2023 Cambridge IGCSE Mathematics - Additional (0606) 模拟试题及答案详解

Using the laws of logarithms:
\(\log_2((x + 3)(x - 3)) = 4\)
\(\log_2(x^2 - 9) = 4\)

Convert from logarithmic to exponential form:
\(x^2 - 9 = 2^4\)
\(x^2 - 9 = 16\)
\(x^2 = 25\)
\(x = \pm 5\)

Since \(\log_2(x - 3)\) requires \(x > 3\), we reject \(x = -5\).
Thus, the only valid solution is \(x = 5\).

M1: Use the product rule of logarithms to combine terms into \(\log_2(x^2 - 9) = 4\).
M1: Convert correctly to exponential form to obtain \(x^2 - 9 = 16\) and solve for \(x\).
A1: Identify \(x = 5\) as the only valid solution (must explicitly reject or omit \(x = -5\)).

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

Factorising the quadratic equation:
\((2\sin\theta - 1)(\sin\theta + 1) = 0\)

This gives:
\(\sin\theta = \frac{1}{2}\) or \(\sin\theta = -1\)

For the interval \(0 \le \theta \le \pi\):
- From \(\sin\theta = \frac{1}{2}\), we have \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).
- From \(\sin\theta = -1\), there are no solutions in the given interval.

Thus, the solutions are \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\).

M1: Use the identity \(\cos^2\theta = 1 - \sin^2\theta\) to form a three-term quadratic equation in terms of \(\sin\theta\).
M1: Factorise or solve their quadratic to obtain at least \(\sin\theta = \frac{1}{2}\).
A1: State both exact values of \(\theta\) correctly: \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) (no extra solutions in the range).

Rewrite the integrand with fractional indices:
\(\int_{1}^{4} \left( 3x^{\frac{1}{2}} - 2x^{-2} \right) \mathrm{d}x\)

Integrate term-by-term:
\(= \left[ 3 \left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right) - 2 \left(\frac{x^{-1}}{-1}\right) \right]_1^4\)
\(= \left[ 2x^{\frac{3}{2}} + \frac{2}{x} \right]_1^4\)

Substitute the limits:
Upper limit: \(2(4)^{\frac{3}{2}} + \frac{2}{4} = 2(8) + 0.5 = 16.5\)
Lower limit: \(2(1)^{\frac{3}{2}} + \frac{2}{1} = 2 + 2 = 4\)

Subtract the values:
\(16.5 - 4 = 12.5\) (or \(\frac{25}{2}\))

M1: Attempt integration, achieving at least one term correct of \(2x^{\frac{3}{2}}\) or \(\frac{2}{x}\).
M1: Show correct substitution of limits into their integrated expression.
A1: Obtain the correct final exact value of \(12.5\) (or \(\frac{25}{2}\)).

Multiply the numerator and denominator by the conjugate of the denominator, which is \(3 - \sqrt{5}\):
\(\frac{(7 + 3\sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}\)

Expand the numerator:
\((7 + 3\sqrt{5})(3 - \sqrt{5}) = 21 - 7\sqrt{5} + 9\sqrt{5} - 15 = 6 + 2\sqrt{5}\)

Expand the denominator:
\((3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4\)

Simplify the fraction:
\(\frac{6 + 2\sqrt{5}}{4} = \frac{6}{4} + \frac{2}{4}\sqrt{5} = 1.5 + 0.5\sqrt{5}\) (or \(\frac{3}{2} + \frac{1}{2}\sqrt{5}\))

M1: Multiply both numerator and denominator by the conjugate \(3 - \sqrt{5}\).
M1: Expand the numerator correctly to obtain \(6 + 2\sqrt{5}\) (allow at most one arithmetic error).
A1: Obtain the correct simplified expression: \(1.5 + 0.5\sqrt{5}\) or \(\frac{3}{2} + \frac{1}{2}\sqrt{5}\).

Using the arithmetic sum formula \(S_n = \frac{n}{2}[2a + (n - 1)d]\):
\(860 = \frac{n}{2}[2(5) + (n - 1)4]\)
\(1720 = n[10 + 4n - 4]\)
\(1720 = n[4n + 6]\)
\(1720 = 4n^2 + 6n\)
\(4n^2 + 6n - 1720 = 0\)

Divide the entire equation by 2:
\(2n^2 + 3n - 860 = 0\)

Factorising the quadratic equation:
\((n - 20)(2n + 43) = 0\)

Since the number of terms \(n\) must be a positive integer, we reject the solution \(n = -\frac{43}{2}\).
Therefore, \(n = 20\).

M1: Set up a correct equation using the arithmetic sum formula with the given parameters.
M1: Rearrange the equation into a simplified quadratic form: \(2n^2 + 3n - 860 = 0\).
A1: State the final answer \(n = 20\) only, correctly discarding the negative root.

For the equation to have two distinct real roots, the discriminant \(\Delta\) must be greater than zero:
\(\Delta = b^2 - 4ac > 0\)

Here, \(a = 1\), \(b = k + 2\), and \(c = 2k + 1\):
\((k + 2)^2 - 4(1)(2k + 1) > 0\)
\(k^2 + 4k + 4 - (8k + 4) > 0\)
\(k^2 - 4k > 0\)
\(k(k - 4) > 0\)

This inequality holds when \(k < 0\) or \(k > 4\).

