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

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

Method 1: Case analysis
Since the left-hand side is a modulus, we must have \(3x + 1 \geqslant 0 \implies x \geqslant -\frac{1}{3}\).

Case 1: \(2x - 5 \geqslant 0 \implies x \geqslant \frac{5}{2}\)
\[ 2x - 5 \leqslant 3x + 1 \implies -x \leqslant 6 \implies x \geqslant -6 \]
Combining with \(x \geqslant \frac{5}{2}\) gives \(x \geqslant \frac{5}{2}\).

Case 2: \(2x - 5 < 0 \implies x < \frac{5}{2}\)
\[ -(2x - 5) \leqslant 3x + 1 \implies -2x + 5 \leqslant 3x + 1 \implies 5x \geqslant 4 \implies x \geqslant \frac{4}{5} \]
Combining with \(x < \frac{5}{2}\) gives \(\frac{4}{5} \leqslant x < \frac{5}{2}\).

Combining both valid cases gives:
\[ x \geqslant \frac{4}{5} \]

Method 2: Squaring both sides
\[ (2x - 5)^2 \leqslant (3x + 1)^2 \quad \text{for } 3x + 1 \geqslant 0 \]
\[ 4x^2 - 20x + 25 \leqslant 9x^2 + 6x + 1 \]
\[ 5x^2 + 26x - 24 \geqslant 0 \]
\[ (5x - 4)(x + 6) \geqslant 0 \]
Critical values: \(x = \frac{4}{5}\), \(x = -6\).
This gives \(x \leqslant -6\) or \(x \geqslant \frac{4}{5}\).
Since \(|2x - 5| \geqslant 0\), we must reject \(x \leqslant -6\) because \(3x + 1 < 0\) for those values.

Hence, the solution is \(x \geqslant \frac{4}{5}\).

M1: Valid attempt to find critical values by equating \(2x - 5 = \pm(3x + 1)\) or squaring to form \(5x^2 + 26x - 24 = 0\).
A1: Obtains critical values \(x = \frac{4}{5}\) (or \(0.8\)) and \(x = -6\).
M1: Realises that \(3x + 1 \geqslant 0\) is required (or tests regions / uses graph) to reject the region \(x \leqslant -6\).
A0.5: Fully correct final inequality \(x \geqslant \frac{4}{5}\) (or \(x \geqslant 0.8\)).

(a)
Complete the square for both \(x\) and \(y\):
\[ (x - 3)^2 - 9 + (y + 4)^2 - 16 - 11 = 0 \]
\[ (x - 3)^2 + (y + 4)^2 = 36 \]
\[ (x - 3)^2 + (y + 4)^2 = 6^2 \]

Therefore:
- Centre is \((3, -4)\)
- Radius is \(r = \sqrt{36} = 6\)

(b)
The line \(y = k\) is a horizontal line.
For \(y = k\) to be a tangent to the circle, its distance from the centre \((3, -4)\) must equal the radius \(6\).
\[ |-4 - k| = 6 \]
\[ -4 - k = 6 \implies k = -10 \]
\[ -4 - k = -6 \implies k = 2 \]

Hence, \(k = 2\) or \(k = -10\).

(a)
M1: Completes the square for both \(x\) and \(y\) terms to obtain \((x - 3)^2 + (y + 4)^2 = C\).
A1: Centre \((3, -4)\) and radius \(6\) (both correct).

(b)
M0.5: Method using \(y\text{-coordinate of centre} \pm \text{radius}\) (i.e. \(-4 \pm 6\)) or setting discriminant \(b^2 - 4ac = 0\).
A1: Both values \(k = 2\) and \(k = -10\) obtained correctly.

(a) Since \( (x - 2) \) is a factor, \( \mathrm{P}(2) = 0 \):
\( 2(2)^3 + a(2)^2 + b(2) + 6 = 0 \)
\( 16 + 4a + 2b + 6 = 0 \implies 4a + 2b = -22 \implies 2a + b = -11 \quad \text{--- (1)} \)

Since dividing by \( (x + 1) \) gives a remainder of \( 6 \), \( \mathrm{P}(-1) = 6 \):
\( 2(-1)^3 + a(-1)^2 + b(-1) + 6 = 6 \)
\( -2 + a - b + 6 = 6 \implies a - b = 2 \quad \text{--- (2)} \)

Adding equations (1) and (2):
\( 3a = -9 \implies a = -3 \)

Substituting \( a = -3 \) into (2):
\( -3 - b = 2 \implies b = -5 \)

(b) Substituting \( a = -3 \) and \( b = -5 \):
\( \mathrm{P}(x) = 2x^3 - 3x^2 - 5x + 6 \)

Dividing \( \mathrm{P}(x) \) by \( (x - 2) \):
\( 2x^3 - 3x^2 - 5x + 6 = (x - 2)(2x^2 + x - 3) \)

Factorising the quadratic quotient:
\( 2x^2 + x - 3 = (2x + 3)(x - 1) \)

Thus, \( \mathrm{P}(x) = (x - 2)(2x + 3)(x - 1) \).

Setting \( \mathrm{P}(x) = 0 \) gives \( x = 2, x = 1, x = -\frac{3}{2} \).

(a)
M1: Substitute \( x = 2 \) into \( \mathrm{P}(x) \) and equate to \( 0 \).
A1: Obtain correct linear equation, e.g. \( 2a + b = -11 \) or \( 4a + 2b = -22 \).
M1: Substitute \( x = -1 \) into \( \mathrm{P}(x) \), equate to \( 6 \), and attempt to solve the simultaneous equations.
A1: Correct values \( a = -3 \) and \( b = -5 \).

(b)
M1: Valid division or inspection to obtain quadratic factor \( 2x^2 + x - 3 \) and attempt full linear factorisation.
A1: Correct factors \( (x - 2)(2x + 3)(x - 1) \) and all three roots \( x = 2, x = 1, x = -\frac{3}{2} \) stated.

(a) Let \( y = \frac{3x + 1}{x - 2} \):
\( y(x - 2) = 3x + 1 \)
\( xy - 2y = 3x + 1 \)
\( xy - 3x = 2y + 1 \)
\( x(y - 3) = 2y + 1 \implies x = \frac{2y + 1}{y - 3} \)

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

(b) For \( x \geq -5 \), \( \sqrt{x + 5} \geq 0 \).
Thus, \( 2\sqrt{x + 5} - 1 \geq -1 \).
Range of \( \mathrm{g} \): \( \mathrm{g}(x) \geq -1 \) (or \( y \geq -1 \)).

(c) \( \mathrm{f}(\mathrm{g}(x)) = 5 \implies \mathrm{g}(x) = \mathrm{f}^{-1}(5) \).

