2024 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

To find the points of intersection, set the two equations equal to each other:
\[x^2 - 3x - 11 = 2x + 3\]
Rearrange to form a quadratic equation equal to zero:
\[x^2 - 3x - 2x - 11 - 3 = 0\]
\[x^2 - 5x - 14 = 0\]
Factorise the quadratic expression:
\[(x - 7)(x + 2) = 0\]
This gives:
\[x = 7 \quad \text{and} \quad x = -2\]

M1 for setting the expressions equal: \(x^2 - 3x - 11 = 2x + 3\)
M1 for writing in the form \(x^2 - 5x - 14 = 0\) (or equivalent)
A1 for both solutions: \(-2\) and \(7\)

Multiply the entire equation by the common denominator \(x(x-1)\):
\[6x - 4(x-1) = 1 \cdot x(x-1)\]
Expand both sides:
\[6x - 4x + 4 = x^2 - x\]
\[2x + 4 = x^2 - x\]
Rearrange into standard quadratic form:
\[x^2 - 3x - 4 = 0\]
Factorise the quadratic equation:
\[(x - 4)(x + 1) = 0\]
This gives:
\[x = 4 \quad \text{or} \quad x = -1\]

M1 for eliminating denominators correctly: \(6x - 4(x-1) = x(x-1)\) (or equivalent)
M1 for rearranging to standard quadratic form: \(x^2 - 3x - 4 = 0\)
A1 for \(x = 4\) and \(x = -1\)

Factorise the numerator:
\[2x^2 - 5x - 3 = (2x + 1)(x - 3)\]
Factorise the denominator using the difference of two squares:
\[4x^2 - 1 = (2x - 1)(2x + 1)\]
Now write as a simplified fraction by cancelling the common factor \((2x + 1)\):
\[\frac{(2x+1)(x-3)}{(2x-1)(2x+1)} = \frac{x-3}{2x-1}\]

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

Find the first and second differences of the sequence:
Terms: \(4, \quad 11, \quad 22, \quad 37, \quad 56\)
First differences: \(7, \quad 11, \quad 15, \quad 19\)
Second differences: \(4, \quad 4, \quad 4\)
Since the second difference is constant, the sequence is quadratic. The coefficient of \(n^2\) is half of the second difference: \(a = \frac{4}{2} = 2\).
Subtract \(2n^2\) from each term:
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 \(n=4\): \(37 - 2(4)^2 = 5\)
The remaining linear sequence is \(2, 3, 4, 5, \dots\), which has the \(n\)-th term of \(n + 1\).
Combining these gives the complete \(n\)-th term: \(2n^2 + n + 1\).

M1 for finding the second difference is 4 (or establishing the \(2n^2\) term)
M1 for attempting to find the linear part by subtracting \(2n^2\) from terms or using simultaneous equations
A1 for the correct expression: \(2n^2 + n + 1\)

Express both bases as powers of 3:
\[27 = 3^3 \implies 27^{2x - 1} = (3^3)^{2x - 1} = 3^{3(2x - 1)} = 3^{6x - 3}\]
\[\frac{1}{9} = 3^{-2} \implies \left(\frac{1}{9}\right)^{x - 4} = (3^{-2})^{x - 4} = 3^{-2(x - 4)} = 3^{-2x + 8}\]
Equating the indices since the bases are equal:
\[6x - 3 = -2x + 8\]
Solve for \(x\):
\[8x = 11\]
\[x = \frac{11}{8} = 1.375\]

M1 for expressing both sides correctly as powers of 3: \(3^{3(2x-1)}\) and \(3^{-2(x-4)}\) (or equivalent)
M1 for equating the exponents: \(3(2x - 1) = -2(x - 4)\) (or equivalent)
A1 for \(\frac{11}{8}\) or \(1.375\)

First, calculate the numerator:
\[(3.2 \times 1.5) \times 10^{6 + (-3)} = 4.8 \times 10^3\]
Next, divide by the denominator:
\[\frac{4.8 \times 10^3}{8.0 \times 10^{-2}} = \left(\frac{4.8}{8.0}\right) \times 10^{3 - (-2)}\]
\[= 0.6 \times 10^5\]
Convert to standard form:
\[= 6 \times 10^4\]

M1 for simplifying the numerator to \(4.8 \times 10^3\) (or \(4800\))
M1 for dividing to find \(60000\) or \(0.6 \times 10^5\)
A1 for \(6 \times 10^4\) (or \(6.0 \times 10^4\))

First, calculate the arc length of the sector:
\[\text{Arc length} = \frac{135}{360} \times 2 \times \pi \times 8\]
\[\text{Arc length} = \frac{3}{8} \times 16\pi = 6\pi \approx 18.85\text{ cm}\]
The perimeter of the sector is the arc length plus two radii:
\[\text{Perimeter} = \text{Arc length} + 2r = 6\pi + 2(8)\]
\[\text{Perimeter} = 18.85 + 16 = 34.85\text{ cm}\]
To 3 significant figures, this is \(34.8\text{ cm}\).

M1 for formula or calculation for arc length: \(\frac{135}{360} \times 2 \times \pi \times 8\) (approx. \(18.8\))
M1 for adding two radii to their arc length: \(\text{arc length} + 16\)
A1 for \(34.8\) (accept answers in the range \([34.8, 34.9]\))

The total number of beads is \(5 + 4 = 9\).
There are two ways to get different colours:
1. First red, then blue:
\[P(\text{Red, then Blue}) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}\]
2. First blue, then red:
\[P(\text{Blue, then Red}) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72}\]
Sum these probabilities to find the total probability of choosing different colours:
\[P(\text{Different}) = \frac{20}{72} + \frac{20}{72} = \frac{40}{72} = \frac{5}{9} \approx 0.556\]

M1 for correct probability of one combined outcome, e.g., \(\frac{5}{9} \times \frac{4}{8}\) or \(\frac{4}{9} \times \frac{5}{8}\) (or \(\frac{20}{72}\))
M1 for adding both valid combinations: \(\left(\frac{5}{9} \times \frac{4}{8}\right) + \left(\frac{4}{9} \times \frac{5}{8}\right)\)
A1 for \(\frac{5}{9}\) (or equivalent fraction, or \(0.556\) or \(0.5555...\))

To find the stationary points, we first differentiate \(y = 2x^3 - 9x^2 - 24x + 7\) with respect to \(x\):
\(\frac{dy}{dx} = 6x^2 - 18x - 24\)

At stationary points, \(\frac{dy}{dx} = 0\):
\(6x^2 - 18x - 24 = 0\)

Divide the entire equation by 6:
\(x^2 - 3x - 4 = 0\)

Factorise the quadratic expression:
\((x - 4)(x + 1) = 0\)

This gives \(x = -1\) and \(x = 4\).

