2024 Cambridge IGCSE Mathematics - Additional (0606) 模擬試題連答案詳解

First, find the second \(x\)-intercept by setting \(y = 0\): \((3x-5)\ln(2x-3) = 0\). This occurs when \(3x-5 = 0\) or \(2x-3 = 1\). Given that one point is \((\frac{5}{3}, 0)\), the other point is found when \(2x - 3 = 1\), which yields \(x = 2\). Thus, the point of tangency is \((2, 0)\). Next, differentiate \(y = (3x-5)\ln(2x-3)\) using the product rule: \(\frac{dy}{dx} = 3\ln(2x-3) + (3x-5)\frac{2}{2x-3}\). Substitute \(x = 2\) to find the gradient: \(\frac{dy}{dx} = 3\ln(1) + (6-5)\frac{2}{1} = 0 + 2 = 2\). The equation of the tangent line is \(y - 0 = 2(x - 2)\), which simplifies to \(y = 2x - 4\).

M1: Set \(\ln(2x-3) = 0\) to find \(x = 2\).
M1: Apply the product rule to differentiate \(y\).
A1: Obtain the correct derivative \(\frac{dy}{dx} = 3\ln(2x-3) + \frac{2(3x-5)}{2x-3}\).
M1: Substitute \(x=2\) into their derivative to find the gradient.
M1: Form the equation of the tangent line.
A1.67: Obtain the final simplified equation \(y = 2x - 4\).

Use the double angle identity \(\cos(2\theta) = 1 - 2\sin^2\theta\). Substitute this into the equation: \(1 - 2\sin^2\theta + 3\sin\theta + 1 = 0\). Rearranging gives: \(2\sin^2\theta - 3\sin\theta - 2 = 0\). Factorising this quadratic equation yields: \((2\sin\theta + 1)(\sin\theta - 2) = 0\). Therefore, \(\sin\theta = -0.5\) or \(\sin\theta = 2\) (which has no solution). For \(\sin\theta = -0.5\) within \(0^\circ \le \theta \le 360^\circ\), the basic angle is \(30^\circ\). Since sine is negative, the solutions lie in the third and fourth quadrants: \(\theta = 180^\circ + 30^\circ = 210^\circ\) and \(\theta = 360^\circ - 30^\circ = 330^\circ\).

M1: Substitute \(\cos(2\theta)\) with \(1 - 2\sin^2\theta\).
M1: Form and attempt to factorise a quadratic in terms of \(\sin\theta\).
A1: Correct quadratic \(2\sin^2\theta - 3\sin\theta - 2 = 0\).
M1: Identify \(\sin\theta = -0.5\) as the only valid equation to solve.
A1: Identify the angle \(210^\circ\).
A1.67: Identify the angle \(330^\circ\).

First, expand \((1 - 3x)^6\) up to the \(x^2\) term using the binomial theorem: \((1 - 3x)^6 = 1 + \binom{6}{1}(-3x) + \binom{6}{2}(-3x)^2 + \dots = 1 - 18x + 15(9x^2) - \dots = 1 - 18x + 135x^2 - \dots\). Now, expand \((2 + kx)(1 - 18x + 135x^2)\) to find the total coefficient of the \(x^2\) term: \(2(135x^2) + (kx)(-18x) = (270 - 18k)x^2\). Set this coefficient equal to 90: \(270 - 18k = 90 \implies 18k = 180 \implies k = 10\).

M1: Correctly expand \((1 - 3x)^6\) to find the \(x\) and \(x^2\) terms.
A1: Obtain \(-18x\) and \(135x^2\).
M1: Correctly set up the expression for the coefficient of \(x^2\) as \(2(135) - 18k\).
A1: Formulate the equation \(270 - 18k = 90\).
A2.67: Correctly solve the linear equation to find \(k = 10\).

The perimeter of the sector is given by \(P = 2r + r\theta\), so \(2r + r\theta = 20 \implies r\theta = 20 - 2r\). The area of the sector is given by \(A = \frac{1}{2}r^2\theta = 24 \implies r^2\theta = 48\). Substitute the expression for \(r\theta\) from the perimeter equation into the area equation: \(r(r\theta) = 48 \implies r(20 - 2r) = 48 \implies 20r - 2r^2 = 48\). Dividing the entire equation by 2 and rearranging into a standard quadratic form gives: \(r^2 - 10r + 24 = 0\). Factorising this equation gives: \((r - 4)(r - 6) = 0\), which yields \(r = 4\) or \(r = 6\). If \(r = 4\): \(4\theta = 20 - 2(4) \implies 4\theta = 12 \implies \theta = 3\). If \(r = 6\): \(6\theta = 20 - 2(6) \implies 6\theta = 8 \implies \theta = \frac{4}{3}\). Thus, the two pairs of values are: \(r = 4, \theta = 3\) and \(r = 6, \theta = \frac{4}{3}\).

