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

1. Find the vector \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} -1 \\ 6 \end{pmatrix} - \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}\)

2. Calculate the magnitude of \(\overrightarrow{AB}\):
\(|\overrightarrow{AB}| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)

3. Find the unit vector by dividing the vector by its magnitude:
\(\text{Unit Vector} = \frac{1}{10}\begin{pmatrix} -6 \\ 8 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} -3 \\ 4 \end{pmatrix}\) or \(\begin{pmatrix} -0.6 \\ 0.8 \end{pmatrix}\).

M1: For attempt to find \(\overrightarrow{AB}\) (e.g. \(\overrightarrow{OB} - \overrightarrow{OA}\))
M1: For dividing their vector \(\overrightarrow{AB}\) by its magnitude
A1: For \(\frac{1}{5}\begin{pmatrix} -3 \\ 4 \end{pmatrix}\) or \(\begin{pmatrix} -0.6 \\ 0.8 \end{pmatrix}\) or equivalent

1. Expand and rearrange the inequality:
\((2x - 3)(x + 4) < 2x - 3\)
\(2x^2 + 8x - 3x - 12 < 2x - 3\)
\(2x^2 + 5x - 12 < 2x - 3\)
\(2x^2 + 3x - 9 < 0\)

2. Factorise the quadratic expression:
\((2x - 3)(x + 3) < 0\)

3. Find the critical values:
\(2x - 3 = 0 \Rightarrow x = 1.5\)
\(x + 3 = 0 \Rightarrow x = -3\)

4. Determine the range for which the quadratic is negative (below the x-axis):
\(-3 < x < 1.5\).

M1: For expanding and collecting terms into a 3-term quadratic inequality of the form \(2x^2 + 3x - 9 < 0\) (allow any inequality sign or \(=\))
M1: For factorising or solving their 3-term quadratic to find critical values
A1: For finding the critical values \(-3\) and \(1.5\)
A1: For \(-3 < x < 1.5\) (or \(-3 < x < \frac{3}{2}\))

1. Identify the centre of the circle from the equation:
Centre \(C = (2, -1)\)

2. Find the gradient of the radius joining \(C(2, -1)\) and the point of tangency \(P(5, 3)\):
\(m_{\text{radius}} = \frac{3 - (-1)}{5 - 2} = \frac{4}{3}\)

3. Calculate the gradient of the tangent, which is perpendicular to the radius:
\(m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{3}{4}\)

4. Use the point-slope form with \(P(5, 3)\) to find the equation of the tangent:
\(y - 3 = -\frac{3}{4}(x - 5)\)
Multiply by 4:
\(4y - 12 = -3x + 15\)
\(3x + 4y = 27\).

B1: For identifying the centre of the circle as \((2, -1)\)
M1: For finding the gradient of the radius using their centre and the point \((5, 3)\)
M1: For using the perpendicular gradient rule to find the tangent gradient and attempting to form the straight-line equation
A1: For \(3x + 4y = 27\) or any equivalent form

1. Apply the product rule of logarithms:
\(\log_3((x + 4)(x - 2)) = 3\)

2. Convert from logarithmic form to exponential form:
\((x + 4)(x - 2) = 3^3\)
\(x^2 + 2x - 8 = 27\)

3. Form a quadratic equation:
\(x^2 + 2x - 35 = 0\)

4. Solve the quadratic equation by factorisation:
\((x + 7)(x - 5) = 0\)
\(x = -7\) or \(x = 5\)

5. Check the validity of the solutions against the original log arguments:
Since the arguments of logarithms must be positive, we require \(x + 4 > 0 \Rightarrow x > -4\) and \(x - 2 > 0 \Rightarrow x > 2\).
Therefore, \(x = -7\) is rejected, and the only valid solution is \(x = 5\).

M1: For correct use of log addition rule to obtain \(\log_3((x+4)(x-2))\)
M1: For converting log equation to index form, \((x+4)(x-2) = 3^3\) or 27
A1: For obtaining the quadratic equation \(x^2 + 2x - 35 = 0\) and solving to find \(x = 5\) and \(x = -7\)
B1: For rejecting \(x = -7\) with justification, leaving \(x = 5\) as the only solution

1. Write the equation as \(y = x^2 (2x + 1)^{1/2}\) and differentiate using the product rule:
Let \(u = x^2 \Rightarrow \frac{\text{d}u}{\text{d}x} = 2x\)
Let \(v = (2x + 1)^{1/2} \Rightarrow \frac{\text{d}v}{\text{d}x} = \frac{1}{2}(2x + 1)^{-1/2} \cdot 2 = (2x + 1)^{-1/2}\)

2. Express \(\frac{\text{d}y}{\text{d}x}\):
\(\frac{\text{d}y}{\text{d}x} = u\frac{\text{d}v}{\text{d}x} + v\frac{\text{d}u}{\text{d}x} = x^2(2x+1)^{-1/2} + 2x(2x+1)^{1/2}\)
\(\frac{\text{d}y}{\text{d}x} = \frac{x^2}{\sqrt{2x+1}} + 2x\sqrt{2x+1}\)

3. Substitute \(x = 4\) into the derivative:
\(\frac{\text{d}y}{\text{d}x}\Big|_{x=4} = \frac{4^2}{\sqrt{2(4)+1}} + 2(4)\sqrt{2(4)+1}\)
\(= \frac{16}{\sqrt{9}} + 8\sqrt{9} = \frac{16}{3} + 8(3) = \frac{16}{3} + 24 = \frac{16 + 72}{3} = \frac{88}{3}\).