In ascending order, the \(x\)-coordinates are \(-1, 4\).

M1 for differentiating to get \(6x^2 - 18x - 24\) (at least two terms correct)
M1 for setting their derivative to 0 and attempting to solve the quadratic equation
A1 for both \(x = -1\) and \(x = 4\) (or -1, 4)

Substitute \(y = x - 3\) into the second equation:
\(x^2 + (x - 3)^2 = 29\)
\(x^2 + x^2 - 6x + 9 = 29\)
\(2x^2 - 6x - 20 = 0\)

Divide the entire equation by 2:
\(x^2 - 3x - 10 = 0\)

Factorise the quadratic equation:
\((x - 5)(x + 2) = 0\)

So \(x = 5\) or \(x = -2\).

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

So the solutions are \(x = 5, y = 2\) and \(x = -2, y = -5\).

M1 for substitution of \(y = x - 3\) to obtain a quadratic in \(x\): \(2x^2 - 6x - 20 = 0\) (or equivalent)
M1 for solving their quadratic to find \(x = 5\) and \(x = -2\)
A1 for both correct pairs of solutions: \(x = 5, y = 2\) and \(x = -2, y = -5\) (or written as coordinates)

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

Factorise the denominator as a difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

Simplify the fraction by cancelling the common factor \((2x + 1)\):
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} =
\frac{x - 3}{2x - 1}\)

M1 for factorising numerator: \((2x + 1)(x - 3)\)
M1 for factorising denominator: \((2x - 1)(2x + 1)\)
A1 for final answer \(\frac{x - 3}{2x - 1}\) or equivalent fraction

The gradient of the given line is \(m_1 = \frac{2}{3}\).

Since the lines are perpendicular, the product of their gradients is \(-1\):
\(m_2 = -\frac{1}{2/3} = -\frac{3}{2} = -1.5\)

The equation of the perpendicular line is \(y = -1.5x + c\).

Substitute the point \((4, 7)\) into this equation:
\(7 = -1.5(4) + c\)
\(7 = -6 + c\)
\(c = 13\)

Therefore, the equation of the line is \(y = -1.5x + 13\).

M1 for finding perpendicular gradient = \(-1.5\) (or \(-\frac{3}{2}\))
M1 for substituting \((4, 7)\) into \(y = (their\ m)x + c\) to find \(c\)
A1 for \(y = -1.5x + 13\) (or \(y = -\frac{3}{2}x + 13\))

First, use the reciprocal property of negative indices:
\(\left( \frac{64x^6}{y^3} \right)^{-\frac{2}{3}} = \left( \frac{y^3}{64x^6} \right)^{\frac{2}{3}}\)

Now, apply the power of \(\frac{2}{3}\) to each component of the fraction:
- For the numerator: \((y^3)^{\frac{2}{3}} = y^{3 \times \frac{2}{3}} = y^2\)
- For the denominator:
\(64^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16\)
\((x^6)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4\)

Combining these, we get:
\(\frac{y^2}{16x^4}\)

M1 for dealing with negative index or fractional index of 64 (e.g. \(16\) or \(\frac{1}{16}\) seen)
M1 for a correct power of \(x\) or \(y\) (e.g. \(x^4\) or \(y^2\) in a relevant position)
A1 for \(\frac{y^2}{16x^4}\) (or \(\frac{1}{16}x^{-4}y^2\))

The total number of balls is \(5 + 3 = 8\).

There are two ways to select balls of different colours:
1) Red first, then Blue:
\(P(\text{Red, Blue}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)

2) Blue first, then Red:
\(P(\text{Blue, Red}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)

Add the two probabilities together:
\(P(\text{different colours}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\)

M1 for calculating the probability of one combination correctly, e.g. \(\frac{5}{8} \times \frac{3}{7}\)
M1 for adding the two appropriate products, e.g. \(\frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7}\)
A1 for \(\frac{15}{28}\) (or equivalent fraction, or \(0.536\) or \(0.5357...\))

The diagonal of the base, \(AC\), is calculated using Pythagoras' theorem in triangle \(ABC\):
\(AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}\)

The angle between the diagonal line \(AG\) and the base \(ABCD\) is the angle \(\angle GAC\).

In the right-angled triangle \(ACG\), \(CG = 5\text{ cm}\) (opposite side) and \(AC = 10\text{ cm}\) (adjacent side):
\(\tan(\angle GAC) = \frac{CG}{AC} = \frac{5}{10} = 0.5\)

\(\angle GAC = \tan^{-1}(0.5) \approx 26.565^\circ\)

To 1 decimal place, the angle is \(26.6^\circ\).

M1 for \(AC^2 = 8^2 + 6^2\) or \(AC = 10\)
M1 for \(\tan(\theta) = \frac{5}{their\ AC}\) (or equivalent trigonometric ratio)
A1 for \(26.6\) (accept \(26.56...\))

The perimeter of a sector of a circle is made up of two radii and the arc length \(L\):
\(P = 2r + L\)
\(25 = 2(9) + L\)
\(25 = 18 + L\)
\(L = 7\text{ cm}\)

Now use the arc length formula to find the sector angle \(x\):
\(L = \frac{x}{360} \times 2\pi r\)
\(7 = \frac{x}{360} \times 2 \times \pi \times 9\)
\(7 = \frac{18\pi x}{360}\)
\(7 = \frac{\pi x}{20}\)
\(x = \frac{140}{\pi} \approx 44.563^\circ\)

To 1 decimal place, the sector angle is \(44.6^\circ\).

