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

We can solve the modulus equation by splitting it into two cases:

**Case 1:**
\(2x - 5 = x + 1\)
\(x = 6\)

**Case 2:**
\(2x - 5 = -(x + 1)\)
\(2x - 5 = -x - 1\)
\(3x = 4\)
\(x = \frac{4}{3}\)

Both values are valid solutions.

M1: For writing two linear equations, e.g., \(2x - 5 = x + 1\) and \(2x - 5 = -x - 1\), or for squaring both sides.
A1: For \(x = 6\).
A1: For \(x = \frac{4}{3}\).

Using the laws of logarithms:

\(\log_3 (x^2) - \log_3(x-2) = 2\)

\(\log_3 \left(\frac{x^2}{x-2}\right) = 2\)

\(\frac{x^2}{x-2} = 3^2 = 9\)

\(x^2 = 9x - 18\)

\(x^2 - 9x + 18 = 0\)

\((x-3)(x-6) = 0\)

\(x=3\) or \(x=6\)

Both values are positive and greater than 2, so both are valid.

M1: For correct use of the power law and subtraction law to obtain \(\log_3\left(\frac{x^2}{x-2}\right) = 2\).
M1: For removing logarithms to obtain \(\frac{x^2}{x-2} = 9\) and forming a quadratic equation.
A1: For both \(x=3\) and \(x=6\) and no other solutions.

First, divide by 2:
\(\cos(2\theta) = \frac{1}{2}\)

Since \(0 \le \theta \le \pi\), we have \(0 \le 2\theta \le 2\pi\).

In this interval, the solutions for \(2\theta\) are:
\(2\theta = \frac{\pi}{3}\) or \(2\theta = \frac{5\pi}{3}\)

Dividing by 2 gives:
\(\theta = \frac{\pi}{6}\) or \(\theta = \frac{5\pi}{6}\)

M1: For simplifying to \(\cos(2\theta) = \frac{1}{2}\).
M1: For identifying at least one correct value of \(2\theta\) (e.g. \(\frac{\pi}{3}\) or \(\frac{5\pi}{3}\)).
A1: For both \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\) and no others in the range.

Using the product rule to differentiate \(y = (2x - 3)\ln x\):

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(2x - 3) \cdot \ln x + (2x - 3) \cdot \frac{\mathrm{d}}{\mathrm{d}x}(\ln x)\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2 \ln x + (2x - 3) \cdot \frac{1}{x}\)

At the point where \(x = 1\):

\(\text{Gradient} = 2 \ln(1) + \frac{2(1) - 3}{1}\)

\(\text{Gradient} = 2(0) + \frac{-1}{1} = -1\)

M1: For an attempt to use the product rule (must have two terms of the form \(a \ln x + \frac{2x-3}{x}\)).
A1: For the correct derivative \(2\ln x + \frac{2x-3}{x}\) oe.
A1: For gradient \(= -1\).

The perimeter of the sector is given by:
\(P = 2r + r\theta = 24 \implies r\theta = 24 - 2r\)

Therefore, \(\theta = \frac{24 - 2r}{r}\).

The area of the sector is given by:
\(A = \frac{1}{2}r^2\theta = 32\)

Substitute the expression for \(\theta\) into the area equation:
\(\frac{1}{2}r^2\left(\frac{24 - 2r}{r}\right) = 32\)

\(\frac{1}{2}r(24 - 2r) = 32\)

\(12r - r^2 = 32\)

\(r^2 - 12r + 32 = 0\)

Factorising the quadratic equation:
\((r - 4)(r - 8) = 0\)

Thus, \(r = 4\) or \(r = 8\).

M1: For writing down the equations for perimeter and area: \(2r + r\theta = 24\) and \(\frac{1}{2}r^2\theta = 32\).
M1: For eliminating \(\theta\) to obtain a quadratic equation in terms of \(r\).
A1: For \(r = 4\) or \(r = 8\).

To find the number of even 4-digit numbers greater than 4000 without repetition, we consider two cases based on the first digit:

**Case 1: The first digit is odd and greater than 4000 (i.e. 5 or 7)**
- There are 2 choices for the first digit (5 or 7).
- The last digit must be even: 2, 4, or 6 (3 choices).
- The remaining 2 middle digits can be filled using any of the remaining 5 digits: \(5 \times 4 = 20\) ways.
- Number of ways for Case 1: \(2 \times 3 \times 20 = 120\).

**Case 2: The first digit is even and greater than 4000 (i.e. 4 or 6)**
- There are 2 choices for the first digit (4 or 6).
- The last digit must be even. Since one even digit is already used in the first position, there are 2 choices left for the last digit.
- The remaining 2 middle digits can be filled using any of the remaining 5 digits: \(5 \times 4 = 20\) ways.
- Number of ways for Case 2: \(2 \times 2 \times 20 = 80\).

**Total number of ways:**
\(120 + 80 = 200\).

M1: For identifying the need to split into cases based on the first digit being odd or even, or attempting a systematic approach.
M1: For correct calculation of one of the cases (either 120 or 80).
A1: For 200.

By the Remainder Theorem, since dividing \(p(x)\) by \(x - 2\) gives a remainder of 18:
\(p(2) = 18\)

Substitute \(x = 2\):
\(2^3 + k(2^2) - 4(2) + 6 = 18\)

\(8 + 4k - 8 + 6 = 18\)

\(4k + 6 = 18 \implies 4k = 12 \implies k = 3\)

So, \(p(x) = x^3 + 3x^2 - 4x + 6\).

