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

Combine the logarithms using the product rule:
\(\log_2((x+3)(x-3)) = 4\)

Convert the logarithmic equation to its exponential form:
\((x+3)(x-3) = 2^4\)
\(x^2 - 9 = 16\)
\(x^2 = 25\)

This gives \(x = 5\) or \(x = -5\).

Since the argument of a logarithm must be strictly positive:
For \(\log_2(x-3)\), we require \(x - 3 > 0 \implies x > 3\).

Therefore, the only valid solution is \(x = 5\).

M1: for applying the product rule of logarithms to obtain \(\log_2((x+3)(x-3)) = 4\).
M1: for converting to exponential form to obtain \(x^2 - 9 = 16\) or equivalent quadratic.
A1: for solving and stating \(x = 5\) as the only valid solution (rejecting \(x = -5\)).

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

To find the coefficient of \(x^2\), we set the exponent of \(x\) to 2:
\(6 - 2r = 2 \implies 2r = 4 \implies r = 2\)

Substitute \(r = 2\) into the general term:
\(\binom{6}{2} 2^{4} (-1)^2 = 15 \times 16 \times 1 = 240\)

Thus, the coefficient of \(x^2\) is 240.

M1: for expressing the general term or setting up the equation \(6 - 2r = 2\) to find \(r = 2\).
M1: for evaluating \(\binom{6}{2} = 15\) and \(2^4 = 16\).
A1: for the correct coefficient of 240.

The committee can have 2, 3, or 4 women.

Case 1: 2 women and 2 men
Number of ways = \(\binom{4}{2} \times \binom{5}{2} = 6 \times 10 = 60\)

Case 2: 3 women and 1 man
Number of ways = \(\binom{4}{3} \times \binom{5}{1} = 4 \times 5 = 20\)

Case 3: 4 women and 0 men
Number of ways = \(\binom{4}{4} \times \binom{5}{0} = 1 \times 1 = 1\)

Total number of different committees = \(60 + 20 + 1 = 81\).

M1: for calculating the combinations of at least one valid case (e.g. \(\binom{4}{2} \times \binom{5}{2} = 60\)).
M1: for a complete method of identifying and summing all three cases (or using subtraction from total: \(\binom{9}{4} - \binom{5}{4} - \binom{4}{1}\binom{5}{3}\)).
A1: for the correct answer 81.

First, find the sum of the two vectors:
\(\mathbf{a} + \mathbf{b} = (3\mathbf{i} - 4\mathbf{j}) + (k\mathbf{i} + 12\mathbf{j}) = (3 + k)\mathbf{i} + 8\mathbf{j}\)

For \(\mathbf{a} + \mathbf{b}\) to be parallel to \(\mathbf{i} + \mathbf{j}\), the ratio of the \(\mathbf{i}\) component to the \(\mathbf{j}\) component must be equal:
\(\frac{3+k}{8} = \frac{1}{1}\)
\(3 + k = 8 \implies k = 5\)

M1: for finding the vector sum in terms of \(k\): \((3+k)\mathbf{i} + 8\mathbf{j}\).
A1: for setting up the proportion \(3+k = 8\) and correctly solving to get \(k = 5\).

Let \(y = \mathrm{f}(x)\):
\(y = \frac{2x+3}{x-1}\)

Multiply both sides by \(x-1\):
\(y(x-1) = 2x + 3\)
\(xy - y = 2x + 3\)

Rearrange to group all terms containing \(x\) on one side:
\(xy - 2x = y + 3\)
\(x(y-2) = y + 3\)

Solve for \(x\):
\(x = \frac{y+3}{y-2}\)

Thus, \(\mathrm{f}^{-1}(x) = \frac{x+3}{x-2}\) for \(x \neq 2\).

M1: for writing \(y = \frac{2x+3}{x-1}\) and expanding to \(xy - y = 2x + 3\).
M1: for collecting terms in \(x\) and factorising: \(x(y-2) = y + 3\).
A1: for obtaining the correct inverse function \(\mathrm{f}^{-1}(x) = \frac{x+3}{x-2}\) (or equivalent).

First, write the curve's equation as \(y = x + 9x^{-1}\).

Differentiate with respect to \(x\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - 9x^{-2} = 1 - \frac{9}{x^2}\)

For a stationary point, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(1 - \frac{9}{x^2} = 0 \implies x^2 = 9\)

Since \(x > 0\), we have \(x = 3\).

Substitute \(x = 3\) back into the original curve equation:
\(y = 3 + \frac{9}{3} = 6\)

Therefore, the stationary point is \((3, 6)\).

M1: for correctly differentiating the curve to find \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{9}{x^2}\).
M1: for setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solving to find \(x = 3\) (condone inclusion of \(x = -3\) at this stage).
A1: for the correct final coordinates \((3, 6)\) only.

Let \( y = 3^x \).
Then the equation \( 3(3^{2x}) - 28(3^x) + 9 = 0 \) becomes:
\( 3y^2 - 28y + 9 = 0 \)

Factorizing the quadratic equation:
\( (3y - 1)(y - 9) = 0 \)

This gives:
\( y = \frac{1}{3} \) or \( y = 9 \)

Since \( y = 3^x \):
For \( y = \frac{1}{3} \):
\( 3^x = 3^{-1} \implies x = -1 \)

For \( y = 9 \):
\( 3^x = 3^2 \implies x = 2 \)

M1: For substituting \( y = 3^x \) to obtain a correct 3-term quadratic equation in \( y \), i.e., \( 3y^2 - 28y + 9 = 0 \).
M1: For factorizing or solving their 3-term quadratic to find two values of \( y \) (\( y = \frac{1}{3} \) and \( y = 9 \)).
A1: For obtaining either \( x = -1 \) or \( x = 2 \).
A1: For obtaining both correct solutions \( x = -1 \) and \( x = 2 \).

