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

First, find the indefinite integral: \(\int \left(2x - x^{-2}\right) \text{d}x = x^2 - \frac{x^{-1}}{-1} + c = x^2 + \frac{1}{x} + c\). Now, evaluate the definite integral with limits 1 and 2: \(\left[ x^2 + \frac{1}{x} \right]_{1}^{2} = \left(2^2 + \frac{1}{2}\right) - \left(1^2 + \frac{1}{1}\right) = \left(4 + \frac{1}{2}\right) - (1 + 1) = \frac{9}{2} - 2 = \frac{5}{2} = 2.5\).

M1: For integrating to obtain \(x^2 + \frac{1}{x}\) (allow one error in sign or coefficient). M1: For substituting limits 1 and 2 correctly into their integrated expression and subtracting. A1: For a correct final answer of \(\frac{5}{2}\) or 2.5.

Using the laws of logarithms: \(\log_3((x + 4)(x - 2)) = 3\). Rewrite in exponential form: \((x + 4)(x - 2) = 3^3\). Expand and simplify: \(x^2 + 2x - 8 = 27 \implies x^2 + 2x - 35 = 0\). Factorise the quadratic equation: \((x + 7)(x - 5) = 0\), which gives \(x = -7\) or \(x = 5\). Since the arguments of the logarithms must be positive, we must have \(x > 2\). Therefore, \(x = -7\) is rejected, leaving the single solution \(x = 5\).

M1: For using logarithm laws to combine terms to get \(\log_3((x+4)(x-2)) = 3\). M1: For converting to quadratic form \(x^2 + 2x - 35 = 0\) and solving. A1: For \(x = 5\) only (must explicitly or implicitly reject \(x = -7\)).

The magnitude of a vector \(\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}\) is given by \(|\mathbf{a}| = \sqrt{x^2 + y^2}\). Thus, \(\sqrt{(k-1)^2 + 5^2} = \sqrt{74}\). Square both sides: \((k-1)^2 + 25 = 74 \implies (k-1)^2 = 49\). Taking square roots: \(k-1 = \pm 7\). This gives two possible values: \(k = 8\) or \(k = -6\). Since \(k\) is specified to be a negative constant, we have \(k = -6\).

M1: For setting up the magnitude equation \((k-1)^2 + 25 = 74\). M1: For solving to find \(k - 1 = \pm 7\) (or expanding and solving \(k^2 - 2k - 48 = 0\)). A1: For \(k = -6\) only (rejecting \(k = 8\)).

Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\) into the equation: \(2(1 - \cos^2 \theta) - \cos \theta - 1 = 0\). Simplify: \(2 - 2\cos^2 \theta - \cos \theta - 1 = 0 \implies 2\cos^2 \theta + \cos \theta - 1 = 0\). Factorise the quadratic equation in terms of \(\cos \theta\): \((2\cos \theta - 1)(\cos \theta + 1) = 0\). This yields: \(\cos \theta = \frac{1}{2}\) or \(\cos \theta = -1\). For the interval \(0^\circ \le \theta \le 180^\circ\): From \(\cos \theta = \frac{1}{2}\), \(\theta = 60^\circ\). From \(\cos \theta = -1\), \(\theta = 180^\circ\).

M1: For using \(\sin^2 \theta = 1 - \cos^2 \theta\) to obtain a quadratic in \(\cos \theta\). M1: For solving the quadratic to find \(\cos \theta = \frac{1}{2}\) or \(\cos \theta = -1\). A1: For both \(\theta = 60^\circ\) and \(\theta = 180^\circ\) and no other solutions in the range.

The sum to infinity is given by \(S_{\infty} = \frac{a}{1-r}\). Substituting the given values: \(8 = \frac{12}{1-r} \implies 8(1-r) = 12 \implies 1-r = 1.5 \implies r = -0.5\). The third term of a geometric progression is given by \(u_3 = a r^2\). Substituting \(a = 12\) and \(r = -0.5\): \(u_3 = 12 \times (-0.5)^2 = 12 \times 0.25 = 3\).

M1: For using \(S_{\infty} = \frac{a}{1-r}\) to form an equation in \(r\). A1: For finding the correct common ratio \(r = -0.5\) (or \(-\frac{1}{2}\)). A1: For finding the correct third term \(3\).

For a quadratic equation to have no real roots, the discriminant must be negative (\(b^2 - 4ac < 0\)). Here, \(a = 1\), \(b = k+3\), and \(c = 2k+6\). So, \((k+3)^2 - 4(1)(2k+6) < 0 \implies k^2 + 6k + 9 - 8k - 24 < 0 \implies k^2 - 2k - 15 < 0\). Factorising the quadratic inequality: \((k-5)(k+3) < 0\). The critical values are \(k = 5\) and \(k = -3\). Since we want the expression to be less than zero, the set of values is \(-3 < k < 5\).