M1 for finding arc length = \(25 - 2 \times 9\) or \(7\)
M1 for \(\frac{x}{360} \times 2 \times \pi \times 9 = their\ 7\)
A1 for \(44.6\) (accept \(44.56...\))

To find the stationary points, we first find the derivative: \(\frac{dy}{dx} = 6x^2 - 18x - 24\). Set the derivative to 0: \(6x^2 - 18x - 24 = 0\) which simplifies to \(x^2 - 3x - 4 = 0\), and factorises to \((x - 4)(x + 1) = 0\). This gives \(x = 4\) or \(x = -1\). To determine which is the minimum, we use the second derivative: \(\frac{d^2y}{dx^2} = 12x - 18\). For \(x = 4\): \(\frac{d^2y}{dx^2} = 12(4) - 18 = 30 > 0\), which confirms a local minimum. Finding the corresponding \(y\)-value: \(y = 2(4)^3 - 9(4)^2 - 24(4) + 7 = 128 - 144 - 96 + 7 = -105\). Thus, the coordinates of the local minimum are \((4, -105)\).

M1 for correctly differentiating to get at least two terms of \(6x^2 - 18x - 24\); M1 for setting \(6x^2 - 18x - 24 = 0\) and solving to find \(x = 4\); A1 for the correct coordinates \((4, -105)\).

Multiply both sides of the equation by the common denominator \((x + 2)(x - 1)\) to get: \(3(x - 1) + 1(x + 2) = 2(x + 2)(x - 1)\). Expanding both sides yields: \(3x - 3 + x + 2 = 2(x^2 + x - 2)\), which simplifies to \(4x - 1 = 2x^2 + 2x - 4\). Rearranging into standard quadratic form gives: \(2x^2 - 2x - 3 = 0\). Applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) gives: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4(2)(-3)}}{2(2)} = \frac{2 \pm \sqrt{28}}{4}\). Calculating the two solutions gives: \(x = \frac{2 + 5.2915}{4} \approx 1.82\) and \(x = \frac{2 - 5.2915}{4} \approx -0.82\).

M1 for clearing fractions correctly to get \(3(x - 1) + (x + 2) = 2(x + 2)(x - 1)\); M1 for obtaining the simplified quadratic equation \(2x^2 - 2x - 3 = 0\); A1 for both correct answers \(1.82\) and \(-0.82\).

First, multiply both sides by \((2 - p)\) to eliminate the fraction: \(T(2 - p) = 3p + 5\). Expand the left-hand side: \(2T - Tp = 3p + 5\). Collect all terms containing \(p\) on one side and the rest on the other: \(2T - 5 = 3p + Tp\). Factorise \(p\) on the right-hand side: \(2T - 5 = p(3 + T)\). Divide both sides by \(3 + T\) to solve for \(p\): \(p = \frac{2T - 5}{T + 3}\).

M1 for multiplying by \(2 - p\) to get \(T(2 - p) = 3p + 5\); M1 for collecting terms containing \(p\) on one side and factoring out \(p\) to get \(p(3 + T) = 2T - 5\); A1 for the correct final answer \(p = \frac{2T - 5}{T + 3}\) or equivalent.

Set the two equations equal to each other to find the intersection points: \(3x - 1 = 2x^2 - 5x + 5\). Rearranging into standard quadratic form gives: \(2x^2 - 8x + 6 = 0\). Dividing the entire equation by 2 yields: \(x^2 - 4x + 3 = 0\). Factorising the quadratic gives: \((x - 1)(x - 3) = 0\), which means \(x = 1\) or \(x = 3\). Substituting these back into the linear equation \(y = 3x - 1\) gives the corresponding \(y\)-values: For \(x = 1\), \(y = 3(1) - 1 = 2\); For \(x = 3\), \(y = 3(3) - 1 = 8\). Thus, the points of intersection are \((1, 2)\) and \((3, 8)\).

M1 for equating equations to form a quadratic, e.g. \(2x^2 - 8x + 6 = 0\); M1 for finding both \(x\)-values: \(x = 1\) and \(x = 3\); A1 for both correct coordinate pairs \((1, 2)\) and \((3, 8)\).

From the first equation, express \(y\) in terms of \(x\): \(y = 5 - x\). Substitute this into the second equation: \(x^2 + (5 - x)^2 = 17\). Expanding the bracket gives: \(x^2 + 25 - 10x + x^2 = 17\), which simplifies to \(2x^2 - 10x + 8 = 0\). Dividing by 2 gives: \(x^2 - 5x + 4 = 0\). Factorising gives: \((x - 1)(x - 4) = 0\), so \(x = 1\) or \(x = 4\). If \(x = 1\), then \(y = 5 - 1 = 4\). If \(x = 4\), then \(y = 5 - 4 = 1\). Thus, the solutions are \(x = 1, y = 4\) and \(x = 4, y = 1\).

M1 for correctly substituting \(y = 5 - x\) into the quadratic equation; M1 for simplifying to a 3-term quadratic and solving to find two values of \(x\) (or \(y\)), e.g. 1 and 4; A1 for the correct solutions: \(x = 1, y = 4\) and \(x = 4, y = 1\).

First, factorise the numerator: \(2x^2 - 7x - 15 = 2x^2 - 10x + 3x - 15 = 2x(x - 5) + 3(x - 5) = (2x + 3)(x - 5)\). Next, factorise the denominator using the difference of two squares: \(x^2 - 25 = (x - 5)(x + 5)\). Now, divide the factorised forms and cancel the common factor \((x - 5)\): \(\frac{(2x + 3)(x - 5)}{(x - 5)(x + 5)} = \frac{2x + 3}{x + 5}\).

M1 for factorising the numerator: \((2x + 3)(x - 5)\); M1 for factorising the denominator: \((x - 5)(x + 5)\); A1 for the correct fully simplified fraction \(\frac{2x + 3}{x + 5}\).