To find the stationary point, we first find the derivative \( \frac{dy}{dx} \) using the quotient rule:
\( \frac{dy}{dx} = \frac{x^2 \frac{d}{dx}(\ln x) - \ln x \frac{d}{dx}(x^2)}{(x^2)^2} \)
\( \frac{dy}{dx} = \frac{x^2 \left(\frac{1}{x}\right) - \ln x (2x)}{x^4} \)
\( \frac{dy}{dx} = \frac{x - 2x\ln x}{x^4} = \frac{1 - 2\ln x}{x^3} \)

For a stationary point, set \( \frac{dy}{dx} = 0 \):
\( 1 - 2\ln x = 0 \implies \ln x = \frac{1}{2} \implies x = e^{1/2} = \sqrt{e} \)

Substitute \( x = e^{1/2} \) back into the equation of the curve to find the \( y \)-coordinate:
\( y = \frac{\ln(e^{1/2})}{(e^{1/2})^2} = \frac{\frac{1}{2}}{e} = \frac{1}{2e} \)

Thus, the coordinates of the stationary point are \( \left(\sqrt{e}, \frac{1}{2e}\right) \).

M1: For a correct attempt to differentiate \( y = \frac{\ln x}{x^2} \) using the quotient rule or product rule.
A1: For a correct unsimplified or simplified derivative, e.g., \( \frac{dy}{dx} = \frac{1 - 2\ln x}{x^3} \).
M1: For setting their \( \frac{dy}{dx} = 0 \) and solving to find a value for \( x \) in terms of \( e \).
A1: For the correct exact \( x \)-coordinate: \( x = e^{1/2} \) (or \( \sqrt{e} \)).
A1: For the correct exact \( y \)-coordinate: \( y = \frac{1}{2e} \).

To find where the line and curve do not intersect, we set their equations equal to each other:
\( x^2 - 3x - 1 = kx - 5 \)

Rearranging into a standard quadratic form:
\( x^2 - (3 + k)x + 4 = 0 \)

For no intersection, the discriminant must be less than zero (\( b^2 - 4ac < 0 \)):
\( [-(3 + k)]^2 - 4(1)(4) < 0 \)
\( (k + 3)^2 - 16 < 0 \)
\( k^2 + 6k + 9 - 16 < 0 \)
\( k^2 + 6k - 7 < 0 \)

Factorizing the inequality:
\( (k + 7)(k - 1) < 0 \)

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

M1: For equating the line and the curve and rearranging into a quadratic equation in \( x \): \( x^2 - (k + 3)x + 4 = 0 \) (or equivalent).
M1: For using the discriminant \( b^2 - 4ac < 0 \) on their quadratic equation to set up an inequality in terms of \( k \).
A1: For identifying the critical values \( k = -7 \) and \( k = 1 \).
A1: For writing the correct range as a single inequality: \( -7 < k < 1 \).

By the Factor Theorem, since \( \mathrm{p}(x) \) is exactly divisible by \( x - 2 \):
\( \mathrm{p}(2) = 0 \)
\( 2(2)^3 - (2)^2 + a(2) + b = 0 \)
\( 16 - 4 + 2a + b = 0 \implies 2a + b = -12 \) --- (Equation 1)

By the Remainder Theorem, since dividing \( \mathrm{p}(x) \) by \( x + 1 \) leaves a remainder of \( 30 \):
\( \mathrm{p}(-1) = 30 \)
\( 2(-1)^3 - (-1)^2 + a(-1) + b = 30 \)
\( -2 - 1 - a + b = 30 \implies -a + b = 33 \) --- (Equation 2)

From Equation 2, we have \( b = a + 33 \).
Substitute this into Equation 1:
\( 2a + (a + 33) = -12 \)
\( 3a + 33 = -12 \)
\( 3a = -45 \implies a = -15 \)

Substitute \( a = -15 \) back into the expression for \( b \):
\( b = -15 + 33 = 18 \)

Therefore, the values are \( a = -15 \) and \( b = 18 \).

M1: For applying the Factor Theorem \( \mathrm{p}(2) = 0 \) to obtain a correct linear equation in \( a \) and \( b \) (e.g., \( 2a + b = -12 \)).
M1: For applying the Remainder Theorem \( \mathrm{p}(-1) = 30 \) to obtain a correct linear equation in \( a \) and \( b \) (e.g., \( -a + b = 33 \)).
M1: For a valid attempt to solve their two simultaneous equations for \( a \) and \( b \).
A1: For \( a = -15 \).
A1: For \( b = 18 \).

Let the first term of the progression be \( a \) and the common ratio be \( r \).
The third term is:
\( u_3 = ar^2 = 12 \) --- (Equation 1)

The sixth term is:
\( u_6 = ar^5 = \frac{32}{9} \) --- (Equation 2)

Dividing Equation 2 by Equation 1:
\( \frac{ar^5}{ar^2} = \frac{\frac{32}{9}}{12} \)
\( r^3 = \frac{32}{108} = \frac{8}{27} \)
Taking the cube root of both sides:
\( r = \frac{2}{3} \)

Substitute \( r = \frac{2}{3} \) back into Equation 1 to find \( a \):
\( a \left(\frac{2}{3}\right)^2 = 12 \)
\( a \left(\frac{4}{9}\right) = 12 \)
\( a = 12 \times \frac{9}{4} = 27 \)

Since \( |r| < 1 \), the sum to infinity \( S_{\infty} \) is defined as:
\( S_{\infty} = \frac{a}{1 - r} = \frac{27}{1 - \frac{2}{3}} = \frac{27}{\frac{1}{3}} = 81 \)

M1: For writing down the two equations in terms of \( a \) and \( r \): \( ar^2 = 12 \) and \( ar^5 = \frac{32}{9} \) (or equivalent).
M1: For dividing the equations and solving to obtain \( r = \frac{2}{3} \).
A1: For finding the first term \( a = 27 \).
A1: For substituting their \( a \) and \( r \) into the sum to infinity formula to obtain \( 81 \).