M1: For setting up the inequality \((k+3)^2 - 4(2k+6) < 0\). M1: For simplifying to \(k^2 - 2k - 15 < 0\) and finding critical values \(k = 5\) and \(k = -3\). A1: For the correct inequality range \(-3 < k < 5\).

Let the 5-digit number be represented by five slots. The first digit must be even. The available even digits are \(\{2, 4, 6\}\), giving 3 choices. The last digit must be odd to make the overall number odd. The available odd digits are \(\{1, 3, 5, 7\}\), giving 4 choices. Since the sets of even and odd digits are completely separate, we can choose the first and last digits independently. This leaves \(7 - 2 = 5\) unused digits for the middle 3 positions. The number of ways to choose and arrange 3 digits from 5 is \(P(5, 3) = 5 \times 4 \times 3 = 60\). The total number of different 5-digit numbers is \(3 \times 60 \times 4 = 720\).

M1: For identifying the number of options for the first digit (3) and the last digit (4), and multiplying them. M1: For calculating the permutations of the remaining digits, \(5 \times 4 \times 3\) or \(P(5, 3)\). A1: For the correct final answer of 720.

(i) Using the product rule on \(y = (2x - 3)e^{1-x}\), we let \(u = 2x - 3\) and \(v = e^{1-x}\). This gives \(u' = 2\) and \(v' = -e^{1-x}\). Thus, \(\frac{dy}{dx} = 2e^{1-x} - (2x-3)e^{1-x} = (5-2x)e^{1-x}\). Setting \(\frac{dy}{dx} = 0\) yields \(5-2x = 0\), which gives \(x = 2.5\). Substituting \(x = 2.5\) into the equation of the curve gives \(y = (2(2.5) - 3)e^{1-2.5} = 2e^{-1.5}\). Hence, the stationary point is indeed \((2.5, 2e^{-1.5})\). (ii) To find the second derivative, we apply the product rule to \(\frac{dy}{dx} = (5-2x)e^{1-x}\), giving \(\frac{d^2y}{dx^2} = -2e^{1-x} - (5-2x)e^{1-x} = (2x-7)e^{1-x}\). Substituting \(x = 2.5\) yields \(\frac{d^2y}{dx^2} = (2(2.5)-7)e^{1-2.5} = -2e^{-1.5}\). Since \(-2e^{-1.5} < 0\), the stationary point is a local maximum. (iii) To find the definite integral, we integrate by parts: \(\int u \, dv = uv - \int v \, du\). Let \(u = 2x-3\) and \(dv = e^{1-x} \, dx\). This gives \(du = 2 \, dx\) and \(v = -e^{1-x}\). Thus, \(\int (2x-3)e^{1-x} \, dx = -(2x-3)e^{1-x} - \int -2e^{1-x} \, dx = (3-2x)e^{1-x} - 2e^{1-x} + c = (1-2x)e^{1-x} + c\). Evaluating this over the limits 2 to 3 gives \([(1-2x)e^{1-x}]_2^3 = -5e^{-2} - (-3e^{-1}) = 3e^{-1} - 5e^{-2}\).

(i) M1 for correct product rule application. A1 for finding the correct first derivative. M1 for setting the first derivative to 0 and solving for x. A1 for substituting back to get y. A1 for writing down the full coordinates. (ii) M1 for finding the second derivative. A1 for evaluating it at x = 2.5. A0.8 for correctly stating it is a maximum with reasoning. (iii) M1 for setting up integration by parts. A1 for the correct expression of the indefinite integral. M1 for correctly substituting limits 2 and 3. A1 for the simplified exact final answer.

(a) We use the double angle formula \(\cos 2\theta = 2\cos^2 \theta - 1\). Substituting this into the equation gives \(2\cos^2 \theta - 1 + 3\cos \theta + 2 = 0\), which simplifies to \(2\cos^2 \theta + 3\cos \theta + 1 = 0\). Factoring this quadratic equation gives \((2\cos \theta + 1)(\cos \theta + 1) = 0\). This yields two cases: \(\cos \theta = -1/2\) or \(\cos \theta = -1\). For the interval \(0 \le \theta \le 2\pi\), \(\cos \theta = -1/2\) gives \(\theta = 2\pi/3\) and \(\theta = 4\pi/3\), and \(\cos \theta = -1\) gives \(\theta = \pi\). Thus, the solutions are \(\theta = 2\pi/3, \pi, 4\pi/3\). (b) We use the identity \(\tan^2 \phi = \sec^2 \phi - 1\). Substituting this into the equation gives \(3(\sec^2 \phi - 1) - 5\sec \phi + 1 = 0\), which simplifies to \(3\sec^2 \phi - 5\sec \phi - 2 = 0\). Factoring this quadratic equation gives \((3\sec \phi + 1)(\sec \phi - 2) = 0\). This gives two cases: \(\sec \phi = -1/3\) or \(\sec \phi = 2\). If \(\sec \phi = -1/3\), then \(\cos \phi = -3\), which has no real solution since the range of cosine is between -1 and 1. If \(\sec \phi = 2\), then \(\cos \phi = 1/2\). For the interval \(0 \le \phi \le 2\pi\), this gives \(\phi = \pi/3\) and \(\phi = 5\pi/3\).

