题目内容
7.若(log23)x-(log53)x≥(log23)-y-(log53)-y,判断x+y和0的关系,并证明.分析 令F(x)=(log23)x-(log53)x,然后根据复合函数的单调性法则确定F(x)的单调性,最后根据单调性解F(x)≥F(-y)即可.
解答 解:令F(x)=(log23)x-(log53)x
∵log23>1,0<log53<1,
∴函数F(x)在R上单调递增,
∵(log23)x-(log53)x≥(log23)-y-(log53)-y,
∴F(x)≥F(-y),
∴x≥-y,
即x+y≥0.
点评 本题主要考查了对数函数的单调性,以及复合函数的单调性和构造法的运用,属于中档题.
练习册系列答案
相关题目
17.在下列各式中错误的个数是( )
①1∈{0,1,2};
②{1}∈{0,1,2};
③{0,1,2}⊆{0,1,2};
④{0,1,2}={2,0,1};
⑤{0,1}⊆{(0,1)};
⑥∅⊆{0}.
①1∈{0,1,2};
②{1}∈{0,1,2};
③{0,1,2}⊆{0,1,2};
④{0,1,2}={2,0,1};
⑤{0,1}⊆{(0,1)};
⑥∅⊆{0}.
| A. | 1 | B. | 2 | C. | 3 | D. | 4 |
18.若关于x的函数$y=2x-\frac{m}{x}$在(1,+∞)上是增函数,则m的取值范围是( )
| A. | [-2,+∞) | B. | [2,+∞) | C. | (-∞,-2] | D. | (-∞,2] |