2024 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

Substitute \(n=2\) and \(n=4\) into the formula: \(u_2 = a(2)^2 + b(2) = 4a + 2b = 14\) which simplifies to \(2a + b = 7\). For \(n=4\), \(u_4 = a(4)^2 + b(4) = 16a + 4b = 44\) which simplifies to \(4a + b = 11\). Subtracting the first simplified equation from the second gives: \((4a + b) - (2a + b) = 11 - 7 \implies 2a = 4 \implies a = 2\). Substitute \(a = 2\) back into \(2a + b = 7\) to obtain \(2(2) + b = 7 \implies b = 3\).

M1 for setting up two simultaneous equations. M1 for solving to find either \(a\) or \(b\). A1 for both correct values.

First differences are \(11-5=6\), \(19-11=8\), \(29-19=10\). Second differences are \(8-6=2\) and \(10-8=2\). Since the second difference is constant, the sequence is quadratic, written as \(an^2 + bn + c\), where \(2a = 2 \implies a = 1\). Subtracting \(n^2\) from each term gives the sequence \(4, 7, 10, 13, \dots\) which is a linear sequence with a common difference of 3, so its term is \(3n + 1\). Combining these results, we get the \(n\)-th term: \(n^2 + 3n + 1\).

M1 for identifying second difference is 2 and finding \(a = 1\). M1 for subtracting \(n^2\) and finding the linear sequence relation \(3n+1\). A1 for the correct final term.

Multiply both sides by \(5 - 2x\) to get \(y(5 - 2x) = 3x + 2\). Expand the left side to get \(5y - 2xy = 3x + 2\). Rearrange the terms to group all terms containing \(x\) on one side: \(5y - 2 = 3x + 2xy\). Factorise \(x\) from the right side: \(5y - 2 = x(3 + 2y)\). Finally, divide by \(3 + 2y\) to isolate \(x\): \(x = \frac{5y - 2}{3 + 2y}\) (or equivalent).

M1 for clearing the fraction. M1 for isolating terms with \(x\) and factorising. A1 for the correct final equation.

Group the terms in pairs: \((12ax - 8bx) - (3ay - 2by)\). Factorise the common factor from each group: \(4x(3a - 2b) - y(3a - 2b)\). Now, factorise out the common bracket \((3a - 2b)\) to get \((4x - y)(3a - 2b)\).

M1 for grouping and factorising two terms, e.g. \(4x(3a-2b)\) or \(-y(3a-2b)\). M1 for identifying both bracket components. A1 for the fully correct factorised form.

The matrix \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) maps the unit base vectors as follows: \(\begin{pmatrix} 1 \\ 0 \end{pmatrix} \to \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix} \to \begin{pmatrix} -1 \\ 0 \end{pmatrix}\). This movement corresponds to a rotation of \(90^\circ\) anticlockwise (or \(270^\circ\) clockwise) around the origin \((0, 0)\).

M1 for identifying the transformation as a rotation. M1 for specifying 90 degrees anticlockwise. A1 for identifying the centre of rotation as the origin.

Substitute \(\mathrm{f}(x)\) into \(\mathrm{g}(x)\) to find the composite function: \(\mathrm{g}(\mathrm{f}(x)) = (2x - 3)^2 + 4\). Set this expression equal to 13: \((2x - 3)^2 + 4 = 13 \implies (2x - 3)^2 = 9\). Taking square roots on both sides gives \(2x - 3 = 3\) or \(2x - 3 = -3\). Solving these linear equations gives \(2x = 6 \implies x = 3\), or \(2x = 0 \implies x = 0\).

M1 for substituting \(2x-3\) into the function \(g\). M1 for simplifying and setting up a solvable quadratic or square root equation. A1 for finding both correct values.

A stretch parallel to the \(y\)-axis with scale factor 3 maps the function \(y = \mathrm{f}(x)\) to \(y = 3\mathrm{f}(x)\). Then, a translation by the vector \(\begin{pmatrix} 0 \\ -4 \end{pmatrix}\) shifts the graph vertically downwards by 4 units, giving the final equation as \(y = 3\mathrm{f}(x) - 4\). Thus, \(\mathrm{g}(x) = 3\mathrm{f}(x) - 4\).

M1 for writing down \(3\mathrm{f}(x)\) to represent the stretch. M1 for showing the translation effect of subtracting 4. A1 for the complete correct expression.

Vertical asymptotes occur where the denominator is zero and the numerator is non-zero. Setting \(x^2 - 9 = 0\) gives \((x-3)(x+3) = 0\), which yields \(x = 3\) and \(x = -3\). Since \(k > 0\), we get \(k = 3\). To find the horizontal asymptote, we analyze the behavior as \(x \to \pm\infty\). Since the degree of the denominator is larger than the numerator, the function values approach 0, meaning the horizontal asymptote is \(y = 0\).

M1 for factorising the denominator to find vertical asymptotes. M1 for evaluating the behavior as \(x \to \infty\) to locate the horizontal asymptote. A1 for both correct values.

Let's find the differences between successive terms of the sequence:
First differences: \(10 - 3 = 7\), \(21 - 10 = 11\), \(36 - 21 = 15\), \(55 - 36 = 19\).
Second differences: \(11 - 7 = 4\), \(15 - 11 = 4\), \(19 - 15 = 4\).
Since the second difference is constant, the sequence is quadratic of the form \(an^2 + bn + c\).
The coefficient \(a\) is given by half of the second difference: \(a = \frac{4}{2} = 2\).
Subtracting \(2n^2\) from each term of the original sequence:
- For \(n = 1\): \(3 - 2(1)^2 = 1\)
- For \(n = 2\): \(10 - 2(2)^2 = 2\)
- For \(n = 3\): \(21 - 2(3)^2 = 3\)
This leaves the linear sequence \(1, 2, 3, \dots\) which has the \(n\)-th term \(n\).
Combining the quadratic and linear parts gives the final expression: \(2n^2 + n\).