(a)
The centre of \( C_1 \) is \( (3, 5) \).
Gradient of the radius from \( (3, 5) \) to \( (6, 9) \):
\( m_{\text{radius}} = \frac{9 - 5}{6 - 3} = \frac{4}{3} \).

Since the tangent is perpendicular to the radius:
\( m_{\text{tangent}} = -\frac{3}{4} \).

Using the point-slope form with \( (6, 9) \):
\( y - 9 = -\frac{3}{4}(x - 6) \)
\( 4y - 36 = -3x + 18 \)
\( 3x + 4y = 54 \).

(b)
The second circle, \( C_2 \), has equation:
\( x^2 + y^2 - 2x + 4y + 4 = 0 \implies (x - 1)^2 + (y + 2)^2 - 1 - 4 + 4 = 0 \implies (x - 1)^2 + (y + 2)^2 = 1 \).

The centre of \( C_2 \) is \( (1, -2) \) and its radius is \( r_2 = 1 \).
The centre of \( C_1 \) is \( (3, 5) \) and its radius is \( r_1 = 5 \).

The distance \( d \) between the centres is:
\( d = \sqrt{(3 - 1)^2 + (5 - (-2))^2} = \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53} \).

Since \( \sqrt{53} > \sqrt{49} = 7 \), we have \( d > 7 \).
The sum of the radii is \( r_1 + r_2 = 5 + 1 = 6 \).
Since \( d > r_1 + r_2 \) (as \( \sqrt{53} > 6 \)), the two circles do not intersect.

(a) [4 marks]
- **B1**: Identifies centre of circle as \( (3, 5) \)
- **M1**: Calculates gradient of radius as \( \frac{4}{3} \) and uses perpendicular gradient rule to get \( -\frac{3}{4} \)
- **M1**: Sets up equation of tangent line using their perpendicular gradient and \( (6, 9) \)
- **A1**: Correctly simplifes to \( 3x + 4y = 54 \) (or equivalent integer form)

(b) [2 marks]
- **M1**: Completes the square to find the centre \( (1, -2) \) and radius \( r = 1 \) of the second circle
- **A1**: Calculates the distance between centres as \( \sqrt{53} \) and explains why this is greater than the sum of the radii (\( 6 \)), concluding they do not intersect

Using the change of base formula, we rewrite \( \frac{3}{\log_x 2} \) as \( 3\log_2 x \).

The equation becomes:
\( 3\log_2 x - \log_2(x + 4) = 3 \)

Using log properties:
\( \log_2(x^3) - \log_2(x + 4) = 3 \)
\( \log_2\left(\frac{x^3}{x + 4}\right) = 3 \)

Converting to exponential form:
\( \frac{x^3}{x + 4} = 2^3 \)
\( \frac{x^3}{x + 4} = 8 \)
\( x^3 = 8(x + 4) \)
\( x^3 - 8x - 32 = 0 \)

By testing small integer values, we find that \( x = 4 \) is a root:
\( 4^3 - 8(4) - 32 = 64 - 32 - 32 = 0 \).

Factoring \( x^3 - 8x - 32 \):
\( (x - 4)(x^2 + 4x + 8) = 0 \).

For the quadratic factor \( x^2 + 4x + 8 \), the discriminant is:
\( \Delta = 4^2 - 4(1)(8) = 16 - 32 = -16 < 0 \).

Thus, there are no other real roots. Since the domain specifies \( x > 1 \), the only solution is:
\( x = 4 \).

- **B1**: Rewrites \( \frac{3}{\log_x 2} \) as \( 3\log_2 x \)
- **M1**: Uses subtraction/power rules of logarithms to combine terms into a single log: \( \log_2\left(\frac{x^3}{x+4}\right) \)
- **M1**: Converts from logarithmic form to exponential form correctly: \( \frac{x^3}{x+4} = 8 \)
- **A1**: Obtains correct cubic equation \( x^3 - 8x - 32 = 0 \) (or equivalent)
- **M1**: Uses factor theorem/algebraic method to find the integer root \( x = 4 \)
- **A1**: Explicitly identifies \( x = 4 \) as the only real solution, demonstrating the quadratic has no real roots

(a)
Since \( 2x - 1 \) is a factor, \( p\left(\frac{1}{2}\right) = 0 \):
\( 2\left(\frac{1}{8}\right) + a\left(\frac{1}{4}\right) - 13\left(\frac{1}{2}\right) + b = 0 \)
\( \frac{1}{4} + \frac{a}{4} - \frac{26}{4} + b = 0 \implies a + 4b = 25 \) --- (Equation 1)

Since \( x + 3 \) is a factor, \( p(-3) = 0 \):
\( 2(-27) + a(9) - 13(-3) + b = 0 \)
\( -54 + 9a + 39 + b = 0 \implies 9a + b = 15 \) --- (Equation 2)

Multiply Equation 2 by 4:
\( 36a + 4b = 60 \)

Subtract Equation 1 from this:
\( 35a = 35 \implies a = 1 \).

Substitute \( a = 1 \) back into Equation 2:
\( 9(1) + b = 15 \implies b = 6 \).