First, rewrite the equation as: \(y = 8x^{-1} + x^2\). Differentiate with respect to \(x\) to find the gradient function: \(\frac{dy}{dx} = -8x^{-2} + 2x = -\frac{8}{x^2} + 2x\). To find the gradient at \(x = 4\), substitute \(x = 4\) into the derivative: \(\frac{dy}{dx} = -\frac{8}{4^2} + 2(4) = -\frac{8}{16} + 8 = -0.5 + 8 = 7.5\).

M1 for writing the first term as \(8x^{-1}\) and attempting differentiation (at least one term correct); M1 for the correct derivative \(-8x^{-2} + 2x\); A1 for substituting \(x = 4\) to obtain \(7.5\) (or equivalent fraction).

(a)(i) When \( x = 0 \), \( y = 0^3 - 4(0)^2 - 2(0) + 8 = 8 \). So the \( y \)-intercept is \( (0, 8) \).
(ii) When \( y = 0 \), \( x^3 - 4x^2 - 2x + 8 = 0 \).
Factoring by grouping: \( x^2(x - 4) - 2(x - 4) = 0 \implies (x^2 - 2)(x - 4) = 0 \).
This gives \( x = 4 \) or \( x = \pm\sqrt{2} \approx \pm 1.414 \).
So the \( x \)-intercepts are \( (4, 0) \), \( (1.41, 0) \), and \( (-1.41, 0) \).

(b) To find the turning points, set \( f'(x) = 0 \):
\( f'(x) = 3x^2 - 8x - 2 = 0 \).
Using the quadratic formula:
\( x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-2)}}{2(3)} = \frac{8 \pm \sqrt{64 + 24}}{6} = \frac{8 \pm \sqrt{88}}{6} \).
\( x_1 = \frac{8 + 9.3808}{6} \approx 2.897 \)
\( x_2 = \frac{8 - 9.3808}{6} \approx -0.230 \).

Substitute these back into \( f(x) \) to find the \( y \)-coordinates:
For \( x \approx 2.897 \):
\( y = (2.897)^3 - 4(2.897)^2 - 2(2.897) + 8 \approx -7.051 \).
For \( x \approx -0.230 \):
\( y = (-0.230)^3 - 4(-0.230)^2 - 2(-0.230) + 8 \approx 8.236 \).
So the turning points are \( (2.90, -7.05) \) and \( (-0.230, 8.24) \).

(c) We are given \( x^3 - 4x^2 - 5x + 10 = 0 \).
We want to write this in terms of \( f(x) = x^3 - 4x^2 - 2x + 8 \):
\( (x^3 - 4x^2 - 2x + 8) - 3x + 2 = 0 \implies f(x) = 3x - 2 \).
So the straight line equation is \( y = 3x - 2 \).

(d) For the equation \( f(x) = k \) to have three distinct real solutions, the horizontal line \( y = k \) must intersect the curve at three distinct points. This occurs when \( k \) lies strictly between the \( y \)-coordinates of the two turning points:
\( -7.05 < k < 8.24 \).

(a)(i) B1: For \( (0, 8) \).
(ii) M1: For attempting to factor the cubic by grouping or finding one root.
A1: For root \( x = 4 \).
A1: For roots \( x = \pm\sqrt{2} \) (or 1.41, -1.41).

(b) M1: For differentiating to get \( 3x^2 - 8x - 2 \).
M1: For setting derivative to 0 and solving using quadratic formula.
A1: For \( x \approx 2.90 \) and \( x \approx -0.230 \).
M1: For substituting at least one of their \( x \)-values into the original cubic function.
A1: For final coordinates \( (2.90, -7.05) \) and \( (-0.230, 8.24) \).

(c) M1: For equating \( x^3 - 4x^2 - 2x + 8 = mx + c \) and comparing coefficients to reach \( mx + c \).
A1: For \( y = 3x - 2 \).

(d) M1: For identifying that \( k \) must lie between the \( y \)-values of the turning points.
A1: For \( -7.05 < k < 8.24 \) (accept answers consistent with their turning points from part b).

(a) Time taken for the first part: \( t_1 = \frac{12}{x} \)
Time taken for the second part: \( t_2 = \frac{15}{x - 2} \)
Total time: \( \frac{12}{x} + \frac{15}{x - 2} = 3.5 \)
Multiply by \( x(x - 2) \) to clear fractions:
\( 12(x - 2) + 15x = 3.5x(x - 2) \)
\( 12x - 24 + 15x = 3.5(x^2 - 2x) \)
\( 27x - 24 = 3.5x^2 - 7x \)
Multiply the entire equation by 2 to clear decimals:
\( 54x - 48 = 7x^2 - 14x \)
Rearrange to standard quadratic form:
\( 7x^2 - 68x + 48 = 0 \) (Showed)

(b) Using the quadratic formula:
\( x = \frac{68 \pm \sqrt{(-68)^2 - 4(7)(48)}}{2(7)} \)
\( x = \frac{68 \pm \sqrt{4624 - 1344}}{14} \)
\( x = \frac{68 \pm \sqrt{3280}}{14} \)
\( x \approx \frac{68 \pm 57.271}{14} \)
\( x_1 = \frac{125.271}{14} \approx 8.948 \implies 8.95 \)
\( x_2 = \frac{10.729}{14} \approx 0.766 \implies 0.77 \)

(c) If \( x = 0.77 \) km/h, the speed for the second part is \( x - 2 = 0.77 - 2 = -1.23 \) km/h, which is impossible. Thus, \( x = 8.95 \) km/h is the only realistic speed.
Time taken for the second part is:
\( t_2 = \frac{15}{8.948 - 2} = \frac{15}{6.948} \approx 2.1589 \) hours.
Converting to minutes: \( 0.1589 \times 60 \approx 9.53 \) minutes.
So, the time is 2 hours and 10 minutes.

(d) Average speed over the entire run:
\( \text{Total Distance} = 12 + 15 = 27 \) km.
\( \text{Total Time} = 3.5 \) hours.
\( \text{Average Speed} = \frac{27}{3.5} = 7.714 \approx 7.71 \) km/h.

(a) M1: For writing an expression for time: \( \frac{12}{x} \) or \( \frac{15}{x-2} \).
M1: For the equation \( \frac{12}{x} + \frac{15}{x-2} = 3.5 \).
M1: For algebraically expanding and attempting to clear fractions.
A1: For successfully showing the given quadratic equation with intermediate steps.

