Higher November 2022 Paper 1 Q16
16
(a) Prove that \[(2m + 1)^2 - (2n - 1)^2 = 4(m + n)(m - n + 1)\] (3)
Sophia says that the result in part (a) shows that the difference of the squares of any two odd numbers must be a multiple of 4
(b) Is Sophia correct?
You must give reasons for your answer. (1)
You must give reasons for your answer. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof | M1 | for expansion of \((2m + 1)^2\) or \((2n - 1)^2\), all 4 terms correct with or without signs (and no additional terms) or 3 out of 4 terms correct with signs, eg \(4m^2 + 2m + 2m + 1\) or \(4n^2 - 2n - 2n + 1\) or for correct expansion of \(4(m + n)(m - n + 1)\) or \((m + n)(m - n + 1)\) eg \(4m^2 - 4mn + 4m + 4mn - 4n^2 + 4n\) oe or \(m^2 - mn + m + mn - n^2 + n\) oe or for \([2m + 1 + 2n - 1][(2m + 1) - (2n - 1)]\) |
| M1 | for correct expression after expansion for \((2m + 1)^2 - (2n - 1)^2\) eg \((4m^2 + 4m + 1) - (4n^2 - 4n + 1)\) or \(4m^2 + 4m + 1 - 4n^2 + 4n - 1\) oe \((= 4m^2 + 4m - 4n^2 + 4n)\) or for \([2m + 1 + 2n - 1][2m + 1 - 2n + 1]\) | |
| C1 | for a complete proof without any errors, eg uses difference of two squares to show that LHS = RHS or expands both sides and shows that LHS = RHS or expands and simplifies LHS and factorises convincingly to get RHS |
Additional guidance
Note that, for example, \(4m + 1\) is regarded as 3 terms in the expansion of \((2m + 1)^2\)
Must see correct expression
\[\begin{aligned} &4m^2 - 4n^2 + 4m + 4n \\ &= 4[(m^2 - n^2) + (m + n)] \\ &= 4[(m + n)(m - n) + (m + n)] \\ &= 4(m + n)(m - n + 1) \end{aligned}\]
| Answer | Mark | Mark scheme |
|---|---|---|
| Yes (supported) | C1 | for yes with explanation, eg \(2m + 1\) and \(2n - 1\) are odd numbers (for any positive integer value of \(m\), \(n\)) and the right-hand side is a multiple of 4 |