M1: Set up the perimeter equation \(2r + r\theta = 20\).
M1: Set up the area equation \(\frac{1}{2}r^2\theta = 24\).
M1: Substitute the first equation into the second to form a quadratic equation in \(r\).
A1: Obtain the simplified quadratic \(r^2 - 10r + 24 = 0\).
A1: Solve for both values of \(r\) to find \(r = 4\) and \(r = 6\).
A1.67: Find the corresponding values of \(\theta\): \(\theta = 3\) and \(\theta = \frac{4}{3}\).

(i) Total number of people is \(6 + 5 = 11\). The number of ways to select 5 people from 11 is \(\binom{11}{5} = \frac{11!}{5!6!} = 462\). (ii) To have more women than men, the composition of the committee of 5 can be: - 3 women and 2 men: \(\binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150\) ways, - 4 women and 1 man: \(\binom{5}{4} \times \binom{6}{1} = 5 \times 6 = 30\) ways, - 5 women and 0 men: \(\binom{5}{5} \times \binom{6}{0} = 1 \times 1 = 1\) way. Total number of ways is \(150 + 30 + 1 = 181\).

M1: Correct use of combination formula to find \(\binom{11}{5}\).
A1: Obtain 462 for part (i).
M1: Identify the three mutually exclusive cases for part (ii).
M2: Calculate combinations for each of the three cases: 150, 30, and 1.
A1.67: Find the sum of the cases to obtain 181.

Equating the line and the curve: \(2x^2 + 3x - 3 = kx - 5 \implies 2x^2 + (3 - k)x + 2 = 0\). For the line and curve not to intersect, the discriminant of this quadratic equation must be strictly negative (\(b^2 - 4ac < 0\)): \((3 - k)^2 - 4(2)(2) < 0 \implies (3 - k)^2 - 16 < 0\). Expanding and simplifying: \(9 - 6k + k^2 - 16 < 0 \implies k^2 - 6k - 7 < 0\). Factorising the quadratic inequality yields: \((k - 7)(k + 1) < 0\). The critical values are \(k = 7\) and \(k = -1\). Since the inequality is less than zero, \(k\) must lie between the critical values: \(-1 < k < 7\).

M1: Equate the equations and group into a standard quadratic in \(x\).
A1: Obtain the correct quadratic equation \(2x^2 + (3-k)x + 2 = 0\).
M1: Use the discriminant condition \(\Delta < 0\).
A1: Simplify to the inequality \(k^2 - 6k - 7 < 0\).
M1: Factorise to find critical values \(k = 7\) and \(k = -1\).
A1.67: Deduce the correct range \(-1 < k < 7\).

(i) To find the inverse function, let \(y = \frac{2x + 1}{x - 3}\). Then rearrange to make \(x\) the subject: \(y(x - 3) = 2x + 1 \implies xy - 3y = 2x + 1 \implies xy - 2x = 3y + 1 \implies x(y - 2) = 3y + 1 \implies x = \frac{3y + 1}{y - 2}\). Thus, \(\mathrm{f}^{-1}(x) = \frac{3x + 1}{x - 2}\). The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\). Since \(\mathrm{f}(x) = \frac{2(x-3) + 7}{x-3} = 2 + \frac{7}{x-3}\), and \(x > 3\), we have \(\frac{7}{x-3} > 0\), meaning \(\mathrm{f}(x) > 2\). Therefore, the domain of \(\mathrm{f}^{-1}\) is \(x > 2\). (ii) To solve \(\mathrm{f}^2(x) = 3 \implies \mathrm{f}(\mathrm{f}(x)) = 3\), apply the inverse function to both sides: \(\mathrm{f}(x) = \mathrm{f}^{-1}(3)\). Substitute 3 into \(\mathrm{f}^{-1}\): \(\mathrm{f}^{-1}(3) = \frac{3(3) + 1}{3 - 2} = 10\). Now solve \(\mathrm{f}(x) = 10 \implies \frac{2x + 1}{x - 3} = 10 \implies 2x + 1 = 10x - 30 \implies 8x = 31 \implies x = 3.875\).

M1: Set \(y = \mathrm{f}(x)\) and rearrange to find \(x\) as a subject.
A1: Obtain \(\mathrm{f}^{-1}(x) = \frac{3x+1}{x-2}\).
M1: State the domain of the inverse function as \(x > 2\).
M1: Apply the property \(\mathrm{f}(x) = \mathrm{f}^{-1}(3)\).
A1: Obtain \(\mathrm{f}(x) = 10\).
A1.67: Solve to find \(x = 3.875\) (or \(\frac{31}{8}\)).

Apply the product law of logarithms: \(\log_3((x - 2)(x + 6)) = 2\). Rewrite the equation in exponential form: \((x - 2)(x + 6) = 3^2 \implies x^2 + 4x - 12 = 9\). Rearranging gives the quadratic equation: \(x^2 + 4x - 21 = 0\). Factorising this quadratic equation yields: \((x - 3)(x + 7) = 0\). This gives the potential solutions \(x = 3\) and \(x = -7\). We must check these solutions against the original logarithmic terms: for \(\log_3(x - 2)\) to be defined, we require \(x - 2 > 0 \implies x > 2\). For \(\log_3(x + 6)\) to be defined, we require \(x + 6 > 0 \implies x > -6\). Therefore, \(x = -7\) is invalid, and the only valid solution is \(x = 3\).