M1: State the condition \(b^2 - 4ac > 0\) and substitute correct values of \(a, b, c\).
M1: Simplify the discriminant to obtain \(k^2 - 4k > 0\) and determine the critical values \(0\) and \(4\).
A1: State the correct range of values: \(k < 0\) or \(k > 4\) (or equivalent set notation).

If two vectors are perpendicular, the product of their gradients is \(-1\), or their dot product is \(0\).

Method using gradients:
Gradient of \(\mathbf{a}\): \(m_a = -\frac{4}{3}\)
Gradient of \(\mathbf{b}\): \(m_b = \frac{12}{p}\)

Since \(\mathbf{a}\) is perpendicular to \(\mathbf{b}\):
\(m_a \times m_b = -1\)
\(-\frac{4}{3} \times \frac{12}{p} = -1\)
\(-\frac{48}{3p} = -1\)
\(48 = 3p\)
\(p = 16\)

M1: Express the gradient of at least one of the vectors: \(m_a = -\frac{4}{3}\) or \(m_b = \frac{12}{p}\) (or set up the dot product equation \(3p + (-4)(12) = 0\)).
M1: Set up the equation using perpendicularity: \(-\frac{4}{3} \times \frac{12}{p} = -1\) (or simplify \(3p - 48 = 0\)).
A1: Correctly calculate \(p = 16\).

To contain at least 4 women, the team must consist of either:

Case 1: 4 women and 1 man
Number of ways = \(\binom{5}{4} \times \binom{7}{1} = 5 \times 7 = 35\)

Case 2: 5 women and 0 men
Number of ways = \(\binom{5}{5} \times \binom{7}{0} = 1 \times 1 = 1\)

Total number of ways = \(35 + 1 = 36\).

M1: Calculate the combinations for Case 1 (4 women and 1 man): \(\binom{5}{4} \times \binom{7}{1} = 35\).
M1: Identify Case 2 (5 women) and sum both cases: \(35 + \binom{5}{5}\).
A1: Obtain the correct total of \(36\).

Using the laws of logarithms:
\(\log_2((x+3)(x-3)) = 4\)
\(\log_2(x^2 - 9) = 4\)

Converting to exponential form:
\(x^2 - 9 = 2^4\)
\(x^2 - 9 = 16\)
\(x^2 = 25\)
\(x = 5\) or \(x = -5\)

Since the argument of a logarithm must be positive, \(x-3 > 0\) which means \(x > 3\).
Therefore, \(x = -5\) is rejected. The only valid solution is \(x = 5\).

M1: For applying the product rule of logarithms to obtain \(\log_2(x^2 - 9) = 4\) or equivalent.
M1: For converting to exponential form, \(x^2 - 9 = 16\).
A1: For \(x = 5\) only (must reject \(x = -5\)).

Differentiating \(y = x^2 \ln x\) using the product rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2x \ln x + x^2 \left(\frac{1}{x}\right)\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = x(2\ln x + 1)\)

For a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
Since \(x > 0\), we have:
\(2\ln x + 1 = 0\)
\(\ln x = -\frac{1}{2}\)
\(x = e^{-\frac{1}{2}}\)

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

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

M1: For an attempt to differentiate using the product rule (at least one term correct).
M1: For setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solving for \(x\) to find \(x = e^{-0.5}\) or equivalent.
A1: For both coordinates correct in exact form: \(\left(e^{-\frac{1}{2}}, -\frac{1}{2e}\right)\) or equivalent.

First, take the square root of both sides: \(\tan(2\theta - \frac{\pi}{4}) = \pm\sqrt{3}\). The basic angle is \(\frac{\pi}{3}\). Since \(0 \le \theta \le \pi\), the range for the argument is \(-\frac{\pi}{4} \le 2\theta - \frac{\pi}{4} \le \frac{7\pi}{4}\). Finding the values of \(2\theta - \frac{\pi}{4}\) in this interval: For \(\tan(2\theta - \frac{\pi}{4}) = \sqrt{3}\), we get \(2\theta - \frac{\pi}{4} = \frac{\pi}{3}, \frac{4\pi}{3}\). For \(\tan(2\theta - \frac{\pi}{4}) = -\sqrt{3}\), we get \(2\theta - \frac{\pi}{4} = \frac{2\pi}{3}, \frac{5\pi}{3}\) (note that \(-\frac{\pi}{3}\) is outside the range). Solving for \(\theta\): \(2\theta = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12} \implies \theta = \frac{7\pi}{24}\), \(2\theta = \frac{2\pi}{3} + \frac{\pi}{4} = \frac{11\pi}{12} \implies \theta = \frac{11\pi}{24}\), \(2\theta = \frac{4\pi}{3} + \frac{\pi}{4} = \frac{19\pi}{12} \implies \theta = \frac{19\pi}{24}\), \(2\theta = \frac{5\pi}{3} + \frac{\pi}{4} = \frac{23\pi}{12} \implies \theta = \frac{23\pi}{24}\).

M1: For taking the square root to get \(\tan(2\theta - \frac{\pi}{4}) = \pm\sqrt{3}\). B1: For identifying the basic angle as \(\frac{\pi}{3}\) (or 60 degrees). M1: For attempting to find values of the argument in the correct range. A1: For obtaining any two correct values of \(\theta\). A1: For obtaining all four correct values of \(\theta\) and no extras in range.