M1: For an attempt to use the product rule with two terms
A1: For correct derivative of \((2x+1)^{1/2}\) yielding \((2x+1)^{-1/2}\)
M1: For substituting \(x = 4\) into their expression for \(\frac{\text{d}y}{\text{d}x}\)
A1: For \(\frac{88}{3}\) or equivalent exact value

1. Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\) to obtain an equation in terms of \(\cos \theta\):
\(2(1 - \cos^2 \theta) - 3\cos \theta = 0\)
\(2 - 2\cos^2 \theta - 3\cos \theta = 0\)
\(2\cos^2 \theta + 3\cos \theta - 2 = 0\)

2. Factorise the quadratic expression:
\((2\cos \theta - 1)(\cos \theta + 2) = 0\)

3. Solve for \(\cos \theta\):
\(2\cos \theta - 1 = 0 \Rightarrow \cos \theta = \frac{1}{2}\)
\(\cos \theta + 2 = 0 \Rightarrow \cos \theta = -2\) (No real solutions since \(-1 \le \cos \theta \le 1\))

4. Find the angles in the range \(0^\circ \le \theta \le 360^\circ\):
With basic angle \(60^\circ\) where cosine is positive (1st and 4th quadrants):
\(\theta = 60^\circ\)
\(\theta = 360^\circ - 60^\circ = 300^\circ\).

M1: For using \(\sin^2 \theta = 1 - \cos^2 \theta\) to form a 3-term quadratic in \(\cos \theta\)
A1: For correct quadratic equation \(2\cos^2 \theta + 3\cos \theta - 2 = 0\)
M1: For solving their quadratic to find \(\cos \theta = \frac{1}{2}\) (and rejecting \(\cos \theta = -2\))
A1: For both \(\theta = 60^\circ\) and \(\theta = 300^\circ\) and no extras

1. Express the terms of the progression in terms of \(a\) and \(r\):
\(u_3 = ar^2 = 12\)
\(u_6 = ar^5 = \frac{32}{9}\)

2. Divide \(u_6\) by \(u_3\) to eliminate \(a\):
\(\frac{ar^5}{ar^2} = \frac{32/9}{12}\)
\(r^3 = \frac{32}{108} = \frac{8}{27}\)

3. Solve for \(r\):
\(r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\)

4. Substitute \(r\) back into the equation for \(u_3\) to find \(a\):
\(a \left(\frac{2}{3}\right)^2 = 12\)
\(a \left(\frac{4}{9}\right) = 12\)
\(a = 12 \times \frac{9}{4} = 27\).

M1: For expressing both terms in terms of \(a\) and \(r\) (i.e. \(ar^2 = 12\) and \(ar^5 = \frac{32}{9}\))
M1: For dividing the equations to find an equation for \(r^3\)
A1: For finding \(r = \frac{2}{3}\)
A1: For finding \(a = 27\)

1. Split the equation into two linear cases based on the absolute value definition:

Case 1: \(3x - 5 = 2x + 1\)
\(3x - 2x = 1 + 5\)
\(x = 6\)

Case 2: \(3x - 5 = -(2x + 1)\)
\(3x - 5 = -2x - 1\)
\(5x = 4\)
\(x = 0.8\) (or \(\frac{4}{5}\))

2. Verify the solutions by substituting them back into the original equation:
For \(x = 6\): \(|3(6) - 5| = |13| = 13\) and \(2(6) + 1 = 13\) (Valid)
For \(x = 0.8\): \(|3(0.8) - 5| = |-2.6| = 2.6\) and \(2(0.8) + 1 = 2.6\) (Valid)

Therefore, the solutions are \(x = 0.8\) and \(x = 6\).

M1: For setting up the two linear equations: \(3x - 5 = 2x + 1\) and \(3x - 5 = -(2x + 1)\)
A1: For solving Case 1 to find \(x = 6\)
A1: For solving Case 2 to find \(x = 0.8\) or \(\frac{4}{5}\)
B1: For validating both solutions (e.g., checking that \(2x + 1 \ge 0\) or evaluating both sides)

First, find the vector \(\vec{AB}\): \(2\vec{AB} = \begin{pmatrix} 12 \\ 10 \end{pmatrix} \implies \vec{AB} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}\). Next, find the position vector of \(B\), which is \(\vec{OB}\): \(\vec{OB} = \vec{OA} + \vec{AB} = \begin{pmatrix} -1 \\ 7 \end{pmatrix} + \begin{pmatrix} 6 \\ 5 \end{pmatrix} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}\). Now, calculate the magnitude of \(\vec{OB}\): \(|\vec{OB}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\). Finally, find the unit vector in the direction of \(\vec{OB}\): \(\hat{OB} = \frac{1}{13}\begin{pmatrix} 5 \\ 12 \end{pmatrix}\).

M1: For finding \(\vec{AB} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}\) and attempting to find \(\vec{OB} = \vec{OA} + \vec{AB}\). A1: For obtaining \(\vec{OB} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}\). A1: For the correct unit vector \(\frac{1}{13}\begin{pmatrix} 5 \\ 12 \end{pmatrix}\) (or equivalent).

From the linear equation, express \(x\) in terms of \(y\): \(x = 10 - 2y\). Substitute this expression for \(x\) into the quadratic equation: \((10 - 2y)^2 + y^2 = 25\). Expand the brackets: \(100 - 40y + 4y^2 + y^2 = 25\), which simplifies to \(5y^2 - 40y + 75 = 0\). Divide the entire equation by 5: \(y^2 - 8y + 15 = 0\). Factorise the quadratic equation: \((y - 3)(y - 5) = 0\). This gives two values for \(y\): \(y = 3\) or \(y = 5\). Find the corresponding values of \(x\): When \(y = 3\), \(x = 10 - 2(3) = 4\). When \(y = 5\), \(x = 10 - 2(5) = 0\). The solutions are \(x = 4, y = 3\) and \(x = 0, y = 5\).

M1: Expresses one variable in terms of the other and substitutes into the quadratic equation. M1: Expands and simplifies to a 3-term quadratic equation in one variable (e.g., \(5y^2 - 40y + 75 = 0\)). A1: Finds correct values of one variable (e.g., \(y = 3, 5\)). M1: Substitutes their values back to find the corresponding values of the other variable. A1: Correct pairs of solutions: \(x = 4, y = 3\) and \(x = 0, y = 5\).