So \( a = 1 \) and \( b = 6 \).

(b)
With \( a = 1, b = 6 \), \( p(x) = 2x^3 + x^2 - 13x + 6 \).
The product of the known factors is:
\( (2x - 1)(x + 3) = 2x^2 + 5x - 3 \).

We divide \( 2x^3 + x^2 - 13x + 6 \) by \( 2x^2 + 5x - 3 \) using polynomial long division:
\( \frac{2x^3 + x^2 - 13x + 6}{2x^2 + 5x - 3} = x - 2 \).

Therefore, the linear factors are:
\( p(x) = (2x - 1)(x + 3)(x - 2) \).

(a) [4 marks]
- **M1**: Applies factor theorem with \( x = 1/2 \) and sets equal to 0
- **M1**: Applies factor theorem with \( x = -3 \) and sets equal to 0
- **A1**: Obtains two correct linear equations in \( a \) and \( b \) (e.g., \( a + 4b = 25 \) and \( 9a + b = 15 \))
- **A1**: Solves the system correctly to get \( a = 1 \) and \( b = 6 \)

(b) [2 marks]
- **M1**: Divides the polynomial by the quadratic factor \( (2x - 1)(x + 3) \) or uses synthetic division/coefficient matching
- **A1**: Expresses correctly as \( (2x - 1)(x + 3)(x - 2) \)

(a)
We differentiate \( y = 16(x+2)^{-1} + x \):
\( \frac{dy}{dx} = -16(x+2)^{-2} + 1 = 1 - \frac{16}{(x+2)^2} \).

For stationary points, set \( \frac{dy}{dx} = 0 \):
\( 1 - \frac{16}{(x+2)^2} = 0 \implies (x+2)^2 = 16 \).

Since \( x > 0 \), we take the positive square root:
\( x + 2 = 4 \implies x = 2 \).

When \( x = 2 \):
\( y = \frac{16}{2+2} + 2 = 4 + 2 = 6 \).

So the coordinates of the stationary point are \( (2, 6) \).

(b)
The area \( A \) under the curve from \( x = 0 \) to \( x = 2 \) is given by:
\( A = \int_0^2 \left( \frac{16}{x+2} + x \right) dx \)

Integrating term by term:
\( A = \left[ 16\ln(x+2) + \frac{x^2}{2} \right]_0^2 \)

Substitute limits:
\( A = \left( 16\ln(2+2) + \frac{2^2}{2} \right) - \left( 16\ln(0+2) + \frac{0^2}{2} \right) \)
\( A = (16\ln 4 + 2) - 16\ln 2 \)
\( A = 16\ln(2^2) + 2 - 16\ln 2 \)
\( A = 32\ln 2 + 2 - 16\ln 2 \)
\( A = 16\ln 2 + 2 \).

(a) [3 marks]
- **M1**: Differentiates term by term to find \( \frac{dy}{dx} = -16(x+2)^{-2} + 1 \)
- **M1**: Equates derivative to 0 and solves for \( x \)
- **A1**: Obtains stationary point \( (2, 6) \) (must have both coordinates)

(b) [3 marks]
- **M1**: Sets up the correct definite integral \( \int_0^2 \left( \frac{16}{x+2} + x \right) dx \)
- **A1**: Integrates correctly to \( \left[ 16\ln(x+2) + \frac{x^2}{2} \right] \)
- **A1**: Substitutes limits correctly and simplifies to the exact form \( 16\ln 2 + 2 \) (or equivalent single logarithm representation)

Let \( y = 2x + \frac{\pi}{6} \).
Since \( 0 \le x \le \pi \), the interval for \( y \) is:
\( 0 \le 2x \le 2\pi \implies \frac{\pi}{6} \le y \le 2\pi + \frac{\pi}{6} = \frac{13\pi}{6} \).

Using the identity \( \sin^2 y = 1 - \cos^2 y \):
\( 4(1 - \cos^2 y) - 8\cos y = 1 \)
\( 4 - 4\cos^2 y - 8\cos y = 1 \)
\( 4\cos^2 y + 8\cos y - 3 = 0 \).

Factoring the quadratic equation:
\( (2\cos y - 1)(2\cos y + 3) = 0 \).

This gives:
\( \cos y = \frac{1}{2} \) or \( \cos y = -\frac{3}{2} \) (which has no solution since \( |\cos y| \le 1 \)).

Now we solve \( \cos y = \frac{1}{2} \) within the interval \( \frac{\pi}{6} \le y \le \frac{13\pi}{6} \):
- First quadrant solution: \( y = \frac{\pi}{3} \) (valid as \( \frac{\pi}{3} \ge \frac{\pi}{6} \))
- Fourth quadrant solution: \( y = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \) (valid as \( \frac{5\pi}{3} \le \frac{13\pi}{6} \))

Now we convert back to \( x \):
1) \( 2x + \frac{\pi}{6} = \frac{\pi}{3} \implies 2x = \frac{\pi}{6} \implies x = \frac{\pi}{12} \)

2) \( 2x + \frac{\pi}{6} = \frac{5\pi}{3} \implies 2x = \frac{10\pi}{6} - \frac{\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2} \implies x = \frac{3\pi}{4} \).

Both values are within the range \( 0 \le x \le \pi \).
So the solutions are \( x = \frac{\pi}{12} \) and \( x = \frac{3\pi}{4} \).