(a) M1 for using the double angle formula correctly. A1 for the correct quadratic equation in terms of cos. M1 for factoring the quadratic. A1 for finding both cos values. A1 for the solutions 2pi/3 and 4pi/3. A1 for the solution pi. (b) M1 for using the Pythagorean identity correctly. A1 for the correct quadratic equation in terms of sec. M1 for factoring the quadratic. A1 for identifying sec = 2 (and rejecting sec = -1/3). A1.8 for finding both correct angles pi/3 and 5pi/3.

(a) The 3rd term is \(a + 2d = 14\). The sum of the first 12 terms is \(S_{12} = \frac{12}{2}(2a + 11d) = 6(2a + 11d) = 378\), which simplifies to \(2a + 11d = 63\). We have a system of simultaneous equations: 1) \(a + 2d = 14\) and 2) \(2a + 11d = 63\). From (1), we have \(2a + 4d = 28\). Subtracting this from (2) gives \(7d = 35\), so \(d = 5\). Substituting \(d = 5\) into (1) gives \(a + 2(5) = 14\), so \(a = 4\). (b)(i) The 1st term of the AP is \(T_1 = a = 4\). The 3rd term is \(T_3 = a + 2d = 14\). The \(k\)-th term is \(T_k = a + (k-1)d = 4 + 5(k-1) = 5k - 1\). Since these three terms form a geometric progression, the common ratio must be equal: \(\frac{14}{4} = \frac{5k-1}{14}\). This gives \(14^2 = 4(5k-1)\), which simplifies to \(196 = 20k - 4\). Thus, \(20k = 200\), so \(k = 10\). (b)(ii) The geometric progression has first term \(A = \frac{1}{a} = \frac{1}{4}\) and common ratio \(R = \frac{1}{d} = \frac{1}{5}\). Since \(|R| = 1/5 < 1\), the sum to infinity exists and is given by \(S_{\infty} = \frac{A}{1-R} = \frac{1/4}{1 - 1/5} = \frac{1/4}{4/5} = \frac{5}{16}\).

(a) M1 for writing down the correct expression for the 3rd term. M1 for writing down the correct expression for the sum of 12 terms. M1 for setting up and solving the simultaneous equations. A1 for obtaining both correct values of a and d. (b)(i) M1 for finding the expression of the k-th term in terms of k. M1 for setting up the ratio equation for geometric progression. M1 for clearing the fraction to form a linear equation in k. A2 for finding the correct value of k = 10. (b)(ii) M1 for substituting the correct values of a and d into the formula for the sum to infinity. A1.8 for the correct simplified fraction 5/16.

(a)(i) Since \(P\) lies on \(OA\) in the ratio \(2:1\), we have \(\overrightarrow{OP} = \frac{2}{3}\overrightarrow{OA} = \frac{2}{3}\mathbf{a}\). (a)(ii) Since \(Q\) lies on \(AB\) in the ratio \(3:2\), we have \(\overrightarrow{AQ} = \frac{3}{5}\overrightarrow{AB} = \frac{3}{5}(\mathbf{b} - \mathbf{a})\). Thus, \(\overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AQ} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a}) = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\). (a)(iii) Using vector subtraction, we have \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \left(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) - \frac{2}{3}\mathbf{a} = \left(\frac{2}{5} - \frac{2}{3}\right)\mathbf{a} + \frac{3}{5}\mathbf{b} = -\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b}\). (b) We express \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\): \(\overrightarrow{OM} = \overrightarrow{OP} + \overrightarrow{PM} = \frac{2}{3}\mathbf{a} + k\overrightarrow{PQ} = \frac{2}{3}\mathbf{a} + k\left(-\frac{4}{15}\mathbf{a} + \frac{3}{5}\mathbf{b}\right) = \left(\frac{2}{3} - \frac{4k}{15}\right)\mathbf{a} + \frac{3k}{5}\mathbf{b}\常规. Since we are given \)\overrightarrow{OM} = m\mathbf{b}\), the coefficient of \(\mathbf{a}\) must be 0: \(\frac{2}{3} - \frac{4k}{15} = 0\), which gives \(\frac{4k}{15} = \frac{2}{3}\), so \(4k = 10\) and hence \(k = 2.5\). Substituting \(k = 2.5\) into the coefficient of \(\mathbf{b}\) gives \(m = \frac{3(2.5)}{5} = 1.5\).