(b) M1: For correct substitution into the quadratic formula.
B1: For \( \sqrt{3280} \) or \( 57.27... \).
A1: For \( x = 8.95 \) (correct to 2 d.p.).
A1: For \( x = 0.77 \) (correct to 2 d.p.).

(c) B1: For choosing \( x = 8.95 \) and giving the reason that \( x = 0.77 \) leads to a negative speed.
M1: For calculating \( \frac{15}{\text{their } x - 2} \).
A1: For 2 hours and 10 minutes.

(d) M1: For \( \frac{27}{3.5} \).
A1: For \( 7.71 \) (or \( 7.714 \)).

(a) Factor the numerator: \( 2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2) \).
Factor the denominator: \( x^2 + 5x - 14 = (x + 7)(x - 2) \).
Simplify by cancelling the common factor \( (x - 2) \):
\( \frac{2(x - 2)(x + 2)}{(x + 7)(x - 2)} = \frac{2(x + 2)}{x + 7} = \frac{2x + 4}{x + 7} \).

(b) Find a common denominator: \( (2x - 1)(x + 3) \).
Express with the common denominator:
\( \frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)} \)
Expand and simplify the numerator:
\( 3x + 9 - 4x + 2 = -x + 11 \).
So, the simplified fraction is \( \frac{11 - x}{(2x - 1)(x + 3)} \).

(c) Multiply both sides by \( (2x + 1) \):
\( y(2x + 1) = 3x - 5 \)
\( 2xy + y = 3x - 5 \)
Group terms containing \( x \) on one side:
\( y + 5 = 3x - 2xy \)
Factor out \( x \):
\( y + 5 = x(3 - 2y) \)
Divide by \( (3 - 2y) \):
\( x = \frac{y + 5}{3 - 2y} \).

(a) M1: For factoring numerator: \( 2(x - 2)(x + 2) \).
M1: For factoring denominator: \( (x + 7)(x - 2) \).
M1: For cancelling common factor of \( (x - 2) \).
A1: For final answer \( \frac{2x + 4}{x + 7} \) (or \( \frac{2(x+2)}{x+7} \)).

(b) M1: For writing over a common denominator: \( (2x - 1)(x + 3) \).
M1: For expanding numerator: \( 3(x + 3) - 2(2x - 1) \).
A1: For correct expansion \( 3x + 9 - 4x + 2 \).
A1: For final answer \( \frac{11 - x}{(2x - 1)(x + 3)} \).

(c) M1: For multiplying by \( (2x + 1) \) correctly.
M1: For expanding to get \( 2xy + y = 3x - 5 \).
M1: For separating terms containing \( x \) on one side.
M1: For factoring out \( x \) from their terms.
A1: For final correct rearranged formula \( x = \frac{y + 5}{3 - 2y} \) (or equivalent).

(a) Volume of cylinder: \( V_{\text{cyl}} = \pi r^2 h = \pi \times 4^2 \times 10 = 160\pi \approx 502.65 \) cm\(^3 \).
Volume of cone: \( V_{\text{cone}} = \frac{1}{3}\pi r^2 H = \frac{1}{3} \times \pi \times 4^2 \times 6 = 32\pi \approx 100.53 \) cm\(^3 \).
Total volume: \( V_{\text{total}} = 160\pi + 32\pi = 192\pi \approx 603.19 \) cm\(^3 \).
To 3 s.f., the total volume is \( 603 \) cm\(^3 \).

(b) Using Pythagoras' theorem for the slant height, \( l \), of the cone:
\( l = \sqrt{r^2 + H^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.211 \) cm.
To 3 s.f., the slant height is \( 7.21 \) cm.

(c) Total Surface Area comprises:
1. Base area of cylinder: \( \pi r^2 = \pi \times 4^2 = 16\pi \approx 50.27 \) cm\(^2 \).
2. Curved surface area of cylinder: \( 2\pi r h = 2\pi \times 4 \times 10 = 80\pi \approx 251.33 \) cm\(^2 \).
3. Curved surface area of cone: \( \pi r l = \pi \times 4 \times \sqrt{52} \approx 90.62 \) cm\(^2 \).
Total Surface Area:
\( A_{\text{total}} = 16\pi + 80\pi + 4\pi\sqrt{52} = 96\pi + 4\pi\sqrt{52} \approx 301.59 + 90.62 = 392.21 \) cm\(^2 \).
To 3 s.f., the total surface area is \( 392 \) cm\(^2 \).

(d) Let the linear scale factor be \( k \).
\( k^3 = \frac{V_{\text{large}}}{V_{\text{small}}} = \frac{2035}{192\pi} \approx \frac{2035}{603.186} \approx 3.3736 \).
\( k = \sqrt[3]{3.3736} \approx 1.4998 \).
Height of the cylinder of the larger toy:
\( h_{\text{large}} = 10 \times k \approx 10 \times 1.4998 = 14.998 \) cm.
To 3 s.f., the height of the cylinder is \( 15.0 \) cm.

(a) M1: For correct volume of cylinder formula: \( \pi \times 4^2 \times 10 \).
M1: For correct volume of cone formula: \( \frac{1}{3} \times \pi \times 4^2 \times 6 \).
A1: For total volume \( 603 \) (or 603.1 - 603.2).

(b) M1: For \( \sqrt{4^2 + 6^2} \).
A1: For \( 7.21 \) (or 7.211).

(c) B1: For area of base \( 16\pi \) (or 50.3).
M1: For curved surface area of cylinder: \( 2\pi \times 4 \times 10 = 80\pi \) (or 251).
M1: For curved surface area of cone: \( \pi \times 4 \times \text{their } l \).
M1: For summing their three calculated areas.
A1: For total area \( 392 \) (or 392.2).

(d) M1: For setting up the ratio of volumes: \( \frac{2035}{603} \) (or \( \frac{2035}{192\pi} \)).
M1: For taking the cube root of the volume ratio to get linear scale factor \( k \).
A1: For \( 15.0 \) (accept 15).