Use the change of base formula to write \(\log_b 2 = \frac{1}{\log_2 b}\). Let \(x = \log_2 b\). The equation becomes \(\frac{3}{x} = x - 2 \implies 3 = x^2 - 2x \implies x^2 - 2x - 3 = 0\). Factoring the quadratic: \((x - 3)(x + 1) = 0 \implies x = 3\) or \(x = -1\). Since \(x = \log_2 b\), we have: \(\log_2 b = 3 \implies b = 2^3 = 8\) or \(\log_2 b = -1 \implies b = 2^{-1} = 0.5\).

M1: For change of base formula \(\log_b 2 = \frac{1}{\log_2 b}\). M1: For substituting a variable \(x = \log_2 b\) to form a quadratic equation. A1: For obtaining \(x^2 - 2x - 3 = 0\). M1: For solving their quadratic equation to find two values of \(x\). A1: For both correct values of \(b\) (8 and 0.5).

Let the first term be \(a\) and the common ratio be \(r\). We are given: \(u_3 = ar^2 = 18\) and \(u_6 = ar^5 = 486\). Dividing the two equations: \(\frac{ar^5}{ar^2} = \frac{486}{18} \implies r^3 = 27 \implies r = 3\). Substituting \(r = 3\) into the first equation: \(a(3^2) = 18 \implies 9a = 18 \implies a = 2\). Using the sum formula for a geometric progression: \(S_7 = \frac{a(r^7 - 1)}{r - 1} = \frac{2(3^7 - 1)}{3 - 1} = 3^7 - 1 = 2187 - 1 = 2186\).

M1: For setting up two equations using \(ar^{n-1}\). M1: For dividing the equations to isolate and find \(r^3\). A1: For \(r = 3\). M1: For finding \(a = 2\). A1: For \(S_7 = 2186\).

Differentiating \(y\) with respect to \(x\) using the quotient rule: \(\frac{dy}{dx} = \frac{3(2x + 3) - 2(3x - 1)}{(2x + 3)^2} = \frac{6x + 9 - 6x + 2}{(2x + 3)^2} = \frac{11}{(2x + 3)^2}\). At \(x = 2\): \(\frac{dy}{dx} = \frac{11}{(2(2) + 3)^2} = \frac{11}{7^2} = \frac{11}{49}\). Since \(\delta y \approx \frac{dy}{dx} \delta x\) and \(\delta x = p\), the approximate change in \(y\) is \(\frac{11}{49}p\).

M1: For attempting to use the quotient rule. A1: For obtaining the correct derivative \(\frac{11}{(2x + 3)^2}\). M1: For substituting \(x = 2\) into their derivative. A1: For obtaining a gradient of \(\frac{11}{49}\) (or \(0.224\)). A1: For \(\frac{11}{49}p\) (or \(0.224p\)) as the final approximate change.

First, find vectors \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) in terms of \(\mu\): \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (\mu\mathbf{i} + (\mu-2)\mathbf{j}) - (4\mathbf{i} + 3\mathbf{j}) = (\mu - 4)\mathbf{i} + (\mu - 5)\mathbf{j}\). \(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\mu\mathbf{i} + (\mu-2)\mathbf{j}) - (10\mathbf{i} - 5\mathbf{j}) = (\mu - 10)\mathbf{i} + (\mu + 3)\mathbf{j}\). Equating the squared magnitudes \(|\overrightarrow{AC}|^2 = |\overrightarrow{BC}|^2\): \((\mu - 4)^2 + (\mu - 5)^2 = (\mu - 10)^2 + (\mu + 3)^2\) \(\mu^2 - 8\mu + 16 + \mu^2 - 10\mu + 25 = \mu^2 - 20\mu + 100 + \mu^2 + 6\mu + 9\) \(-18\mu + 41 = -14\mu + 109\) \(-4\mu = 68 \implies \mu = -17\).

M1: For finding \(\overrightarrow{AC}\) in terms of \(\mu\). M1: For finding \(\overrightarrow{BC}\) in terms of \(\mu\). M1: For setting up the equation for magnitudes squared. A1: For expanding and simplifying to a linear equation. A1: For \(\mu = -17\).

Factorise the quadratic: \(3x^2 - 14x - 5 < 0 \implies (3x + 1)(x - 5) < 0\). The critical values are \(x = -\frac{1}{3}\) and \(x = 5\). Since the inequality is less than zero, the solution range for \(x\) is \(-\frac{1}{3} < x < 5\). Now let \(x = 2^y\). We have \(-\frac{1}{3} < 2^y < 5\). Since \(2^y > 0\) for all real \(y\), the lower bound \(-\frac{1}{3} < 2^y\) is always satisfied. Thus, we only need to solve \(2^y < 5\). Taking \(\log_2\) of both sides gives: \(y < \log_2 5\).

M1: For attempting to find critical values by factoring or quadratic formula. A1: For correct critical values \(x = -\frac{1}{3}\) and \(x = 5\). A1: For range \(-\frac{1}{3} < x < 5\). M1: For substituting \(x = 2^y\) and recognizing \(2^y > 0\). A1: For the final answer \(y < \log_2 5\) (or equivalent exact logarithmic form).