M1 for finding the second differences of 4 or using a system of simultaneous equations.
A1 for identifying the quadratic coefficient \(a = 2\).
A0.5 for the fully correct expression \(2n^2 + n\).

First, factorise the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\).
Next, factorise the denominator as a difference of two squares: \(4x^2 - 1 = (2x + 1)(2x - 1)\).
Now, simplify the fraction by dividing the numerator and the denominator by their common factor, \(2x + 1\):
\[\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)} = \frac{x - 3}{2x - 1}\]

M1 for factorising the numerator to \( (2x+1)(x-3) \).
M1 for factorising the denominator to \( (2x+1)(2x-1) \).
A0.5 for cancelling the common factor to obtain the final simplified form \( \frac{x-3}{2x-1} \).

The sequence is geometric with first term \(a = 6\) and common ratio \(r = \frac{18}{6} = 3\).
The general expression for the \(n\)-th term of a geometric sequence is \(u_n = a \cdot r^{n-1}\).
Substituting our values: \[u_n = 6 \times 3^{n-1}\]
This can be simplified as: \[u_n = 2 \times 3 \times 3^{n-1} = 2 \times 3^n\]

M1 for identifying the common ratio \( r = 3 \).
M1 for setting up the term as \( 6 \times 3^{n-1} \) or equivalent.
A0.5 for simplifying to \( 2 \times 3^n \) or writing \( 6 \times 3^{n-1} \) correctly.

We can determine the transformation by examining how the unit basis vectors \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\) map under matrix \(\mathbf{M}\):
\[\mathbf{M} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}\]
\[\mathbf{M} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \end{pmatrix}\]
This maps any point \((x, y)\) to \((-y, -x)\), which represents a reflection in the line \(y = -x\).

M1 for identifying the transformation as a reflection.
A1.5 for the correct line of reflection \( y = -x \) (M1 for line, A0.5 for the full combined statement).

Let \(y = f(x)\), so: \[y = \frac{3x}{x-2}\]
Rearrange the equation to make \(x\) the subject:
\[y(x - 2) = 3x\]
\[xy - 2y = 3x\]
Collect terms involving \(x\) on one side:
\[xy - 3x = 2y\]
Factorise \(x\):
\[x(y - 3) = 2y\]
\[x = \frac{2y}{y - 3}\]
Replacing \(y\) with \(x\) gives the inverse function: \[f^{-1}(x) = \frac{2x}{x - 3}\]

M1 for writing \( y(x-2) = 3x \) or equivalent first step.
M1 for grouping the terms in \( x \) on one side and factorising.
A0.5 for obtaining the correct final expression \( \frac{2x}{x-3} \).

A horizontal shift by \(c\) units to the right is represented by replacing \(x\) with \(x - c\). Here, we have \(x - 4\), which is a horizontal shift of \(+4\) units.
A vertical shift of \(d\) units upwards is represented by adding \(d\) to the function. Here, we add \(2\), which is a vertical shift of \(+2\) units.
Therefore, the transformation is a translation by vector \(\begin{pmatrix} 4 \\ 2 \end{pmatrix}\).

M1 for identifying either a horizontal shift of \( +4 \) or a vertical shift of \( +2 \).
A1.5 for writing the translation correctly as the column vector \( \begin{pmatrix} 4 \\ 2 \end{pmatrix} \).

Let's compare the terms of the sequence with the standard cubic numbers \(n^3\):
- For \(n = 1\): \(1^3 = 1\), term is \(2 = 1 + 1\)
- For \(n = 2\): \(2^3 = 8\), term is \(9 = 8 + 1\)
- For \(n = 3\): \(3^3 = 27\), term is \(28 = 27 + 1\)
- For \(n = 4\): \(4^3 = 64\), term is \(65 = 64 + 1\)
- For \(n = 5\): \(5^3 = 125\), term is \(126 = 125 + 1\)
We can see that each term is exactly 1 more than the corresponding cubic number. Therefore, the \(n\)-th term is \(n^3 + 1\).

M1.5 for identifying the connection to cubic numbers (e.g., matching with sequence \( 1, 8, 27, 64, \dots \)).
A1 for the correct final expression \( n^3 + 1 \).

First, square both sides of the equation to eliminate the square root:
\[p^2 = \frac{w+5}{w-2}\]
Multiply both sides by \((w-2)\) to clear the fraction:
\[p^2(w - 2) = w + 5\]
Expand the brackets:
\[p^2 w - 2p^2 = w + 5\]
Collect all terms containing \(w\) on one side and other terms on the opposite side:
\[p^2 w - w = 2p^2 + 5\]
Factorise \(w\) from the left side:
\[w(p^2 - 1) = 2p^2 + 5\]
Divide both sides by \((p^2 - 1)\) to solve for \(w\):
\[w = \frac{2p^2 + 5}{p^2 - 1}\]

M1 for squaring both sides of the equation.
M1 for gathering all terms with \( w \) on one side and factorising.
A0.5 for the fully correct final expression \( w = \frac{2p^2 + 5}{p^2 - 1} \) or equivalent.

(a)(i) For Sequence \(A\), find the differences: first differences are \(7, 11, 15, \dots\) and second differences are constant at \(4\). The coefficient of \(n^2\) is \(4 / 2 = 2\). Subtracting \(2n^2\) from each term gives: \(4-2(1)=2\), \(11-8=3\), \(22-18=4\), \(37-32=5\). This linear sequence is \(n+1\). Thus, the \(n\)th term of \(A\) is \(2n^2 + n + 1\).
(ii) For Sequence \(B\), first differences are \(8, 20, 38, \dots\), second differences are \(12, 18, \dots\), third differences are constant at \(6\). The coefficient of \(n^3\) is \(6/6 = 1\). Subtracting \(n^3\) from each term gives: \(3-1=2\), \(11-8=3\), \(31-27=4\), \(69-64=5\). This linear sequence is \(n+1\). Thus, the \(n\)th term of \(B\) is \(n^3 + n + 1\).