(a)(i) B1 for the correct vector. (a)(ii) M1 for expressing OQ using OA and AB. A1 for correct substitution of vectors. A1 for simplified correct expression. (a)(iii) M1 for correct vector subtraction formula. A1 for simplified correct expression. (b) M1 for writing OM as OP + kPQ. A1 for substituting the vectors. M1 for grouping the coefficients of a and b. M1 for setting the coefficient of a to 0. A1 for finding k = 2.5. A0.8 for finding m = 1.5.

(a) To find \(\mathrm{g}^{-1}(x)\), let \(y = \frac{2x+5}{x-1}\). Then \(y(x-1) = 2x+5\), which gives \(yx - y = 2x + 5\). Rearranging terms to make \(x\) the subject: \(yx - 2x = y + 5 \implies x(y-2) = y + 5\), so \(x = \frac{y+5}{y-2}\). Thus, \(\mathrm{g}^{-1}(x) = \frac{x+5}{x-2}\). The domain of \(\mathrm{g}^{-1}(x)\) is the range of \(\mathrm{g}(x)\), which is \(x \ne 2\). (b) To solve \(\mathrm{fg}(x) = 13\), we substitute \(\mathrm{g}(x)\) into \(\mathrm{f}(x)\): \(\mathrm{f}(\mathrm{g}(x)) = 3\left(\frac{2x+5}{x-1}\right) - 2 = 13\). This simplifies to \(3\left(\frac{2x+5}{x-1}\right) = 15 \implies \frac{2x+5}{x-1} = 5\). Thus, \(2x + 5 = 5(x-1) \implies 2x + 5 = 5x - 5 \implies 3x = 10 \implies x = 10/3\). (c) Setting \(\mathrm{g}(x) = x\) gives \(\frac{2x+5}{x-1} = x\), which simplifies to \(2x + 5 = x(x-1) \implies 2x + 5 = x^2 - x\). This yields the quadratic equation \(x^2 - 3x - 5 = 0\). The discriminant of this equation is \(\Delta = (-3)^2 - 4(1)(-5) = 9 + 20 = 29\). Since \(\Delta > 0\), the equation has two real roots. Using the quadratic formula, the roots are given by \(x = \frac{-(-3) \pm \sqrt{29}}{2} = \frac{3 \pm \sqrt{29}}{2}\). Here, \(a = 3\) and \(b = 29\) are indeed integers.

(a) M1 for setting y = g(x) and rearranging to multiply by the denominator. M1 for collecting x terms together on one side. A1 for the correct expression of the inverse function. A1 for stating the correct domain. (b) M1 for substituting g(x) into f(x) correctly. A1 for simplifying the equation to a linear equation. A1.8 for the correct solution x = 10/3. (c) M1 for setting up the equation g(x) = x and obtaining the correct quadratic equation. M1 for evaluating the discriminant to prove there are two real roots. A1 for applying the quadratic formula correctly. A1 for obtaining the final exact values.

First, find the coordinate of the point where \(x = 1\):
\(y = 3\ln(2(1) - 1) = 3\ln(1) = 0\).

Next, differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = 3 \cdot \frac{2}{2x - 1} = \frac{6}{2x - 1}\).

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

The gradient of the normal, \(m_n\), is the negative reciprocal of the tangent gradient:
\(m_n = -\frac{1}{6}\).

Use the point-slope form to find the equation of the normal:
\(y - 0 = -\frac{1}{6}(x - 1) \implies y = -\frac{1}{6}x + \frac{1}{6}\).

M1 for correctly differentiating the function to get \(\frac{dy}{dx} = \frac{6}{2x - 1}\).
M1 for finding the gradient of the normal, \(m_n = -\frac{1}{6}\).
A1.25 for the correct final equation \(y = -\frac{1}{6}x + \frac{1}{6}\) or equivalent.

Use the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to rewrite the equation:
\(4(1 - \cos^2 \theta) + 5 \cos \theta - 5 = 0\)
\(4 - 4\cos^2 \theta + 5\cos \theta - 5 = 0\)
\(-4\cos^2 \theta + 5\cos \theta - 1 = 0\)

Multiply by \(-1\):
\(4\cos^2 \theta - 5\cos \theta + 1 = 0\).

Factorise the quadratic equation:
\((4\cos \theta - 1)(\cos \theta - 1) = 0\).

This gives two cases:
1) \(\cos \theta = 1 \implies \theta = 0^\circ, 360^\circ\)
2) \(\cos \theta = \frac{1}{4} \implies \theta = \cos^{-1}(0.25) \approx 75.5^\circ\), and \(\theta = 360^\circ - 75.5^\circ = 284.5^\circ\).

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

M1 for applying the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to obtain a quadratic equation in terms of \(\cos \theta\).
M1 for factorising or solving the quadratic equation to find \(\cos \theta = \frac{1}{4}\) and \(\cos \theta = 1\).
A1.25 for all four correct angles: \(0^\circ, 75.5^\circ, 284.5^\circ, 360^\circ\) (accept answers rounded to 1 decimal place; deduct 0.5 marks for each missing or extra solution outside the range).

