2024 Cambridge IGCSE International Mathematics (0607) 模拟试题及答案详解

First, find the differences between consecutive terms:
First differences: \(6, 8, 10, 12, \dots\)
Second differences: \(2, 2, 2, \dots\)

Since the second difference is constant at 2, the sequence is quadratic. The coefficient of the \(n^2\) term is \(2 \div 2 = 1\).
Subtract \(1n^2\) from each term in the sequence:
- For \(n = 1\): \(3 - 1^2 = 2\)
- For \(n = 2\): \(9 - 2^2 = 5\)
- For \(n = 3\): \(17 - 3^2 = 8\)
- For \(n = 4\): \(27 - 4^2 = 11\)

This gives a linear sequence: \(2, 5, 8, 11, \dots\), which has a first term of 2 and a common difference of 3. The formula for this linear part is \(3n - 1\).

Combining the quadratic and linear parts, the \(n\)-th term of the sequence is \(n^2 + 3n - 1\).

M1 for finding second difference of 2 (so \(n^2\) term is established).
M1 for subtracting \(n^2\) from the sequence to get the linear sequence \(2, 5, 8, \dots\) or finding the linear term \(3n - 1\).
A0.5 for \(n^2 + 3n - 1\) (or equivalent).

Multiply both sides by the denominator \(x-3\):
\(w(x-3) = x+2\)

Expand the brackets:
\(wx - 3w = x+2\)

Collect all terms containing \(x\) on one side and other terms on the opposite side:
\(wx - x = 3w + 2\)

Factorise \(x\) on the left-hand side:
\(x(w-1) = 3w + 2\)

Divide both sides by \(w-1\) to isolate \(x\):
\(x = \frac{3w+2}{w-1}\)

M1 for multiplying by the denominator to get \(w(x-3) = x+2\).
M1 for collecting all terms with \(x\) on one side and factorising to get \(x(w-1) = 3w+2\).
A0.5 for \(x = \frac{3w+2}{w-1}\) or \(x = \frac{-3w-2}{1-w}\).

Set \(y = f(x)\):
\(y = \frac{4}{x+3}\)

Swap \(x\) and \(y\):
\(x = \frac{4}{y+3}\)

Now solve for \(y\):
\(x(y+3) = 4\)
\(y+3 = \frac{4}{x}\)
\(y = \frac{4}{x} - 3\)

Therefore, the inverse function is:
\(f^{-1}(x) = \frac{4}{x} - 3\) or \(f^{-1}(x) = \frac{4-3x}{x}\)

M1 for setting up the equation, e.g., \(x = \frac{4}{y+3}\) or \(y(x+3) = 4\).
M1 for isolating \(y\) (or \(x\)) correctly, e.g., \(y+3 = \frac{4}{x}\).
A0.5 for \(f^{-1}(x) = \frac{4}{x} - 3\) or \(\frac{4-3x}{x}\).

1. The factor 3 multiplying the function represents a vertical stretch. Specifically, it is a stretch parallel to the \(y\)-axis with a scale factor of 3.

2. The term \((x-2)\) inside the function represents a horizontal translation. Specifically, it is a translation by 2 units to the right, which can be written as a translation with vector \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\).

B1 for stretch parallel to the \(y\)-axis, scale factor 3 (accept vertical stretch, scale factor 3).
B1 for translation by \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\) (accept 2 units to the right).
A0.5 for a fully correct response using correct mathematical terminology.

Express both base numbers, 27 and 9, as powers of 3:
\(27 = 3^3\)
\(9 = 3^2\)

Substitute these bases back into the equation:
\((3^3)^{2x-1} = (3^2)^{x+2}\)

Apply the power of a power rule:
\(3^{3(2x-1)} = 3^{2(x+2)} Since the bases are identical, equate the exponents: \)3(2x-1) = 2(x+2)\)
\(6x - 3 = 2x + 4\)

Subtract \(2x\) from both sides and add 3 to both sides:
\(4x = 7\)
\(x = \frac{7}{4} = 1.75\)

M1 for expressing both bases as powers of 3, e.g., \((3^3)^{2x-1}\) and \((3^2)^{x+2}\).
M1 for equating the exponents to form the linear equation \(6x - 3 = 2x + 4\) (or equivalent).
A0.5 for \(x = \frac{7}{4}\) or \(1.75\).

Multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{7} + \sqrt{2}\):

\(\frac{10}{\sqrt{7} - \sqrt{2}} \times \frac{\sqrt{7} + \sqrt{2}}{\sqrt{7} + \sqrt{2}} = \frac{10(\sqrt{7} + \sqrt{2})}{(\sqrt{7})^2 - (\sqrt{2})^2}\)

\(= \frac{10(\sqrt{7} + \sqrt{2})}{7 - 2}\)

\(= \frac{10(\sqrt{7} + \sqrt{2})}{5}\)

Simplify by dividing 10 by 5:
\(= 2(\sqrt{7} + \sqrt{2}) = 2\sqrt{7} + 2\sqrt{2}\)

M1 for multiplying numerator and denominator by \(\sqrt{7} + \sqrt{2}\).
M1 for correctly simplifying the denominator to 5.
A0.5 for \(2\sqrt{7} + 2\sqrt{2}\) or \(2(\sqrt{7} + \sqrt{2})\).

