Higher November 2022 Paper 4 Q9
9 Here are two pieces of work.
For each one, describe the error in the method and give the correct answer.
(a)
Question:
Rearrange \(y = 3x + 17\) to make \(x\) the subject.
Solution:
\(y = 3x + 17\)
\(y + 17 = 3x\)
\(x = \dfrac{y + 17}{3}\)
Error is ……………
Correct answer …………… [2]
(b)
Question:
Rearrange \(A = 4x^2\) to make \(x\) the subject, where \(x \gt 0\).
Solution:
\(A = 4x^2\)
\(\sqrt{A} = \sqrt{4x^2}\)
\(\sqrt{A} = 4x\)
\(x = \dfrac{\sqrt{A}}{4}\)
Error is ……………
Correct answer …………… [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| it should be −17 | 1 | accept any correct explanation | |
| \(x = \frac{y - 17}{3}\) | 1 | allow correct equation if written in the question just missing “\(x\) = ” | |
Appendix: exemplar responses for Q9(a)
| Response | Mark |
|---|---|
| it should be −17 | 1 |
| the sign of 17 is wrong | 1 |
| the error is adding 17 | 1 |
| he has added 17 | 1 |
| the sign on the right should not be plus | 1 |
| he has to do the inverse of +17 | 1 |
| the 17 maintained a positive sign on the other side | 1 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| it should be 2\(x\) | 1 | Do not accept ± 2\(x\) or “the error is that they missed the ±” | |
| \(x = \frac{\sqrt{A}}{2}\) or \(x = \sqrt{\frac{A}{4}}\) | 1 | Do not penalise twice missing the “\(x\) =”. allow correct equation if written in the question just missing “\(x\) = ” | |
Appendix: exemplar responses for Q9(b)
| Response | Mark |
|---|---|
| it should be 2\(x\) | 1 |
| It should be \(\sqrt{4}\) | 1 |
| take the root of 4 | 1 |
| include 4 in the square root | 1 |
| only A is rooted | 1(BOD) |
| they just square rooted A | 1(BOD) |
| they should have divided by 4 first | 0 |