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

(a) Equating the curve and the line equations:
\[12x - 3x^2 = 9\]
\[3x^2 - 12x + 9 = 0\]
\[x^2 - 4x + 3 = 0\]
\[(x-1)(x-3) = 0 \implies x=1, \ x=3\]

(b) The area is given by:
\[\text{Area} = \int_{1}^{3} (12x - 3x^2 - 9) \,dx\]
\[= \left[ 6x^2 - x^3 - 9x \right]_1^3\]
Evaluating at the upper limit \(x=3\):
\[6(9) - 27 - 9(3) = 54 - 27 - 27 = 0\]
Evaluating at the lower limit \(x=1\):
\[6(1) - 1^3 - 9(1) = 6 - 1 - 9 = -4\]
\[\text{Area} = 0 - (-4) = 4\]

(a)
M1: For equating curve and line equation and attempting to solve the resulting 3-term quadratic equation.
A1: For both \(x = 1\) and \(x = 3\) correctly found.

(b)
M1: For correct integration limits used on \(\int (12x - 3x^2 - 9) \,dx\) or \(\int (9 - (12x - 3x^2)) \,dx\).
A1: For correct integration: \(6x^2 - x^3 - 9x\) (ignore signs/order for this mark).
M1: For correct substitution of their limits from part (a).
A1: For \(4\) (must be positive).

(a) Using the quotient rule to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{(2x-1) \cdot \frac{2}{2x-1} - \ln(2x-1) \cdot 2}{(2x-1)^2} = \frac{2 - 2\ln(2x-1)}{(2x-1)^2}\]
At a stationary point, \(\frac{dy}{dx} = 0\):
\[2 - 2\ln(2x-1) = 0 \implies \ln(2x-1) = 1 \implies 2x - 1 = e \implies x = \frac{e+1}{2}\]
Substituting \(x = \frac{e+1}{2}\) back into the equation of the curve:
\[y = \frac{\ln(e)}{e} = \frac{1}{e}\]
So, the stationary point is \(\left(\frac{e+1}{2}, \frac{1}{e}\right)\).

(b) Examining the first derivative:
For \(x < \frac{e+1}{2}\), \(2x-1 < e \implies \ln(2x-1) < 1 \implies \frac{dy}{dx} > 0\).
For \(x > \frac{e+1}{2}\), \(2x-1 > e \implies \ln(2x-1) > 1 \implies \frac{dy}{dx} < 0\).
Since the gradient changes from positive to negative, the stationary point is a local maximum.

(a)
M1: For attempting to use the quotient rule (or product rule).
A1: For correct differentiation to obtain \(\frac{2 - 2\ln(2x-1)}{(2x-1)^2}\) or equivalent.
M1: For setting their derivative equal to 0 and solving for \(\ln(2x-1)\).
A1: For the correct exact \(x\)-coordinate: \(x = \frac{e+1}{2}\).
A1: For the correct exact \(y\)-coordinate: \(y = \frac{1}{e}\).

(b)
M1: For a valid method to determine the nature of the stationary point (e.g. first derivative test or second derivative).
A1: For correct conclusion (maximum) with supporting reasoning.

(a) Expanding \((1 + ax)^n\):
\[(1 + ax)^n = 1 + \binom{n}{1}(ax) + \binom{n}{2}(ax)^2 + \dots = 1 + nax + \frac{n(n-1)}{2} a^2 x^2\]

(b) From the given coefficients:
1) \(na = -24 \implies a = -\frac{24}{n}\)
2) \(\frac{n(n-1)}{2} a^2 = 252\)

Substituting \(a = -\frac{24}{n}\) into the second equation:
\[\frac{n(n-1)}{2} \left(-\frac{24}{n}\right)^2 = 252\]
\[\frac{n(n-1)}{2} \cdot \frac{576}{n^2} = 252\]
\[\frac{288(n-1)}{n} = 252\]
\[288n - 288 = 252n \implies 36n = 288 \implies n = 8\]

Using \(n = 8\) to find \(a\):
\[a = -\frac{24}{8} = -3\]

(a)
B1: For the first two terms: \(1 + nax\).
B1: For the third term: \rac{n(n-1)}{2} a^2 x^2\.

(b)
B1: For identifying the two equations: \(na = -24\) and \(\frac{n(n-1)}{2} a^2 = 252\).
M1: For substituting \(a = -\frac{24}{n}\) (or equivalent substitution) to obtain an equation in \(n\) only.
A1: For \(n = 8\).
A1: For \(a = -3\).

