题目内容
11.函数f(x)=$\sqrt{4-|x|}$+lg$\frac{{{x^2}-5x+6}}{x-3}$的定义域为( )A. | (2,3) | B. | (2,4] | C. | (2,3)∪(3,4] | D. | (-1,3)∪(3,6] |
分析 根据函数成立的条件进行求解即可.
解答 解:要使函数有意义,则$\left\{\begin{array}{l}{4-|x|≥0}\\{\frac{{x}^{2}-5x+6}{x-3}>0}\end{array}\right.$,
即$\left\{\begin{array}{l}{-4≤x≤4}\\{\frac{(x-2)(x-3)}{x-3}>0}\end{array}\right.$,
$\frac{(x-2)(x-3)}{x-3}$>0等价为①$\left\{\begin{array}{l}{x>3}\\{(x-2)(x-3)>0}\end{array}\right.$即$\left\{\begin{array}{l}{x>3}\\{x>3或x<2}\end{array}\right.$,即x>3,
②$\left\{\begin{array}{l}{x<3}\\{(x-2)(x-3)<0}\end{array}\right.$,即$\left\{\begin{array}{l}{x<3}\\{2<x<3}\end{array}\right.$,此时2<x<3,
即2<x<3或x>3,
∵-4≤x≤4,
∴解得3<x≤4且2<x<3,
即函数的定义域为(2,3)∪(3,4],
故选:C
点评 本题主要考查函数的定义域的求解,要求熟练掌握常见函数成立的条件.
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