Since the two vectors are perpendicular, their scalar (dot) product is equal to 0:
\(\mathbf{a} \cdot \mathbf{b} = 0\)
\(k(k - 9) + 4(5) = 0\)
\(k^2 - 9k + 20 = 0\).

Factorise the quadratic equation:
\((k - 4)(k - 5) = 0\).

Thus, the possible values of \(k\) are:
\(k = 4\) or \(k = 5\).

M1 for setting up the scalar product equation \(\mathbf{a} \cdot \mathbf{b} = 0\).
M1 for expanding and formulating the quadratic equation \(k^2 - 9k + 20 = 0\).
A1.25 for obtaining the correct values \(k = 4\) and \(k = 5\).

Let the first term be \(a\) and the common ratio be \(r\).

The sum of the first three terms is given by:
\(a(1 + r + r^2) = 35\)

The sum to infinity is given by:
\(\frac{a}{1 - r} = 40 \implies a = 40(1 - r)\)

Substitute \(a = 40(1 - r)\) into the first equation:
\(40(1 - r)(1 + r + r^2) = 35\)

Since \((1 - r)(1 + r + r^2) = 1 - r^3\), we have:
\(40(1 - r^3) = 35\)
\(1 - r^3 = \frac{35}{40} = \frac{7}{8}\)
\(r^3 = 1 - \frac{7}{8} = \frac{1}{8}\).

Taking the cube root of both sides:
\(r = \frac{1}{2}\).

Substitute \(r = \frac{1}{2}\) back to find \(a\):
\(a = 40\left(1 - \frac{1}{2}\right) = 20\).

M1 for setting up the equations for the sum of three terms and the sum to infinity.
M1 for substituting \(a\) and reducing to a simplified equation in \(r\) (e.g., \(r^3 = \frac{1}{8}\)).
A1.25 for both correct values: \(a = 20\) and \(r = \frac{1}{2}\) (or \(r = 0.5\)).

To have more women than men on a team of 5, the combination of (women, men) can be:

1) 3 women and 2 men:
Number of ways = \(\binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150\)

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

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

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

M1 for identifying the three valid scenarios: (3W, 2M), (4W, 1M), and (5W, 0M).
M1 for calculating the number of combinations for at least two of the scenarios correctly.
A1.25 for the correct total of 181.

First, find the expression for \(\mathrm{f}\mathrm{g}(x)\):
\(\mathrm{f}(\mathrm{g}(x)) = \mathrm{f}\left(\frac{x+2}{5}\right) = 3\left(\frac{x+2}{5}\right) - 1 = \frac{3x + 6 - 5}{5} = \frac{3x + 1}{5}\).

Next, find the inverse function \(\mathrm{g}^{-1}(x)\):
Let \(y = \frac{x+2}{5} \implies 5y = x + 2 \implies x = 5y - 2\).
So, \(\mathrm{g}^{-1}(x) = 5x - 2\).

Now set \(\mathrm{f}\mathrm{g}(x) = \mathrm{g}^{-1}(x)\):
\(\frac{3x + 1}{5} = 5x - 2\).

Multiply by 5:
\(3x + 1 = 25x - 10\).

Rearrange to solve for \(x\):
\(22x = 11 \implies x = 0.5\).

M1 for finding the composite function \(\mathrm{f}\mathrm{g}(x) = \frac{3x+1}{5}\) or the inverse function \(\mathrm{g}^{-1}(x) = 5x - 2\).
M1 for equating \(\mathrm{f}\mathrm{g}(x) = \mathrm{g}^{-1}(x)\) and forming a linear equation.
A1.25 for the correct value \(x = 0.5\) (or \(x = \frac{1}{2}\)).

Rewrite the equation using indices rules:
\(3 \cdot (3^x)^2 - 10(3^x) + 3 = 0\).

Let \(u = 3^x\). The equation becomes:
\(3u^2 - 10u + 3 = 0\).

Factorise the quadratic equation:
\((3u - 1)(u - 3) = 0\).

This gives:
\(u = \frac{1}{3}\) or \(u = 3\).

Now substitute back \(u = 3^x\):
1) \(3^x = \frac{1}{3} = 3^{-1} \implies x = -1\)
2) \(3^x = 3^1 \implies x = 1\).

Therefore, the solutions are \(x = -1\) or \(x = 1\).

M1 for expressing the equation as a quadratic in terms of \(3^x\), e.g., \(3u^2 - 10u + 3 = 0\).
M1 for solving the quadratic equation to find \(3^x = \frac{1}{3}\) and \(3^x = 3\).
A1.25 for obtaining both correct answers \(x = -1\) and \(x = 1\).