M1: Apply the addition law of logarithms to combine terms.
M1: Convert from logarithmic to exponential form to obtain \((x - 2)(x + 6) = 9\).
A1: Obtain the correct quadratic equation \(x^2 + 4x - 21 = 0\).
M1: Solve the quadratic equation to get \(x = 3\) and \(x = -7\).
A1.67: Reject the extraneous root \(x = -7\) and state \(x = 3\) as the final answer.

At \(x = 2\):
\(y = (2(2) - 3)\sqrt{4(2) + 1} = 1 \times \sqrt{9} = 3\).
So the coordinates of the point of contact are \((2, 3)\).

To find the gradient of the tangent, we differentiate \(y\) with respect to \(x\) using the product rule:
Let \(u = 2x - 3 \implies \frac{du}{dx} = 2\)
Let \(v = (4x + 1)^{1/2} \implies \frac{dv}{dx} = \frac{1}{2}(4x + 1)^{-1/2} \times 4 = \frac{2}{\sqrt{4x + 1}}

Applying the product rule:
\)\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)
\(\frac{dy}{dx} = (2x - 3) \cdot \frac{2}{\sqrt{4x + 1}} + 2\sqrt{4x + 1}\)

At \(x = 2\):
\(\frac{dy}{dx} = (2(2) - 3) \cdot \frac{2}{\sqrt{4(2) + 1}} + 2\sqrt{4(2) + 1} = 1 \cdot \frac{2}{3} + 2(3) = \frac{20}{3}\).

The gradient of the tangent at \(x = 2\) is \(\frac{20}{3}\).
The gradient of the normal is the negative reciprocal:
\(m = -\frac{3}{20}\).

Using the point-slope form for the normal at \((2, 3)\):
\(y - 3 = -\frac{3}{20}(x - 2)\)
\(20(y - 3) = -3(x - 2)\)
\(20y - 60 = -3x + 6\)
\(3x + 20y = 66\).

M1: Attempts to find the \(y\)-coordinate when \(x = 2\) (obtaining \(y=3\)).
M1: Applies the product rule (or quotient rule) to find \(\frac{dy}{dx}\).
A1: Obtains a correct derivative expression, e.g. \(2\sqrt{4x + 1} + \frac{2(2x - 3)}{\sqrt{4x + 1}}\).
M1: Substitutes \(x = 2\) into their derivative to find the tangent gradient, then finds the normal gradient (negative reciprocal).
A1: Obtains a normal gradient of \(-\frac{3}{20}\).
A1: For the correct final equation \(3x + 20y = 66\) (or any integer multiple).

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

Multiply by \(-1\) to form a standard quadratic equation:
\(4\cos^2 x + 5\cos x + 1 = 0\)

Factorising the quadratic equation:
\((4\cos x + 1)(\cos x + 1) = 0\)

This yields two possible cases:
1) \(\cos x = -1\)
Within the interval \(0^\circ \le x \le 360^\circ\):
\(x = 180^\circ\)

2) \(\cos x = -0.25\)
The basic angle \(\alpha = \cos^{-1}(0.25) \approx 75.52^\circ\).
Since cosine is negative, the angle lies in the second and third quadrants:
\(x = 180^\circ - 75.52^\circ = 104.48^\circ \approx 104.5^\circ\)
\(x = 180^\circ + 75.52^\circ = 255.52^\circ \approx 255.5^\circ\)

Thus, the solutions in the given interval are \(104.5^\circ\), \(180^\circ\), and \(255.5^\circ\).

M1: Uses the identity \(\sin^2 x = 1 - \cos^2 x\) to express the entire equation in terms of \(\cos x\).
M1: Obtains a 3-term quadratic equation in \(\cos x\), equivalent to \(4\cos^2 x + 5\cos x + 1 = 0\).
M1: Solves the quadratic to get \(\cos x = -1\) or \(\cos x = -0.25\).
A1: Obtains the solution \(180^\circ\).
A1: Obtains the solution \(104.5^\circ\) (accept \(104.48^\circ\) or better).
A1: Obtains the solution \(255.5^\circ\) (accept \(255.52^\circ\) or better).

**(i)**
The terms of the arithmetic progression (AP) are:
- First term: \(T_1 = a\)
- Third term: \(T_3 = a + 2d\)
- Eleventh term: \(T_{11} = a + 10d\)

These terms form a geometric progression (GP). Therefore, the common ratio is equal between successive terms:
\(\frac{a + 2d}{a} = \frac{a + 10d}{a + 2d}

Cross-multiplying:
\)(a + 2d)^2 = a(a + 10d)\)
\(a^2 + 4ad + 4d^2 = a^2 + 10ad\)
\(4d^2 = 6ad\)

Since \(d \ne 0\), dividing by \(2d\) yields:
\(2d = 3a \implies d = 1.5a\) [Shown]

**(ii)**
The sum of the first 8 terms of the AP is:
\(S_8 = \frac{8}{2}[2a + (8 - 1)d] = 4(2a + 7d) = 200\)
\(2a + 7d = 50\)

Substitute \(d = 1.5a\):
\(2a + 7(1.5a) = 50\)
\(12.5a = 50 \implies a = 4\)

Since \(a = 4\), \(d = 1.5(4) = 6\).