(b) Observing Sequence \(C\), the terms can be written as: \(\frac{4}{3}, \frac{11}{11}, \frac{22}{31}, \frac{37}{69}\). The numerators correspond to Sequence \(A\) and the denominators to Sequence \(B\). Therefore, the \(n\)th term of Sequence \(C\) is \(\frac{2n^2 + n + 1}{n^3 + n + 1}\).

(c) Solve \(2n^2 + n + 1 = 301 \Rightarrow 2n^2 + n - 300 = 0\). Using the quadratic formula: \(n = \frac{-1 \pm \sqrt{1 - 4(2)(-300)}}{4} = \frac{-1 \pm \sqrt{2401}}{4} = \frac{-1 \pm 49}{4}\). Since \(n\) must be a positive integer, \(n = \frac{48}{4} = 12\).

(a)(i) M1 for recognizing a quadratic sequence (second difference is constant). M1 for finding the coefficient of \(n^2\) is 2. A1 for complete correct expression \(2n^2 + n + 1\).
(ii) M1 for recognizing a cubic sequence (third difference is constant). M1 for finding the coefficient of \(n^3\) is 1. A1 for complete correct expression \(n^3 + n + 1\).
(b) B1.9 for correct division of expressions from (a)(i) and (a)(ii).
(c) M1 for setting up the equation \(2n^2 + n + 1 = 301\). M1 for solving the quadratic equation (by factoring, formula, or calculator). A1 for selecting the positive integer solution \(n = 12\).

(a) The sum of the first two terms is \(a + ar = a(1+r) = 15\). The sum of the first four terms is \(a(1+r+r^2+r^3) = a(1+r)(1+r^2) = 85\). Substituting the first equation into the second gives: \(15(1+r^2) = 85 \Rightarrow 1+r^2 = \frac{85}{15} = \frac{17}{3}\). Hence, \(r^2 = \frac{17}{3} - 1 = \frac{14}{3}\).

(b) Since \(r^2 = \frac{14}{3}\) and \(r > 0\), we have \(r = \sqrt{\frac{14}{3}} = \frac{\sqrt{42}}{3}\). Substituting this back into \(a(1+r) = 15\) yields: \(a\left(1 + \frac{\sqrt{42}}{3}\right) = 15 \Rightarrow a\left(\frac{3+\sqrt{42}}{3}\right) = 15 \Rightarrow a = \frac{45}{3+\sqrt{42}}\). Rationalizing the denominator: \(a = \frac{45(3-\sqrt{42})}{9-42} = \frac{45(3-\sqrt{42})}{-33} = \frac{15(\sqrt{42}-3)}{11}\).

(c) The 5th term of the sequence is \(u_5 = a r^4\). Since \(r^2 = \frac{14}{3}\), \(r^4 = \left(\frac{14}{3}\right)^2 = \frac{196}{9}\). Therefore, \(u_5 = \frac{15(\sqrt{42}-3)}{11} \times \frac{196}{9} = \frac{980(\sqrt{42}-3)}{33} \approx 103.35\), which is \(103\) to 3 significant figures.

(a) M1 for expressing \(u_1 + u_2 = a(1+r) = 15\). M1 for expressing \(u_1+u_2+u_3+u_4 = a(1+r)(1+r^2) = 85\). M1 for algebraic substitution leading to \(15(1+r^2) = 85\). A1 for final proof of \(r^2 = \frac{14}{3}\).
(b) M1 for finding \(r = \sqrt{14/3}\). M1 for setting up \(a = \frac{15}{1+r}\). A1.9 for simplifying to the exact rationalized form \\frac{15(\\sqrt{42}-3)}{11}\\.
(c) M1 for using \(u_5 = ar^4\). M1 for calculating \(r^4 = 196/9\). A1 for correct evaluation of \(103\) (3 s.f.).

(a)(i) The white tiles are the perfect squares: \(1, 4, 9, 16, \dots\) which is \(n^2\).
(ii) The grey tiles increase by 4 each time: \(8, 12, 16, 20, \dots\) which is an arithmetic sequence with a common difference of 4 and a first term of 8. Thus, the formula is \(8 + 4(n-1) = 4n + 4\).
(iii) The total number of tiles is the sum of white and grey tiles: \(n^2 + 4n + 4 = (n+2)^2\).

(b)(i) For Pattern \(k\), the total number of tiles is \(k^2 + 4k + 4\), and the number of grey tiles is \(4k + 4\). Set up the equation: \(k^2 + 4k + 4 = 5(4k + 4) \Rightarrow k^2 + 4k + 4 = 20k + 20\). Subtracting terms gives: \(k^2 - 16k - 16 = 0\).

(b)(ii) Solve \(k^2 - 16k - 16 = 0\) using the quadratic formula: \(k = \frac{16 \pm \sqrt{(-16)^2 - 4(1)(-16)}}{2} = \frac{16 \pm \sqrt{256 + 64}}{2} = \frac{16 \pm \sqrt{320}}{2} = 8 \pm \sqrt{80}\). Since \(k\) must be positive, \(k = 8 + 8.944 = 16.944\). To the nearest integer, \(k = 17\).