Using the inverse function:
\( \mathrm{f}^{-1}(5) = \frac{2(5) + 1}{5 - 3} = \frac{11}{2} \)

So, \( \mathrm{g}(x) = \frac{11}{2} \):
\( 2\sqrt{x + 5} - 1 = \frac{11}{2} \)
\( 2\sqrt{x + 5} = \frac{13}{2} \)
\( \sqrt{x + 5} = \frac{13}{4} \)

Squaring both sides:
\( x + 5 = \frac{169}{16} \)
\( x = \frac{169}{16} - 5 = \frac{169 - 80}{16} = \frac{89}{16} \).

(a)
M1: Set \( y = \frac{3x + 1}{x - 2} \), multiply by \( (x - 2) \), collect terms in \( x \), and isolate \( x \).
A1: \( \mathrm{f}^{-1}(x) = \frac{2x + 1}{x - 3} \) (condone use of \( y \) if correctly written at final step).

(b)
B1: \( \mathrm{g}(x) \geq -1 \) or \( y \geq -1 \) or \( [-1, \infty) \).

(c)
M1: Set up equation \( \mathrm{g}(x) = \mathrm{f}^{-1}(5) \) or substitute \( \mathrm{g}(x) \) directly into \( \mathrm{f} \) and equate to \( 5 \).
A1: Obtain \( \mathrm{g}(x) = \frac{11}{2} \) (or equivalent single radical equation).
M1: Correctly isolate \( \sqrt{x + 5} \) and square both sides.
A1: \( x = \frac{89}{16} \) (or \( 5.5625 \) or \( 5\frac{9}{16} \)).

(a) Using the power and quotient laws of logarithms:
\( \log_3((x - 1)^2) - \log_3(2x - 5) = 1 \)
\( \log_3\left(\frac{(x - 1)^2}{2x - 5}\right) = 1 \)

Converting to exponential form:
\( \frac{(x - 1)^2}{2x - 5} = 3^1 = 3 \)
\( (x - 1)^2 = 3(2x - 5) \)
\( x^2 - 2x + 1 = 6x - 15 \)
\( x^2 - 8x + 16 = 0 \)
\( (x - 4)^2 = 0 \implies x = 4 \)

Check: \( x - 1 = 3 > 0 \) and \( 2x - 5 = 3 > 0 \), so \( x = 4 \) is valid.

(b) Express the exponential equation in base 2:
\( (2^2)^{2x - y} = 2^5 \implies 2^{4x - 2y} = 2^5 \)
\( 4x - 2y = 5 \quad \text{--- (1)} \)

Express the logarithmic equation in exponential form:
\( x + 3y = 2^3 = 8 \implies x = 8 - 3y \quad \text{--- (2)} \)

Substitute (2) into (1):
\( 4(8 - 3y) - 2y = 5 \)
\( 32 - 12y - 2y = 5 \)
\( 32 - 14y = 5 \implies 14y = 27 \implies y = \frac{27}{14} \)

Substitute \( y = \frac{27}{14} \) into (2):
\( x = 8 - 3\left(\frac{27}{14}\right) = \frac{112 - 81}{14} = \frac{31}{14} \).

(a)
M1: Apply power and quotient laws to write as single logarithm: \( \log_3\left(\frac{(x - 1)^2}{2x - 5}\right) \).
M1: Remove logarithms correctly to form quadratic equation: \( \frac{(x - 1)^2}{2x - 5} = 3 \).
A1: Obtain \( x^2 - 8x + 16 = 0 \).
A1: \( x = 4 \) (must not include extraneous solutions).

(b)
B1: \( 4x - 2y = 5 \) derived from index laws in base 2.
B1: \( x + 3y = 8 \) derived from definition of logarithm.
M1: Valid method to solve the linear simultaneous equations (elimination or substitution).
A1: Exact answers \( x = \frac{31}{14} \) and \( y = \frac{27}{14} \).

(a) Case 1: \( 3x - 5 = 2x + 1 \)
\( x = 6 \)
Check: \( |3(6) - 5| = 13 \) and \( 2(6) + 1 = 13 \) (valid).

Case 2: \( 3x - 5 = -(2x + 1) \)
\( 3x - 5 = -2x - 1 \)
\( 5x = 4 \implies x = \frac{4}{5} \)
Check: \( |3(4/5) - 5| = |-13/5| = \frac{13}{5} \) and \( 2(4/5) + 1 = \frac{13}{5} \) (valid).

Thus, the solutions are \( x = \frac{4}{5} \) and \( x = 6 \).

(b) Rewrite \( 2^{2y+1} \) as \( 2 \cdot (2^y)^2 \):
\( 2(2^y)^2 - 17(2^y) + 8 = 0 \)

Using \( u = 2^y \):
\( 2u^2 - 17u + 8 = 0 \)
\( (2u - 1)(u - 8) = 0 \)
\( u = \frac{1}{2} \quad \text{or} \quad u = 8 \)

Returning to \( y \):
\( 2^y = \frac{1}{2} = 2^{-1} \implies y = -1 \)
\( 2^y = 8 = 2^3 \implies y = 3 \)

Thus, \( y = -1 \) and \( y = 3 \).

(a)
M1: Set up two linear equations: \( 3x - 5 = 2x + 1 \) and \( 3x - 5 = -(2x + 1) \) (or square both sides).
A1: Find one correct solution: \( x = 6 \) or \( x = \frac{4}{5} \).
A1: Find second correct solution with valid verification/no invalid extra solutions.

(b)
M1: Express equation as a quadratic in \( u \): \( 2u^2 - 17u + 8 = 0 \).
A1: Solve quadratic to obtain \( u = \frac{1}{2} \) and \( u = 8 \).
A1: Solve \( 2^y = \frac{1}{2} \) and \( 2^y = 8 \) to get \( y = -1 \) and \( y = 3 \).

(a)(i) Using the product and power laws of logarithms:
\( \log_a(a^3 b^2) = \log_a(a^3) + \log_a(b^2) = 3\log_a a + 2\log_a b = 3(1) + 2p = 3 + 2p \).

(a)(ii) Using the power law and change of base formula:
\( \log_b(a^2) = 2\log_b a = 2 \cdot \frac{1}{\log_a b} = \frac{2}{p} \).

(b) Using the change of base law, \( \log_x 2 = \frac{1}{\log_2 x} \):
\( \log_2 x + \frac{6}{\log_2 x} = 5 \)

Let \( u = \log_2 x \):
\( u + \frac{6}{u} = 5 \implies u^2 - 5u + 6 = 0 \)
\( (u - 2)(u - 3) = 0 \implies u = 2 \quad \text{or} \quad u = 3 \)

Since \( u = \log_2 x \):
\( \log_2 x = 2 \implies x = 2^2 = 4 \)
\( \log_2 x = 3 \implies x = 2^3 = 8 \).