To find the remainder when \(p(x)\) is divided by \(x + 1\), we calculate \(p(-1)\):
\(p(-1) = (-1)^3 + 3(-1)^2 - 4(-1) + 6\)

\(p(-1) = -1 + 3 + 4 + 6 = 12\)

M1: For using \(p(2) = 18\) to find \(k\).
A1: For \(k = 3\).
A1: For finding the remainder is 12 (by evaluating \(p(-1)\) with their value of \(k\)).

(a) When \(x = \frac{1}{2}\), \(y = 4e^{2(\frac{1}{2})-1} = 4e^0 = 4\).

Differentiating \(y = 4e^{2x-1}\) with respect to \(x\) gives:

\(\frac{dy}{dx} = 8e^{2x-1}\)

At \(x = \frac{1}{2}\), the gradient of the tangent is:

\(m = 8e^0 = 8\)

The equation of the tangent is:

\(y - 4 = 8\left(x - \frac{1}{2}\right) \implies y = 8x\)

(b) The region is bounded by the curve \(y = 4e^{2x-1}\), the tangent \(y = 8x\), and the line \(x = 0\). The curve and the tangent meet at \(x = \frac{1}{2}\).

Since the curve is convex, it lies above the tangent line for the interval \(0 \le x \le \frac{1}{2}\).

Thus, the area \(A\) is given by:

\(A = \int_{0}^{1/2} \left(4e^{2x-1} - 8x\right) dx\)

Integrating term-by-term:

\(A = \left[ 2e^{2x-1} - 4x^2 \right]_{0}^{1/2}\)

Substitute the upper limit \(x = \frac{1}{2}\):

\(2e^0 - 4\left(\frac{1}{2}\right)^2 = 2 - 1 = 1\)

Substitute the lower limit \(x = 0\):

\(2e^{-1} - 4(0)^2 = \frac{2}{e}\)

Therefore, the exact area is:

\(A = 1 - \frac{2}{e}\)

(a)
- M1: For attempt to differentiate, must see \(ke^{2x-1}\) where \(k \ne 4\)
- A1: Correct derivative \(\frac{dy}{dx} = 8e^{2x-1}\)
- M1: For finding the coordinate \(y = 4\) and substituting \(x = \frac{1}{2}\) into their derivative to find the gradient
- A1: Correct equation of the tangent \(y = 8x\) or equivalent

(b)
- M1: For identifying the limits of integration as \(0\) and \(\frac{1}{2}\)
- M1: For setting up the integral of the difference of the functions: \(\int (4e^{2x-1} - 8x) dx\) or equivalent
- A1: Correct integration of \(4e^{2x-1}\) to get \(2e^{2x-1}\)
- A1: Correct integration of \(-8x\) to get \(-4x^2\)
- M1: For substituting the limits \(\frac{1}{2}\) and \(0\) into their integrated expression
- A1: Correct evaluation of the upper limit value as \(1\) (or \(2 - 1\))
- A1: Correct evaluation of the lower limit value as \(2e^{-1}\) (or \(\frac{2}{e}\))
- A1: Correct final exact area of \(1 - \frac{2}{e}\) or \(\frac{e-2}{e}\)

(a) From the first equation:

\(\log_3(x - 2y) = 2\)

Using the definition of logarithms, this can be written in exponential form as:

\(x - 2y = 3^2\)

\(x - 2y = 9\)

(b) From the second equation:

\(3^{2x} \cdot 9^{y} = 27^{x-y+2}\)

Express all terms as powers of base 3:

\(3^{2x} \cdot \left(3^2\right)^{y} = \left(3^3\right)^{x-y+2}\)

\(3^{2x} \cdot 3^{2y} = 3^{3(x-y+2)}\)

Using the laws of indices to simplify both sides:

\(3^{2x + 2y} = 3^{3x - 3y + 6}\)

Equating the exponents:

\(2x + 2y = 3x - 3y + 6\)

Rearranging to group \(x\) and \(y\) terms:

\(3x - 2x - 3y - 2y + 6 = 0\)

\(x - 5y + 6 = 0 \implies x - 5y = -6\)

(c) We now have the linear simultaneous equations:

1) \(x - 2y = 9 \implies x = 2y + 9\)

2) \(x - 5y = -6\)

Substitute equation 1 into equation 2:

\((2y + 9) - 5y = -6\)

\(-3y + 9 = -6\)

\(-3y = -15\)

\(y = 5\)

Substitute \(y = 5\) back into equation 1:

\(x = 2(5) + 9\)

\(x = 19\)

Check validity: \(x - 2y = 19 - 2(5) = 9 > 0\), so the logarithm is defined.

Thus, \(x = 19\) and \(y = 5\).

