4MA0 Higher June 2017 Paper 4 Q2
2 \(M = 2t^2 - 7t\)
(a) Work out the value of \(M\) when \(t = -3\) (2)
(b) Solve \(4(x + 3) = 9x - 10\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
\(y\) is an integer.
\(-2 \lt y \leqslant 3\)
(c) Write down all the possible values of \(y\). (2)
| Scheme | Marks |
|---|---|
| \(2 \times (-3)^2 - 7 \times (-3)\) oe e.g. 2(9) –(–21) or 2×9 + 21 or 18 + 21 | M1 |
| 39 | A1 |
| (2) |
Notes
M1: Brackets must be round \((-3)^2\)
| Scheme | Marks |
|---|---|
| \(4x + 12 = 9x - 10\) or \(x + 3 = \dfrac{9x}{4} - \dfrac{10}{4}\) oe | M1 |
| \(12 + 10 = 9x - 4x\) or \(-9x + 4x = -12 - 10\) or \(22 = 5x\) or \(-5x = -22\) or \(3 + 2.5 = 2.25x - x\) or \(1.25x = 5.5\) | M1 |
| 4.4 | A1 |
| (3) |
Notes
M1: for \(4x + 12\) (may not be in an equation) or for dividing RHS by 4
M1: (ft from \(4x + b = 9x - 10\), \(b \neq 0\))
for all terms in \(x\) isolated on one side and numbers on other side
A1: for 4.4 oe eg. \(\dfrac{22}{5}\), \(4\dfrac{2}{5}\) dep on at least M1
| Scheme | Marks |
|---|---|
| −1, 0, 1, 2, 3 | B2 |
| (2) | |
| (7 marks) |
Notes
B2: B1 for −2, −1, 0, 1, 2 or list with one error or omission: e.g. −2, −1, 0, 1, 2, 3 ; −1, 0, 1, 2 ; −1, 1, 2, 3