Foundation January 2023 Paper 2 Q24
24
(a) Write down the value of \((m + 2)^0\) where \(m\) is a positive integer. (1)
(b) Simplify \((3a^2b^4)^3\) (2)
(c) Factorise fully \(14x^2y^4 + 21x^3y^2\) (2)
The diagram shows a straight line drawn on a grid.
(d) Write down an equation of the line. (2)
| Scheme | Marks |
|---|---|
| 1 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(27a^6b^{12}\) | B2 |
| (2) |
Notes
B2: (B1 for 2 of 3 parts in a product)
| Scheme | Marks |
|---|---|
| \(7x^2y^2(2y^2 + 3x)\) | B2 |
| (2) |
Notes
B2: B1 for a correct factorisation with at least 2 factors outside (eg \(7x\), \(x^2\), \(xy\), etc) eg \(7x(2xy^4 + 3x^2y^2)\) eg \(x^2y^2(14y^2 + 21x)\) or for the correct common factor with just one mistake inside the bracket eg \(7x^2y^2(2y + 3x)\) which is missing the squared on the \(y\) term
| Scheme | Marks |
|---|---|
| \(y = mx + 4\) where \(m \neq 0\) oe (eg \(y = 2x + 4\)) or \(y = -2x + c\) or \(y + 2x = c\) oe or \(-2x + 4\) or \(\mathrm{f}(x) = -2x + 4\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(y = -2x + 4\) | A1 |
| (2) | |
| (7 marks) |
Notes
A1: oe eg \(y + 2x = 4\)