2025 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

First differences: \(6, 10, 14\). Second differences: \(4, 4\). Since the second difference is constant, the sequence is quadratic with a term in \(a n^2\) where \(a = \frac{4}{2} = 2\). Subtracting \(2n^2\) from each term: \(3 - 2(1)^2 = 1\), \(9 - 2(2)^2 = 1\), \(19 - 2(3)^2 = 1\), \(33 - 2(4)^2 = 1\). The remaining sequence is constant at \(1\). Therefore, the \(n\)-th term is \(2n^2 + 1\).

M1 for finding first differences \(6, 10, 14\) and second difference \(4\), implying a term in \(2n^2\)
M1 for subtracting \(2n^2\) from the terms of the sequence
A1 for the correct final answer \(2n^2 + 1\)

Factorise the numerator: \(6x^2 + xy - 2y^2 = 6x^2 + 4xy - 3xy - 2y^2 = 2x(3x + 2y) - y(3x + 2y) = (2x - y)(3x + 2y)\). Factorise the denominator (difference of two squares): \(4x^2 - y^2 = (2x - y)(2x + y)\). Cancel the common factor \(2x - y\) to simplify: \(\frac{(2x - y)(3x + 2y)}{(2x - y)(2x + y)} = \frac{3x + 2y}{2x + y}\).

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

Let \(y = \frac{3x+1}{x-2}\). Multiply both sides by \(x-2\): \(y(x-2) = 3x+1 \implies xy - 2y = 3x + 1\). Rearrange to collect the \(x\) terms on one side: \(xy - 3x = 2y + 1\). Factorise \(x\): \(x(y - 3) = 2y + 1\). Divide by \(y-3\): \(x = \frac{2y+1}{y-3}\). Replace \(y\) with \(x\) to obtain the inverse function: \(f^{-1}(x) = \frac{2x+1}{x-3}\).

M1 for setting \(y = f(x)\) and multiplying by \(x-2\)
M1 for correctly rearranging to group and factorise the \(x\) terms
A1 for the correct final inverse function \(\frac{2x+1}{x-3}\)

Use the cosine rule: \(PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(PQR)\). Substitute the given values: \(PR^2 = 5^2 + 8^2 - 2(5)(8)\cos(60^\circ)\). Since \(\cos(60^\circ) = 0.5\), we get \(PR^2 = 25 + 64 - 80 \times 0.5 = 89 - 40 = 49\). Thus, \(PR = \sqrt{49} = 7\text{ cm}\).

M1 for substituting correct values into the cosine rule formula
M1 for using \(\cos(60^\circ) = 0.5\) to find \(PR^2 = 49\)
A1 for \(7\)

Substitute \(M_0 = 80\), \(t = 15\) and \(M = 10\) into the decay formula: \(10 = 80 \times 2^{-15k}\). Divide both sides by 80: \(\frac{10}{80} = 2^{-15k} \implies \frac{1}{8} = 2^{-15k}\). Write both sides with a base of 2: \(2^{-3} = 2^{-15k}\). Equate the powers: \(-3 = -15k \implies k = \frac{3}{15} = \frac{1}{5} = 0.2\).

M1 for substituting the given values correctly into the equation
M1 for simplifying the equation to the form \(2^{-3} = 2^{-15k}\) or similar
A1 for the correct value \(0.2\) or \(\frac{1}{5}\)

Multiply the numerator and denominator by the conjugate of the denominator, which is \(3 + \sqrt{3}\): \(\frac{6(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} = \frac{6(3 + \sqrt{3})}{3^2 - (\sqrt{3})^2} = \frac{6(3 + \sqrt{3})}{9 - 3} = \frac{6(3 + \sqrt{3})}{6} = 3 + \sqrt{3}\).

M1 for multiplying both numerator and denominator by \(3 + \sqrt{3}\)
M1 for simplifying the denominator to \(6\)
A1 for the correct simplified form \(3 + \sqrt{3}\)

Group the terms of the polynomial: \((2x^3 - 3x^2) - (8x - 12) = x^2(2x - 3) - 4(2x - 3)\). Factorise out the common bracket \((2x - 3)\): \((x^2 - 4)(2x - 3)\). Apply the difference of two squares to factorise \(x^2 - 4\): \((x - 2)(x + 2)(2x - 3)\).

M1 for grouping and factorising the first stage to \(x^2(2x - 3) - 4(2x - 3)\) or similar
M1 for identifying the quadratic factor \((x^2 - 4)\)
A1 for completely factorising to \((x - 2)(x + 2)(2x - 3)\)

A quadratic function with vertex \((h, k)\) can be written as \(y = a(x - h)^2 + k\). Here, the vertex is \((1, 8)\), so \(y = a(x - 1)^2 + 8\). Substitute the point \((3, 0)\) to find \(a\): \(0 = a(3 - 1)^2 + 8 \implies 0 = 4a + 8 \implies a = -2\). Expand the equation: \(y = -2(x - 1)^2 + 8 = -2(x^2 - 2x + 1) + 8 = -2x^2 + 4x - 2 + 8 = -2x^2 + 4x + 6\).