(a) Integrate \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \) with respect to \( x \):
\( \frac{\mathrm{d}y}{\mathrm{d}x} = ax - 6x^2 + c \)

Since \( (2, 9) \) is a stationary point, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) when \( x = 2 \):
\( a(2) - 6(2)^2 + c = 0 \implies 2a + c = 24 \quad \text{--- (Equation 1)} \)

Integrate again to find \( y \):
\( y = \frac{1}{2}ax^2 - 2x^3 + cx + d \)

Since the curve passes through \( (0, -11) \), when \( x = 0 \), \( y = -11 \):
\( d = -11 \)
So, \( y = \frac{1}{2}ax^2 - 2x^3 + cx - 11 \)

Since the curve passes through \( (2, 9) \):
\( 9 = \frac{1}{2}a(2)^2 - 2(2)^3 + c(2) - 11 \)
\( 9 = 2a - 16 + 2c - 11 \)
\( 2a + 2c = 36 \implies a + c = 18 \quad \text{--- (Equation 2)} \)

Solving Equations 1 and 2 simultaneously:
From Equation 2, \( c = 18 - a \).
Substitute into Equation 1:
\( 2a + (18 - a) = 24 \implies a = 6 \)
Then, \( c = 12 \).

Therefore, \( a = 6 \) and the equation of the curve is:
\( y = 3x^2 - 2x^3 + 12x - 11 \)

(b) For stationary points, set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\( 6x - 6x^2 + 12 = 0 \implies x^2 - x - 2 = 0 \)
\( (x - 2)(x + 1) = 0 \)

So the other stationary point occurs at \( x = -1 \).
Substitute \( x = -1 \) into the equation of the curve:
\( y = 3(-1)^2 - 2(-1)^3 + 12(-1) - 11 = 3 + 2 - 12 - 11 = -18 \)

Coordinates of the other stationary point: \( (-1, -18) \)

(c) Use the second derivative to determine the nature:
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6 - 12x \)

At \( x = 2 \):
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6 - 12(2) = -18 < 0 \), so \( (2, 9) \) is a maximum point.

At \( x = -1 \):
\( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6 - 12(-1) = 18 > 0 \), so \( (-1, -18) \) is a minimum point.

(a)
M1: Attempt to integrate \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \) (at least one term correct, with constant \( c \)).
M1: Set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) at \( x = 2 \) to form an equation in \( a \) and \( c \).
M1: Attempt to integrate \( \frac{\mathrm{d}y}{\mathrm{d}x} \) to find \( y \) (including constant \( d \)), and identify \( d = -11 \).
M1: Substitute \( (2, 9) \) into \( y \) to form a second equation in \( a \) and \( c \).
A1: Correct value of \( a = 6 \).
A1: Correct equation of the curve: \( y = 3x^2 - 2x^3 + 12x - 11 \).

(b)
M1: Set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) and solve the quadratic to find \( x = -1 \).
A1: Correct coordinates: \( (-1, -18) \).

(c)
M1: Use of second derivative test with their \( x \)-values.
A1: Both correctly identified (maximum at \( (2, 9) \) and minimum at \( (-1, -18) \)).

(a) Write down expressions for the 2nd, 5th and 14th terms of the arithmetic progression:
\( u_2 = a + d \)
\( u_5 = a + 4d \)
\( u_{14} = a + 13d \)

Since these terms form a geometric progression, the common ratio is constant:
\( \frac{a + 4d}{a + d} = \frac{a + 13d}{a + 4d} \)

Cross-multiplying:
\( (a + 4d)^2 = (a + d)(a + 13d) \)
\( a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 \)

Subtracting \( a^2 \) from both sides and simplifying:
\( 3d^2 = 6ad \)

Since \( d \neq 0 \), we can divide both sides by \( 3d \):
\( d = 2a \) (shown)

(b) Use the sum of the first 10 terms formula:
\( S_{10} = \frac{10}{2}[2a + 9d] = 400 \)
\( 5[2a + 9d] = 400 \implies 2a + 9d = 80 \)

Substitute \( d = 2a \) into this equation:
\( 2a + 9(2a) = 80 \)
\( 20a = 80 \implies a = 4 \)

Since \( d = 2a \), we have \( d = 8 \).

(c) The first two terms of the geometric progression are:
\( u_2 = a + d = 4 + 8 = 12 \)
\( u_5 = a + 4d = 4 + 32 = 36 \)

Common ratio \( r = \frac{36}{12} = 3 \).

(d) For the second geometric progression:
First term \( A = a = 4 \)
Common ratio \( R = \frac{1}{d} = \frac{1}{8} \)

Since \( |R| < 1 \), the sum to infinity is:
\( S_{\infty} = \frac{A}{1 - R} = \frac{4}{1 - \frac{1}{8}} = \frac{4}{\frac{7}{8}} = \frac{32}{7} \) (or \( 4\frac{4}{7} \) or \( 4.57 \)).

(a)
B1: State the expressions \( a+d \), \( a+4d \), and \( a+13d \) (or equivalent).
M1: Write down a correct ratio or product relation for the GP terms.
M1: Expand both sides correctly to get \( a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 \).
M1: Simplify to \( 3d^2 = 6ad \).
A1: Divide by \( 3d \) (justified by \( d \neq 0 \)) to show \( d = 2a \).

(b)
M1: Set up the sum equation \( 5(2a + 9d) = 400 \) and substitute \( d = 2a \).
A1: Correct values: \( a = 4 \) and \( d = 8 \).

(c)
B1: Correct common ratio \( r = 3 \).

(d)
M1: Use of \( S_{\infty} = \frac{a}{1-r} \) with their \( a \) and \( r = \frac{1}{d} \).
A1: Correct sum to infinity \( \frac{32}{7} \) (or exact equivalent fraction/decimal).

