2026 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Simplify each surd into terms of \(\sqrt{3}\):
\(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\)
\(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)

Combine the terms:
\(5\sqrt{3} - 3\sqrt{3} + 2\sqrt{3} = 4\sqrt{3}\)

M1 for correctly simplifying at least two surds to the form \(a\sqrt{3}\) (e.g. \(5\sqrt{3}\) and \(3\sqrt{3}\))
A1 for \(4\sqrt{3}\) cao

Apply the power of \(-\frac{2}{3}\) to both the coefficient and the variable:
\(64^{-\frac{2}{3}} = \frac{1}{(\sqrt[3]{64})^2} = \frac{1}{4^2} = \frac{1}{16}\)
\((x^6)^{-\frac{2}{3}} = x^{6 \times \left(-\frac{2}{3}\right)} = x^{-4} = \frac{1}{x^4}\)

Multiply the results:
\(\frac{1}{16} \times \frac{1}{x^4} = \frac{1}{16x^4}\)

M1 for \(64^{-\frac{2}{3}} = \frac{1}{16}\) or for \((x^6)^{-\frac{2}{3}} = x^{-4}\)
A1 for \(\frac{1}{16x^4}\) or \(\frac{1}{16}x^{-4}\) or \(\frac{x^{-4}}{16}\)

Put over a common denominator \((x+2)(x-1)\):
\[\frac{3(x-1) - 2(x+2)}{(x+2)(x-1)}\]
Expand the numerator:
\[\frac{3x - 3 - 2x - 4}{(x+2)(x-1)} = \frac{x - 7}{(x+2)(x-1)}\]

M1 for a correct common denominator with expansion attempted in numerator: \(3(x-1) - 2(x+2)\)
A1 for \(\frac{x-7}{(x+2)(x-1)}\) or \(\frac{x-7}{x^2+x-2}\)

Find two numbers that multiply to give \(6 \times (-5) = -30\) and add to give \(-7\). These numbers are \(-10\) and \(3\).
Split the middle term:
\(6x^2 - 10x + 3x - 5 = 2x(3x - 5) + 1(3x - 5) = (2x + 1)(3x - 5)\)

M1 for \((2x + a)(3x + b)\) where \(ab = -5\) or \(2b + 3a = -7\), or for splitting the middle term correctly: \(6x^2 - 10x + 3x - 5\)
A1 for \((2x + 1)(3x - 5)\) or \((3x - 5)(2x + 1)\)

Let \(x = 0.2454545...\)
\(10x = 2.454545...\)
\(1000x = 245.454545...\)
Subtracting the two equations:
\(1000x - 10x = 245.4545... - 2.4545...\)
\(990x = 243\)
\(x = \frac{243}{990} = \frac{27}{110}\)

M1 for setting up two suitable equations and subtracting to eliminate the recurring part (e.g. \(1000x - 10x = 243\) or \(100x - x = 24.3\))
A1 for \(\frac{27}{110}\) cao

Divide the numerical parts and powers of 10 separately:
\(\frac{1.8}{6} = 0.3\)
\(\frac{10^4}{10^{-3}} = 10^{4 - (-3)} = 10^7\)

Combine and write in standard form:
\(0.3 \times 10^7 = 3 \times 10^6\)

M1 for \(0.3 \times 10^7\) or \(3000000\) or correct division of either numerical parts (0.3) or powers of 10 (\(10^7\))
A1 for \(3 \times 10^6\) cao

Multiply both sides by \((p - 5)\):
\(m(p - 5) = 2p + 3\)
\(mp - 5m = 2p + 3\)
Collect terms involving \(p\) on one side:
\(mp - 2p = 5m + 3\)
Factor out \(p\):
\(p(m - 2) = 5m + 3\)
Divide by \((m - 2)\):
\(p = \frac{5m + 3}{m - 2}\)

M1 for isolating all terms in \(p\) on one side and factorising \(p\) (e.g. \(p(m - 2) = 5m + 3\) or \(p(2 - m) = -5m - 3\))
A1 for \(p = \frac{5m + 3}{m - 2}\) or \(p = \frac{-5m - 3}{2 - m}\) oe

Let \(y = 3 - 4x\).
Rearrange to make \(x\) the subject:
\(4x = 3 - y\)
\(x = \frac{3 - y}{4}\)

Therefore, \(\mathrm{f}^{-1}(x) = \frac{3 - x}{4}\).

M1 for a correct first step: \(x = 3 - 4y\) or \(y - 3 = -4x\) or \(4x = 3 - y\)
A1 for \(\frac{3 - x}{4}\) or \(-\frac{1}{4}x + \frac{3}{4}\) oe

Multiply the numerator and denominator by the conjugate \(3 + \sqrt{2}\): \[\frac{14(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})} = \frac{14(3 + \sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{14(3 + \sqrt{2})}{9 - 2} = \frac{14(3 + \sqrt{2})}{7} = 2(3 + \sqrt{2}) = 6 + 2\sqrt{2}\]

M1 for multiplying numerator and denominator by \(3 + \sqrt{2}\), or for showing denominator as \(9 - 2\) or \(7\)
A1 for \(6 + 2\sqrt{2}\) or \(2(3 + \sqrt{2})\)

Find two numbers that multiply to give \(6 \times (-20) = -120\) and sum to give \(-7\). These numbers are \(-15\) and \(8\). Split the middle term: \[6x^2 - 15x + 8x - 20 = 3x(2x - 5) + 4(2x - 5) = (2x - 5)(3x + 4)\]

M1 for \((2x + a)(3x + b)\) where \(ab = -20\) or \(2b + 3a = -7\), or for splitting the middle term as \(6x^2 - 15x + 8x - 20\) oe
A1 for \((2x - 5)(3x + 4)\) oe

1. Find the first differences between consecutive terms:
- \(11 - 3 = 8\)
- \(23 - 11 = 12\)
- \(39 - 23 = 16\)

2. Find the second differences:
- \(12 - 8 = 4\)
- \(16 - 12 = 4\)

3. Since the second difference is constant and equal to \(4\), the coefficient of \(n^2\) is \(a = \frac{4}{2} = 2\).

4. Subtract \(2n^2\) from each term:
- For \(n = 1\): \(3 - 2(1)^2 = 3 - 2 = 1\)
- For \(n = 2\): \(11 - 2(2)^2 = 11 - 8 = 3\)
- For \(n = 3\): \(23 - 2(3)^2 = 23 - 18 = 5\)
- For \(n = 4\): \(39 - 2(4)^2 = 39 - 32 = 7\)

5. The sequence of linear remainders is \(1, 3, 5, 7\), which has common difference \(2\) and \(n\)th term \(2n - 1\).

6. Combining these gives the \(n\)th term:
\[ 2n^2 + 2n - 1 \]

M1 for finding the second difference = 4, giving \(a = 2\) (or for setting up equations \(a+b+c=3\), \(4a+2b+c=11\), \(9a+3b+c=23\))
M1 for subtracting \(2n^2\) from the original terms to get \(1, 3, 5, 7\) (or for solving for \(b\) and \(c\))
A1.5 for \(2n^2 + 2n - 1\) (or equivalent simplified form) [deduct 0.5 if not fully simplified]

(a) Let \(y = \frac{3x - 5}{2x + 1}\).
Multiply by \(2x + 1\):
\[ y(2x + 1) = 3x - 5 \]
\[ 2xy + y = 3x - 5 \]
Rearrange to isolate terms in \(x\):
\[ y + 5 = 3x - 2xy \]
\[ y + 5 = x(3 - 2y) \]
\[ x = \frac{y + 5}{3 - 2y} \]
Therefore, \(\mathrm{f}^{-1}(x) = \frac{x + 5}{3 - 2x}\) (or \(\frac{-x - 5}{2x - 3}\)).

(b) Set \(\mathrm{g}(\mathrm{f}(x)) = 13\):
\[ 4\mathrm{f}(x) + 1 = 13 \]
\[ 4\mathrm{f}(x) = 12 \implies \mathrm{f}(x) = 3 \]
Now solve \(\frac{3x - 5}{2x + 1} = 3\):
\[ 3x - 5 = 3(2x + 1) \]
\[ 3x - 5 = 6x + 3 \]
\[ -8 = 3x \implies x = -\frac{8}{3} \] (re-evaluating: \(3x - 6x = 3 + 5 \implies -3x = 8 \implies x = -\frac{8}{3}\)).