(a) Total counters = 12.
Probability of both Red:
\( P(\text{Red}, \text{Red}) = \frac{6}{12} \times \frac{5}{11} = \frac{30}{132} \).
Probability of both Blue:
\( P(\text{Blue}, \text{Blue}) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132} \).
Probability of both Green:
\( P(\text{Green}, \text{Green}) = \frac{2}{12} \times \frac{1}{11} = \frac{2}{132} \).
Probability of same color:
\( P(\text{Same}) = \frac{30 + 12 + 2}{132} = \frac{44}{132} = \frac{1}{3} \approx 0.333 \).

(b) The complement event is "neither counter is red".
There are 6 counters that are not red (4 blue + 2 green).
\( P(\text{No Red}) = \frac{6}{12} \times \frac{5}{11} = \frac{30}{132} \).
\( P(\text{At least one Red}) = 1 - P(\text{No Red}) = 1 - \frac{30}{132} = \frac{102}{132} = \frac{17}{22} \approx 0.773 \).

(c) We want to find the conditional probability \( P(\text{First Red} \mid \text{Second Blue}) = \frac{P(\text{First Red and Second Blue})}{P(\text{Second Blue})} \).
\( P(\text{First Red and Second Blue}) = P(\text{Red}, \text{Blue}) = \frac{6}{12} \times \frac{4}{11} = \frac{24}{132} \).
To find \( P(\text{Second Blue}) \), we sum all possibilities where the second is blue:
\( P(\text{Second Blue}) = P(\text{Red}, \text{Blue}) + P(\text{Blue}, \text{Blue}) + P(\text{Green}, \text{Blue}) \)
\( P(\text{Second Blue}) = \frac{6}{12} \times \frac{4}{11} + \frac{4}{12} \times \frac{3}{11} + \frac{2}{12} \times \frac{4}{11} = \frac{24 + 12 + 8}{132} = \frac{44}{132} \).
Thus, the conditional probability is:
\( P(\text{First Red} \mid \text{Second Blue}) = \frac{24/132}{44/132} = \frac{24}{44} = \frac{6}{11} \approx 0.545 \).

(d) If the first two counters drawn were red, 4 red counters remain in the bag out of a total of 10 remaining counters.
\( P(\text{Third is Red}) = \frac{4}{10} = \frac{2}{5} = 0.4 \).

(a) M1: For calculating any correct joint probability (e.g., Red/Red: \( \frac{30}{132} \)).
M1: For summing three correct same-color scenarios: \( P(\text{RR}) + P(\text{BB}) + P(\text{GG}) \).
A1: For unsimplified fraction \( \frac{44}{132} \) (or equivalent fraction).
A1: For final answer \( \frac{1}{3} \) (or 0.333).

(b) M1: For attempting to use the complement method or listing out options (RR, RB, RG, BR, GR).
M1: For correct probability \( \frac{30}{132} \) of getting no red counters.
A1: For \( \frac{17}{22} \) (or 0.773).

(c) M1: For finding \( P(\text{Red, Blue}) = \frac{24}{132} \).
M1: For identifying and summing the three paths for Second Blue: \( \frac{44}{132} \).
M1: For dividing \( \frac{P(\text{Red, Blue})}{P(\text{Second Blue})} \).
A1: For \( \frac{6}{11} \) (or 0.545).

(d) M1: For establishing remaining counters (4 red out of 10 total).
A1: For \( \frac{2}{5} \) (or 0.4).

(a) The maximum value of \( \sin(2x) \) is \( 1 \).
This occurs when \( 2x = 90^\circ \implies x = 45^\circ \).
At \( x = 45^\circ \), \( y = 3(1) - 1 = 2 \).
So, the Maximum point is \( (45^\circ, 2) \).

The minimum value of \( \sin(2x) \) is \( -1 \).
This occurs when \( 2x = 270^\circ \implies x = 135^\circ \).
At \( x = 135^\circ \), \( y = 3(-1) - 1 = -4 \).
So, the Minimum point is \( (135^\circ, -4) \).

(b) Set \( 3\sin(2x) - 1 = 0 \implies \sin(2x) = \frac{1}{3} \).
Find the principal angle: \( 2x = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ \).
Since \( 0^\circ \le x \le 180^\circ \), the range for \( 2x \) is \( 0^\circ \le 2x \le 360^\circ \).
The values of \( 2x \) are:
\( 2x = 19.47^\circ \) and \( 2x = 180^\circ - 19.47^\circ = 160.53^\circ \).
Dividing by 2 gives:
\( x = 9.735^\circ \approx 9.7^\circ \)
\( x = 80.265^\circ \approx 80.3^\circ \).

(c) The three transformations are:
1. A stretch parallel to the \( y \)-axis (vertical stretch) with a scale factor of 3.
2. A stretch parallel to the \( x \)-axis (horizontal stretch) with a scale factor of \( 0.5 \) (or \( \frac{1}{2} \)).
3. A translation of 1 unit down, represented by the vector \( \begin{pmatrix} 0 \\ -1 \end{pmatrix} \).

(a) B1: For finding \( y_{\text{max}} = 2 \) and \( y_{\text{min}} = -4 \).
M1: For setting \( 2x = 90 \) and \( 2x = 270 \) to find \( x \)-coordinates.
A1: For Max point \( (45, 2) \).
A1: For Min point \( (135, -4) \).

(b) M1: For \( \sin(2x) = \frac{1}{3} \).
M1: For finding the first angle \( 2x = 19.5 \) (or 19.47).
M1: For finding the second angle \( 2x = 160.5 \).
A1: For \( x = 9.7^\circ \).
A1: For \( x = 80.3^\circ \).

(c) B1: For vertical stretch, scale factor 3.
B1: For horizontal stretch, scale factor 0.5 (or \( \frac{1}{2} \)).
B2: For translation of \( \begin{pmatrix} 0 \\ -1 \end{pmatrix} \) (B1 for 'translation' and B1 for vector or '1 unit down').