Using the identity \(\tan^2(2\theta) = \sec^2(2\theta) - 1\):
\[3(\sec^2(2\theta) - 1) - 4\sec(2\theta) - 1 = 0\]
\[3\sec^2(2\theta) - 4\sec(2\theta) - 4 = 0\]
Let \(u = \sec(2\theta)\):
\[3u^2 - 4u - 4 = 0 \implies (3u + 2)(u - 2) = 0\]
This gives \(u = -\frac{2}{3}\) or \(u = 2\).

If \(\sec(2\theta) = -\frac{2}{3} \implies \cos(2\theta) = -\frac{3}{2}\), which has no real solutions.

If \(\sec(2\theta) = 2 \implies \cos(2\theta) = \frac{1}{2}\).
Since \(0^\circ \le \theta \le 180^\circ \implies 0^\circ \le 2\theta \le 360^\circ\):
\[2\theta = 60^\circ \implies \theta = 30^\circ\]
\[2\theta = 300^\circ \implies \theta = 150^\circ\]

M1: For using the identity \(\tan^2(2\theta) = \sec^2(2\theta) - 1\) to express the equation in terms of \(\sec(2\theta)\) only.
A1: For obtaining the correct quadratic equation \(3\sec^2(2\theta) - 4\sec(2\theta) - 4 = 0\).
M1: For factorising or solving their quadratic equation to find values for \(\sec(2\theta)\).
A1: For obtaining \(\sec(2\theta) = 2\) and \(\sec(2\theta) = -\frac{2}{3}\).
B1: For rejecting \(\sec(2\theta) = -\frac{2}{3}\) (or showing it has no solution).
M1: For solving \(\cos(2\theta) = \frac{1}{2}\) in the correct range for \(2\theta\).
A1: For \(\theta = 30^\circ, \ 150^\circ\) (with no extra solutions in range).

(a) Since \(M\) is the midpoint of \(OB\), we have \(\overrightarrow{OM} = \frac{1}{2}\mathbf{b}\).
Using the ratio theorem on \(AM\) with ratio \(2 : 1\):
\[\overrightarrow{ON} = \frac{1 \cdot \overrightarrow{OA} + 2 \cdot \overrightarrow{OM}}{1 + 2} = \frac{\mathbf{a} + 2\left(\frac{1}{2}\mathbf{b}\right)}{3} = \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\]

(b) From \(\overrightarrow{OP} = k\overrightarrow{ON}\):
\[\overrightarrow{OP} = \frac{k}{3}\mathbf{a} + \frac{k}{3}\mathbf{b}\]

From \(\overrightarrow{AP} = m\overrightarrow{AB}\):
\[\overrightarrow{OP} - \overrightarrow{OA} = m(\overrightarrow{OB} - \overrightarrow{OA}) \implies \overrightarrow{OP} = (1-m)\mathbf{a} + m\mathbf{b}\]

Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\):
\[\frac{k}{3} = 1-m \quad \text{and} \quad \frac{k}{3} = m\]

Thus:
\[1-m = m \implies 2m = 1 \implies m = \frac{1}{2}\]
\[\frac{k}{3} = \frac{1}{2} \implies k = \frac{3}{2}\]

(a)
B1: For stating \(\overrightarrow{OM} = \frac{1}{2}\mathbf{b}\) (soi).
M1: For a valid vector pathway for \(\overrightarrow{ON}\) (e.g. \(\overrightarrow{OA} + \frac{2}{3}\overrightarrow{AM}\) or using ratio theorem).
A1: For \(\frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\) (or equivalent).

(b)
B1: For writing \(\overrightarrow{OP} = \frac{k}{3}\mathbf{a} + \frac{k}{3}\mathbf{b}\).
M1: For writing \(\overrightarrow{OP} = (1-m)\mathbf{a} + m\mathbf{b}\) (or equivalent in terms of \(m\)).
M1: For equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up two equations.
A1: For both \(k = \frac{3}{2}\) and \(m = \frac{1}{2}\).

(a) Completing the square for both \(x\) and \(y\):
\[(x - 3)^2 - 9 + (y - 4)^2 - 16 + 15 = 0\]
\[(x - 3)^2 + (y - 4)^2 = 10\]
So, the centre \(C\) is \((3, 4)\) and the radius is \(\sqrt{10}\).

(b) The gradient of the radius \(CP\) joining the centre \((3, 4)\) and \(P(4, 7)\) is:
\[m_{CP} = \frac{7 - 4}{4 - 3} = 3\]
Since the tangent is perpendicular to the radius:
\[m_{\text{tangent}} = -\frac{1}{3}\]
Using the point-slope formula with \(P(4, 7)\):
\[y - 7 = -\frac{1}{3}(x - 4) \implies 3y - 21 = -x + 4 \implies x + 3y = 25\]

(a)
M1: For attempting to complete the square for both \(x\) and \(y\).
A1: For correct centre \((3, 4)\).
A1: For correct radius \(\sqrt{10}\).