- **M1**: Uses the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to form a quadratic in \( \cos\left(2x + \frac{\pi}{6}\right) \)
- **M1**: Factors or solves their quadratic to find \( \cos\left(2x + \frac{\pi}{6}\right) = \frac{1}{2} \) (must discard \( -1.5 \))
- **M1**: Determines correct interval of angles \( \left[ \frac{\pi}{6}, \frac{13\pi}{6} \right] \)
- **A1**: Finds the solutions \( 2x + \frac{\pi}{6} = \frac{\pi}{3} \) and \( 2x + \frac{\pi}{6} = \frac{5\pi}{3} \)
- **A1**: Solves for first correct value of \( x = \frac{\pi}{12} \)
- **A1**: Solves for second correct value of \( x = \frac{3\pi}{4} \) with no extra/spurious solutions in range

(a) (i)
Since \( C \) lies on \( OB \) produced such that \( OB : BC = 2 : 3 \), we have:
\( \overrightarrow{BC} = \frac{3}{2}\overrightarrow{OB} = \frac{3}{2}\mathbf{b} \).
Thus, \( \overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC} = \mathbf{b} + \frac{3}{2}\mathbf{b} = \frac{5}{2}\mathbf{b} \).

(ii)
Since \( D \) lies on \( AC \) such that \( AD : DC = 1 : 2 \), \( D \) divides \( AC \) in the ratio \( 1:2 \).
Using the ratio theorem:
\( \overrightarrow{OD} = \frac{2\overrightarrow{OA} + 1\overrightarrow{OC}}{1 + 2} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\left(\frac{5}{2}\mathbf{b}\right) = \frac{2}{3}\mathbf{a} + \frac{5}{6}\mathbf{b} \).

(b)
Since \( E \) lies on \( OD \) produced:
\( \overrightarrow{OE} = \lambda \overrightarrow{OD} = \lambda \left( \frac{2}{3}\mathbf{a} + \frac{5}{6}\mathbf{b} \right) = \frac{2\lambda}{3}\mathbf{a} + \frac{5\lambda}{6}\mathbf{b} \).

We are given:
\( \overrightarrow{OE} = k\mathbf{a} + \frac{5}{2}\mathbf{b} \).

Comparing the coefficients of \( \mathbf{b} \):
\( \frac{5\lambda}{6} = \frac{5}{2} \implies \frac{\lambda}{6} = \frac{1}{2} \implies \lambda = 3 \).

Comparing the coefficients of \( \mathbf{a} \):
\( k = \frac{2\lambda}{3} \).

Since \( \lambda = 3 \):
\( k = \frac{2(3)}{3} = 2 \).

(a) (i) [1 mark]
- **B1**: For \( \frac{5}{2}\mathbf{b} \) (or equivalent)

(a) (ii) [2 marks]
- **M1**: Uses a correct vector route or ratio theorem for \( \overrightarrow{OD} \) in terms of \( \mathbf{a} \) and \( \overrightarrow{OC} \)
- **A1**: Simplifies to \( \frac{2}{3}\mathbf{a} + \frac{5}{6}\mathbf{b} \)

(b) [3 marks]
- **M1**: Sets up equation \( \overrightarrow{OE} = \lambda \overrightarrow{OD} \) with an unknown parameter \( \lambda \)
- **M1**: Equates coefficients of \( \mathbf{b} \) to solve for \( \lambda = 3 \)
- **A1**: Finds \( k = 2 \)

We can write the inequality as \( 3x^2 - x - 2 > 0 \). Factorizing gives \( (3x + 2)(x - 1) > 0 \). The critical values are \( x = -\frac{2}{3} \) and \( x = 1 \). Since the inequality is strictly greater than zero, the solution is \( x < -\frac{2}{3} \) or \( x > 1 \).

M1: For factorizing or finding the critical values \( -\frac{2}{3} \) and \( 1 \). A1: For the correct solution \( x < -\frac{2}{3} \text{ or } x > 1 \).

The magnitude of the vector is \( |\mathbf{a}| = \sqrt{15^2 + (-8)^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \). The unit vector in the direction of \(\mathbf{a}\) is given by \( \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{1}{17}(15\mathbf{i} - 8\mathbf{j}) \).

M1: For calculating the magnitude of \(\mathbf{a}\) as 17. A1: For the correct unit vector \( \frac{1}{17}(15\mathbf{i} - 8\mathbf{j}) \) or equivalent.

The number of ways to choose 2 men from 6 is \( \binom{6}{2} = 15 \). The number of ways to choose 2 women from 5 is \( \binom{5}{2} = 10 \). The total number of different ways is \( 15 \times 10 = 150 \).

M1: For calculating \( \binom{6}{2} \times \binom{5}{2} \) (or seeing 15 and 10). A1: For 150.

Using the sum to infinity formula, \( S_{\infty} = \frac{a}{1-r} \). Substituting the given values: \( 18 = \frac{12}{1-r} \). This simplifies to \( 1-r = \frac{12}{18} = \frac{2}{3} \), which gives \( r = \frac{1}{3} \).

M1: For substituting correctly into the sum to infinity formula, e.g., \( 18 = \frac{12}{1-r} \). A1: For the correct common ratio \( \frac{1}{3} \) (or equivalent).

Rewrite the logarithmic equation in exponential form: \( 2x - 5 = 3^2 \). This gives \( 2x - 5 = 9 \), so \( 2x = 14 \), which results in \( x = 7 \).

M1: For converting correctly to exponential form, i.e., \( 2x - 5 = 3^2 \) or \( 2x - 5 = 9 \). A1: For \( x = 7 \).

Rewrite the integral as \( \int_1^4 3x^{-\frac{1}{2}} \text{d}x \). Integrating gives \( \left[ 6x^{\frac{1}{2}} \right]_1^4 = \left[ 6\sqrt{x} \right]_1^4 \). Substituting the limits: \( 6\sqrt{4} - 6\sqrt{1} = 12 - 6 = 6 \).

