2024 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

Express both equations with a common base for each system:
From the first equation:
\( 3^{x-1} \times (3^2)^y = 3^3 \)
\( 3^{x-1+2y} = 3^3 \)
\( x + 2y - 1 = 3 \implies x + 2y = 4 \) [1]

From the second equation:
\( 2^{2x} \div (2^2)^{y+1} = 2^{-3} \)
\( 2^{2x - 2y - 2} = 2^{-3} \)
\( 2x - 2y - 2 = -3 \implies 2x - 2y = -1 \) [2]

Solve [1] and [2] simultaneously:
Add the two equations:
\( 3x = 3 \implies x = 1 \)

Substitute \( x = 1 \) into [1]:
\( 1 + 2y = 4 \implies 2y = 3 \implies y = 1.5 \)

M1: For expressing both equations as linear equations in \(x\) and \(y\) (e.g., \(x+2y=4\) and \(2x-2y=-1\)).
M1: For a valid method to solve their simultaneous linear equations.
A1: For both correct values \(x=1\) and \(y=1.5\) (or equivalent fractions).

The perimeter \(P\) of the sector is given by:
\( P = 2r + r\theta = 30 \implies r\theta = 30 - 2r \)

The area \(A\) of the sector is given by:
\( A = \frac{1}{2}r^2\theta = 50 \)

Substitute the expression for \(r\theta\) into the area formula:
\( \frac{1}{2}r(r\theta) = 50 \)

\( \frac{1}{2}r(30 - 2r) = 50 \)

\( 15r - r^2 = 50 \)

\( r^2 - 15r + 50 = 0 \)

Factorising the quadratic equation:
\( (r-5)(r-10) = 0 \)

Thus, \( r = 5 \) or \( r = 10 \).

M1: For writing down the two correct equations for perimeter and area: \(2r + r\theta = 30\) and \(\frac{1}{2}r^2\theta = 50\).
M1: For eliminating \(\theta\) to form a quadratic equation in \(r\) (e.g., \(r^2 - 15r + 50 = 0\)).
A1: For finding both correct values of \(r\) (\(r = 5\) and \(r = 10\)).

Use the change of base formula to write the logarithm in base 2:
\( \log_4(x-1) = \frac{\log_2(x-1)}{\log_2 4} = \frac{1}{2}\log_2(x-1) = \log_2\sqrt{x-1} \)

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

Apply the division law of logarithms:
\( \log_2\left(\frac{x+3}{\sqrt{x-1}}\right) = 2 \)

Convert from logarithmic to exponential form:
\( \frac{x+3}{\sqrt{x-1}} = 2^2 = 4 \)

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

Square both sides:
\( (x+3)^2 = 16(x-1) \)

\( x^2 + 6x + 9 = 16x - 16 \)

\( x^2 - 10x + 25 = 0 \)

\( (x-5)^2 = 0 \implies x = 5 \)

M1: For using the change of base rule to write \(\log_4(x-1) = \frac{1}{2}\log_2(x-1)\) or equivalent.
M1: For applying laws of logarithms correctly to obtain \(x+3 = 4\sqrt{x-1}\) or \((x+3)^2 = 16(x-1)\).
A1: For \(x = 5\).

Let the first term be \(a\) and the common ratio be \(r\).
\( u_3 = ar^2 = 12 \)
\( u_6 = ar^5 = -96 \)

Dividing the sixth term by the third term:
\( \frac{ar^5}{ar^2} = \frac{-96}{12} \)

\( r^3 = -8 \implies r = -2 \)

Substitute \(r = -2\) back to find \(a\):
\( a(-2)^2 = 12 \implies 4a = 12 \implies a = 3 \)

Now calculate the sum of the first 8 terms:
\( S_8 = \frac{a(1 - r^8)}{1 - r} = \frac{3(1 - (-2)^8)}{1 - (-2)} = \frac{3(1 - 256)}{3} = 1 - 256 = -255 \)

M1: For setting up the equations \(ar^2 = 12\) and \(ar^5 = -96\) and solving for \(r\).
M1: For finding the correct value of \(a = 3\) and using the geometric series sum formula with \(n=8\).
A1: For correct final answer of \(-255\).

By the Factor Theorem, since \( x-2 \) is a factor of \( \mathrm{p}(x) \):
\( \mathrm{p}(2) = 0 \)
\( 2(2)^3 + a(2)^2 - 7(2) + b = 0 \)
\( 16 + 4a - 14 + b = 0 \implies 4a + b = -2 \) [1]

By the Remainder Theorem, since dividing by \( x+1 \) leaves a remainder of \( -15 \):
\( \mathrm{p}(-1) = -15 \)
\( 2(-1)^3 + a(-1)^2 - 7(-1) + b = -15 \)
\( -2 + a + 7 + b = -15 \implies a + b = -20 \) [2]

Subtract equation [2] from equation [1]:
\( 3a = 18 \implies a = 6 \)

Substitute \( a = 6 \) into equation [2]:
\( 6 + b = -20 \implies b = -26 \)

M1: For applying the factor theorem to get \(4a + b = -2\) or equivalent.
M1: For applying the remainder theorem to get \(a + b = -20\) or equivalent and attempting to solve simultaneously.
A1: For both correct values \(a = 6\) and \(b = -26\).