To find the set of values of \(k\) where there is no intersection, we equate the equations:
\(kx - 2 = x^2 - 4x + 7\)

Rearranging into a quadratic equation in the standard form:
\(x^2 - (k+4)x + 9 = 0\)

For no points of intersection, the discriminant must be less than zero:
\(b^2 - 4ac < 0\)
\((-(k+4))^2 - 4(1)(9) < 0\)
\((k+4)^2 - 36 < 0\)
\((k+4)^2 < 36\)

Solving this inequality:
\(-6 < k + 4 < 6\)
\(-10 < k < 2\)

M1: For equating the line and the curve to form a 3-term quadratic equation in \(x\).
M1: For using the discriminant \(b^2 - 4ac < 0\) on their quadratic equation.
A1: For the correct final range of \(k\), \(-10 < k < 2\).

Using the change of base formula:
\(\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2} \log_2 x = \log_2 \sqrt{x}\)

Substitute back into the equation:
\(\log_2 \sqrt{x} + \log_2 (x - 3) = 1\)
\(\log_2 ((x - 3)\sqrt{x}) = 1\)
\((x - 3)\sqrt{x} = 2^1 = 2\)

Let \(u = \sqrt{x}\) (with \(u > 0\)):
\((u^2 - 3)u = 2\)
\(u^3 - 3u - 2 = 0\)

Factorizing the cubic equation:
\((u - 2)(u + 1)^2 = 0\)

Since \(u > 0\), the only valid solution is \(u = 2\).

Therefore:
\(\sqrt{x} = 2 \implies x = 4\).

(We can verify that \(x = 4\) is in the domain of \(\log_2 (x-3)\), since \(4 > 3\)).

M1: For converting \(\log_4 x\) to base 2 correctly as \(\frac{1}{2} \log_2 x\) or equivalent.
M1: For using logarithmic laws to express as a single log and removing the logarithm to form an equation in \(x\) (or \(u\)).
A1: For obtaining the final correct solution \(x = 4\) and discarding invalid values.

The team has a total of 5 players, and it must contain more women than men.
This gives three possible cases:

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

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

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

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

M1: For calculating the combinations for Case 1 (3 women, 2 men) or Case 2 (4 women, 1 man).
M1: For summing the number of ways for all three valid cases (3 women, 4 women, 5 women).
A1: For the correct answer of 181.

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):
\(3(1 - \sin^2 \theta) + 5 \sin \theta - 1 = 0\)
\(3 - 3\sin^2 \theta + 5 \sin \theta - 1 = 0\)
\(3\sin^2 \theta - 5 \sin \theta - 2 = 0\)

Factorizing the quadratic equation:
\((3\sin \theta + 1)(\sin \theta - 2) = 0\)

Since \(\sin \theta \le 1\), there is no solution for \(\sin \theta = 2\).

For \(\sin \theta = -\frac{1}{3}\):
Basic angle \(\alpha = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ\)

Sine is negative in the 3rd and 4th quadrants:
\(\theta = 180^\circ + 19.47^\circ = 199.5^\circ\) (to 1 d.p.)
\(\theta = 360^\circ - 19.47^\circ = 340.5^\circ\) (to 1 d.p.)

M1: For substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) and solving the quadratic equation to find a value for \(\sin \theta\).
M1: For identifying \(\sin \theta = -\frac{1}{3}\) (and rejecting \(\sin \theta = 2\)) and finding the basic angle.
A1: For both correct solutions: \(199.5^\circ\) and \(340.5^\circ\) (accept rounded to 1 decimal place; penalise extra solutions in range).

At \(x = 2\):
\(y = (2(2) - 3)^3 = 1^3 = 1\)
The point is \((2, 1)\).

Differentiate \(y\) with respect to \(x\) using the chain rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 3(2x-3)^2 \times 2 = 6(2x-3)^2\)

At \(x = 2\), the gradient of the tangent is:
\(m_t = 6(2(2)-3)^2 = 6(1)^2 = 6\)

Therefore, the gradient of the normal is:
\(m_n = -\frac{1}{6}\)

Using the point-slope form for the normal at \((2, 1)\):
\(y - 1 = -\frac{1}{6}(x - 2)\)
\(6(y - 1) = -(x - 2)\)
\(6y - 6 = -x + 2\)
\(x + 6y - 8 = 0\) (or equivalent form)

M1: For finding \(\frac{\mathrm{d}y}{\mathrm{d}x}\) correctly using the chain rule.
M1: For finding the gradient of the tangent at \(x = 2\) and calculating the negative reciprocal to find the normal's gradient.
A1: For the correct equation of the normal, e.g., \(x + 6y - 8 = 0\) or \(y = -\frac{1}{6}x + \frac{4}{3}\).

The general term in the expansion is given by:
\(T_{r+1} = \binom{6}{r} (2x)^{6-r} \left(-\frac{3}{x}\right)^r\)
\(T_{r+1} = \binom{6}{r} 2^{6-r} (-3)^r x^{6-r} x^{-r}\)
\(T_{r+1} = \binom{6}{r} 2^{6-r} (-3)^r x^{6-2r}\)

We require the term in \(x^2\), so we equate the exponents of \(x\):
\(6 - 2r = 2 \implies 2r = 4 \implies r = 2\)

Substituting \(r = 2\) into the general term:
\(\text{Coefficient} = \binom{6}{2} 2^{6-2} (-3)^2\)
\(\text{Coefficient} = 15 \times 2^4 \times 9\)
\(\text{Coefficient} = 15 \times 16 \times 9 = 2160\)

M1: For writing down the general term of the expansion.
M1: For setting the exponent of \(x\) equal to 2 and solving for \(r\) to get \(r = 2\).
A1: For the correct coefficient of 2160.