M1 for setting up the vertex form of the equation: \(y = a(x - 1)^2 + 8\)
M1 for substituting \((3, 0)\) and finding \(a = -2\)
M1 for expanding their equation to general form
A1 for the correct final function \(y = -2x^2 + 4x + 6\)

First, factorise the numerator and denominator:
Numerator:
$$2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)$$

Denominator:
$$x^2 - 9 = (x - 3)(x + 3)$$

Substitute these back into the fraction:
$$\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}$$

M1 for factorising the numerator correctly as $(2x+1)(x-3)$
M1 for factorising the denominator correctly as $(x-3)(x+3)$
A1 for the correct final simplified fraction $\frac{2x+1}{x+3}$

Use the given terms to set up a system of equations:
For $n = 1$:
$$u_1 = a(1)^2 + b(1) = a + b = 3$$

For $n = 2$:
$$u_2 = a(2)^2 + b(2) = 4a + 2b = 10 \implies 2a + b = 5$$

Subtract the first equation from the second equation:
$$(2a + b) - (a + b) = 5 - 3 \implies a = 2$$

Substitute $a = 2$ back into the first equation:
$$2 + b = 3 \implies b = 1$$

So, the $n$-th term is $u_n = 2n^2 + n$.

For $n = 5$:
$$u_5 = 2(5)^2 + 5 = 2(25) + 5 = 50 + 5 = 55$$

M1 for setting up the equations $a + b = 3$ and $4a + 2b = 10$
M1 for solving the system of equations to find $a = 2$ and $b = 1$
A1 for finding $u_5 = 55$

Using the Cosine Rule:
$$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)$$

Substitute the given values:
$$BC^2 = 6^2 + 8^2 - 2 \cdot 6 \cdot 8 \cdot \cos(60^\circ)$$
$$BC^2 = 36 + 64 - 96 \cdot 0.5$$
$$BC^2 = 100 - 48 = 52$$
$$BC = \sqrt{52} = 2\sqrt{13}\text{ cm}$$

M1 for correct substitution into the cosine rule: $6^2 + 8^2 - 2(6)(8)\cos(60^\circ)$
M1 for simplifying to $100 - 48$ or $52$
A1 for $2\sqrt{13}$ or $\sqrt{52}$

The general formula for exponential decay in this scenario is:
$$P(t) = P_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{3}}$$

Substitute $P_0 = 8000$ and $t = 12$:
$$P(12) = 8000 \cdot \left(\frac{1}{2}\right)^{\frac{12}{3}}$$
$$P(12) = 8000 \cdot \left(\frac{1}{2}\right)^4$$
$$P(12) = 8000 \cdot \frac{1}{16} = 500$$

M1 for identifying that the population halves 4 times (since $12 / 3 = 4$)
M1 for calculating $8000 \times (0.5)^4$ or $8000 / 16$
A1 for $500$

Let $y = \frac{3x+2}{x-1}$. Rearrange the equation to make $x$ the subject:
$$y(x - 1) = 3x + 2$$
$$yx - y = 3x + 2$$
$$yx - 3x = y + 2$$
$$x(y - 3) = y + 2$$
$$x = \frac{y+2}{y-3}$$

Therefore, the inverse function is:
$$f^{-1}(x) = \frac{x+2}{x-3}$$

M1 for setting $y = f(x)$ and clearing the fraction: $y(x-1) = 3x+2$
M1 for grouping $x$ terms together: $x(y-3) = y+2$
A1 for the correct inverse function expression $\frac{x+2}{x-3}$

First expand both parts of the expression:
$$(2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$$
$$(x - 2)^2 = x^2 - 4x + 4$$

Now subtract the second expanded expression from the first:
$$(2x^2 + 5x - 12) - (x^2 - 4x + 4) = 2x^2 + 5x - 12 - x^2 + 4x - 4$$
$$= x^2 + 9x - 16$$

M1 for correct expansion of $(2x-3)(x+4)$ to $2x^2 + 5x - 12$
M1 for correct expansion of $(x-2)^2$ to $x^2 - 4x + 4$
A1 for the correct fully simplified expression $x^2 + 9x - 16$

Express both sides as powers of the base 3:
$$9 = 3^2 \quad \text{and} \quad 27 = 3^3$$
$$(3^2)^{x-1} = (3^3)^{2x+3}$$
$$3^{2(x-1)} = 3^{3(2x+3)}$$

Equate the exponents:
$$2(x - 1) = 3(2x + 3)$$
$$2x - 2 = 6x + 9$$
$$-11 = 4x$$
$$x = -\frac{11}{4}$$

M1 for expressing both sides with base 3: $3^{2(x-1)} = 3^{3(2x+3)}$
M1 for equating exponents and forming the linear equation: $2x - 2 = 6x + 9$
A1 for $x = -\frac{11}{4}$ or $-2.75$