First, list the prime numbers between 1 and 10:
\(P = \{2, 3, 5, 7\}\)

Next, list the factors of 24 that lie between 1 and 10:
\(F = \{1, 2, 3, 4, 6, 8\}\)

Identify the numbers that are in both sets (prime numbers and factors of 24):
\(P \cap F = \{2, 3\}\)

There are 2 numbers that satisfy both conditions out of 10 possible outcomes.

The probability is \(\frac{2}{10} = \frac{1}{5}\).

M1 for listing prime numbers \(\{2, 3, 5, 7\}\) or factors of 24 \(\{1, 2, 3, 4, 6, 8\}\) up to 10.
M1 for identifying the overlap \(\{2, 3\}\).
A0.5 for \(\frac{1}{5}\) (accept \(\frac{2}{10}\) only if working is fully correct but fraction is unsimplified, though simplest form is requested so A0.5 is for simplified form).

First, calculate the gradient of the original line passing through \((2, 5)\) and \((6, -3)\):
\(m_1 = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2\)

The gradient of a perpendicular line, \(m_2\), is the negative reciprocal of \(m_1\):
\(m_2 = -\frac{1}{-2} = \frac{1}{2}\)

Use the equation of a straight line, \(y - y_1 = m(x - x_1)\), with the point \((4, 1)\):
\(y - 1 = \frac{1}{2}(x - 4)\)
\(y - 1 = \frac{1}{2}x - 2\)
\(y = \frac{1}{2}x - 1\)

M1 for finding the gradient of the original line, \(m_1 = -2\).
M1 for finding the perpendicular gradient, \(m_2 = \frac{1}{2}\), and substituting the point \((4, 1)\) into a linear equation.
A0.5 for \(y = \frac{1}{2}x - 1\) (or equivalent form like \(y = 0.5x - 1\)).

First differences: \(7, 11, 15, 19, \dots\)
Second differences: \(4, 4, 4, \dots\)
Since the second difference is constant, the sequence is quadratic of the form \(an^2 + bn + c\).
\(2a = 4 \implies a = 2\).
Subtracting \(2n^2\) from the terms of the sequence:
- For \(n = 1\): \(3 - 2(1)^2 = 1\)
- For \(n = 2\): \(10 - 2(2)^2 = 2\)
- For \(n = 3\): \(21 - 2(3)^2 = 3\)
The remaining sequence is \(1, 2, 3, \dots\) which is simply \(n\).
Thus, the \(n\)th term is \(2n^2 + n\).

M1 for finding \(2n^2\) as the quadratic term
A1 for the fully correct expression \(2n^2 + n\)

Factorise the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)

Factorise the denominator:
\(x^2 - 9 = (x - 3)(x + 3)\)

Cancel the common factor of \((x - 3)\):
\[\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}\]

M1 for factorising the numerator to \((2x + 1)(x - 3)\)
M1 for factorising the denominator to \((x - 3)(x + 3)\)
A1 for final simplified expression \(\frac{2x + 1}{x + 3}\)

Express \(\frac{1}{125}\) as a power of 5:
\(\frac{1}{125} = 5^{-3}\)

Equate exponents:
\(3x - 1 = -3\)

Solve for \(x\):
\(3x = -2 \implies x = -\frac{2}{3}\)

M1 for writing \(\frac{1}{125}\) as \(5^{-3}\)
M1 for equating exponents to form \(3x - 1 = -3\)
A1 for \(x = -\frac{2}{3}\) or equivalent

Multiply out the terms:
\(4 \times 2 + 4 \times 3\sqrt{3} - \sqrt{3} \times 2 - \sqrt{3} \times 3\sqrt{3}\)

Simplify each term:
\(8 + 12\sqrt{3} - 2\sqrt{3} - 3 \times 3\)
\(8 + 10\sqrt{3} - 9\)

Combine like terms:
\(-1 + 10\sqrt{3}\)

M1 for expansion of at least three terms correctly
A1 for combining integer terms to \(-1\) or surd terms to \(10\sqrt{3}\)
A1 for fully correct simplified expression \(-1 + 10\sqrt{3}\)

Calculate the vector \(2\mathbf{a} - \mathbf{b}\):
\[2\mathbf{a} - \mathbf{b} = 2\begin{pmatrix} -1 \\ 7 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} -2 \\ 14 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} -5 \\ 15 \end{pmatrix}\]

Calculate the magnitude:
\[|2\mathbf{a} - \mathbf{b}| = \sqrt{(-5)^2 + 15^2} = \sqrt{25 + 225} = \sqrt{250}\]

Simplify the surd:
\[\sqrt{250} = \sqrt{25 \times 10} = 5\sqrt{10}\]

M1 for finding vector \(2\mathbf{a} - \mathbf{b} = \begin{pmatrix} -5 \\ 15 \end{pmatrix}\)
M1 for using Pythagoras' theorem to find magnitude \(\sqrt{(-5)^2 + 15^2}\)
A1 for the final answer \(5\sqrt{10}\)

The two possible combinations for different colours are Red then Blue (RB) or Blue then Red (BR).

