June 2023 Paper 1 Q2
2
Solve the equation \(2^{2y} - 7 \times 2^y - 8 = 0\). [4]
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{3 + 2\sqrt{x} + 3 - 2\sqrt{x}}{\left(3 - 2\sqrt{x}\right)\left(3 + 2\sqrt{x}\right)}\) | M1 | 1.1 |
| \(\dfrac{6}{9 - 4x}\) | A1 | 2.1 |
| [2] | ||
| (ii) \(\dfrac{6}{9 - 4x} = 2\) \(6 = 18 - 8x\) \(8x = 12\) | M1 | 1.1a |
| \(x = \frac{3}{2}\) | A1 | 1.1 |
| [2] |
Notes
(a)(i)
M1: Attempt to rewrite fractions using correct common denominator
Common denominator could just appear as \(9 - 4x\)
Must include correct attempt at numerators as well
A1: Obtain correct simplified fraction
No need to state values for \(a\), \(b\) and \(c\) explicitly
www – if middle terms shown for expansion of denominator, then these must be correct
ISW any further attempt to ‘simplify’
SC B1 for answer only, with no method shown
(a)(ii)
M1: 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 eg \(6 = 18 - 4x\), but allow sign error when combining constant terms
A1: Obtain \(x = \frac{3}{2}\)
aef, but fractions must be simplified
| Scheme | Marks | AO |
|---|---|---|
| DR \(\left(2^y - 8\right)\left(2^y + 1\right)\) | M1 | 3.1a |
| \(2^y = 8,\ 2^y = -1\) | A1 | 2.1 |
| \(y = \log_2 8 = 3\) | M1 | 1.1 |
| \(y = 3\) only; \(2^y = -1\) has no solutions as \(2^y \gt 0\) for all \(y\) | A1 | 2.3 |
| [4] |
Notes
M1: 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 \(x - p = \pm\sqrt{q}\)
A1: Obtain two correct roots (could still be in terms of eg \(u\) 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)
M1: Attempt to solve \(2^y = k\), where \(k \gt 0\)
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
A1: Obtain \(y = 3\), having rejected \(2^y = -1\) with some reasoning
Must have some reason, eg ‘\(2^y\) 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