First find the vector \( \overrightarrow{AB} \):
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \)

\( \overrightarrow{AB} = (p\mathbf{i} - 2\mathbf{j}) - (3\mathbf{i} + 4\mathbf{j}) = (p-3)\mathbf{i} - 6\mathbf{j} \)

Now apply the magnitude formula:
\( |\overrightarrow{AB}| = \sqrt{(p-3)^2 + (-6)^2} = 10 \)

Square both sides:
\( (p-3)^2 + 36 = 100 \)

\( (p-3)^2 = 64 \)

Take the square root of both sides:
\( p - 3 = \pm 8 \)

This gives two cases:
\( p - 3 = 8 \implies p = 11 \)

\( p - 3 = -8 \implies p = -5 \)

M1: For finding an expression for \( \overrightarrow{AB} \) in terms of \( p \), i.e., \( (p-3)\mathbf{i} - 6\mathbf{j} \).
M1: For setting up the magnitude equation \((p-3)^2 + (-6)^2 = 10^2\) and attempting to solve for \(p\).
A1: For both correct values \(p = 11\) and \(p = -5\).

(a) Using the product rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 \cdot \sqrt{2x + 5} + (x - 2) \cdot \frac{1}{2\sqrt{2x + 5}} \cdot 2\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{2x + 5} + \frac{x - 2}{\sqrt{2x + 5}} = \frac{2x + 5 + x - 2}{\sqrt{2x + 5}} = \frac{3x + 3}{\sqrt{2x + 5}}\).
So \(a = 3\) and \(b = 3\).

(b) At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies 3x + 3 = 0 \implies x = -1\).
When \(x = -1\), \(y = (-1 - 2)\sqrt{2(-1) + 5} = -3\sqrt{3}\).
Stationary point coordinates are \((-1, -3\sqrt{3})\).

(c) The curve crosses the \(y\)-axis when \(x = 0\).
When \(x = 0\), \(y = (0 - 2)\sqrt{5} = -2\sqrt{5}\).
The gradient of the tangent at \(x = 0\) is \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{\sqrt{5}}\).
Hence, the gradient of the normal is \(-\frac{\sqrt{5}}{3}\).
The equation of the normal is:
\(y - (-2\sqrt{5}) = -\frac{\sqrt{5}}{3}(x - 0) \implies y = -\frac{\sqrt{5}}{3}x - 2\sqrt{5}\).

(d) We have \(\frac{\mathrm{d}}{\mathrm{d}x} \left( (x - 2)\sqrt{2x + 5} \right) = \frac{3x + 3}{\sqrt{2x + 5}}\).
So, \(\frac{3x + 1}{\sqrt{2x + 5}} = \frac{3x + 3 - 2}{\sqrt{2x + 5}} = \frac{3x + 3}{\sqrt{2x + 5}} - \frac{2}{\sqrt{2x + 5}}\).
Integrating both sides:
\(\int \frac{3x + 1}{\sqrt{2x + 5}} \, \mathrm{d}x = \int \frac{3x + 3}{\sqrt{2x + 5}} \, \mathrm{d}x - 2 \int (2x + 5)^{-1/2} \, \mathrm{d}x\)
\(= (x - 2)\sqrt{2x + 5} - 2 \left[ \frac{(2x + 5)^{1/2}}{\frac{1}{2} \cdot 2} \right] + C\)
\(= (x - 2)\sqrt{2x + 5} - 2\sqrt{2x + 5} + C\)
\(= (x - 4)\sqrt{2x + 5} + C\).

(a)
M1: For application of product rule, with at least one term correct.
A1: Correct unsimplified expression for derivative.
M1: For finding a common denominator and simplifying.
A1: For correct final answer \(\frac{3x + 3}{\sqrt{2x + 5}}\).

(b)
M1: Equating their numerator from part (a) to 0 and solving for \(x\).
A1: For \(x = -1\).
A1: For \(y = -3\sqrt{3}\).

(c)
B1: For finding \(y = -2\sqrt{5}\) when \(x = 0\).
M1: Substituting \(x = 0\) into their derivative to find the tangent gradient, and finding the negative reciprocal to get the normal gradient.
A1: Correct normal gradient \(-\frac{\sqrt{5}}{3}\).
A1: Correct equation of the normal, e.g., \(y = -\frac{\sqrt{5}}{3}x - 2\sqrt{5}\) or equivalent.

(d)
M1: Recognising that \(3x + 1 = 3x + 3 - 2\) or similar split of the fraction.
A1: Correct split: \(\frac{3x + 3}{\sqrt{2x + 5}} - \frac{2}{\sqrt{2x + 5}}\).
B1: For \(\int \frac{3x + 3}{\sqrt{2x + 5}} \, \mathrm{d}x = (x - 2)\sqrt{2x + 5}\).
M1: For correct integration of \(2(2x + 5)^{-1/2}\) to get \(2\sqrt{2x + 5}\).
A1: For the fully simplified correct answer: \((x - 4)\sqrt{2x + 5} + C\) (condone missing \(+ C\)).

(a)(i) \(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(= \frac{n}{2}[2\ln(k) + (n-1)\ln(2)] = n\ln(k) + \frac{n(n-1)}{2}\ln(2)\)
\(= \ln(k^n) + \ln\left(2^{\frac{n(n-1)}{2}}\right) = \ln\left(2^{\frac{n(n-1)}{2}} k^n\right)\).