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

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

Substitute \(\theta\) into the area equation:
\(\frac{1}{2}r^2 \left(\frac{24 - 2r}{r}\right) = 32\)
\(\frac{1}{2}r(24 - 2r) = 32\)
\(12r - r^2 = 32\).

Rearrange into a standard quadratic equation:
\(r^2 - 12r + 32 = 0\).

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

Thus, the possible values of \(r\) are \(r = 4\) or \(r = 8\).

M1 for writing down equations for both perimeter and area of the sector.
M1 for substituting one equation into the other to form a quadratic equation in \(r\).
A1.25 for finding both correct values of \(r = 4\) and \(r = 8\).

(a) Using the product rule with \(u = x - 2\) and \(v = (2x + 5)^{1/2}\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 1\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(2x + 5)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x + 5}}\).
Using \(\frac{\mathrm{d}y}{\mathrm{d}x} = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x-2}{\sqrt{2x+5}} + \sqrt{2x+5}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x-2) + (2x+5)}{\sqrt{2x+5}} = \frac{3x+3}{\sqrt{2x+5}}\).

(b) For a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\).
\(\frac{3x+3}{\sqrt{2x+5}} = 0 \Rightarrow 3x + 3 = 0 \Rightarrow x = -1\).
When \(x = -1\):
\(y = (-1 - 2)\sqrt{2(-1) + 5} = -3\sqrt{3}\).
The stationary point is \((-1, -3\sqrt{3})\).

(c) The curve crosses the positive \(x\)-axis when \(y = 0\) and \(x > 0\).
\((x-2)\sqrt{2x+5} = 0 \Rightarrow x = 2\).
So point \(P\) is \((2, 0)\).
At \(x = 2\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3(2)+3}{\sqrt{2(2)+5}} = \frac{9}{\sqrt{9}} = 3\).
The equation of the tangent is:
\(y - 0 = 3(x - 2) \Rightarrow y = 3x - 6\).

(d) From part (a), \(\frac{\mathrm{d}}{\mathrm{d}x} \left[ (x-2)\sqrt{2x+5} \right] = \frac{3(x+1)}{\sqrt{2x+5}}\).
Integrating both sides:
\((x-2)\sqrt{2x+5} = 3 \int \frac{x+1}{\sqrt{2x+5}} \, \mathrm{d}x\).
Thus, \(\int \frac{x+1}{\sqrt{2x+5}} \, \mathrm{d}x = \frac{1}{3}(x-2)\sqrt{2x+5} + C\).

Part (a) [4 marks]:
M1: Attempt to apply product rule to \(y = (x-2)\sqrt{2x+5}\).
A1: Correct derivative of \(\sqrt{2x+5}\), which is \(\frac{1}{\sqrt{2x+5}}\).
M1: Correct algebraic steps to combine terms over a common denominator.
A1: Fully correct completion showing \(\frac{3x+3}{\sqrt{2x+5}}\).

Part (b) [2 marks]:
M1: Setting their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solving for \(x\).
A1: Correct stationary point coordinates \((-1, -3\sqrt{3})\) or exact equivalent.

Part (c) [3 marks]:
B1: Finding the correct crossing point \(P(2,0)\).
M1: Finding the gradient of the tangent at \(x=2\) and attempting to find the line equation.
A1: Correct equation \(y = 3x - 6\) or equivalent.

Part (d) [2 marks]:
M1: Realising that integration is the reverse of differentiation and attempting to divide their part (a) result by 3.
A1: Correct answer \(\frac{1}{3}(x-2)\sqrt{2x+5} + C\) (accept without \(C\)).

(a) In the right-angled triangle \(OPR\):
\(PR = 12\sin\theta\) and \(OR = 12\cos\theta\).
Since the radius of the sector is \(12\text{ cm}\), we have \(OQ = 12\text{ cm}\).
Therefore, \(QR = OQ - OR = 12 - 12\cos\theta = 12(1 - \cos\theta)\).
The arc length \(PQ = r\theta = 12\theta\).
The perimeter of the shaded region is:
\(P = \text{Arc } PQ + QR + RP\)
\(P = 12\theta + 12(1 - \cos\theta) + 12\sin\theta = 12(\theta + \sin\theta + 1 - \cos\theta)\).

(b) Area of shaded region = Area of sector \(OPQ\) - Area of right-angled triangle \(OPR\).
Area of sector \(OPQ = \frac{1}{2} r^2 \theta = \frac{1}{2} (12^2) \left(\frac{\pi}{6}\right) = 12\pi\).
Area of triangle \(OPR = \frac{1}{2} \times OR \times PR = \frac{1}{2} (12\cos\theta)(12\sin\theta) = 72\sin\theta\cos\theta\).
For \(\theta = \frac{\pi}{6}\):
\(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) and \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\).
Area of triangle \(OPR = 72 \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) = 18\sqrt{3}\).
Thus, the exact area of the shaded region is \(12\pi - 18\sqrt{3}\) \(\text{cm}^2\).

