2026 Cambridge IGCSE Mathematics - Additional (0606) 模擬試題連答案詳解

(a) Since \(x \geqslant -5\), \(\sqrt{x+5} \geqslant 0\). Therefore, the minimum value of \(\mathrm{f}(x)\) occurs at \(x = -5\), where \(\mathrm{f}(-5) = 2(0) - 3 = -3\).
Hence, the range is \(\mathrm{f}(x) \geqslant -3\).

(b) Let \(y = 2\sqrt{x+5} - 3\).
Rearranging to make \(x\) the subject:
\(y + 3 = 2\sqrt{x+5}\)
\(\sqrt{x+5} = \frac{y+3}{2}\)
\(x + 5 = \left(\frac{y+3}{2}\right)^2 = \frac{(y+3)^2}{4}\)
\(x = \frac{(y+3)^2}{4} - 5\)

Therefore, \(\mathrm{f}^{-1}(x) = \frac{(x+3)^2}{4} - 5\) (or \(\frac{x^2+6x-11}{4}\)).
The domain of \(\mathrm{f}^{-1}\) is equal to the range of \(\mathrm{f}\), which is \(x \geqslant -3\).

(a) B1: Correct range \(\mathrm{f}(x) \geqslant -3\) or \(y \geqslant -3\) or \([-3, \infty)\) (accept \(\mathrm{f} \geqslant -3\); do not accept \(x \geqslant -3\)).

(b)
M1: Correct algebraic steps to isolate \(\sqrt{x+5}\) and square both sides (allow one sign error).
A1: Correct expression for \(\mathrm{f}^{-1}(x) = \frac{(x+3)^2}{4} - 5\) or \(\frac{x^2+6x-11}{4}\) in terms of \(x\).
B1: Correct domain \(x \geqslant -3\) for \(\mathrm{f}^{-1}\).

Method 1: Squaring both sides (valid for \(2x + 1 \geqslant 0\))
\((3x - 4)^2 \leqslant (2x + 1)^2\)
\(9x^2 - 24x + 16 \leqslant 4x^2 + 4x + 1\)
\(5x^2 - 28x + 15 \leqslant 0\)
Factorising the quadratic:
\((5x - 3)(x - 5) \leqslant 0\)
Critical values are \(x = \frac{3}{5}\) and \(x = 5\).
Since the inequality is \(\leqslant 0\), the solution set is \(\frac{3}{5} \leqslant x \leqslant 5\).
Note that for all \(x\) in this interval, \(2x + 1 > 0\), so all solutions are valid.

Method 2: Linear inequalities
\(-(2x + 1) \leqslant 3x - 4 \leqslant 2x + 1\)
From \(3x - 4 \leqslant 2x + 1 \implies x \leqslant 5\).
From \(-2x - 1 \leqslant 3x - 4 \implies 5x \geqslant 3 \implies x \geqslant \frac{3}{5}\).
Combining these gives \(\frac{3}{5} \leqslant x \leqslant 5\).

M1: Attempt to solve by squaring both sides to obtain a 3-term quadratic or setting up two linear equations/inequalities \(3x - 4 = 2x + 1\) and \(3x - 4 = -(2x + 1)\).
A1: Finding the two critical values \(x = \frac{3}{5}\) and \(x = 5\).
M1: Choosing the correct inside region for their critical values.
A1: Correct final inequality \(\frac{3}{5} \leqslant x \leqslant 5\) (or \(0.6 \leqslant x \leqslant 5\)).

Applying the addition and subtraction laws of logarithms on the left-hand side:
\(\log_2 \left( \frac{(x + 3)(x - 2)}{x + 1} \right) = 2\)

Converting the logarithmic equation into exponential form:
\(\frac{(x + 3)(x - 2)}{x + 1} = 2^2\)
\(\frac{x^2 + x - 6}{x + 1} = 4\)

Multiplying by \((x + 1)\):
\(x^2 + x - 6 = 4(x + 1)\)
\(x^2 + x - 6 = 4x + 4\)
\(x^2 - 3x - 10 = 0\)

Factorising the quadratic equation:
\((x - 5)(x + 2) = 0\)
This gives \(x = 5\) or \(x = -2\).

Checking for extraneous solutions:
For \(x = -2\), \(\log_2(x - 2) = \log_2(-4)\), which is undefined as the argument of a logarithm must be strictly positive.
For \(x = 5\), all arguments \(x + 3 = 8 > 0\), \(x - 2 = 3 > 0\), and \(x + 1 = 6 > 0\) are valid.

Thus, the only valid solution is \(x = 5\).

M1: Correct use of logarithm laws to combine terms into a single logarithm \(\log_2 \left( \frac{(x + 3)(x - 2)}{x + 1} \right)\).
M1: Correctly removing the logarithm to obtain \(\frac{(x + 3)(x - 2)}{x + 1} = 2^2\) or \(4\).
M1: Expanding and rearranging to form a 3-term quadratic equation \(x^2 - 3x - 10 = 0\) and attempting to solve it.
A1: \(x = 5\) only (must explicitly discard or omit \(x = -2\)).

Using the Factor Theorem, since \((x - 2)\) is a factor, \(\mathrm{P}(2) = 0\):
\(\mathrm{P}(2) = 2(2)^3 + a(2)^2 + b(2) - 12 = 0\)
\(2(8) + 4a + 2b - 12 = 0\)
\(16 + 4a + 2b - 12 = 0\)
\(4a + 2b + 4 = 0\)
\(2a + b = -2\) --- (Equation 1)

Using the Remainder Theorem, since the remainder when divided by \((x + 1)\) is \(-18\), \(\mathrm{P}(-1) = -18\):
\(\mathrm{P}(-1) = 2(-1)^3 + a(-1)^2 + b(-1) - 12 = -18\)
\(-2 + a - b - 12 = -18\)
\(a - b - 14 = -18\)
\(a - b = -4\) --- (Equation 2)

Adding Equation 1 and Equation 2:
\((2a + b) + (a - b) = -2 + (-4)\)
\(3a = -6 \implies a = -2\)

Substituting \(a = -2\) into Equation 2:
\(-2 - b = -4 \implies b = 2\)

Therefore, \(a = -2\) and \(b = 2\).

M1: Setting \(\mathrm{P}(2) = 0\) and forming a linear equation in \(a\) and \(b\) (e.g. \(4a + 2b = -4\) or \(2a + b = -2\)).
M1: Setting \(\mathrm{P}(-1) = -18\) and forming a second linear equation in \(a\) and \(b\) (e.g. \(a - b = -4\)).
M1: Valid method to solve their two simultaneous linear equations for \(a\) or \(b\).
A1: Both correct values \(a = -2\) and \(b = 2\).