(a)(ii) For \(n = 10\):
\(S_{10} = \ln\left(2^{45} k^{10}\right) = 45\ln(2) + 10\ln(k)\).
We are given \(S_{10} = 35\ln(2)\).
So, \(45\ln(2) + 10\ln(k) = 35\ln(2) \implies 10\ln(k) = -10\ln(2) \implies \ln(k) = -\ln(2) = \ln\left(\frac{1}{2}\right)\).
Thus, \(k = \frac{1}{2}\).

(b)(i) First term \(a = 1\), common ratio \(r = \cos^2 \theta\).
Since \(0 < \theta < \frac{\pi}{2}\), \(0 < r < 1\), so sum to infinity exists:
\(S_{\infty} = \frac{a}{1-r} = \frac{1}{1 - \cos^2 \theta} = \frac{1}{\sin^2 \theta} = \csc^2 \theta\).

(b)(ii) \(S_{\infty} = \frac{4}{3} \implies \sin^2 \theta = \frac{3}{4} \implies \sin \theta = \frac{\sqrt{3}}{2}\) (since \(0 < \theta < \frac{\pi}{2}\)).
Thus, \(\theta = \frac{\pi}{3}\).

(b)(iii) For \(\theta = \frac{\pi}{3}\), \(r = \cos^2\left(\frac{\pi}{3}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).
\(S_5 = \frac{a(1 - r^5)}{1-r} = \frac{1\left(1 - (1/4)^5\right)}{1 - 1/4} = \frac{1 - 1/1024}{3/4} = \frac{1023}{1024} \times \frac{4}{3} = \frac{341}{256} \approx 1.33\) (correct to 3 significant figures).

(a)(i)
M1: Use of AP sum formula with \(a = \ln(k)\) and \(d = \ln(2)\).
M1: Use of log laws (power law and addition law) to combine terms.
A1: Obtains the correct shown expression clearly.

(a)(ii)
M1: Sets up the equation for \(S_{10}\) with \(n=10\).
M1: Equates to \(35\ln(2)\) and solves for \(\ln(k)\).
A1: Correct exact value \(k = \frac{1}{2}\).

(b)(i)
M1: Use of \(S_{\infty} = \frac{a}{1-r}\) with \(a = 1\) and \(r = \cos^2 \theta\).
A1: Simplifies to \(\frac{1}{\sin^2 \theta}\) or \(\csc^2 \theta\).

(b)(ii)
M1: Equates \(S_{\infty}\) to \(\frac{4}{3}\) to find \(\sin^2 \theta\).
A1: Finds \(\sin \theta = \frac{\sqrt{3}}{2}\) (accepts \(\pm\) but must select positive).
A1: Exact value \(\theta = \frac{\pi}{3}\) (or \(60^\circ\)).

(b)(iii)
M1: Finds \(r = \frac{1}{4}\).
M1: Use of GP sum formula for \(S_5\).
A1: Correct unsimplified or exact fraction \(\frac{341}{256}\).
A1: Correct to 3 sig figs: \(1.33\).

(a) Since \(\mathrm{e}^{3x} > 0\) for all real \(x\), we have \(2\mathrm{e}^{3x} - 5 > -5\).
Thus, the range of \(\mathrm{f}\) is \(\mathrm{f}(x) > -5\).

(b) Let \(y = 2\mathrm{e}^{3x} - 5\).
\(y + 5 = 2\mathrm{e}^{3x} \implies \mathrm{e}^{3x} = \frac{y + 5}{2} \implies 3x = \ln\left(\frac{y + 5}{2}\right)\).
Thus, \(\mathrm{f}^{-1}(x) = \frac{1}{3}\ln\left(\frac{x + 5}{2}\right)\).
Its domain is the range of \(\mathrm{f}\), which is \(x > -5\).

(c) \(\mathrm{fg}(x) = \mathrm{f}(\ln(x + 4)) = 2\mathrm{e}^{3\ln(x + 4)} - 5 = 2(x + 4)^3 - 5\).
We set this equal to 11:
\(2(x + 4)^3 - 5 = 11 \implies 2(x + 4)^3 = 16 \implies (x + 4)^3 = 8\).
Taking the cube root of both sides gives:
\(x + 4 = 2 \implies x = -2\).

(d) \(\mathrm{f}(x) - 9\mathrm{e}^{-3x} = 12 \implies 2\mathrm{e}^{3x} - 5 - 9\mathrm{e}^{-3x} = 12 \implies 2\mathrm{e}^{3x} - 17 - 9\mathrm{e}^{-3x} = 0\).
Let \(u = \mathrm{e}^{3x}\). Since \(\mathrm{e}^{3x} > 0\), we must have \(u > 0\).
Multiply the entire equation by \(u\):
\(2u^2 - 17u - 9 = 0\).
Factorising this quadratic equation:
\((2u + 1)(u - 9) = 0\).
This gives \(u = -\frac{1}{2}\) (which is rejected as \(u > 0\)) or \(u = 9\).
Therefore, \(\mathrm{e}^{3x} = 9\).
Taking the natural logarithm of both sides:
\(3x = \ln(9) \implies x = \frac{1}{3}\ln(9)\).
Thus, \(a = 3\) and \(b = 9\).

