite expression \(e^{2\ln(x-3)} + 3\).
M1: For using log properties to simplify \(2\ln(x-3)\) to \\ln(x-3)^2\).
A1: For the final simplified expression \((x-3)^2 + 3\) (or \(x^2 - 6x + 12\)).
B1: For stating the domain as \(x > 3\).

(a) The perimeter \(P\) of the sector consists of two radii and the arc length:
\(P = 2r + r\theta = 32\) --- (Equation 1)

The area \(A\) of the sector is given by:
\(A = \frac{1}{2}r^2\theta = 48 \implies r^2\theta = 96\) --- (Equation 2)

From Equation 1, express \(\theta\) in terms of \(r\):
\(r\theta = 32 - 2r \implies \theta = \frac{32}{r} - 2\)

Substitute this expression for \(\theta\) into Equation 2:
\(r^2\left(\frac{32}{r} - 2\right) = 96\)
\(32r - 2r^2 = 96\)

Rearranging to standard quadratic form:
\(2r^2 - 32r + 96 = 0\)
Dividing through by 2:
\(r^2 - 16r + 48 = 0\) (as required).

(b) Solve the quadratic equation by factorising:
\((r - 12)(r - 4) = 0\)

This gives \(r = 12\) or \(r = 4\).

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

For \(r = 4\):
\(\theta = \frac{32}{4} - 2 = 8 - 2 = 6\) radians.

(a)
B1: For the perimeter equation \(2r + r\theta = 32\).
B1: For the area equation \(\frac{1}{2}r^2\theta = 48\).
M1: For substituting one equation into the other to eliminate \(\theta\) (or \(r\)).
A1: For correctly showing \(r^2 - 16r + 48 = 0\).

(b)
M1: For solving the quadratic equation to find both values of \(r\).
A1: For the pair \(r = 12, \theta = \frac{2}{3}\).
A1: For the pair \(r = 4, \theta = 6\).

(a) Let \(M\) be the midpoint of the chord \(AB\).
\(AM = \frac{1}{2} AB = 3\sqrt{3}\text{ cm}\).
In the right-angled triangle \(AOM\):
\(\sin(\angle AOM) = \frac{AM}{OA} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}\).
Therefore, \(\angle AOM = \frac{\pi}{3}\) radians.
Thus, \(\angle AOB = 2 \times \angle AOM = \frac{2\pi}{3}\) radians.

(b) First, find the radius \(r\) of the second circle with centre \(C\).
Since \(\angle ACB = \frac{\pi}{3}\) radians, in the right-angled triangle \(ACM\):
\(\angle ACM = \frac{\pi}{6}\) radians.
\(\sin(\angle ACM) = \frac{AM}{AC} \implies \sin\left(\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{r}\)
\(\frac{1}{2} = \frac{3\sqrt{3}}{r} \implies r = 6\sqrt{3}\text{ cm}\).

Now, calculate the area of the segment of the first circle:
\(Area_1 = \frac{1}{2} R^2 (\theta - \sin \theta) = \frac{1}{2} (6^2) \left(\frac{2\pi}{3} - \sin\left(\frac{2\pi}{3}\right)\right)\)
\(Area_1 = 18 \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = 12\pi - 9\sqrt{3}\).

Calculate the area of the segment of the second circle:
\(Area_2 = \frac{1}{2} r^2 (\phi - \sin \phi) = \frac{1}{2} (6\sqrt{3})^2 \left(\frac{\pi}{3} - \sin\left(\frac{\pi}{3}\right)\right)\)
\(Area_2 = 54 \left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) = 18\pi - 27\sqrt{3}\).

Since both centres lie on the same side of the chord, the enclosed area between the two arcs is the difference between the two segments:
\(Area_{enclosed} = Area_1 - Area_2 = (12\pi - 9\sqrt{3}) - (18\pi - 27\sqrt{3}) = 18\sqrt{3} - 6\pi\text{ cm}^2\).

(a)
- M1: For attempting to find the angle using trigonometry, e.g., \(\sin(\angle AOM) = \frac{3\sqrt{3}}{6}\) or using the cosine rule on triangle \(AOB\).
- A1: Obtaining \(\frac{2\pi}{3}\) exactly.