(a) Starting with the left-hand side and combining over a common denominator:
\[ \begin{aligned} \text{LHS} &= \frac{\sin \theta(1 + \cos \theta) - \sin \theta(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} \[6pt] &= \frac{\sin \theta + \sin \theta \cos \theta - \sin \theta + \sin \theta \cos \theta}{1 - \cos^2 \theta} \[6pt] &= \frac{2\sin \theta \cos \theta}{\sin^2 \theta} \[6pt] &= \frac{2\cos \theta}{\sin \theta} \[6pt] &= 2 \cot \theta = \text{RHS} \end{aligned} \]

(b) Using the identity with \(\theta = 2x\):
\[ 2\cot 2x = 2\sqrt{3} \implies \cot 2x = \sqrt{3} \implies \tan 2x = \frac{1}{\sqrt{3}} \]
For \(0 \leqslant x \leqslant \pi\), the domain for \(2x\) is \(0 \leqslant 2x \leqslant 2\pi\).
\[ 2x = \frac{\pi}{6} \quad \text{or} \quad 2x = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \]
Dividing by 2 gives:
\[ x = \frac{\pi}{12}, \quad x = \frac{7\pi}{12} \]

(a)
M1: Puts fractions over a common denominator \((1-\cos\theta)(1+\cos\theta)\) with correct numerator expansion.
M1: Applies \(1 - \cos^2 \theta = \sin^2 \theta\) in denominator and simplifies numerator to \(2\sin\theta\cos\theta\).
A1: Cancels \(\sin \theta\) and deduces \(2\cot \theta\) with complete, clear working shown.

(b)
B1: Obtains \(\cot 2x = \sqrt{3}\) or \(\tan 2x = \frac{1}{\sqrt{3}}\).
M1: Finds at least one valid value for \(2x\) in radians (e.g., \(\frac{\pi}{6}\) or \(\frac{7\pi}{6}\)).
A1: Both exact answers \(x = \frac{\pi}{12}\) and \(x = \frac{7\pi}{12}\) with no extra values in the range.

(a)(i) When \(x = 0\):
\[ y = 3\mathrm{e}^0 - 5 = 3(1) - 5 = -2 \]
So the \(y\)-intercept is \((0, -2)\).

(a)(ii) When \(y = 0\):
\[ 3\mathrm{e}^{2x} - 5 = 0 \implies 3\mathrm{e}^{2x} = 5 \implies \mathrm{e}^{2x} = \frac{5}{3} \]
Taking natural logarithms:
\[ 2x = \ln\left(\frac{5}{3}\right) \implies x = \frac{1}{2}\ln\left(\frac{5}{3}\right) \]
So the \(x\)-intercept is \(\left(\frac{1}{2}\ln\left(\frac{5}{3}\right), 0\right)\).

(b) As \(x \to -\infty\), \(3\mathrm{e}^{2x} \to 0\), so the horizontal asymptote is \(y = -5\).
The graph is an increasing exponential curve starting just above \(y = -5\) in the third quadrant, crossing the \(y\)-axis at \((0, -2)\), crossing the \(x\)-axis at \(\left(\frac{1}{2}\ln\left(\frac{5}{3}\right), 0\right)\), and increasing rapidly into the first quadrant.

(a)(i)
B1: \((0, -2)\) or \(y = -2\).

(a)(ii)
M1: Sets \(y = 0\) and solves for \(x\) using logarithms (e.g. \(2x = \ln(5/3)\)).
A1: Correct exact coordinates \(\left(\frac{1}{2}\ln\left(\frac{5}{3}\right), 0\right)\) or \(\left(\ln\sqrt{\frac{5}{3}}, 0\right)\).

(b)
B1: Straight dashed line or explicit statement for asymptote \(y = -5\).
B1: Correct shape: strictly increasing curve, approaching asymptote as \(x \to -\infty\).
B1: Both intercept coordinates clearly indicated/labelled on the sketch.

(a) The expression \(|2x - 5| = 0\) when \(x = \frac{5}{2}\), giving vertex \(\left(\frac{5}{2}, -3\right)\).
- \(y\)-intercept: when \(x = 0\), \(y = |0 - 5| - 3 = 5 - 3 = 2\), so \((0, 2)\).
- \(x\)-intercepts: when \(y = 0\), \(|2x - 5| - 3 = 0 \implies |2x - 5| = 3\).
\(2x - 5 = 3 \implies 2x = 8 \implies x = 4\)
\(2x - 5 = -3 \implies 2x = 2 \implies x = 1\)
So intercepts are \((1, 0)\) and \((4, 0)\).
The graph is a V-shape with vertex at \(\left(\frac{5}{2}, -3\right)\), crossing the \(x\)-axis at \((1, 0)\) and \((4, 0)\), and the \(y\)-axis at \((0, 2)\).

(b) The inequality is \(|2x - 5| - 3 \leqslant x + 1 \iff |2x - 5| \leqslant x + 4\).
Setting the linear boundaries:
1) \(2x - 5 = x + 4 \implies x = 9\)
2) \(-(2x - 5) = x + 4 \implies -2x + 5 = x + 4 \implies 3x = 1 \implies x = \frac{1}{3}\)
From the graph, the V-shape lies below or on the line \(y = x + 1\) between these two critical values.
Thus, \(\frac{1}{3} \leqslant x \leqslant 9\).

(a)
B1: Correct V-shape with minimum point.
B1: Vertex \(\left(\frac{5}{2}, -3\right)\) labelled.
B1: All axis intercepts labelled: \((0, 2)\), \((1, 0)\), and \((4, 0)\).

(b)
M1: Solves \(2x - 5 = x + 4\) and \(-(2x - 5) = x + 4\) (or squares both sides).
A1: Critical values \(x = \frac{1}{3}\) and \(x = 9\).
A1: Correct closed inequality \(\frac{1}{3} \leqslant x \leqslant 9\) (or equivalent set notation).

(a) Using the product rule \(\frac{\mathrm{d}}{\mathrm{d}x}(uv) = u'v + uv'\):
Let \(u = x^2\) and \(v = \ln(3x)\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{3}{3x} = \frac{1}{x}\).
\[ \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left( x^2 \ln(3x) \right) &= 2x \ln(3x) + x^2 \left(\frac{1}{x}\right) \[4pt] &= 2x \ln(3x) + x \[4pt] &= x\left(1 + 2\ln(3x)\right) \end{aligned} \]

(b) At a stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ x\left(1 + 2\ln(3x)\right) = 0 \]
Since \(x > 0\), \(x \neq 0\):
\[ 1 + 2\ln(3x) = 0 \implies \ln(3x) = -\frac{1}{2} \implies 3x = \mathrm{e}^{-1/2} \implies x = \frac{1}{3\sqrt{\mathrm{e}}} \]
Finding the \(y\)-coordinate:
\[ y = \left(\frac{1}{3\sqrt{\mathrm{e}}}\right)^2 \ln\left(3 \cdot \frac{1}{3\sqrt{\mathrm{e}}}\right) = \frac{1}{9\mathrm{e}} \left(-\frac{1}{2}\right) = -\frac{1}{18\mathrm{e}} \]
To determine the nature, find the second derivative:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}\left(x + 2x\ln(3x)\right) = 1 + 2\ln(3x) + 2x\left(\frac{1}{x}\right) = 3 + 2\ln(3x) \]
At \(x = \frac{1}{3\sqrt{\mathrm{e}}}\), \(\ln(3x) = -\frac{1}{2}\):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 3 + 2\left(-\frac{1}{2}\right) = 3 - 1 = 2 > 0 \]
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\), the stationary point is a local minimum.