(a)
- M1: For converting logarithmic equation to exponential form, i.e., \(x - 2y = 3^2\) or better
- A1: Correctly showing \(x - 2y = 9\)

(b)
- M1: For expressing \(9^y\) as \(3^{2y}\) or \(27^{x-y+2}\) as \(3^{3(x-y+2)}\)
- A1: For correct expression with all bases equal to 3: \(3^{2x} \cdot 3^{2y} = 3^{3(x-y+2)}\)
- M1: For applying index law of multiplication to LHS: \(3^{2x+2y}\)
- M1: For equating exponents: \(2x + 2y = 3(x - y + 2)\)
- A1: Fully correct simplification to \(x - 5y = -6\) (or equivalent)

(c)
- M1: For a valid method to solve the two linear equations, e.g., substitution or elimination
- A1: For eliminating one variable to get an equation in a single variable, e.g., \(-3y = -15\) or \(-3x = -57\)
- A1: For finding either \(y = 5\) or \(x = 19\)
- A1: For finding the other variable's correct value
- B1: For verifying that the solution satisfies the logarithmic constraint \(x - 2y > 0\)

(a) To show the identity, we start with the Left Hand Side (LHS):

\(LHS = \frac{\cos\theta}{1-\sin\theta} - \tan\theta\)

Express \(\tan\theta\) as \(\frac{\sin\theta}{\cos\theta}\):

\(LHS = \frac{\cos\theta}{1-\sin\theta} - \frac{\sin\theta}{\cos\theta}\)

Find a common denominator, which is \(\cos\theta(1-\sin\theta)\):

\(LHS = \frac{\cos^2\theta - \sin\theta(1-\sin\theta)}{\cos\theta(1-\sin\theta)}\)

Expand the numerator:

\(LHS = \frac{\cos^2\theta - \sin\theta + \sin^2\theta}{\cos\theta(1-\sin\theta)}\)

Since \(\sin^2\theta + \cos^2\theta = 1\):

\(LHS = \frac{1 - \sin\theta}{\cos\theta(1-\sin\theta)}\)

Cancel the common factor of \(1 - \sin\theta\) from the numerator and denominator (given \(\sin\theta \ne 1\)):

\(LHS = \frac{1}{\cos\theta}\)

Using the definition of secant, \(\sec\theta = \frac{1}{\cos\theta}\):

\(LHS = \sec\theta = RHS\) (proved)

(b) Using the identity proven in part (a) with \(\theta = 2\phi\), the equation:

\(\frac{\cos(2\phi)}{1-\sin(2\phi)} - \tan(2\phi) = 2\)

simplifies directly to:

\(\sec(2\phi) = 2\)

Since \(\sec(2\phi) = \frac{1}{\cos(2\phi)}\), this is equivalent to:

\(\cos(2\phi) = \frac{1}{2}\)

We are given the range for \(\phi\) as:

\(0 \le \phi \le \pi \implies 0 \le 2\phi \le 2\pi\)

In the interval \(0 \le 2\phi \le 2\pi\), the solutions for \(\cos(2\phi) = \frac{1}{2}\) are:

\(2\phi = \frac{\pi}{3}\) or \(2\phi = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\)

Dividing by 2 to find \(\phi\):

\(\phi = \frac{\pi}{6}\) or \(\phi = \frac{5\pi}{6}\)

Both solutions lie within the range \(0 \le \phi \le \pi\).

(a)
- M1: For writing \(\tan\theta\) as \(\frac{\sin\theta}{\cos\theta}\)
- M1: For writing the expression with a common denominator of \(\cos\theta(1-\sin\theta)\)
- A1: For correct numerator expansion to \(\cos^2\theta - \sin\theta + \sin^2\theta\)
- A1: For correct use of \(\sin^2\theta + \cos^2\theta = 1\) and simplification to \(\sec\theta\) with all intermediate steps shown clearly

(b)
- M1: For using the identity from part (a) to write the equation as \(\sec(2\phi) = 2\)
- A1: For converting to \(\cos(2\phi) = \frac{1}{2}\)
- M1: For identifying the correct interval for \(2\phi\), which is \(0 \le 2\phi \le 2\pi\)
- M1: For finding one correct value of \(2\phi\) (e.g., \(\frac{\pi}{3}\))
- A1: For finding the second correct value of \(2\phi\) (e.g., \(\frac{5\pi}{3}\))
- M1: For dividing their angles by 2 to solve for \(\phi\)
- A1: For \(\phi = \frac{\pi}{6}\)
- A1: For \(\phi = \frac{5\pi}{6}\) and no other solutions in the given range

(a) To find \(\vec{PQ}\):

\(\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 8 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 8 - 2 \\ 3 - (-1) \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}\)

(b) Since \(P\), \(Q\), and \(R\) lie on a straight line, the vectors \(\vec{PQ}\) and \(\vec{PR}\) are parallel (collinear).

First, express \(\vec{PR}\) in terms of \(m\):

\(\vec{PR} = \vec{OR} - \vec{OP} = \begin{pmatrix} 14 \\ m \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 12 \\ m + 1 \end{pmatrix}\)

Since \(\vec{PR}\) is parallel to \(\vec{PQ}\), there exists a scalar \(k\) such that:

\(\vec{PR} = k \vec{PQ}\)

\(\begin{pmatrix} 12 \\ m + 1 \end{pmatrix} = k \begin{pmatrix} 6 \\ 4 \end{pmatrix}\)

Equating the \(x\)-components:

\(12 = 6k \implies k = 2\)

Equating the \(y\)-components:

\(m + 1 = 4k\)

Substitute \(k = 2\) into the equation:

\(m + 1 = 4(2) = 8\)

\(m = 7\)