(a)
B1: Correct range: \(\mathrm{f}(x) > -5\) (or \(y > -5\)). Do not accept \(y \ge -5\).

(b)
M1: Attempts to make \(x\) (or \(y\)) the subject of \(y = 2\mathrm{e}^{3x} - 5\).
M1: Correctly introduces logarithms.
A1: Correct expression for \(\mathrm{f}^{-1}(x)\).
B1: Correct domain \(x > -5\) (must match range of \(f\)).

(c)
M1: Correctly writes \(\mathrm{fg}(x)\) as \(2\mathrm{e}^{3\ln(x + 4)} - 5\).
A1: Simplifies \(\mathrm{fg}(x)\) to \(2(x + 4)^3 - 5\).
M1: Equates to 11 and solves for \(x\).
A1: Correct value \(x = -2\).

(d)
M1: Writes the equation in terms of \(\mathrm{e}^{3x}\): \(2\mathrm{e}^{3x} - 17 - 9\mathrm{e}^{-3x} = 0\).
M1: Substitutes \(u = \mathrm{e}^{3x}\) to form a quadratic equation.
A1: Correct quadratic equation \(2u^2 - 17u - 9 = 0\).
M1: Solves the quadratic equation to get two values of \(u\).
A1: Identifies \(u = 9\) as the only valid solution and rejects \(u = -1/2\).
M1: Solves \(\mathrm{e}^{3x} = 9\) using logarithms.
A1: Correct final answer: \(x = \frac{1}{3}\ln(9)\) (or equivalent form with \(a = 3, b = 9\)).

(a) Using the formula \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\):
For particle \(P\):
\(\mathbf{r}_P = (3\mathbf{i} - 4\mathbf{j}) + t(2\mathbf{i} + 5\mathbf{j}) = (3 + 2t)\mathbf{i} + (-4 + 5t)\mathbf{j}\).
For particle \(Q\):
\(\mathbf{r}_Q = (-9\mathbf{i} + 5\mathbf{j}) + t(6\mathbf{i} + 2\mathbf{j}) = (-9 + 6t)\mathbf{i} + (5 + 2t)\mathbf{j}\).

(b) For a collision to occur, there must be a common time \(t\) such that \(\mathbf{r}_P = \mathbf{r}_Q\).
Equating the \(\mathbf{i}\) components:
\(3 + 2t = -9 + 6t \implies 4t = 12 \implies t = 3\).
Now we check if this value of \(t\) also satisfies the equality of the \(\mathbf{j}\) components:
For \(P\): \(y_P = -4 + 5(3) = 11\).
For \(Q\): \(y_Q = 5 + 2(3) = 11\).
Since both components are equal at \(t = 3\), the particles collide at \(t = 3\) seconds.
The position vector of the point of collision is \(9\mathbf{i} + 11\mathbf{j}\).

(c) The velocity of particle \(P\) is \(2\mathbf{i} + 5\mathbf{j}\).
Speed \(= |\mathbf{v}_P| = \sqrt{2^2 + 5^2} = \sqrt{29} \approx 5.39 \text{ m/s}\).

(d) At \(t = 2\):
\(\mathbf{r}_P = (3 + 4)\mathbf{i} + (-4 + 10)\mathbf{j} = 7\mathbf{i} + 6\mathbf{j}\).
\(\mathbf{r}_Q = (-9 + 12)\mathbf{i} + (5 + 4)\mathbf{j} = 3\mathbf{i} + 9\mathbf{j}\).
The relative position vector \(\overrightarrow{PQ} = \mathbf{r}_Q - \mathbf{r}_P = (3 - 7)\mathbf{i} + (9 - 6)\mathbf{j} = -4\mathbf{i} + 3\mathbf{j}\).
Distance \(= |\overrightarrow{PQ}| = \sqrt{(-4)^2 + 3^2} = \sqrt{25} = 5 \text{ metres}\).

(e) At any time \(t\), the vector from \(P\) to \(Q\) is:
\(\overrightarrow{PQ} = \mathbf{r}_Q - \mathbf{r}_P = [(-9 + 6t) - (3 + 2t)]\mathbf{i} + [(5 + 2t) - (-4 + 5t)]\mathbf{j}\)
\(= (4t - 12)\mathbf{i} + (9 - 3t)\mathbf{j} = (t - 3)(4\mathbf{i} - 3\mathbf{j})\).
Thus, the distance \(d\) at time \(t\) is:
\(d = |t - 3|\sqrt{4^2 + (-3)^2} = 5|t - 3|\).
We want \(d = 10 \implies 5|t - 3| = 10 \implies |t - 3| = 2\).
This gives two solutions:
\(t - 3 = 2 \implies t = 5\)
\(t - 3 = -2 \implies t = 1\).
Both values are positive, so \(t = 1\) or \(t = 5\).

(a)
B1: Correct vector for \(P\).
B1: Correct vector for \(Q\).

(b)
M1: Equates the \(\mathbf{i}\) components and solves for \(t\).
A1: Correct value \(t = 3\).
M1: Substitutes \(t = 3\) into both \(\mathbf{j}\) components to verify they are equal.
A1: Confirms collision at \(t = 3\) with correct reasoning.
A1: Correct position vector \(9\mathbf{i} + 11\mathbf{j}\).