(a)(i)
M1: Apply addition rule: \( \log_a(a^3) + \log_a(b^2) \) or power rule.
A1: \( 3 + 2p \).

(a)(ii)
M1: Apply power rule and/or change of base \( \log_b a = \frac{1}{\log_a b} \).
A1: \( \frac{2}{p} \).

(b)
M1: Use \( \log_x 2 = \frac{1}{\log_2 x} \) to set up and solve quadratic equation in \( \log_2 x \).
A1: Both correct values \( x = 4 \) and \( x = 8 \).

(a) Completing the square:
\( \mathrm{f}(x) = 2(x^2 - 6x) + 13 \)
\( = 2\left((x - 3)^2 - 9\right) + 13 \)
\( = 2(x - 3)^2 - 18 + 13 \)
\( = 2(x - 3)^2 - 5 \)
So \( a = 2, b = 3, c = -5 \).

(b) For \( \mathrm{f} \) to have an inverse, \( \mathrm{f} \) must be a one-one function.
The line of symmetry / vertex of the parabola is at \( x = 3 \).
Thus, the least value of \( k \) is \( 3 \).

(c) Let \( y = 2(x - 3)^2 - 5 \) for \( x \geq 3 \):
\( y + 5 = 2(x - 3)^2 \)
\( (x - 3)^2 = \frac{y + 5}{2} \)

Since \( x \geq 3 \), \( x - 3 \geq 0 \), so take the positive square root:
\( x - 3 = \sqrt{\frac{y + 5}{2}} \)
\( x = 3 + \sqrt{\frac{y + 5}{2}} \)

Therefore, \( \mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x + 5}{2}} \).

The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \).
Since the minimum value of \( \mathrm{f}(x) \) is \( -5 \), the range of \( \mathrm{f} \) is \( \mathrm{f}(x) \geq -5 \).
Hence, the domain of \( \mathrm{f}^{-1} \) is \( x \geq -5 \).

(a)
M1: Complete the square by factoring out 2: \( 2(x - 3)^2 + c \).
A1: Fully correct expression: \( 2(x - 3)^2 - 5 \).

(b)
B1: \( k = 3 \).

(c)
M1: Set \( y = 2(x - 3)^2 - 5 \) and rearrange to isolate \( (x - 3)^2 \).
M1: Take the positive square root (justified by \( x \geq 3 \)) and solve for \( x \).
A1: \( \mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x + 5}{2}} \) (must be in terms of \( x \)).
B1: State correct domain of \( \mathrm{f}^{-1} \): \( x \geq -5 \) (or \( [-5, \infty) \)).

(a) Using the product rule \(u = 2x - 1\) and \(v = \mathrm{e}^{3x}\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 2\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = 3\mathrm{e}^{3x}\).
\(\frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x} = 3(2x - 1)\mathrm{e}^{3x} + 2\mathrm{e}^{3x} = (6x - 3 + 2)\mathrm{e}^{3x} = (6x - 1)\mathrm{e}^{3x}\).

(b) At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\((6x - 1)\mathrm{e}^{3x} = 0\).
Since \(\mathrm{e}^{3x} \neq 0\) for all real \(x\), \(6x - 1 = 0 \implies x = \frac{1}{6}\).
Substitute \(x = \frac{1}{6}\) into the equation of the curve:
\(y = \left(2\left(\frac{1}{6}\right) - 1\right)\mathrm{e}^{3(1/6)} = \left(\frac{1}{3} - 1\right)\mathrm{e}^{1/2} = -\frac{2}{3}\mathrm{e}^{1/2} = -\frac{2}{3}\sqrt{\mathrm{e}}\).
To determine the nature, find \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6\mathrm{e}^{3x} + 3(6x - 1)\mathrm{e}^{3x} = (18x + 3)\mathrm{e}^{3x}\).
At \(x = \frac{1}{6}\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \left(18\left(\frac{1}{6}\right) + 3\right)\mathrm{e}^{1/2} = 6\sqrt{\mathrm{e}} > 0\).
Hence, the stationary point is a local minimum.

(c) When \(x = 0\):
\(y = (2(0) - 1)\mathrm{e}^0 = -1\).
Gradient \(m = \frac{\mathrm{d}y}{\mathrm{d}x}\Big|_{x=0} = (6(0) - 1)\mathrm{e}^0 = -1\).
Equation of the tangent: \(y - (-1) = -1(x - 0) \implies y = -x - 1\).

(a)
M1: Applies product rule correctly to differentiate \((2x-1)\mathrm{e}^{3x}\)
A1: Correct unsimplified derivative \(2\mathrm{e}^{3x} + 3(2x-1)\mathrm{e}^{3x}\)
A1: Fully simplified form \((6x - 1)\mathrm{e}^{3x}\) with \(a = 6, b = -1\)

(b)
M1: Sets derivative equal to 0 and solves for \(x\) (gets \(x = \frac{1}{6}\))
A1: Obtains exact \(y\)-coordinate \(-\frac{2}{3}\sqrt{\mathrm{e}}\) or \(-\frac{2}{3}\mathrm{e}^{1/2}\)
A1: Correctly justifies nature as a minimum using \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) or sign test of first derivative

(c)
M1: Finds gradient at \(x = 0\) and attempts equation of straight line through \((0, -1)\)
A1: \(y = -x - 1\) (or \(y + x + 1 = 0\))

(a) Using \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\):
\(\mathbf{r}_A = (2\mathbf{i} + 15\mathbf{j}) + t(3\mathbf{i} - \mathbf{j}) = (2 + 3t)\mathbf{i} + (15 - t)\mathbf{j}\)
\(\mathbf{r}_B = (-14\mathbf{i} - 5\mathbf{j}) + t(7\mathbf{i} + 3\mathbf{j}) = (-14 + 7t)\mathbf{i} + (-5 + 3t)\mathbf{j}\)

(b) Boat \(B\) is due south of boat \(A\) when their \(\mathbf{i}\)-components (east-west positions) are equal and the \(\mathbf{j}\)-component of \(B\) is less than that of \(A\):
\(2 + 3t = -14 + 7t\)
\(4t = 16 \implies t = 4\).
Check \(\mathbf{j}\)-components at \(t = 4\):
\(y_A = 15 - 4 = 11\)
\(y_B = -5 + 3(4) = 7\)
Since \(7 < 11\), \(B\) is indeed due south of \(A\) at \(t = 4\).