(a)
M1: Applies product rule to \(x^2 \ln(3x)\), differentiating \(\ln(3x)\) correctly to \(\frac{1}{x}\).
A1: Fully simplifies and factors out \(x\) to obtain the printed form \(x(1 + 2\ln(3x))\).

(b)
M1: Equates derivative to 0 and solves \(1 + 2\ln(3x) = 0\) for \(x\).
A1: Exact \(x\)-coordinate \(x = \frac{1}{3\sqrt{\mathrm{e}}}\) (or \(\frac{1}{3}\mathrm{e}^{-1/2}\)).
A1: Exact \(y\)-coordinate \(y = -\frac{1}{18\mathrm{e}}\) (or \(-\frac{1}{18}\mathrm{e}^{-1}\)).
B1: Finds \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 3 + 2\ln(3x)\) (or uses sign test) and correctly concludes it is a Minimum since \(2 > 0\).

(a) Completing the square for \(x\) and \(y\):
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0 \implies (x - 3)^2 + (y + 2)^2 = 25 \]
Centre \(M = (3, -2)\), radius \(r = \sqrt{25} = 5\).

(b)(i) Gradient of the normal radius \(MP\):
\[ m_{MP} = \frac{1 - (-2)}{7 - 3} = \frac{3}{4} \]
Since tangent \(L\) is perpendicular to \(MP\), the gradient of \(L\) is \(m_L = -\frac{4}{3}\).
Equation of \(L\):
\[ y - 1 = -\frac{4}{3}(x - 7) \implies 3(y - 1) = -4(x - 7) \implies 3y - 3 = -4x + 28 \]
Rearranging gives:
\[ 4x + 3y - 31 = 0 \]

(b)(ii) Point \(R\) is the \(y\)-intercept of \(L\):
Setting \(x = 0\) in \(4(0) + 3y - 31 = 0 \implies y = \frac{31}{3}\).
So \(R = \left(0, \frac{31}{3}\right)\).
The vertices of triangle \(OMR\) are \(O(0, 0)\), \(M(3, -2)\), and \(R\left(0, \frac{31}{3}\right)\).
Taking the base along the \(y\)-axis from \(O(0, 0)\) to \(R\left(0, \frac{31}{3}\right)\), the length of the base is \(\frac{31}{3}\).
The perpendicular height is the horizontal distance from the \(y\)-axis to \(M(3, -2)\), which is \(|3| = 3\).
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{31}{3} \times 3 = \frac{31}{2} = 15.5 \]

(a)
B1: Centre \(M(3, -2)\).
B1: Radius \(r = 5\).

(b)(i)
M1: Finds gradient of radius \(MP = \frac{3}{4}\) and uses perpendicular gradient condition \(m = -\frac{4}{3}\).
A1: Correctly forms equation of line and rearranges to given form \(4x + 3y - 31 = 0\) with full working.

(b)(ii)
B1: Identifies \(R\left(0, \frac{31}{3}\right)\).
M1: Uses a valid method for triangle area, e.g. \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \left(\frac{31}{3}\right)(3)\) or matrix determinant method.
A1: Exact area \(\frac{31}{2}\) or \(15.5\).

(a) Rewriting the curve in index form: \(y = 6x^{\frac{1}{2}} - x^{\frac{3}{2}}\).

Differentiating with respect to \(x\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 6 \cdot \frac{1}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = 3x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}}\).

At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(3x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = 0 \implies \frac{3}{\sqrt{x}} = \frac{3\sqrt{x}}{2} \implies 3x = 6 \implies x = 2\).

When \(x = 2\):
\(y = 6\sqrt{2} - 2\sqrt{2} = 4\sqrt{2}\).
So the stationary point is \((2, 4\sqrt{2})\).

(b) When \(x = 1\):
\(y = 6\sqrt{1} - 1\sqrt{1} = 5\).
Gradient of the tangent at \(x = 1\):
\(m = 3(1)^{-\frac{1}{2}} - \frac{3}{2}(1)^{\frac{1}{2}} = 3 - \frac{3}{2} = \frac{3}{2}\).

Equation of the tangent:
\(y - 5 = \frac{3}{2}(x - 1) \implies y = \frac{3}{2}x + \frac{7}{2}\).

At the \(y\)-axis, \(x = 0 \implies y = \frac{7}{2}\).
Thus, \(T\left(0, \frac{7}{2}\right)\).

(c) The area \(A\) is given by:
\(A = \int_0^2 (6x^{\frac{1}{2}} - x^{\frac{3}{2}})\,\mathrm{d}x = \left[ 6 \cdot \frac{2}{3}x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}} \right]_0^2 = \left[ 4x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}} \right]_0^2\).

Evaluating at limits:
At \(x = 2\): \(4(2^{\frac{3}{2}}) - \frac{2}{5}(2^{\frac{5}{2}}) = 4(2\sqrt{2}) - \frac{2}{5}(4\sqrt{2}) = 8\sqrt{2} - \frac{8\sqrt{2}}{5} = \frac{32\sqrt{2}}{5}\).
At \(x = 0\): \(0\).
Therefore, the exact area is \(\frac{32\sqrt{2}}{5}\).

(a)
M1: Differentiates \(y = 6x^{1/2} - x^{3/2}\) to achieve at least one correct term in powers of \(x\).
A1: Correct derivative \(3x^{-1/2} - \frac{3}{2}x^{1/2}\) and equates to 0 to obtain \(x = 2\).
A1: Correct coordinates \((2, 4\sqrt{2})\).

(b)
M1: Evaluates derivative at \(x = 1\) to obtain gradient \(m = \frac{3}{2}\) and finds \(y(1) = 5\).
M1: Forms equation of straight line through \((1, 5)\) with gradient \(\frac{3}{2}\).
A1: Obtains \(T\left(0, \frac{7}{2}\right)\) or \(x = 0, y = \frac{7}{2}\).

(c)
M1: Integrates terms correctly to obtain \(kx^{3/2} - cx^{5/2}\).
A1: Correct antiderivative \(4x^{3/2} - \frac{2}{5}x^{5/2}\).
A1: Fully simplified exact answer \(\frac{32\sqrt{2}}{5}\) nfww.