\(P(RB) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)
\(P(BR) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)

Sum of probabilities:
\(P(\text{different colours}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56}\)

Simplify the fraction:
\(\frac{30}{56} = \frac{15}{28}\)

M1 for finding probability of one combination (e.g., \(\frac{15}{56}\))
M1 for adding both combinations (e.g., \(\frac{15}{56} + \frac{15}{56}\))
A1 for \(\frac{15}{28}\) in its simplest form

First, find the composite function \(f(g(x))\):
\[f(g(x)) = 2\left(1 + \frac{6}{x}\right) - 5 = 2 + \frac{12}{x} - 5 = \frac{12}{x} - 3\]

Set this equal to 1:
\[\frac{12}{x} - 3 = 1\]
\[\frac{12}{x} = 4\]
\[x = 3\]

M1 for finding composite function expression \(2\left(1 + \frac{6}{x}\right) - 5\) or setting up \(g(x) = f^{-1}(1)\)
M1 for setting up equation \(\frac{12}{x} = 4\) or equivalent
A1 for \(x = 3\)

Using log laws, combine the terms on the left side:
\(3 \log_a 2 = \log_a(2^3) = \log_a 8\)

\(\log_a 8 + \log_a 5 - \log_a 10 = \log_a\left(\frac{8 \times 5}{10}\right)\)
\(= \log_a\left(\frac{40}{10}\right) = \log_a 4\)

The equation becomes:
\(\log_a 4 = 2\)

Convert to exponential form:
\(a^2 = 4\)

Since base \(a > 0\), \(a = 2\).

M1 for using power law: \(3 \log_a 2 = \log_a 8\)
M1 for combining the terms correctly to obtain \(\log_a 4 = 2\)
A1 for \(a = 2\)

(a) Continuing the patterns, the next terms are: Sequence P has a second difference of 2, so the next term is \(29 + 12 = 41\). Sequence Q has a common difference of 4, so the next term is \(15 + 4 = 19\). Sequence R has term values 2, 4, 8, 14, where differences are 2, 4, 6, so the next difference is 8, making the next term \(14 + 8 = 22\).
(b) Sequence Q is an arithmetic progression with first term \(a = 3\) and common difference \(d = 4\). The \(n\)-th term is \(3 + (n-1)4 = 4n - 1\).
(c) Sequence P is quadratic: \(an^2 + bn + c\). The second differences are constant at 2, so \(2a = 2 \Rightarrow a = 1\). Subtracting \(n^2\) from each term gives the sequence 4, 7, 10, 13, ..., which is linear with formula \(3n + 1\). Thus, the \(n\)-th term of Sequence P is \(n^2 + 3n + 1\).
(d) The \(n\)-th term of Sequence R is the difference between P and Q: \(P_n - Q_n = (n^2 + 3n + 1) - (4n - 1) = n^2 - n + 2\).

Part (a): [3 marks] - B1 for each correct next term.
Part (b): [2 marks] - M1 for recognizing arithmetic sequence with difference 4, A1 for \(4n - 1\).
Part (c): [3 marks] - M1 for showing constant second difference is 2, M1 for setting up linear system or subtracting \(n^2\), A1 for \(n^2 + 3n + 1\).
Part (d): [3 marks] - M1 for expressing \(R_n = P_n - Q_n\), M1 for correct substitution, A1 for complete algebraic proof leading to \(n^2 - n + 2\).

(a) Sketch shows a curve in the second quadrant going to infinity as \(x \to 0^-\), having a minimum, and a curve crossing the x-axis in the first/fourth quadrant going to negative infinity as \(x \to 0^+\).
(b) The function is undefined at \(x = 0\), so the vertical asymptote is \(x = 0\).
(c) Using the minimum feature on a GDC, the coordinates of the local minimum are approximately \((-1.14, 3.93)\).
(d) Set \(x^2 - \frac{3}{x} = 2x + 1\). Using GDC to find the intersection of the curve and the line yields \(x \approx 2.76\).
(e) From the graph, there is only one intersection with a horizontal line \(y = k\) when \(k\) is below the local minimum y-coordinate. Thus, \(k < 3.93\).

Part (a): [3 marks] - B1 for correct shape in negative x region, B1 for correct shape in positive x region, B1 for asymptotic behavior near y-axis.
Part (b): [1 mark] - B1 for \(x = 0\).
Part (c): [2 marks] - B1 for correct x-coordinate \(-1.14\) (3 s.f.), B1 for correct y-coordinate \(3.93\) (3 s.f.).
Part (d): [3 marks] - M1 for setting up equation \(x^2 - \frac{3}{x} = 2x + 1\), M1 for rearranging or using solver, A1 for \(2.76\) (accept 2.75 to 2.77).
Part (e): [2 marks] - M1 for identifying the local minimum boundary, A1 for \(k < 3.93\).

(a) Multiply coordinates of P, Q, and R by \(-2\): \(P'(-2, -4)\), \(Q'(-8, -4)\), \(R'(-8, -8)\).
(b) The mapping is \((x, y) \to (y, x)\). This is a reflection in the line \(y = x\).
(c) Translating the rotation center to origin: \((x-1, y-1)\). Rotating anticlockwise by \(90^\circ\): \((-(y-1), x-1)\). Translating back: \((2-y, x)\). Applying this: \(P(1,2) \to (0, 1)\), \(Q(4,2) \to (0, 4)\), \(R(4,4) \to (-2, 4)\).
(d) Add the vector to the coordinates of Q: \((4 + (-3), 2 + 5) = (1, 7)\).