(c)
M1: Use of Pythagoras to find the magnitude of \(\mathbf{v}_P\).
A1: \(\sqrt{29}\) (or awrt \(5.39\)).

(d)
M1: Substitution of \(t = 2\) into both position vectors.
M1: Finds the magnitude of the difference vector.
A1: Correct distance \(5\) m.

(e)
M1: Expresses distance in terms of \(t\): \(\sqrt{(4t-12)^2 + (9-3t)^2}\) or \(5|t-3|\).
M1: Sets up the equation \(5|t-3| = 10\) (or equivalent quadratic equation) and solves for \(t\).
A1: Correct values \(t = 1\) and \(t = 5\).

Using the Factor Theorem, since \(\mathrm{p}(x)\) is divisible by \(x - 2\), we have:
\(\mathrm{p}(2) = 0\)
\(3(2)^3 + a(2)^2 + b(2) - 12 = 0\)
\(24 + 4a + 2b - 12 = 0\)
\(4a + 2b = -12\)
\(2a + b = -6\) --- (Equation 1)

Using the Remainder Theorem, since dividing \(\mathrm{p}(x)\) by \(x + 1\) gives a remainder of \(-15\), we have:
\(\mathrm{p}(-1) = -15\)
\(3(-1)^3 + a(-1)^2 + b(-1) - 12 = -15\)
\(-3 + a - b - 12 = -15\)
\(a - b = 0\)
\(a = b\) --- (Equation 2)

Substituting Equation 2 into Equation 1:
\(2a + a = -6 \implies 3a = -6 \implies a = -2\)

Since \(a = b\), we also have \(b = -2\).

B1: For setting up the equation \(2a + b = -6\) using \(\mathrm{p}(2) = 0\)
B1: For setting up the equation \(a - b = 0\) using \(\mathrm{p}(-1) = -15\)
M1: For attempting to solve the simultaneous equations to find \(a\) or \(b\)
A1: For both correct values: \(a = -2\) and \(b = -2\)

First, change the base of the second logarithm to 3:
\(\log_9(x - 2) = \frac{\log_3(x - 2)}{\log_3 9} = \frac{1}{2}\log_3(x - 2)\)

Substitute this back into the equation:
\(\log_3(2x - 3) - \frac{1}{2}\log_3(x - 2) = 1\)

Multiply by 2:
\(2\log_3(2x - 3) - \log_3(x - 2) = 2\)
\(\log_3(2x - 3)^2 - \log_3(x - 2) = 2\)
\(\log_3 \left( \frac{(2x-3)^2}{x-2} \right) = 2\)

Convert to exponential form:
\(\frac{(2x-3)^2}{x-2} = 3^2\)
\(\frac{4x^2 - 12x + 9}{x-2} = 9\)
\(4x^2 - 12x + 9 = 9(x - 2)\)
\(4x^2 - 12x + 9 = 9x - 18\)
\(4x^2 - 21x + 27 = 0\)

Factor the quadratic equation:
\((4x - 9)(x - 3) = 0\)

Thus, \(x = 2.25\) or \(x = 3\).
Both values are valid since they make the arguments inside the logarithms positive.

B1: Correctly apply the change of base formula to obtain \(\frac{1}{2}\log_3(x-2)\)
M1: Correctly apply log laws to combine the terms into a single logarithm
M1: Correctly remove logs to form a quadratic equation and attempt to solve it
A1: For both correct solutions: \(x = 3\) and \(x = 2.25\) (or \(\frac{9}{4}\))

From the first equation, express \(y\) in terms of \(x\):
\(y = 5 - 2x\)

Substitute this expression into the second equation:
\(x^2 - x(5 - 2x) + 2(5 - 2x)^2 = 14\)

Expand the terms:
\(x^2 - 5x + 2x^2 + 2(25 - 20x + 4x^2) = 14\)
\(3x^2 - 5x + 50 - 40x + 8x^2 = 14\)
\(11x^2 - 45x + 36 = 0\)

Factorise the quadratic equation:
\((x - 3)(11x - 12) = 0\)

This gives \(x = 3\) or \(x = \frac{12}{11}\).

If \(x = 3\), then:
\(y = 5 - 2(3) = -1\)

If \(x = \frac{12}{11}\), then:
\(y = 5 - 2\left(\frac{12}{11}\right) = \frac{31}{11}\).

M1: Express \(y\) in terms of \(x\) (or vice versa) correctly
M1: Substitute into the quadratic equation and expand
M1: Form a correct 3-term quadratic equation in one variable and attempt to solve it
A1: Both pairs of correct coordinates: \((3, -1)\) and \((\frac{12}{11}, \frac{31}{11})\)

The perimeter of the sector is given by:
\(P = 2r + r\theta = 30 \implies r\theta = 30 - 2r\) --- (Equation 1)

The area of the sector is given by:
\(A = \frac{1}{2}r^2\theta = 50 \implies r^2\theta = 100\) --- (Equation 2)

Substitute Equation 1 into Equation 2:
\(r(r\theta) = 100\)
\(r(30 - 2r) = 100\)
\(30r - 2r^2 = 100\)
\(2r^2 - 30r + 100 = 0\)
\(r^2 - 15r + 50 = 0\)

Factorise the quadratic equation:
\((r - 5)(r - 10) = 0\)

This gives \(r = 5\) or \(r = 10\).