(b)
M1: For finding the gradient of the radius \(CP\).
M1: For using \(m_1 m_2 = -1\) to find the gradient of the tangent, and attempting to form the equation of the line.
A1: For \(x + 3y = 25\) or any equivalent form with integer coefficients.

(a) Since \(2x - 1\) is a factor, \(p(1/2) = 0\):
\[2\left(\frac{1}{8}\right) + a\left(\frac{1}{4}\right) + b\left(\frac{1}{2}\right) + 6 = 0 \implies 1 + a + 2b + 24 = 0 \implies a + 2b = -25\]

Using the Remainder Theorem, \(p(-1) = 12\):
\[2(-1)^3 + a(-1)^2 + b(-1) + 6 = 12 \implies -2 + a - b + 6 = 12 \implies a - b = 8\]

We solve the simultaneous equations:
1) \(a + 2b = -25\)
2) \(a - b = 8 \implies a = b + 8\)

Substituting (2) into (1):
\[b + 8 + 2b = -25 \implies 3b = -33 \implies b = -11\]
Then \(a = -11 + 8 = -3\).

(b) The polynomial is \(p(x) = 2x^3 - 3x^2 - 11x + 6\).
Since \(2x - 1\) is a factor, we can divide \(p(x)\) by \((2x - 1)\):
\[2x^3 - 3x^2 - 11x + 6 = (2x - 1)(x^2 - x - 6)\]
Factorising the quadratic factor \(x^2 - x - 6\):
\[x^2 - x - 6 = (x + 2)(x - 3)\]
So, \(p(x) = (2x - 1)(x + 2)(x - 3)\).

(a)
M1: For substituting \(x = \frac{1}{2}\) into \(p(x)\) and setting to 0.
M1: For substituting \(x = -1\) into \(p(x)\) and setting to 12.
A1: For obtaining correct simplified equations: \(a + 2b = -25\) and \(a - b = 8\).
A1: For \(a = -3\) and \(b = -11\) (both correct).

(b)
M1: For attempting to divide \(p(x)\) by \((2x - 1)\) to obtain a quadratic factor.
A1: For correct quadratic factor \(x^2 - x - 6\).
A1: For completely factorised form: \((2x - 1)(x + 2)(x - 3)\).

(a) Since \(e^{2x} > 0\) for all \(x \in \mathbb{R}\), we have \(3e^{2x} - 1 > -1\).
So, the range of \(f\) is \(f(x) > -1\).

(b) Let \(y = 3e^{2x} - 1\).
\[y + 1 = 3e^{2x} \implies e^{2x} = \frac{y+1}{3}\]
\[2x = \ln\left(\frac{y+1}{3}\right) \implies x = \frac{1}{2}\ln\left(\frac{y+1}{3}\right)\]
So, \(f^{-1}(x) = \frac{1}{2}\ln\left(\frac{x+1}{3}\right)\).

The domain of \(f^{-1}\) is the range of \(f\), which is \(x > -1\).

(c) Finding the composite function:
\[gf(x) = g(3e^{2x} - 1) = \ln(3e^{2x} - 1 + 2) = \ln(3e^{2x} + 1)\]

Solving \(gf(x) = \ln 5\):
\[\ln(3e^{2x} + 1) = \ln 5 \implies 3e^{2x} + 1 = 5\]
\[3e^{2x} = 4 \implies e^{2x} = \frac{4}{3} \implies 2x = \ln\left(\frac{4}{3}\right)\]
\[x = \frac{1}{2}\ln\left(\frac{4}{3}\right)\]

(a)
B1: For \(f(x) > -1\) or \(y > -1\) (or interval notation \((-1, \infty)\)).

(b)
M1: For attempting to make \(x\) the subject of \(y = 3e^{2x} - 1\).
A1: For correct expression: \(f^{-1}(x) = \frac{1}{2}\ln\left(\frac{x+1}{3}\right)\) or equivalent.
B1: For stating correct domain: \(x > -1\).

(c)
M1: For setting up the equation \(\ln(3e^{2x} + 1) = \ln 5\) and attempting to solve for \(e^{2x}\).
A1: For correct exact value: \(x = \frac{1}{2}\ln\left(\frac{4}{3}\right)\) or equivalent.