Using the vertex form of a quadratic equation:
$$y = a(x - h)^2 + k$$

Substitute the vertex $(1, 4)$:
$$y = a(x - 1)^2 + 4$$

Since the curve passes through $(0, 3)$, substitute $x = 0$ and $y = 3$:
$$3 = a(0 - 1)^2 + 4$$
$$3 = a(1) + 4 \implies a = -1$$

Now expand the equation to the standard form $y = ax^2 + bx + c$:
$$y = -1(x - 1)^2 + 4$$
$$y = -(x^2 - 2x + 1) + 4$$
$$y = -x^2 + 2x - 1 + 4$$
$$y = -x^2 + 2x + 3$$

Therefore, $a = -1$, $b = 2$, and $c = 3$.

M1 for writing down the vertex form: $y = a(x-1)^2 + 4$
M1 for substituting $(0,3)$ to find $a = -1$
A1 for expanding correctly to obtain $a = -1$, $b = 2$, $c = 3$

First, factorise the numerator:
$$3x^2 - 5x - 2 = (3x + 1)(x - 2)$$

Next, factorise the denominator using the difference of two squares:
$$9x^2 - 1 = (3x - 1)(3x + 1)$$

Substitute these back into the fraction:
$$\frac{(3x + 1)(x - 2)}{(3x - 1)(3x + 1)}$$

Cancel the common factor $(3x + 1)$ from the numerator and denominator:
$$\frac{x - 2}{3x - 1}$$

M1 for factorising the numerator into $(3x + 1)(x - 2)$
M1 for factorising the denominator into $(3x - 1)(3x + 1)$
A1 for the correct final simplified fraction

Find the first and second differences of the sequence:

Sequence: $4, \quad 11, \quad 22, \quad 37, \quad 56$
First differences: $7, \quad 11, \quad 15, \quad 19$
Second differences: $4, \quad 4, \quad 4$

Since the second differences are constant, the sequence is quadratic of the form $an^2 + bn + c$.
The coefficient $a$ is half of the second difference:
$$a = \frac{4}{2} = 2$$

Subtract $2n^2$ from each term of the sequence to find the linear part:
For $n = 1$: $4 - 2(1)^2 = 2$
For $n = 2$: $11 - 2(2)^2 = 3$
For $n = 3$: $22 - 2(3)^2 = 4$
For $n = 4$: $37 - 2(4)^2 = 5$

This linear sequence is $2, 3, 4, 5, \dots$, which has the $n$th term $n + 1$.

Combining these, the $n$th term of the original sequence is:
$$2n^2 + n + 1$$

M1 for finding second differences are constant and equal to $4$ (implied by $2n^2$)
M1 for subtracting $2n^2$ to find the linear sequence $n + 1$ or attempting to solve equations for $b$ and $c$
A1 for the correct $n$th term

Express all terms with a base of 2:
$$8 = 2^3 \implies 8^{x - 2} = (2^3)^{x - 2} = 2^{3(x - 2)}$$
$$4 = 2^2 \implies 4^{x + 3} = (2^2)^{x + 3} = 2^{2(x + 3)}$$
$$32 = 2^5$$

Rewrite the equation using base 2:
$$2^{3(x - 2)} = \frac{2^{2(x + 3)}}{2^5}$$

Apply the laws of indices to simplify the right-hand side:
$$2^{3x - 6} = 2^{2x + 6 - 5}$$
$$2^{3x - 6} = 2^{2x + 1}$$

Since the bases are equal, equate the exponents:
$$3x - 6 = 2x + 1$$
$$x = 7$$

M1 for expressing the terms correctly with a common base of 2
M1 for applying index laws correctly to obtain the linear equation $3x - 6 = 2x + 1$ (or equivalent)
A1 for the correct value of $x$

(a) Substitute t = 6 into the equation: V = 1500 * 1.052^6 ≈ 2034.39. To the nearest dollar, the value is $2034. (b) Set up the inequality: 1500 * 1.052^t > 2500, which simplifies to 1.052^t > 1.6667. Taking logarithms on both sides: t * ln(1.052) > ln(1.6667), which gives t > 10.07. Since t must be a complete number of years, the first such year is 11.

(a) M1 for 1500 * 1.052^6. A1 for 2034. (b) M1 for 1500 * 1.052^t = 2500. M1 for t ≈ 10.1 or trial values for t = 10 and t = 11. A1 for 11.

(a) Using the cosine rule: AC^2 = 8.4^2 + 11.2^2 - 2 * 8.4 * 11.2 * cos(62) ≈ 70.56 + 125.44 - 188.16 * 0.46947 ≈ 107.66. Thus, AC = √107.66 ≈ 10.38 cm, which rounds to 10.4 cm to 3 significant figures. (b) Using the area formula: Area = 0.5 * 8.4 * 11.2 * sin(62) ≈ 47.04 * 0.88295 ≈ 41.53 cm², which rounds to 41.5 cm² to 3 significant figures.

