题目内容
9.已知函数f(x)=lg(2-x),那么f(x)的定义域是( )| A. | R | B. | {x|x>2} | C. | {x|x<2} | D. | {x|0<x<2} |
分析 直接利用对数的真数大于0,求解即可.
解答 解:函数f(x)=lg(2-x)有意义,可得2-x>0,解得x<2.
函数的定义域为:(-∞,2).
故选:C.
点评 本题考查函数的定义域的求法,考查计算能力.
练习册系列答案
相关题目
18.下列关系中正确的是( )
| A. | ${(\frac{1}{2})}^{\frac{2}{3}}$<${(\frac{1}{5})}^{\frac{2}{3}}$<${(\frac{1}{2})}^{\frac{1}{3}}$ | B. | ${(\frac{1}{2})}^{\frac{1}{3}}$<${(\frac{1}{2})}^{\frac{2}{3}}$<${(\frac{1}{5})}^{\frac{2}{3}}$ | ||
| C. | ${(\frac{1}{5})}^{\frac{2}{3}}$<${(\frac{1}{2})}^{\frac{1}{3}}$<${(\frac{1}{2})}^{\frac{2}{3}}$ | D. | ${(\frac{1}{5})}^{\frac{2}{3}}$<${(\frac{1}{2})}^{\frac{2}{3}}$<${(\frac{1}{2})}^{\frac{1}{3}}$ |
19.${∫}_{0}^{2π}$|cosx|dx等于( )
| A. | 0 | B. | 1 | C. | 2 | D. | 4 |