(a) To find the equation of the curve, we integrate the gradient function: \(y = \int 4(2x+1)^{-\frac{1}{2}} \mathrm{d}x\). Integrating gives \(y = 4 \times \frac{(2x+1)^{\frac{1}{2}}}{\frac{1}{2} \times 2} + C = 4\sqrt{2x+1} + C\). Substituting the point \((4, 15)\) to find \(C\), we get \(15 = 4\sqrt{2(4)+1} + C\), which simplifies to \(15 = 12 + C\), so \(C = 3\). Thus, the equation of the curve is \(y = 4\sqrt{2x+1} + 3\). (b) Since the gradient of the normal is \(-\frac{5}{4}\), the gradient of the tangent is \(\frac{4}{5}\). Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4}{5}\) gives \(\frac{4}{\sqrt{2x+1}} = \frac{4}{5}\), which simplifies to \(\sqrt{2x+1} = 5\). Squaring both sides, we get \(2x+1 = 25\), so \(x = 12\). Substituting \(x = 12\) into the curve equation gives \(y = 4\sqrt{2(12)+1} + 3 = 23\). Therefore, the coordinates of the point are \((12, 23)\).

(a) M1: Attempt to integrate obtaining \(k\sqrt{2x+1}\). A1: Correct integration to obtain \(4\sqrt{2x+1} (+ C)\). M1: Substitute \(x=4, y=15\) to find \(C\). A1: Correct equation \(y = 4\sqrt{2x+1} + 3\). (b) M1: State gradient of tangent is \(\frac{4}{5}\) and set \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4}{5}\). A1: Solve to get \(x=12\). A1: Find \(y=23\) and state coordinates \((12, 23)\).

(a) Let the first term be \(a\) and the common ratio be \(r\). We have \(ar^2 = 12\) and \(ar^5 = \frac{32}{9}\). Dividing the two equations gives \(r^3 = \frac{32/9}{12} = \frac{8}{27}\), so \(r = \frac{2}{3}\). Substituting this back gives \(a\left(\frac{4}{9}\right) = 12\), so \(a = 27\). (b) The sum to infinity is \(S_{\infty} = \frac{a}{1-r} = \frac{27}{1 - 2/3} = 81\). (c) The sum of the first 4 terms is \(S_4 = \frac{27(1 - (2/3)^4)}{1 - 2/3} = 81(1 - 16/81) = 65\). The difference is \(S_{\infty} - S_4 = 81 - 65 = 16\). Alternatively, the difference is \(\frac{ar^4}{1-r} = 81 \times \frac{16}{81} = 16\).

(a) M1: Set up equations and eliminate \(a\) to find \(r^3\). A1: Correct common ratio \(r = \frac{2}{3}\). A1: Correct first term \(a = 27\). (b) M1: Use sum to infinity formula with their \(a\) and \(r\). A1: Correct sum to infinity \(81\). (c) M1: Use a correct method to find the difference. A1: Correct difference \(16\).

(a) By the factor theorem, \(\mathrm{p}(2) = 0\), which gives \(2(8) + k(4) - 17(2) + m = 0 \implies 4k + m = 18\). By the remainder theorem, \(\mathrm{p}(-1) = 12\), which gives \(2(-1) + k(1) - 17(-1) + m = 12 \implies k + m = -3\). Subtracting the equations gives \(3k = 21 \implies k = 7\). Then \(7 + m = -3 \implies m = -10\). (b) The polynomial is \(\mathrm{p}(x) = 2x^3 + 7x^2 - 17x - 10\). Dividing by \(x-2\) gives \((x-2)(2x^2 + 11x + 5)\). Factorising the quadratic yields \((2x+1)(x+5)\). Thus, the completely factorised form is \(\mathrm{p}(x) = (x-2)(2x+1)(x+5)\).

(a) M1: Use factor theorem to obtain \(4k + m = 18\). M1: Use remainder theorem to obtain \(k + m = -3\). M1: Solve simultaneous equations. A1: Correct values \(k = 7\) and \(m = -10\). (b) M1: Attempt to divide \(\mathrm{p}(x)\) by \(x-2\). A1: Correct quadratic factor \(2x^2 + 11x + 5\). A1: Correct complete factorisation \((x-2)(2x+1)(x+5)\).

(a) Completing the square for \(x\) and \(y\) gives \((x-3)^2 - 9 + (y+4)^2 - 16 - 15 = 0 \implies (x-3)^2 + (y+4)^2 = 40\). The centre is \((3, -4)\) and the radius is \(\sqrt{40}\) (or \(2\sqrt{10}\)). (b) Substituting \((5, 2)\) into the LHS of the equation gives \(5^2 + 2^2 - 6(5) + 8(2) - 15 = 25 + 4 - 30 + 16 - 15 = 0\), which equals the RHS, so \(P\) lies on the circle. (c) The gradient of the radius from \(C(3, -4)\) to \(P(5, 2)\) is \(m = \frac{2 - (-4)}{5 - 3} = \frac{6}{2} = 3\). The gradient of the tangent is perpendicular, which is \(-\frac{1}{3}\). The equation of the tangent is \(y - 2 = -\frac{1}{3}(x - 5)\), which simplifies to \(x + 3y = 11\).