Use the laws of logarithms to combine the terms on the left-hand side: \(\log_3((x - 2)(x + 4)) = 3\). Convert from logarithmic form to exponential form: \((x - 2)(x + 4) = 3^3 \implies x^2 + 2x - 8 = 27\). Rearrange into a quadratic equation set to zero: \(x^2 + 2x - 35 = 0\). Factorise the quadratic: \((x + 7)(x - 5) = 0\). This gives two potential solutions: \(x = -7\) or \(x = 5\). Check for extraneous solutions: For \(\log_3(x - 2)\) to be defined, we must have \(x - 2 > 0 \implies x > 2\). Therefore, \(x = -7\) is invalid, and the only solution is \(x = 5\).

M1: For correctly combining logarithms to get \(\log_3((x - 2)(x + 4)) = 3\). M1: For removing logarithms to get \((x - 2)(x + 4) = 27\) and forming a quadratic equation. A1: For finding the roots \(x = 5\) and \(x = -7\). A1: For rejecting \(x = -7\) with a valid reason to give the final answer \(x = 5\) only.

(a) Midpoint of PQ is \((\frac{1+5}{2}, \frac{4+2}{2}) = (3, 3)\). The gradient of PQ is \(\frac{2-4}{5-1} = -\frac{1}{2}\). Therefore, the gradient of the perpendicular bisector is \(2\). The equation is \(y - 3 = 2(x - 3)\), which simplifies to \(y = 2x - 3\). (b) Since the center lies on both \(y = 2x - 3\) and \(y = x + 2\), we solve these simultaneously: \(2x - 3 = x + 2\), which gives \(x = 5\) and \(y = 7\). Thus, the center is \((5, 7)\). The radius \(r\) is the distance from the center to \(P(1, 4)\): \(r^2 = (5-1)^2 + (7-4)^2 = 16 + 9 = 25\), so \(r = 5\). The equation of the circle is \((x - 5)^2 + (y - 7)^2 = 25\).

M1 for finding midpoint of PQ. A1 for gradient of perpendicular bisector. A1 for showing the equation y = 2x - 3. M1 for setting up simultaneous equations to find the center. A1 for correct center (5, 7). M1 for finding the radius or radius squared. A1 for the correct final circle equation.

(a) Using the product rule and chain rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 \cdot \sqrt{2x + 5} + (x - 2) \cdot \frac{1}{2\sqrt{2x + 5}} \cdot 2 = \sqrt{2x + 5} + \frac{x - 2}{\sqrt{2x + 5}} = \frac{2x + 5 + x - 2}{\sqrt{2x + 5}} = \frac{3x + 3}{\sqrt{2x + 5}}\). Thus \(a = 3\) and \(b = 3\). (b) At \(x = 2\), \(y = 0\). The gradient of the tangent is \(\frac{3(2) + 3}{\sqrt{2(2) + 5}} = \frac{9}{3} = 3\). The gradient of the normal is \(-\frac{1}{3}\). The equation of the normal is \(y - 0 = -\frac{1}{3}(x - 2)\), which simplifies to \(y = -\frac{1}{3}x + \frac{2}{3}\).

M1 for applying product rule. M1 for correct derivative of \(\sqrt{2x+5}\). A1 for combining into a single fraction. A1 for identifying a=3 and b=3. B1 for finding y = 0 at x = 2. M1 for finding normal gradient -1/3. A1 for correct equation of the normal.

(a) Substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) into \(2\cos^2 \theta + 3\sin \theta - 3 = 0\): \(2(1 - \sin^2 \theta) + 3\sin \theta - 3 = 0 \Rightarrow 2 - 2\sin^2 \theta + 3\sin \theta - 3 = 0 \Rightarrow -2\sin^2 \theta + 3\sin \theta - 1 = 0 \Rightarrow 2\sin^2 \theta - 3\sin \theta + 1 = 0\). (b) Let \(\theta = 2x\). Since \(0^\circ \le x \le 180^\circ\), \(0^\circ \le \theta \le 360^\circ\). Solve \(2\sin^2 \theta - 3\sin \theta + 1 = 0 \Rightarrow (2\sin \theta - 1)(\sin \theta - 1) = 0\). So \(\sin \theta = \frac{1}{2}\) or \(\sin \theta = 1\). For \(0^\circ \le \theta \le 360^\circ\), \(\theta = 30^\circ, 150^\circ\) or \(\theta = 90^\circ\). Thus, \(2x = 30^\circ, 90^\circ, 150^\circ \Rightarrow x = 15^\circ, 45^\circ, 75^\circ\).

M1 for using identity \(\cos^2 \theta = 1 - \sin^2 \theta\). A1 for showing the required quadratic form. M1 for factoring the quadratic in \(\sin(2x)\). A1 for finding the values \(\sin(2x) = 1/2\) and \(\sin(2x) = 1\). M1 for finding the values of \(2x\). A2 for all three correct angles of x (deduct 1 mark for any extra or missing solution).

(a) We are given \(S_2 = a + ar = a(1 + r) = 15\) and \(S_\infty = \frac{a}{1-r} = 27\). From the second equation, \(a = 27(1 - r)\). Substituting this into the first equation: \(27(1 - r)(1 + r) = 15 \Rightarrow 27(1 - r^2) = 15 \Rightarrow 1 - r^2 = \frac{15}{27} = \frac{5}{9} \Rightarrow r^2 = 1 - \frac{5}{9} = \frac{4}{9}\). (b) Since \(r^2 = \frac{4}{9}\), we have \(r = \frac{2}{3}\) or \(r = -\frac{2}{3}\) (both values have absolute value less than 1, so they are valid for an infinite sum to exist). If \(r = \frac{2}{3}\), \(a = 27(1 - \frac{2}{3}) = 9\). If \(r = -\frac{2}{3}\), \(a = 27(1 + \frac{2}{3}) = 45\).