If \(r = 5\), substitute into Equation 1:
\(5\theta = 30 - 2(5) \implies 5\theta = 20 \implies \theta = 4\)

If \(r = 10\), substitute into Equation 1:
\(10\theta = 30 - 2(10) \implies 10\theta = 10 \implies \theta = 1\).

B1: State perimeter equation \(2r + r\theta = 30\) (or equivalent)
B1: State area equation \(\frac{1}{2}r^2\theta = 50\) (or equivalent)
M1: Eliminate \(\theta\) to form a quadratic equation in \(r\) and solve
A1: Obtain both correct pairs: \(r = 5, \theta = 4\) and \(r = 10, \theta = 1\)

First, find the point of intersection of \(y = \frac{6}{x + 2}\) and \(y = x + 1\):
\(\frac{6}{x + 2} = x + 1\)
\(6 = (x + 2)(x + 1)\)
\(6 = x^2 + 3x + 2\)
\(x^2 + 3x - 4 = 0\)
\((x - 1)(x + 4) = 0\)

Since \(x > 0\), we choose \(x = 1\).
When \(x = 1\), \(y = 1 + 1 = 2\). So the intersection point is \((1, 2)\).

Now, differentiate \(y = 6(x + 2)^{-1}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = -6(x + 2)^{-2} = -\frac{6}{(x + 2)^2}\)

At the point \(x = 1\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{6}{(1 + 2)^2} = -\frac{6}{9} = -\frac{2}{3}\)

The gradient of the tangent is \(-\frac{2}{3}\), so the gradient of the normal is:
\(m = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}\)

The equation of the normal at \((1, 2)\) is:
\(y - 2 = \frac{3}{2}(x - 1)\)
\(y - 2 = \frac{3}{2}x - \frac{3}{2}\)
\(y = \frac{3}{2}x + \frac{1}{2}\)

B1: Find the correct intersection point \((1, 2)\)
M1: Correctly differentiate to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
M1: Calculate the gradient of the normal, \(m_n = -\frac{1}{m_t}\), at \(x = 1\)
A1: Find the correct normal equation: \(y = 1.5x + 0.5\) (or any equivalent form)

Find the vector \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\)
\(\overrightarrow{AB} = ((k + 2)\mathbf{i} + 9\mathbf{j}) - (4\mathbf{i} + k\mathbf{j})\)
\(\overrightarrow{AB} = (k - 2)\mathbf{i} + (9 - k)\mathbf{j}\)

Since \(\overrightarrow{AB}\) is parallel to \(\mathbf{i} - 2\mathbf{j}\), the ratio of the \(\mathbf{j}\) component to the \(\mathbf{i}\) component must be equal:
\(\frac{9 - k}{k - 2} = \frac{-2}{1}\)

Solve for \(k\):
\(9 - k = -2(k - 2)\)
\(9 - k = -2k + 4\)
\(k = 4 - 9\)
\(k = -5\)

M1: Attempt to find \(\overrightarrow{AB}\) by subtracting vectors
A1: Correct vector \(\overrightarrow{AB} = (k - 2)\mathbf{i} + (9 - k)\mathbf{j}\)
M1: Set up a correct ratio equation using the parallel condition
A1: Correct value of \(k = -5\)

**(a)(i)**
The number of ways to select 6 people from 14 is:
$$\binom{14}{6} = \frac{14!}{6!(14-6)!} = 3003$$

**(a)(ii)**
Since there must be more women than men, the possible compositions of the committee of 6 are:
- 4 women and 2 men:
$$\binom{6}{4} \times \binom{8}{2} = 15 \times 28 = 420$$
- 5 women and 1 man:
$$\binom{6}{5} \times \binom{8}{1} = 6 \times 8 = 48$$
- 6 women and 0 men:
$$\binom{6}{6} \times \binom{8}{0} = 1 \times 1 = 1$$

Total number of committees:
$$420 + 48 + 1 = 469$$

**(b)**
Treat the 6 women as a single unit. There are 8 men and 1 unit of women, giving 9 units in total to arrange, which can be done in $9!$ ways.
Within the unit, the 6 women can be arranged in $6!$ ways.
Total number of arrangements:
$$9! \times 6! = 362\,880 \times 720 = 261\,273\,600$$

**(a)(i)**
B1: For 3003

**(a)(ii)**
M1: For considering 3 cases (4W 2M, 5W 1M, 6W 0M)
M1: For correct calculation of combinations of at least two cases
A1: For individual correct values (420, 48, 1)
A1: For final answer of 469

**(b)**
M1: For treating women as a single unit (e.g. seeing $9!$)
M1: For arranging women within the unit (e.g. seeing $6!$)
A1: For final answer of 261273600 (or $9! \times 6!$)

**(a)**
To find the intersection points, equate the two equations:
$$\frac{3}{x} + 1 = 5 - x$$
Multiply by $x$:
$$3 + x = 5x - x^2$$
$$x^2 - 4x + 3 = 0$$
$$(x-1)(x-3) = 0$$
So, $x = 1$ and $x = 3$.
When $x = 1$, $y = 5 - 1 = 4$. So, $A = (1, 4)$.
When $x = 3$, $y = 5 - 3 = 2$. So, $B = (3, 2)$.