(a) M1: Attempt to complete the square for both \(x\) and \(y\). A1: Correct centre \((3, -4)\) and radius \(\sqrt{40}\). (b) B1: Show substitute \((5, 2)\) gives 0. (c) M1: Find gradient of radius \(CP\). M1: Find perpendicular gradient of tangent. M1: Find equation of line using their gradient and point \(P\). A1: Correct equation \(x + 3y = 11\) (or any equivalent simplified form).

(a) Squaring both sides:
\((3x - 2)^2 \le (x + 4)^2\)
\(9x^2 - 12x + 4 \le x^2 + 8x + 16\)
\(8x^2 - 20x - 12 \le 0\)
Divide by 4:
\(2x^2 - 5x - 3 \le 0\)
\((2x + 1)(x - 3) \le 0\)
Critical values are \(x = -\frac{1}{2}\) and \(x = 3\).
Hence, \(-\frac{1}{2} \le x \le 3\).

(b) Substitute the line into the curve:
\(mx - 5 = x^2 - 4x + 4\)
\(x^2 - (m + 4)x + 9 = 0\)
For no intersection, the discriminant must be less than 0:
\(D = (m + 4)^2 - 4(1)(9) < 0\)
\((m + 4)^2 - 36 < 0\)
\((m + 4)^2 < 36\)
\(-6 < m + 4 < 6\)
\(-10 < m < 2\)

(a)
M1: Attempt to square both sides or solve separate linear equations
M1: Factorise or solve their quadratic to find critical values
A1: Identify critical values \(-\frac{1}{2}\) and \(3\)
A1: Correct final inequality \(-\frac{1}{2} \le x \le 3\)

(b)
M1: Equate line and curve and write as a three-term quadratic
M1: Use \(b^2 - 4ac < 0\)
A1: Obtain critical values \(m = -10\) and \(m = 2\)
A1: Correct range \(-10 < m < 2\)

(a) Let \(y = \ln(2x - 3)\)
\(e^y = 2x - 3\)
\(2x = e^y + 3\)
\(x = \frac{e^y + 3}{2}\)
Thus, \(f^{-1}(x) = \frac{e^x + 3}{2}\).
The domain of \(f^{-1}\) is the range of \(f\). Since \(f(x)\) maps to all real numbers, the domain is \(x \in \mathbb{R}\).

(b) \(gf(x) = g(\ln(2x - 3)) = e^{2\ln(2x - 3)} + 1 = 17\)
\(e^{\ln((2x - 3)^2)} + 1 = 17\)
\((2x - 3)^2 + 1 = 17\)
\((2x - 3)^2 = 16\)
\(2x - 3 = 4\) or \(2x - 3 = -4\)
If \(2x - 3 = 4 \implies 2x = 7 \implies x = 3.5\)
If \(2x - 3 = -4 \implies 2x = -1 \implies x = -0.5\)
Since the domain of \(f\) requires \(x > 1.5\), we reject \(x = -0.5\).
Hence, the only solution is \(x = 3.5\).

(a)
M1: Attempt to make \(x\) the subject of \(y = \ln(2x - 3)\)
M1: Correct use of exponential to eliminate logarithm
A1: Correct formula \(f^{-1}(x) = \frac{e^x + 3}{2}\)
B1: Correct domain \(x \in \mathbb{R}\)

(b)
M1: Form the composite function \(gf(x)\)
M1: Use \(e^{2\ln A} = A^2\) to simplify
A1: Solve the quadratic or linear equations to find \(x = 3.5\) and \(x = -0.5\)
A1: Reject \(x = -0.5\) and state final answer \(x = 3.5\) only

(a) Completing the square for \(x^2 - 6x\) and \(y^2 + 8y\):
\((x - 3)^2 - 9 + (y + 4)^2 - 16 = 0\)
\((x - 3)^2 + (y + 4)^2 = 25\)
Centre is \((3, -4)\) and radius is \(\sqrt{25} = 5\).

(b) Substitute \(y = x - 2\) into the circle's equation:
\(x^2 + (x - 2)^2 - 6x + 8(x - 2) = 0\)
\(x^2 + x^2 - 4x + 4 - 6x + 8x - 16 = 0\)
\(2x^2 - 2x - 12 = 0\)
\(x^2 - x - 6 = 0\)
\((x - 3)(x + 2) = 0\)
Intersection points are at \(x = 3\) and \(x = -2\).
When \(x = 3\), \(y = 3 - 2 = 1\), so \(A = (3, 1)\).
When \(x = -2\), \(y = -2 - 2 = -4\), so \(B = (-2, -4)\).
Length of chord \(AB = \sqrt{(3 - (-2))^2 + (1 - (-4))^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}\).