(a)(i) B1 for \(n^2\).
(ii) M1 for identifying linear sequence with difference 4. A1 for \(4n + 4\).
(iii) M1 for adding expressions. A1 for \(n^2 + 4n + 4\).
(b)(i) M1 for setting up the equation \(T_k = 5 \times G_k\). M1 for substituting expressions. M1 for expanding \(5(4k + 4)\). A0.9 for arriving at \(k^2 - 16k - 16 = 0\).
(ii) M1 for applying the quadratic formula or using a GDC solver. A1 for \(17\) (rounded from \(16.94\)).

(a) First, factor all expressions:
- \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)
- \(4x^2 - 1 = (2x + 1)(2x - 1)\)
- \(x^2 - 6x + 9 = (x - 3)^2\)

Rewrite the division as multiplication by the reciprocal of the second fraction:
\( \frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)} \times \frac{(2x + 1)(x - 3)}{(x - 3)^2} \)

Cancel common terms:
- Cancel \(x-3\) and \(x-3\) in numerators with \((x-3)^2\) in the denominator.
- Cancel \(2x+1\) in the first numerator with \(2x+1\) in the first denominator.

This leaves: \( \frac{2x + 1}{2x - 1} \).

(b) Multiply both sides by the denominator:
\( w(2 - 5\sqrt{p}) = 3\sqrt{p} + q \Rightarrow 2w - 5w\sqrt{p} = 3\sqrt{p} + q \)

Rearrange to group all terms with \(\sqrt{p}\) on one side:
\( 2w - q = 3\sqrt{p} + 5w\sqrt{p} \Rightarrow 2w - q = \sqrt{p}(3 + 5w) \)

Divide both sides by \(3 + 5w\):
\( \sqrt{p} = \frac{2w - q}{3 + 5w} \)

Square both sides to solve for \(p\):
\( p = \left(\frac{2w - q}{5w + 3}\right)^2 \)

(a) M1 for factoring the quadratic trinomial \((2x+1)(x-3)\). M1 for factoring the difference of squares \((2x-1)(2x+1)\). M1 for factoring the perfect square trinomial \((x-3)^2\). M1 for correctly inverting the second fraction and multiplying. A1 for final simplified fraction \( \frac{2x + 1}{2x - 1} \).
(b) M1 for clearing the fraction to obtain \(2w - 5w\sqrt{p} = 3\sqrt{p} + q\). M1 for isolating terms containing \(\sqrt{p}\). M1 for factoring out \(\sqrt{p}\) on one side. M1 for dividing by the factor. A1.9 for squaring both sides to get \(p = \left(\frac{2w - q}{5w + 3}\right)^2\).

(a) An open box has 5 faces: the bottom base, and 4 side faces.
- Area of bottom base: \(x(x+5) = x^2 + 5x\).
- Area of 2 sides of size \(x \times (2x-3)\): \(2x(2x-3) = 4x^2 - 6x\).
- Area of 2 sides of size \((x+5) \times (2x-3)\): \(2(x+5)(2x-3) = 2(2x^2 + 7x - 15) = 4x^2 + 14x - 30\).

Summing these areas:
\( A = (x^2 + 5x) + (4x^2 - 6x) + (4x^2 + 14x - 30) = 9x^2 + 13x - 30\).

(b)(i) Given \(A = 90\):
\( 9x^2 + 13x - 30 = 90 \Rightarrow 9x^2 + 13x - 120 = 0\).
Solve using the quadratic formula:
\( x = \frac{-13 \pm \sqrt{13^2 - 4(9)(-120)}}{18} = \frac{-13 \pm \sqrt{169 + 4320}}{18} = \frac{-13 \pm \sqrt{4489}}{18} = \frac{-13 \pm 67}{18}\).
Since \(x\) must be positive, \(x = \frac{54}{18} = 3\).

(b)(ii) The dimensions of the box when \(x = 3\) are:
- Width = \(3\) cm
- Length = \(3 + 5 = 8\) cm
- Height = \(2(3) - 3 = 3\) cm

Volume \(V = \text{length} \times \text{width} \times \text{height} = 8 \times 3 \times 3 = 72\) \(\text{cm}^3\).

(a) M1 for finding bottom area \(x^2 + 5x\). M1 for finding side areas \(4x^2-6x\) and \(4x^2+14x-30\). M1 for summing the 5 face areas. A1 for arriving correctly at the given expression \(9x^2+13x-30\).
(b)(i) M1 for setting up the equation \(9x^2+13x-30 = 90\). M1 for rearranging to standard form \(9x^2+13x-120=0\). A1.9 for finding positive root \(x = 3\).
(b)(ii) M1 for substituting \(x = 3\) to find dimensions: 3, 8, and 3. A1 for volume \(72\).

(a) Under a reflection in \(y = -x\), the mapping is \((x, y) \rightarrow (-y, -x)\). Thus:
- \(A(1, 2) \rightarrow A_1(-2, -1)\)
- \(B(3, 2) \rightarrow B_1(-2, -3)\)
- \(C(2, 5) \rightarrow C_1(-5, -2)\)

(b) Under translation by \(\begin{pmatrix} 4 \\ 4 \end{pmatrix}\), add 4 to both coordinates:
- \(A_1(-2, -1) \rightarrow A_2(2, 3)\)
- \(B_1(-2, -3) \rightarrow B_2(2, 1)\)
- \(C_1(-5, -2) \rightarrow C_2(-1, 2)\)

(c) To find the single transformation mapping \(T\) to \(T_2\):
Observe the coordinates of \(T\) and \(T_2\):
- \(A(1,2) \rightarrow A_2(2,3)\)
- \(B(3,2) \rightarrow B_2(2,1)\)
- \(C(2,5) \rightarrow C_2(-1,2)\)

Notice that the orientation of the shape is reversed, so it is a reflection. Let the line of reflection be \(y = mx + c\). The midpoints of the lines connecting corresponding vertices are:
- Midpoint of \(AA_2\) is \(\left(\frac{1+2}{2}, \frac{2+3}{2}\right) = (1.5, 2.5)\)
- Midpoint of \(BB_2\) is \(\left(\frac{3+2}{2}, \frac{2+1}{2}\right) = (2.5, 1.5)\)
- Midpoint of \(CC_2\) is \(\left(\frac{2-1}{2}, \frac{5+2}{2}\right) = (0.5, 3.5)\)

These midpoints all lie on the straight line \(x + y = 4\), which is \(y = -x + 4\). The gradient of the segment \(AA_2\) is \(\frac{3-2}{2-1} = 1\), which is perpendicular to the line of gradient \(-1\).
Thus, the single transformation is a reflection in the line \(y = -x + 4\).