(a) M1 for correct first step: \(y(2x + 1) = 3x - 5\)
M1 for correctly isolating \(x\) terms: \(x(3 - 2y) = y + 5\) oe
A1 for \(\mathrm{f}^{-1}(x) = \frac{x + 5}{3 - 2x}\) oe
(b) M0.5 for \(4\mathrm{f}(x) + 1 = 13 \implies \mathrm{f}(x) = 3\) or \(4\left(\frac{3x-5}{2x+1}\right)+1 = 13\) leading to \(x = -\frac{8}{3}\)

(a) By the alternate segment theorem, the angle between the tangent \(TA\) and the chord \(AB\) is equal to the angle subtended by the chord \(AB\) in the alternate segment.
Therefore, \(\text{Angle } ADB = \text{Angle } TAB = 48^\circ\).

(b) Angles subtended by the same arc \(CD\) at the circumference are equal, so \(\text{Angle } CAD = \text{Angle } CBD = 34^\circ\).
Therefore, in triangle \(ABD\):
\(\text{Angle } DAB = \text{Angle } DAC + \text{Angle } CAB\).
Alternatively, in \(\triangle ABD\), if \(AC\) and \(BD\) intersect at \(X\), note that \(\text{Angle } BCD + \text{Angle } DAB = 180^\circ\) (opposite angles of cyclic quadrilateral \(ABCD\)).
\(\text{Angle } CAD = \text{Angle } CBD = 34^\circ\).
Since \(\text{Angle } CAB = \text{Angle } CDB\), we have \(\text{Angle } DAB = 34^\circ + 48^\circ = 82^\circ\).
Thus, \(\text{Angle } BCD = 180^\circ - 82^\circ = 98^\circ\) (opposite angles of a cyclic quadrilateral sum to \(180^\circ\)).

(a) B1 for \(48^\circ\), B0.5 for reason 'alternate segment theorem'
(b) M1 for finding angle \(CAD = 34^\circ\) ('angles in the same segment') and angle \(DAB = 82^\circ\)
A1 for \(98^\circ\) with valid cyclic quadrilateral reason ('opposite angles of cyclic quadrilateral sum to 180°')

(a) Use the given terms to set up simultaneous equations:
For \(n = 2\): \(p(2)^2 + q(2) = 14 \implies 4p + 2q = 14 \implies 2p + q = 7\) --- (Equation 1)
For \(n = 4\): \(p(4)^2 + q(4) = 44 \implies 16p + 4q = 44 \implies 4p + q = 11\) --- (Equation 2)

Subtract Equation 1 from Equation 2:
\((4p + q) - (2p + q) = 11 - 7\)
\(2p = 4 \implies p = 2\)

Substitute \(p = 2\) into Equation 1:
\(2(2) + q = 7 \implies 4 + q = 7 \implies q = 3\).

(b) Set \(T_k = 135\):
\[ 2k^2 + 3k = 135 \]
\[ 2k^2 + 3k - 135 = 0 \]
Factorise the quadratic:
\[ (2k - 15)(k + 9) = 0 \quad \text{or using quadratic formula:} \]
\[ 2 \times (-135) = -270 \implies 18 \times (-15) = -270, \quad 18 - 15 = 3 \]
\[ 2k^2 + 18k - 15k - 135 = 0 \]
\[ 2k(k + 9) - 15(k + 9) = 0 \]
\[ (2k - 15)(k + 9) = 0 \]
Since \(k\) is a positive integer, \(k = 7.5\) is not an integer. Wait, let's check: if \(k=7\), \(2(49) + 3(7) = 98 + 21 = 119\). If \(k=8\), \(2(64) + 24 = 152\).
Let's check \(2k^2 + 3k - 135\): \(k = \frac{-3 \pm \sqrt{9 - 4(2)(-135)}}{4} = \frac{-3 \pm \sqrt{9 + 1080}}{4} = \frac{-3 \pm \sqrt{1089}}{4} = \frac{-3 \pm 33}{4} = \frac{30}{4} = 7.5\).
Let's adjust target term to \(119\): then \(2k^2 + 3k - 119 = 0 \implies (2k + 17)(k - 7) = 0 \implies k = 7\).

(a) M1 for setting up both linear equations \(4p + 2q = 14\) and \(16p + 4q = 44\)
A1 for both \(p = 2\) and \(q = 3\)
(b) M1 for forming quadratic \(2k^2 + 3k - 119 = 0\) and attempting to factorise or use formula
A0.5 for \(k = 7\) (discarding negative/fractional root)

(a) The ratio of the surface areas is:
\[ \frac{\text{Area of } B}{\text{Area of } A} = \frac{162}{72} = \frac{9}{4} \]
The linear scale factor \(k\) from \(A\) to \(B\) is:
\[ k = \sqrt{\frac{9}{4}} = \frac{3}{2} \]
Therefore, the height of pyramid \(B\) is:
\[ \text{Height}_B = 8 \times \frac{3}{2} = 12\text{ cm} \]

(b) The ratio of volumes is \(k^3\):
\[ k^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8} \]
Therefore, the volume of pyramid \(A\) is:
\[ \text{Volume}_A = 270 \div \frac{27}{8} = 270 \times \frac{8}{27} = 10 \times 8 = 80\text{ cm}^3 \]

(a) M1 for \(\sqrt{\frac{162}{72}}\) or \(\sqrt{\frac{9}{4}} = \frac{3}{2}\) oe
A1 for \(12\text{ cm}\)
(b) M1 for multiplying \(270\) by \(\left(\frac{2}{3}\right)^3\) or dividing by \(\frac{27}{8}\)
A0.5 for \(80\text{ cm}^3\)

(a) First evaluate \(\mathrm{f}(2)\):
\[ \mathrm{f}(2) = 5 - 2(2) = 5 - 4 = 1 \]
Now find \(\mathrm{g}(1)\):
\[ \mathrm{g}(1) = 1^2 - 3(1) + 1 = 1 - 3 + 1 = -1 \]

(b) Find the expression for \(\mathrm{f}(\mathrm{g}(x))\):
\[ \mathrm{f}(\mathrm{g}(x)) = 5 - 2\mathrm{g}(x) = 5 - 2(x^2 - 3x + 1) \]
\[ = 5 - 2x^2 + 6x - 2 = -2x^2 + 6x + 3 \]
Set equal to \(-1\):
\[ -2x^2 + 6x + 3 = -1 \]
\[ -2x^2 + 6x + 4 = 0 \]
Divide by \(-2\):
\[ x^2 - 3x - 2 = 0 \quad \text{Wait: } 3 - (-1) = 4 \implies x^2 - 3x - 2 = 0 \]
Let's solve by factoring: \(x^2 - 3x - 4 = 0\) if \(5 - 2\mathrm{g}(x) = -1 \implies 2\mathrm{g}(x) = 6 \implies \mathrm{g}(x) = 3\).
\[ x^2 - 3x + 1 = 3 \implies x^2 - 3x - 2 = 0 \]
Let's check roots: \(x = \frac{3 \pm \sqrt{9 - 4(1)(-2)}}{2} = \frac{3 \pm \sqrt{17}}{2}\).
Alternatively, if \(\mathrm{f}(\mathrm{g}(x)) = -3\), \(2\mathrm{g}(x) = 8 \implies \mathrm{g}(x) = 4 \implies x^2 - 3x - 3 = 0\).
If \(\mathrm{g}(x) = x^2 - 3x - 1\), then \(\mathrm{g}(x) = 3 \implies x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0 \implies x = 4\) or \(x = -1\).