B1 for expressing \(S_2 = a(1+r) = 15\). B1 for expressing \(S_\infty = \frac{a}{1-r} = 27\). M1 for substituting one equation into the other. A1 for completing the algebra to show \(r^2 = \frac{4}{9}\). B1 for stating the two possible values of r. M1 for substituting r to find a. A1 for both correct values of a.

(a) Convert \(\log_9(x+4)\) to base 3: \(\log_9(x+4) = \frac{\log_3(x+4)}{\log_3 9} = \frac{1}{2}\log_3(x+4)\). The equation becomes: \(\log_3(x-2) + \frac{1}{2}\log_3(x+4) = 2 \Rightarrow 2\log_3(x-2) + \log_3(x+4) = 4 \Rightarrow \log_3((x-2)^2(x+4)) = 4 \Rightarrow (x-2)^2(x+4) = 3^4 = 81\). Expand and simplify: \((x^2 - 4x + 4)(x + 4) = 81 \Rightarrow x^3 + 4x^2 - 4x^2 - 16x + 4x + 16 = 81 \Rightarrow x^3 - 12x - 65 = 0\). (b) Since \(x = 5\) gives \(5^3 - 12(5) - 65 = 125 - 60 - 65 = 0\), \(x = 5\) is a root. Factorizing: \(x^3 - 12x - 65 = (x - 5)(x^2 + 5x + 13) = 0\). The quadratic factor \(x^2 + 5x + 13 = 0\) has discriminant \(D = 25 - 4(13) = -27 < 0\), so it has no real roots. Therefore, \(x = 5\) is the only real solution. (Note: \(x = 5\) is valid since \(x > 2\) is required for \(\log_3(x-2)\) to be defined).

M1 for change of base to 3. M1 for combining logarithms using log laws. A1 for removing logarithms to get a cubic. A1 for correct expansion to \(x^3 - 12x - 65 = 0\). B1 for showing x = 5 is a root. M1 for factorizing into linear and quadratic terms. A1 for using discriminant to show the quadratic has no real roots.

(a) If they do not intersect, then the equation \(kx - 1 = x^2 + (2k - 3)x + 3\) has no real roots. Rearranging gives: \(x^2 + (2k - 3 - k)x + 4 = 0 \Rightarrow x^2 + (k - 3)x + 4 = 0\). For no real roots, the discriminant must be less than zero: \(D = (k - 3)^2 - 4(1)(4) < 0 \Rightarrow k^2 - 6k + 9 - 16 < 0 \Rightarrow k^2 - 6k - 7 < 0\). (b) Factorize \(k^2 - 6k - 7 < 0 \Rightarrow (k - 7)(k + 1) < 0\). The critical values are \(k = -1\) and \(k = 7\). For the inequality to be less than zero, \(k\) must lie between these values: \(-1 < k < 7\).

M1 for setting equations equal to each other. A1 for correct simplified quadratic equation. M1 for setting discriminant \(D < 0\). A1 for showing \(k^2 - 6k - 7 < 0\). M1 for finding critical values of k. A1 for correct factorisation or inequality setup. A1 for correct final range -1 < k < 7.

(a) Substitute \(y = 2x - 5\) into \(x^2 - xy + y^2 = 7\): \(x^2 - x(2x - 5) + (2x - 5)^2 = 7 \Rightarrow x^2 - 2x^2 + 5x + 4x^2 - 20x + 25 = 7 \Rightarrow 3x^2 - 15x + 18 = 0\). Divide both sides by 3 to get the required quadratic: \(x^2 - 5x + 6 = 0\). (b) Solve \(x^2 - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0\). Thus \(x = 2\) or \(x = 3\). If \(x = 2\), \(y = 2(2) - 5 = -1\). If \(x = 3\), \(y = 2(3) - 5 = 1\). So the solutions are \((2, -1)\) and \((3, 1)\).

M1 for correct substitution of y. M1 for expansion of \((2x - 5)^2\). A1 for correct grouping of terms to get \(3x^2 - 15x + 18 = 0\). A1 for division by 3 to reach the final form. M1 for factorising the quadratic to find x = 2, 3. A1 for finding correct corresponding y values. A1 for writing solutions clearly as pairs.

(a) Set the curve equal to the line: \(x^2 - 4x + 5 = 5 \Rightarrow x^2 - 4x = 0 \Rightarrow x(x - 4) = 0\). So \(x = 0\) or \(x = 4\). When \(x = 0\), \(y = 5\). When \(x = 4\), \(y = 5\). The points of intersection are \((0, 5)\) and \((4, 5)\). (b) The area is given by the integral of the upper line minus the lower curve from \(x = 0\) to \(x = 4\): \(\text{Area} = \int_{0}^{4} (5 - (x^2 - 4x + 5)) \mathrm{d}x = \int_{0}^{4} (4x - x^2) \mathrm{d}x = [2x^2 - \frac{x^3}{3}]_{0}^{4} = (2(4)^2 - \frac{4^3}{3}) - 0 = (32 - \frac{64}{3}) = \frac{96 - 64}{3} = \frac{32}{3}\).

M1 for setting equations equal and finding x-values. A1 for giving full coordinates of the intersection points. M1 for setting up the correct definite integral of (line - curve). A1 for correct integration to find \(2x^2 - \frac{x^3}{3}\). M1 for correct substitution of limits. A1 for simplifying to the final answer of \(\frac{32}{3}\).

For perpendicular vectors, the scalar product is 0.

