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1. notscience
'13.4.28 9:19 AM (98.217.xxx.116)m=12?
2. notscience
'13.4.28 9:41 AM (98.217.xxx.116)We are given
sqrt(a) - 2 < m, (1)
and
m < sqrt(a) + 1. (2).
From (1), we have m+2 > 0, and thus m-1 > - 3, that is, m-1 >= -2.
We will consider Cases (I) m-1>=0, and (II) 0 > m-1 >= -2 separately.3. notscience
'13.4.28 9:52 AM (98.217.xxx.116)Bare in mind that a is nonnegative, since a is an integer inside square root.
4. notscience
'13.4.28 9:53 AM (98.217.xxx.116)(I) m-1 >= 0
Note that (1) is equivalent to a < (m+2)^2.
Also note that (2) is equivalent to (m-1)^2 < a, since m-1 >=0.
Thus the condition [(1) and (2)] is equivalent to
(m-1)^2 < a < (m+2)^2.
Since these inequalities hold for 74 integers,
76 > (m+2)^2 - (m-1)^2 >= 74.
Hence
76 > 6m + 3 >= 74.
Therefore m=12.5. notscience
'13.4.28 9:54 AM (98.217.xxx.116)(II) 0 > m-1 >= -2
In this case, we have 3 > m+2 >= 1. Hence from (1), we have sqrt(a) < 3. So a < 9. This inequality does not hold for 74 nonnegative integers.6. notscience
'13.4.28 9:55 AM (98.217.xxx.116)By the arguments in Cases (I) and (II), m=12.
7. 준
'13.4.28 11:16 AM (59.7.xxx.194)^^감사합니다~~~