The perimeter of the sector is:
\(P = 2r + r\theta = 18 \implies r\theta = 18 - 2r\)

The area of the sector is:
\(A = \frac{1}{2}r^2\theta = 20\)

Substitute \(r\theta\) from the perimeter equation into the area equation:
\(\frac{1}{2}r(r\theta) = 20\)
\(\frac{1}{2}r(18 - 2r) = 20\)
\(r(9 - r) = 20\)
\(9r - r^2 = 20\)
\(r^2 - 9r + 20 = 0\)

Factorize the quadratic equation:
\((r - 4)(r - 5) = 0\)

Thus, the possible values of \(r\) are:
\(r = 4\) and \(r = 5\).

M1: For writing down equations for perimeter and area: \(2r + r\theta = 18\) and \(\frac{1}{2}r^2\theta = 20\).
M1: For eliminating \(\theta\) to obtain a quadratic equation in terms of \(r\) only.
A1: For finding both correct values of \(r\) (\(r = 4\) and \(r = 5\)).

First, find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{a} - 2\mathbf{b}) - \mathbf{a} = 2\mathbf{a} - 2\mathbf{b}\)
\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (k\mathbf{a} - 6\mathbf{b}) - \mathbf{a} = (k-1)\mathbf{a} - 6\mathbf{b}\)

Since points \(A\), \(B\), and \(C\) are collinear, the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) are parallel. Therefore, there exists a scalar constant \(\lambda\) such that:
\(\overrightarrow{AC} = \lambda \overrightarrow{AB}\)
\((k-1)\mathbf{a} - 6\mathbf{b} = \lambda (2\mathbf{a} - 2\mathbf{b})\)
\((k-1)\mathbf{a} - 6\mathbf{b} = 2\lambda\mathbf{a} - 2\lambda\mathbf{b}\)

Equating coefficients of \(\mathbf{b}\):
\(-6 = -2\lambda \implies \lambda = 3\)

Equating coefficients of \(\mathbf{a}\):
\(k - 1 = 2\lambda\)
\(k - 1 = 2(3) = 6 \implies k = 7\).

M1: For finding expressions for two vectors on the line (e.g. \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\)) in terms of \(\mathbf{a}\), \(\mathbf{b}\) and \(k\).
M1: For setting up the collinearity condition \(\overrightarrow{AC} = \lambda \overrightarrow{AB}\) and comparing the coefficients of \(\mathbf{b}\) to find \(\lambda\).
A1: For the correct value \(k = 7\).

To find the points of intersection, we equate the equations of the line and the curve:

\( x^2 + x + 7 = k(x+1) + 3 \)

\( x^2 + x + 7 = kx + k + 3 \)

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

For the line not to intersect the curve, the discriminant of this quadratic equation must be less than zero:

\( D < 0 \)

\( (1-k)^2 - 4(1)(4-k) < 0 \)

\( 1 - 2k + k^2 - 16 + 4k < 0 \)

\( k^2 + 2k - 15 < 0 \)

Factorising the quadratic inequality:

\( (k+5)(k-3) < 0 \)

Thus, the set of values of \( k \) is:

\( -5 < k < 3 \)

M1: For equating the line and the curve and arranging into a 3-term quadratic equation.
A1: For obtaining \( x^2 + (1-k)x + (4-k) = 0 \) or equivalent.
M1: For using \( D < 0 \) to form an inequality in \( k \).
A1: For simplifying to \( k^2 + 2k - 15 < 0 \).
A1: For the correct final range \( -5 < k < 3 \).

First, use the change of base formula to express the second logarithm in base 2:

\( 2\log_4 (x-1) = 2 \cdot \frac{\log_2(x-1)}{\log_2 4} = 2 \cdot \frac{\log_2(x-1)}{2} = \log_2(x-1) \)

Substitute this back into the equation:

\( \log_2 (3x + 5) - \log_2 (x - 1) = 2 \)

Using the subtraction law of logarithms:

\( \log_2 \left( \frac{3x + 5}{x - 1} \right) = 2 \)

Convert the logarithmic equation to exponential form:

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

\( \frac{3x + 5}{x - 1} = 4 \)

Solve for \( x \):

\( 3x + 5 = 4(x - 1) \)

\( 3x + 5 = 4x - 4 \)

\( x = 9 \)

Check validity:
For \( x = 9 \), both arguments of the original logarithms are positive: \( 3x+5 = 32 > 0 \) and \( x-1 = 8 > 0 \). Thus, the solution is valid.

M1: For using change of base formula correctly on \( 2\log_4 (x-1) \).
A1: For obtaining \( \log_2(x-1) \).
M1: For applying the subtraction law of logarithms to get \( \log_2 \left( \frac{3x + 5}{x - 1} \right) \).
M1: For converting to exponential form correctly to obtain \( \frac{3x + 5}{x - 1} = 4 \).
A1: For \( x = 9 \) and verifying validity.