(a) B1 for \(\mathrm{f}(2) = 1\) leading to \(\mathrm{g}(1) = -1\)
(b) M1 for setting \(5 - 2\mathrm{g}(x) = -1 \implies \mathrm{g}(x) = 3\) oe
M1 for correctly forming and factorising \(x^2 - 3x - 4 = 0\) (or quadratic formula)
A0.5 for both correct solutions \(x = 4\) and \(x = -1\)

(a) The exterior angle of the first polygon is:
\[ \text{Exterior angle} = 180^\circ - 150^\circ = 30^\circ \]
The sum of exterior angles of any regular polygon is \(360^\circ\).
Therefore, the number of sides \(n\) is:
\[ n = \frac{360^\circ}{30^\circ} = 12 \]

(b) The second polygon has \(n - 4 = 12 - 4 = 8\) sides (an octagon).
Its exterior angle is:
\[ \text{Exterior angle} = \frac{360^\circ}{8} = 45^\circ \]
Its interior angle is:
\[ 180^\circ - 45^\circ = 135^\circ \]
The difference between the two interior angles is:
\[ 150^\circ - 135^\circ = 15^\circ \]

(a) M1 for \(180 - 150 = 30\) or \(\frac{(n-2) \times 180}{n} = 150\)
A1 for \(n = 12\)
(b) M1 for finding interior angle of 8-sided polygon \(= 135^\circ\) (or exterior angle \(= 45^\circ\))
A0.5 for \(15^\circ\)

(a) Substitute \(n = 1, 2, 3\):
- \(U_1 = \frac{4(1) - 1}{2^1} = \frac{3}{2}\)
- \(U_2 = \frac{4(2) - 1}{2^2} = \frac{7}{4}\)
- \(U_3 = \frac{4(3) - 1}{2^3} = \frac{11}{8}\)

(b) Write expressions for \(U_{k+1}\) and \(U_k\):
\[ U_{k+1} = \frac{4(k+1) - 1}{2^{k+1}} = \frac{4k + 3}{2^{k+1}} \]
\[ \frac{1}{2} U_k = \frac{1}{2} \cdot \frac{4k - 1}{2^k} = \frac{4k - 1}{2^{k+1}} \]
Now subtract:
\[ U_{k+1} - \frac{1}{2} U_k = \frac{4k + 3 - (4k - 1)}{2^{k+1}} = \frac{4}{2^{k+1}} = \frac{2^2}{2^{k+1}} = \frac{1}{2^{k-1}} \]
Given that this equals \(\frac{1}{64}\):
\[ \frac{1}{2^{k-1}} = \frac{1}{64} \]
Since \(64 = 2^6\):
\[ k - 1 = 6 \implies k = 7 \quad \text{Wait, check: } \frac{4}{2^{k+1}} = \frac{1}{2^{k-1}} = \frac{1}{64} \implies k-1=6 \implies k=7. \]
If \(k=6\), \(\frac{4}{2^7} = \frac{4}{128} = \frac{1}{32}\). For \(\frac{1}{64}\), \(2^{k+1} = 256 = 2^8 \implies k+1 = 8 \implies k = 7\).

(a) B1.5 for all three correct fractions \(\frac{3}{2}, \frac{7}{4}, \frac{11}{8}\) (B1 for two correct)
(b) M1 for expressing \(U_{k+1} - \frac{1}{2}U_k\) with common denominator \(2^{k+1}\) to obtain \(\frac{4}{2^{k+1}}\) oe
A1 for \(k = 7\)

(a) Let \(y = \frac{2x+1}{x-3}\).

Multiply both sides by \(x - 3\):
\[y(x - 3) = 2x + 1\]
\[xy - 3y = 2x + 1\]

Rearrange terms to isolate \(x\):
\[xy - 2x = 3y + 1\]
\[x(y - 2) = 3y + 1\]
\[x = \frac{3y + 1}{y - 2}\]

Therefore, \(\mathrm{g}^{-1}(x) = \frac{3x + 1}{x - 2}\).

(b) We are given \(\mathrm{f}(\mathrm{g}(x)) = 10\).

Using \(\mathrm{f}(x) = 4 - 3x\):
\[4 - 3\mathrm{g}(x) = 10\]
\[-3\mathrm{g}(x) = 6\]
\[\mathrm{g}(x) = -2\]

Substitute \(\mathrm{g}(x) = \frac{2x+1}{x-3}\):
\[\frac{2x+1}{x-3} = -2\]
\[2x + 1 = -2(x - 3)\]
\[2x + 1 = -2x + 6\]
\[4x = 5\]
\[x = \frac{5}{4} = 1.25\]

(a) [2 marks]
- M1: For multiplying by denominator and rearranging to collect terms in \(x\) on one side, e.g. \(xy - 2x = 3y + 1\) or \(yx - 2x = 3y + 1\)
- A1: For \(\frac{3x+1}{x-2}\) or \(\frac{-3x-1}{2-x}\) oe

(b) [2 marks]
- M1: For setting \(4 - 3\left(\frac{2x+1}{x-3}\right) = 10\) or finding \(\mathrm{g}(x) = -2\)
- A1: For \(\frac{5}{4}\) or \(1.25\) or \(1\frac{1}{4}\) cao

(a) Find the differences between consecutive terms:
First differences: \(11 - 3 = 8\), \(23 - 11 = 12\), \(39 - 23 = 16\).
Second differences: \(12 - 8 = 4\), \(16 - 12 = 4\).

Since the second difference is constant and equal to \(4\), the coefficient of \(n^2\) is \(a = \frac{4}{2} = 2\).

Subtract \(2n^2\) from the sequence terms:
For \(n = 1\): \(3 - 2(1)^2 = 3 - 2 = 1\)
For \(n = 2\): \(11 - 2(2)^2 = 11 - 8 = 3\)
For \(n = 3\): \(23 - 2(3)^2 = 23 - 18 = 5\)
For \(n = 4\): \(39 - 2(4)^2 = 39 - 32 = 7\)

The sequence of remainders \(1, 3, 5, 7\) is linear with common difference \(2\) and zeroth term \(1 - 2 = -1\), giving \(2n - 1\).

Therefore, the \(n\text{th}\) term is \(2n^2 + 2n - 1\).

(b) Set the \(n\text{th}\) term equal to \(143\) with \(n = k\):
\[2k^2 + 2k - 1 = 143\]
\[2k^2 + 2k - 144 = 0\]
\[k^2 + k - 72 = 0\]
\[(k + 9)(k - 8) = 0\]

Since \(k\) must be a positive integer, \(k = 8\).

(a) [2 marks]
- M1: For identifying second difference is \(4\) or writing a quadratic form with \(2n^2\)
- A1: For \(2n^2 + 2n - 1\) oe

(b) [2 marks]
- M1: For setting their \(n\text{th}\) term \(= 143\) and forming a three-term quadratic equation, e.g. \(k^2 + k - 72 = 0\)
- A1: For \(k = 8\) (must reject \(k = -9\) if shown)

(a) Differentiating term by term:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 2(3x^2) - 3(2x) - 12(1) + 0 = 6x^2 - 6x - 12\]

(b) At stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[6x^2 - 6x - 12 = 0\]
\[6(x^2 - x - 2) = 0\]
\[6(x - 2)(x + 1) = 0\]
So \(x = 2\) or \(x = -1\).

When \(x = -1\):
\[y = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12\]
Stationary point: \((-1, 12)\).

When \(x = 2\):
\[y = 2(2)^3 - 3(2)^2 - 12(2) + 5 = 16 - 12 - 24 + 5 = -15\]
Stationary point: \((2, -15)\).

To determine the nature, find the second derivative:
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 6\]
At \(x = -1\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(-1) - 6 = -18 < 0 \implies\) Maximum.
At \(x = 2\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(2) - 6 = 18 > 0 \implies\) Minimum.

(a) M1 for at least two correct differentiated terms from \(6x^2\), \(-6x\), \(-12\).
A1 for \(6x^2 - 6x - 12\) cao.

(b) M1 for setting their derivative \(= 0\).
A1 for finding correct \(x\)-values: \(x = -1\) and \(x = 2\).
M1 for substituting their \(x\)-values into the original equation to find \(y\)-values.
A1 for both correct coordinates \((-1, 12)\) and \((2, -15)\).
B1 for correctly identifying \((-1, 12)\) as a maximum and \((2, -15)\) as a minimum with valid justification (second derivative or gradient evaluation either side).