(c) For \(\theta = 0.8\):
\(P = 12(0.8 +
\sin(0.8) + 1 - \cos(0.8))\).
Using a calculator in radian mode:
\(\sin(0.8) \approx 0.71736\)
\(\cos(0.8) \approx 0.69671\)
\(P \approx 12(0.8 + 0.71736 + 1 - 0.69671) = 12(1.82065) \approx 21.848\).
To 1 decimal place, the perimeter is \(21.8\) cm.

Part (a) [4 marks]:
B1: Finding \(PR = 12\sin\theta\) or equivalent.
B1: Finding \(OR = 12\cos\theta\) or equivalent.
M1: Subtracting \(OR\) from \(OQ\) to find \(QR = 12 - 12\cos\theta\).
A1: Combining arc length \(12\theta\) and lines to get the given perimeter formula clearly.

Part (b) [4 marks]:
M1: Correct formula for area of sector, \(\frac{1}{2}r^2\theta\), and substitution of \(r=12\), \(\theta = \frac{\pi}{6}\).
M1: Correct expression for area of triangle, e.g. \(\frac{1}{2}(12\cos\theta)(12\sin\theta)\).
M1: Substitution of exact values for \(\sin(\pi/6)\) and \(\cos(\pi/6)\).
A1: Correct exact area \(12\pi - 18\sqrt{3}\) or exact equivalent.

Part (c) [3 marks]:
M1: Substituting \(\theta = 0.8\) into the perimeter formula.
A1: Evaluation of trigonometric terms in radian mode.
A1: Correct answer of \(21.8\) (accept \(21.8\) or \(21.9\) depending on rounding).

(a) Number of ways to choose 6 members from 15 people (7 teachers + 8 students):
\(\binom{15}{6} = 5005\).

(b) Since the committee has 6 members, the options for (teachers, students) are:
- 2 teachers and 4 students:
\(\binom{7}{2} \times \binom{8}{4} = 21 \times 70 = 1470\)
- 3 teachers and 3 students:
\(\binom{7}{3} \times \binom{8}{3} = 35 \times 56 = 1960\)
- 4 teachers and 2 students:
\(\binom{7}{4} \times \binom{8}{2} = 35 \times 28 = 980\)

Total number of ways = \(1470 + 1960 + 980 = 4410\).

(c)(i) There are two possible patterns for alternate seating:
Pattern 1 (T S T S T S): \(3! \times 3! = 6 \times 6 = 36\) ways.
Pattern 2 (S T S T S T): \(3! \times 3! = 36\) ways.
Total arrangements = \(36 + 36 = 72\).

(c)(ii) Total unrestricted seating arrangements for 6 people = \(6! = 720\).
Number of arrangements where Mr. Jones and Sam sit together:
Treat them as one unit. There are 5 units to arrange: \(5! = 120\) ways.
Within the unit, they can be arranged in \(2! = 2\) ways.
Arrangements together = \(120 \times 2 = 240\).
Arrangements not together = \(720 - 240 = 480\).

Part (a) [2 marks]:
M1: Attempting to use combinations \(\binom{15}{6}\).
A1: Correct value 5005.

Part (b) [4 marks]:
M1: Realising three cases (2T/4S, 3T/3S, 4T/2S) and attempting combinations for at least one case.
M1: Correct product of combinations for all three cases.
M1: Summing the three cases.
A1: Correct answer 4410.

Part (c)(i) [2 marks]:
M1: Realising there are 2 patterns and attempting to calculate \(3! \times 3! \times 2\).
A1: Correct answer 72.

Part (c)(ii) [3 marks]:
M1: Calculating total unrestricted arrangements as \(6! = 720\).
M1: Attempting to calculate the together arrangements as \(2 \times 5! = 240\) and subtracting from total.
A1: Correct answer 480.

(a) LHS = \(\frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta}\)
Express with a common denominator:
LHS = \(\frac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)}\)
LHS = \(\frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1 + \cos\theta)}\)
Since \(\sin^2\theta + \cos^2\theta = 1\):
LHS = \(\frac{1 + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)}\)
LHS = \(\frac{2 + 2\cos\theta}{\sin\theta(1 + \cos\theta)}\)
LHS = \(\frac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)}\)
LHS = \(\frac{2}{\sin\theta} = 2\csc\theta\) = RHS.

(b) \(3\tan^2 x - 5\sec x + 1 = 0\)
Using the identity \(\tan^2 x = \sec^2 x - 1\):
\(3(\sec^2 x - 1) - 5\sec x + 1 = 0\)
\(3\sec^2 x - 5\sec x - 2 = 0\).
Let \(u = \sec x\):
\(3u^2 - 5u - 2 = 0 \Rightarrow (3u + 1)(u - 2) = 0\).
So, \(\sec x = -\frac{1}{3}\) or \(\sec x = 2\).