Now, for the geometric progression:
- First term of GP, \(G_1 = a = 4\)
- Second term of GP, \(G_2 = a + 2d = 4 + 2(6) = 16\)
- Common ratio, \(r = \frac{16}{4} = 4\)

The sum of the first 5 terms of the GP is:
\(S_5 = \frac{G_1(r^5 - 1)}{r - 1} = \frac{4(4^5 - 1)}{4 - 1} = \frac{4(1024 - 1)}{3} = \frac{4 \times 1023}{3} = 1364\).

M1: Equates the common ratio of GP terms using AP expressions: \(\frac{a+2d}{a} = \frac{a+10d}{a+2d}\).
A1: Conducts correct algebraic reduction to obtain \(d = 1.5a\).
M1: Applies the AP sum formula for \(S_8 = 200\) and substitutes \(d = 1.5a\).
A1: Obtains \(a = 4\) and \(d = 6\).
M1: Finds the common ratio \(r = 4\) and applies the geometric progression sum formula for \(n = 5\).
A1: Obtains the final answer of 1364.

First, use the change of base formula to express \(\log_4 (2x - 7)\) in base 2:
\(\log_4 (2x - 7) = \frac{\log_2 (2x - 7)}{\log_2 4} = \frac{\log_2 (2x - 7)}{2}\)

Substitute this back into the original equation:
\(\log_2 (x - 2) - \frac{1}{2}\log_2 (2x - 7) = 1\)

Multiply the entire equation by 2 to clear fractions:
\(2\log_2 (x - 2) - \log_2 (2x - 7) = 2\)

Use logarithmic laws to rewrite the left side as a single logarithm:
\(\log_2 (x - 2)^2 - \log_2 (2x - 7) = 2\)
\(\log_2 \frac{(x - 2)^2}{2x - 7} = 2\)

Convert from logarithmic form to exponential form:
\(\frac{(x - 2)^2}{2x - 7} = 2^2 = 4\)

Solve the resulting quadratic equation:
\((x - 2)^2 = 4(2x - 7)\)
\(x^2 - 4x + 4 = 8x - 28\)
\(x^2 - 12x + 32 = 0\)

Factorise the quadratic expression:
\((x - 4)(x - 8) = 0\)

This gives candidate solutions:
\(x = 4\) or \(x = 8\)

Check for validity (both logarithmic arguments must be strictly positive):
- For \(x = 4\): \(x-2 = 2 > 0\) and \(2x-7 = 1 > 0\) (valid).
- For \(x = 8\): \(x-2 = 6 > 0\) and \(2x-7 = 9 > 0\) (valid).

Both values are valid solutions.

M1: Applies change of base formula correctly to obtain \(\frac{1}{2}\log_2 (2x - 7)\).
M1: Applies power law to get \(\log_2 (x - 2)^2\).
M1: Uses subtraction law of logarithms to combine terms into a single logarithm.
M1: Successfully converts the logarithmic equation to an algebraic equation: \(\frac{(x - 2)^2}{2x - 7} = 4\).
A1: Simplifies to the correct quadratic form \(x^2 - 12x + 32 = 0\).
A1: Obtains both correct and validated solutions: \(x = 4\) and \(x = 8\).

Substitute \(\cos(2\theta) = 1 - 2\sin^2\theta\) into the given equation:
\(3(1 - 2\sin^2\theta) + 7\sin\theta - 5 = 0\)
\(3 - 6\sin^2\theta + 7\sin\theta - 5 = 0\)
\(-6\sin^2\theta + 7\sin\theta - 2 = 0\)
\(6\sin^2\theta - 7\sin\theta + 2 = 0\)

Factorise the quadratic equation in terms of \(\sin\theta\):
\((3\sin\theta - 2)(2\sin\theta - 1) = 0\)

This gives:
\(\sin\theta = \frac{2}{3}\) or \(\sin\theta = \frac{1}{2}\)

For \(\sin\theta = \frac{1}{2}\):
\(\theta = 30^\circ\) or \(\theta = 180^\circ - 30^\circ = 150^\circ\)

For \(\sin\theta = \frac{2}{3}\):
\(\theta = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.8^\circ\) or \(\theta = 180^\circ - 41.8^\circ = 138.2^\circ\)

Thus, the solutions in the given range are \(\theta = 30^\circ, 41.8^\circ, 138.2^\circ, 150^\circ\).