(a) M1 for substituting correctly into the cosine rule. A1 for AC^2 ≈ 107.7 or better. A1 for 10.4 (or 10.37 to 10.38). (b) M1 for substituting into the area formula 0.5 * a * b * sin(C). A1 for 41.5 (or 41.53 to 41.54).

(a) Substitute n = 5 into the formula for Sequence A: u_5 = 3(5^2) - 2 = 73. (b) The 2nd term of Sequence A is u_2 = 3(2^2) - 2 = 10. The 3rd term of Sequence B is v_3 = k * 2^3 = 8k. Since v_3 = 4 * u_2, we have 8k = 4 * 10, which simplifies to 8k = 40, so k = 5.

(a) M1 for 3(5^2) - 2. A1 for 73. (b) M1 for finding u_2 = 10. M1 for setting up the equation k * 2^3 = 4 * 10 (or their u_2). A1 for 5.

(a) First calculate f(3) = (2(3)+3)/(3-1) = 9/2 = 4.5. Then find ff(3) = f(4.5) = (2(4.5)+3)/(4.5-1) = 12/3.5 = 24/7 ≈ 3.43. (b) Let y = (2x+3)/(x-1). Then y(x-1) = 2x+3, which expands to xy - y = 2x+3. Rearranging to isolate x: xy - 2x = y+3, which factors to x(y-2) = y+3. Thus, x = (y+3)/(y-2), and the inverse function is f^-1(x) = (x+3)/(x-2).

(a) M1 for finding f(3) = 4.5 or 9/2. A1 for 24/7 or 3.43. (b) M1 for clearing fractions: y(x-1) = 2x+3. M1 for grouping and factoring x terms: x(y-2) = y+3. A1 for (x+3)/(x-2).

(a) Factor out the highest common factor, 3x: 15x^2 - 12xy = 3x(5x - 4y). (b) Use the difference of two squares formula, x^2 - y^2 = (x-y)(x+y): 4a^2 - 25b^2 = (2a - 5b)(2a + 5b). (c) Factorise the quadratic expression: 2x^2 + 5x - 3 = (2x - 1)(x + 3).

(a) M1 for factoring out x or 3. A1 for 3x(5x - 4y). (b) B1 for (2a - 5b)(2a + 5b). (c) M1 for (2x+p)(x+q) where pq=-3 or p+2q=5. A1 for (2x - 1)(x + 3).

(a) p * q = (4.5 * 10^7) * (1.2 * 10^-4) = (4.5 * 1.2) * 10^(7 - 4) = 5.4 * 10^3. (b) First, calculate p/q = (4.5 * 10^7)/(1.2 * 10^-4) = 3.75 * 10^11. Then, take the square root: √(3.75 * 10^11) = √(37.5 * 10^10) = √37.5 * 10^5 ≈ 6.12 * 10^5.

(a) M1 for 4.5 * 1.2 or for standard index manipulation. A1 for 5.4 * 10^3. (b) M1 for p/q = 3.75 * 10^11. M1 for attempting the square root of their division result. A1 for 6.12 * 10^5 (accept answers in range 6.12 * 10^5 to 6.124 * 10^5).

(a) Set f(x) = 0: x^3 - 3x^2 - 4x + 12 = 0. Factor by grouping: x^2(x-3) - 4(x-3) = 0, which gives (x^2 - 4)(x-3) = 0, so (x-2)(x+2)(x-3) = 0. Thus, x = -2, 2, 3. (b) Use a GDC to find the local minimum, or differentiate: f'(x) = 3x^2 - 6x - 4. Set f'(x) = 0, solving the quadratic equation gives x ≈ 2.53 (local minimum) or x ≈ -0.53 (local maximum). Substitute x ≈ 2.53 back into f(x) to find the y-coordinate: y ≈ (2.53)^3 - 3(2.53)^2 - 4(2.53) + 12 ≈ -1.13.

(a) M1 for setting f(x) = 0 and attempting to factorise or solve. A1 for -2, 2, 3. (b) M1 for 3x^2 - 6x - 4 = 0 (or using GDC to find the minimum point). A1 for x ≈ 2.53. A1 for y ≈ -1.13.

Use the vertex form of a quadratic: f(x) = a(x - h)^2 + k. Substituting the vertex (2, -5) gives f(x) = a(x - 2)^2 - 5. Substitute the point (4, 7) into this equation: 7 = a(4 - 2)^2 - 5, which simplifies to 7 = 4a - 5, so 4a = 12 and a = 3. Now expand the function: f(x) = 3(x - 2)^2 - 5 = 3(x^2 - 4x + 4) - 5 = 3x^2 - 12x + 12 - 5 = 3x^2 - 12x + 7. Thus, a = 3, b = -12, and c = 7.

M1 for writing vertex form a(x - 2)^2 - 5. M1 for substituting (4, 7) into vertex form. A1 for a = 3. M1 for expanding 3(x - 2)^2 - 5. A1 for b = -12 and c = 7.