Part (a): [3 marks] - B1 for each correct vertex coordinate set.
Part (b): [3 marks] - B1 for 'Reflection', B2 for 'line \(y = x\)' (B1 if only 'line' omitted but correct axis defined).
Part (c): [3 marks] - B1 for each correct vertex of Triangle W.
Part (d): [2 marks] - M1 for translation calculation, A1 for \((1, 7)\).

(a) Square both sides: \( u^2 = \frac{w - 2v}{v + 1} \).
Multiply by \((v+1)\): \( u^2 v + u^2 = w - 2v \).
Group \(v\) terms: \( u^2 v + 2v = w - u^2 \).
Factor out \(v\): \( v(u^2 + 2) = w - u^2 \).
Divide: \( v = \frac{w - u^2}{u^2 + 2} \).
(b) Factor the numerator: \( 3x^2 - 14x - 5 = (3x+1)(x-5) \).
Factor the denominator: \( 9x^2 - 1 = (3x-1)(3x+1) \).
Divide out common factor \((3x+1)\): \( \frac{x-5}{3x-1} \).
(c) Multiply terms by common denominator \((2x-3)(x+2)\):
\( 4(x+2) - 3(2x-3) = (2x-3)(x+2) \).
Expand: \( 4x + 8 - 6x + 9 = 2x^2 + x - 6 \).
Simplify: \( -2x + 17 = 2x^2 + x - 6 \Rightarrow 2x^2 + 3x - 23 = 0 \).
Use quadratic formula: \( x = \frac{-3 \pm \sqrt{9 - 4(2)(-23)}}{4} = \frac{-3 \pm \sqrt{193}}{4} \).
So \(x \approx 2.72\) or \(x \approx -4.22\).

Part (a): [4 marks] - M1 for squaring, M1 for cross-multiplying, M1 for collecting \(v\) terms, A1 for \(v = \frac{w - u^2}{u^2 + 2}\).
Part (b): [3 marks] - M1 for factoring numerator, M1 for factoring denominator, A1 for \(\frac{x-5}{3x-1}\).
Part (c): [4 marks] - M1 for eliminating denominators, M1 for simplifying to a quadratic equation, A1 for one correct solution, A1 for second correct solution.

(a) \( g(11) = \frac{11-1}{5} = 2 \). Then \( f(2) = 2(2) + 3 = 7 \).
(b) Let \( y = \frac{x-1}{5} \Rightarrow 5y = x-1 \Rightarrow x = 5y + 1 \). So \( g^{-1}(x) = 5x + 1 \).
(c) \( f(f(x)) = 2(2x+3) + 3 = 4x + 6 + 3 = 4x + 9 \).
(d) \( h(f(x)) = 3^{2x+3} = \frac{1}{27} = 3^{-3} \). Therefore, \( 2x+3 = -3 \Rightarrow 2x = -6 \Rightarrow x = -3 \).
(e) \( h(2) = 3^2 = 9 \). Set \( \frac{x-1}{5} = 9 \Rightarrow x-1 = 45 \Rightarrow x = 46 \).

Part (a): [2 marks] - M1 for evaluating \(g(11)\), A1 for 7.
Part (b): [2 marks] - M1 for valid method to change subject, A1 for \(5x + 1\).
Part (c): [2 marks] - M1 for substituting \(f(x)\) into itself, A1 for \(4x+9\).
Part (d): [3 marks] - M1 for equating bases (e.g. expressing 1/27 as base 3), M1 for setting exponents equal, A1 for \(x = -3\).
Part (e): [2 marks] - M1 for establishing \(g(x) = 9\), A1 for \(x = 46\).

(a) Row A increases by 4 each term: 17 + 4 = 21. Row B doubles each term: 24 * 2 = 48.
(b) Row A is an arithmetic sequence with first term 5 and difference 4: \(5 + (n-1)4 = 4n+1\).
(c) Row B is geometric with first term 3 and common ratio 2: \(3 \times 2^{n-1}\).
(d) Row C is the product of Row A and Row B: \( (4n+1) \times 3 \times 2^{n-1} = 3(4n+1)2^{n-1}\).
(e) Put \(n=10\) in Row C: \( 3(4(10)+1) \times 2^9 = 3(41) \times 512 = 123 \times 512 = 62976 \).

Part (a): [2 marks] - B1 for Row A next term, B1 for Row B next term.
Part (b): [2 marks] - M1 for linear pattern with difference 4, A1 for \(4n+1\).
Part (c): [2 marks] - M1 for geometric pattern with ratio 2, A1 for \(3 \times 2^{n-1}\).
Part (d): [2 marks] - M1 for multiplying Row A and Row B expressions, A1 for \(3(4n+1)2^{n-1}\).
Part (e): [2 marks] - M1 for substituting \(n=10\), A1 for 62976.

(a) Substitute \(y = 3x - 2\) into the second equation:
\( 2x^2 + (3x-2)^2 = 19 \)
\( 2x^2 + 9x^2 - 12x + 4 = 19 \)
\( 11x^2 - 12x - 15 = 0 \).
Use quadratic formula: \( x = \frac{12 \pm \sqrt{144 - 4(11)(-15)}}{22} = \frac{12 \pm \sqrt{804}}{22} \).
\( x_1 \approx 1.83 \Rightarrow y_1 = 3(1.83) - 2 \approx 3.50 \).
\( x_2 \approx -0.743 \Rightarrow y_2 = 3(-0.743) - 2 \approx -4.23 \).
(b) Let \( u = 2^x \). The equation becomes:
\( u^2 - 5u + 4 = 0 \Rightarrow (u-4)(u-1) = 0 \).
Thus \(u = 4\) or \(u = 1\).
If \( 2^x = 4 \Rightarrow x = 2 \).
If \( 2^x = 1 \Rightarrow x = 0 \).