M1: Use correct double angle identity \(\cos(2\theta) = 1 - 2\sin^2\theta\) to obtain a quadratic in terms of \(\sin\theta\).
M1: Correctly factorise or solve the quadratic equation to find \(\sin\theta = \frac{1}{2}\) and \(\sin\theta = \frac{2}{3}\).
A1: Find \(\theta = 30^\circ\) and \(\theta = 150^\circ\).
A1: Find \(\theta = 41.8^\circ\) and \(\theta = 138.2^\circ\) (rounded to 1 d.p.).

Equate the line and the curve to find their intersection points:
\(2x^2 + 3x + 3 = kx - 5\)

Rearrange into standard quadratic form:
\(2x^2 + (3 - k)x + 8 = 0\)

For the line and curve not to intersect, the discriminant must be strictly less than zero (\(b^2 - 4ac < 0\)):
\((3 - k)^2 - 4(2)(8) < 0\)

Simplify and solve the inequality:
\((3 - k)^2 - 64 < 0\)

\((3 - k)^2 < 64\)

\(-8 < 3 - k < 8\)

Subtract 3 from all parts:

\(-11 < -k < 5\)

Multiply by \(-1\) and reverse the inequality signs:

\(-5 < k < 11\)

M1: Equate the equations and express as a three-term quadratic in \(x\).
A1: Correct quadratic equation \(2x^2 + (3 - k)x + 8 = 0\).
M1: Apply discriminant condition \(b^2 - 4ac < 0\).
A1: Correct expansion and inequality \((3 - k)^2 - 64 < 0\).
M1: Solve the quadratic inequality to find critical values.
A1: Correct final range of values: \(-5 < k < 11\).

Let the first term be \(a\) and the common ratio be \(r\).
The sum to infinity is:
\(\frac{a}{1-r} = 12 \implies a = 12(1-r)\) [Equation 1]

The second term is:
\(ar = 2.25 = \frac{9}{4}\) [Equation 2]

Substitute Equation 1 into Equation 2:
\(12(1-r)r = \frac{9}{4}\)

\(12r - 12r^2 = \frac{9}{4}\)

Multiply the entire equation by 4 to clear the fraction:

\(48r - 48r^2 = 9 \implies 48r^2 - 48r + 9 = 0\)

Divide by 3:

\(16r^2 - 16r + 3 = 0\)

Factorise the quadratic:

\((4r - 1)(4r - 3) = 0\)

Thus, \(r = 0.25\) or \(r = 0.75\).

If \(r = 0.25\):
\(a = 12(1 - 0.25) = 12(0.75) = 9\)

If \(r = 0.75\):
\(a = 12(1 - 0.75) = 12(0.25) = 3\)

The possible solutions are \(a = 9, r = 0.25\) and \(a = 3, r = 0.75\).

M1: Write correct formulas for the sum to infinity and second term in terms of \(a\) and \(r\).
A1: Obtain the simultaneous equations \(\frac{a}{1-r} = 12\) and \(ar = 2.25\).
M1: Eliminate \(a\) or \(r\) to form a quadratic equation in one variable.
A1: Obtain the simplified quadratic \(16r^2 - 16r + 3 = 0\) (or equivalent for \(a\)).
M1: Factorise or solve the quadratic to find two values for \(r\) (or \(a\)).
A1: State both pairs of answers clearly: \(a = 9, r = 0.25\) and \(a = 3, r = 0.75\).

Let \(Y = \lg y\) and \(X = x^2\).
The points given are \((X_1, Y_1) = (2, 5)\) and \((X_2, Y_2) = (6, 13)\).

Find the gradient \(m\) of the straight line:
\(m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\)

Using the line equation \(Y - Y_1 = m(X - X_1)\):
\(Y - 5 = 2(X - 2)\)

\(Y = 2X + 1\)

Substitute back \(Y = \lg y\) and \(X = x^2\):
\(\lg y = 2x^2 + 1\)

Express in exponential form (base 10):
\(y = 10^{2x^2 + 1}\)

Using the laws of indices to write in the requested form:
\(y = 10^1 \times 10^{2x^2}\)

\(y = 10 \times (10^2)^{x^2}\)

\(y = 10 \times 100^{x^2}\)

So, \(A = 10\) and \(B = 100\).

M1: Calculate the gradient of the straight line using \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
A1: Obtain gradient \(m = 2\).
M1: State the equation of the line in the form \(\lg y = m x^2 + c\) and solve for \(c\).
A1: Obtain \(\lg y = 2x^2 + 1\).
M1: Convert logarithmic equation to exponential form base 10.
A1: Correctly express in the final form \(y = 10 \times 100^{x^2}\) with \(A = 10\) and \(B = 100\).

Given \(y = (2x - 3)e^{x^2}\), we use the product rule to differentiate with respect to \(x\).
Let \(u = 2x - 3 \implies \frac{du}{dx} = 2\).
Let \(v = e^{x^2} \implies \frac{dv}{dx} = 2x e^{x^2}\).