(a) A3 for all 3 coordinates correct. (A2 for 2 coordinates correct, A1 for 1 coordinate correct).
(b) A2.9 for all 3 coordinates correct. (A1.9 for 2 coordinates correct, A1 for 1 coordinate correct).
(c) M1 for identifying the transformation as a reflection. M2 for finding the midpoints of corresponding points. M1 for verifying perpendicularity. A1 for specifying the correct line equation \(y = -x + 4\) (or \(x + y = 4\)).

(a) Let \(y = 2x - 3 \Rightarrow y + 3 = 2x \Rightarrow x = \frac{y+3}{2}\). Thus, \(f^{-1}(x) = \frac{x+3}{2}\).

(b) \(g(f(x)) = g(2x - 3) = (2x - 3)^2 - 6 = (4x^2 - 12x + 9) - 6 = 4x^2 - 12x + 3\).

(c) First, find \(f(g(x)) = f(x^2 - 6) = 2(x^2 - 6) - 3 = 2x^2 - 12 - 3 = 2x^2 - 15\).
Now set \(f(g(x)) = g(f(x))\):
\(2x^2 - 15 = 4x^2 - 12x + 3\)
\(0 = 2x^2 - 12x + 18\).
Divide the entire equation by 2:
\(x^2 - 6x + 9 = 0\).

(d) Solve \(x^2 - 6x + 9 = 0 \Rightarrow (x - 3)^2 = 0 \Rightarrow x = 3\).

(a) M1 for writing \(y = 2x - 3\) and attempting to make \(x\) the subject. M1 for swapping variables. A1 for \(\frac{x+3}{2}\).
(b) M1 for substituting \(f(x)\) into \(g(x)\). M1 for expanding \((2x-3)^2\). A1 for simplified quadratic \(4x^2 - 12x + 3\).
(c) M1 for finding \(f(g(x)) = 2x^2 - 15\). M1 for equating \(2x^2 - 15 = 4x^2 - 12x + 3\). A1.9 for correctly simplifying to the given equation \(x^2 - 6x + 9 = 0\).
(d) B1 for \(x = 3\).

(a) Plotting the function on the calculator, we observe a W-shaped quartic curve with:
- \(y\)-intercept at \((0, -2)\).
- First local minimum at approx \((-0.47, -3.44)\).
- Local maximum at approx \((1.13, 2.06)\).
- Second local minimum at approx \((2.35, -0.62)\).

(b)(i) The \(x\)-intercepts are found by setting \(f(x) = 0\). Using the GDC root solver, the roots are:
\(x = -1.00\), \(x = 0.38\), \(x = 2.00\), and \(x = 2.62\).

(b)(ii) For the horizontal line \(y = k\) to intersect the curve \(y = f(x)\) at exactly 4 distinct points, \(k\) must lie strictly between the value of the higher local minimum and the local maximum:
- Higher local minimum: \(y \approx -0.62\)
- Local maximum: \(y \approx 2.06\)

Thus, the range is \(-0.62 < k < 2.06\).

(a) G1 for a correct quartic W-shape. G1 for correctly indicating the \(y\)-intercept at \(-2\). G1 for first local minimum at \((-0.47, -3.44)\). G1 for local maximum at \((1.13, 2.06)\). G1 for second local minimum at \((2.35, -0.62)\).
(b)(i) A2.9 for all four intercepts correct: \(-1\), \(0.38\), \(2\), and \(2.62\) (deduct 1 mark for each missing or incorrect intercept).
(b)(ii) M1 for realizing the boundary values are the local maximum and the higher local minimum. A1 for lower limit \(-0.62\). A1 for upper limit \(2.06\) with correct inequalities.

\( \textbf{Part (a)} \)
Let \(u_n = an^2 + bn + c\).
The first differences are: \(12-5=7\), \(25-12=13\), \(44-25=19\), \(69-44=25\).
The second differences are: \(13-7=6\), \(19-13=6\), \(25-19=6\).
Since the second difference is constant, \(2a = 6 \implies a = 3\).
Subtracting \(3n^2\) from the terms:
For \(n=1\): \(5 - 3(1)^2 = 2\)
For \(n=2\): \(12 - 3(2)^2 = 0\)
For \(n=3\): \(25 - 3(3)^2 = -2\)
For \(n=4\): \(44 - 3(4)^2 = -4\)
This linear sequence has the form \(bn + c\).
The common difference is \(-2\), so \(b = -2\).
When \(n=1\), \(-2(1) + c = 2 \implies c = 4\).
Thus, the \(n\)-th term of Sequence A is \(u_n = 3n^2 - 2n + 4\).

\( \textbf{Part (b)} \)
Sequence B is geometric with first term \(a = 2\) and common ratio \(r = 3\).
The \(n\)-th term is \(w_n = a r^{n-1} = 2 \times 3^{n-1}\).

\( \textbf{Part (c)(i)} \)
\(v_n = u_n - w_n = (3n^2 - 2n + 4) - (2 \times 3^{n-1}) = 3n^2 - 2n + 4 - 2 \times 3^{n-1}\).