Since the vertex is \((2, -5)\), the quadratic function can be written in vertex form as:
\(f(x) = a(x - 2)^2 - 5\)

Since it passes through the point \((5, 13)\), substitute \(x = 5\) and \(f(x) = 13\):
\(13 = a(5 - 2)^2 - 5\)
\(18 = a(3)^2\)
\(18 = 9a\)
\(a = 2\)

Substitute \(a = 2\) back into the vertex form:
\(f(x) = 2(x - 2)^2 - 5\)

Expand and simplify:
\(f(x) = 2(x^2 - 4x + 4) - 5\)
\(f(x) = 2x^2 - 8x + 8 - 5\)
\(f(x) = 2x^2 - 8x + 3\)

M1 for setting up vertex form: \(a(x - 2)^2 - 5\)
M1 for substituting \((5, 13)\) to find \(a\)
A1 for \(a = 2\)
M1 for expanding \(2(x - 2)^2 - 5\)
A1 for \(2x^2 - 8x + 3\)

First, find the length of the side \(AC\) using the Cosine Rule:
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\)
\(AC^2 = 8.4^2 + 11.5^2 - 2 \cdot 8.4 \cdot 11.5 \cdot \cos(58^\circ)\)
\(AC^2 = 70.56 + 132.25 - 193.2 \cdot \cos(58^\circ)\)
\(AC^2 \approx 202.81 - 102.38 = 100.43\)
\(AC \approx 10.02\text{ cm}\)

Next, use the Sine Rule to calculate angle \(ACB\):
\(\frac{\sin(ACB)}{AB} = \frac{\sin(ABC)}{AC}\)
\(\frac{\sin(ACB)}{8.4} = \frac{\sin(58^\circ)}{10.02}\)
\(\sin(ACB) = \frac{8.4 \cdot \sin(58^\circ)}{10.02} \approx 0.7109\)
\(ACB = \sin^{-1}(0.7109) \approx 45.3^\circ\)

M1 for correct substitution into Cosine Rule: \(8.4^2 + 11.5^2 - 2 \cdot 8.4 \cdot 11.5 \cdot \cos(58^\circ)\)
A1 for \(AC \approx 10.0\text{ cm}\) (or \(10.02\))
M1 for correct substitution into Sine Rule: \(\frac{\sin(ACB)}{8.4} = \frac{\sin(58^\circ)}{\text{their } AC}\)
M1 for \(\sin(ACB) \approx 0.71\) or \(\cos(ACB) \approx 0.70\)
A1 for \(45.3^\circ\) (accept range \(45.2^\circ\) to \(45.4^\circ\))

Find the differences between consecutive terms:
First differences: \(11 - 4 = 7\), \(22 - 11 = 11\), \(37 - 22 = 15\), \(56 - 37 = 19\)
Second differences: \(11 - 7 = 4\), \(15 - 11 = 4\), \(19 - 15 = 4\)

Since the second differences are constant, the sequence is quadratic and has the form \(an^2 + bn + c\).
The coefficient \(a = \frac{\text{second difference}}{2} = \frac{4}{2} = 2\).

Subtract \(2n^2\) from each term of the sequence to find the linear part:
For \(n=1\): \(4 - 2(1)^2 = 2\)
For \(n=2\): \(11 - 2(2)^2 = 3\)
For \(n=3\): \(22 - 2(3)^2 = 4\)
For \(n=4\): \(37 - 2(4)^2 = 5\)
For \(n=5\): \(56 - 2(5)^2 = 6\)

The resulting sequence is \(2, 3, 4, 5, 6, \dots\), which is represented by \(n + 1\).
Therefore, the \(n\)-th term is \(2n^2 + n + 1\).

M1 for first differences: \(7, 11, 15, 19\)
M1 for second difference: \(4\)
M1 for \(a = 2\) (coefficient of \(n^2\))
M1 for subtracting \(2n^2\) or finding the linear sequence \(n + 1\)
A1 for \(2n^2 + n + 1\)

Factorise each numerator and denominator expression:
1. \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\)
2. \(4x^2 - 1 = (2x - 1)(2x + 1)\) (difference of two squares)
3. \(2x^2 - x = x(2x - 1)\)
4. \(x^2 + 3x = x(x + 3)\)

Substitute the factorised forms back into the expression:
\(\frac{(2x - 1)(x + 3)}{(2x - 1)(2x + 1)} \times \frac{x(2x - 1)}{x(x + 3)} Cancel the common factors \)(2x - 1)\), \((x + 3)\), and \(x\):
- \((x + 3)\) cancels from numerator and denominator.
- \((2x - 1)\) cancels from numerator and denominator.
- \(x\) cancels from numerator and denominator.