(c) With \(m = 7\), the vector \(\vec{PR}\) is:

\(\vec{PR} = \begin{pmatrix} 12 \\ 7 + 1 \end{pmatrix} = \begin{pmatrix} 12 \\ 8 \end{pmatrix}\)

To find the unit vector in the direction of \(\vec{PR}\), we first find its magnitude \(|\vec{PR}|\):

\(|\vec{PR}| = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208}\)

Simplify the surd:

\(\sqrt{208} = \sqrt{16 \times 13} = 4\sqrt{13}\)

Now, the unit vector is:

\(\hat{u} = \frac{1}{|\vec{PR}|} \vec{PR} = \frac{1}{4\sqrt{13}} \begin{pmatrix} 12 \\ 8 \end{pmatrix} = \frac{1}{\sqrt{13}} \begin{pmatrix} 3 \\ 2 \end{pmatrix}\)

(a)
- M1: For subtractive combination of position vectors, i.e., \(\vec{OQ} - \vec{OP}\)
- A1: Correct column vector \(\begin{pmatrix} 6 \\ 4 \end{pmatrix}\)

(b)
- M1: For finding \(\vec{PR} = \begin{pmatrix} 12 \\ m + 1 \end{pmatrix}\) (or \(\vec{QR}\))
- M1: For setting up collinearity condition, e.g., \(\vec{PR} = k\vec{PQ}\)
- A1: For finding the scale factor \(k = 2\) (or equivalent ratio)
- M1: For setting up the equation for \(m\): \(m + 1 = 4k\) (or equivalent)
- A1: Correct value of \(m = 7\) with no errors seen

(c)
- M1: For substituting their \(m\) to get a numerical vector \(\vec{PR} = \begin{pmatrix} 12 \\ 8 \end{pmatrix}\)
- M1: For calculating the magnitude \(\sqrt{12^2 + 8^2}\)
- A1: Correct simplified magnitude \(4\sqrt{13}\) (or \(\sqrt{208}\))
- M1: For dividing their vector \(\vec{PR}\) by their magnitude
- A1: Correct unit vector \(\frac{1}{\sqrt{13}}\begin{pmatrix} 3 \\ 2 \end{pmatrix}\) or \(\begin{pmatrix} 3/\sqrt{13} \\ 2/\sqrt{13} \end{pmatrix}\) or equivalent

(a) (i) There are 10 people in total, and we need to choose 5 of them with no restrictions. This is a combination problem:

\(\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252\)

(ii) The committee of 5 must contain exactly 3 students. Therefore, the remaining 2 members must be teachers.
Choose 3 students from the 6 available:

\(\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\)

Choose 2 teachers from the 4 available:

\(\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6\)

Total number of ways:

\(20 \times 6 = 120\)

(iii) The committee of 5 must contain at least 1 teacher. This is easiest to solve using the subtraction method:

\(\text{Ways with at least 1 teacher} = \text{Total ways} - \text{Ways with 0 teachers (all students)}\)

Choose 5 students from the 6 available (and 0 teachers):

\(\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6\)

Total ways from part (a)(i) is 252.

\(\text{Ways with at least 1 teacher} = 252 - 6 = 246\)

(b) We arrange 10 people in a line such that the 4 teachers must stand together.

Treat the 4 teachers as a single block. This block plus the 6 individual students gives 7 units to arrange in a line:
Number of ways to arrange 7 units is:

\(7! = 5040\)

Within the block, the 4 teachers can be arranged among themselves in:

\(4! = 24\) ways.

Using the multiplication principle, the total number of arrangements is:

\(7! \times 4! = 5040 \times 24 = 120960\)

(a) (i)
- M1: For \(\binom{10}{5}\) written out in factorial or product form
- A1: Correct answer 252

(a) (ii)
- M1: For either \(\binom{6}{3}\) or \(\binom{4}{2}\) evaluated correctly
- M1: For multiplying their \(\binom{6}{3}\) and \(\binom{4}{2}\)
- A1: Correct answer 120

(a) (iii)
- M1: For identifying that the case of 0 teachers is \(\binom{6}{5}\) or 6
- M1: For subtracting their 0-teacher case from their total (252)
- A1: Correct answer 246

(b)
- M1: For treating the 4 teachers as 1 unit, leading to 7 units to arrange
- A1: For \(7!\) (or 5040) seen
- M1: For multiplying their arrangement of units by \(4!\) (or 24)
- A1: Correct answer 120960

The perimeter of a sector is given by \(P = 2r + r\theta\).
Given \(P = 36\), we have:
\(2r + r\theta = 36 \implies r\theta = 36 - 2r\)

The area of a sector is given by \(A = \frac{1}{2}r^2\theta\).
Given \(A = 80\), we have:
\(\frac{1}{2}r(r\theta) = 80\)

Substitute the expression for \(r\theta\) from the perimeter equation into the area equation:
\(\frac{1}{2}r(36 - 2r) = 80\)
\(r(18 - r) = 80\)
\(18r - r^2 = 80\)
\(r^2 - 18r + 80 = 0\)

Factorising the quadratic equation:
\((r - 8)(r - 10) = 0\)

This gives the two possible values of \(r\):
\(r = 8\) or \(r = 10\).