Part (a): [6 marks] - M1 for substitution, M1 for expansion of bracket, M1 for simplifying to a 3-term quadratic, A1 for both x values correct, M1 for finding y values, A1 for final coordinate pairs correct to 3 s.f.
Part (b): [5 marks] - M1 for variable substitution (e.g. \(u = 2^x\)), M1 for quadratic equation setup, A1 for finding values of \(u\) (1 and 4), M1 for solving exponential equations, A1 for \(x=0\) and \(x=2\).

(a) \( S_1 = 3(1^2) - 1 = 2 \).
\( S_2 = 3(2^2) - 2 = 10 \).
\( S_3 = 3(3^2) - 3 = 24 \).
(b) \( u_1 = S_1 = 2 \).
\( u_2 = S_2 - S_1 = 10 - 2 = 8 \).
\( u_3 = S_3 - S_2 = 24 - 10 = 14 \).
(c) The terms 2, 8, 14 form an arithmetic sequence with common difference 6. The formula is \( u_n = 2 + (n-1)6 = 6n - 4 \).
Alternatively: \( u_n = S_n - S_{n-1} = (3n^2 - n) - (3(n-1)^2 - (n-1)) = 6n - 4 \).
(d) Set \( 6n - 4 = 296 \Rightarrow 6n = 300 \Rightarrow n = 50 \).

Part (a): [2 marks] - B1 for at least two correct, B2 for all three correct.
Part (b): [3 marks] - B1 for each correct term.
Part (c): [3 marks] - M1 for recognizing arithmetic progression or doing \(S_n - S_{n-1}\), A2 for \(6n-4\) (A1 for \(6n + k\)).
Part (d): [3 marks] - M1 for setting their expression equal to 296, M1 for solving, A1 for \(n = 50\).

(a) For Sequence \(P\), the terms increase by successive odd numbers: \(+5, +7, +9, +11, \dots\). The next term is \(30 + 13 = 43\).

For Sequence \(Q\), the terms double each time: \(48 \times 2 = 96\).

(b)(i) First differences: \(5, 7, 9, 11, \dots\)

Second differences: \(2, 2, 2, \dots\)

Since the second difference is constant, the \(n\)-th term is quadratic: \(an^2 + bn + c\) where \(a = 2 / 2 = 1\).

Let the term be \(n^2 + bn + c\).

For \(n = 1\): \(1^2 + b + c = -2 \implies b + c = -3\)

For \(n = 2\): \(2^2 + 2b + c = 3 \implies 2b + c = -1\)

Subtracting the first equation from the second:

\(b = 2\)

Substitute \(b = 2\) back: \(2 + c = -3 \implies c = -5\).

Thus, the expression is \(n^2 + 2n - 5\).

(ii) Sequence \(Q\) is geometric with first term \(a = 3\) and common ratio \(r = 2\).

The \(n\)-th term is \(3 \times 2^{n-1}\).

(iii) First differences: \(6, 10, 14, 18, \dots\)

Second differences: \(4, 4, 4, \dots\)

Thus, the quadratic term has \(a = 4 / 2 = 2\).

Let the term be \(2n^2 + bn + c\).

For \(n = 1\): \(2(1)^2 + b + c = 5 \implies b + c = 3\)

For \(n = 2\): \(2(2)^2 + 2b + c = 11 \implies 2b + c = 3\)

Subtracting: \(b = 0 \implies c = 3\).

Thus, the expression is \(2n^2 + 3\).

(c) Set \(n^2 + 2n - 5 = 435\):

\(n^2 + 2n - 440 = 0\)

\((n + 22)(n - 20) = 0\)

Since \(n > 0\), \(n = 20\).

So, it is the 20th term.

(a) [2 marks] B1 for 43, B1 for 96.

(b)(i) [3 marks] M1 for finding second difference of 2. M1 for setting up simultaneous equations or using formula methods. A1 for \(n^2 + 2n - 5\).

(b)(ii) [2 marks] M1 for recognizing geometric pattern of common ratio 2. A1 for \(3 \times 2^{n-1}\) or equivalent.

(b)(iii) [2 marks] M1 for finding second difference of 4 or recognizing quadratic term is \(2n^2\). A1 for \(2n^2 + 3\).

(c) [2 marks] M1 for setting their quadratic expression equal to 435 and solving. A1 for 20.

(a) Using a graphic display calculator (GDC) to plot \(y = x^3 - 3x^2 - x + 4\):

(i) The local maximum is located at \((-0.155, 4.08)\) (correct to 3 significant figures).

(ii) The local minimum is located at \((2.15, -2.08)\) (correct to 3 significant figures).

(b) The roots of \(f(x) = 0\) are the \(x\)-intercepts of the graph. From the GDC solver, they are:

\(x ≈ -1.11\), \(x ≈ 1.25\), \(x ≈ 2.86\) (each to 3 s.f.).