(a) Completing the square for the circle equation:
\((x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0\)
\((x - 3)^2 + (y + 2)^2 = 25\)
Therefore, the centre is \((3, -2)\) and the radius is \(r = \sqrt{25} = 5\).

(b)(i) The perpendicular distance \(d\) from the centre \((3, -2)\) to the line \(3x - 4y + k = 0\) must equal the radius \(5\):
\(d = \frac{|3(3) - 4(-2) + k|}{\sqrt{3^2 + (-4)^2}} = 5\)
\(\frac{|9 + 8 + k|}{\sqrt{9 + 16}} = 5\)
\(\frac{|17 + k|}{5} = 5 \implies |17 + k| = 25\)

Setting up the two cases:
\(17 + k = 25 \implies k = 8\)
\(17 + k = -25 \implies k = -42\)
Thus, the values of \(k\) are \(8\) and \(-42\).

(ii) For \(k = 8\), the tangent equation is \(3x - 4y + 8 = 0 \implies y = \frac{3}{4}x + 2\).
The gradient of this tangent is \(\frac{3}{4}\).
The normal to the circle at the point of contact passes through the centre \((3, -2)\) and has gradient \(-\frac{4}{3}\):
\(y - (-2) = -\frac{4}{3}(x - 3) \implies y + 2 = -\frac{4}{3}x + 4 \implies y = -\frac{4}{3}x + 2\).

Equating the tangent and normal to find the intersection:
\(\frac{3}{4}x + 2 = -\frac{4}{3}x + 2\)
\(\left(\frac{3}{4} + \frac{4}{3}\right)x = 0 \implies \frac{25}{12}x = 0 \implies x = 0\).

Substituting \(x = 0\) into \(y = \frac{3}{4}(0) + 2\) gives \(y = 2\).
So the point of contact is \((0, 2)\).

(a)
B1: Centre \((3, -2)\).
B1: Radius \(r = 5\).

(b)(i)
M1: Identifies condition for tangency (e.g. perpendicular distance from centre to line equals radius, or discriminant \(b^2 - 4ac = 0\) from substitution).
M1: Sets up correct equation: \(\frac{|3(3) - 4(-2) + k|}{\sqrt{3^2 + (-4)^2}} = 5\) or correct quadratic in \(x\) / \(y\).
M1: Simplifies to \(|17 + k| = 25\) or \(k^2 + 34k - 336 = 0\).
A1: Clearly shows \(k = 8\) and \(k = -42\) with fully justified working.

(b)(ii)
M1: Finds gradient of normal \(-\frac{4}{3}\) and writes equation of normal through \((3, -2)\).
M1: Solves simultaneous linear equations of normal and tangent \(3x - 4y + 8 = 0\).
A1: Obtains \((0, 2)\).

(a) Differentiating \(\mathrm{f}(x) = 4\cos 2x + 8\sin x\):
\(\mathrm{f}'(x) = 4(-2\sin 2x) + 8\cos x = -8\sin 2x + 8\cos x\).

Using the double-angle identity \(\sin 2x = 2\sin x \cos x\):
\(\mathrm{f}'(x) = -8(2\sin x \cos x) + 8\cos x = -16\sin x \cos x + 8\cos x = 8\cos x(1 - 2\sin x)\).

(b) At stationary points, \(\mathrm{f}'(x) = 0\):
\(8\cos x(1 - 2\sin x) = 0\).

Case 1: \(\cos x = 0\).
For \(0 \le x \le \pi\), \(x = \frac{\pi}{2}\).

Case 2: \(1 - 2\sin x = 0 \implies \sin x = \frac{1}{2}\).
For \(0 \le x \le \pi\), the solutions are \(x = \frac{\pi}{6}\) and \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\).

Therefore, the exact \(x\)-coordinates of the stationary points are \(x = \frac{\pi}{6}\), \(x = \frac{\pi}{2}\), and \(x = \frac{5\pi}{6}\).

(c) Differentiating \(\mathrm{f}'(x) = -8\sin 2x + 8\cos x\):
\(\mathrm{f}''(x) = -16\cos 2x - 8\sin x\).

Evaluating \(\mathrm{f}''(x)\) at each stationary point:

1. At \(x = \frac{\pi}{6}\):
\(\mathrm{f}''\left(\frac{\pi}{6}\right) = -16\cos\left(\frac{\pi}{3}\right) - 8\sin\left(\frac{\pi}{6}\right) = -16\left(\frac{1}{2}\right) - 8\left(\frac{1}{2}\right) = -8 - 4 = -12 < 0\).
Hence, \(x = \frac{\pi}{6}\) is a local maximum.

2. At \(x = \frac{\pi}{2}\):
\(\mathrm{f}''\left(\frac{\pi}{2}\right) = -16\cos(\pi) - 8\sin\left(\frac{\pi}{2}\right) = -16(-1) - 8(1) = 16 - 8 = 8 > 0\).
Hence, \(x = \frac{\pi}{2}\) is a local minimum.

3. At \(x = \frac{5\pi}{6}\):
\(\mathrm{f}''\left(\frac{5\pi}{6}\right) = -16\cos\left(\frac{5\pi}{3}\right) - 8\sin\left(\frac{5\pi}{6}\right) = -16\left(\frac{1}{2}\right) - 8\left(\frac{1}{2}\right) = -8 - 4 = -12 < 0\).
Hence, \(x = \frac{5\pi}{6}\) is a local maximum.

(a)
M1: Applies chain rule to obtain \(-8\sin 2x + 8\cos x\).
A1: Uses \(\sin 2x = 2\sin x \cos x\) and fully factors out \(8\cos x\) to establish the given form.

(b)
M1: Sets factors to zero: \(\cos x = 0\) and \(\sin x = \frac{1}{2}\).
A1: Obtains \(x = \frac{\pi}{2}\) and \(x = \frac{\pi}{6}\).
A1: Obtains \(x = \frac{5\pi}{6}\) with no extra values in range.

(c)
M1: Correctly differentiates to find \(\mathrm{f}''(x) = -16\cos 2x - 8\sin x\).
M1: Evaluates \(\mathrm{f}''(x)\) at the candidate points.
A1: Correctly classifies all three stationary points with supporting numerical values of \(\mathrm{f}''(x)\).