M1: For writing down the equations for perimeter and area: \(2r + r\theta = 36\) and \(\frac{1}{2}r^2\theta = 80\), and attempting to eliminate \(\theta\).
M1: For forming a quadratic equation in \(r\): \(r^2 - 18r + 80 = 0\) (or equivalent).
A1: For both \(r = 8\) and \(r = 10\).

Using the Factor Theorem, since \(x - 2\) is a factor of \(\text{p}(x)\), we have \(\text{p}(2) = 0\):
\(2(2)^3 + a(2)^2 + b(2) - 6 = 0\)
\(16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5\) (Equation 1)

Using the Remainder Theorem, since the remainder when \(\text{p}(x)\) is divided by \(x + 1\) is \(-12\), we have \(\text{p}(-1) = -12\):
\(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -12\)
\(-2 + a - b - 6 = -12 \implies a - b = -4\) (Equation 2)

Solving Equations 1 and 2 simultaneously:
From Equation 2, \(b = a + 4\).
Substitute this into Equation 1:
\(2a + (a + 4) = -5\)
\(3a = -9 \implies a = -3\)

Substitute \(a = -3\) back into the expression for \(b\):
\(b = -3 + 4 = 1\).

Therefore, \(a = -3\) and \(b = 1\).

M1: For using \(\text{p}(2) = 0\) to obtain \(2a + b = -5\) (or equivalent).
M1: For using \(\text{p}(-1) = -12\) to obtain \(a - b = -4\) (or equivalent).
A1: For \(a = -3\) and \(b = 1\).

For the quadratic equation to have real roots, its discriminant must be greater than or equal to zero (\(b^2 - 4ac \ge 0\)).

For the equation \(x^2 + (k - 2)x + k + 1 = 0\), we have:
\(a = 1\)
\(b = k - 2\)
\(c = k + 1\)

Calculate the discriminant:
\(D = (k - 2)^2 - 4(1)(k + 1) \ge 0\)
\(k^2 - 4k + 4 - 4k - 4 \ge 0\)
\(k^2 - 8k \ge 0\)
\(k(k - 8) \ge 0\)

The critical values are \(k = 0\) and \(k = 8\).
Since we require \(k(k - 8) \ge 0\), the set of values of \(k\) is:
\(k \le 0\) or \(k \ge 8\).

M1: For using \(b^2 - 4ac \ge 0\) with correct substitution of coefficients.
M1: For simplifying to \(k^2 - 8k \ge 0\) and finding critical values \(0\) and \(8\).
A1: For the correct range \(k \le 0\) or \(k \ge 8\) (allow equivalent inequality or interval notation).

Let \(Y = y^2\) and \(X = \ln x\).
Since the graph of \(Y\) against \(X\) is a straight line, we can write its equation in the form \(Y = mX + c\).

The line passes through the points \((2, 5)\) and \((6, 17)\).

First, find the gradient \(m\):
\(m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3\)

Now use one of the points, say \((2, 5)\), to find the vertical intercept \(c\):
\(5 = 3(2) + c\)
\(5 = 6 + c \implies c = -1\)

So the linear equation is:
\(Y = 3X - 1\)

Substituting back \(Y = y^2\) and \(X = \ln x\):
\(y^2 = 3\ln x - 1\).

M1: For finding the gradient of the line: \(m = 3\).
M1: For using a point to find the vertical intercept: \(c = -1\), or writing the linear equation \(Y = 3X - 1\).
A1: For the correct final relation \(y^2 = 3\ln x - 1\) (or equivalent).

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

Cross-multiply to solve for \(k\):
\(k^2 = (k + 4)(2k - 15)\)
\(k^2 = 2k^2 - 15k + 8k - 60\)
\(k^2 = 2k^2 - 7k - 60\)
\(k^2 - 7k - 60 = 0\)

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

This gives two possible values of \(k\): \(k = 12\) or \(k = -5\).

Now, test each value of \(k\) to check if the geometric progression has a sum to infinity (which requires \(|r| < 1\)):
- If \(k = 12\):
\(r = \frac{12}{12 + 4} = \frac{12}{16} = 0.75\). Since \(|0.75| < 1\), a sum to infinity exists.
- If \(k = -5\):
\(r = \frac{-5}{-5 + 4} = 5\). Since \(|5| \ge 1\), a sum to infinity does not exist.

Therefore, the common ratio is \(r = 0.75\) (or \(\frac{3}{4}\)).

M1: For setting up the ratio equation: \(\frac{k}{k + 4} = \frac{2k - 15}{k}\) and cross-multiplying to form a quadratic in \(k\).
M1: For solving the quadratic equation to get \(k = 12\) or \(k = -5\).
A1: For identifying \(r = 0.75\) (or \(3/4\)) as the only valid common ratio.

Using the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(4(1 - \cos^2 \theta) + 5\cos \theta - 5 = 0\)
\(4 - 4\cos^2 \theta + 5\cos \theta - 5 = 0\)
\(4\cos^2 \theta - 5\cos \theta + 1 = 0\)

Factorising this quadratic equation in \(\cos \theta\):
\((4\cos \theta - 1)(\cos \theta - 1) = 0\)

This yields two possible values for \(\cos \theta\):
1) \(\cos \theta = 1\)
2) \(\cos \theta = 0.25\)

Solving for \(\theta\) in the interval \(0^\circ \le \theta \le 360^\circ\):
- From \(\cos \theta = 1\):
\(\theta = 0^\circ, 360^\circ\)

- From \(\cos \theta = 0.25\):
\(\theta = \cos^{-1}(0.25) \approx 75.5^\circ\) (to 1 decimal place)
The other solution in the range is \(360^\circ - 75.5^\circ = 284.5^\circ\).