(c) Plotting both \(y = f(x)\) and \(y = 0.5x^2 + 1\) on the GDC, the intersection points represent solutions to \(f(x) = g(x)\):

\(x ≈ -0.942\), \(x ≈ 0.903\), \(x ≈ 3.54\) (each to 3 s.f.).

(d) For \(f(x) = k\) to have three distinct real solutions, the horizontal line \(y = k\) must cross the local maximum and minimum loops. Thus, \(k\) must be strictly between the \(y\)-value of the local minimum and the local maximum:

\(-2.08 < k < 4.08\).

(a)(i) [2 marks] B1 for \(x\)-coordinate \(-0.155\) (or accurate to 2 s.f.), B1 for \(y\)-coordinate \(4.08\).

(a)(ii) [2 marks] B1 for \(x\)-coordinate \(2.15\), B1 for \(y\)-coordinate \(-2.08\).

(b) [2 marks] B2 for all three solutions correct: \(-1.11\), \(1.25\), \(2.86\). (B1 for any two correct solutions, allow 1 d.p. equivalent if consistent).

(c) [3 marks] B1 for each correct intersection value to 3 s.f.: \(-0.942\), \(0.903\), \(3.54\).

(d) [2 marks] M1 for identifying bounds from local extrema. A1 for the correct range \(-2.08 < k < 4.08\) (allow FT from their local max/min values).

(a) The transformation \(y = h(x - 3) + 2\) represents a horizontal translation of 3 units to the right, and a vertical translation of 2 units upwards.

(i) New coordinates of \(A\): \((2 + 3, 6 + 2) = (5, 8)\).

(ii) New coordinates of \(B\): \((-1 + 3, -3 + 2) = (2, -1)\).

(b) The transformation \(y = -2h(x)\) stretches the graph vertically by a scale factor of 2, and reflects it in the \(x\)-axis (multiply \(y\)-coordinates by \(-2\)).

(i) Image of point \(A\): \((2, 6 \times -2) = (2, -12)\).

(ii) Image of point \(B\): \((-1, -3 \times -2) = (-1, 6)\).

(c)(i) The transformation \(y = h(2x)\) is a horizontal stretch with scale factor \(\frac{1}{2}\) (parallel to the \(x\)-axis).

(ii) The transformation \(y = h(-x)\) is a reflection in the \(y\)-axis (or reflection in the line \(x = 0\)).

(a)(i) [2 marks] B1 for \(x = 5\), B1 for \(y = 8\).

(a)(ii) [2 marks] B1 for \(x = 2\), B1 for \(y = -1\).

(b)(i) [2 marks] B1 for \(x = 2\), B1 for \(y = -12\).

(b)(ii) [2 marks] B1 for \(x = -1\), B1 for \(y = 6\).

(c)(i) [2 marks] B1 for 'stretch', B1 for 'scale factor \(\frac{1}{2}\) parallel to the \(x\)-axis' (or horizontal stretch).

(c)(ii) [1 mark] B1 for 'reflection in the \(y\)-axis' (or reflection in \(x = 0\)).

(a) \(S_2(4) = 1\times 2 + 2\times 3 + 3\times 4 + 4\times 5 = 2 + 6 + 12 + 20 = 40\).

(b) Using the formula with \(n = 4\):
\(S_2(4) = \frac{4(4+1)(4+2)}{3} = \frac{4 \times 5 \times 6}{3} = \frac{120}{3} = 40\).
This matches the calculated value.

M1 for writing out the sum: \(2 + 6 + 12 + 20\) or equivalent
A1 for 40
M1 for substituting \(n=4\) into the given formula: \(\frac{4 \times 5 \times 6}{3}\)
A1 for showing that the calculation equals 40 and matches part (a)

(a) For \(n = 3\):
\(S_3(3) = 1\times 2\times 3 + 2\times 3\times 4 + 3\times 4\times 5 = 6 + 24 + 60 = 90\).

(b) For \(n = 4\):
\(S_3(4) = S_3(3) + 4\times 5\times 6 = 90 + 120 = 240\).

B1 for writing the calculation for \(S_3(3)\): \(6 + 24 + 60\) or equivalent
A1 for \(S_3(3) = 90\)
M1 for writing the calculation for \(S_3(4)\): \(90 + 4\times 5\times 6\) or equivalent
A2 for \(S_3(4) = 240\)

(a) If \(S_3(n) = \frac{n(n+1)(n+2)(n+a)}{b}\):
Using \(n = 1\):
\(S_3(1) = \frac{1(2)(3)(1+a)}{b} = \frac{6(1+a)}{b} = 6 \implies 1+a = b\).
Using \(n = 2\):
\(S_3(2) = \frac{2(3)(4)(2+a)}{b} = \frac{24(2+a)}{b} = 30 \implies \frac{4(2+a)}{b} = 5 \implies 8 + 4a = 5b\).
Substitute \(b = 1+a\):
\(8 + 4a = 5(1+a) \implies 8 + 4a = 5 + 5a \implies a = 3\).
Then \(b = 4\).

(b) For \(n = 10\):
\(S_3(10) = \frac{10 \times 11 \times 12 \times 13}{4} = 10 \times 11 \times 3 \times 13 = 4290\).