(c) At \(t = 4\), \(\mathbf{r}_A = 14\mathbf{i} + 11\mathbf{j}\) and \(\mathbf{r}_B = 14\mathbf{i} + 7\mathbf{j}\).
Vector \(\vec{BA} = \mathbf{r}_A - \mathbf{r}_B = (14 - 14)\mathbf{i} + (11 - 7)\mathbf{j} = 4\mathbf{j}\).
Distance \(= |4\mathbf{j}| = 4\text{ km}\).

(d) For a collision to occur, there must exist a single time \(t\) such that \(\mathbf{r}_A = \mathbf{r}_B\).
Equating the \(\mathbf{i}\)-components gives \(2 + 3t = -14 + 7t \implies t = 4\).
However, at \(t = 4\), the \(\mathbf{j}\)-components are \(y_A = 11\) and \(y_B = 7\).
Since \(y_A \neq y_B\) at \(t = 4\), there is no value of \(t\) for which both components are simultaneously equal. Hence the boats do not collide.

(a)
B1: Correct expression for \(\mathbf{r}_A\)
B1: Correct expression for \(\mathbf{r}_B\)

(b)
M1: Equates the \(\mathbf{i}\)-components of \(\mathbf{r}_A\) and \(\mathbf{r}_B\)
A1: \(t = 4\)

(c)
M1: Calculates the difference between the \(\mathbf{j}\)-components at their value of \(t\)
A1: \(4\text{ km}\)

(d)
M1: Considers the condition for collision (equating position vectors or evaluating \(\mathbf{j}\)-components at \(t = 4\))
A1: Fully explains that at \(t = 4\), \(y_A = 11 \neq 7 = y_B\), so \(\mathbf{r}_A \neq \mathbf{r}_B\) for any \(t\)

(a) Let \(u = \sin 2x\) and \(v = 2 + \cos 2x\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2\cos 2x\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = -2\sin 2x\).
Applying the quotient rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{(2 + \cos 2x)(2\cos 2x) - (\sin 2x)(-2\sin 2x)}{(2 + \cos 2x)^2}\)
\(= \frac{4\cos 2x + 2\cos^2 2x + 2\sin^2 2x}{(2 + \cos 2x)^2}\)
Using the identity \(\sin^2 2x + \cos^2 2x = 1\):
\(= \frac{4\cos 2x + 2(1)}{(2 + \cos 2x)^2} = \frac{2 + 4\cos 2x}{(2 + \cos 2x)^2}\).
Thus, \(a = 2\) and \(b = 4\).

(b) From part (a), \(\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\sin 2x}{2 + \cos 2x}\right) = \frac{2(1 + 2\cos 2x)}{(2 + \cos 2x)^2}\).
Therefore, \(\frac{1 + 2\cos 2x}{(2 + \cos 2x)^2} = \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\sin 2x}{2 + \cos 2x}\right)\).
\(\int_0^{\frac{\pi}{4}} \frac{1 + 2\cos 2x}{(2 + \cos 2x)^2}\,\mathrm{d}x = \frac{1}{2} \left[ \frac{\sin 2x}{2 + \cos 2x} \right]_0^{\frac{\pi}{4}}\)
At the upper limit \(x = \frac{\pi}{4}\):
\(\frac{\sin\left(2 \cdot \frac{\pi}{4}\right)}{2 + \cos\left(2 \cdot \frac{\pi}{4}\right)} = \frac{\sin\left(\frac{\pi}{2}\right)}{2 + \cos\left(\frac{\pi}{2}\right)} = \frac{1}{2 + 0} = \frac{1}{2}\).
At the lower limit \(x = 0\):
\(\frac{\sin 0}{2 + \cos 0} = \frac{0}{2 + 1} = 0\).
Hence, the value of the integral is \(\frac{1}{2} \left(\frac{1}{2} - 0\right) = \frac{1}{4}\).

(c) Using the chain rule for connected rates of change: \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\).
When \(x = \frac{\pi}{6}\), \(2x = \frac{\pi}{3}\), so \(\cos 2x = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\).
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2 + 4\left(\frac{1}{2}\right)}{\left(2 + \frac{1}{2}\right)^2} = \frac{2 + 2}{\left(\frac{5}{2}\right)^2} = \frac{4}{\frac{25}{4}} = \frac{16}{25}\).
Given \(\frac{\mathrm{d}x}{\mathrm{d}t} = 0.5 = \frac{1}{2}\):
\(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{16}{25} \times \frac{1}{2} = \frac{8}{25}\text{ units s}^{-1}\) (or \(0.32\)).

(a)
M1: Correct structure of quotient rule applied to \(\frac{\sin 2x}{2 + \cos 2x}\)
A1: Correct differentiation of numerator and denominator including chain rule factor of 2
M1: Uses identity \(\sin^2 2x + \cos^2 2x = 1\) in the numerator
A1: Correct final form with \(a = 2\) and \(b = 4\)

(b)
M1: Recognises integral as \(\frac{1}{2}\left[\frac{\sin 2x}{2 + \cos 2x}\right]\)
M1: Correct substitution of limits \(x = \frac{\pi}{4}\) and \(x = 0\)
A1: Correct exact fraction \(\frac{1}{4}\)

(c)
M1: Applies chain rule \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\) with \(x = \frac{\pi}{6}\)
A1: \(\frac{8}{25}\) or \(0.32\)

(a) Velocity \(v = \int a\,\mathrm{d}t = \int (6 - 4\mathrm{e}^{-2t})\,\mathrm{d}t = 6t + \frac{-4}{-2}\mathrm{e}^{-2t} + C = 6t + 2\mathrm{e}^{-2t} + C\).
When \(t = 0\), \(v = -5\):
\(-5 = 6(0) + 2\mathrm{e}^0 + C \implies -5 = 2 + C \implies C = -7\).
Thus, \(v = 6t + 2\mathrm{e}^{-2t} - 7\).

(b) Displacement \(s = \int v\,\mathrm{d}t = \int (6t + 2\mathrm{e}^{-2t} - 7)\,\mathrm{d}t = 3t^2 - \mathrm{e}^{-2t} - 7t + D\).
When \(t = 0\), the particle is at \(O\), so \(s = 0\):
\(0 = 3(0)^2 - \mathrm{e}^0 - 7(0) + D \implies 0 = -1 + D \implies D = 1\).
Thus, \(s = 3t^2 - 7t + 1 - \mathrm{e}^{-2t}\).

(c) Set \(a = 4\):
\(6 - 4\mathrm{e}^{-2t} = 4\)
\(4\mathrm{e}^{-2t} = 2\)
\(\mathrm{e}^{-2t} = \frac{1}{2} \implies \mathrm{e}^{2t} = 2\)
Taking the natural logarithm of both sides:
\(2t = \ln 2 \implies t = \frac{1}{2}\ln 2\).