To solve \(|3x - 5| \leqslant 2x + 1\):

Method 1: Algebraic definition
The inequality is equivalent to the compound inequality:
\(-(2x + 1) \leqslant 3x - 5 \leqslant 2x + 1\)

1. Left inequality:
\(-2x - 1 \leqslant 3x - 5\)
\(4 \leqslant 5x\)
\(x \geqslant \frac{4}{5}\)

2. Right inequality:
\(3x - 5 \leqslant 2x + 1\)
\(x \leqslant 6\)

Combining both conditions (and noting that \(2x + 1 \geqslant 0\) for \(x \geqslant -0.5\), which holds):
\(\frac{4}{5} \leqslant x \leqslant 6\)

Method 2: Squaring both sides
\((3x - 5)^2 \leqslant (2x + 1)^2\) (for \(2x + 1 \geqslant 0\))
\(9x^2 - 30x + 25 \leqslant 4x^2 + 4x + 1\)
\(5x^2 - 34x + 24 \leqslant 0\)
\((5x - 4)(x - 6) \leqslant 0\)
Critical values are \(x = \frac{4}{5}\) and \(x = 6\).
Hence, \(\frac{4}{5} \leqslant x \leqslant 6\).

M1: Valid method to set up boundary equations or critical values (e.g. squaring both sides to form a quadratic, or splitting into two linear inequalities \(3x - 5 = \pm(2x + 1)\)).
A1: Correct critical values \(x = \frac{4}{5}\) (or \(0.8\)) and \(x = 6\).
M1: Appropriate test of region / inequality manipulation consistent with \(\leqslant\).
A1: Fully correct inequality \(\frac{4}{5} \leqslant x \leqslant 6\) or \(0.8 \leqslant x \leqslant 6\).

(a) Taking the natural logarithm of both sides of \(y = A \mathrm{e}^{kx}\):
\(\ln y = \ln(A \mathrm{e}^{kx}) = \ln A + \ln(\mathrm{e}^{kx})\)
\(\ln y = kx + \ln A\)

This is in linear form \(Y = mX + c\), where the gradient is \(m = k\) and the intercept is \(c = \ln A\).

Using the given points \((1, 3.2)\) and \((4, 8.6)\):
\(k = \frac{8.6 - 3.2}{4 - 1} = \frac{5.4}{3} = 1.8\)

(b) Substitute \(k = 1.8\) and the point \((1, 3.2)\) into the line equation:
\(3.2 = 1.8(1) + \ln A\)
\(\ln A = 3.2 - 1.8 = 1.4\)
\(A = \mathrm{e}^{1.4} \approx 4.0552\)

To 3 significant figures, \(A = 4.06\).

(a)
M1: Converts equation to linear form \(\ln y = kx + \ln A\) and calculates gradient \(\frac{8.6 - 3.2}{4 - 1}\).
A1: Obtains \(k = 1.8\) or \(\frac{9}{5}\).

(b)
M1: Correctly calculates \(\ln A = 1.4\) and applies inverse exponential \(A = \mathrm{e}^{1.4}\).
A1: \(A = 4.06\) (must be given to 3 s.f.).

(a) Let \(u = 2x - 3\) and \(v = \mathrm{e}^{2x}\).
Then \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = 2\) and \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = 2\mathrm{e}^{2x}\).

Using the product rule:
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = u \dfrac{\mathrm{d}v}{\mathrm{d}x} + v \dfrac{\mathrm{d}u}{\mathrm{d}x}\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = (2x - 3)(2\mathrm{e}^{2x}) + 2\mathrm{e}^{2x}\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{2x}(2x - 3 + 1)\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{2x}(2x - 2) = 4(x - 1)\mathrm{e}^{2x}\)

(b) For stationary points, \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(4(x - 1)\mathrm{e}^{2x} = 0\)
Since \(\mathrm{e}^{2x} > 0\) for all real \(x\), \(x - 1 = 0 \implies x = 1\).
When \(x = 1\), \(y = (2(1) - 3)\mathrm{e}^{2(1)} = -1\mathrm{e}^2 = -\mathrm{e}^2\).

To determine the nature, find \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\):
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4\mathrm{e}^{2x} + 8(x - 1)\mathrm{e}^{2x} = 4(2x - 1)\mathrm{e}^{2x}\)
At \(x = 1\), \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4(1)\mathrm{e}^2 = 4\mathrm{e}^2 > 0\).
Therefore, the stationary point is a local minimum.

(a)
M1: Correct application of the product rule: \(2\mathrm{e}^{2x} + 2(2x - 3)\mathrm{e}^{2x}\).
A1: Factorises completely and clearly obtains the given result \(4(x - 1)\mathrm{e}^{2x}\) with no skipped steps.

(b)
B1: Exact coordinates \((1, -\mathrm{e}^2)\).
B1: Minimum, fully justified by evaluating \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) at \(x = 1\) (or a clear first derivative sign change analysis).

(a) Start from the left-hand side:
\(\text{LHS} = \dfrac{\cos \theta(1 + \sin \theta) - \cos \theta(1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)}\)
\(\text{LHS} = \dfrac{\cos \theta + \cos \theta \sin \theta - \cos \theta + \cos \theta \sin \theta}{1 - \sin^2 \theta}\)
\(\text{LHS} = \dfrac{2\cos \theta \sin \theta}{\cos^2 \theta}\)
\(\text{LHS} = \dfrac{2\sin \theta}{\cos \theta} = 2\tan \theta = \text{RHS}\)

(b) Using the identity from part (a):
\(2\tan \theta = 3\sqrt{3}\)
\(\tan \theta = \dfrac{3\sqrt{3}}{2} \approx 2.5981\)

Basic acute angle:
\(\alpha = \tan^{-1}\left(\dfrac{3\sqrt{3}}{2}\right) \approx 68.962^\circ\)

Since \(\tan \theta > 0\), \(\theta\) lies in the first and third quadrants for \(0^\circ \leqslant \theta \leqslant 360^\circ\):
\(\theta = 68.962^\circ \approx 69.0^\circ\)
\(\theta = 180^\circ + 68.962^\circ = 248.962^\circ \approx 249.0^\circ\)

(a)
M1: Combines fractions over a common denominator \((1 - \sin \theta)(1 + \sin \theta)\) and uses \(1 - \sin^2 \theta = \cos^2 \theta\).
A1: Fully simplified working leading convincingly to \(2\tan \theta\).

(b)
M1: Sets \(2\tan \theta = 3\sqrt{3} \implies \tan \theta = \frac{3\sqrt{3}}{2}\) and finds the principal value.
A1: \(\theta = 69.0^\circ\) and \(\theta = 249.0^\circ\) (both correct to 1 decimal place, no extra solutions in range).

(a) Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \):
\[ 3(1 + \tan^2 \theta) - 5\tan \theta = 5 \]
\[ 3 + 3\tan^2 \theta - 5\tan \theta - 5 = 0 \]
\[ 3\tan^2 \theta - 5\tan \theta - 2 = 0 \]

(b) Let \( \theta = 2x - 20^\circ \).
From part (a), \( (3\tan \theta + 1)(\tan \theta - 2) = 0 \).
So \( \tan \theta = 2 \) or \( \tan \theta = -\frac{1}{3} \).

Given \( 0^\circ \leqslant x \leqslant 180^\circ \), the interval for \( \theta \) is \( -20^\circ \leqslant \theta \leqslant 340^\circ \).

