**(i)** translation of 3 units in negative y-direction | B1 | State translation.

**State or imply 3 units in negative y-direction** | B1 | 2 | Independent of first B1. Statement needs to clearly intend a vertical downwards move of 3, without ambiguity or contradiction, such as '3 down', '-3 in the y direction' etc or vector notation. B0 if direction unclear, such as 'in the y-axis' (could be along or towards) or 'along the y-axis' (unless direction made clear). Allow '3' or '3 units' but not '3 places', '3 spaces', '3 squares', '3 coordinates' or mention of (scale) factor of 3. If both a valid statement and an ambiguous statement are made eg '3 units down on the y-axis' then still award B1. Ignore irrelevant statements, such as where the y-intercept is, whether correct or incorrect.

Give B0D on double negatives eg 'down the y-axis by -3 units' unless clearly wrong or contradictory eg 'negative y-direction by $\begin{pmatrix} -3 \\ 0 \end{pmatrix}$'.

**(ii)** $y = -2$ | B1 | 1 | State or imply $y = -2$. Just stating -2 is enough. B0 for final answer of $2^0 - 3$ or $1 - 3$.

(-2, 0) is B0 unless -2 already seen or implied as y-coordinate.

**(iii)** $2^x = 3$, $x = \log_2 3$ | M1 | Attempt to solve $2^x - 3 = 0$. Rearrange to $2^x = 3$, introduce logarithms (could be no base or any base as long as consistent) and then attempt expression for $x$.

M0 for $x = \log_2 2$. M1 A0 for alternative, correct, log expressions such as $\log_3 2$ or $1/\log_3 2$.

Decimal equivalent of 1.58 can get M1 A0.

$x = \log_3 (y + 3)$ is M0 (unless $y$ then becomes 0).

**State $\log_2 3$** | A1 | 2 | Doesn't need to be $x = ...$. Change of base is not on the specification, but is a valid method and can gain both marks. Allow if base not initially specified, but then both logs become base 2.

NB $x - \log_2 3 = 0$ leading to correct answer, can get full marks as there is no incorrect statement seen.

**(iv)** $2^p = 65$, $\log 2^p = \log 65$, $p\log 2 = \log 65$, $p = 6.02$ | M1* | Rearrange equation and introduce logs (or $\log_2$). Must first rearrange to $2^p = k$, with $k$ from attempt at $62 \pm 3$, before introducing logs.

**Drop power and attempt to solve** | M1d* | Dependent on first M1. If taking logs to any other base, or no base, or $\log_2$ on both sides then need to drop power of $p$ and attempt to solve using a sound algebraic method ie $p = \frac{\log k}{\log 2}$.

**Obtain 6.02, or better** | A1 | 3 | Decimal answer reqd, if more than 3sf it must be in range [6.022, 6.023]. Answer only, or trial and improvement, is 0/3 as no evidence of using logs as requested.

**(v)** $0.5 \times 0.5 \left[ 2^3 - 3 + 2(2^{3.5} - 3) + 2^4 - 3 \right] = 8.66$ | M1 | Attempt y-values at $x = 3, 3.5, 4$. M0 if other y-values also found (unless not used in trap rule). Allow M1 for using incorrect function as long as still clearly y-values that are intended to be the original function eg $2x - 3$ or $2^{(y-3)}$.

**Attempt correct trapezium rule** | M1 | Must be correct structure ie $0.5 \times 0.5 \times (y_0 + 2y_1 + y_2)$. Must be finding area from 3 to 4, so using eg $x = 0, 0.5, 1$ is M0. Allow if still in terms of $y_i$ etc as long as these have been clearly defined elsewhere. Using x-values in trapezium rule is M0, even if labelled y-values. Allow a different number of strips (except 1) as long as their $h$ is consistent with this, and the limits are still 3 and 4.

**Obtain 8.66, or better** | A1 | 3 | If final answer given to more than 3sf, allow answers in range [8.655, 8.657].

Exact answer from integrating $e^{\ln 2 - 3}$ is 0/3.

Answer only is 0/3.

Attempting integration before using trapezium rule is 0/3.

Using two separate trapezia is fine.

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