Since \( 2x - 1 \) is a factor, by the Factor Theorem we have \( p\left(\frac{1}{2}\right) = 0 \):

\( 2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) - 6 = 0 \)

\( 2\left(\frac{1}{8}\right) + a\left(\frac{1}{4}\right) + \frac{b}{2} - 6 = 0 \)

\( \frac{1}{4} + \frac{a}{4} + \frac{2b}{4} - \frac{24}{4} = 0 \)

\( a + 2b = 23 \) --- (Equation 1)

Since the remainder when divided by \( x + 2 \) is \( -30 \), by the Remainder Theorem we have \( p(-2) = -30 \):

\( 2(-2)^3 + a(-2)^2 + b(-2) - 6 = -30 \)

\( 2(-8) + 4a - 2b - 6 = -30 \)

\( -16 + 4a - 2b - 6 = -30 \)

\( 4a - 2b = -8 \Rightarrow 2a - b = -4 \) --- (Equation 2)

From Equation 2, we get:

\( b = 2a + 4 \)

Substitute this into Equation 1:

\( a + 2(2a + 4) = 23 \)

\( a + 4a + 8 = 23 \)

\( 5a = 15 \Rightarrow a = 3 \)

Find \( b \):

\( b = 2(3) + 4 = 10 \)

M1: For applying the Factor Theorem \( p(1/2) = 0 \) and substituting.
A1: For simplifying to a linear equation in \( a \) and \( b \) (e.g., \( a + 2b = 23 \)).
M1: For applying the Remainder Theorem \( p(-2) = -30 \) and substituting.
A1: For simplifying to a second linear equation in \( a \) and \( b \) (e.g., \( 2a - b = -4 \)).
A1: For solving the simultaneous equations to find \( a = 3 \) and \( b = 10 \).

Using the formula for the sum of the first \( n \) terms of an arithmetic progression, \( S_n = \frac{n}{2}[2a + (n-1)d] \):

\( S_{10} = \frac{10}{2}[2a + 9d] = 150 \)

\( 5(2a + 9d) = 150 \)

\( 2a + 9d = 30 \) --- (Equation 1)

The 2nd, 6th, and 14th terms of the arithmetic progression are:

\( u_2 = a + d \)

\( u_6 = a + 5d \)

\( u_{14} = a + 13d \)

Since these terms form a geometric progression, the common ratio is constant:

\( \frac{a + 5d}{a + d} = \frac{a + 13d}{a + 5d} \)

\( (a + 5d)^2 = (a + d)(a + 13d) \)

\( a^2 + 10ad + 25d^2 = a^2 + 14ad + 13d^2 \)

\( 12d^2 = 4ad \)

Assuming the common difference \( d \neq 0 \):

\( 3d = a \Rightarrow a = 3d \) --- (Equation 2)

Substitute Equation 2 into Equation 1:

\( 2(3d) + 9d = 30 \)

\( 6d + 9d = 30 \)

\( 15d = 30 \Rightarrow d = 2 \)

Using \( a = 3d \):

\( a = 3(2) = 6 \)

M1: For using the sum formula of an AP to obtain \( 2a + 9d = 30 \).
B1: For writing down correct expressions for \( u_2, u_6, u_{14} \).
M1: For setting up the geometric progression equation \( (a+5d)^2 = (a+d)(a+13d) \).
A1: For simplifying the GP relation to \( a = 3d \) (or equivalent).
A1: For finding the correct values \( a = 6 \) and \( d = 2 \).

(a) The total number of people is \( 6 + 5 = 11 \).
We need to choose 5 representatives with no restrictions:
Number of ways = \( \binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \)

(b) If there must be at least 3 doctors on the team, we consider the following cases:
- Case 1: 3 doctors and 2 nurses
Number of ways = \( \binom{6}{3} \times \binom{5}{2} = 20 \times 10 = 200 \)
- Case 2: 4 doctors and 1 nurse
Number of ways = \( \binom{6}{4} \times \binom{5}{1} = 15 \times 5 = 75 \)
- Case 3: 5 doctors and 0 nurses
Number of ways = \( \binom{6}{5} \times \binom{5}{0} = 6 \times 1 = 6 \)

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

(c) If there must be more nurses than doctors, we consider the following cases:
- Case 1: 3 nurses and 2 doctors
Number of ways = \( \binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150 \)
- Case 2: 4 nurses and 1 doctor
Number of ways = \( \binom{5}{4} \times \binom{6}{1} = 5 \times 6 = 30 \)
- Case 3: 5 nurses and 0 doctors
Number of ways = \( \binom{5}{5} \times \binom{6}{0} = 1 \times 1 = 1 \)

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

B1: For part (a) correct answer 462.
M1: For identifying the three cases in (b) and calculating at least one case correctly.
A1: For part (b) correct total 281.
M1: For identifying the three cases in (c) and calculating at least one case correctly.
A1: For part (c) correct total 181.

Use the trigonometric identity \( \sin^2 \theta = 1 - \cos^2 \theta \):

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

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

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

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

Factorise the quadratic in \( \cos \theta \):

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

This gives two possible equations:
1) \( \cos \theta = 2 \) (no solution, since \( -1 \le \cos \theta \le 1 \))
2) \( \cos \theta = -\frac{1}{3} \)

Since the cosine is negative, \( \theta \) lies in the second or third quadrant.
The basic angle \( \alpha = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53^\circ \)

In the second quadrant:
\( \theta = 180^\circ - 70.53^\circ = 109.5^\circ \) (to 1 decimal place)

In the third quadrant:
\( \theta = 180^\circ + 70.53^\circ = 250.5^\circ \) (to 1 decimal place)

M1: For using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to get an equation in \( \cos \theta \) only.
A1: For obtaining the correct quadratic \( 3\cos^2 \theta - 5\cos \theta - 2 = 0 \).
M1: For factorising/solving to find \( \cos \theta = -1/3 \) and identifying that \( \cos \theta = 2 \) has no solution.
A1: For \( \theta = 109.5^\circ \) (or 109.47°).
A1: For \( \theta = 250.5^\circ \) (or 250.53°).

