题目内容
14.已知m∈R,函数f(x)=$\left\{\begin{array}{l}{|3x+1|,x<0}\\{lo{g}_{3}x,x>0}\end{array}\right.$,g(x)=x2-2x+2m-1,若函数y=f(g(x))-m有6个零点,则实数m的取值范围是( )A. | (0,$\frac{5}{7}$) | B. | ($\frac{3}{7}$,$\frac{5}{7}$) | C. | (0,$\frac{3}{7}$) | D. | ($\frac{2}{7}$,1) |
分析 由于函数f(x)=$\left\{\begin{array}{l}|3x+1|,x<0\\ lo{g}_{3}x,x>0\end{array}\right.$,g(x)=x2-2x+2m-1.可得当g(x)=(x-1)2+2m-2<0,当g(x)=(x-1)2+2m-2>0,求出函数y=f(g(x))-m的表达式.再对m分类讨论,利用直线y=m与函数y=f(g(x))图象的交点必须是6个即可得出.
解答 解:∵函数f(x)=$\left\{\begin{array}{l}|3x+1|,x<0\\ lo{g}_{3}x,x>0\end{array}\right.$,g(x)=x2-2x+2m-1.
∴当g(x)=(x-1)2+2m-2<0,即(x-1)2<2-2m时,
则y=f(g(x))=|3g(x)+1|=|3(x-1)2+6m-5|.
当g(x)=(x-1)2+2m-2>0,即(x-1)2>2-2m时,则y=f(g(x))=${log}_{3}[{(x-1)}^{2}+2m-2]$.
①当m≥1时,y=m只与y=f(g(x))=${log}_{3}[{(x-1)}^{2}+2m-2]$的图象有两个交点,不满足题意,应该舍去.
②当m<1时,y=m与y=f(g(x))=${log}_{3}[{(x-1)}^{2}+2m-2]$的图象有两个交点,需要直线y=m与函数y=f(g(x))=|3g(x)+1|=|3(x-1)2+6m-5|的图象有四个交点时才满足题意.
∴0<m<5-6m,又m<1,解得0<m<$\frac{5}{7}$.
综上可得:m的取值范围:(0,$\frac{5}{7}$).
故选:A.
点评 本题考查了分段函数的图象与性质、含绝对值函数的图象、对数函数的图象、函数图象的交点的与函数零点的关系,考查了推理能力与计算能力,考查了数形结合的思想方法,考查了推理能力与计算能力,属于难题.
A. | -6 | B. | $\frac{11}{10}$ | C. | $\frac{9}{10}$ | D. | -9 |
A. | α+β<$\frac{π}{2}$ | B. | α+β=$\frac{π}{2}$ | C. | α+β>$\frac{π}{2}$ | D. | α>β |