M1: For integrating to obtain the term \( k\sqrt{x} \) where \( k \ne 3 \). A1: For 6.

Let \( y = 2x^3 - 1 \). Rearranging to make \( x \) the subject: \( y + 1 = 2x^3 \implies x^3 = \frac{y+1}{2} \implies x = \sqrt[3]{\frac{y+1}{2}} \). Thus, \( \text{f}^{-1}(x) = \sqrt[3]{\frac{x+1}{2}} \).

M1: For a valid attempt to make \( x \) the subject of the equation, obtaining at least \( x^3 = \frac{y+1}{2} \) (or equivalent with swapped variables). A1: For \( \sqrt[3]{\frac{x+1}{2}} \) (or equivalent).

Using the identity \( \sin^2\phi + \cos^2\phi = 1 \), we have \( \cos^2\phi = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169} \). Since \(\phi\) is obtuse (in the second quadrant), \( \cos\phi \) must be negative, so \( \cos\phi = -\frac{12}{13} \).

M1: For using \( \sin^2\phi + \cos^2\phi = 1 \) or constructing a correct triangle to find the magnitude of the cosine is \( \frac{12}{13} \). A1: For \( -\frac{12}{13} \).

First, find the critical values by setting the quadratic expression equal to zero:

\((2x + 3)(3x - 5) = 0\)

This gives the critical values:
\(x = -\frac{3}{2}\) and \(x = \frac{5}{3}\).

Since the inequality is of the form \(P(x) \leqslant 0\) for a quadratic with a positive leading coefficient, the solution is the region between the two critical values.

Thus, the solution is \(-\frac{3}{2} \leqslant x \leqslant \frac{5}{3}\) (or \(-1.5 \leqslant x \leqslant 1.67\)).

B1: For both correct critical values, \(x = -1.5\) and \(x = \frac{5}{3}\) (or 1.67) obtained.
B1: For the correct final inequality range \(-\frac{3}{2} \leqslant x \leqslant \frac{5}{3}\).

(a) Since \(\mathrm{e}^{2x} > 0\) for all real \(x\), we have \(3\mathrm{e}^{2x} - 5 > -5\). Thus, the range of \(\mathrm{f}\) is \(\mathrm{f}(x) > -5\). (b) Let \(y = 3\mathrm{e}^{2x} - 5\). Then \(y + 5 = 3\mathrm{e}^{2x}\), which gives \(\mathrm{e}^{2x} = \frac{y+5}{3}\). Taking the natural logarithm of both sides: \(2x = \ln\left(\frac{y+5}{3}\right)\), so \(x = \frac{1}{2}\ln\left(\frac{y+5}{3}\right)\). Replacing \(y\) with \(x\) gives \(\mathrm{f}^{-1}(x) = \frac{1}{2}\ln\left(\frac{x+5}{3}\right)\).

Range of \(\mathrm{f}\): B1: For \(\mathrm{f}(x) > -5\) (or equivalent). Expression for \(\mathrm{f}^{-1}(x)\): M1: For setting \(y = 3\mathrm{e}^{2x} - 5\) and attempting to make \(\mathrm{e}^{2x}\) or \(2x\) the subject. M1: For correct use of logarithms to eliminate the exponential. A1: For \(\mathrm{f}^{-1}(x) = \frac{1}{2}\ln\left(\frac{x+5}{3}\right)\) (or equivalent expression in terms of \(x\)).

Apply the quotient rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 \cdot \frac{3}{3x-1} - 2x\ln(3x-1)}{x^4}\). Simplifying this gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x}{x^3(3x-1)} - \frac{2\ln(3x-1)}{x^3} = \frac{3}{x(3x-1)} - \frac{2\ln(3x-1)}{x^3}\). Substituting \(x = 1\) yields \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{1(2)} - \frac{2\ln 2}{1^3} = \frac{3}{2} - 2\ln 2\).

M1: For attempting to apply the quotient rule (or product rule on \(x^{-2}\ln(3x-1)\)), with at least one term correctly differentiated. A1: For a correct derivative expression, e.g. \(\frac{x^2 \left(\frac{3}{3x-1}\right) - 2x\ln(3x-1)}{x^4}\) or equivalent. M1: For substituting \(x = 1\) into their derivative. A1: For \(\frac{3}{2} - 2\ln 2\) or equivalent exact form.

Let \(u = 3^x\). Then the equation becomes \(3u^2 - 10u + 3 = 0\). Factorising the quadratic gives \((3u - 1)(u - 3) = 0\), so \(u = \frac{1}{3}\) or \(u = 3\). Substituting back \(u = 3^x\): If \(3^x = \frac{1}{3}\), then \(x = -1\). If \(3^x = 3\), then \(x = 1\).

M1: For using a substitution like \(u = 3^x\) to obtain a quadratic equation \(3u^2 - 10u + 3 = 0\) (or equivalent). A1: For correct factorisation or solution of the quadratic, giving \(u = \frac{1}{3}\) and \(u = 3\). M1: For setting \(3^x = \text{their } u\) and solving for \(x\). A1: For both \(x = -1\) and \(x = 1\).

Let \(a\) be the first term and \(d\) be the common difference. Using \(S_n = \frac{n}{2}[2a + (n-1)d]\): \(S_4 = 2(2a + 3d) = 34 \implies 2a + 3d = 17\) (1), and \(S_8 = 4(2a + 7d) = 116 \implies 2a + 7d = 29\) (2). Subtracting (1) from (2) gives \(4d = 12 \implies d = 3\). Substituting \(d = 3\) into (1) gives \(2a + 9 = 17 \implies 2a = 8 \implies a = 4\).