\( \textbf{Part (c)(ii)} \)
We need to solve \(3n^2 - 2n + 4 - 2 \times 3^{n-1} = -4194\).
Using a GDC to find the intersection of \(y = 3x^2 - 2x + 4 - 2 \times 3^{x-1}\) and \(y = -4194\), we find the integer solution is \(x = 8\).
Thus, \(n = 8\).

\( \textbf{Part (c)(iii)} \)
Substituting \(n = 10\):
\(v_{10} = 3(10)^2 - 2(10) + 4 - 2 \times 3^9\)
\(v_{10} = 300 - 20 + 4 - 2(19683)\)
\(v_{10} = 284 - 39366 = -39082\).

(a) M1 for attempting to find second differences (or setting up equations in terms of a, b, c)
A1 for \(a = 3\)
A1 for \(3n^2 - 2n + 4\)

(b) M1 for identifying a geometric sequence or showing ratio is 3
A1 for \(2 \times 3^{n-1}\) or equivalent

(c)(i) B1 for substituting both expressions into \(v_n = u_n - w_n\) and completing the show that.

(c)(ii) M1 for setting up the equation \(3n^2 - 2n + 4 - 2 \times 3^{n-1} = -4194\)
M1 for using GDC to find intersection or table of values
A1 for \(n = 8\)

(c)(iii) M1 for substituting \(n = 10\) into their expression
A1 for \(-39082\)

\( \textbf{Part (a)} \)
The sketch should show a curve with two branches separated by a vertical asymptote at \(x=0\).
For \(x < 0\), the curve increases from negative values, crosses the x-axis, and goes to \(+\infty\) as \(x \to 0^-\).
For \(x > 0\), as \(x \to 0^+\), the curve goes to \(-\infty\), rises to a local maximum at \(x \approx 1.23\), falls to a local minimum at \(x \approx 2.50\), and then rises to \(+\infty\).

\( \textbf{Part (b)} \)
Since \(f(x) \to \pm\infty\) as \(x \to 0\) due to the term \(-\frac{8}{x}\), the vertical asymptote is \(x = 0\).

\( \textbf{Part (c)} \)
Using the GDC's minimum solver on the interval \(x > 0\):
\(x \approx 2.50\) (or precisely \(2.495\)), \(y \approx -7.57\).
Thus, the local minimum is \((2.50, -7.57)\).

\( \textbf{Part (d)} \)
Using the GDC's maximum solver on the interval \(0 < x < 2\):
\(x \approx 1.23\) (or precisely \(1.229\)), \(y \approx -5.70\).
Thus, the local maximum is \((1.23, -5.70)\).

\( \textbf{Part (e)} \)
To solve \(f(x) \le 5\), we find the intersections of \(y = f(x)\) and \(y = 5\).
This is equivalent to solving \(x^3 - 4x^2 - \frac{8}{x} + 5 = 5 \implies x^4 - 4x^3 - 8 = 0\).
The GDC gives the intersection points:
\(x \approx -1.16\) and \(x \approx 4.11\).
Looking at the graph, \(f(x) \le 5\) for:
\(x \le -1.16\) and \(0 < x \le 4.11\).

(a) G3 for a sketch showing:
- Correct shape for \(x < 0\) branch with no turning points, crossing the negative x-axis (1 mark)
- Correct shape for \(x > 0\) branch showing one local maximum and one local minimum (1 mark)
- Asymptotic behavior to the y-axis (1 mark)

(b) B1 for \(x = 0\)

(c) B1 for \(x\)-coordinate \(2.50\) (accept \(2.49\) to \(2.50\))
B1 for \(y\)-coordinate \(-7.57\) (accept \(-7.58\) to \(-7.57\))

(d) B1 for \(x\)-coordinate \(1.23\)
B1 for \(y\)-coordinate \(-5.70\)

(e) M1 for finding the critical values \(x \approx -1.16\) and \(x \approx 4.11\)
A1 for \(x \le -1.16\)
A1 for \(0 < x \le 4.11\) (do not accept \(x \le 4.11\) without the lower bound 0)

\( \textbf{Part (a)(i)} \)
By completing the square or differentiation:
\(f'(x) = 2x - 6 = 0 \implies x = 3\).
\(f(3) = 3^2 - 6(3) + 4 = 9 - 18 + 4 = -5\).
The coordinates of the local minimum are \((3, -5)\).

\( \textbf{Part (a)(ii)} \)
Set \(f(x) = 0 \implies x^2 - 6x + 4 = 0\).
Using the quadratic formula:
\(x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(4)}}{2} = \frac{6 \pm \sqrt{20}}{2} = 3 \pm \sqrt{5}\).
Thus, \(x \approx 0.764\) and \(x \approx 5.24\).
The coordinates of the x-intercepts are \((0.764, 0)\) and \((5.24, 0)\).

\( \textbf{Part (b)} \)
A translation of \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\) corresponds to mapping \(f(x)\) to \(f(x + 2) + 3\).
\(g(x) = (x+2)^2 - 6(x+2) + 4 + 3\)
\(g(x) = x^2 + 4x + 4 - 6x - 12 + 7\)
\(g(x) = x^2 - 2x - 1\).

\( \textbf{Part (c)(i)} \)
A stretch parallel to the \(y\)-axis with scale factor 2 maps \(f(x) \to 2f(x)\).
A reflection in the \(x\)-axis maps \(y \to -y\), i.e., \(2f(x) \to -2f(x)\).
Thus, \(h(x) = -2f(x) = -2(x^2 - 6x + 4) = -2x^2 + 12x - 8\).