**(b)**
For $1 \le x \le 3$, the line lies above the curve.
$$\text{Area} = \int_{1}^{3} \left( (5 - x) - \left(\frac{3}{x} + 1\right) \right) dx$$
$$= \int_{1}^{3} \left( 4 - x - \frac{3}{x} \right) dx$$
$$= \left[ 4x - \frac{x^2}{2} - 3\ln x \right]_{1}^{3}$$
$$= \left( 4(3) - \frac{3^2}{2} - 3\ln 3 \right) - \left( 4(1) - \frac{1^2}{2} - 3\ln 1 \right)$$
$$= \left( 12 - 4.5 - 3\ln 3 \right) - \left( 4 - 0.5 - 0 \right)$$
$$= (7.5 - 3\ln 3) - 3.5$$
$$= 4 - 3\ln 3$$
Hence, $a = 4$ and $b = 3$.

**(a)**
M1: For equating curve and line equations to form a quadratic equation
M1: For solving the quadratic equation to find $x = 1$ and $x = 3$
A1: For correct coordinates $(1, 4)$ and $(3, 2)$

**(b)**
M1: For setting up the integral $\int (y_{\text{line}} - y_{\text{curve}}) dx$ with limits 1 and 3
A1: For correct integration of $4 - x - \frac{3}{x}$ to obtain $4x - \frac{x^2}{2} - 3\ln x$
M1: For substituting limits 3 and 1 into their integrated expression
A1: For correct evaluation of the terms with limit 3 (i.e. $7.5 - 3\ln 3$)
A1: For correct evaluation of the terms with limit 1 (i.e. $3.5$)
A1: For final exact area $4 - 3\ln 3$

**(a)**
Using logarithm base-change formula, $\log_{25} x = \frac{\log_5 x}{\log_5 25} = \frac{1}{2}\log_5 x$.
Also, $\log_5(x^3) = 3\log_5 x$.
Substitute these into the equation:
$$3\log_5 x - \frac{1}{2}\log_5 x = 10$$
$$\frac{5}{2}\log_5 x = 10$$
$$\log_5 x = 4$$
$$x = 5^4 = 625$$

**(b)**
Let $u = e^{y+1}$.
The equation becomes:
$$2u - 5 - \frac{3}{u} = 0$$
Multiply by $u$:
$$2u^2 - 5u - 3 = 0$$
$$(2u + 1)(u - 3) = 0$$
Since $u = e^{y+1} > 0$, we reject $u = -0.5$.
Thus, $u = 3 \implies e^{y+1} = 3$.
Taking natural logs:
$$y + 1 = \ln 3 \implies y = \ln 3 - 1$$

**(a)**
M1: For using $\log_{25} x = \frac{1}{2}\log_5 x$ or equivalent base change
M1: For using power rule $\log_5(x^3) = 3\log_5 x$
A1: For obtaining $\frac{5}{2}\log_5 x = 10$ or equivalent
A1: For $x = 625$

**(b)**
M1: For substitution $u = e^{y+1}$ (or similar) to form a quadratic equation
M1: For correct method to solve their quadratic equation
A1: For $u = 3$ (and rejecting $u = -0.5$)
M1: For equating $e^{y+1} = 3$ and taking logs
A1: For final exact answer $y = \ln 3 - 1$

**(a)**
Since $x-2$ is a factor of $f(x)$, by the factor theorem:
$$f(2) = 0 \implies 2(2)^3 + p(2)^2 + q(2) + 6 = 0$$
$$16 + 4p + 2q + 6 = 0 \implies 4p + 2q = -22 \implies 2p + q = -11 \quad (1)$$

Since dividing by $x+1$ gives a remainder of $18$, by the remainder theorem:
$$f(-1) = 18 \implies 2(-1)^3 + p(-1)^2 + q(-1) + 6 = 18$$
$$-2 + p - q + 6 = 18 \implies p - q + 4 = 18 \implies p - q = 14 \quad (2)$$

Adding (1) and (2):
$$(2p + q) + (p - q) = -11 + 14$$
$$3p = 3 \implies p = 1$$

Substitute $p = 1$ into (2):
$$1 - q = 14 \implies q = -13$$

**(b)**
The polynomial is $f(x) = 2x^3 + x^2 - 13x + 6$.
Divide $f(x)$ by $x-2$:
$$2x^3 + x^2 - 13x + 6 = (x-2)(2x^2 + ax + b)$$
By comparing coefficients:
- Constant term: $-2b = 6 \implies b = -3$
- $x^2$ term: $-4 + a = 1 \implies a = 5$

Thus, $2x^3 + x^2 - 13x + 6 = (x-2)(2x^2 + 5x - 3)$.
Factorise the quadratic:
$$2x^2 + 5x - 3 = (2x - 1)(x + 3)$$
So completely factorised, $f(x) = (x-2)(2x-1)(x+3)$.