Therefore, the solutions are \(\theta = 0^\circ, 75.5^\circ, 284.5^\circ, 360^\circ\).

M1: For using \(\sin^2 \theta = 1 - \cos^2 \theta\) to obtain a quadratic equation in \(\cos \theta\): \(4̂\cos^2 \theta - 5\cos \theta + 1 = 0\).
M1: For solving the quadratic to get \(\cos \theta = 1\) and \(\cos \theta = 0.25\) and finding at least one correct non-trivial angle (e.g., \(75.5^\circ\)).
A1: For all four correct values: \(\theta = 0^\circ, 75.5^\circ, 284.5^\circ, 360^\circ\).

The team of 5 must contain more women than men. The possible compositions of the team are:
- Case 1: 3 women and 2 men
Number of ways = \(\binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150\)

- Case 2: 4 women and 1 man
Number of ways = \(\binom{5}{4} \times \binom{6}{1} = 5 \times 6 = 30\)

- Case 3: 5 women and 0 men
Number of ways = \(\binom{5}{5} \times \binom{6}{0} = 1 \times 1 = 1\)

Total number of different teams = \(150 + 30 + 1 = 181\).

M1: For identifying the three cases (3W 2M, 4W 1M, 5W 0M) and showing intent to calculate using combinations.
M1: For calculating the correct number of combinations for at least two of the cases (e.g., 150, 30, or 1).
A1: For the correct total of 181.

Write the equation of the curve in index form:
\(y = (2x - 3)(x + 1)^{\frac{1}{2}}\)

Differentiate with respect to \(x\) using the product rule:
\(\frac{\text{d}y}{\text{d}x} = \frac{\text{d}}{\text{d}x}(2x - 3) \cdot (x + 1)^{\frac{1}{2}} + (2x - 3) \cdot \frac{\text{d}}{\text{d}x}(x + 1)^{\frac{1}{2}}\)
\(\frac{\text{d}y}{\text{d}x} = 2(x + 1)^{\frac{1}{2}} + (2x - 3) \cdot \frac{1}{2}(x + 1)^{-\frac{1}{2}}\)
\(\frac{\text{d}y}{\text{d}x} = 2\sqrt{x + 1} + \frac{2x - 3}{2\sqrt{x + 1}}\)

For a stationary point, set \(\frac{\text{d}y}{\text{d}x} = 0\):
\(2\sqrt{x + 1} + \frac{2x - 3}{2\sqrt{x + 1}} = 0\)

Multiply the entire equation by \(2\sqrt{x + 1}\) to eliminate the fraction:
\(4(x + 1) + (2x - 3) = 0\)
\(4x + 4 + 2x - 3 = 0\)
\(6x + 1 = 0 \implies x = -\frac{1}{6}\)

Therefore, the \(x\)-coordinate of the stationary point is \(x = -\frac{1}{6}\) (or \(-0.167\)).

M1: For applying the product rule to differentiate \(y = (2x - 3)(x + 1)^{1/2}\).
A1: For correct derivative: \(\frac{\text{d}y}{\text{d}x} = 2\sqrt{x + 1} + \frac{2x - 3}{2\sqrt{x + 1}}\).
M1: For setting \(\frac{\text{d}y}{\text{d}x} = 0\) and attempting to solve for \(x\).
A1: For finding \(x = -\frac{1}{6}\) (or equivalent, such as \(-0.167\)).

(a) Given \(y = 8 + 2x - x^2\), differentiating with respect to \(x\) gives \(\frac{dy}{dx} = 2 - 2x\). At the point \(P(3, 5)\), the gradient of the tangent is \(\frac{dy}{dx} = 2 - 2(3) = -4\). Using the equation of a straight line, the tangent is \(y - 5 = -4(x - 3)\), which simplifies to \(y = -4x + 17\).

(b) The region is bounded by the tangent line \(y = -4x + 17\), the curve \(y = 8 + 2x - x^2\), and the y-axis (\(x = 0\)), up to the point of tangency at \(x = 3\). Since the curve is concave down, the tangent lies above the curve in this interval. The area of the region is given by \(\int_{0}^{3} [(-4x + 17) - (8 + 2x - x^2)] dx = \int_{0}^{3} (x^2 - 6x + 9) dx = \left[ \frac{x^3}{3} - 3x^2 + 9x \right]_{0}^{3} = \left( \frac{27}{3} - 27 + 27 \right) - 0 = 9\).

(a) M1: For differentiating to get \(\frac{dy}{dx} = 2 - 2x\).
A1: For correct gradient of \(-4\) at \(x = 3\).
M1: For using \(y - y_1 = m(x - x_1)\) with their gradient and point \(P\).
A1: For obtaining \(y = -4x + 17\) or equivalent.

(b) M1: For setting up the integral \(\int (y_{\text{line}} - y_{\text{curve}}) dx\).
A1: For correct integrand \(x^2 - 6x + 9\) (or negative if subtracted in reverse order, but must resolve to positive area).
M1: For integrating their 3-term quadratic expression, at least two terms correct.
A1: For correct integration: \(\frac{x^3}{3} - 3x^2 + 9x\).
M1: For substituting limits \(3\) and \(0\) into their integrated expression.
A1: For showing all steps clearly to arrive at the given area of 9.