(a)
M1: Attempt to complete the square on both quadratic expressions
A1: Correct centre \((3, -4)\)
A1: Correct radius \(5\)

(b)
M1: Substitute line into circle equation
M1: Simplify to a three-term quadratic and solve for \(x\)
A1: Obtain correct coordinates \((3, 1)\) and \((-2, -4)\)
M1: Apply distance formula to find chord length
A1: Correct exact length \(5\sqrt{2}\)

(a) \(\text{LHS} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta(1 + \cos \theta)}\)
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\(= \frac{1 + 1 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2(1 + \cos \theta)}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2}{\sin \theta} = 2\csc \theta = \text{RHS}\).

(b) Using the identity with \(\theta = 2x\):
\(2\csc 2x = 4\)
\(\csc 2x = 2\)
\(\sin 2x = \frac{1}{2}\)
Since \(0^\circ \le x \le 180^\circ\), we have \(0^\circ \le 2x \le 360^\circ\).
\(2x = 30^\circ\) or \(2x = 150^\circ\)
\(x = 15^\circ\) or \(x = 75^\circ\).

(a)
M1: Place over a common denominator
M1: Expand numerator correctly
M1: Use \(\sin^2\theta + \cos^2\theta = 1\) to simplify numerator
A1: Correctly factorise numerator and cancel \(1 + \cos\theta\) to achieve \(2\csc\theta\)

(b)
M1: Use part (a) to write \(2\csc 2x = 4\)
M1: Rewrite as \(\sin 2x = \frac{1}{2}\)
A1: Find one correct value of \(x\) (either \(15^\circ\) or \(75^\circ\))
A1: Obtain both \(15^\circ\) and \(75^\circ\) with no extras in range

(a) The area of sector \(OAB = \frac{1}{2} r^2 \theta\).
In the right-angled triangle \(OCA\), we have:
\(AC = r\sin\theta\)
\(OC = r\cos\theta\)
Area of triangle \(OCA = \frac{1}{2} (OC)(AC) = \frac{1}{2} (r\cos\theta)(r\sin\theta) = \frac{1}{2} r^2 \sin\theta\cos\theta\).
Area of the shaded region \(= \text{Area of sector } OAB - \text{Area of triangle } OCA\)
\(= \frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin\theta\cos\theta = \frac{1}{2} r^2 (\theta - \sin\theta\cos\theta)\).

(b) The perimeter consists of the arc \(AB\), the segment \(AC\), and the segment \(CB\).
Arc \(AB = r\theta = 8\left(\frac{\pi}{3}\right) = \frac{8\pi}{3}\).
Segment \(AC = r\sin\theta = 8\sin\left(\frac{\pi}{3}\right) = 8\left(\frac{\sqrt{3}}{2}\right) = 4\sqrt{3}\).
Segment \(CB = OB - OC = r - r\cos\theta = 8 - 8\cos\left(\frac{\pi}{3}\right) = 8 - 8\left(\frac{1}{2}\right) = 4\).
Perimeter \(= \text{Arc } AB + AC + CB = \frac{8\pi}{3} + 4\sqrt{3} + 4\).

(a)
B1: State the area of sector as \(\frac{1}{2} r^2 \theta\)
M1: Express \(AC\) and \(OC\) in terms of \(r\) and \(\theta\)
M1: Obtain the area of triangle \(OCA\) as \(\frac{1}{2} r^2 \sin\theta\cos\theta\)
A1: Show subtraction to get the required shaded area

(b)
B1: Calculate arc length \(AB = \frac{8\pi}{3}\)
M1: Calculate length of \(AC\) using trigonometry
M1: Calculate length of \(CB\) as \(r - r\cos\theta\)
A1: State the exact perimeter as \(\frac{8\pi}{3} + 4\sqrt{3} + 4\)

(a) (i) Total number of people is \(7 + 6 = 13\).
Number of ways to choose 5 is \(\binom{13}{5} = 1287\).

(ii) For more women than men, the possible selections of (women, men) are:
- (3W, 2M): \(\binom{6}{3} \times \binom{7}{2} = 20 \times 21 = 420\)
- (4W, 1M): \(\binom{6}{4} \times \binom{7}{1} = 15 \times 7 = 105\)
- (5W, 0M): \(\binom{6}{5} \times \binom{7}{0} = 6 \times 1 = 6\)
Total ways = \(420 + 105 + 6 = 531\).

(b) (i) The letters of `SECRET` are S, E, C, R, E, T (6 letters containing two Es).
Number of arrangements = \(\frac{6!}{2!} = \frac{720}{2} = 360\).

