Intermediate June 2018 Unit 1 Q12
12.
(a) Expand and simplify the following expression. [4]\[x(5x - 2) - 3(x^2 - 2x + 7)\]
(b) Solve \(\dfrac{22 - f}{3} = 6\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| \(5x^2 - 2x - 3x^2 + 6x - 21\) | B2 | B1 for sight of \(5x^2 - 2x\). B1 for sight of \(-\,3x^2 + 6x - 21\). Brackets must be removed. Allow both of the above B marks even if not part of a single expression. FT for B2 if at least two x2 terms AND at least two x terms to be simplified. FT for B1 if at least two x2 terms OR at least two x terms to be simplified. |
| \(= 2x^2 + 4x - 21\) | B2 | If B2 not awarded, allow B1 for correct collection of ‘x2 terms’ (2x2) OR B1 for correct collection of ‘x terms’ (+4x). This 2nd B2 (or B1) is for their final answer. Any compensating errors leading to a ‘correct’ answer is B0. Penalise −1 for any attempt to equate their expression to zero (and attempting to solve) OR Incorrectly factorising. |
| Scheme | Marks | Comments |
|---|---|---|
| 22 – f = 3 × 6 or equivalent. | M1 | |
| 22 – 18 = f OR −f = 18 – 22 | A1 | |
| f = 4 | A1 | C.A.O. Accept 4 = f. M1A1A0 for –f = −4. Mark final answer. Allow all 3 marks for \(\dfrac{22 - 4}{3} = 6\) with no further work. Allow 2 marks for \(\dfrac{22 - 4}{3} = 6\) followed by ‘f ≠ 4. If no marks gained. Allow SC1 for an unsupported f = −4. Allow SC1 for sight of 18 from 3 × 6. |