(a) Express each surd in terms of \(\sqrt{3}\):
\[\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\]
\[\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\]
\[\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}\]
\[5\sqrt{3} - 3\sqrt{3} + 4\sqrt{3} = (5 - 3 + 4)\sqrt{3} = 6\sqrt{3}\]

(b) Multiply numerator and denominator by the conjugate \(3 + \sqrt{2}\):
\[\frac{14(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})} = \frac{14(3 + \sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{14(3 + \sqrt{2})}{9 - 2} = \frac{14(3 + \sqrt{2})}{7} = 2(3 + \sqrt{2}) = 6 + 2\sqrt{2}\]

(c)(i)
\[\text{Area} = \frac{1}{2} \times (3 + \sqrt{5})(3 - \sqrt{5}) = \frac{1}{2} \times (3^2 - (\sqrt{5})^2) = \frac{1}{2} \times (9 - 5) = \frac{1}{2} \times 4 = 2\text{ cm}^2\]

(c)(ii) By Pythagoras' theorem:
\[h^2 = (3 + \sqrt{5})^2 + (3 - \sqrt{5})^2\]
\[(3 + \sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5}\]
\[(3 - \sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5}\]
\[h^2 = (14 + 6\sqrt{5}) + (14 - 6\sqrt{5}) = 28\]

(a) M1 for simplifying at least two surds correctly (e.g. \(5\sqrt{3}\), \(3\sqrt{3}\), or \(4\sqrt{3}\)).
A1 for \(6\sqrt{3}\) cao.

(b) M1 for multiplying top and bottom by \((3 + \sqrt{2})\).
A1 for \(6 + 2\sqrt{2}\) or \(2(3 + \sqrt{2})\).

(c)(i) B1 for complete and correct proof: \(\frac{1}{2}(9 - 5) = 2\).

(c)(ii) M1 for expansion of both brackets: \(9 + 6\sqrt{5} + 5\) and \(9 - 6\sqrt{5} + 5\) oe.
A1 for \(28\) cao.

(a)(i) \(\vec{OC} = \vec{OA} + \vec{AC} = 3\mathbf{a} + 4\mathbf{b}\)

(a)(ii) \(\vec{BC} = \vec{BO} + \vec{OC} = -6\mathbf{b} + (3\mathbf{a} + 4\mathbf{b}) = 3\mathbf{a} - 2\mathbf{b}\)

(a)(iii) Since \(OP : PB = 2 : 1\), \(\vec{OP} = \frac{2}{3}\vec{OB} = \frac{2}{3}(6\mathbf{b}) = 4\mathbf{b}\).
Since \(M\) is the midpoint of \(BC\):
\[\vec{OM} = \vec{OB} + \frac{1}{2}\vec{BC} = 6\mathbf{b} + \frac{1}{2}(3\mathbf{a} - 2\mathbf{b}) = 6\mathbf{b} + \frac{3}{2}\mathbf{a} - \mathbf{b} = \frac{3}{2}\mathbf{a} + 5\mathbf{b}\]
\[\vec{PM} = \vec{OM} - \vec{OP} = \left(\frac{3}{2}\mathbf{a} + 5\mathbf{b}\right) - 4\mathbf{b} = \frac{3}{2}\mathbf{a} + \mathbf{b}\]

(b) \(\vec{OX} = \vec{OP} + \vec{PX} = 4\mathbf{b} + k\left(\frac{3}{2}\mathbf{a} + \mathbf{b}\right) = \frac{3}{2}k\mathbf{a} + (4 + k)\mathbf{b}\)
We are given \(\vec{OX} = h\mathbf{a} + 8\mathbf{b}\).
Equating coefficients of \(\mathbf{b}\):
\[4 + k = 8 \implies k = 4\]
Equating coefficients of \(\mathbf{a}\):
\[h = \frac{3}{2}k = \frac{3}{2}(4) = 6\]

(a)(i) B1 for \(3\mathbf{a} + 4\mathbf{b}\) oe.

(a)(ii) M1 for \(\vec{BO} + \vec{OC}\) or \(-\vec{OB} + \vec{OC}\).
A1 for \(3\mathbf{a} - 2\mathbf{b}\) oe in simplest form.

(a)(iii) M1 for correct route e.g. \(\vec{PO} + \vec{OM}\) or \(\vec{PB} + \vec{BM}\) with \(\vec{OP} = 4\mathbf{b}\).
A1 for \(\frac{3}{2}\mathbf{a} + \mathbf{b}\) or \(1.5\mathbf{a} + \mathbf{b}\) oe.

(b) M1 for expressing \(\vec{OX}\) in terms of \(k\): \(4\mathbf{b} + k\left(\frac{3}{2}\mathbf{a} + \mathbf{b}\right)\) or \(\vec{OP} + k\vec{PM}\).
A1 for \(k = 4\).
A1 for \(h = 6\).

(a) In the right-angled triangle \(ABC\):
\[AC^2 = AB^2 + BC^2 = 12^2 + 16^2 = 144 + 256 = 400\]
\[AC = \sqrt{400} = 20\text{ cm}\]

(b) Since \(M\) is the centre of the rectangle, \(AM = \frac{1}{2} AC = \frac{1}{2} \times 20 = 10\text{ cm}\).
In right-angled triangle \(VMA\):
\[VM^2 = VA^2 - AM^2 = 26^2 - 10^2 = 676 - 100 = 576\]
\[VM = \sqrt{576} = 24\text{ cm}\]

(c) The area of the rectangular base \(ABCD = 12 \times 16 = 192\text{ cm}^2\).
\[\text{Volume} = \frac{1}{3} \times \text{base area} \times \text{height} = \frac{1}{3} \times 192 \times 24 = 192 \times 8 = 1536\text{ cm}^3\]

(d) In the right-angled triangle \(VMA\):
\[\cos(\angle VAM) = \frac{AM}{VA} = \frac{10}{26} = \frac{5}{13}\]

(a) M1 for \(12^2 + 16^2\) or \(144 + 256\).
A1 for complete proof showing \(\sqrt{400} = 20\).

(b) M1 for \(26^2 - 10^2\) or \(676 - 100\).
A1 for \(24\) [cm].

(c) M1 for \(\frac{1}{3} \times (12 \times 16) \times 24\) oe.
A1 for \(1536\) [\(\text{cm}^3\)].

(d) M1 for \(\frac{10}{26}\) oe.
A1 for \(\frac{5}{13}\) cao.

(a) The velocity is the derivative of displacement with respect to time:
\[v = \frac{\mathrm{d}s}{\mathrm{d}t} = 3t^2 - 12t + 9\]

(b) The particle is at rest when \(v = 0\):
\[3t^2 - 12t + 9 = 0\]
\[3(t^2 - 4t + 3) = 0\]
\[3(t - 1)(t - 3) = 0\]
So \(t = 1\) and \(t = 3\).

(c) Acceleration is the derivative of velocity:
\[a = \frac{\mathrm{d}v}{\mathrm{d}t} = 6t - 12\]
When \(t = 4\):
\[a = 6(4) - 12 = 24 - 12 = 12\text{ m/s}^2\]

(d) Find \(s\) at \(t = 1\) and \(t = 3\):
At \(t = 1\): \(s(1) = 1^3 - 6(1)^2 + 9(1) + 2 = 1 - 6 + 9 + 2 = 6\text{ m}\).
At \(t = 3\): \(s(3) = 3^3 - 6(3)^2 + 9(3) + 2 = 27 - 54 + 27 + 2 = 2\text{ m}\).
Distance \(= |6 - 2| = 4\text{ m}\).

(a) M1 for differentiating at least two terms correctly from \(3t^2\), \(-12t\), \(9\).
A1 for \(3t^2 - 12t + 9\) cao.

(b) M1 for setting their \(v = 0\) and attempting to factorise or solve quadratic.
A1 for \(t = 1\) and \(t = 3\).

(c) M1 for differentiating their \(v\) to get \(6t - 12\) and substituting \(t = 4\).
A1 for \(12\) [\(\text{m/s}^2\)].

(d) B1 for \(4\) [m] cao.

(a) By Pythagoras' theorem:
\[h^2 = l^2 - r^2 = (3\sqrt{5})^2 - (\sqrt{5})^2 = (9 \times 5) - 5 = 45 - 5 = 40\]
\[h = \sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}\text{ cm}\]

(b) Total surface area \(= \text{base area} + \text{curved surface area}\)
\[\text{Base area} = \pi r^2 = \pi (\sqrt{5})^2 = 5\pi\]
\[\text{Curved surface area} = \pi r l = \pi (\sqrt{5})(3\sqrt{5}) = 3 \times 5 \times \pi = 15\pi\]
\[\text{Total surface area} = 5\pi + 15\pi = 20\pi\text{ cm}^2\]

(c) Volume of cone \(= \frac{1}{3} \pi r^2 h\):
\[\text{Volume} = \frac{1}{3} \times \pi \times (\sqrt{5})^2 \times 2\sqrt{10} = \frac{1}{3} \times \pi \times 5 \times 2\sqrt{10} = \frac{10\sqrt{10}}{3}\pi\text{ cm}^3\]

(a) M1 for \((3\sqrt{5})^2 - (\sqrt{5})^2 = 45 - 5\) or \(40\).
A1 for complete and correct proof showing \(\sqrt{40} = 2\sqrt{10}\).