(ii) Let the two Es be next to each other as a single unit `EE`.
Number of arrangements of S, C, R, T, (EE) is \(5! = 120\).
Number of arrangements with the two Es not next to each other = \(360 - 120 = 240\).

(a)
(i) B1: Correct calculation of \(1287\)
(ii) M1: Identify three cases (3W,2M), (4W,1M), (5W,0M) and attempt combinations
M1: Perform calculations for the cases (e.g. at least one case fully correct)
A1: Correct total of \(531\)

(b)
(i) B1: Show division by \(2!\)
A1: Correct answer of \(360\)
(ii) M1: Calculate arrangements with Es together (e.g., \(5!\) or \(120\))
A1: Correct answer of \(240\)

(a) Arithmetic Progression terms:
\(u_3 = a + 2d\)
\(u_{11} = a + 10d\)
Geometric Progression terms:
\(g_2 = ar\)
\(g_3 = ar^2\)
From the given conditions:
\(a + 2d = ar \implies 2d = a(r - 1)\) (Equation 1)
\(a + 10d = ar^2 \implies 10d = a(r^2 - 1)\) (Equation 2)
Dividing Equation 2 by Equation 1:
\(\frac{10d}{2d} = \frac{a(r^2 - 1)}{a(r - 1)}\)
\(5 = \frac{(r - 1)(r + 1)}{r - 1}\)
Since \(d \ne 0 \implies r \ne 1\), we can cancel \(r - 1\):
\(5 = r + 1 \implies r = 4\).
Substitute \(r = 4\) back into Equation 1:
\(2d = a(4 - 1) = 3a \implies d = 1.5a\).

(b) Sum of first 4 terms of AP:
\(S_4 = \frac{4}{2} [2a + 3d] = 2(2a + 3d) = 26\)
\(2a + 3d = 13\)
Substitute \(d = 1.5a\):
\(2a + 3(1.5a) = 13\)
\(6.5a = 13 \implies a = 2\)
Then \(d = 1.5(2) = 3\).

(a)
M1: Formulate expressions for \(u_3, u_{11}\) and \(g_2, g_3\)
M1: Establish two equations relating \(a, d, r\)
M1: Attempt to eliminate \(d\) by dividing the equations
A1: Simplify correctly to show \(r = 4\)
A1: Obtain \(d = 1.5a\) or equivalent relation

(b)
M1: Write the sum of AP formula for \(S_4 = 26\)
M1: Substitute \(d = 1.5a\) into their sum equation
A1: Find correct values \(a = 2\) and \(d = 3\)

(a) Use the quotient rule with \(u = x^2 - 3\) and \(v = x - 2\):
\(\frac{du}{dx} = 2x\)
\(\frac{dv}{dx} = 1\)
\(\frac{dy}{dx} = \frac{2x(x - 2) - (x^2 - 3)(1)}{(x - 2)^2}\)
\(\frac{dy}{dx} = \frac{2x^2 - 4x - x^2 + 3}{(x - 2)^2} = \frac{x^2 - 4x + 3}{(x - 2)^2}\)
For stationary points, set \(\frac{dy}{dx} = 0\):
\(x^2 - 4x + 3 = 0\)
\((x - 1)(x - 3) = 0\)
Thus, stationary points occur at \(x = 1\) and \(x = 3\).

(b) Using the second derivative test:
Rewrite \(\frac{dy}{dx} = \frac{(x-2)^2 - 1}{(x-2)^2} = 1 - (x-2)^{-2}\).
\(\frac{d^2y}{dx^2} = 2(x-2)^{-3} = \frac{2}{(x-2)^3}\).
At \(x = 1\), \(\frac{d^2y}{dx^2} = \frac{2}{(1-2)^3} = -2 < 0\), so \(x = 1\) is a local maximum.
At \(x = 3\), \(\frac{d^2y}{dx^2} = \frac{2}{(3-2)^3} = 2 > 0\), so \(x = 3\) is a local minimum.