Case 1: \( \tan \theta = 2 \)
\( \theta = \tan^{-1}(2) \approx 63.435^\circ \) or \( 63.435^\circ + 180^\circ = 243.435^\circ \).
- \( 2x - 20 = 63.435 \implies 2x = 83.435 \implies x \approx 41.7^\circ \)
- \( 2x - 20 = 243.435 \implies 2x = 263.435 \implies x \approx 131.7^\circ \)

Case 2: \( \tan \theta = -\frac{1}{3} \)
\( \theta = \tan^{-1}\left(-\frac{1}{3}\right) \approx -18.435^\circ \) or \( -18.435^\circ + 180^\circ = 161.565^\circ \).
- \( 2x - 20 = -18.435 \implies 2x = 1.565 \implies x \approx 0.8^\circ \) (or \( 0.78^\circ \) to 1 d.p. is \( 0.8^\circ \))
Wait: \( 2x = 20 - 18.4349 = 1.5651 \implies x = 0.78^\circ \approx 0.8^\circ \).
- \( 2x - 20 = 161.565 \implies 2x = 181.565 \implies x \approx 90.8^\circ \).

(a)
M1: Substitute \( \sec^2 \theta = 1 + \tan^2 \theta \) into the given equation.
A1: Fully correct algebraic steps leading to \( 3\tan^2 \theta - 5\tan \theta - 2 = 0 \) with no errors.

(b)
M1: Factorise or use quadratic formula to solve for \( \tan(2x - 20^\circ) \), obtaining \( \tan(2x - 20^\circ) = 2 \) or \( -\frac{1}{3} \).
M1: Valid attempt to find reference angles and values for \( 2x - 20^\circ \) in correct quadrants.
A1: Any two correct solutions for \( x \).
A1: Remaining two correct solutions for \( x \) (i.e. \( x = 0.8^\circ, 41.7^\circ, 90.8^\circ, 131.7^\circ \)).
A1: Deduct 1 mark if extra values inside the domain are given or if rounding to 1 d.p. is not followed.

(a) For an arithmetic progression, \( a = 1200 \), \( n = 10 \), and \( S_{10} = 16500 \).
\[ S_{10} = \frac{10}{2} [2(1200) + (10 - 1)d] = 16500 \]
\[ 5(2400 + 9d) = 16500 \]
\[ 2400 + 9d = 3300 \]
\[ 9d = 900 \implies d = 100 \]

(b) For a geometric progression, \( a = 800 \) and common ratio \( r = 1 + 0.15 = 1.15 \).
Total production over \( n \) years is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1} = \frac{800(1.15^n - 1)}{1.15 - 1} = \frac{800(1.15^n - 1)}{0.15} \]
We require \( S_n > 25000 \):
\[ \frac{800(1.15^n - 1)}{0.15} > 25000 \]
\[ 1.15^n - 1 > \frac{25000 \times 0.15}{800} = 4.6875 \]
\[ 1.15^n > 5.6875 \]
Taking natural logarithms on both sides:
\[ n \ln(1.15) > \ln(5.6875) \]
\[ n > \frac{\ln(5.6875)}{\ln(1.15)} \approx \frac{1.7383}{0.13976} \approx 12.44 \]
Since \( n \) must be an integer, the first year is year 13. Wait, let us check: for \( n = 12 \), \( S_{12} = \frac{800(1.15^{12} - 1)}{0.15} = 23201.3 \). For \( n = 13 \), \( S_{13} = \frac{800(1.15^{13} - 1)}{0.15} = 27481.5 > 25000 \).
Thus, the smallest integer is \( n = 13 \).

(a)
M1: Use \( S_n = \frac{n}{2}(2a + (n-1)d) \) with \( n=10 \) and \( a=1200 \).
M1: Set expression equal to 16500 and solve for \( d \).
A1: \( d = 100 \)

(b)
M1: Identify GP with \( a = 800 \) and \( r = 1.15 \).
M1: Set \( \frac{800(1.15^n - 1)}{1.15 - 1} > 25000 \) or \( = 25000 \).
M1: Correctly apply logarithms to solve \( 1.15^n > 5.6875 \).
A1: Obtain \( n = 13 \) (or Year 13) with supporting working.

(a) Particle is at rest when \( v = 0 \):
\[ 6t - 2t^2 = 0 \implies 2t(3 - t) = 0 \]
So \( t = 0 \) or \( t = 3 \). The first time after \( t = 0 \) is \( t = 3 \).
Acceleration \( a = \frac{\mathrm{d}v}{\mathrm{d}t} = 6 - 4t \).
At \( t = 3 \):
\[ a = 6 - 4(3) = -6 \text{ m s}^{-2} \]

(b) Displacement \( s(t) = \int (6t - 2t^2) \, \mathrm{d}t = 3t^2 - \frac{2}{3}t^3 \) (taking \( s(0) = 0 \)).
The particle changes direction at \( t = 3 \).
At \( t = 3 \):
\[ s(3) = 3(3)^2 - \frac{2}{3}(3)^3 = 27 - 18 = 9 \text{ m} \]
At \( t = 4 \):
\[ s(4) = 3(4)^2 - \frac{2}{3}(4)^3 = 48 - \frac{128}{3} = \frac{16}{3} \approx 5.333 \text{ m} \]
Distance travelled from \( t = 0 \) to \( t = 3 \) is \( |9 - 0| = 9 \text{ m} \).
Distance travelled from \( t = 3 \) to \( t = 4 \) is \( |s(4) - s(3)| = \left|\frac{16}{3} - 9\right| = \frac{11}{3} \approx 3.667 \text{ m} \).

Total distance \( = 9 + \frac{11}{3} = \frac{38}{3} \approx 12.7 \text{ m} \) (or \( 12\frac{2}{3} \text{ m} \)).

(a)
M1: Set \( v = 0 \) and solve to find \( t = 3 \).
M1: Differentiate \( v \) to find \( a = 6 - 4t \).
A1: Obtain \( a = -6 \text{ m s}^{-2} \).

(b)
M1: Integrate \( v \) to obtain displacement \( s(t) = 3t^2 - \frac{2}{3}t^3 (+ c) \).
M1: Calculate displacement at the turning point \( t = 3 \) (i.e. \( 9 \)) and at \( t = 4 \) (i.e. \( \frac{16}{3} \)).
M1: Correctly calculate the sum of individual distances \( |s(3) - s(0)| + |s(4) - s(3)| \).
A1: \( \frac{38}{3} \) or \( 12.7 \text{ m} \) (3 s.f.).

(a) The maximum occurs when \( \cos\left(\frac{\pi t}{6}\right) = 1 \):
\( h_{\max} = 5 + 3(1) = 8 \text{ m} \).
The minimum occurs when \( \cos\left(\frac{\pi t}{6}\right) = -1 \):
\( h_{\min} = 5 + 3(-1) = 2 \text{ m} \).