M1 for setting up equations using \(n=1\) and/or \(n=2\)
A1 for finding \(a = 3\)
A1 for finding \(b = 4\)
M1 for substituting \(n = 10\) into their formula
A1 for \(4290\) (allow follow-through from incorrect values of \(a\) and \(b\))

(a) Following the pattern:
\(S_1(n) = \frac{n(n+1)}{2}\)
\(S_2(n) = \frac{n(n+1)(n+2)}{3}\)
\(S_3(n) = \frac{n(n+1)(n+2)(n+3)}{4}\)
In general, \(S_k(n) = \frac{n(n+1)(n+2)\dots(n+k)}{k+1}\).

(b) For \(k=4\):
\(S_4(n) = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}\).

(c) For \(n=8\):
\(S_4(8) = \frac{8 \times 9 \times 10 \times 11 \times 12}{5} = 19008\).

M1 for identifying the general pattern in the denominator \(k+1\)
A1 for the expression \(S_k(n) = \frac{n(n+1)\dots(n+k)}{k+1}\) or equivalent
B1 for \(S_4(n) = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}\)
M1 for substituting \(n=8\) into their expression for \(S_4(n)\)
A2 for 19008

(a) Substituting \(v_f = 100\) and \(k_j = 120\) into the given formula:
\(v = 100 \left(1 - \frac{k}{120}\right) = 100 - \frac{5}{6}k\).

(b) When \(k = 30\):
\(v = 100 - \frac{5}{6}(30) = 100 - 25 = 75\text{ km/h}\).

M1 for substituting \(v_f = 100\) and \(k_j = 120\) into the formula
A1 for \(v = 100 - \frac{5}{6}k\) or any equivalent simplified form
M1 for substituting \(k=30\) into their formula
A1 for \(75\)

(a) \(q = k \times v = k \left(100 - \frac{5}{6}k\right) = 100k - \frac{5}{6}k^2\).

(b) The equation \(q = 100k - \frac{5}{6}k^2\) is a quadratic curve opening downwards.
The maximum flow occurs at the vertex of this parabola.
The axis of symmetry is at \(k = -\frac{b}{2a} = -\frac{100}{2(-5/6)} = 60\).
So, \(k = 60\text{ vehicles/km}\) maximizes traffic flow.

M1 for multiplying their expression for \(v\) by \(k\)
A1 for \(q = 100k - \frac{5}{6}k^2\) or equivalent
M1 for identifying that the maximum of a quadratic occurs at its vertex
M1 for calculation of vertex position: \(\frac{-100}{2 \times (-5/6)}\) or equivalent
A1 for \(60\)

(a) Substituting the optimal density \(k = 60\) into the flow equation:
\(q_{\text{max}} = 100(60) - \frac{5}{6}(60)^2 = 6000 - 3000 = 3000\text{ vehicles/hour}\).

(b) Find the speed \(v\) when \(k = 60\):
\(v = 100 - \frac{5}{6}(60) = 50\text{ km/h}\).
Alternatively, using \(v = \frac{q}{k} = \frac{3000}{60} = 50\text{ km/h}\).

M1 for substituting \(k = 60\) into their expression for \(q\)
A1 for 3000
M1 for substituting \(k = 60\) into their expression for \(v\), or using \(v = q_{\text{max}} / 60\)
A2 for 50 (M1 for partial correct calculation)

(a) The transition point \(k_c\) occurs when the two speed functions are equal:
\(100\left(1 - \frac{k_c}{120}\right) = 80 \implies 1 - \frac{k_c}{120} = 0.8 \implies \frac{k_c}{120} = 0.2 \implies k_c = 24\).

(b) The traffic flow under this model is \(q = k \times v_{\text{new}}(k)\).
For \(0 \le k \le 24\), \(q = 80k\), which has a maximum value of \(80 \times 24 = 1920\text{ vehicles/hour}\).
For \(24 < k \le 120\), \(q = 100k - \frac{5}{6}k^2\), which has its overall peak at \(k = 60\) (from Question 6).
Since \(k = 60\) lies in the interval \(24 < k \le 120\), the maximum traffic flow remains \(q = 3000\text{ vehicles/hour}\).

M1 for setting \(100\left(1 - \frac{k_c}{120}\right) = 80\)
A1 for showing \(k_c = 24\) clearly
M1 for checking maximum flow for \(0 \le k \le 24\): \(q = 80 \times 24 = 1920\)
M1 for noting that the peak flow at \(k = 60\) (value \(3000\)) lies within \(24 < k \le 120\)
A2 for concluding \(q_{\text{max}} = 3000\text{ vehicles/hour}\) with complete reasoning

(a) To find the 6th pentagonal number, we substitute \(n = 6\) into the given formula:
\[P_6 = \frac{6(3 \times 6 - 1)}{2} = \frac{6(17)}{2} = 3 \times 17 = 51\]

(b) The sum of the first 5 pentagonal numbers is:
\[S_5 = 1 + 5 + 12 + 22 + 35 = 75\]
Adding the 6th pentagonal number, \(P_6\), we get the sum of the first 6 pentagonal numbers:
\[S_6 = S_5 + P_6 = 75 + 51 = 126\]

For part (a):
M1 for substituting \(n=6\) into the formula: \(\frac{6(18-1)}{2}\)
A1 for 51

For part (b):
M1 for adding their part (a) to 75 (sum of first 5 pentagonal numbers)
A1 for 126

We extend the difference table by keeping the third difference constant at 3:
- The 2nd differences for the next steps are:
\(13 + 3 = 16\) (for \(n=5\) to \(6\))
\(16 + 3 = 19\) (for \(n=6\) to \(7\))
- Using these 2nd differences, we find the 1st differences:
\(35 + 16 = 51\) (for \(n=5\) to \(6\))
\(51 + 19 = 70\) (for \(n=6\) to \(7\))
- Finally, we calculate the values of \(S_n\):
\(S_6 = S_5 + 51 = 75 + 51 = 126\)
\(S_7 = S_6 + 70 = 126 + 70 = 196\)

B1 for identifying that the third difference remains constant at 3.
M1 for finding the next two second differences: 16 and 19.
M1 for finding the next two first differences: 51 and 70.
M1 for calculating \(S_6 = 126\).
A1 for \(S_7 = 196\).

