Higher June 2024 Paper 2 Q15
15
(a) Make \(g\) the subject of \(\quad e = \sqrt{\dfrac{7g + 5}{11 + 2g}}\) (4)
(b) Solve the inequality \(\quad 3y^2 + 4y - 32 \gt 0\)
Show your working clearly. (3)
Show your working clearly. (3)
| Scheme | Marks |
|---|---|
| \(e^2 = \dfrac{7g + 5}{11 + 2g}\) | M1 |
| \(11e^2 + 2e^2g = 7g + 5\) | M1 |
| eg \(2e^2g - 7g = 5 - 11e^2\) or \(11e^2 - 5 = 7g - 2e^2g\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(g = \dfrac{5 - 11e^2}{2e^2 - 7}\) | A1 |
| (4) |
Notes
M1: for removing square root
M1: For multiplying by denominator and expanding in a correct equation
M1: For gathering terms in \(g\) on one side and other terms the other side in a correct equation.
A1: or \(g = \dfrac{11e^2 - 5}{7 - 2e^2}\) oe eg \(g = \dfrac{\dfrac{5}{e^2} - 11}{2 - \dfrac{7}{e^2}}\) or \(g = \left(\dfrac{5 - 11e^2}{e^2 - 3.5}\right) \div 2\)
etc
| Scheme | Marks |
|---|---|
| \((3y - 8)(y + 4)\) | M1 |
| \(y = \dfrac{8}{3}\), \(y = -4\) | A1 |
working required Answer: \(y \lt -4\), \(y \gt \dfrac{8}{3}\) | A1 |
| (3) | |
| (7 marks) |
Notes
M1: For correct factorisation or correct use of quadratic formula \(\dfrac{-4 \pm \sqrt{4^2 - 4 \times 3 \times -32}}{2 \times 3}\) or as far as \(\dfrac{-4 \pm \sqrt{400}}{6}\)
\(\left(y - \dfrac{8}{3}\right)(y + 4)\) is not valid factorisation, unless preceded by division of quadratic by 3, so no marks
A1: dep on M1 for correct critical values (allow 2.6 or better or 2.7 )
A1: oe dep on M1 (allow use of \(x\) rather than \(y\))
or \((-\infty, -4), \left(\dfrac{8}{3}, (+)\infty\right)\) or \((-\infty, -4) \cup \left(\dfrac{8}{3}, (+)\infty\right)\) oe