(a) LHS \(= \frac{\cos x(1 + \sin x) - \cos x(1 - \sin x)}{(1 - \sin x)(1 + \sin x)} = \frac{\cos x + \cos x \sin x - \cos x + \cos x \sin x}{1 - \sin^2 x} = \frac{2 \cos x \sin x}{\cos^2 x} = 2 \left(\frac{\sin x}{\cos x}\right) = 2 \tan x = \text{RHS}\).

(b) Let \(\alpha = 2\theta - 10^\circ\). The equation becomes \(2 \tan \alpha = 4 \sin \alpha\). This simplifies to \(\frac{2 \sin \alpha}{\cos \alpha} - 4 \sin \alpha = 0 \implies 2 \sin \alpha \left(\frac{1}{\cos \alpha} - 2\right) = 0\). This gives two cases:
1) \(\sin \alpha = 0\)
2) \(\cos \alpha = 0.5\).
Since \(0^\circ \le \theta \le 180^\circ\), we have \(0^\circ \le 2\theta \le 360^\circ \implies -10^\circ \le 2\theta - 10^\circ \le 350^\circ\), so \(-10^\circ \le \alpha \le 350^\circ\).
For Case 1: \(\sin \alpha = 0 \implies \alpha = 0^\circ, 180^\circ\).
- \(2\theta - 10^\circ = 0^\circ \implies \theta = 5^\circ\)
- \(2\theta - 10^\circ = 180^\circ \implies \theta = 95^\circ\).
For Case 2: \(\cos \alpha = 0.5 \implies \alpha = 60^\circ, 300^\circ\).
- \(2\theta - 10^\circ = 60^\circ \implies \theta = 35^\circ\)
- \(2\theta - 10^\circ = 300^\circ \implies \theta = 155^\circ\).
Thus, the solutions are \(\theta = 5^\circ, 35^\circ, 95^\circ, 155^\circ\).

(a) M1: For writing over a common denominator \((1 - \sin x)(1 + \sin x)\) or \(1 - \sin^2 x\).
A1: For correct expansion of numerator to get \(2 \cos x \sin x\).
M1: For using \(1 - \sin^2 x = \cos^2 x\) in the denominator and simplifying.
A1: For fully correct proof showing all steps.

(b) M1: For recognizing the identity and rewriting LHS as \(2 \tan(2\theta - 10^\circ)\).
M1: For separating into two trigonometric equations: \(\sin(2\theta - 10^\circ) = 0\) and \(\cos(2\theta - 10^\circ) = 0.5\).
A1: For finding either set of values for \(2\theta - 10^\circ\) (e.g. \(0^\circ, 180^\circ\) or \(60^\circ, 300^\circ\)).
M1: For a correct attempt to find values for \(\theta\) from their values for \(\alpha\).
A1: For any two correct values of \(\theta\).
A2: For all four correct values of \(\theta\) (\(5^\circ, 35^\circ, 95^\circ, 155^\circ\)) and no extra values in range.

(a) The equation of a circle with center \((h, k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). Substituting the center \((3, -1)\) and radius \(5\) gives \((x - 3)^2 + (y + 1)^2 = 25\).

(b) To find the intersection points, substitute \(y = 2x - 2\) into the circle's equation:
\((x - 3)^2 + (2x - 2 + 1)^2 = 25 \implies (x - 3)^2 + (2x - 1)^2 = 25\)
\(x^2 - 6x + 9 + 4x^2 - 4x + 1 = 25 \implies 5x^2 - 10x + 10 = 25\)
\(5x^2 - 10x - 15 = 0 \implies x^2 - 2x - 3 = 0\)
\((x - 3)(x + 1) = 0 \implies x = 3\) or \(x = -1\).
Substitute \(x\) values back into the line equation \(y = 2x - 2\):
- When \(x = 3\), \(y = 2(3) - 2 = 4\), giving the point \((3, 4)\).
- When \(x = -1\), \(y = 2(-1) - 2 = -4\), giving the point \((-1, -4)\).
Thus, the coordinates are \((3, 4)\) and \((-1, -4)\).

(c) The exact length of chord \(AB\) is the distance between \((3, 4)\) and \((-1, -4)\):
Length \(= \sqrt{(3 - (-1))^2 + (4 - (-4))^2} = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}\).

(a) B1: For LHS correct: \((x - 3)^2 + (y + 1)^2\).
B1: For RHS correct: \(25\).

(b) M1: For substituting \(y = 2x - 2\) into their circle equation.
A1: For obtaining a correct 3-term quadratic equation, e.g., \(5x^2 - 10x - 15 = 0\) or \(x^2 - 2x - 3 = 0\).
M1: For a valid method to solve their 3-term quadratic equation (factorisation or quadratic formula).
A1: For \(x = 3\) and \(x = -1\).
M1: For substituting their \(x\) values into the linear equation to find corresponding \(y\) values.
A1: For both points \((3, 4)\) and \((-1, -4)\) correctly paired.

