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MC卷 第 39-45題
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MC卷 第 39-45題 [IMG]http://i293.photobucket.com/albums/mm67/zaza520/39-45.jpg[/IMG]
最佳解答:
(39) Comparing the log values of the 4 nos. 3000 log 500 = 3000 (2 + log 5) = 6000 + 3000 log 5 2500 log 2000 = 2500 (3 + log 2) = 7500 + 2500 log 2 2000 log 2500 = 2000 (3 + log 2.5) = 6000 + 2000 log 2.5 500 log 3000 = 500 (3 + log 3) = 1500 + 500 log 3 So we have 2500 log 2000 being the greatest, ans = B (40) Since f(x) is divisible by x - 1, f(1) = 0 So for f(2x + 1), it will become zero when x = -1/2 and hence it is divisible by x + 1/2 or 2x + 1. (41) The digit 7 occupies the 164 digit, up to 169 digit for A So the ans should be A. (42) q2 - 12q + 6 = 2q - 7 q2 - 14q + 13 = 0 (q - 13)(q - 1) = 0 q = 1 or 13 p = -5 or 19 (43) y >= 4 means that it should be above the line y = 4 (including the line) x + y = 8 means that it should be above the line 2x + y = 8 (including the line) So region II satisfies all the conditions. (44) From the given, we can find that the common difference is 6, so The sum up to the 28th term is 1624, i.e. (a1 + a1 + 27 x 6) x (28/2) = 1624 2a1 + 162 = 116 a1 = -23 (45) First term = 4 Common ratio = -1/2 So with the sum to infinity, we have: S = a/(1 - R) = 4/(1 + 1/2) = 8/3
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