To determine which expression must be an odd integer given that [tex]\(m+I\)[/tex] is an even integer, let us analyze each option step-by-step.
1. Option E: [tex]\(m - 1\)[/tex]:
- Given that [tex]\(m + I\)[/tex] is even, let [tex]\(m + I = 2k\)[/tex] for some integer [tex]\(k\)[/tex].
- Substituting [tex]\(m = 2k - I\)[/tex], we get:
[tex]\[
m - 1 = (2k - I) - 1
\][/tex]
- The parity (odd or even nature) of [tex]\(m - 1\)[/tex] depends on [tex]\(I\)[/tex]. Since [tex]\(I\)[/tex] is not specified, [tex]\(m - 1\)[/tex] can either be odd or even depending on the value of [tex]\(I\)[/tex].
2. Option F: [tex]\(2m - 2\)[/tex]:
- Using [tex]\(m = 2k - I\)[/tex], we calculate:
[tex]\[
2m - 2 = 2(2k - I) - 2 = 4k - 2I - 2
\][/tex]
- Since [tex]\(4k - 2I - 2\)[/tex] is always even (product and difference of even integers), [tex]\(2m - 2\)[/tex] is always even and therefore cannot be an odd integer.
3. Option G: [tex]\(2m + 1\)[/tex]:
- Substituting [tex]\(m = 2k - I\)[/tex], compute:
[tex]\[
2m + 1 = 2(2k - I) + 1 = 4k - 2I + 1
\][/tex]
- [tex]\(4k - 2I\)[/tex] is an even number since [tex]\(4k\)[/tex] and [tex]\(-2I\)[/tex] are both even. Adding 1 to an even number gives us an odd number.
- Therefore, [tex]\(2m + 1\)[/tex] is always odd.
4. Option H: [tex]\(2m + 2\)[/tex]:
- Substituting [tex]\(m = 2k - I\)[/tex], we get:
[tex]\[
2m + 2 = 2(2k - I) + 2 = 4k - 2I + 2
\][/tex]
- Since [tex]\(4k - 2I + 2\)[/tex] is always even (sum of even numbers), [tex]\(2m + 2\)[/tex] is always even and hence cannot be an odd integer.
Among the provided options, only [tex]\(2m + 1\)[/tex] must be an odd integer.
Therefore, the correct answer is:
[tex]\[ \boxed{G} \][/tex]