(a) The angle \( \angle BAC \) is the difference in bearings:
\( \angle BAC = 130^\circ - 55^\circ = 75^\circ \).
We use the Cosine Rule to find \( BC \):
\( BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC) \)
\( BC^2 = 18^2 + 25^2 - 2(18)(25)\cos(75^\circ) \)
\( BC^2 = 324 + 625 - 900\cos(75^\circ) \)
\( BC^2 = 949 - 900(0.258819) = 949 - 232.937 = 716.063 \).
\( BC = \sqrt{716.063} \approx 26.759 \) km.
To 3 s.f., the distance is \( 26.8 \) km.

(b) First, use the Sine Rule to find the angle \( \angle ABC \):
\( \frac{\sin(\angle ABC)}{AC} = \frac{\sin(\angle BAC)}{BC} \)
\( \sin(\angle ABC) = \frac{25 \times \sin(75^\circ)}{26.759} \approx \frac{24.148}{26.759} \approx 0.90242 \).
\( \angle ABC = \sin^{-1}(0.90242) \approx 64.48^\circ \).
Now find the bearing. Draw a North line at \( B \).
Since the bearing of \( B \) from \( A \) is \( 055^\circ \), the line \( BA \) has a back-bearing of \( 55^\circ + 180^\circ = 235^\circ \).
Because \( C \) is further south-east, looking from \( B \) to \( A \) (direction \( 235^\circ \)), we turn counter-clockwise by angle \( \angle ABC \) to look towards \( C \).
So, the bearing of \( C \) from \( B \) is:
\( 235^\circ - 64.48^\circ = 170.52^\circ \approx 170.5^\circ \).

(c) Area of triangle \( ABC \):
\( \text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle BAC) \)
\( \text{Area} = \frac{1}{2} \times 18 \times 25 \times \sin(75^\circ) = 225 \times 0.96593 \approx 217.33 \) km\(^2 \).
To 3 s.f., the area is \( 217 \) km\(^2 \).

(d) The shortest distance \( d \) from \( A \) to \( BC \) is the perpendicular height of triangle \( ABC \) when \( BC \) is the base:
\( \text{Area} = \frac{1}{2} \times BC \times d \)
\( 217.33 = \frac{1}{2} \times 26.759 \times d \)
\( d = \frac{2 \times 217.33}{26.759} \approx 16.24 \) km.
To 3 s.f., the shortest distance is \( 16.2 \) km.

(a) B1: For \( \angle BAC = 75^\circ \).
M1: For substituting into Cosine Rule correctly.
A1: For \( BC^2 = 716.06 \).
A1: For \( BC = 26.8 \) km (accept 26.75-26.8).

(b) M1: For substituting into Sine Rule correctly.
A1: For \( \sin(\angle ABC) \approx 0.902 \).
A1: For \( \angle ABC = 64.5^\circ \) (or 64.48).
M1: For correct bearing calculation step: \( 235 - \text{their } 64.5 \).
A1: For bearing \( 170.5^\circ \).

(c) M1: For correct area formula: \( \frac{1}{2} \times 18 \times 25 \times \sin(75^\circ) \).
A1: For \( 217 \) km\(^2 \) (or 217.3).

(d) M1: For setting up equation \( \frac{1}{2} \times BC \times d = \text{Area} \).
A1: For \( 16.2 \) km (or 16.24).

(a) Express \( y \) in terms of \( x \) from the first equation:
\( y = 7 - 2x \).
Substitute this into the second equation:
\( 3x^2 - (7 - 2x)^2 = 8 \)
Expand the squared term:
\( (7 - 2x)^2 = 49 - 28x + 4x^2 \).
Now substitute this expansion back into the equation:
\( 3x^2 - (49 - 28x + 4x^2) = 8 \)
Remove parentheses by distributing the negative sign:
\( 3x^2 - 49 + 28x - 4x^2 = 8 \)
Simplify terms:
\( -x^2 + 28x - 49 = 8 \)
Move all terms to one side to set to zero:
\( -x^2 + 28x - 57 = 0 \)
Multiply by \(-1\) to get the required form:
\( x^2 - 28x + 57 = 0 \) (Showed)

(b) Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{28 \pm \sqrt{(-28)^2 - 4(1)(57)}}{2} \)
\( x = \frac{28 \pm \sqrt{784 - 228}}{2} = \frac{28 \pm \sqrt{556}}{2} \)
\( \sqrt{556} \approx 23.5796 \)
\( x_1 = \frac{28 + 23.5796}{2} \approx 25.7898 \implies 25.8 \) (to 3 s.f.)
\( x_2 = \frac{28 - 23.5796}{2} \approx 2.2102 \implies 2.21 \) (to 3 s.f.)

(c) Substitute the \( x \) values back into \( y = 7 - 2x \):
For \( x_1 = 25.7898 \):
\( y_1 = 7 - 2(25.7898) = 7 - 51.5796 = -44.5796 \approx -44.6 \).
For \( x_2 = 2.2102 \):
\( y_2 = 7 - 2(2.2102) = 7 - 4.4204 = 2.5796 \approx 2.58 \).
So the solutions are \( (25.8, -44.6) \) and \( (2.21, 2.58) \).

(a) M1: For writing \( y = 7 - 2x \) or equivalent expression.
M1: For substitution of their expression into \( 3x^2 - y^2 = 8 \).
A1: For expanding \( (7-2x)^2 \) correctly as \( 49 - 28x + 4x^2 \).
A1: For simplifying to the target equation \( x^2 - 28x + 57 = 0 \) with all steps shown.

(b) M1: For correct substitution into the quadratic formula.
A1: For \( \sqrt{556} \) or \( 23.579... \).
A1: For \( x = 25.8 \).
A1: For \( x = 2.21 \).
B1: For answers rounded to 3 significant figures.

(c) M1: For attempting to substitute at least one of their \( x \) values into a linear equation to find \( y \).
A1: For \( y_1 \approx -44.6 \).
A1: For \( y_2 \approx 2.58 \).
A1: For writing final answers as coordinate pairs: \( (25.8, -44.6) \) and \( (2.21, 2.58) \).

Let's solve each part systematically:

(a)(i) Speed is given by \(v = \frac{d}{t}\), so time is \(t = \frac{d}{v}\).
For the outward journey, distance is 36 km and average speed is \(x\) km/h.
Time taken = \(\frac{36}{x}\) hours.