\(\mathbf{p} \cdot \mathbf{q} = (k-7)k + 2(6) = 0\)

\(k^2 - 7k + 12 = 0\)

\((k-3)(k-4) = 0\)

\(k = 3\) or \(k = 4\)

M1: For setting up the scalar product equal to zero: \((k-7)k + 2(6) = 0\)
M1: For forming a correct 3-term quadratic equation: \(k^2 - 7k + 12 = 0\)
A1: For both correct values of \(k\): \(k = 3, k = 4\)

Equate the line and the curve:

\(mx - 5 = x^2 - 3x - 1 \Rightarrow x^2 - (m+3)x + 4 = 0\)

For no intersection, the discriminant \(\Delta < 0\):

\(\Delta = (m+3)^2 - 4(1)(4) < 0\)

\((m+3)^2 - 16 < 0\)

\((m+3-4)(m+3+4) < 0 \Rightarrow (m-1)(m+7) < 0\)

Thus, \(-7 < m < 1\).

M1: For equating the expressions and forming a quadratic equation: \(x^2 - (m+3)x + 4 = 0\)
M1: For using \(\Delta < 0\) to obtain \((m+3)^2 - 16 < 0\) or equivalent
A1: For correct set of values: \(-7 < m < 1\) (allow equivalent notation)

Replace \(\tan^2 \theta\) with \(\sec^2 \theta - 1\):

\(3(\sec^2 \theta - 1) - 5 \sec \theta + 1 = 0\)

\(3 \sec^2 \theta - 5 \sec \theta - 2 = 0\)

\((3 \sec \theta + 1)(\sec \theta - 2) = 0\)

Since \(\sec \theta = -1/3\) has no solution (as \(\cos \theta = -3\)), we solve \(\sec \theta = 2 \Rightarrow \cos \theta = 0.5\).

\(\theta = 60^\circ\) or \(\theta = 360^\circ - 60^\circ = 300^\circ\).

M1: For using the identity \(\tan^2 \theta = \sec^2 \theta - 1\) to obtain a quadratic in \(\sec \theta\)
M1: For solving the quadratic to get \(\sec \theta = 2\) (or \(\cos \theta = 0.5\))
A1: For both correct angles: \(\theta = 60^\circ, 300^\circ\)

Using the common ratio property of a geometric progression:

\(\frac{x+2}{x-1} = \frac{3x}{x+2}\)

\((x+2)^2 = 3x(x-1)\)

\(x^2 + 4x + 4 = 3x^2 - 3x\)

\(2x^2 - 7x - 4 = 0\)

\((2x+1)(x-4) = 0\)

Since all terms are positive, \(x > 1\), so we choose \(x = 4\).

M1: For setting up the ratio equation: \(\frac{x+2}{x-1} = \frac{3x}{x+2}\)
M1: For solving the quadratic equation \(2x^2 - 7x - 4 = 0\) to find \(x = 4\) and \(x = -0.5\)
A1: For identifying \(x = 4\) as the only valid solution since terms must be positive

Use the laws of logarithms:

\(\lg(x^2) - \lg(x+4) = \lg 2\)

\(\lg\left(\frac{x^2}{x+4}\right) = \lg 2\)

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

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

\((x-4)(x+2) = 0\)

Since \(x > 0\), \(x = 4\).

M1: For applying logarithm laws correctly to obtain \(\frac{x^2}{x+4} = 2\)
A1: For solving and obtaining the single valid solution \(x = 4\) (must reject \(x = -2\))

Identify the three mutually exclusive cases for the committee of 5 with at least 3 women:

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

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

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

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

M1: For finding the number of ways for at least two correct cases (e.g., \(10 \times 15\) or \(5 \times 6\))
M1: For sum of all three correct combinations: \(\binom{5}{3}\binom{6}{2} + \binom{5}{4}\binom{6}{1} + \binom{5}{5}\binom{6}{0}\)
A1: For correct final answer of 181

Let \(Y = \ln y\) and \(X = x^2\). The points on the straight-line graph are \((2, 5)\) and \((6, 13)\).

Gradient \(m = \frac{13 - 5}{6 - 2} = 2\).

Using the line equation \(Y - Y_1 = m(X - X_1)\):

\(\ln y - 5 = 2(x^2 - 2)\)

\(\ln y = 2x^2 + 1\).

M1: For finding the gradient \(m = 2\) and attempting to write the linear equation
A1: For correct final equation: \(\ln y = 2x^2 + 1\)

Differentiate \(y = x^2 \ln x\) using the product rule:

\(\frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x} = x(2\ln x + 1)\)

At the stationary point, \(\frac{dy}{dx} = 0\):

\(x(2\ln x + 1) = 0\)

Since \(x > 0\), we must have:

\(2\ln x + 1 = 0 \Rightarrow \ln x = -0.5 \Rightarrow x = e^{-0.5} = \frac{1}{\sqrt{e}}\)

Substitute \(x\) back into the original equation to find \(y\):

\(y = (e^{-0.5})^2 \ln(e^{-0.5}) = e^{-1}(-0.5) = -\frac{1}{2e}\)

So the coordinates are \(\left(e^{-0.5}, -\frac{1}{2e}\right)\).

M1: For differentiating correctly using the product rule: \(\frac{dy}{dx} = 2x \ln x + x\)
M1: For setting \(\frac{dy}{dx} = 0\) and solving to find \(x = e^{-0.5}\)
A1: For correct coordinates \(\left(e^{-0.5}, -\frac{1}{2e}\right)\) or equivalent exact form

The team can have either 3 nurses and 1 doctor, or 4 nurses and 0 doctors.