(a)
M1: Correct use of the quotient rule (allow minor slip in numerator)
A1: Correct unsimplified numerator \(2x(x-2) - (x^2-3)\)
A1: Simplify numerator to \(x^2 - 4x + 3\)
M1: Set derivative equal to zero and solve the quadratic
A1: Confirm stationary points at \(x=1\) and \(x=3\)

(b)
M1: Attempt to find the second derivative or evaluate the sign of first derivative around \(x=1\) and \(x=3\)
A1: Show \(x = 1\) is a maximum with correct supporting working
A1: Show \(x = 3\) is a minimum with correct supporting working

(a) Starting with the logarithmic equation:
\(2\log_3 x - \log_3(x + 6) = 1\)

Apply the power law of logarithms:
\(\log_3(x^2) - \log_3(x + 6) = 1\)

Apply the quotient law of logarithms:
\(\log_3\left(\frac{x^2}{x + 6}\right) = 1\)

Convert the logarithmic equation to its exponential form:
\(\frac{x^2}{x + 6} = 3^1\)
\(\frac{x^2}{x + 6} = 3\)

Solve the quadratic equation:
\(x^2 = 3(x + 6)\)
\(x^2 = 3x + 18\)
\(x^2 - 3x - 18 = 0\)
\((x - 6)(x + 3) = 0\)

This gives \(x = 6\) or \(x = -3\).
Since the term \(\log_3 x\) requires \(x > 0\), we must reject \(x = -3\).
Thus, the only valid solution is \(x = 6\).

(b) Starting with the exponential equation:
\(2^{2y+2} - 9(2^y) + 2 = 0\)

Rewrite \(2^{2y+2}\) as \(2^2 \cdot (2^y)^2\):
\(4(2^y)^2 - 9(2^y) + 2 = 0\)

Let \(u = 2^y\). The equation becomes:
\(4u^2 - 9u + 2 = 0\)

Factor the quadratic equation:
\((4u - 1)(u - 2) = 0\)

This gives \(u = \frac{1}{4}\) or \(u = 2\).

Substitute back \(u = 2^y\):
Case 1: \(2^y = \frac{1}{4} = 2^{-2} \implies y = -2\)
Case 2: \(2^y = 2 = 2^1 \implies y = 1\)

Both values are valid since there are no restrictions on the exponent \(y\).
Thus, \(y = -2\) or \(y = 1\).

(a)
M1: For applying the power law and quotient law of logarithms correctly to obtain \(\log_3\left(\frac{x^2}{x+6}\right) = 1\).
M1: For converting to exponential form correctly to obtain \(\frac{x^2}{x+6} = 3\).
A1: For solving the quadratic equation to get \(x = 6\) and \(x = -3\).
A1: For stating \(x = 6\) only and rejecting \(x = -3\).

(b)
M1: For using index laws to express the equation as a quadratic in terms of \(2^y\), i.e., \(4(2^y)^2 - 9(2^y) + 2 = 0\).
M1: For solving the quadratic equation in terms of \(2^y\) to find \(2^y = \frac{1}{4}\) or \(2^y = 2\).
A1: For obtaining one correct value of \(y\) (either \(y = -2\) or \(y = 1\)).
A1: For obtaining both correct values: \(y = -2\) and \(y = 1\).

(a)(i) Since there are no restrictions, we choose 5 students from the total of 11 students:
Number of ways = \(\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462\)

(a)(ii) For there to be more girls than boys on a 5-student committee, we can have the following scenarios:
- Case 1: 3 girls and 2 boys
Number of ways = \(\binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200\)
- Case 2: 4 girls and 1 boy
Number of ways = \(\binom{6}{4} \times \binom{5}{1} = 15 \times 5 = 75\)
- Case 3: 5 girls and 0 boys
Number of ways = \(\binom{6}{5} \times \binom{5}{0} = 6 \times 1 = 6\)

Total number of ways = \(200 + 75 + 6 = 281\)

(b)(i) The word 'KEEPER' contains 6 letters in total, with the letter 'E' repeating 3 times, while 'K', 'P', and 'R' appear once.
Number of unrestricted arrangements = \(\frac{6!}{3!} = \frac{720}{6} = 120\)

(b)(ii) To find the number of arrangements where the three 'E's are not all next to each other, we can use the subtraction method:
First, find the number of arrangements where the three 'E's are all next to each other.
Treat the group of three 'E's (EEE) as a single block. This leaves us with 4 entities to arrange: (EEE), K, P, and R.
Number of arrangements where 'EEE' are together = \(4! = 24\)

Arrangements where 'EEE' are not all next to each other = Total arrangements - Arrangements where 'EEE' are together
\(= 120 - 24 = 96\)

(a)(i)
B1: For 462

(a)(ii)
M1: For identifying the three valid cases: (3G, 2B), (4G, 1B), and (5G, 0B).
M1: For summing the products of the combinations for the three cases: \(\binom{6}{3}\binom{5}{2} + \binom{6}{4}\binom{5}{1} + \binom{6}{5}\binom{5}{0}\).
A1: For 281

(b)(i)
M1: For identifying repeating letters and writing \(\frac{6!}{3!}\).
A1: For 120

(b)(ii)
M1: For treating 'EEE' as a single block and finding \(4! = 24\).
A1: For subtracting the 'together' arrangements from the total to get 96.