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

M1 for factorising \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\)
M1 for factorising \(4x^2 - 1 = (2x - 1)(2x + 1)\)
M1 for factorising both \(2x^2 - x = x(2x - 1)\) and \(x^2 + 3x = x(x + 3)\)
M1 for canceling common factors
A1 for \(\frac{2x - 1}{2x + 1}\) or \((2x - 1)(2x + 1)^{-1}\)

First, find the growth factor \(k\):
\(2343 = 1200 \times k^3\)
\(k^3 = \frac{2343}{1200} = 1.9525\)
\(k = \sqrt[3]{1.9525} \approx 1.25\)

Now set up the equation for \(B = 10\,000\):
\(10\,000 = 1200 \times (1.25)^t\)
\((1.25)^t = \frac{10\,000}{1200} = 8.3333\)

Take logarithms on both sides to solve for \(t\):
\(t \log(1.25) = \log(8.3333)\)
\(t = \frac{\log(8.3333)}{\log(1.25)} \approx 9.50\) hours.

(If using the more precise value \(k \approx 1.2499\), then \(t = \frac{\log(8.3333)}{\log(1.2499)} \approx 9.51\) hours.)
Correct to 1 decimal place, the time is \(9.5\) hours.

M1 for \(2343 = 1200 \times k^3\)
A1 for \(k = 1.25\) (or \(1.2499\dots\))
M1 for \(10000 = 1200 \times (\text{their } k)^t\)
M1 for \(t = \frac{\log(8.3333)}{\log(\text{their } k)}\)
A1 for \(9.5\) (accept \(9.5\) or \(9.51\))

1. Vertical asymptote:
The denominator cannot be zero, so we solve:
\(2x + 4 = 0 \implies x = -2\)
Thus, \(p = -2\).

2. Horizontal asymptote:
As \(x \to \pm\infty\), terms without \(x\) become negligible:
\(y = \frac{3}{2} = 1.5\)
Thus, \(q = 1.5\).

3. \(x\)-axis intercept:
Set \(f(x) = 0\):
\(3x - 5 = 0 \implies x = \frac{5}{3} \approx 1.67\)
Thus, \(r = 1.67\) (or \(\frac{5}{3}\)).

4. \(y\)-axis intercept:
Set \(x = 0\):
\(f(0) = \frac{3(0) - 5}{2(0) + 4} = -\frac{5}{4} = -1.25\)
Thus, \(s = -1.25\) (or \(-\frac{5}{4}\)).

B1 for \(p = -2\)
B1 for \(q = 1.5\) (or \(\frac{3}{2}\))
M1 for setting numerator equal to zero to find \(r\)
A1 for \(r = 1.67\) (or \(\frac{5}{3}\))
B1 for \(s = -1.25\) (or \(-\frac{5}{4}\))

(a) To find the inverse function, let \(y = 3x - 2\).
Swap \(x\) and \(y\):
\(x = 3y - 2\)
Solve for \(y\):
\(3y = x + 2 \implies y = \frac{x + 2}{3}\)
Therefore, \(f^{-1}(x) = \frac{x + 2}{3}\).

(b) To find \(g(f(x))\), substitute \(f(x)\) into \(g(x)\):
\(g(f(x)) = \frac{5}{(3x - 2) - 1}\)
Simplify the denominator:
\(g(f(x)) = \frac{5}{3x - 3}\) (or \(\frac{5}{3(x - 1)}\))

M1 for attempting to find inverse (e.g. swap \(x\) and \(y\))
A1 for \(f^{-1}(x) = \frac{x + 2}{3}\) oe
M1 for substituting \(3x - 2\) into \(g(x)\) to get \(\frac{5}{(3x - 2) - 1}\)
M1 for simplifying the denominator
A1 for \(\frac{5}{3x - 3}\) oe

(a) (i) On a grid of width 10, adjacent squares to 25 are:
- Left: $25 - 1 = 24$
- Right: $25 + 1 = 26$
- Up: $25 - 10 = 15$
- Down: $25 + 10 = 35$
So, the five numbers are 15, 24, 25, 26, 35.
(ii) Sum = $15 + 24 + 25 + 26 + 35 = 125$.

(b) Let the center be $C$. On a grid of width 10, the numbers adjacent to $C$ are:
- Left: $C - 1$
- Right: $C + 1$
- Up: $C - 10$
- Down: $C + 10$
Adding all 5 squares:
Sum = $(C - 10) + (C - 1) + C + (C + 1) + (C + 10) = 5C$.

(c) We are given Sum = 385. Since Sum = $5C$, we set up:
$5C = 385 \implies C = 77$.

(a) (i) B1 for listing: 15, 24, 25, 26, 35 (any order).
(ii) B1 for 125.
(b) M1 for expressing adjacent squares as $C-1, C+1, C-10, C+10$.
M1 for setting up the sum equation.
A1 for simplifying to $5C$.
(c) M1 for $5C = 385$.
A1 for $C = 77$.

(a) For $k = 3$ and $C = 50$:
- Horizontal arm extends 3 units left and right: 47, 48, 49, 50, 51, 52, 53
- Vertical arm extends 3 units up and down (multiples of 10): 20, 30, 40, 50, 60, 70, 80
- Sum = $(4 \times 3 + 1) \times 50 = 13 \times 50 = 650$.