Case 1: 3 nurses and 1 doctor
\(\binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60\)

Case 2: 4 nurses and 0 doctors
\(\binom{5}{4} \times \binom{6}{0} = 5 \times 1 = 5\)

Total number of ways = 60 + 5 = 65

M1: for attempting to calculate at least one of the combinations, e.g. \(\binom{5}{3} \times \binom{6}{1}\) or \(\binom{5}{4}\)
M1: for adding the two mutually exclusive cases
A1: for 65

First, calculate \(\mathbf{q} - \mathbf{p}\):
\(\mathbf{q} - \mathbf{p} = (10\mathbf{i} + 12\mathbf{j}) - (2\mathbf{i} - 3\mathbf{j}) = 8\mathbf{i} + 15\mathbf{j}\)

Next, calculate the magnitude of the vector:
\(|8\mathbf{i} + 15\mathbf{j}| = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\)

Finally, find the unit vector:
\(\frac{1}{17}(8\mathbf{i} + 15\mathbf{j})\)

M1: for finding the vector difference \(8\mathbf{i} + 15\mathbf{j}\)
M1: for finding the magnitude 17 and dividing their vector by it
A1: for \(\frac{1}{17}(8\mathbf{i} + 15\mathbf{j})\) or equivalent

The given line is \(2x - 5y = 10\), which can be written as \(y = \frac{2}{5}x - 2\).
The gradient of this line is \(\frac{2}{5}\).

Since the required line is perpendicular, its gradient is \(m = -\frac{5}{2}\).

Using the point-slope form with \((4, -3)\):
\(y - (-3) = -\frac{5}{2}(x - 4)\)
\(y + 3 = -\frac{5}{2}x + 10\)
Multiply by 2 to clear fractions:
\(2y + 6 = -5x + 20\)
Rearrange into the form \(ax + by = c\):
\(5x + 2y = 14\)

M1: for finding the perpendicular gradient of \(-\frac{5}{2}\)
A1: for the correct equation in the specified form, e.g. \(5x + 2y = 14\) or any non-zero integer multiple

For the quadratic equation to have no real roots, the discriminant must be less than 0.
\(\Delta = b^2 - 4ac < 0\)
Here, \(a = 1\), \(b = k-2\), and \(c = k+1\).
\(\Delta = (k-2)^2 - 4(1)(k+1) < 0\)
\(k^2 - 4k + 4 - 4k - 4 < 0\)
\(k^2 - 8k < 0\)
\(k(k-8) < 0\)

Thus, \(0 < k < 8\).

M1: for using \(b^2 - 4ac < 0\) with correct values
A1: for simplifying to \(k^2 - 8k < 0\)
A1: for the correct range \(0 < k < 8\)

The equation of the circle with center \((4, 2)\) and radius \(\sqrt{10}\) is given by:
\((x-4)^2 + (y-2)^2 = 10\)

Substitute the equation of the line \(y = 5 - 2x\) into the circle's equation:
\((x-4)^2 + (5 - 2x - 2)^2 = 10\)
\((x-4)^2 + (3 - 2x)^2 = 10\)

Expand both squared terms:
\((x^2 - 8x + 16) + (9 - 12x + 4x^2) = 10\)
\(5x^2 - 20x + 25 = 10\)
\(5x^2 - 20x + 15 = 0\)

Divide the entire equation by 5:
\(x^2 - 4x + 3 = 0\)

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

This gives the x-coordinates of the intersection points:
\(x = 1\) or \(x = 3\)

Find the corresponding y-coordinates using \(y = 5 - 2x\):
For \(x = 1\): \(y = 5 - 2(1) = 3\)
For \(x = 3\): \(y = 5 - 2(3) = -1\)

Therefore, the coordinates of the points of intersection are \((1, 3)\) and \((3, -1)\).

M1: For writing down the correct equation of the circle: \((x-4)^2 + (y-2)^2 = 10\).
M1: For substituting the linear equation into the circle equation.
M1: For expanding and simplifying to a 3-term quadratic equation \(5x^2 - 20x + 15 = 0\) (or equivalent).
A1: For correct x-values: \(x=1\) and \(x=3\).
A1: For both correct coordinates: \((1, 3)\) and \((3, -1)\).

Start by changing the base of the first term to base 2:
\(\log_4(5y - 1) = \frac{\log_2(5y - 1)}{\log_2 4} = \frac{\log_2(5y - 1)}{2}\)

Substitute this back into the original equation:
\(\frac{1}{2}\log_2(5y - 1) - \log_2(y) = 1\)

Multiply the entire equation by 2:
\(\log_2(5y - 1) - 2\log_2(y) = 2\)

Apply the power law of logarithms to the second term:
\(\log_2(5y - 1) - \log_2(y^2) = 2\)

Apply the division law of logarithms:
\(\log_2\left(\frac{5y - 1}{y^2}\right) = 2\)

Convert from logarithmic to exponential form:
\(\frac{5y - 1}{y^2} = 2^2\)

\(\frac{5y - 1}{y^2} = 4\)

Multiply by \(y^2\) and rearrange into a standard quadratic form:
\(5y - 1 = 4y^2\)

\(4y^2 - 5y + 1 = 0\)

Factorise the quadratic equation:
\((4y - 1)(y - 1) = 0\)

This yields the solutions:
\(y = \frac{1}{4} = 0.25\) or \(y = 1\)

Both values satisfy the original log constraints (\(5y - 1 > 0\) and \(y > 0\)).

M1: For applying the change of base formula to express \(\log_4(5y - 1)\) in terms of base 2.
M1: For applying log laws (power law or division law) to combine the terms.
M1: For removing logarithms correctly to get an equation of the form \(\frac{5y-1}{y^2} = 4\).
M1: For rearranging into a solvable 3-term quadratic equation.
A1: For finding the values \(y = 0.25\) and \(y = 1\).
A1: For confirming both values are valid within the domain of the logarithmic function.

(a) For a team of 5 with at least 3 women, we consider the following mutually exclusive cases:

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

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

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

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

(b) Total unrestricted ways to choose 5 members from 11 people (6 men + 5 women) is:
Total = \(\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462\)

Now find the number of ways where both the particular man and the particular woman are selected. This means we have already chosen these 2 specific people, so we need to choose the remaining 3 members from the remaining 9 people:
Ways with both selected = \(\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\)

Therefore, the number of ways where they cannot both be on the team is:
Required ways = \(462 - 84 = 378\)

(a)
M1: For identifying and writing down the combinations for the three cases (3W/2M, 4W/1M, 5W).
M1: For calculating the values of the combinations correctly for at least two cases.
A1: For the correct total of 181.