\( \textbf{Part (c)(ii)} \)
The local minimum of \(f(x)\) was at \((3, -5)\).
The stretch maps this point to \((3, -10)\).
The reflection in the \(x\)-axis maps this point to \((3, 10)\).
Since the reflection turns a minimum into a maximum, the local maximum of \(h(x)\) is at \((3, 10)\).

(a)(i) M1 for differentiation or completing the square to find \(x=3\).
A1 for \((3, -5)\).

(a)(ii) M1 for setting \(x^2 - 6x + 4 = 0\) and using formula or GDC.
A1 for \((0.764, 0)\) and \((5.24, 0)\) (or 3 s.f. equivalents).

(b) M1 for writing \(f(x+2) + 3\).
M1 for algebraic expansion: \((x+2)^2 - 6(x+2) + 7\).
A1 for \(x^2 - 2x - 1\).

(c)(i) M1 for applying stretch to get \(2(x^2-6x+4)\) or reflection to get \(-(x^2-6x+4)\).
A1 for completing the steps to show \(-2x^2 + 12x - 8\) correctly.

(c)(ii) M1 for applying the transformations to the vertex \((3, -5)\) (either step shown).
A1 for \((3, 10)\).

Following the alternating pattern in blocks of four terms:
- For \(n = 1, 2\), \(x_n = 1\).
- For \(n = 3, 4\), \(x_n = -1\).
- For \(n = 5, 6\), \(x_n = 2\).
- For \(n = 7, 8\), \(x_n = -2\).

Following this rule:
- For \(n = 9, 10\), \(x_n = 3\).
- For \(n = 11, 12\), \(x_n = -3\).

Therefore, the values of \(x_9, x_{10}, x_{11}, x_{12}\) are 3, 3, -3, -3 respectively.

M1 for recognizing that the values increase in magnitude every 4 terms with alternating signs.
A1 for 3, 3.
A1 for -3, -3.

Let us analyze the terms for \(y_{2k}\):
- For \(k=1\) (odd): \(y_2 = 1\)
- For \(k=2\) (even): \(y_4 = -1\)
- For \(k=3\) (odd): \(y_6 = 2\)
- For \(k=4\) (even): \(y_8 = -2\)
- For \(k=5\) (odd): \(y_{10} = 3\)
- For \(k=6\) (even): \(y_{12} = -3\)

We see that:
- When \(k\) is odd, the sequence of values is \(1, 2, 3, \dots\), which can be written as \(\frac{k+1}{2}\).
- When \(k\) is even, the sequence of values is \(-1, -2, -3, \dots\), which can be written as \(-\frac{k}{2}\).

M1 for identifying the pattern of positive and negative terms.
A1 for finding the correct expression \(\frac{k+1}{2}\) for odd \(k\).
A1 for finding the correct expression \(-\frac{k}{2}\) for even \(k\).

(i) Comparing the term number \(m\) with the value \(d_{4m}^2\):
- For \(m = 1\): \(d_4^2 = 2 = 2(1)^2\)
- For \(m = 2\): \(d_8^2 = 8 = 2(2)^2\)
- For \(m = 3\): \(d_{12}^2 = 18 = 2(3)^2\)
Thus, the formula is \(d_{4m}^2 = 2m^2\).

(ii) For the 40th vertex, \(n = 40\), which corresponds to \(4m = 40 \implies m = 10\).
Using the formula, \(d_{40}^2 = 2(10)^2 = 2(100) = 200\).

M1 for finding the relationship between \(m\) and the squared distance (multiples of squares).
A1 for the formula \(2m^2\).
M1 for substituting \(m = 10\) into their formula.
A1 for the final answer of 200.

We set up a system of equations using the values for \(m = 1, 2, 3\):
1) For \(m = 1\): \(a(1)^2 + b(1) + c = 1 \implies a + b + c = 1\)
2) For \(m = 2\): \(a(2)^2 + b(2) + c = 5 \implies 4a + 2b + c = 5\)
3) For \(m = 3\): \(a(3)^2 + b(3) + c = 13 \implies 9a + 3b + c = 13\)

Subtracting equation (1) from (2):
\(3a + b = 4\) (Equation 4)

Subtracting equation (2) from (3):
\(5a + b = 8\) (Equation 5)

Subtracting Equation (4) from (5):
\(2a = 4 \implies a = 2\).

Substitute \(a = 2\) into Equation (4):
\(3(2) + b = 4 \implies 6 + b = 4 \implies b = -2\).

Substitute \(a = 2\) and \(b = -2\) into equation (1):
\(2 - 2 + c = 1 \implies c = 1\).

Thus, \(a = 2\), \(b = -2\), and \(c = 1\).

M1 for setting up at least two correct equations in terms of \(a\), \(b\), and \(c\).
M1 for attempting to find the second differences (which is 4) or solving the system of equations.
A1 for finding \(a = 2\).
A1 for finding \(b = -2\) and \(c = 1\).

As \(t \to \infty\), \(e^{-0.15 t} \to 0\).
Therefore, \(P(t) \to \frac{A}{1 + B(0)} = A\).
Since the population stabilizes at 1500 beetles, we have \(A = 1500\).

At \(t = 0\), \(P(0) = 120\).
Substituting these values into the model:
\(120 = \frac{1500}{1 + B e^{-0.15(0)}}\)
\(120 = \frac{1500}{1 + B}\)
\(1 + B = \frac{1500}{120}\)
\(1 + B = 12.5\)
\(B = 11.5\).

M1 for explaining why \(A = 1500\) as \(t \to \infty\).
M1 for substituting \(t = 0\) and \(P(0) = 120\) into the model.
A1 for obtaining \(1 + B = 12.5\).
A1 for the correct value \(B = 11.5\).