(b) The horizontal arm contains the numbers from $C - k$ to $C + k$. The sum of these $2k + 1$ numbers is:
Sum(horizontal) = $(2k + 1)C$
The vertical arm contains the numbers $C - 10i$ and $C + 10i$ for $i = 1, \dots, k$, plus $C$. The sum of these numbers is:
Sum(vertical) = $(2k + 1)C$
The total sum of the cross is the sum of both arms minus the center $C$ (which is counted twice):
Total Sum = $(2k + 1)C + (2k + 1)C - C = (4k + 1)C$.

(c) For $k = 4$, the sum is $(4(4) + 1)C = 17C$.
We are given $17C = 935 \implies C = 55$.

(a) B1 for horizontal arm: 47, 48, 49, 50, 51, 52, 53.
B1 for vertical arm: 20, 30, 40, 50, 60, 70, 80.
B1 for sum: 650.
(b) M1 for showing horizontal sum is $(2k+1)C$.
M1 for showing vertical sum is $(2k+1)C$.
A0.5 for subtracting the center $C$ once to obtain $(4k+1)C$.
(c) M1 for setting $17C = 935$.
A1 for $C = 55$.

(a) On a grid of width $W$, the horizontal arm is unaffected by the width, so its sum is still:
Sum(horizontal) = $(2k+1)C$
The vertical arm consists of the center $C$, and for each $i = 1, \dots, k$, the numbers $C - iW$ and $C + iW$.
The sum of each symmetric vertical pair is:
$(C - iW) + (C + iW) = 2C$
Since there are $k$ such pairs, the sum of all vertical squares is:
Sum(vertical) = $2kC + C = (2k+1)C$
Adding horizontal and vertical sums and subtracting $C$:
Total Sum = $(2k+1)C + (2k+1)C - C = (4k+1)C$.
This shows the sum is independent of $W$.

(b) (i) For a cross of size $k = 2$ with center $C = 42$:
- The horizontal arm has largest value $42 + 2 = 44$.
- The vertical arm has largest value $42 + 2W$.
Since $W \ge 1$, we have $42 + 2W \ge 44$, so the largest value overall is $42 + 2W$.
(ii) Given $42 + 2W = 82$:
$2W = 40 \implies W = 20$.

(a) M1 for horizontal arm sum is $(2k+1)C$.
M1 for expressing the vertical arm elements as $C - iW$ and $C + iW$.
M1 for summing the vertical arm to $(2k+1)C$.
A0.5 for final sum $(4k+1)C$.
(b) (i) B1.5 for $42 + 2W$.
(ii) M1.5 for setting up $42 + 2W = 82$.
A1 for $W = 20$.

(a) The 8 surrounding squares are:
- Horizontal: $C - 1$, $C + 1$
- Vertical: $C - W$, $C + W$
- Diagonal 1: $C - (W + 1)$, $C + (W + 1)$
- Diagonal 2: $C - (W - 1)$, $C + (W - 1)$

(b) Sum = $(C - 1) + (C + 1) + (C - W) + (C + W) + (C - W - 1) + (C + W + 1) + (C - W + 1) + (C + W - 1) + C$
Grouping the symmetric pairs around $C$:
$= (C - 1 + C + 1) + (C - W + C + W) + (C - W - 1 + C + W + 1) + (C - W + 1 + C + W - 1) + C$
$= 2C + 2C + 2C + 2C + C$
$= 9C$.

(c) A Double Cross has 8 arms of length $k$, which can be grouped into 4 opposite pairs.
For any direction offset $D$ (where $D$ can be $1$, $W$, $W+1$, or $W-1$):
An opposite pair of arms consists of squares of the form $C - iD$ and $C + iD$ for $i = 1, \dots, k$.
The sum of each opposite pair is:
$\sum_{i=1}^k (C - iD + C + iD) = 2kC$
Since there are 4 such opposite pairs, the sum of all 8 arms is:
$4 \times 2kC = 8kC$
Adding the center $C$ once, the total sum is:
$(8k + 1)C$.

(a) B1 for horizontal and vertical squares: $C-1, C+1, C-W, C+W$.
B1 for diagonal squares: $C-W-1, C+W+1, C-W+1, C+W-1$.
(b) M1 for writing the sum of all 9 squares.
M1 for pairing symmetric opposites to show the offset cancels out.
A1 for final simplification to $9C$.
(c) M1 for noting that opposite arms of length $k$ have a sum of $2kC$.
M1 for identifying 4 such pairs of opposite arms.
A0.5 for adding the center $C$ to obtain the final expression $(8k+1)C$.