(a)
M1: Integrates \(a\) with respect to \(t\), obtaining at least one correct term \(6t\) or \(2\mathrm{e}^{-2t}\) and includes constant \(C\)
M1: Substitutes \(t = 0, v = -5\) to find \(C\)
A1: \(v = 6t + 2\mathrm{e}^{-2t} - 7\)

(b)
M1: Integrates their \(v\) expression with respect to \(t\) and includes constant of integration \(D\)
M1: Uses initial condition \(t = 0, s = 0\) to evaluate \(D\)
A1: \(s = 3t^2 - 7t + 1 - \mathrm{e}^{-2t}\) (or equivalent simplified form)

(c)
M1: Sets \(6 - 4\mathrm{e}^{-2t} = 4\) and rearranges to \(\mathrm{e}^{-2t} = \frac{1}{2}\) or \(\mathrm{e}^{2t} = 2\)
M1: Takes natural logarithms correctly to isolate \(t\)
A1: \(t = \frac{1}{2}\ln 2\)

(a) The general term in the binomial expansion of \(\left(2x + \frac{k}{x^2}\right)^6\) is given by:
\[ T_{r+1} = \binom{6}{r} (2x)^{6-r} \left(\frac{k}{x^2}\right)^r = \binom{6}{r} 2^{6-r} k^r x^{6-3r} \]
For the term independent of \(x\), we set the power of \(x\) to zero:
\[ 6 - 3r = 0 \implies r = 2 \]
The term is:
\[ \binom{6}{2} 2^{6-2} k^2 = 15 \times 16 \times k^2 = 240k^2 \]
Given that this term equals \(2160\):
\[ 240k^2 = 2160 \implies k^2 = 9 \]
Since \(k > 0\), we have \(k = 3\).

(b) With \(k = 3\), the expansion is \((3 - x^3)\left(2x + \frac{3}{x^2}\right)^6\).
In the expansion of \(\left(2x + \frac{3}{x^2}\right)^6\), the power of \(x\) is \(6 - 3r\).
To obtain an \(x^3\) term when multiplied by \((3 - x^3)\):
1. The \(x^3\) term from \(\left(2x + \frac{3}{x^2}\right)^6\) multiplied by \(3\):
\(6 - 3r = 3 \implies r = 1\).
Coefficient: \(\binom{6}{1} 2^5 (3)^1 = 6 \times 32 \times 3 = 576\).
2. The constant term (\(x^0\)) from \(\left(2x + \frac{3}{x^2}\right)^6\) multiplied by \(-x^3\):
\(r = 2\), which gives the constant term \(2160\).

Combining these:
\[ \text{Coefficient of } x^3 = 3(576) - 1(2160) = 1728 - 2160 = -432 \]

(a)
M1: Identifies general term and sets power of \(x\) to 0, finding \(r = 2\).
M1: Forms equation \(\binom{6}{2} 2^4 k^2 = 2160\) or \(240k^2 = 2160\).
A1: \(k = 3\) (reject \(k = -3\) since \(k > 0\)).

(b)
B1: Finds coefficient of \(x^3\) term in \(\left(2x + \frac{3}{x^2}\right)^6\) as \(576\).
M1: Correctly combines terms: \(3 \times 576 - 1 \times 2160\).
A1: \(-432\).

(a) The number of ways to choose and arrange 5 distinct digits from 6 available digits is:
\[ {}^6P_5 = \frac{6!}{(6-5)!} = 720 \]

(b) For a 5-digit number to be greater than \(50\,000\), the first digit must be \(5, 7,\) or \(8\).
For the number to be even, the last digit must be \(2, 4,\) or \(8\).
Since \(8\) is in both sets, we split into two mutually exclusive cases:

Case 1: First digit is odd (\(5\) or \(7\))
- Number of choices for the 1st digit: \(2\) (either 5 or 7).
- Number of choices for the 5th digit: \(3\) (2, 4, or 8).
- Number of choices for the remaining 3 middle digits from the 4 remaining digits: \({}^4P_3 = 4 \times 3 \times 2 = 24\).
- Subtotal for Case 1: \(2 \times 3 \times 24 = 144\).

Case 2: First digit is \(8\)
- Number of choices for the 1st digit: \(1\) (only 8).
- Number of choices for the 5th digit: \(2\) (either 2 or 4).
- Number of choices for the remaining 3 middle digits from the 4 remaining digits: \({}^4P_3 = 24\).
- Subtotal for Case 2: \(1 \times 2 \times 24 = 48\).

Total number of required numbers = \(144 + 48 = 192\).

(a)
B1: \(720\) or \({}^6P_5\).

(b)
M1: Identifies correct restriction for first digit (\(5, 7, 8\)) and last digit (\(2, 4, 8\)), splitting into valid cases to avoid double counting.
M1: Calculates permutations for both cases correctly (e.g. \(2 \times 3 \times 24\) and \(1 \times 2 \times 24\)).
A1: \(192\).

(a) Taking the natural logarithm of both sides of \(y = A \mathrm{e}^{kx}\):
\[ \ln y = \ln(A \mathrm{e}^{kx}) = \ln A + kx \]
This is in linear form \(Y = mX + c\), where \(Y = \ln y\), \(X = x\), gradient \(m = k\), and \(Y\)-intercept \(c = \ln A\).
The gradient \(k\) is:
\[ k = \frac{5.6 - 3.1}{7 - 2} = \frac{2.5}{5} = 0.5 \]

(b) Using the point \((2, 3.1)\) in the line equation:
\[ 3.1 = \ln A + 0.5(2) \]
\[ 3.1 = \ln A + 1.0 \implies \ln A = 2.1 \]
\[ A = \mathrm{e}^{2.1} \approx 8.1661699... \]
Correct to 3 significant figures, \(A = 8.17\).

(a)
M1: Expresses relation in linear form \(\ln y = kx + \ln A\) and calculates gradient \(\frac{5.6 - 3.1}{7 - 2}\).
A1: \(k = 0.5\) or \(\frac{1}{2}\).

(b)
M1: Substitutes \(k\) and one given point to find \(\ln A\), and evaluates \(A = \mathrm{e}^{2.1}\).
A1: \(8.17\) (accept exact \(\mathrm{e}^{2.1}\) or rounding to 3 s.f.).

(a) For \(y = |4 - 2x|\):
- Vertex on the \(x\)-axis where \(4 - 2x = 0 \implies x = 2\), so point is \((2, 0)\).
- \(y\)-intercept at \(x = 0 \implies y = |4| = 4\), so point is \((0, 4)\).
- The graph is a V-shaped curve with vertex at \((2, 0)\).

For \(y = x + 1\):
- Straight line with \(y\)-intercept \((0, 1)\) and \(x\)-intercept \((-1, 0)\).