(b)
- B1: Finding the radius of the second circle, \(r = 6\sqrt{3}\text{ cm}\), using valid trigonometry.
- M1: Attempting to calculate the area of the segment of the first circle.
- A1: Obtaining \(12\pi - 9\sqrt{3}\).
- M1: Attempting to calculate the area of the segment of the second circle.
- A1: Obtaining \(18\pi - 27\sqrt{3}\).
- A1: Finding the difference \(Area_1 - Area_2\) to obtain the final exact area \(18\sqrt{3} - 6\pi\).

Using the change of base rule:
\(\log_4(x - 1) = \frac{\log_2(x - 1)}{\log_2(4)} = \frac{1}{2}\log_2(x - 1) = \log_2\sqrt{x - 1}\).

Substitute this back into the equation:
\(\log_2(x + 2) - \log_2\sqrt{x - 1} = 2\).

Apply the division rule for logarithms:
\(\log_2\left(\frac{x + 2}{\sqrt{x - 1}}\right) = 2\).

Convert from logarithmic to exponential form:
\(\frac{x + 2}{\sqrt{x - 1}} = 2^2 = 4\).

Multiply by \(\sqrt{x - 1}\):
\(x + 2 = 4\sqrt{x - 1}\).

Square both sides:
\((x + 2)^2 = 16(x - 1)\)
\(x^2 + 4x + 4 = 16x - 16\)
\(x^2 - 12x + 20 = 0\).

Factorise the quadratic equation:
\((x - 2)(x - 10) = 0\).

This gives \(x = 2\) or \(x = 10\).

Check for validity in the original equation:
For \(x = 2\), \(x+2 = 4 > 0\) and \(x-1 = 1 > 0\) (valid).
For \(x = 10\), \(x+2 = 12 > 0\) and \(x-1 = 9 > 0\) (valid).

- B1: Expressing \(\log_4(x - 1)\) as \(\frac{1}{2}\log_2(x - 1)\) or \(\log_2\sqrt{x - 1}\).
- M1: Applying subtraction log laws correctly to obtain a single logarithmic expression: \(\log_2\left(\frac{x + 2}{\sqrt{x - 1}}\right) = 2\).
- M1: Removing logarithms to form \(\frac{x + 2}{\sqrt{x - 1}} = 4\).
- M1: Squaring both sides to eliminate the square root.
- A1: Forming the correct quadratic equation \(x^2 - 12x + 20 = 0\).
- M1: Attempting to solve their 3-term quadratic equation.
- A1: Finding both correct solutions \(x = 2\) and \(x = 10\), confirming both are valid.

Using the trigonometric identity \(\tan^2(2x) = \sec^2(2x) - 1\), substitute this into the equation:
\(2(\sec^2(2x) - 1) + 3\sec(2x) = 0\)
\(2\sec^2(2x) + 3\sec(2x) - 2 = 0\).

Let \(u = \sec(2x)\). The equation becomes a quadratic in \(u\):
\(2u^2 + 3u - 2 = 0\)
\((2u - 1)(u + 2) = 0\).

This gives two cases:
1) \(\sec(2x) = \frac{1}{2} \implies \cos(2x) = 2\), which has no real solutions since the range of cosine is \([-1, 1]\).
2) \(\sec(2x) = -2 \implies \cos(2x) = -\frac{1}{2}\).

Since \(0 \le x \le \pi\), the range for the angle \(2x\) is \(0 \le 2x \le 2\pi\).
Solving \(\cos(2x) = -\frac{1}{2}\) within this interval:
\(2x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\)
\(2x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\).

Divide by 2 to find \(x\):
\(x = \frac{\pi}{3}\)
\(x = \frac{2\pi}{3}\).

- M1: Substituting \(\tan^2(2x) = \sec^2(2x) - 1\) into the given equation.
- A1: Forming the correct quadratic equation \(2\sec^2(2x) + 3\sec(2x) - 2 = 0\).
- M1: Attempting to factorise or solve their quadratic in \(\sec(2x)\).
- B1: Stating or implying that \(\sec(2x) = \frac{1}{2}\) has no real solutions.
- M1: Setting up \(\cos(2x) = -\frac{1}{2}\) and identifying the correct quadrant angles for \(2x\).
- A1: Finding the values for \(2x\) as \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\).
- A1: Obtaining the final correct values \(x = \frac{\pi}{3}\) and \(x = \frac{2\pi}{3}\) only.