Using the product rule:
\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\)
\(\frac{dy}{dx} = (2x - 3)(2x e^{x^2}) + 2e^{x^2}\)
\(\frac{dy}{dx} = e^{x^2} [2x(2x - 3) + 2]\)
\(\frac{dy}{dx} = e^{x^2} [4x^2 - 6x + 2]\)

For stationary points, set \(\frac{dy}{dx} = 0\):
\(e^{x^2} [4x^2 - 6x + 2] = 0\)

Since \(e^{x^2} \neq 0\) for all real \(x\), we solve:
\(4x^2 - 6x + 2 = 0\)

Divide by 2:
\(2x^2 - 3x + 1 = 0\)

Factorise:
\((2x - 1)(x - 1) = 0\)

Thus, \(x = 0.5\) or \(x = 1\).

Find the corresponding \(y\)-coordinates:
For \(x = 0.5\):
\(y = (2(0.5) - 3)e^{(0.5)^2} = -2e^{0.25}\)

For \(x = 1\):
\(y = (2(1) - 3)e^{1^2} = -1e^1 = -e\)

So the stationary points are \((0.5, -2e^{0.25})\) and \((1, -e)\).

M1: State and apply product rule correctly.
A1: Obtain correct derivative \(\frac{dy}{dx} = 2e^{x^2} + 2x(2x - 3)e^{x^2}\).
M1: Set derivative to 0 and factor out \(e^{x^2}\) to get quadratic in \(x\).
A1: Find correct \(x\)-values: \(x = 0.5\) and \(x = 1\).
M1: Substitute both \(x\)-values back into original equation to find \(y\)-coordinates.
A1: Correct exact coordinates: \((0.5, -2e^{0.25})\) (or equivalent) and \((1, -e)\).

(i) Total number of people is \(8 + 6 = 14\).
We choose 6 from 14:
\(\binom{14}{6} = 3003\)

(ii) 'At least 4 women' means we can choose 4, 5, or 6 women:
Case 1: 4 women and 2 men:
\(\binom{6}{4} \times \binom{8}{2} = 15 \times 28 = 420\)
Case 2: 5 women and 1 man:
\(\binom{6}{5} \times \binom{8}{1} = 6 \times 8 = 48\)
Case 3: 6 women and 0 men:
\(\binom{6}{6} \times \binom{8}{0} = 1 \times 1 = 1\)

Total number of ways = \(420 + 48 + 1 = 469\)

(iii) Let the particular man be \(M_1\) and the particular woman be \(W_1\).
We can find the number of ways where they are BOTH on the committee, then subtract this from the total number of unrestricted ways:
If both \(M_1\) and \(W_1\) are chosen, we must choose the remaining 4 committee members from the remaining 12 people (14 total - 2 specific people):
\(\binom{12}{4} = 495\)

Number of ways where both are not on the committee together:
\(3003 - 495 = 2508\)

(i) M1: Identify use of combinations \(\binom{14}{6}\).
A1: Correct answer 3003.

(ii) M1: Calculate combinations for at least one of the three valid cases.
M1: Sum all three mutually exclusive cases: 4 women, 5 women, and 6 women.
A1: Correct answer 469.

(iii) M1: Apply subtraction method (Total - cases with both present) or sum valid alternative scenarios.
A1: Correct answer 2508.

(i) To find the inverse of \(f(x)\), we first complete the square for \(f(x)\):
\(f(x) = 2(x^2 - 2x) + 5 = 2[(x-1)^2 - 1] + 5 = 2(x-1)^2 + 3\)

Let \(y = 2(x-1)^2 + 3\):
\(y - 3 = 2(x-1)^2\)

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

Since the domain of \(f\) is \(x \ge 1\), we choose the positive square root:
\(x - 1 = \sqrt{\frac{y-3}{2}}\)

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

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

(ii) The domain of \(f^{-1}\) is the range of \(f\).
Since \(f(x) = 2(x-1)^2 + 3\) for \(x \ge 1\), the minimum value of \(f(x)\) is at \(x = 1\), which gives \(f(1) = 3\).
So, the range of \(f\) is \(f(x) \ge 3\).
Thus, the domain of \(f^{-1}\) is \(x \ge 3\).

(iii) To find \(gf(3)\):
First, calculate \(f(3)\):
\(f(3) = 2(3)^2 - 4(3) + 5 = 18 - 12 + 5 = 11\)

Now, calculate \(g(11)\):
\(g(11) = \ln(11 + 1) = \ln 12\).

(i) M1: Complete the square for the quadratic function \(f(x)\).
M1: Rearrange the equation to express \(x\) in terms of \(y\).
A1: Correct formula for \(f^{-1}(x)\) with positive root chosen.

(ii) B1: State correct domain \(x \ge 3\) (allow any correct notation).

(iii) M1: Calculate \(f(3) = 11\) and substitute into \(g(x)\).
A1: Find correct exact value \\ln 12\" (do not accept decimal approximation)."},

(i) The graph of \(y = |3x - 6|\) is a V-shaped curve:
- The vertex is at \((2, 0)\).
- When \(x = 0\), \(y = |-6| = 6\), so the y-intercept is \((0, 6)\).