Case 1: \(\sec x = -\frac{1}{3} \Rightarrow \cos x = -3\).
Since \(-1 \le \cos x \le 1\), there are no real solutions for this case.

Case 2: \(\sec x = 2 \Rightarrow \cos x = \frac{1}{2}\).
For \(0^\circ \le x \le 360^\circ\):
\(x = 60^\circ\) or \(x = 360^\circ - 60^\circ = 300^\circ\).

Part (a) [4 marks]:
M1: Attempt to write over a common denominator.
M1: Expand \((1 + \cos\theta)^2\) correctly and use \(\sin^2\theta + \cos^2\theta = 1\).
M1: Factorise the numerator to obtain \(2(1 + \cos\theta)\).
A1: Correct simplification to \(2\csc\theta\) showing all steps clearly.

Part (b) [7 marks]:
M1: Using the identity \(\tan^2 x = \sec^2 x - 1\) to form a quadratic equation in \(\sec x\).
A1: Correct quadratic equation \(3\sec^2 x - 5\sec x - 2 = 0\).
M1: Attempting to factorise or solve their 3-term quadratic equation.
A1: Finding \(\sec x = 2\) and \(\sec x = -1/3\) (or \(\cos x = 1/2\) and \(\cos x = -3\)).
B1: Stating that \(\cos x = -3\) has no solutions.
M1: Finding at least one correct angle for \(\cos x = 1/2\).
A1: Both \(x = 60^\circ\) and \(x = 300^\circ\) and no other solutions in the range.

(a)(i) Since \(OQ : QB = 3 : 2\), we have \(\overrightarrow{OQ} = \frac{3}{5}\mathbf{b}\).
\(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{3}{5}\mathbf{b}\).

(a)(ii) Since \(OP : PA = 2 : 1\), we have \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
\(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).

(b) Using \(\overrightarrow{AX} = \lambda \overrightarrow{AQ}\):
\(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + \lambda\left(-\mathbf{a} + \frac{3}{5}\mathbf{b}\right) = (1 - \lambda)\mathbf{a} + \frac{3}{5}\lambda\mathbf{b}\).

Using \(\overrightarrow{BX} = \mu \overrightarrow{BP}\):
\(\overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + \mu\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\mu\mathbf{a} + (1 - \mu)\mathbf{b}\).

Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
1) \(1 - \lambda = \frac{2}{3}\mu\)
2) \(\frac{3}{5}\lambda = 1 - \mu\)

From (2), we get \(\mu = 1 - \frac{3}{5}\lambda\).
Substitute this into (1):
\(1 - \lambda = \frac{2}{3}\left(1 - \frac{3}{5}\lambda\right)\)
\(1 - \lambda = \frac{2}{3} - \frac{2}{5}\lambda \Rightarrow \frac{1}{3} = \frac{3}{5}\lambda \Rightarrow \lambda = \frac{5}{9}\).

Substitute \(\lambda = \frac{5}{9}\) back to find \(\mu\):
\(\mu = 1 - \frac{3}{5}\left(\frac{5}{9}\right) = 1 - \frac{1}{3} = \frac{2}{3}\).

(c) Substitute \(\lambda = \frac{5}{9}\) into the expression for \(\overrightarrow{OX}\):
\(\overrightarrow{OX} = \left(1 - \frac{5}{9}\right)\mathbf{a} + \frac{3}{5}\left(\frac{5}{9}\right)\mathbf{b} = \frac{4}{9}\mathbf{a} + \frac{1}{3}\mathbf{b}\).

Part (a)(i) [1 mark]:
B1: Correct expression \(-\mathbf{a} + \frac{3}{5}\mathbf{b}\).

Part (a)(ii) [1 mark]:
B1: Correct expression \(\frac{2}{3}\mathbf{a} - \mathbf{b}\).

Part (b) [7 marks]:
M1: Expressing \(\overrightarrow{OX}\) as \(\overrightarrow{OA} + \lambda \overrightarrow{AQ}\).
A1: Obtaining \(\overrightarrow{OX} = (1 - \lambda)\mathbf{a} + \frac{3}{5}\lambda\mathbf{b}\).
M1: Expressing \(\overrightarrow{OX}\) as \(\overrightarrow{OB} + \mu \overrightarrow{BP}\).
A1: Obtaining \(\overrightarrow{OX} = \frac{2}{3}\mu\mathbf{a} + (1 - \mu)\mathbf{b}\).
M1: Equating coefficients to form two simultaneous equations.
A1: Correct value \(\lambda = \frac{5}{9}\).
A1: Correct value \(\mu = \frac{2}{3}\).

Part (c) [2 marks]:
M1: Substituting their \(\lambda\) or \(\mu\) into an expression for \(\overrightarrow{OX}\).
A1: Correct final vector \(\frac{4}{9}\mathbf{a} + \frac{1}{3}\mathbf{b}\).