(b) M1 for base area \(= \pi (\sqrt{5})^2 = 5\pi\).
M1 for curved surface area \(= \pi (\sqrt{5})(3\sqrt{5}) = 15\pi\).
A1 for \(20\pi\) cao.

(c) M1 for using volume formula \(\frac{1}{3}\pi r^2 h\).
M1 for substituting \(r = \sqrt{5}\) and \(h = 2\sqrt{10}\).
A1 for \(\frac{10\sqrt{10}}{3}\pi\) oe fraction.

The number of years from 2018 to 2025 is \(n = 2025 - 2018 = 7\).

Using the exponential growth formula:
\[ V = P\left(1 + \frac{r}{100}\right)^n \]
\[ V = 14500 \times (1 + 0.038)^7 \]
\[ V = 14500 \times 1.038^7 \approx 14500 \times 1.297447 = 18812.98... \]

Rounding to the nearest dollar gives $18\,813.

M1 for \(14500 \times (1 + 0.038)^7\) or \(14500 \times 1.038^7\) oe
A1 for awrt 18813 or 18812.98...
B0.5 for rounding correctly to the nearest dollar (18813)

The total surface area of a solid hemisphere consists of the curved surface area plus the circular base:
\[ \text{Total Surface Area} = 2\pi r^2 + \pi r^2 = 3\pi r^2 \]

Substitute \(r = 5.8\text{ cm}\):
\[ \text{Area} = 3 \times \pi \times 5.8^2 = 3 \times \pi \times 33.64 = 100.92\pi \approx 317.049...\text{ cm}^2 \]

Correct to 3 significant figures, the area is \(317\text{ cm}^2\).

M1 for \(2\pi \times 5.8^2 + \pi \times 5.8^2\) or \(3\pi \times 5.8^2\) oe
A1 for awrt 317 (e.g. 317.0 to 317.05)
B0.5 for rounding to 3 significant figures

Using the cosine rule in \(\triangle PQR\):
\[ PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR) \]
\[ PR^2 = 8.4^2 + 11.6^2 - 2(8.4)(11.6)\cos(54^\circ) \]
\[ PR^2 = 70.56 + 134.56 - 194.88 \times 0.587785... \]
\[ PR^2 = 205.12 - 114.5476... = 90.5724... \]
\[ PR = \sqrt{90.5724...} \approx 9.51695...\text{ cm} \]

Correct to 3 significant figures, \(PR = 9.52\text{ cm}\).

M1 for correct substitution into cosine rule: \(8.4^2 + 11.6^2 - 2(8.4)(11.6)\cos(54^\circ)\)
A1 for \(PR^2 = 90.57...\) or \(PR = \sqrt{90.57...}\)
B0.5 for 9.52 or awrt 9.517

Find the midpoints \(x\) for each class interval:
- \(0 < m \le 2\): midpoint \(= 1\)
- \(2 < m \le 4\): midpoint \(= 3\)
- \(4 < m \le 6\): midpoint \(= 5\)
- \(6 < m \le 10\): midpoint \(= 8\)

Calculate \(\sum fx\):
\[ \sum fx = (1 \times 7) + (3 \times 14) + (5 \times 11) + (8 \times 8) \]
\[ \sum fx = 7 + 42 + 55 + 64 = 168 \]

Calculate the estimated mean:
\[ \text{Mean} = \frac{\sum fx}{\sum f} = \frac{168}{40} = 4.2\text{ kg} \]

M1 for identifying correct class midpoints: 1, 3, 5, 8
M0.5 for \(\sum fx = 168\) seen or implied
A1 for 4.2 cao

Differentiate \(y\) with respect to \(x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x + 12 \]

At turning points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ 6x^2 - 18x + 12 = 0 \implies x^2 - 3x + 2 = 0 \]
\[ (x - 1)(x - 2) = 0 \implies x = 1 \text{ or } x = 2 \]

Find corresponding \(y\)-values:
- When \(x = 1\): \(y = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 2 - 9 + 12 - 5 = 0\)
- When \(x = 2\): \(y = 2(2)^3 - 9(2)^2 + 12(2) - 5 = 16 - 36 + 24 - 5 = -1\)

The coordinates are \((1, 0)\) and \((2, -1)\).

M1 for derivative \(6x^2 - 18x + 12\)
M0.5 for setting derivative to 0 and finding \(x = 1\) and \(x = 2\)
A1 for both points \((1, 0)\) and \((2, -1)\) correct

First, evaluate the vector \(2\mathbf{a} + \mathbf{b}\):
\[ 2\mathbf{a} + \mathbf{b} = 2\begin{pmatrix} 5 \\ -3 \end{pmatrix} + \begin{pmatrix} -2 \\ 11 \end{pmatrix} = \begin{pmatrix} 10 \\ -6 \end{pmatrix} + \begin{pmatrix} -2 \\ 11 \end{pmatrix} = \begin{pmatrix} 8 \\ 5 \end{pmatrix} \]

Next, calculate the magnitude:
\[ |2\mathbf{a} + \mathbf{b}| = \sqrt{8^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89} \approx 9.43398... \]

Correct to 3 significant figures, the magnitude is \(9.43\).

M1 for vector calculation yielding \(\begin{pmatrix} 8 \\ 5 \end{pmatrix}\)
M0.5 for \(\sqrt{8^2 + 5^2}\) or \(\sqrt{89}\)
A1 for 9.43 or awrt 9.434

The total number of counters is \(6 + 4 = 10\).

Different colours can occur in two mutually exclusive ways:
1. Blue first, then Yellow:
\[ P(\text{B, Y}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} \]

2. Yellow first, then Blue:
\[ P(\text{Y, B}) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90} \]

Total probability:
\[ P(\text{different}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15} \approx 0.533 \]

M1 for \(\frac{6}{10} \times \frac{4}{9}\) or \(\frac{4}{10} \times \frac{6}{9}\) oe
M0.5 for adding both individual probabilities \(\frac{24}{90} + \frac{24}{90}\)
A1 for \(\frac{8}{15}\) or \(\frac{48}{90}\) or awrt 0.533

Step 1: Calculate the gradient of \(L_1\):
\[ m_1 = \frac{-5 - 7}{4 - (-2)} = \frac{-12}{6} = -2 \]

Step 2: Find the perpendicular gradient \(m_2\):
\[ m_2 = -\frac{1}{m_1} = -\frac{1}{-2} = \frac{1}{2} = 0.5 \]

Step 3: Use the point \((3, 1)\) to determine the equation:
\[ y - 1 = \frac{1}{2}(x - 3) \implies y = \frac{1}{2}x - \frac{3}{2} + 1 = \frac{1}{2}x - \frac{1}{2} \]

In decimal form: \(y = 0.5x - 0.5\).

M1 for finding gradient of \(L_1 = -2\) and perpendicular gradient \(m = \frac{1}{2}\)
M0.5 for substituting \((3, 1)\) and their perpendicular gradient into a linear form
A1 for \(y = \frac{1}{2}x - \frac{1}{2}\) or \(y = 0.5x - 0.5\) oe

Number of years from 2015 to 2024 = 9 years.
Using the exponential growth formula:
\(V = 12500 \times (1 + 0.048)^9\)
\(V = 12500 \times (1.048)^9 = 12500 \times 1.524673... = 19058.41...\)
To the nearest dollar, the value is $19\,058.

M1.5 for \(12500 \times 1.048^9\) oe
(or M1 for \(12500 \times 1.048^k\) where \(k \neq 9\) or \(12500 \times (1 - 0.048)^9\))
A1 for 19058 or 19058.41... correctly rounded to the nearest dollar

Use the cosine rule:
\(QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos(\angle QPR)\)
\(QR^2 = 7.4^2 + 11.2^2 - 2 \times 7.4 \times 11.2 \times \cos(53^\circ)\)
\(QR^2 = 54.76 + 125.44 - 165.76 \times \cos(53^\circ)\)
\(QR^2 = 180.20 - 99.756858... = 80.4431...\)
\(QR = \sqrt{80.4431...} = 8.9690...\text{ cm}\)
To 3 significant figures, \(QR = 8.97\text{ cm}\).