The graph of \(y = 2x + 1\) is a straight line:
- When \(x = 0\), \(y = 1\), so the y-intercept is \((0, 1)\).
- When \(y = 0\), \(x = -0.5\), so the x-intercept is \((-0.5, 0)\).

(ii) Solve the modular equation \(|3x - 6| = 2x + 1\):
Case 1: \(3x - 6 = 2x + 1\)
\(x = 7\)

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

Both solutions are valid as they yield positive values for \(2x + 1\) (which are 15 and 3 respectively).
Thus, the solutions are \(x = 1\) and \(x = 7\).

(i) B1: Correctly shaped V-graph for \(y = |3x - 6|\) with vertex at \((2, 0)\) and y-intercept at \((0, 6)\).
B1: Correct straight line \(y = 2x + 1\) passing through \((0, 1)\) and \((-0.5, 0)\).
B1: Clearly label all four axis intersection coordinates on the diagram.

(ii) M1: Set up two algebraic equations to solve the modulus: \(3x - 6 = 2x + 1\) and \(-(3x - 6) = 2x + 1\).
A1: Obtain \(x = 7\).
A1: Obtain \(x = 1\).

To find the area of the region, we integrate the function from \(x = 0\) to \(x = \frac{\pi}{6}\):

\[\text{Area} = \int_{0}^{\frac{\pi}{6}} (3\sin(2x) + 2\cos(x)) \,\mathrm{d}x\]

Find the antiderivative:
\[\int (3\sin(2x) + 2\cos(x)) \,\mathrm{d}x = -\frac{3}{2}\cos(2x) + 2\sin(x)\]

Now, apply the limits of integration:
\[\left[ -\frac{3}{2}\cos(2x) + 2\sin(x) \right]_{0}^{\frac{\pi}{6}}\]

At the upper limit \(x = \frac{\pi}{6}\):
\[-\frac{3}{2}\cos\left(2 \cdot \frac{\pi}{6}\right) + 2\sin\left(\frac{\pi}{6}\right) = -\frac{3}{2}\cos\left(\frac{\pi}{3}\right) + 2\sin\left(\frac{\pi}{6}\right)\]
\[= -\frac{3}{2}\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) = -\frac{3}{4} + 1 = \frac{1}{4}\]

At the lower limit \(x = 0\):
\[-\frac{3}{2}\cos(0) + 2\sin(0) = -\frac{3}{2}(1) + 2(0) = -\frac{3}{2}\]

Subtract the lower limit value from the upper limit value:
\[\text{Area} = \frac{1}{4} - \left(-\frac{3}{2}\right) = \frac{1}{4} + \frac{6}{4} = \frac{7}{4}\]

Therefore, the exact area of the region is \(\frac{7}{4}\) (or \(1.75\)).

M1: Integrates \(3\sin(2x)\) to obtain \(k\cos(2x)\) where \(k \neq 3\).
M1: Integrates \(2\cos(x)\) to obtain \(2\sin(x)\).
A1: Correct integration to obtain \(-\frac{3}{2}\cos(2x) + 2\sin(x)\).
M1: Correctly substitutes the limits \(x = \frac{\pi}{6}\) and \(x = 0\) into their integrated expression.
A1: Correct calculation of the upper limit value as \(\frac{1}{4}\) and the lower limit value as \(-\frac{3}{2}\).
A1: Subtracts correctly to find the final exact area of \(\frac{7}{4}\) (or \(1.75\)).

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

(a) We are given:
\[u_3 = ar^2 = 12\]
\[u_6 = ar^5 = \frac{32}{9}\]

Dividing the equation for \(u_6\) by the equation for \(u_3\):
\[\frac{ar^5}{ar^2} = \frac{\frac{32}{9}}{12}\]
\[r^3 = \frac{32}{108} = \frac{8}{27}\]

Taking the cube root of both sides:
\[r = \frac{2}{3}\]

Now, substitute \(r = \frac{2}{3}\) back into the equation for \(u_3\):
\[a\left(\frac{2}{3}\right)^2 = 12\]
\[a\left(\frac{4}{9}\right) = 12\]
\[a = 12 \times \frac{9}{4} = 27\]

So, the common ratio is \(\frac{2}{3}\) and the first term is \(27\).

(b) The sum to infinity is given by \(S_{\infty} = \frac{a}{1-r}\):
\[S_{\infty} = \frac{27}{1 - \frac{2}{3}} = \frac{27}{\frac{1}{3}} = 81\]

Part (a):
M1: Writes down the correct formula for \(u_3\) and \(u_6\) in terms of \(a\) and \(r\).
M1: Eliminates \(a\) to obtain an equation in terms of \(r^3\).
A1: Finds the correct common ratio \(r = \frac{2}{3}\) (or decimal equivalent to at least 3 s.f.).
A1: Finds the correct first term \(a = 27\).

Part (b):
M1: Applies the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\) with their values of \(a\) and \(r\) (where \(|r| < 1\)).
A1: Obtains the correct sum to infinity of \(81\).