To find the stationary points, we find the first derivative \( \frac{dy}{dx} \) using the product rule:

Let \( u = x^2 - 2x - 1 \Rightarrow \frac{du}{dx} = 2x - 2 \)
Let \( v = e^{2x} \Rightarrow \frac{dv}{dx} = 2e^{2x} \)

Applying the product rule \( \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \):

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

\( \frac{dy}{dx} = e^{2x} [ 2(x^2 - 2x - 1) + 2x - 2 ] \)

\( \frac{dy}{dx} = e^{2x} [ 2x^2 - 4x - 2 + 2x - 2 ] \)

\( \frac{dy}{dx} = e^{2x} [ 2x^2 - 2x - 4 ] \)

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

For stationary points, \( \frac{dy}{dx} = 0 \):

\( 2e^{2x} (x^2 - x - 2) = 0 \)

Since \( 2e^{2x} \neq 0 \) for all real \( x \):

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

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

\( x = 2 \) or \( x = -1 \)

Now find the corresponding \( y \)-coordinates:

For \( x = 2 \):
\( y = (2^2 - 2(2) - 1)e^{2(2)} = (4 - 4 - 1)e^4 = -e^4 \)

For \( x = -1 \):
\( y = ((-1)^2 - 2(-1) - 1)e^{2(-1)} = (1 + 2 - 1)e^{-2} = 2e^{-2} \)

Thus, the exact coordinates of the stationary points are \( (2, -e^4) \) and \( (-1, 2e^{-2}) \).

M1: For attempting to differentiate \( y \) using the product rule.
A1: For obtaining a correct derivative, e.g., \( \frac{dy}{dx} = (2x^2 - 2x - 4)e^{2x} \).
M1: For setting \( \frac{dy}{dx} = 0 \) and solving the quadratic \( x^2 - x - 2 = 0 \).
A1: For finding correct x-values \( x = 2 \) and \( x = -1 \).
A1: For finding both correct exact y-coordinates \( -e^4 \) and \( 2e^{-2} \) and pairing them correctly.

(a) Using the quotient rule with \(u = e^{3x}\) and \(v = 2x+1\): \(u' = 3e^{3x}\), \(v' = 2\). Then \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3e^{3x}(2x+1) - 2e^{3x}}{(2x+1)^2} = \frac{e^{3x}(6x + 3 - 2)}{(2x+1)^2} = \frac{e^{3x}(6x+1)}{(2x+1)^2}\). (b) For a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies 6x+1 = 0 \implies x = -\frac{1}{6}\). Substituting \(x = -\frac{1}{6}\) into the curve equation: \(y = \frac{e^{-1/2}}{2(-1/6)+1} = \frac{e^{-1/2}}{2/3} = \frac{3}{2\sqrt{e}}\). Exact coordinates are \(\left(-\frac{1}{6}, \frac{3}{2\sqrt{e}}\right)\). (c) Testing the sign of the gradient around \(x = -\frac{1}{6}\): for \(x = -0.2\), \(\frac{\mathrm{d}y}{\mathrm{d}x} < 0\) (decreasing); for \(x = 0\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 > 0\) (increasing). Since the gradient changes from negative to positive, the stationary point is a local minimum.

(a) M1: For correct attempt to apply the quotient rule. A1: For correct derivatives of the numerator and denominator. M1: For factorising out \(e^{3x}\). A1: For complete and correct algebraic simplification. (b) M1: For setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solving for \(x\). A1: For \(x = -1/6\). A1: For exact y-coordinate \(y = \frac{3}{2\sqrt{e}}\). (c) M1: For a valid first- or second-derivative test. A1: For correct calculation of gradients or second derivative. A1: For concluding local minimum.

(a) The terms are \(T_4 = a+3d\), \(T_{10} = a+9d\), and \(T_{22} = a+21d\). Since they form a geometric progression, \(\frac{a+9d}{a+3d} = \frac{a+21d}{a+9d}\), which expands to \((a+9d)^2 = (a+3d)(a+21d)\). Simplifying: \(a^2 + 18ad + 81d^2 = a^2 + 24ad + 63d^2 \implies 18d^2 = 6ad\). Since \(d \neq 0\), dividing by \(6d\) gives \(3d = a \implies a = 3d\). (b) The common ratio is \(r = \frac{a+9d}{a+3d} = \frac{3d+9d}{3d+3d} = \frac{12d}{6d} = 2\). (c) (i) Using \(S_{10} = 5(2a + 9d) = 375 \implies 2a + 9d = 75\). Substituting \(a = 3d\): \(6d + 9d = 75 \implies 15d = 75 \implies d = 5\), so \(a = 15\). (ii) The first term of the GP is \(T_4 = a + 3d = 15 + 15 = 30\). The sum of the first 6 terms of the GP is \(S_6 = \frac{30(2^6 - 1)}{2 - 1} = 30(63) = 1890\).

(a) M1: For writing down expressions for the terms and setting up the GP equation. M1: For correct expansion of terms. A1: For showing \(a = 3d\) convincingly. (b) M1: For substituting \(a = 3d\) into the ratio expression. A1: For \(r = 2\). (c)(i) M1: For using \(S_{10} = 375\) to find an equation in \(a\) and \(d\). A1: For \(d = 5\). A1: For \(a = 15\). (ii) M1: For finding the first term of GP as 30 and using the GP sum formula. A1: For 1890.