(b)
M1: For calculating the total unrestricted ways \(\binom{11}{5} = 462\) and subtracting the restricted ways \(\binom{9}{3} = 84\) (or equivalent correct casework).
A1: For the correct final answer of 378.

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

Expand and simplify:
\(3 + 3\tan^2 x - 5\tan x - 5 = 0\)

\(3\tan^2 x - 5\tan x - 2 = 0\)

Factorise the quadratic equation in terms of \(\tan x\):
\((3\tan x + 1)(\tan x - 2) = 0\)

This gives:
\(\tan x = 2\) or \(\tan x = -\frac{1}{3}\)

Solve for \(x\) in the range \(0 \le x \le \pi\):

Case 1: \(\tan x = 2\)
Since 2 is positive, \(x\) lies in the first quadrant:
\(x = \arctan(2) \approx 1.107\) radians.

Case 2: \(\tan x = -\frac{1}{3}\)
Since \(-\frac{1}{3}\) is negative, \(x\) lies in the second quadrant:
\(x = \pi - \arctan\left(\frac{1}{3}\right) \approx 3.1416 - 0.3218 = 2.820\) radians.

Correct to 2 decimal places, the solutions are \(x = 1.11\) and \(x = 2.82\).

M1: For using the identity \(\sec^2 x = 1 + \tan^2 x\) to write the equation in terms of \(\tan x\) only.
M1: For simplifying to a 3-term quadratic equation \(3\tan^2 x - 5\tan x - 2 = 0\).
M1: For solving the quadratic equation to get \(\tan x = 2\) and \(\tan x = -1/3\).
M1: For attempting to find a radian angle in either quadrant within the interval \([0, \pi]\).
A1: For \(x = 1.11\) (allow awrt 1.11).
A1: For \(x = 2.82\) (allow awrt 2.82).

(a) To find \(\mathrm{f}^{-1}(x)\), let \(y = \frac{3x+1}{x-2}\) and solve for \(x\):
\(y(x-2) = 3x+1\)
\(xy - 2y = 3x+1\)
\(xy - 3x = 2y+1\)
\(x(y-3) = 2y+1\)
\(x = \frac{2y+1}{y-3}\)

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

To find the domain of \(\mathrm{f}^{-1}(x)\), we find the range of \(\mathrm{f}(x)\) for \(x > 2\):
\(\mathrm{f}(x) = \frac{3(x-2) + 7}{x-2} = 3 + \frac{7}{x-2}\)
Since \(x > 2\), \(x-2 > 0\), which implies \(\frac{7}{x-2} > 0\).
Therefore, \(\mathrm{f}(x) > 3\).

Since the domain of the inverse function is the range of the original function, the domain of \(\mathrm{f}^{-1}(x)\) is \(x > 3\).

(b) First find \(\mathrm{f}(3)\):
\(\mathrm{f}(3) = \frac{3(3)+1}{3-2} = \frac{10}{1} = 10\)

Now find \(\mathrm{ff}(3) = \mathrm{f}(10)\):
\(\mathrm{f}(10) = \frac{3(10)+1}{10-2} = \frac{31}{8} = 3.875\).

(a)
M1: For attempting to rearrange the equation \(y = \frac{3x+1}{x-2}\) to make \(x\) the subject.
A1: For obtaining \(\mathrm{f}^{-1}(x) = \frac{2x+1}{x-3}\) (or equivalent form).
B1: For stating the domain correctly as \(x > 3\).

(b)
M1: For evaluating \(\mathrm{f}(3) = 10\) and attempting to substitute this result back into \(\mathrm{f}(x)\).
A1: For the final answer \(3.875\) (or \(\frac{31}{8}\)).

Let the geometric progression have first term \(a\) and common ratio \(r\).

The sum of the first three terms is given by:
\(a(1 + r + r^2) = 26\) --- (Equation 1)

The sum to infinity is given by:
\(\frac{a}{1-r} = 27\) --- (Equation 2)

From Equation 2, we can express \(a\) in terms of \(r\):
\(a = 27(1-r)\)

Substitute this expression for \(a\) into Equation 1:
\(27(1-r)(1 + r + r^2) = 26\)

Since \((1-r)(1+r+r^2) = 1 - r^3\) (the difference of cubes identity), the equation simplifies to:
\(27(1 - r^3) = 26\)

Divide both sides by 27:

\(1 - r^3 = \frac{26}{27}\)

\(r^3 = 1 - \frac{26}{27}\)

\(r^3 = \frac{1}{27}\)

Taking the cube root of both sides gives the common ratio:
\(r = \frac{1}{3}\)

Now, substitute \(r = \frac{1}{3}\) back into the expression for \(a\):
\(a = 27\left(1 - \frac{1}{3}\right) = 27\left(\frac{2}{3}\right) = 18\)

Thus, the first term is \(a = 18\) and the common ratio is \(r = \frac{1}{3}\).

M1: For using the sum of the first three terms to write \(a(1+r+r^2) = 26\) (or equivalent).
M1: For using the sum to infinity formula to write \(\frac{a}{1-r} = 27\).
M1: For substituting \(a = 27(1-r)\) into the first equation to obtain an equation in terms of \(r\) only.
A1: For simplifying the equation correctly to \(27(1-r^3) = 26\) (or equivalent).
A1: For solving to find \(r = \frac{1}{3}\).
A1: For evaluating \(a = 18\).