(b) To solve \(|4 - 2x| = x + 1\):
Case 1: \(4 - 2x = x + 1\)
\[ 3x = 3 \implies x = 1 \]
Check: \(|4 - 2(1)| = 2\) and \(1 + 1 = 2\) (valid).

Case 2: \(-(4 - 2x) = x + 1 \implies 2x - 4 = x + 1\)
\[ x = 5 \]
Check: \(|4 - 2(5)| = |-6| = 6\) and \(5 + 1 = 6\) (valid).

So the solutions are \(x = 1\) and \(x = 5\).

(c) From the sketch, \(|4 - 2x| \leqslant x + 1\) corresponds to the region where the V-shaped graph lies below or on the line \(y = x + 1\).
This occurs between the two intersection points:
\[ 1 \leqslant x \leqslant 5 \]

(a)
B1: Correct V-shape graph for \(y = |4 - 2x|\) with vertex at \((2, 0)\) and \(y\)-intercept \((0, 4)\) labelled.
B1: Correct straight line \(y = x + 1\) with intercepts \((0, 1)\) and \((-1, 0)\) labelled.

(b)
M1: Sets up two linear equations \(4 - 2x = x + 1\) and \(4 - 2x = -(x + 1)\) (or squares both sides).
A1: Obtains \(x = 1\) and \(x = 5\).

(c)
B1: \(1 \leqslant x \leqslant 5\) (must be in inequality format).

(a)(i) Using the binomial expansion:
\(\left(1 + \frac{x}{4}\right)^{12} = 1 + \binom{12}{1}\left(\frac{x}{4}\right) + \binom{12}{2}\left(\frac{x}{4}\right)^2 + \dots\)
\(= 1 + 12\left(\frac{x}{4}\right) + 66\left(\frac{x^2}{16}\right) + \dots\)
\(= 1 + 3x + \frac{33}{8}x^2 + \dots\)

Multiply by \((2 - kx)\):
\((2 - kx)\left(1 + 3x + \frac{33}{8}x^2\right) = 2 + 6x + \frac{33}{4}x^2 - kx - 3kx^2 + \dots\)
\(= 2 + (6 - k)x + \left(\frac{33}{4} - 3k\right)x^2 + \dots\)

Equating the coefficient of \(x\):
\(6 - k = 5 \implies k = 1\).

(a)(ii) Equating the coefficient of \(x^2\):
\(a = \frac{33}{4} - 3k = \frac{33}{4} - 3(1) = \frac{33}{4} - \frac{12}{4} = \frac{21}{4} = 5.25\).

(b) The terms of the arithmetic progression are:
\(u_1 = p\)
\(u_5 = p + 4d\)
\(u_{17} = p + 16d\)

Since \(u_1, u_5, u_{17}\) form a geometric progression:
\(\frac{u_5}{u_1} = \frac{u_{17}}{u_5}\)
\(\frac{p + 4d}{p} = \frac{p + 16d}{p + 4d}\)
\((p + 4d)^2 = p(p + 16d)\)
\(p^2 + 8pd + 16d^2 = p^2 + 16pd\)
\(16d^2 = 8pd\)

Since \(d \ne 0\), dividing both sides by \(8d\) gives:
\(2d = p\), hence \(p = 2d\).

(a)(i)
M1: For term in x: \(\binom{12}{1}\frac{1}{4} = 3\) or full binomial expansion up to \(x^2\) seen.
M1: For equating coefficient of \(x\): \(2(3) - k = 5\) or \(6 - k = 5\).
A1: \(k = 1\).

(a)(ii)
M1: For coefficient of \(x^2\) expression: \(2\left(\frac{33}{8}\right) - 3k\) or \(\frac{33}{4} - 3(1)\).
A1: \(a = \frac{21}{4}\) or \(5.25\).

(b)
B1: For writing correct expressions for the three terms: \(p\), \(p + 4d\), \(p + 16d\).
M1: For setting up the condition for GP: \((p + 4d)^2 = p(p + 16d)\) or \(\frac{p+4d}{p} = \frac{p+16d}{p+4d}\).
A1: For correct expansion \(p^2 + 8pd + 16d^2 = p^2 + 16pd\) leading to \(16d^2 = 8pd\) and correctly concluding \(p = 2d\).

(a) Taking the natural logarithm of both sides of \(y\sqrt{x} = A b^x\):
\(\ln(y\sqrt{x}) = \ln(A b^x) = \ln A + \ln(b^x) = (\ln b)x + \ln A\)
Thus, \(m = \ln b\) and \(c = \ln A\).

(b) Gradient \(m = \frac{3.46 - 1.84}{5 - 2} = \frac{1.62}{3} = 0.54\).
Vertical intercept \(c\):
Using \((2, 1.84)\):
\(1.84 = 0.54(2) + c \implies c = 1.84 - 1.08 = 0.76\).

(c) From part (a):
\(\ln b = 0.54 \implies b = e^{0.54} \approx 1.7160\dots \approx 1.72\)
\(\ln A = 0.76 \implies A = e^{0.76} \approx 2.1383\dots \approx 2.14\).

(a)
M1: For taking logarithms correctly using laws of logs: \(\ln A + x\ln b\).
A1: For identifying \(m = \ln b\) and \(c = \ln A\) (or written clearly as \(\ln(y\sqrt{x}) = (\ln b)x + \ln A\)).

(b)
B1: Gradient \(= 0.54\).
B1: Vertical intercept \(= 0.76\).

(c)
M1: For using \(b = e^m\) or \(A = e^c\).
A1: \(b = 1.72\) (accept 1.716 to 3 sf).
A1: \(A = 2.14\) (accept 2.138 to 3 sf).

(a)(i) In right-angled triangle \(OCA\), \(\sin\theta = \frac{AC}{OA} = \frac{AC}{r}\), so \(AC = r\sin\theta\).

(a)(ii) In right-angled triangle \(OCA\), \(\cos\theta = \frac{OC}{OA} = \frac{OC}{r}\), so \(OC = r\cos\theta\).

(b)(i) For \(r = 12\) and \(\theta = \frac{\pi}{3}\):
\(AC = 12\sin\left(\frac{\pi}{3}\right) = 12\left(\frac{\sqrt{3}}{2}\right) = 6\sqrt{3}\)
\(OC = 12\cos\left(\frac{\pi}{3}\right) = 12\left(\frac{1}{2}\right) = 6\)
\(BC = OB - OC = 12 - 6 = 6\)
\(\text{Arc } AB = r\theta = 12\left(\frac{\pi}{3}\right) = 4\pi\)

\(\text{Perimeter} = \text{Arc } AB + BC + CA = 4\pi + 6 + 6\sqrt{3}\text{ cm}\).