(b) Set \( h = 6.5 \):
\[ 5 + 3\cos\left(\frac{\pi t}{6}\right) = 6.5 \]
\[ 3\cos\left(\frac{\pi t}{6}\right) = 1.5 \]
\[ \cos\left(\frac{\pi t}{6}\right) = 0.5 \]
For \( 0 \leqslant t \leqslant 12 \), \( 0 \leqslant \frac{\pi t}{6} \leqslant 2\pi \).
\[ \frac{\pi t}{6} = \frac{\pi}{3} \implies t = 2 \]
\[ \frac{\pi t}{6} = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \implies t = 10 \]

(c) The depth \( h \geqslant 6.5 \) when \( \cos\left(\frac{\pi t}{6}\right) \geqslant 0.5 \).
This occurs for \( 0 \leqslant \frac{\pi t}{6} \leqslant \frac{\pi}{3} \) and \( \frac{5\pi}{3} \leqslant \frac{\pi t}{6} \leqslant 2\pi \).
- Interval 1: \( 0 \leqslant t \leqslant 2 \), duration = 2 hours.
- Interval 2: \( 10 \leqslant t \leqslant 12 \), duration = 2 hours.
Total time \( = 2 + 2 = 4 \text{ hours} \).

(a)
B1: Maximum depth \( = 8 \text{ m} \).
B1: Minimum depth \( = 2 \text{ m} \).

(b)
M1: Equate \( 5 + 3\cos\left(\frac{\pi t}{6}\right) = 6.5 \) to get \( \cos\left(\frac{\pi t}{6}\right) = 0.5 \).
M1: Find first angle \( \frac{\pi t}{6} = \frac{\pi}{3} \) to get \( t = 2 \).
A1: Find second solution \( t = 10 \).

(c)
M1: Identify the time intervals \( 0 \leqslant t \leqslant 2 \) and \( 10 \leqslant t \leqslant 12 \) (or equivalent graphical understanding).
A1: Total time = 4 hours.

(a) The second term is \( u_2 = ar = 18 \implies a = \frac{18}{r} \).
The sum to infinity is \( S_\infty = \frac{a}{1 - r} = 72 \).
Substituting \( a = \frac{18}{r} \):
\[ \frac{18}{r(1 - r)} = 72 \]
\[ 18 = 72r(1 - r) \]
\[ 18 = 72r - 72r^2 \]
Dividing by 18:
\[ 1 = 4r - 4r^2 \]
\[ 4r^2 - 4r + 1 = 0 \]
Factorising:
\[ (2r - 1)^2 = 0 \implies r = \frac{1}{2} \]
Then \( a = \frac{18}{1/2} = 36 \).

(b) We are given \( S_\infty - S_n < 0.01 \).
Since \( S_\infty - S_n = \frac{a}{1-r} - \frac{a(1 - r^n)}{1 - r} = \frac{a r^n}{1 - r} = S_\infty r^n \):
\[ 72 \left(\frac{1}{2}\right)^n < 0.01 \]
\[ \left(\frac{1}{2}\right)^n < \frac{0.01}{72} = \frac{1}{7200} \]
\[ 2^n > 7200 \]
Taking logarithms:
\[ n \ln 2 > \ln 7200 \]
\[ n > \frac{\ln 7200}{\ln 2} \approx \frac{8.8818}{0.6931} \approx 12.81 \]
Since \( n \) is an integer, the least value of \( n \) is 13.

(a)
M1: State \( ar = 18 \) and \( \frac{a}{1-r} = 72 \).
M1: Substitute \( a = \frac{18}{r} \) into \( S_\infty \) formula.
A1: Correctly obtain the quadratic equation \( 4r^2 - 4r + 1 = 0 \).
A1: Solve to find \( r = \frac{1}{2} \) and \( a = 36 \).

(b)
M1: Form the inequality \( 72\left(\frac{1}{2}\right)^n < 0.01 \) and apply logarithms.
A1: \( n = 13 \).

(a) The volume of the cylinder is given by:
\[ V = \pi r^2 h = 288\pi \implies h = \frac{288}{r^2} \]
The cost consists of the top and bottom circular bases and the curved vertical side:
\[ \text{Cost of top and bottom} = 2 \times (\pi r^2) \times 2 = 4\pi r^2 \]
\[ \text{Cost of curved side} = (2\pi r h) \times 3 = 6\pi r h \]
Substitute \(h = \frac{288}{r^2}\) into the cost equation:
\[ C = 4\pi r^2 + 6\pi r\left(\frac{288}{r^2}\right) = 4\pi r^2 + \frac{1728\pi}{r} \]

(b) Differentiate \(C\) with respect to \(r\):
\[ \frac{\mathrm{d}C}{\mathrm{d}r} = 8\pi r - \frac{1728\pi}{r^2} \]
For a stationary value, set \(\frac{\mathrm{d}C}{\mathrm{d}r} = 0\):
\[ 8\pi r = \frac{1728\pi}{r^2} \implies 8r^3 = 1728 \implies r^3 = 216 \implies r = 6 \]

(c) Calculate the stationary cost when \(r = 6\):
\[ C = 4\pi(6^2) + \frac{1728\pi}{6} = 144\pi + 288\pi = 432\pi \]
Find the second derivative:
\[ \frac{\mathrm{d}^2C}{\mathrm{d}r^2} = 8\pi + \frac{3456\pi}{r^3} \]
At \(r = 6\):
\[ \frac{\mathrm{d}^2C}{\mathrm{d}r^2} = 8\pi + \frac{3456\pi}{216} = 8\pi + 16\pi = 24\pi \]
Since \(\frac{\mathrm{d}^2C}{\mathrm{d}r^2} = 24\pi > 0\), the cost is a minimum.

(a)
M1: Expresses \(h\) in terms of \(r\) using \(\pi r^2 h = 288\pi\) to get \(h = \frac{288}{r^2}\).
M1: Forms cost expression using \(2 \times 2\pi r^2 + 3 \times 2\pi r h\).
A1: Obtains \(C = 4\pi r^2 + \frac{1728\pi}{r}\) with completely correct unsimplified working shown.

(b)
M1: Differentiates \(C\) to obtain form \(A r + B r^{-2}\).
A1: Fully correct derivative \(\frac{\mathrm{d}C}{\mathrm{d}r} = 8\pi r - 1728\pi r^{-2}\).
A1: Equates derivative to zero and solves to find \(r = 6\).

(c)
B1: Obtains exact stationary cost \(432\pi\) (or \(1357\) / \(1360\)).
M1: Finds \(\frac{\mathrm{d}^2C}{\mathrm{d}r^2} = 8\pi + 3456\pi r^{-3}\) and evaluates at \(r = 6\).
A1: Evaluates to \(24\pi\) (or \(75.4\)) and concludes it is a minimum with valid justification (\(>0\)).