(a) To find \(\mathrm{f}^{-1}(x)\), let \(y = 2\ln(3x - 1)\):
\[\frac{y}{2} = \ln(3x - 1)\]

Take the exponential of both sides:
\[\mathrm{e}^{\frac{y}{2}} = 3x - 1\]
\[3x = \mathrm{e}^{\frac{y}{2}} + 1\]
\[x = \frac{\mathrm{e}^{\frac{y}{2}} + 1}{3}\]

Thus, the inverse function is:
\[\mathrm{f}^{-1}(x) = \frac{\mathrm{e}^{\frac{x}{2}} + 1}{3}\]

(b) The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\). Since the range of \(\mathrm{f}\) is all real numbers, the domain of \(\mathrm{f}^{-1}\) is \(x \in \mathbb{R}\).
The range of \(\mathrm{f}^{-1}\) is the domain of \(\mathrm{f}\). Thus, the range is \(\mathrm{f}^{-1}(x) > \frac{1}{3}\).

(c) Solve \(2\ln(3x - 1) = \ln(8x + 1)\):
\[\ln((3x - 1)^2) = \ln(8x + 1)\]
\[(3x - 1)^2 = 8x + 1\]
\[9x^2 - 6x + 1 = 8x + 1\]
\[9x^2 - 14x = 0\]
\[x(9x - 14) = 0\]

This gives \(x = 0\) or \(x = \frac{14}{9}\).
However, for \(\mathrm{f}(x)\) to be defined, we must have \(x > \frac{1}{3}\). Therefore, \(x = 0\) is rejected.

The only solution is \(x = \frac{14}{9}\) (or \(1.56\) to 3 significant figures).

Part (a):
M1: Sets \(y = 2\ln(3x-1)\) and makes progress by dividing by 2 and exponentiating.
A1: Correct expression for \(\mathrm{f}^{-1}(x) = \frac{\mathrm{e}^{x/2} + 1}{3}\) (or equivalent form).

Part (b):
B1: States the correct domain of \(\mathrm{f}^{-1}\) is \(x \in \mathbb{R}\) (or equivalent).
B1: States the correct range of \(\mathrm{f}^{-1}\) is \(\mathrm{f}^{-1}(x) > \frac{1}{3}\).

Part (c):
M1: Uses laws of logarithms to write \(2\ln(3x-1)\) as \(\ln(3x-1)^2\).
M1: Obtains and solves the quadratic equation \(9x^2 - 14x = 0\).
A1: Correctly identifies \(x = \frac{14}{9}\) and rejects \(x = 0\) with valid justification based on the domain.

We begin by using the trigonometric identity \(\sin^2\theta = 1 - \cos^2\theta\) to rewrite the equation in terms of \(\cos\theta\):
\[4(1 - \cos^2\theta) - 5\cos\theta - 5 = 0\]
\[4 - 4\cos^2\theta - 5\cos\theta - 5 = 0\]
\[-4\cos^2\theta - 5\cos\theta - 1 = 0\]

Multiply by \(-1\) to obtain a standard quadratic form:
\[4\cos^2\theta + 5\cos\theta + 1 = 0\]

Factorize the quadratic equation:
\[(4\cos\theta + 1)(\cos\theta + 1) = 0\]

This gives two possible cases:
1) \(\cos\theta = -1\)
2) \(\cos\theta = -\frac{1}{4}\)

Case 1: \(\cos\theta = -1\)
For \(0^\circ \le \theta \le 360^\circ\):
\[\theta = 180^\circ\]

Case 2: \(\cos\theta = -0.25\)
Find the basic angle \(\alpha\):
\[\alpha = \cos^{-1}(0.25) \approx 75.52^\circ\]

Since \(\cos\theta\) is negative, \(\theta\) lies in the second and third quadrants:
In the second quadrant:
\[\theta = 180^\circ - 75.52^\circ = 104.48^\circ \approx 104.5^\circ\]

In the third quadrant:
\[\theta = 180^\circ + 75.52^\circ = 255.48^\circ \approx 255.5^\circ\]

Thus, the solutions in the range \(0^\circ \le \theta \le 360^\circ\) are:
\[\theta = 104.5^\circ, 180^\circ, 255.5^\circ\]

M1: Uses the identity \(\sin^2\theta = 1 - \cos^2\theta\) to form a quadratic equation in terms of \(\cos\theta\).
A1: Obtains the correct simplified quadratic equation \(4\cos^2\theta + 5\cos\theta + 1 = 0\).
M1: Factorizes or solves their quadratic equation to find values for \(\cos\theta\).
A1: Identifies \(\cos\theta = -1\) and \(\cos\theta = -\frac{1}{4}\).
B1: Finds \(\theta = 180^\circ\).
M1: Finds one correct quadrant angle for \(\cos\theta = -0.25\) (e.g. \(104.5^\circ\) or \(255.5^\circ\)).
A1: Obtains all three correct angles: \(104.5^\circ\), \(180^\circ\), and \(255.5^\circ\) (rounded correctly to 1 decimal place, with no extra incorrect angles in the range).