3 and 4 are incorrect, because the middle line is the median and the median of A is bigger than B

\( 20^2 + 21^2 = 29^2 \)

The smallest triple is \( 3^2 + 4^2 = 5^2 \). If this is not known, it can easily be worked out by writing down all the square numbers until you see 2 that will add to make another one.

Sum of the integers in both equations = \( 20 + 21 + 29 + 3 + 4 + 5 = 82 \)

Product of the two largest numbers minus the 2 smallest numbers from both equations = \( (29 + 21) + (3 + 4) = 57 \)

HCF of 57 and 82:

82 factors: 1, 2, 41, 82
57 factors: 1, 3, 19, 57

HCF = 1

We use algebra to write each type of number. A number must always be a multiple of 3, 1 more than or 1 less than a multiple of 3. Therefore, we can test all numbers using the values n = 3k, n = 3k + 1, n = 3k -1.

Candidate should recognise the shape of the 3lnx graph and understand that a negative gradient will only cut the graph once for any value.

Since \( x^2 = 2x + 5 \rightarrow x = \pm \sqrt{2x + 5} \), there is one incorrect solution, coming from \( x = 1 – \sqrt{6} \), caused by an error made in line 2.

The fact that it has rained and been hot for 3 days is not important, the probability is still the same for each day. The easiest way to see the answer is to draw a probability table.

From this we can see the probability of it being hot will be 1.0 – 0.42 = 0.58
the probability of it raining and being hot must be 0.58 – 0.15 = 0.43
the probability of it raining and not being hot must be 0.55 – 0.43 = 0.12
the probability of it not raining 1.0 – 0.55 = 0.45
the probability it is not raining and not hot must therefore be 0.3

line 2, the symbol should be → not ↔, because if x is negative, the first line is not possible.
• line 3 the symbol should be ↔ not →
• line 4, the solution should be ±√125, or the arrow should just be → not ↔.

Candidate must identify that any stretch or reflection of the reciprocal graph will only fill 2 quadrants, eliminating options A and C. Likewise with the ln x graph, hence options D, E and F are eliminated. For option G, the minimum y coordinate is usually -1, which would be translated to 1 (by a translation 6 upwards, and then squashed by a factor of 5 in the vertical direction).

K: A : J
• for biscuit boxes:
• 3𝑥 ∶ 12𝑥 ∶ 𝑥 = 3: 12: 1
• for biscuits, the ratio remains the same:
• 12𝑥 = 𝑥 + 33, 𝑥 = 3
• therefore, there were 36 + 9 + 3 = 48 biscuits in total, 8 boxes.
In a month 32 boxes

\( f(g(x)) = f(4x – 2) = \frac{x – 3}{2} \)

Find the transformation that takes \( (4x – 2) \) to \( \frac{x – 3}{2} \)

divide by 8, then \( -1.25 \)

therefore \( f(x) = \frac{x}{8} – 1.25 \)

so \( f(x^2 – 2) = \frac{x^2 – 2}{8} – 1.25 = \frac{x^2 – 12}{8} \)

\( \frac{dy}{dx} = 48x^3 + 15x^2 + 14x + 3 \)

Where \( x = 2 \), gradient = \( 475 \) and \( y = 267 \)

The equation of the tangent is \( y = 475x – 683 \)

\( A_1 = x \)

\( A_2 = 3x + \frac{1}{3} \)

\( A_3 = 3\left(3x + \frac{1}{3}\right) + \frac{1}{3} = 9x + \frac{4}{3} \)

\( A_4 = 3\left(9x + \frac{4}{3}\right) + \frac{1}{3} = 27x + \frac{13}{3} \)

\( A_5 = 3\left(27x + \frac{13}{3}\right) + \frac{1}{3} = 81x + \frac{40}{3} \)

\( A_6 = 3\left(81x + \frac{40}{3}\right) + \frac{1}{3} = 243x + \frac{121}{3} \)

\( A_6 + A_5 + A_4 + A_3 = 243x + \frac{121}{3} + 81x + \frac{40}{3} + 27x + \frac{13}{3} + 9x + \frac{4}{3} \)

\( = 360x + \frac{178}{3} \)

\( 360x + \frac{178}{3} = 30 \)

\( 1080x + 178 = 90 \)

\( 1080x = -88 \)

\( x = -\frac{88}{1080} \)

\( y \) intercept is at \( 19 \)

the gradient is positive at first, so positive \( x^3 \)

A is a negative cubic graph, C is a quadratic, D and E are reciprocal graphs

Split each number into its factors:

\( \frac{12^{a-b} \times 36^{a+b}}{9^{3a+b} \times 16^a} = \frac{(3 \times 4)^{a-b} \times (3^2 \times 4)^{a+b}}{(3^2)^{3a+b} \times (4^2)^a} \)

\( = \frac{3^{a-b} \times 4^{a-b} \times 3^{2a+2b} \times 4^{a+b}}{3^{6a+2b} \times 4^{2a}} \)

\( = \frac{3^{3a+b} \times 4^{2a}}{3^{6a+2b} \times 4^{2a}} \)

\( = \frac{1}{3^{3a+b}} \)

\( V = Ak^t \)

When \( t = 0 \), \( V = A = 3000 \)

When \( t = 2 \), \( 9000 = 3000k^2 \)

\( 3 = k^2 \)

\( k = \sqrt{3} \)

when \( V = 270000 \),

\( 270000 = 3000\sqrt{3}^t \)

\( 90 = 3^{\frac{t}{2}} \)

\( 8100 = 3^t \)

\( t = \log_3 8100 = 2\log_3 90 \)

The year will be \( 2000 + t \), so is \( 2000 + 2\log_3 90 \)

\( \int_{2}^{4} \frac{2x^3 + 5\sqrt{x}}{x^4} \, dx \)

Separate out variables

\( = \int_{2}^{4} \left( 2x^{-1} + 5x^{-\frac{7}{2}} \right) dx \)

\( = \left[ 2\ln x – 2x^{-\frac{5}{2}} \right]_{2}^{4} \)

\( = \left( 2\ln 4 – 2(4^{-\frac{5}{2}}) \right) – \left( 2\ln 2 – 2(2^{-\frac{5}{2}}) \right) \)

\( = \left( \ln 16 – 2\left(\frac{1}{32}\right) \right) – \left( \ln 4 – 2^{-\frac{3}{2}} \right) \)

\( = \ln 4 – \frac{1}{16} + 2^{-\frac{3}{2}} \)

Sum of the first 10 terms = \( 5(2a + 9d) \)

Sum of the first 16 terms = \( 8(2a + 15d) \)

\( 10a + 45d = 16a + 120d \)

\( -6a = 75d \)

\( a = -\frac{25}{2} d \)

\( 2x + y = 2a \) can be rearranged to \( y = 2a – 2x \)

Sub into equation I:

\( 2x^2 + 3x(2a – 2x) = 4 \)

\( 2x^2 + 6ax – 6x^2 = 4 \)

\( -4x^2 + 6ax – 4 = 0 \)

Using the discriminant,

\( b^2 – 4ac = 0 \) if there is only one real root

\( (6a)^2 – (4 \times -4 \times -4) = 0 \)

\( 36a^2 – 64 = 0 \)

\( a^2 = \frac{16}{9} \)

\( a = \frac{4}{3} \) or \( -\frac{4}{3} \)