## Question 2:

### Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{3 + 2\sqrt{x} + 3 - 2\sqrt{x}}{(3 - 2\sqrt{x})(3 + 2\sqrt{x})}$ | M1 (AO 1.1) | Attempt to rewrite fractions using correct common denominator. Common denominator could just appear as $9 - 4x$. Must include correct attempt at numerators as well. |
| $\dfrac{6}{9 - 4x}$ | A1 (AO 2.1) | Obtain correct simplified fraction. No need to state values for $a$, $b$ and $c$ explicitly. www – if middle terms shown for expansion of denominator, these must be correct. ISW any further attempt to 'simplify'. SC B1 for answer only, with no method shown. |

**[2 marks]**

### Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{6}{9-4x} = 2$, so $6 = 18 - 8x$, $8x = 12$ | M1 (AO 1.1a) | Attempt to solve equation – as far as clearing the fraction and combining constant terms. M1 for using their fraction, as long as of correct form. Correct method to clear fraction, so M0 for e.g. $6 = 18 - 4x$, but allow sign error when combining constant terms. |
| $x = \dfrac{3}{2}$ | A1 (AO 1.1) | Obtain $x = \dfrac{3}{2}$. aef, but fractions must be simplified. |

**[2 marks]**

### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| **DR** $(2^y - 8)(2^y + 1)$ | M1 (AO 3.1a) | Attempt to solve disguised quadratic in $2^y$. If factorising then expansion should give $x^2$ and one other term correct. Quadratic formula should be correct – allow one slip when substituting as long as general formula already seen as correct. Completing the square needs to go as far as $x - p = \pm\sqrt{q}$. |
| $2^y = 8,\ 2^y = -1$ | A1 (AO 2.1) | Obtain two correct roots (could still be in terms of e.g. $u$ if substitution used). SC If no method shown then award B1 in place of M1A1 for both correct roots (final two marks can still be awarded). |
| $y = \log_2 8 = 3$ | M1 (AO 1.1) | Attempt to solve $2^y = k$, where $k > 0$. May just see $y = 3$, with no explicit use of $\log_2$. Allow BOD if attempt at solving $2^y = -1$ still present. If $k \neq 8$ then solution method must be seen, even if $k$ is a power of 2. |
| $y = 3$ only; $2^y = -1$ has no solutions as $2^y > 0$ for all $y$ | A1 (AO 2.3) | Obtain $y = 3$, having rejected $2^y = -1$ with some reasoning. Must have some reason, e.g. '$2^y$ is always positive', '$2^y$ cannot be negative', 'cannot take log of a negative number', 'not defined', 'not real', 'no solutions'. A0 for 'math error', 'does not work', 'not possible'. SC If no method at all shown then allow B1 for $y = 3$, with no other solutions. |

**[4 marks]**