(c) M1: For using the distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) with their coordinates for \(A\) and \(B\).
A1: For \(\sqrt{80}\) or equivalent exact unsimplified form.
A1: For simplifying to the exact form \(4\sqrt{5}\).

(a) Total people = \(6 + 8 = 14\). Choosing 5 people with no restrictions is \(\binom{14}{5} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002\).

(b) The committee must have at least 3 professional photographers (P) out of 5 members, so the remaining are amateur photographers (A):
- Case 1: 3 P and 2 A \(\implies \binom{6}{3} \times \binom{8}{2} = 20 \times 28 = 560\).
- Case 2: 4 P and 1 A \(\implies \binom{6}{4} \times \binom{8}{1} = 15 \times 8 = 120\).
- Case 3: 5 P and 0 A \(\implies \binom{6}{5} \times \binom{8}{0} = 6 \times 1 = 6\).
Total combinations \(= 560 + 120 + 6 = 686\).

(c) (i) There are 3 P and 2 A. Treat the 3 P as a single block. This gives \(1 \text{ block} + 2 \text{ amateurs} = 3\) entities to arrange in a line, which can be done in \(3! = 6\) ways. Within the block, the 3 P can be arranged in \(3! = 6\) ways. Total ways \(= 6 \times 6 = 36\).

(ii) There are 4 P and 1 A. Treat the 4 P as a single block. This gives \(1 \text{ block} + 1 \text{ amateur} = 2\) entities to arrange in a line, which can be done in \(2! = 2\) ways. Within the block, the 4 P can be arranged in \(4! = 24\) ways. Total ways \(= 2 \times 24 = 48\).

(a) M1: For writing or calculating \(\binom{14}{5}\).
A1: For 2002.

(b) M1: For identifying the three cases: 3P/2A, 4P/1A, and 5P/0A.
M1: For at least two correct product terms (e.g., \(\binom{6}{3} \times \binom{8}{2}\) and \(\binom{6}{4} \times \binom{8}{1}\)).
A1: For obtaining the correct counts for the individual cases: 560, 120, and 6.
A1: For the correct sum of 686.

(c)(i) M1: For treating the 3 professionals as a single unit, yielding 3 units to arrange (\(3!\)).
M1: For multiplying by the internal arrangements of professionals (\(3!\)).
A1: For 36.

(c)(ii) M1: For arranging 2 units (\(2!\)) and multiplying by the internal arrangements of 4 professionals (\(4!\)).
A1: For 48.

(a) The terms of the arithmetic progression are \(T_1 = a\), \(T_5 = a + 4d\), and \(T_{13} = a + 12d\). The terms of the geometric progression are \(G_1 = a\), \(G_2 = ar\), and \(G_3 = ar^2\). Since \(T_1 = G_1 = a\), we have:
\(T_5 = G_2 \implies a + 4d = ar \implies r = \frac{a+4d}{a} = 1 + \frac{4d}{a}\).
Also, \(T_{13} = G_3 \implies a + 12d = ar^2\).
Substituting \(r\) into this equation:
\(a + 12d = a \left(1 + \frac{4d}{a}\right)^2 \implies a + 12d = a \left(1 + \frac{8d}{a} + \frac{16d^2}{a^2}\right)\)
\(a + 12d = a + 8d + \frac{16d^2}{a}\)
Subtracting \(a + 8d\) from both sides:
\(4d = \frac{16d^2}{a}\).
Since \(d \ne 0\), we can divide by \(4d\):
\(1 = \frac{4d}{a} \implies d = \frac{1}{4}a\).

(b) Substituting \(d = \frac{1}{4}a\) into the expression for \(r\):
\(r = 1 + \frac{4(\frac{1}{4}a)}{a} = 1 + 1 = 2\).

(c) The 10th term of the arithmetic progression is \(T_{10} = a + 9d = 39\).
Substituting \(d = \frac{1}{4}a\):
\(a + 9\left(\frac{1}{4}a\right) = 39 \implies \frac{13}{4}a = 39 \implies a = 39 \times \frac{4}{13} = 12\).
So, the geometric progression has first term \(a = 12\) and common ratio \(r = 2\).
The sum of the first 5 terms of the geometric progression is:
\(S_5 = \frac{a(r^5 - 1)}{r - 1} = \frac{12(2^5 - 1)}{2 - 1} = 12(32 - 1) = 12 \times 31 = 372\).

(a) B1: For expressing terms of AP in terms of \(a\) and \(d\) (\(a + 4d\) and \(a + 12d\)).
M1: For equating to GP terms to write equations in \(a, d, r\) (e.g., \(a+4d=ar\) and \(a+12d=ar^2\)).
M1: For eliminating \(r\) to obtain an equation in \(a\) and \(d\).
A1: For simplifying to \(4d = \frac{16d^2}{a}\) or equivalent.
A1: For dividing by \(d\) (justified by \(d \ne 0\)) and concluding \(d = \frac{1}{4}a\).

(b) M1: For substituting their relationship between \(a\) and \(d\) into their expression for \(r\).
A1: For showing the work to evaluate \(r\).
A1: For \(r = 2\).

(c) M1: For using \(a + 9d = 39\) with their relationship from part (a) to find \(a\) (and \(d\)).
A1: For finding \(a = 12\).
A1: For using the GP sum formula with \(a = 12\), \(r = 2\), and \(n = 5\) to find \(S_5 = 372\).