(b)(ii) \(\text{Area of sector } OAB = \frac{1}{2}r^2\theta = \frac{1}{2}(12^2)\left(\frac{\pi}{3}\right) = \frac{1}{2}(144)\left(\frac{\pi}{3}\right) = 24\pi\)
\(\text{Area of triangle } OCA = \frac{1}{2} \times OC \times AC = \frac{1}{2}(6)(6\sqrt{3}) = 18\sqrt{3}\)

\(\text{Shaded Area} = \text{Area of sector } OAB - \text{Area of triangle } OCA = 24\pi - 18\sqrt{3}\)
Thus, \(a = 24\) and \(b = 18\).

(a)(i)
B1: \(AC = r\sin\theta\)

(a)(ii)
B1: \(OC = r\cos\theta\)

(b)(i)
B1: For \(\text{Arc } AB = 4\pi\) or \(12\left(\frac{\pi}{3}\right)\).
M1: For finding \(BC = 12 - 6 = 6\) and \(AC = 6\sqrt{3}\).
A1: For fully correct sum showing \(4\pi + 6 + 6\sqrt{3}\).

(b)(ii)
M1: For area of sector \(OAB = \frac{1}{2}(12^2)\left(\frac{\pi}{3}\right) = 24\pi\).
M1: For area of triangle \(OCA = \frac{1}{2}(6)(6\sqrt{3}) = 18\sqrt{3}\) or \(\frac{1}{2}(12)(6)\sin(\frac{\pi}{3})\).
A1: \(24\pi - 18\sqrt{3}\) (with \(a = 24, b = 18\)).

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

So \(\tan 2x = 2\) or \(\tan 2x = -\frac{1}{3}\).

Since \(0^{\circ} \le x \le 180^{\circ}\), the interval for \(2x\) is \(0^{\circ} \le 2x \le 360^{\circ}\).

For \(\tan 2x = 2\):
\(2x = 63.435^{\circ}, 180^{\circ} + 63.435^{\circ} = 243.435^{\circ}\)
\(x = 31.7^{\circ}, 121.7^{\circ}\)

For \(\tan 2x = -\frac{1}{3}\):
Basic angle \(= \tan^{-1}\left(\frac{1}{3}\right) = 18.435^{\circ}\)
\(2x = 180^{\circ} - 18.435^{\circ} = 161.565^{\circ}\)
\(2x = 360^{\circ} - 18.435^{\circ} = 341.565^{\circ}\)
\(x = 80.8^{\circ}, 170.8^{\circ}\)

Thus, \(x = 31.7^{\circ}, 80.8^{\circ}, 121.7^{\circ}, 170.8^{\circ}\).

(b) Using \(\sec\theta = \frac{1}{\cos\theta}\):
\(4\sin\left(y + \frac{\pi}{6}\right)\cos\left(y + \frac{\pi}{6}\right) = \sqrt{3}\)
\(2\left[2\sin\left(y + \frac{\pi}{6}\right)\cos\left(y + \frac{\pi}{6}\right)\right] = \sqrt{3}\)
\(2\sin\left(2y + \frac{\pi}{3}\right) = \sqrt{3}\)
\(\sin\left(2y + \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)

For \(0 \le y \le 2\pi\), the domain for \(\theta = 2y + \frac{\pi}{3}\) is \(\frac{\pi}{3} \le 2y + \frac{\pi}{3} \le \frac{13\pi}{3}\).

\(2y + \frac{\pi}{3} = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}, \frac{13\pi}{3}\)
\(2y = 0, \frac{\pi}{3}, 2\pi, \frac{7\pi}{3}, 4\pi\)
\(y = 0, \frac{\pi}{6}, \pi, \frac{7\pi}{6}, 2\pi\).

(a)
M1: For using \(\sec^2 2x = 1 + \tan^2 2x\) to form a quadratic in \(\tan 2x\).
M1: For solving the 3-term quadratic to obtain \(\tan 2x = 2\) and \(\tan 2x = -\frac{1}{3}\).
A1: For any two correct solutions for \(x\) (e.g. \(31.7^{\circ}\) and \(121.7^{\circ}\)).
A1: For all four correct solutions: \(x = 31.7^{\circ}, 80.8^{\circ}, 121.7^{\circ}, 170.8^{\circ}\) (and no extras in range).

(b)
M1: For multiplying by \(\cos\left(y + \frac{\pi}{6}\right)\) and applying double angle identity \(2\sin\theta\cos\theta = \sin 2\theta\) to get \(\sin\left(2y + \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\).
A1: For obtaining at least three correct values of \(y\).
A1: For all five correct values: \(y = 0, \frac{\pi}{6}, \pi, \frac{7\pi}{6}, 2\pi\).

(a) Set \(y = 0\):
\[5 - \frac{8}{2x+1} = 0 \implies 2x+1 = \frac{8}{5} = 1.6 \implies 2x = 0.6 \implies x = 0.3 = \frac{3}{10}\]
So \(P\left(\frac{3}{10}, 0\right)\).

(b) At \(x = 1\):
\[y = 5 - \frac{8}{2(1)+1} = 5 - \frac{8}{3} = \frac{7}{3}\]
Differentiating \(y = 5 - 8(2x+1)^{-1}\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -8(-1)(2x+1)^{-2} \cdot 2 = \frac{16}{(2x+1)^2}\]
At \(x = 1\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{16}{(3)^2} = \frac{16}{9}\]
Equation of the tangent at \(Q\left(1, \frac{7}{3}\right)\):
\[y - \frac{7}{3} = \frac{16}{9}(x - 1) \implies 9y - 21 = 16x - 16 \implies 16x - 9y + 5 = 0\]

(c) Set \(y = 0\) in the tangent equation:
\[16x + 5 = 0 \implies x = -\frac{5}{16}\]
So \(R\left(-\frac{5}{16}, 0\right)\).