M1: For using the sum formula to write an equation in \(a\) and \(d\) for \(S_4\) or \(S_8\). A1: For both correct simplified equations: \(2a + 3d = 17\) and \(2a + 7d = 29\) (or equivalent). M1: For a valid method of solving the simultaneous equations to find \(a\) and \(d\). A1: For \(a = 4\) and \(d = 3\).

(a) The magnitude of \(\mathbf{p}\) is \(\sqrt{12^2 + k^2} = 15\), which simplifies to \(144 + k^2 = 225 \implies k^2 = 81\). Since \(k < 0\), we have \(k = -9\). (b) The unit vector in the direction of \(\mathbf{p}\) is \(\frac{\mathbf{p}}{|\mathbf{p}|} = \frac{12\mathbf{i} - 9\mathbf{j}}{15} = \frac{4}{5}\mathbf{i} - \frac{3}{5}\mathbf{j}\).

(a) M1: For setting up the equation \(12^2 + k^2 = 15^2\) or equivalent. A1: For \(k = -9\). (b) M1: For dividing their vector \(\mathbf{p}\) by 15. A1: For \(\frac{4}{5}\mathbf{i} - \frac{3}{5}\mathbf{j}\) or equivalent.

The committee of 5 must contain at least 3 women. Case 1: Exactly 3 women and 2 men. Number of ways = \(\binom{4}{3} \times \binom{6}{2} = 4 \times 15 = 60\). Case 2: Exactly 4 women and 1 man. Number of ways = \(\binom{4}{4} \times \binom{6}{1} = 1 \times 6 = 6\). Since there are only 4 women in total, we cannot have 5 women on the committee. Total number of ways = \(60 + 6 = 66\).

M1: For calculating the number of ways for 3 women and 2 men: \(\binom{4}{3} \times \binom{6}{2}\). A1: For obtaining 60. M1: For calculating the number of ways for 4 women and 1 man: \(\binom{4}{4} \times \binom{6}{1}\) and adding the two cases. A1: For 66.

1. Gradient of \(L_1\) is \(m_1 = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\). 2. Since \(L_2\) is perpendicular to \(L_1\), its gradient is \(m_2 = -\frac{1}{2}\). 3. Midpoint of \(AB\) is \(M = \left(\frac{2 + 6}{2}, \frac{5 + 13}{2}\right) = (4, 9)\). 4. Equation of \(L_2\) is \(y - 9 = -\frac{1}{2}(x - 4) \implies y - 9 = -\frac{1}{2}x + 2 \implies y = -\frac{1}{2}x + 11\).

M1: For finding the gradient of \(L_1\) and stating the perpendicular gradient \(m_2 = -\frac{1}{2}\). M1: For finding the coordinates of the midpoint of \(AB\), which is \((4, 9)\). M1: For substituting their perpendicular gradient and midpoint into a line equation formula. A1: For \(y = -\frac{1}{2}x + 11\) or equivalent.

For the line to be a tangent to the curve, equating them must result in a quadratic equation with a discriminant of 0: \(2x + c = x^2 - 4x + 12 \implies x^2 - 6x + (12 - c) = 0\). Setting the discriminant \(\Delta = 0\): \((-6)^2 - 4(1)(12 - c) = 0 \implies 36 - 48 + 4c = 0 \implies -12 + 4c = 0 \implies c = 3\).

M1: For equating the line and the curve: \(2x + c = x^2 - 4x + 12\). M1: For rearranging into a standard quadratic equation: \(x^2 - 6x + (12 - c) = 0\). M1: For setting the discriminant of their quadratic equal to 0. A1: For \(c = 3\).

Let \(Y = \frac{1}{y}\) and \(X = \ln x\).

The straight-line graph passes through the points \((1, 3)\) and \((4, 9)\).

First, find the gradient, \(m\), of the line:
\(m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2\)

Using the straight-line equation \(Y - Y_1 = m(X - X_1)\) with the point \((1, 3)\):
\(Y - 3 = 2(X - 1)\)
\(Y = 2X + 1\)

Now, substitute \(Y = \frac{1}{y}\) and \(X = \ln x\) back into the equation:
\(\frac{1}{y} = 2\ln x + 1\)

Solve for \(y\) to express it in terms of \(x\):
\(y = \frac{1}{2\ln x + 1}\)

**M1**: For an attempt to find the gradient of the line: \(m = \frac{9-3}{4-1}\) (or equivalent)
**M1**: For an attempt to write the equation of the line using their gradient and one of the points, in the form \(\frac{1}{y} = m\ln x + c\)
**A1**: For obtaining \(\frac{1}{y} = 2\ln x + 1\) (or any equivalent unsimplified form)
**A1**: For the correct final expression \(y = \frac{1}{2\ln x + 1}\) (or equivalent, such as \(y = \frac{1}{\ln(x^2) + 1}\))

(a) Integrating \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find \(y\):

\(y = \int \left(4x^3 - 4x^{-2}\right) \mathrm{d}x\)

\(y = x^4 + \frac{4}{x} + c\)

Using the point \((2, 20)\):

\(20 = 2^4 + \frac{4}{2} + c\)

\(20 = 16 + 2 + c\)

\(c = 2\)

So, the equation of the curve is \(y = x^4 + \frac{4}{x} + 2\).

(b) At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):

\(4x^3 - \frac{4}{x^2} = 0\)

\(4x^3 = \frac{4}{x^2}\)

\(x^5 = 1 \implies x = 1\) (since \(x > 0\))

Substituting \(x = 1\) into the curve equation:

\(y = 1^4 + \frac{4}{1} + 2 = 7\)

So, the coordinates of the stationary point are \((1, 7)\).