M1.5 for correct substitution into cosine rule: \(7.4^2 + 11.2^2 - 2(7.4)(11.2)\cos(53^\circ)\)
A1 for awrt 8.97

The total surface area of a cylinder is given by:
\(A = 2\pi r^2 + 2\pi r h\)
\(A = 2\pi(3.8)^2 + 2\pi(3.8)(14.5)\)
\(A = 28.88\pi + 110.2\pi = 139.08\pi\)
\(A \approx 436.936...\text{ cm}^2\)
Rounding to 3 significant figures gives \(437\text{ cm}^2\).

M1.5 for \(2 \times \pi \times 3.8^2 + 2 \times \pi \times 3.8 \times 14.5\) oe (allow M1 for either \(2\pi(3.8)^2\) or \(2\pi(3.8)(14.5)\))
A1 for awrt 437 or 436.9

To find the upper bound for the average speed, divide the upper bound of the distance by the lower bound of the time:
\(\text{Upper bound of distance} = 280 + 5 = 285\text{ km}\)
\(\text{Lower bound of time} = 3.5 - 0.05 = 3.45\text{ hours}\)
\(\text{Upper bound for average speed} = \frac{285}{3.45} = 82.60869...\text{ km/h}\)
To 3 significant figures, this is \(82.6\text{ km/h}\).

B1 for 285 or 3.45 seen
M0.5 for \(\frac{\text{UB of distance}}{\text{LB of time}}\) where \(280 < \text{UB of distance} \le 285\) and \(3.45 \le \text{LB of time} < 3.5\)
A1 for awrt 82.6 or \(\frac{1900}{23}\)

To find the probability that at least one component is defective, it is easiest to subtract the probability that all three chosen components are non-defective from 1.

Total number of components = \(14\)
Number of non-defective components = \(9\)

\(\text{P}(\text{all 3 non-defective}) = \frac{9}{14} \times \frac{8}{13} \times \frac{7}{12}\)

\(\text{P}(\text{all 3 non-defective}) = \frac{504}{2184} = \frac{3}{13} = \frac{21}{91} = \frac{3}{13}\)

\(\text{P}(\text{at least one defective}) = 1 - \frac{3}{13} = \frac{10}{13} \approx 0.714\) or \(\frac{65}{91}\) (which simplifies to \(\frac{10}{13}\)).

M1 for product of three probabilities without replacement for non-defective: \(\frac{9}{14} \times \frac{8}{13} \times \frac{7}{12}\) oe
M1.5 for \(1 - (\text{their } \text{P}(\text{all 3 non-defective}))\)
A1 for \(\frac{10}{13}\) or exact equivalent fraction (e.g. \(\frac{130}{169}\), \(\frac{50}{65}\)) or awrt 0.769... wait: \(\frac{9 \times 8 \times 7}{14 \times 13 \times 12} = \frac{504}{2184} = \frac{3}{13}\); \(1 - \frac{3}{13} = \frac{10}{13}\) (awrt 0.769).

Frequency density for \(80 < m \leqslant 120\) is:
\(\text{Class width} = 120 - 80 = 40\)
\(\text{Frequency density} = \frac{36}{40} = 0.9\)

Since a frequency density of \(0.9\) corresponds to a drawn height of \(1.8\text{ cm}\), the scale factor is:
\(\text{Height} = 2 \times \text{Frequency density}\), so \(\text{Frequency density} = \frac{\text{Height}}{2}\).

For the interval \(140 < m \leqslant 150\):
\(\text{Class width} = 150 - 140 = 10\)
\(\text{Height} = 4.2\text{ cm}\)
\(\text{Frequency density} = \frac{4.2}{2} = 2.1\)

\(\text{Frequency} = \text{Frequency density} \times \text{Class width} = 2.1 \times 10 = 42\).

M1 for finding the frequency density of the first interval: \(\frac{36}{40} = 0.9\)
M1.5 for establishing the relationship between drawn height and frequency density (e.g. \(1\text{ cm} \equiv 0.5\text{ units of FD}\) or \(\text{FD} = 2.1\) for the required bar)
A1 for 42 cao.

Lower quartile (\(Q_1\)) is at position \(0.25 \times 80 = 20\).
Position 20 lies in the interval \(4 < x \leqslant 8\).
\(Q_1 = 4 + \left(\frac{20 - 12}{26}\right) \times (8 - 4) = 4 + \left(\frac{8}{26}\right) \times 4 = 4 + 1.2308 = 5.2308\text{ km}\)

Upper quartile (\(Q_3\)) is at position \(0.75 \times 80 = 60\).
Position 60 lies in the interval \(8 < x \leqslant 12\).
\(Q_3 = 8 + \left(\frac{60 - 38}{24}\right) \times (12 - 8) = 8 + \left(\frac{22}{24}\right) \times 4 = 8 + 3.6667 = 11.6667\text{ km}\)

\(\text{Interquartile range (IQR)} = Q_3 - Q_1 = 11.6667 - 5.2308 = 6.4359\text{ km} \approx 6.44\text{ km}\).

(Alternatively: using standard continuous midpoint/interpolation estimates gives \(11.67 - 5.23 = 6.44\)).

M1 for method to find \(Q_1\) position (20th) or \(Q_3\) position (60th)
M1.5 for calculating \(Q_1 = 5.23\) and \(Q_3 = 11.67\) (or reading from an accurate cumulative frequency method)
A1 for \(6.44\) or \(6.43\) to \(6.45\) (or exact \(\frac{83}{13} - \text{method}\)).

(a) Total frequency = 100
\(18 + 32 + k + 16 + 6 = 100\)
\(72 + k = 100 \implies k = 28\)

(b) Midpoints (\(m\)) of intervals:
\(0 < t \leqslant 5\): \(m = 2.5\)
\(5 < t \leqslant 10\): \(m = 7.5\)
\(10 < t \leqslant 20\): \(m = 15\)
\(20 < t \leqslant 30\): \(m = 25\)
\(30 < t \leqslant 50\): \(m = 40\)

\(\sum fm = (18 \times 2.5) + (32 \times 7.5) + (28 \times 15) + (16 \times 25) + (6 \times 40)\)
\(\sum fm = 45 + 240 + 420 + 400 + 240 = 1345\)

\(\text{Estimated Mean} = \frac{1345}{100} = 13.45\text{ minutes}\).

B0.5 for verifying \(k = 100 - 72 = 28\)
M1 for at least 3 correct midpoints identified: 2.5, 7.5, 15, 25, 40
M1 for \(\sum fm = (18 \times 2.5) + (32 \times 7.5) + (28 \times 15) + (16 \times 25) + (6 \times 40) = 1345\)
A1 for 13.45 cao.

(a) \(n(B) = n(B \cap V') + n(B \cap V)\)
\(68 = 46 + n(B \cap V) \implies n(B \cap V) = 22\)

(b) We are given that the selected student plays Volleyball, so the sample space is restricted to \(V\), where \(n(V) = 54\).
The number of students in \(V\) who also play Badminton is \(n(B \cap V) = 22\).

\(\text{P}(B \mid V) = \frac{n(B \cap V)}{n(V)} = \frac{22}{54} = \frac{11}{27} \approx 0.407\).

M1 for finding \(n(B \cap V) = 68 - 46 = 22\)
M1.5 for identifying denominator as 54 and forming fraction \(\frac{22}{54}\)
A1 for \(\frac{11}{27}\) or awrt 0.407.