Since \(S_0 = 0\), we have:
\[d = 0\]
Now we substitute \(n = 1, 2, 3\) into \(S_n = an^3 + bn^2 + cn\):
1. For \(n = 1\):
\[a + b + c = 1 \quad \text{(Equation 1)}\]
2. For \(n = 2\):
\[8a + 4b + 2c = 6 \implies 4a + 2b + c = 3 \quad \text{(Equation 2)}\]
3. For \(n = 3\):
\[27a + 9b + 3c = 18 \implies 9a + 3b + c = 6 \quad \text{(Equation 3)}\]

Subtracting Equation 1 from Equation 2:
\[3a + b = 2 \quad \text{(Equation 4)}\]
Subtracting Equation 2 from Equation 3:
\[5a + b = 3 \quad \text{(Equation 5)}\]
Subtracting Equation 4 from Equation 5:
\[2a = 1 \implies a = \frac{1}{2}\]
Substitute \(a = \frac{1}{2}\) into Equation 4:
\[3\left(\frac{1}{2}\right) + b = 2 \implies b = \frac{1}{2}\]
Substitute \(a = \frac{1}{2}\) and \(b = \frac{1}{2}\) into Equation 1:
\[\frac{1}{2} + \frac{1}{2} + c = 1 \implies c = 0\]
Thus, the formula is:
\[S_n = \frac{1}{2}n^3 + \frac{1}{2}n^2 = \frac{n^2(n+1)}{2}\]

B1 for stating \(d = 0\).
M1 for setting up a system of three equations in \(a, b, c\).
M1 for using elimination or substitution to solve for \(a\) and \(b\).
A1 for finding \(a = \frac{1}{2}\), \(b = \frac{1}{2}\), and \(c = 0\).
A1 for showing that the formula simplifies to \(S_n = \frac{n^2(n+1)}{2}\).

(a) Substituting \(t = 0\):
\[E(0) = r = 12\]
Substituting \(t = 5\):
\[E(5) = 25p + 5q + 12 = 14.5 \implies 25p + 5q = 2.5 \implies 5p + q = 0.5 \quad \text{(Equation 1)}\]
Substituting \(t = 10\):
\[E(10) = 100p + 10q + 12 = 16.0 \implies 100p + 10q = 4 \implies 10p + q = 0.4 \quad \text{(Equation 2)}\]

Subtract Equation 1 from Equation 2:
\[5p = -0.1 \implies p = -0.02\]
Substitute \(p = -0.02\) into Equation 1:
\[5(-0.02) + q = 0.5 \implies -0.1 + q = 0.5 \implies q = 0.6\]
So the model is \(E(t) = -0.02t^2 + 0.6t + 12\).

(b) The peak of a quadratic function \(E(t) = pt^2 + qt + r\) occurs at the vertex:
\[t = -\frac{q}{2p} = -\frac{0.6}{2(-0.02)} = \frac{0.6}{0.04} = 15\]
Since \(t\) is the number of years since 2010, the peak occurs in the year:
\[2010 + 15 = 2025\]

For part (a):
B1 for \(r = 12\).
M1 for substituting \(t=5, 10\) to get two linear equations in \(p\) and \(q\).
A1 for \(p = -0.02\) and \(q = 0.6\).

For part (b):
M1 for finding the vertex time using \(t = -\frac{q}{2p}\).
A1 for year 2025.

(a) Substituting \(x = 5\) and \(y = 90.39\):
\[90.39 = 100 \cdot b^5\]
\[b^5 = 0.9039\]
\[b = (0.9039)^{1/5} \approx 0.980007\]
Thus, \(b = 0.98\) correct to 2 decimal places.

(b) Setting \(y = 50\) and \(b = 0.98\):
\[50 = 100 \cdot (0.98)^x\]
\[0.5 = (0.98)^x\]
Taking natural logarithms on both sides:
\[\ln(0.5) = x \ln(0.98)\]
\[x = \frac{\ln(0.5)}{\ln(0.98)} \approx \frac{-0.69315}{-0.02020} \approx 34.31\text{ years}\]
Correct to the nearest year, it takes 34 years.

For part (a):
M1 for substituting \(x = 5, y = 90.39\) and setting up \(b^5 = 0.9039\).
A1 for evaluating \(b \approx 0.98\).

For part (b):
M1 for setting up \(50 = 100 \cdot (0.98)^x\) (or \((0.98)^x = 0.5\)).
M1 for using logarithms to solve for \(x\): \(x = \frac{\log(0.5)}{\log(0.98)}\).
A1 for 34 years (accept 34).