(d) Area of the triangle formed by the tangent, the vertical line \(x = 1\), and the \(x\)-axis:
\[\text{Base} = 1 - \left(-\frac{5}{16}\right) = \frac{21}{16}\]
\[\text{Height} = \frac{7}{3}\]
\[\text{Area}_{\text{triangle}} = \frac{1}{2} \times \frac{21}{16} \times \frac{7}{3} = \frac{49}{32}\]
Area under the curve from \(x = \frac{3}{10}\) to \(x = 1\):
\[\int_{3/10}^1 \left(5 - \frac{8}{2x+1}\right) \mathrm{d}x = \left[ 5x - 4\ln(2x+1) \right]_{3/10}^1\]
\[= (5 - 4\ln 3) - \left(5(0.3) - 4\ln(1.6)\right)\]
\[= 5 - 4\ln 3 - 1.5 + 4\ln\left(\frac{8}{5}\right) = 3.5 - 4\ln\left(\frac{15}{8}\right) = \frac{7}{2} - 4\ln\left(\frac{15}{8}\right)\]
Area of the shaded region:
\[\text{Area} = \text{Area}_{\text{triangle}} - \text{Area}_{\text{under curve}} = \frac{49}{32} - \left(\frac{7}{2} - 4\ln\left(\frac{15}{8}\right)\right) = 4\ln\left(\frac{15}{8}\right) - \frac{63}{32}\]

(a) B1: Correct coordinates \((0.3, 0)\) or \(\left(\frac{3}{10}, 0\right)\)

(b) B1: Correct \(y\)-coordinate \(\frac{7}{3}\)
M1: Correct method to differentiate \(5 - 8(2x+1)^{-1}\) to obtain form \(k(2x+1)^{-2}\)
A1: Correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{16}{(2x+1)^2}\) and gradient \(\frac{16}{9}\)
A1: Correct equation in required integer form: \(16x - 9y + 5 = 0\) (or \(-16x + 9y - 5 = 0\))

(c) B1: Correct coordinates \(\left(-\frac{5}{16}, 0\right)\) or \((-0.3125, 0)\) (allow FT from their tangent)

(d) M1: Valid attempt to find area of triangle using \(\frac{1}{2} \times \text{base} \times \text{height}\) giving \(\frac{49}{32}\)
M1: Correct integration of the curve to obtain \(5x - 4\ln(2x+1)\)
M1: Correct substitution of limits \(\frac{3}{10}\) and \(1\) into integrated expression
A1: Correct exact area \(4\ln\left(\frac{15}{8}\right) - \frac{63}{32}\) (or exact equivalent such as \(4\ln 1.875 - 1.96875\))

(a) Differentiate \(y = 6x^{1/2} - x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{-1/2} - 1 = \frac{3}{\sqrt{x}} - 1\]
Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[\frac{3}{\sqrt{x}} = 1 \implies \sqrt{x} = 3 \implies x = 9\]
When \(x = 9\), \(y = 6\sqrt{9} - 9 = 18 - 9 = 9\).
Stationary point is \((9, 9)\).

(b) At \(x = 4\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{\sqrt{4}} - 1 = \frac{3}{2} - 1 = \frac{1}{2}\]
Equation of tangent at \(P(4, 8)\):
\[y - 8 = \frac{1}{2}(x - 4) \implies y = \frac{1}{2}x + 6\]

(c) When \(x = 0\), \(y = 6\), so \(T(0, 6)\).

(d) The region extends from \(x = 0\) to \(x = 4\), with the tangent above the curve:
\[\text{Area} = \int_0^4 \left( \left(\frac{1}{2}x + 6\right) - (6x^{1/2} - x) \right) \mathrm{d}x\]
\[= \int_0^4 \left( \frac{3}{2}x + 6 - 6x^{1/2} \right) \mathrm{d}x\]
\[= \left[ \frac{3}{4}x^2 + 6x - 4x^{3/2} \right]_0^4\]
\[= \left( \frac{3}{4}(16) + 6(4) - 4(8) \right) - 0\]
\[= 12 + 24 - 32 = 4\]

(a) M1: Differentiating to obtain \(a x^{-1/2} - 1\)
A1: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{-1/2} - 1\) and setting to 0 to find \(x = 9\)
A1: Correct coordinates \((9, 9)\)

(b) M1: Substituting \(x = 4\) into their derivative to find gradient
A1: Correct gradient \(m = \frac{1}{2}\)
A1: Correct equation of tangent, e.g. \(y = \frac{1}{2}x + 6\) or \(x - 2y + 12 = 0\)

(c) B1: Correct coordinates \((0, 6)\)

(d) M1: Setting up the integral \(\int_0^4 (y_{\text{tangent}} - y_{\text{curve}}) \mathrm{d}x\) with correct limits 0 and 4
M1: Integrating correctly to obtain \(\left[ \frac{3}{4}x^2 + 6x - 4x^{3/2} \right]\) (or equivalent separate integrals)
A1: Correct final area \(4\)

(a) Differentiating \(y = x^3 - 3x^2 + 4\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 6x = 3x(x - 2)\]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies x = 0\) or \(x = 2\).
When \(x = 0\), \(y = 4\).
When \(x = 2\), \(y = 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0\).
Stationary points are \((0, 4)\) and \((2, 0)\).
Second derivative:
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x - 6\]
At \(x = 0\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -6 < 0 \implies (0, 4)\) is a local maximum.
At \(x = 2\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6 > 0 \implies (2, 0)\) is a local minimum.

(b) At \(x = 3\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3(3^2) - 6(3) = 27 - 18 = 9\]
Gradient of the normal \(m_{\text{normal}} = -\frac{1}{9}\).
Equation of the normal at \(P(3, 4)\):
\[y - 4 = -\frac{1}{9}(x - 3) \implies 9y - 36 = -x + 3 \implies x + 9y - 39 = 0\]

(c) The line \(y = 4\) intersects the curve where:
\[x^3 - 3x^2 + 4 = 4 \implies x^2(x - 3) = 0 \implies x = 0 \text{ and } x = 3\]
For \(0 \le x \le 3\), the line \(y = 4\) lies above the curve:
\[\text{Area} = \int_0^3 \left( 4 - (x^3 - 3x^2 + 4) \right) \mathrm{d}x = \int_0^3 (3x^2 - x^3) \mathrm{d}x\]
\[= \left[ x^3 - \frac{x^4}{4} \right]_0^3 = \left( 27 - \frac{81}{4} \right) - 0 = \frac{27}{4} = 6.75\]

(a) M1: Differentiating to obtain \(3x^2 - 6x\) and equating to 0
A1: Finding both correct points \((0, 4)\) and \((2, 0)\)
M1: Valid test for nature using second derivative (or sign of first derivative)
A1: Correctly identifies \((0, 4)\) as maximum and \((2, 0)\) as minimum with working shown

(b) M1: Finding gradient of curve at \(x = 3\) and calculating perpendicular gradient \(-\frac{1}{m}\)
A1: Using \(m_{\text{normal}} = -\frac{1}{9}\) with point \((3, 4)\)
A1: Correct integer form \(x + 9y - 39 = 0\) (or \(-x - 9y + 39 = 0\))

(c) M1: Solving \(x^3 - 3x^2 + 4 = 4\) to find upper limit \(x = 3\)
M1: Integrating \((3x^2 - x^3)\) to obtain \(x^3 - \frac{x^4}{4}\)
A1: Correct area \(6.75\) or \(\frac{27}{4}\)