We set \(P(t) = 750\):
\(750 = \frac{1500}{1 + 11.5 e^{-0.15 t}}\)

Divide both sides by 750:
\(1 = \frac{2}{1 + 11.5 e^{-0.15 t}}\)
\(1 + 11.5 e^{-0.15 t} = 2\)
\(11.5 e^{-0.15 t} = 1\)
\(e^{-0.15 t} = \frac{1}{11.5}\)

Take the natural logarithm of both sides:
\(-0.15 t = \ln\left(\frac{1}{11.5}\right)\)
\(-0.15 t = -\ln(11.5)\)
\(t = \frac{\ln(11.5)}{0.15}\)

Using a calculator:
\(t \approx \frac{2.4423}{0.15} \approx 16.282\) months.
Rounding to 1 decimal place gives \(16.3\) months.

M1 for setting up the equation \(750 = \frac{1500}{1 + 11.5 e^{-0.15 t}}\).
M1 for isolating the exponential term, e.g., \(e^{-0.15 t} = \frac{1}{11.5}\).
M1 for taking natural logarithms to solve for \(t\).
A1 for \(16.3\) (accept 16.28).

Using a graphic display calculator (GDC), we sketch the function \(y = \frac{2587.5 e^{-0.15 x}}{(1 + 11.5 e^{-0.15 x})^2}\).
Using the maximum finder function on the calculator:
- The maximum y-value (rate of growth) is found to be exactly \(56.25\) beetles per month.
Rounding to the nearest whole number gives 56.
- The x-value (time \(t\)) at this maximum is approximately \(16.282\) months.
Rounding to 1 decimal place gives \(16.3\) months.

M1 for sketching the function on GDC with appropriate window dimensions.
A1 for finding the maximum rate of growth as 56.25 or 56.
M1 for locating the x-coordinate of the peak.
A1 for \(16.3\) months (or 16.28).

(i) For a steady state, set \(G(P_s) = 0\):
\(0.15 P_s \left(1 - \frac{P_s}{1500}\right) - h P_s = 0\)

Since \(P_s \neq 0\), we can divide both sides by \(P_s\):
\(0.15 \left(1 - \frac{P_s}{1500}\right) - h = 0\)
\(0.15 - \frac{0.15 P_s}{1500} = h\)
\(0.15 - \frac{P_s}{10000} = h\)
\(\frac{P_s}{10000} = 0.15 - h\)
\(P_s = 10000(0.15 - h) = 1500 - 10000h\).

(ii) Set \(P_s = 900\):
\(900 = 1500 - 10000h\)
\(10000h = 1500 - 900\)
\(10000h = 600\)
\(h = 0.06\).

M1 for setting \(G(P) = 0\) and dividing by \(P\).
A1 for the correct expression \(P_s = 1500 - 10000h\) or equivalent.
M1 for substituting \(P_s = 900\) into their expression.
A1 for finding the correct predation rate \(h = 0.06\).

The area \(A_n\) can be found by adding the area of the next square \(s_n^2\) to the previous rectangle's area:
\(A_5 = A_4 + s_5^2 = 15 + 5^2 = 15 + 25 = 40\)
\(A_6 = A_5 + s_6^2 = 40 + 8^2 = 40 + 64 = 104\)

M1 for \(A_5 = 15 + 5^2\) or equivalent
A1 for \(A_5 = 40\)
M1 for \(A_6 = 40 + 8^2\) or equivalent
A1.6 for \(A_6 = 104\)

Since the dimensions of the \(n\)-th rectangle \(R_n\) are \(s_n\) and \(s_{n+1}\), the area \(A_n\) is the product of these dimensions:
\(A_n = s_n \times s_{n+1} = s_n s_{n+1}\)

M2 for recognizing that the dimensions of \(R_n\) are \(s_n\) and \(s_{n+1}\)
A2.6 for the correct formula \(A_n = s_n s_{n+1}\)

Substitute \(A_n = s_n s_{n+1}\) into the equation:
\(A_n + s_{n+1}^2 = s_n s_{n+1} + s_{n+1}^2\)
Factorise by taking out the common term \(s_{n+1}\):
\(s_{n+1}(s_n + s_{n+1})\)
Thus, the expression in the bracket is \(s_n + s_{n+1}\).

M2 for factorising \(s_{n+1}\) from \(s_n s_{n+1} + s_{n+1}^2\)
A2.6 for the correct expression \(s_n + s_{n+1}\)

Substitute \(t = 0\):
\(P(0) = a \cdot b^0 = a = 200\)

Substitute \(t = 3\) and \(a = 200\):
\(P(3) = 200 \cdot b^3 = 1600\)
\(b^3 = 8\)
\(b = 2\)

So, \(a = 200\) and \(b = 2\).

M1 for setting \(a = 200\) using the initial condition
M2 for setting up \(200 \cdot b^3 = 1600\) and solving for \(b\)
A1.6 for both correct values: \(a = 200\) and \(b = 2\)

Set \(N(t) = 5000\):
\(\frac{10000}{1 + 49 \cdot 2^{-t}} = 5000\)

Divide both sides by 5000:
\(2 = 1 + 49 \cdot 2^{-t}\)
\(1 = 49 \cdot 2^{-t}\)
\(2^t = 49\)

Take natural logarithms (or log base 2) of both sides:
\(t = \frac{\ln(49)}{\ln(2)} \approx 5.6147\) hours

Rounding to 2 decimal places gives \(t = 5.61\) hours.

M1 for substituting values into the formula: \(\frac{10000}{1 + 49 \cdot 2^{-t}} = 5000\)
M1 for simplifying to \(2^t = 49\) or \(2^{-t} = \frac{1}{49}\)
M1 for taking logarithms to solve: \(t = \frac{\ln(49)}{\ln(2)}\) or equivalent GDC method
A1.6 for \(5.61\) (accept \(5.61\) to \(5.62\))