(a) As \(t\) becomes very large, \((0.92)^t \to 0\). Thus, \(T \to 20\) °C. The room temperature is \(20\) °C.
(b) When \(t = 15\), \(T = 65(0.92)^{15} + 20 \approx 65(0.2863) + 20 = 18.61 + 20 = 38.61 \approx 38.6\) °C.
(c) (i) \(65(0.92)^t + 20 = 35\)
(ii) \(65(0.92)^t = 15 \implies 0.92^t = \frac{15}{65} \approx 0.23077\)
Taking logarithms on both sides:
\(t \log(0.92) = \log(0.23077) \implies t = \frac{\log(0.23077)}{\log(0.92)} \approx 17.6\) minutes.
(d) As \(t \to \infty\), the term \(65(0.92)^t\) approaches \(0\), meaning the temperature of the liquid approaches the room temperature of \(20\) °C but never drops below it.

(a) B1 for 20 (accept 20 °C).
(b) M1 for substituting t = 15 into the equation, i.e., 65(0.92)^15 + 20.
A1 for 38.6 (accept 38.61).
(c) (i) B1 for 65(0.92)^t + 20 = 35 or equivalent.
(ii) M1 for simplifying to 0.92^t = 15/65 (or 3/13).
M1 for taking logarithms to solve for t, e.g., t = log(3/13) / log(0.92).
A1 for 17.6 (accept 17.58 to 17.60).
(d) B1 for recognizing that (0.92)^t approaches 0.
B1 for stating that the temperature approaches 20 °C.
B1 for explaining that it will never drop below 20 °C.

(a) When \(t = 0\), \(h = 2.5\) metres. Thus, the height of the launch platform is \(2.5\) metres.
(b) The coefficient of \(t\) represents the initial velocity, which is \(24.5\) m/s.
(c) (i) Substitute \(t = 3\) into the equation:
\(h = -4.9(3)^2 + 24.5(3) + 2.5 = -44.1 + 73.5 + 2.5 = 31.9\) metres.
(ii) The maximum height occurs at the vertex, where \(t = -\frac{b}{2a} = -\frac{24.5}{2(-4.9)} = 2.5\) seconds.
Substitute \(t = 2.5\) into the formula:
\(h = -4.9(2.5)^2 + 24.5(2.5) + 2.5 = -30.625 + 61.25 + 2.5 = 33.125\) metres.
(d) Set \(h = 0\):
\(-4.9t^2 + 24.5t + 2.5 = 0\)
Using the quadratic formula:
\(t = \frac{-24.5 \pm \sqrt{24.5^2 - 4(-4.9)(2.5)}}{2(-4.9)} = \frac{-24.5 \pm \sqrt{649.25}}{-9.8}\)
Since \(t\) must be positive:
\(t = \frac{-24.5 - 25.480}{-9.8} \approx 5.10\) seconds.

(a) B1 for 2.5 (accept 2.5 m).
(b) B1 for 24.5 (accept 24.5 m/s).
(c) (i) M1 for substituting t = 3, i.e., -4.9(3)^2 + 24.5(3) + 2.5.
A1 for 31.9.
(ii) M1 for finding the time of maximum height, t = 2.5.
M1 for substituting their t = 2.5 into the equation.
A1 for 33.125 (accept 33.1 or 33.13).
(d) M1 for setting the equation to 0, i.e., -4.9t^2 + 24.5t + 2.5 = 0.
M1 for attempting to solve the quadratic equation using the quadratic formula or a graphics calculator.
A1 for 5.10.

(a) The amplitude \(|a|\) of the fluctuation is half the difference between the maximum and minimum values:
\(|a| = \frac{28 - 12}{2} = 8\)
Since January (\(x=1\)) is the coldest month (minimum temperature), and \(\cos(0) = 1\), the cosine wave must be inverted. Therefore, \(a = -8\).
The vertical shift \(c\) represents the average of the maximum and minimum values:
\(c = \frac{28 + 12}{2} = 20\).
(b) The period of the cosine function is 12 months.
Thus, \(12b = 360^\circ \implies b = 30^\circ\).
(c) The complete model is:
\(T = -8 \cos(30^\circ(x - 1)) + 20\)
(d) (i) For April (\(x = 4\)):
\(T = -8 \cos(30^\circ(4 - 1)) + 20 = -8 \cos(90^\circ) + 20 = -8(0) + 20 = 20\) °C.
(ii) For October (\(x = 10\)):
\(T = -8 \cos(30^\circ(10 - 1)) + 20 = -8 \cos(270^\circ) + 20 = -8(0) + 20 = 20\) °C.

(a) M1 for calculating the amplitude (28 - 12) / 2 = 8.
M1 for explaining why a = -8 (since January, x = 1, is the minimum).
A1 for showing c = (28 + 12) / 2 = 20.
(b) M1 for setting up the equation 12b = 360.
A1 for b = 30 (or b = 30°).
(c) B1 for T = -8 cos(30(x - 1)) + 20 or equivalent.
(d) (i) M1 for substituting x = 4 into their model.
A1 for 20 (accept 20 °C).
(ii) M1 for substituting x = 10 into their model.
A1 for 20 (accept 20 °C).