(a) [4 marks]

M1: Attempt to integrate \(\frac{\mathrm{d}y}{\mathrm{d}x}\), obtaining at least one term correct.

A1: Correct integration to get \(y = x^4 + \frac{4}{x} + c\) (or equivalent form).

M1: Substituting \((2, 20)\) into their integrated expression to find \(c\).

A1: Finding \(c = 2\) and stating the correct final equation \(y = x^4 + \frac{4}{x} + 2\).

(b) [2 marks]

M1: Setting their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solving for \(x\).

A1: Finding \(x = 1\) and \(y = 7\), yielding the coordinates \((1, 7)\).

(a) Treat the 4 boys as a single block.

The number of objects to arrange is the 6 girls plus 1 block of boys, which is 7 objects.

These 7 objects can be arranged in \(7!\) ways.

Within the block, the 4 boys can be arranged in \(4!\) ways.

Total number of ways = \(7! \times 4! = 5040 \times 24 = 120\,960\).

(b) The committee must contain at least 2 girls. The possible cases are:

- 2 girls and 2 boys: \(\binom{6}{2} \times \binom{4}{2} = 15 \times 6 = 90\)

- 3 girls and 1 boy: \(\binom{6}{3} \times \binom{4}{1} = 20 \times 4 = 80\)

- 4 girls and 0 boys: \(\binom{6}{4} \times \binom{4}{0} = 15 \times 1 = 15\)

Total number of different committees = \(90 + 80 + 15 = 185\).

(a) [3 marks]

M1: For considering 7 blocks/items to arrange, i.e., \(7!\).

M1: For considering internal arrangement of 4 boys, i.e., \(4!\).

A1: Correct final answer \(120\,960\).

(b) [3 marks]

M1: For identifying the three valid scenarios (2G 2B, 3G 1B, 4G 0B) and expressing them as combinations.

M1: For calculating the number of ways for at least two of the scenarios correctly.

A1: Correct final answer \(185\).

(a) Rewrite the equation as:

\(2 \cdot (2^x)^2 - 15(2^x) - 8 = 0\)

Let \(u = 2^x\).

\(2u^2 - 15u - 8 = 0\)

\((2u + 1)(u - 8) = 0\)

\(u = -\frac{1}{2}\) or \(u = 8\)

Since \(2^x > 0\) for all real \(x\), \(2^x = -\frac{1}{2}\) has no solution.

Thus, \(2^x = 8 \implies x = 3\).

(b) Using the subtraction rule for logarithms:

\(\log_3\left(\frac{y + 5}{y - 1}\right) = 2\)

Converting to exponential form:

\(\frac{y + 5}{y - 1} = 3^2\)

\(\frac{y + 5}{y - 1} = 9\)

\(y + 5 = 9(y - 1)\)

\(y + 5 = 9y - 9\)

\(8y = 14 \implies y = 1.75\) (or \(\frac{7}{4}\)).

(a) [4 marks]

M1: Rewriting the equation as a quadratic in terms of \(2^x\).

M1: Correct factorisation of their quadratic equation.

A1: Stating \(2^x = 8\) and rejecting the negative root.

A1: Correct final answer \(x = 3\).

(b) [3 marks]

M1: Using the division property of logarithms to obtain \(\log_3\left(\frac{y + 5}{y - 1}\right) = 2\).

M1: Removing logarithms to get \(\frac{y + 5}{y - 1} = 9\) and attempting to solve for \(y\).

A1: Stating the correct final answer \(y = 1.75\) (or \(\frac{7}{4}\)).

(a) Since \(AC : CB = 1 : 2\), the position vector of \(C\) can be found using:

\(\overrightarrow{OC} = \mathbf{a} + \frac{1}{3}\overrightarrow{AB} = \mathbf{a} + \frac{1}{3}(\mathbf{b} - \mathbf{a}) = \frac{2\mathbf{a} + \mathbf{b}}{3}\)

\(\overrightarrow{OC} = \frac{2(\mathbf{i} + 2\mathbf{j}) + (7\mathbf{i} + 8\mathbf{j})}{3}\)

\(\overrightarrow{OC} = \frac{(2\mathbf{i} + 4\mathbf{j}) + (7\mathbf{i} + 8\mathbf{j})}{3} = \frac{9\mathbf{i} + 12\mathbf{j}}{3}\)

\(\overrightarrow{OC} = 3\mathbf{i} + 4\mathbf{j}\).

(b) First, find the magnitude of \(\overrightarrow{OC}\):

\(|\overrightarrow{OC}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\).

The unit vector in the direction of \(\overrightarrow{OC}\) is:

\(\frac{\overrightarrow{OC}}{|\overrightarrow{OC}|} = \frac{3\mathbf{i} + 4\mathbf{j}}{5} = \frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j}\) (or \(0.6\mathbf{i} + 0.8\mathbf{j}\)).

(a) [3 marks]

M1: For writing a correct vector expression for \(\overrightarrow{OC}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

M1: Substituting \(\mathbf{a} = \mathbf{i} + 2\mathbf{j}\) and \(\mathbf{b} = 7\mathbf{i} + 8\mathbf{j}\) into their expression.

A1: Finding the correct position vector \(\overrightarrow{OC} = 3\mathbf{i} + 4\mathbf{j}\).

(b) [4 marks]

M1: For attempting to find the magnitude of their \(\overrightarrow{OC}\).

A1: Finding the correct magnitude \(5\).

M1: Dividing their \(\overrightarrow{OC}\) by its magnitude.

A1: Correct final unit vector \(\frac{3}{5}\mathbf{i} + \frac{4}{5}\mathbf{j}\) (or equivalent).