**(c)**
To solve $f(x) = 0$:
$$(x-2)(2x-1)(x+3) = 0$$
So, $x = 2$, $x = \frac{1}{2}$ (or $0.5$), $x = -3$.

**(a)**
M1: For applying $f(2) = 0$ to get $2p + q = -11$
M1: For applying $f(-1) = 18$ to get $p - q = 14$
M1: For solving simultaneous equations to find $p$ or $q$
A1: For $p = 1$ and $q = -13$

**(b)**
M1: For division by $x-2$ or equivalent method to find quadratic factor
A1: For $2x^2 + 5x - 3$
A1: For $(x-2)(2x-1)(x+3)$

**(c)**
B2: For all three correct solutions: $x = 2$, $x = 0.5$, and $x = -3$. (B1 for any one or two correct solutions)

**(a)**
Using the trigonometric identity $\sec^2 \theta = 1 + \tan^2 \theta$:
$$3(1 + \tan^2 \theta) + 5\tan \theta = 5$$
$$3 + 3\tan^2 \theta + 5\tan \theta = 5$$
Subtract 5 from both sides:
$$3\tan^2 \theta + 5\tan \theta - 2 = 0$$
This is in the required form.

**(b)**
Factorise the quadratic in $\tan \theta$:
$$(3\tan \theta - 1)(\tan \theta + 2) = 0$$
Thus, $\tan \theta = \frac{1}{3}$ or $\tan \theta = -2$.

**Case 1**: $\tan \theta = \frac{1}{3}$
Basic angle $\alpha = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.43^\circ$.
Since $\tan \theta > 0$, in the range $-180^\circ \le \theta \le 180^\circ$:
$$\theta = 18.4^\circ$$
$$\theta = 18.4^\circ - 180^\circ = -161.6^\circ$$

**Case 2**: $\tan \theta = -2$
Basic angle $\alpha = \tan^{-1}(2) \approx 63.43^\circ$.
Since $\tan \theta < 0$, in the range $-180^\circ \le \theta \le 180^\circ$:
$$\theta = -63.4^\circ$$
$$\theta = 180^\circ - 63.4^\circ = 116.6^\circ$$

So the solutions are $\theta = -161.6^\circ, -63.4^\circ, 18.4^\circ, 116.6^\circ$.

**(a)**
M1: For using $\sec^2 \theta = 1 + \tan^2 \theta$
M1: For expanding the brackets correctly to get $3 + 3\tan^2 \theta + 5\tan \theta = 5$
A1: For correct simplification to $3\tan^2 \theta + 5\tan \theta - 2 = 0$

**(b)**
M1: For attempt to solve the quadratic equation to get $\tan \theta = \frac{1}{3}$ and $\tan \theta = -2$
A1: For obtaining basic angle $18.4^\circ$ (or $18.43^\circ$)
A1: For obtaining basic angle $63.4^\circ$ (or $63.43^\circ$)
A1: For any two correct values of $\theta$ from $-161.6^\circ, -63.4^\circ, 18.4^\circ, 116.6^\circ$
A1: For a third correct value of $\theta$
A1: For the final correct value of $\theta$ and no extra angles in the range

**(a)**
We have $y = (2x-3)(4x+1)^{1/2}$.
Using the product rule:
$$\frac{dy}{dx} = (2)(4x+1)^{1/2} + (2x-3) \cdot \frac{1}{2}(4x+1)^{-1/2} \cdot 4$$
$$= 2\sqrt{4x+1} + \frac{2(2x-3)}{\sqrt{4x+1}}$$
$$= \frac{2(4x+1) + 2(2x-3)}{\sqrt{4x+1}}$$
$$= \frac{8x + 2 + 4x - 6}{\sqrt{4x+1}}$$
$$= \frac{12x - 4}{\sqrt{4x+1}} \quad \left(\text{or } \frac{4(3x-1)}{\sqrt{4x+1}}\right)$$

**(b)**
At the stationary point, $\frac{dy}{dx} = 0$:
$$\frac{12x - 4}{\sqrt{4x+1}} = 0 \implies 12x - 4 = 0$$
$$x = \frac{4}{12} = \frac{1}{3}$$

**(c)**
We are given $\frac{dx}{dt} = 0.5$. We want to find $\frac{dy}{dt}$ when $x = 2$.
First, calculate $\frac{dy}{dx}$ at $x = 2$:
$$\frac{dy}{dx} = \frac{12(2) - 4}{\sqrt{4(2)+1}} = \frac{24 - 4}{\sqrt{9}} = \frac{20}{3}$$

Using the chain rule:
$$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$$
$$\frac{dy}{dt} = \frac{20}{3} \times 0.5 = \frac{10}{3} \approx 3.33 \text{ units/s}$$

**(a)**
M1: For applying the product rule to $(2x-3)(4x+1)^{1/2}$
A1: For $2(4x+1)^{1/2} + 2(2x-3)(4x+1)^{-1/2}$ oe
A1: For simplifying to $\frac{12x-4}{\sqrt{4x+1}}$ oe

**(b)**
M1: For setting their numerator $12x - 4 = 0$
M1: For solving for $x$
A1: For $x = \frac{1}{3}$ (or $0.333$)

**(c)**
M1: For substituting $x = 2$ into their $\frac{dy}{dx}$
A1: For obtaining $\frac{dy}{dx} = \frac{20}{3}$
M1: For using $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$ with $\frac{dx}{dt} = 0.5$
A1: For $\frac{10}{3}$ or $3.33$