论文部分内容阅读
1两种解法都正确吗问题设函数f(x)=|lgx|,若a≠b,且f(a)=f(b),求a+b的取值范围.解法1由已知不妨设a1.因为f(a)=f(b),所以lga=lgb.所以-lga=lgb,lga+lgb=0.所以lgab=0,ab=1.所以a+b≥2(ab)~(1/2)=21=2.因为a≠b,所以上式取不到“=”号.所以a+b的取值范围为(2,+∞).反思这是很多数学参考资料中的解答.仔细思考这种解法严密吗?(a+b)取不到2就能得出(a+b)的取值范围为(2,+∞)吗?大于2的一切实数都能取得到吗?
1 The two solutions are correct? The problem set function f (x) = | lgx |, if a ≠ b, and f (a) = f (b), find the value range of a + Let a 1. Since f (a) = f (b), lga = lgb. So -lga = lgb, lga + lgb = 0. So lgab = 0, ab = 1, so a + b≥2 (ab) ~ (1/2) = 21 = 2. Because a ≠ b, the above formula can not get “=” sign, so the range of a + b is (2, + ∞). Reflections This is the answer to many mathematical references. Do you think that this solution is tight? (A + b) Taking less than 2 we get the value of (a + b) 2, + ∞)? All real numbers greater than 2 can be achieved?