First, find the points of intersection by setting the curve equal to the line:
\(3\sqrt{x} = x + 2\)

Square both sides of the equation:
\(9x = (x+2)^2\)

\(9x = x^2 + 4x + 4\)

Rearrange into a standard quadratic equation:
\(x^2 - 5x + 4 = 0\)

Factorise the quadratic:
\((x-1)(x-4) = 0\)

So, the intersection points occur at \(x = 1\) and \(x = 4\).

To find the area of the region, integrate the difference between the upper function (the curve) and the lower function (the line) from \(x = 1\) to \(x = 4\):

\(\text{Area} = \int_{1}^{4} (3\sqrt{x} - (x+2)) \,\mathrm{d}x\)

\(\text{Area} = \int_{1}^{4} (3x^{1/2} - x - 2) \,\mathrm{d}x\)

Integrate each term with respect to \(x\):

\(\int (3x^{1/2} - x - 2) \,\mathrm{d}x = \left[ 2x^{3/2} - \frac{1}{2}x^2 - 2x \right]_{1}^{4}\)

Evaluate the definite integral at the upper limit \(x = 4\):

\(\left( 2(4)^{3/2} - \frac{1}{2}(4)^2 - 2(4) \right) = 2(8) - 8 - 8 = 16 - 16 = 0\)

Evaluate the definite integral at the lower limit \(x = 1\):

\(\left( 2(1)^{3/2} - \frac{1}{2}(1)^2 - 2(1) \right) = 2 - 0.5 - 2 = -0.5\)

Calculate the total area:

\(\text{Area} = 0 - (-0.5) = 0.5\).

M1: For setting up the intersection equation \(3\sqrt{x} = x + 2\) and squaring both sides.
A1: For solving to find the limits of integration: \(x=1\) and \(x=4\).
M1: For attempting to integrate the expression \(3x^{1/2} - x - 2\) (power of at least one term increased correctly).
A1: For the correct integrated expression: \(2x^{3/2} - \frac{1}{2}x^2 - 2x\).
M1: For substituting the limits \(4\) and \(1\) into their integrated expression.
A1: For obtaining the correct area of \(0.5\) (or \(1/2\)).

The point \(C\) divides the line segment \(AB\) internally in the ratio \(2:1\). By the section formula (ratio theorem), the position vector of \(C\) is given by:
\(\mathbf{c} = \frac{1\mathbf{a} + 2\mathbf{b}}{1+2} = \frac{\mathbf{a} + 2\mathbf{b}}{3}\)

Multiply both sides by 3:
\(3\mathbf{c} = \mathbf{a} + 2\mathbf{b}\)

Substitute the given vectors into this relation:
\(3(5\mathbf{i} + 2\mathbf{j}) = (3\mathbf{i} + \mu\mathbf{j}) + 2(\lambda\mathbf{i} - 2\mathbf{j})\)

Simplify both sides:
\(15\mathbf{i} + 6\mathbf{j} = 3\mathbf{i} + \mu\mathbf{j} + 2\lambda\mathbf{i} - 4\mathbf{j}\)

\(15\mathbf{i} + 6\mathbf{j} = (3 + 2\lambda)\mathbf{i} + (\mu - 4)\mathbf{j}\)

Equate the corresponding \(\mathbf{i}\) and \(\mathbf{j}\) components:

For the \(\mathbf{i}\) component:
\(15 = 3 + 2\lambda\)

\(2\lambda = 12\)

\(\lambda = 6\)

For the \(\mathbf{j}\) component:
\(6 = \mu - 4\)

\(\mu = 10\)

So, \(\lambda = 6\) and \(\mu = 10\).

M1: For using the ratio theorem or vector division to express \(\mathbf{c}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), such as \(\mathbf{c} = \frac{\mathbf{a} + 2\mathbf{b}}{3}\) (or setting up \(\overrightarrow{AC} = 2\overrightarrow{CB}\)).
M1: For substituting the vector expressions into their vector equation.
M1: For equating the \(\mathbf{i}\) and \(\mathbf{j}\) components to obtain two separate scalar equations in terms of \(\lambda\) and \(\mu\).
A1: For finding \(\lambda = 6\).
A1: For finding \(\mu = 10\).

(a) First, we select exactly 2 rock songs from the 6 available rock songs:
\[\binom{6}{2} = \frac{6 \times 5}{2} = 15\text{ ways}\]
Next, we need to select the remaining 3 tracks from the 5 pop songs and 4 classical pieces such that we have at least 2 pop songs.
There are two mutually exclusive cases for the remaining 3 tracks:
Case 1: Exactly 2 pop songs and 1 classical piece:
\[\binom{5}{2} \times \binom{4}{1} = 10 \times 4 = 40\text{ ways}\]
Case 2: Exactly 3 pop songs and 0 classical pieces:
\[\binom{5}{3} \times \binom{4}{0} = 10 \times 1 = 10\text{ ways}\]
Total ways to select the remaining 3 tracks: \(40 + 10 = 50\) ways.

Therefore, the total number of ways to make the selection is:
\[15 \times 50 = 750\]

(b) To arrange the 15 tracks such that the 4 classical pieces are next to each other, we treat the 4 classical pieces as a single block.
This leaves us with 6 rock songs + 5 pop songs + 1 block of classical pieces = 12 items to arrange.
These 12 items can be arranged in \(12!\) ways.
Within the classical block, the 4 classical pieces can be arranged in \(4!\) ways.

Thus, the total number of arrangements is:
\[12! \times 4! = 24 \times 12!\]

Comparing this to the form \(a \times 12!\), we find:
\[a = 24\]

(a)
- M1: For calculating the number of ways to choose rock songs: \(\binom{6}{2} = 15\)
- M1: For identifying and summing the two cases for the remaining songs: \(\binom{5}{2}\binom{4}{1} + \binom{5}{3}\binom{4}{0} = 50\)
- A1: Correct final answer of 750

(b)
- M1: For realizing the block method and writing the product of arrangements, e.g., showing \(12! \times 4!\) or \(12! \times 24\)
- A1: Correct value of \(a = 24\)