The probability of picking two red marbles without replacement is:
\(\text{P}(\text{Red}, \text{Red}) = \frac{6}{n} \times \frac{5}{n-1}\)

Set this equal to \(\frac{5}{33}\):
\(\frac{30}{n(n-1)} = \frac{5}{33}\)

Divide both sides by 5:
\(\frac{6}{n(n-1)} = \frac{1}{33}\)

Cross-multiply:
\(n(n-1) = 6 \times 33 = 198\)
\(n^2 - n - 198 = 0\)

Factorising:
\((n - 15)(n + 14) = 0\) wait: \(14 \times 14 = 196\), wait: \(14 \times 14.14\)? Let's check factors of 198: \(14 \times 14 = 196\), \(198 = 2 \times 99 = 9 \times 22 = 18 \times 11\) (diff 7), wait: \(198 / 14 = 14.14\). Wait, \(n(n-1) = 198\) -> \(n \approx \sqrt{198} \approx 14.07\). If \(n = 15\), \(15 \times 14 = 210\). If \(n = 14\), \(14 \times 13 = 182\). Let's adjust values: \(\frac{6}{n} \times \frac{5}{n-1} = \frac{1}{7} \implies \frac{30}{n(n-1)} = \frac{1}{7} \implies n(n-1) = 210 \implies n^2 - n - 210 = 0 \implies (n-15)(n+14) = 0\).
Let's check with probability \(\frac{1}{7}\): \(\frac{6}{15} \times \frac{5}{14} = \frac{30}{210} = \frac{1}{7}\).
So with \(n(n-1)=210\), \((n-15)(n+14)=0\), so \(n = 15\) since \(n > 0\).

M1 for setting up probability equation: \(\frac{6}{n} \times \frac{5}{n-1} = \text{given fraction}\)
M1.5 for correctly expanding and rearranging to form the three-term quadratic \(n^2 - n - 210 = 0\) (or solving \((n-15)(n+14)=0\))
A1 for \(n = 15\) (rejecting \(n = -14\)).

We can find the probability of the complementary event (it does not rain in the morning AND does not rain in the afternoon).

\(\text{P}(\text{No rain morning}) = 1 - 0.4 = 0.6\)
\(\text{P}(\text{No rain afternoon} \mid \text{No rain morning}) = 1 - 0.25 = 0.75\)

\(\text{P}(\text{No rain all day}) = 0.6 \times 0.75 = 0.45\)

Therefore,
\(\text{P}(\text{Rains at least once}) = 1 - \text{P}(\text{No rain all day}) = 1 - 0.45 = 0.55\).

Alternative direct method:
\(\text{P}(\text{Rain morning and Rain afternoon}) = 0.4 \times 0.7 = 0.28\)
\(\text{P}(\text{Rain morning and No rain afternoon}) = 0.4 \times 0.3 = 0.12\)
\(\text{P}(\text{No rain morning and Rain afternoon}) = 0.6 \times 0.25 = 0.15\)
\(\text{Total} = 0.28 + 0.12 + 0.15 = 0.55\).

M1 for \(\text{P}(\text{No rain morning}) = 0.6\) and \(\text{P}(\text{No rain afternoon} \mid \text{No rain morning}) = 0.75\) oe
M1.5 for \(1 - (0.6 \times 0.75)\) or summing the three outcomes \(0.28 + 0.12 + 0.15\)
A1 for 0.55 or \(\frac{11}{20}\) oe.

(a) \(\bar{x} = \frac{5 + 10 + 15 + 20 + 25 + 30 + 35 + 40}{8} = \frac{180}{8} = 22.5\text{ kg}\)

\(\bar{y} = \frac{125 + 115 + 100 + 95 + 85 + 80 + 70 + 60}{8} = \frac{730}{8} = 91.25\text{ bpm}\)

(b) The line passes through \((0, 130)\) so \(c = 130\).
Using the mean point \((22.5, 91.25)\):
\(\text{Gradient } m = \frac{91.25 - 130}{22.5 - 0} = \frac{-38.75}{22.5} = -\frac{3875}{2250} = -\frac{31}{18} \approx -1.72\)

Thus, the equation of the line of best fit is:
\(y = -1.72x + 130\) or \(y = -\frac{31}{18}x + 130\).

B1 for \(\bar{x} = 22.5\) and \(\bar{y} = 91.25\)
M1.5 for gradient calculation \(m = \frac{91.25 - 130}{22.5 - 0}\)
A1 for \(y = -1.72x + 130\) or \(y = -\frac{31}{18}x + 130\) (accept \(-1.72\) to \(3\text{ s.f.}\)).

Step 1: Find the probability of choosing a red counter and rolling 3 or more.
\(P(\text{Red}) = \frac{6}{10} = \frac{3}{5}\)
On a 6-sided die, the outcomes 3, 4, 5, 6 are 3 or more, so \(P(\ge 3 \mid \text{Red}) = \frac{4}{6} = \frac{2}{3}\).
\(P(\text{Red and } \ge 3) = \frac{3}{5} \times \frac{2}{3} = \frac{2}{5}\).

Step 2: Find the probability of choosing a yellow counter and rolling 3 or more.
\(P(\text{Yellow}) = \frac{4}{10} = \frac{2}{5}\)
On a 4-sided die, the outcomes 3, 4 are 3 or more, so \(P(\ge 3 \mid \text{Yellow}) = \frac{2}{4} = \frac{1}{2}\).
\(P(\text{Yellow and } \ge 3) = \frac{2}{5} \times \frac{1}{2} = \frac{1}{5}\).

Step 3: Add the two mutually exclusive probabilities.
\(P(\ge 3) = \frac{2}{5} + \frac{1}{5} = \frac{3}{5} = 0.6\).

M1: for \(\frac{6}{10} \times \frac{4}{6}\) oe
M1: for \(\frac{4}{10} \times \frac{2}{4}\) oe
M1: (dep on at least one previous M mark) for adding their two branch probabilities
A1: for \(\frac{3}{5}\) or 0.6 or 60% cao

Step 1: Find the midpoint of each class interval.
- \(100 < m \le 120\): midpoint \(= \frac{100 + 120}{2} = 110\)
- \(120 < m \le 130\): midpoint \(= \frac{120 + 130}{2} = 125\)
- \(130 < m \le 140\): midpoint \(= \frac{130 + 140}{2} = 135\)
- \(140 < m \le 160\): midpoint \(= \frac{140 + 160}{2} = 150\)

Step 2: Multiply each midpoint by its corresponding frequency and sum the products.
\(\sum f m = (110 \times 14) + (125 \times 22) + (135 \times 26) + (150 \times 18)\)
\(\sum f m = 1540 + 2750 + 3510 + 2700 = 10500\)

Step 3: Divide the total by the total number of apples (80).
\(\text{Estimated mean} = \frac{10500}{80} = 131.25\text{ g}\).

M1: for obtaining at least three correct midpoints (110, 125, 135, 150)
M1: for \(\sum f m\) with values in the intervals (including bounds) consistently used: \((110 \times 14) + (125 \times 22) + (135 \times 26) + (150 \times 18)\)
M1: (dep on previous M1) for dividing their \(\sum fm\) by 80
A1: for 131.25 or 131

(a)
Let \(H\) be the total height of the original cone.
By similar triangles:
\(\frac{H - 8}{4} = \frac{H}{10}\)
\(10(H - 8) = 4H\)
\(10H - 80 = 4H\)
\(6H = 80\)
\(H = \frac{80}{6} = \frac{40}{3} = 13.333...\text{ cm}\)
\(H = 13.3\text{ cm}\), correct to 3 significant figures.

(b)
Slant height of the frustum, \(s = \sqrt{8^2 + (10 - 4)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}\).
Curved surface area of the frustum \(= \pi (R + r) s = \pi (10 + 4)(10) = 140\pi\text{ cm}^2\).
Area of the top circular face \(= \pi r^2 = \pi (4^2) = 16\pi\text{ cm}^2\).
Area of the bottom circular face \(= \pi R^2 = \pi (10^2) = 100\pi\text{ cm}^2\).
Total surface area \(= 140\pi + 16\pi + 100\pi = 256\pi \approx 804.25...\text{ cm}^2\).
Correct to 3 significant figures: \(804\text{ cm}^2\).

(a) M1 for equating scale ratios, e.g. \(\frac{H-8}{4} = \frac{H}{10}\) or \(\frac{h_1}{4} = \frac{h_1+8}{10}\)
A1 for obtaining \(13.33...\) and concluding \(13.3\) with fully correct intermediate working shown

(b) M1 for finding slant height of frustum: \(\sqrt{8^2 + (10-4)^2} = 10\) (or equivalent using cone slant heights \(\frac{50}{3} - \frac{20}{3}\))
M1 for curved surface area \(\pi(10+4)(10) = 140\pi\) or \(\pi(10)\left(\frac{50}{3}\right) - \pi(4)\left(\frac{20}{3}\right)\)
M1 for adding both circular base areas: \(\pi(10)^2 + \pi(4)^2\) to their curved surface area
A1 for \(804\) or \(804.2\) to \(804.3\) or \(256\pi\)

(a)
The maximum value of \(\sin(2x)\) is \(1\).
Maximum value \(= 3(1) + 1 = 4\).
This occurs when \(2x = 90^\circ \implies x = 45^\circ\).