(a) When \(x = 0\):
\[ y = 8 - \frac{18}{3} = 2 \implies P(0, 2) \]
Differentiate \(y = 8 - 18(2x+3)^{-1}\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = -18(-1)(2x+3)^{-2}(2) = \frac{36}{(2x+3)^2} \]
At \(x = 0\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{36}{3^2} = 4 \]
Equation of the tangent:
\[ y - 2 = 4(x - 0) \implies y = 4x + 2 \]

(b) For point \(A\) (tangent intersects \(x\)-axis, \(y = 0\)):
\[ 4x + 2 = 0 \implies x = -\frac{1}{2} \implies A\left(-\frac{1}{2}, 0\right) \]
For point \(B\) (curve intersects \(x\)-axis, \(y = 0\)):
\[ 8 - \frac{18}{2x+3} = 0 \implies 2x+3 = \frac{18}{8} = \frac{9}{4} \implies 2x = -\frac{3}{4} \implies x = -\frac{3}{8} \implies B\left(-\frac{3}{8}, 0\right) \]

(c) The region lies between \(x = -\frac{1}{2}\) and \(x = 0\).
Area under tangent from \(x = -\frac{1}{2}\) to \(x = 0\):
\[ \text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2} \times 2 = \frac{1}{2} \]
Area under curve from \(x = -\frac{3}{8}\) to \(x = 0\):
\[ \int_{-\frac{3}{8}}^{0} \left(8 - \frac{18}{2x+3}\right) \mathrm{d}x = \left[ 8x - 9\ln(2x+3) \right]_{-\frac{3}{8}}^{0} \]
At \(x = 0\):
\[ 0 - 9\ln(3) = -9\ln 3 \]
At \(x = -\frac{3}{8}\):
\[ 8\left(-\frac{3}{8}\right) - 9\ln\left(2\left(-\frac{3}{8}\right)+3\right) = -3 - 9\ln\left(\frac{9}{4}\right) \]
Integral value:
\[ (-9\ln 3) - \left(-3 - 9\ln\left(\frac{9}{4}\right)\right) = 3 - 9\ln 3 + 9\ln\left(\frac{9}{4}\right) = 3 + 9\ln\left(\frac{9/4}{3}\right) = 3 + 9\ln\left(\frac{3}{4}\right) \]
The required area is the area under the tangent minus the area under the curve:
\[ \text{Area} = \frac{1}{2} - \left(3 + 9\ln\left(\frac{3}{4}\right)\right) = -\frac{5}{2} - 9\ln\left(\frac{3}{4}\right) = 9\ln\left(\frac{4}{3}\right) - \frac{5}{2} \]

(a)
M1: Differentiates using chain rule to find \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{k}{(2x+3)^2}\).
A1: Evaluates gradient as \(4\) and finds \(P(0, 2)\).
A1: Obtains tangent equation \(y = 4x + 2\) or \(y - 2 = 4x\).

(b)
B1: Finds \(A\left(-\frac{1}{2}, 0\right)\).
B1: Sets \(8 - \frac{18}{2x+3} = 0\) and correctly finds \(B\left(-\frac{3}{8}, 0\right)\).

(c)
B1: Area under tangent \(= \frac{1}{2}\) (via triangle formula or integration \(\int_{-1/2}^0 (4x+2)\mathrm{d}x\)).
M1: Integrates curve function to get \(8x - 9\ln(2x+3)\).
M1: Applies limits \(-\frac{3}{8}\) to \(0\) correctly to the integrated function.
A1: Subtracts curve area from tangent area to obtain exact answer \(9\ln\left(\frac{4}{3}\right) - \frac{5}{2}\) or simplified equivalent (e.g. \(18\ln 2 - 9\ln 3 - 2.5\)).

(a) By similar triangles:
\[ \frac{r}{h} = \frac{8}{24} = \frac{1}{3} \implies r = \frac{h}{3} \]
The volume of a cone is \(V = \frac{1}{3}\pi r^2 h\).
Substituting \(r = \frac{h}{3}\):
\[ V = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h = \frac{1}{3}\pi \left(\frac{h^2}{9}\right) h = \frac{\pi}{27} h^3 \]

(b) Differentiate \(V\) with respect to \(h\):
\[ \frac{\mathrm{d}V}{\mathrm{d}h} = \frac{3\pi}{27} h^2 = \frac{\pi}{9} h^2 \]
By the chain rule:
\[ \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \div \frac{\mathrm{d}V}{\mathrm{d}h} = \frac{12\pi}{\frac{\pi}{9}h^2} = \frac{108}{h^2} \]
When \(h = 6\):
\[ \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{108}{6^2} = \frac{108}{36} = 3\text{ cm s}^{-1} \]

(c) The area of the water surface is \(S = \pi r^2\).
Since \(r = \frac{h}{3}\):
\[ S = \pi \left(\frac{h}{3}\right)^2 = \frac{\pi}{9} h^2 \]
Differentiating \(S\) with respect to \(h\):
\[ \frac{\mathrm{d}S}{\mathrm{d}h} = \frac{2\pi}{9} h \]
By the chain rule:
\[ \frac{\mathrm{d}S}{\mathrm{d}t} = \frac{\mathrm{d}S}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t} \]
At \(h = 6\), \(\frac{\mathrm{d}h}{\mathrm{d}t} = 3\):
\[ \frac{\mathrm{d}S}{\mathrm{d}t} = \left(\frac{2\pi}{9} \times 6\right) \times 3 = \frac{12\pi}{9} \times 3 = 4\pi\text{ cm}^2\text{ s}^{-1} \]

(a)
M1: Uses similar triangles to express \(r\) in terms of \(h\) as \(r = \frac{1}{3}h\).
A1: Substitutes into \(V = \frac{1}{3}\pi r^2 h\) to obtain \(V = \frac{\pi}{27}h^3\) with clear working.

(b)
M1: Differentiates \(V\) with respect to \(h\) to find \(\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{\pi}{9}h^2\).
M1: Applies chain rule \(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \times \frac{\mathrm{d}h}{\mathrm{d}V}\).
A1: Obtains \(\frac{\mathrm{d}h}{\mathrm{d}t} = 3\text{ cm s}^{-1}\).

(c)
M1: Expresses \(S\) in terms of \(h\) as \(S = \frac{\pi}{9}h^2\) (or in terms of \(r\) and finds \(\frac{\mathrm{d}r}{\mathrm{d}t}\)).
M1: Uses chain rule \(\frac{\mathrm{d}S}{\mathrm{d}t} = \frac{\mathrm{d}S}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t}\) (or \(\frac{\mathrm{d}S}{\mathrm{d}t} = 2\pi r \frac{\mathrm{d}r}{\mathrm{d}t}\)).
A1: Obtains exact value \(4\pi\) (or \(12.6\)).