(a)(ii) On the return journey, the speed is 3 km/h slower, so speed = \(x - 3\) km/h.
Time taken = \(\frac{36}{x - 3}\) hours.

(b) The return journey takes 40 minutes longer than the outward journey.
Since 40 minutes = \(\frac{40}{60} = \frac{2}{3}\) hours:
\(\frac{36}{x - 3} - \frac{36}{x} = \frac{2}{3}\)

Divide the entire equation by 2:
\(\frac{18}{x - 3} - \frac{18}{x} = \frac{1}{3}\)

Multiply through by the common denominator \(3x(x - 3)\):
\(3x(18) - 3(x - 3)(18) = x(x - 3)\)
\(54x - 54(x - 3) = x^2 - 3x\)
\(54x - 54x + 162 = x^2 - 3x\)
\(162 = x^2 - 3x\)
\(x^2 - 3x - 162 = 0\) (as required).

(c) Solve \(x^2 - 3x - 162 = 0\):
Factorising the quadratic equation:
\((x - 18)(x + 9) = 0\)
This gives \(x = 18\) or \(x = -9\).
Since speed must be positive, the speed of the outward journey is \(18\) km/h.

(d) For Bilal:
Original speed = \(\frac{36}{t}\) km/h.
If his speed is 2 km/h faster, his new speed is \(\frac{36}{t} + 2\) km/h.
His new time is \(t - 1.5\) hours.
Since distance = speed \(\times\) time:
\(\left(\frac{36}{t} + 2\right)(t - 1.5) = 36\)

Expand the brackets:
\(36 - \frac{54}{t} + 2t - 3 = 36\)
\(2t - 3 - \frac{54}{t} = 0\)

Multiply by \(t\):
\(2t^2 - 3t - 54 = 0\) (as required).

Solve \(2t^2 - 3t - 54 = 0\) by factorising:
\((2t + 9)(t - 6) = 0\)
This gives \(t = 6\) or \(t = -4.5\).
Since time must be positive, \(t = 6\).

(a)(i) [1 mark] B1 for \(\frac{36}{x}\)
(a)(ii) [1 mark] B1 for \(\frac{36}{x - 3}\)
(b) [4 marks] M1 for setting up algebraic equation \(\frac{36}{x - 3} - \frac{36}{x} = \frac{40}{60}\) (or equivalent)
M1 for eliminating denominators, e.g., \(36x - 36(x-3) = \frac{2}{3}x(x-3)\)
M1 for expanding brackets: \(36x - 36x + 108 = \frac{2}{3}x^2 - 2x\)
A1 for fully correct simplification leading to \(x^2 - 3x - 162 = 0\) with no errors seen.
(c) [3 marks] M1 for factorising \((x - 18)(x + 9)\) or correct substitution into quadratic formula.
A1 for \(x = 18\) and \(x = -9\) as algebraic solutions.
A1 for selecting positive value \(18\) as final answer.
(d) [4 marks] M1 for expression of new speed \(\frac{36}{t} + 2\) or new time \(t - 1.5\).
M1 for setting up the equation \((\frac{36}{t} + 2)(t - 1.5) = 36\) or equivalent.
M1 for simplifying to \(2t^2 - 3t - 54 = 0\) and attempting to solve (by factorising or formula).
A1 for \(t = 6\) (rejecting \(-4.5\)).

(a) Let us calculate the missing values in the table:
When \(x = 1\):
\(y = 1^2 - \frac{4}{1} = 1 - 4 = -3\).
So, \(p = -3\).

When \(x = 2\):
\(y = 2^2 - \frac{4}{2} = 4 - 2 = 2\).
So, \(q = 2\).

When \(x = 3\):
\(y = 3^2 - \frac{4}{3} = 9 - 1.333... = 7.67\) (to 3 s.f.).
So, \(r = 7.67\).

(b) The curve crosses the \(x\)-axis when \(y = 0\):
\(x^2 - \frac{4}{x} = 0\)
\(x^2 = \frac{4}{x}\)
\(x^3 = 4\)
\(x = \sqrt[3]{4} \approx 1.5874...\)
To 3 significant figures, \(x = 1.59\).

(c) Rewrite \(y\) as \(y = x^2 - 4x^{-1}\).
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = 2x - 4(-1)x^{-2} = 2x + 4x^{-2} = 2x + \frac{4}{x^2}\).

(d) To find the gradient of the curve at \(x = 2\), substitute \(x = 2\) into the derivative:
\(\text{Gradient} = 2(2) + \frac{4}{2^2} = 4 + \frac{4}{4} = 4 + 1 = 5\).

(e) The tangent is a straight line \(y = mx + c\) with gradient \(m = 5\).
The point of tangency is \((2, 2)\) since \(x = 2\) and \(y = 2\).
Substitute \((2, 2)\) and \(m = 5\) into \(y - y_1 = m(x - x_1)\):
\(y - 2 = 5(x - 2)\)
\(y - 2 = 5x - 10\)
\(y = 5x - 8\).

(a) [3 marks] B1 for \(p = -3\).
B1 for \(q = 2\).
B1 for \(r = 7.67\) (or \(7.7\) or \(7\frac{2}{3}\)).

(b) [2 marks] M1 for setting \(x^2 - \frac{4}{x} = 0\) and obtaining \(x^3 = 4\).
A1 for \(1.59\) (or \(1.587...\)).

(c) [2 marks] M1 for \(2x\) or \(4/x^2\) (or \(4x^{-2}\)).
A1 for fully correct derivative: \(2x + \frac{4}{x^2}\).

(d) [2 marks] M1 for substituting \(x = 2\) into their derivative.
A1 for gradient = \(5\).

(e) [4 marks] M1 for using their gradient (5) as \(m\) in \(y = mx + c\).
M1 for finding the point coordinates \((2, 2)\) by using \(x=2\) in the original function.
M1 for substituting \((2, 2)\) and \(m=5\) into \(y - y_1 = m(x - x_1)\) or equivalent.
A1 for \(y = 5x - 8\).