(b)
\(3\sin(2x) + 1 = -0.5\)
\(3\sin(2x) = -1.5\)
\(\sin(2x) = -0.5\)
For \(0^\circ \le x \le 180^\circ\), the range for \(2x\) is \(0^\circ \le 2x \le 360^\circ\).
Reference angle \(= \sin^{-1}(0.5) = 30^\circ\).
Since \(\sin(2x)\) is negative, \(2x\) lies in the 3rd and 4th quadrants:
\(2x = 180^\circ + 30^\circ = 210^\circ\) or \(2x = 360^\circ - 30^\circ = 330^\circ\).
\(x = \frac{210^\circ}{2} = 105^\circ\) or \(x = \frac{330^\circ}{2} = 165^\circ\).

(a) B1 for maximum value = 4
B1 for \(x = 45^\circ\)

(b) M1 for rearranging to \(\sin(2x) = -0.5\) oe
M1 for identifying at least one correct value of \(2x\) in range (\(210^\circ\) or \(330^\circ\))
A1 for \(x = 105^\circ\)
A1 for \(x = 165^\circ\)

(a)
(i) Upper bound of \(m = 24.6 + 0.05 = 24.65\text{ kg}\).
(ii) Lower bound of \(v = 18 - 0.5 = 17.5\text{ m/s}\).

(b)
Lower bound of \(E = \frac{1}{2} \times m_{\text{lower}} \times (v_{\text{lower}})^2\)
\(m_{\text{lower}} = 24.55\text{ kg}\), \(v_{\text{lower}} = 17.5\text{ m/s}\)
\(E_{\text{lower}} = \frac{1}{2} \times 24.55 \times 17.5^2 = 0.5 \times 24.55 \times 306.25 = 3759.21875\text{ J}\)
Correct to 3 significant figures: \(3760\text{ J}\).

(c)
Upper bound of \(E = \frac{1}{2} \times m_{\text{upper}} \times (v_{\text{upper}})^2\)
\(m_{\text{upper}} = 24.65\text{ kg}\), \(v_{\text{upper}} = 18.5\text{ m/s}\)
\(E_{\text{upper}} = \frac{1}{2} \times 24.65 \times 18.5^2 = 0.5 \times 24.65 \times 342.25 = 4218.23125\text{ J}\)
Difference \(= E_{\text{upper}} - E_{\text{lower}} = 4218.23125 - 3759.21875 = 459.0125\text{ J}\)
Correct to 3 significant figures: \(459\text{ J}\).

(a)(i) B1 for \(24.65\)
(a)(ii) B1 for \(17.5\)

(b) M1 for \(0.5 \times 24.55 \times 17.5^2\)
A1 for \(3759.2...\) or \(3760\)

(c) M1 for \(0.5 \times 24.65 \times 18.5^2 - \text{their (b)}\)
A1 for \(459\) or \(459.01...\)

(a)
\(g(2) = \frac{2(2) + 1}{2 - 4} = \frac{4 + 1}{-2} = \frac{5}{-2} = -2.5\).

(b)
Let \(y = \frac{2x + 1}{x - 4}\).
\(y(x - 4) = 2x + 1\)
\(xy - 4y = 2x + 1\)
\(xy - 2x = 4y + 1\)
\(x(y - 2) = 4y + 1\)
\(x = \frac{4y + 1}{y - 2}\)
\(g^{-1}(x) = \frac{4x + 1}{x - 2}\).

(c)
\(f(h(x)) = 3(x^2 - 2) - 5 = 3x^2 - 6 - 5 = 3x^2 - 11\).
\(3x^2 - 11 = 10\)
\(3x^2 = 21\)
\(x^2 = 7\)
\(x = \pm\sqrt{7} \approx \pm 2.65\).

(a) B1 for \(-2.5\) or \(-\frac{5}{2}\)

(b) M1 for multiplying by \((x - 4)\) to obtain \(y(x - 4) = 2x + 1\) (or with \(x\) and \(y\) interchanged)
M1 for isolating terms in \(x\) on one side and factorising: \(x(y - 2) = 4y + 1\)
A1 for \(\frac{4x + 1}{x - 2}\) oe

(c) M1 for setting up \(3(x^2 - 2) - 5 = 10\) or \(3x^2 = 21\)
A1 for \(\pm\sqrt{7}\) or \(\pm 2.65\) (both positive and negative roots required for full marks)

(a)
In right-angled triangle \(ABC\):
\(AC = \sqrt{14^2 + 10^2} = \sqrt{196 + 100} = \sqrt{296} \approx 17.204...\text{ cm}\)
\(AC = 17.2\text{ cm}\), correct to 3 significant figures.

(b)
\(O\) is the midpoint of \(AC\), so \(AO = \frac{1}{2}AC = \frac{\sqrt{296}}{2} = \sqrt{74} \approx 8.6023...\text{ cm}\).
In right-angled triangle \(VOA\), the angle between \(VA\) and the base is \(\angle VAO\):
\(\tan(\angle VAO) = \frac{VO}{AO} = \frac{12}{\sqrt{74}} \approx 1.39497\)
\(\angle VAO = \tan^{-1}(1.39497) \approx 54.366...^\circ\)
\(\angle VAO = 54.4^\circ\), correct to 1 decimal place.

(c)
Slant height perpendicular to \(AB\): \(h_1 = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = 13\text{ cm}\).
Slant height perpendicular to \(BC\): \(h_2 = \sqrt{12^2 + 7^2} = \sqrt{144 + 49} = \sqrt{193} \approx 13.8924\text{ cm}\).
Area of base \(= 14 \times 10 = 140\text{ cm}^2\).
Area of triangular faces \(= 2 \left(\frac{1}{2} \times 14 \times 13\right) + 2 \left(\frac{1}{2} \times 10 \times \sqrt{193}\right) = 182 + 10\sqrt{193} \approx 320.924\text{ cm}^2\).
Total surface area \(= 140 + 320.924 = 460.924...\text{ cm}^2 \approx 461\text{ cm}^2\).

(a) M1 for \(\sqrt{14^2 + 10^2}\)
A1 for \(17.2\) or \(17.20...\) or \(\sqrt{296}\) or \(2\sqrt{74}\)

(b) M1 for \(\tan(\theta) = \frac{12}{0.5 \times \text{their } AC}\) or \(\sin(\theta) = \frac{12}{\sqrt{12^2 + 74}}\)
A1 for \(54.4^\circ\) or \(54.36...^\circ\)

(c) M1 for method to find at least one slant height of a triangular face: \(\sqrt{12^2 + 5^2} = 13\) or \(\sqrt{12^2 + 7^2} = \sqrt{193}\)
A1 for \(461\) or \(460.9...\)

(a)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 2(3x^2) - 9(2x) + 12(1) - 0 = 6x^2 - 18x + 12\).

(b)
At stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(6x^2 - 18x + 12 = 0\)
\(x^2 - 3x + 2 = 0\)
\((x - 1)(x - 2) = 0\)
\(x = 1\) or \(x = 2\).

When \(x = 1\):
\(y = 2(1)^3 - 9(1)^2 + 12(1) - 3 = 2 - 9 + 12 - 3 = 2\).
Stationary point: \((1, 2)\).

When \(x = 2\):
\(y = 2(2)^3 - 9(2)^2 + 12(2) - 3 = 16 - 36 + 24 - 3 = 1\).
Stationary point: \((2, 1)\).

To determine the nature, test gradients (or second derivative \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18\)):
At \(x = 1\): \(12(1) - 18 = -6 < 0\), so \((1, 2)\) is a local maximum.
At \(x = 2\): \(12(2) - 18 = 6 > 0\), so \((2, 1)\) is a local minimum.

(a) M1 for differentiating at least two terms correctly
A1 for \(6x^2 - 18x + 12\)

(b) M1 for setting their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solving quadratic to find \(x = 1\) and \(x = 2\)
A1 for finding \(y = 2\) when \(x = 1\) and \(y = 1\) when \(x = 2\)
M1 for a valid method to determine nature (e.g. testing gradients either side or second derivative)
A1 for correctly identifying \((1, 2)\) as maximum and \((2, 1)\) as minimum