题目内容
7.已知f(x)=(m2-m-1)x-5m-3,求m为何值时.(1)f(x)是正比例函数;
(2)f(x)是反比例函数;
(3)f(x)是二次函数;
(4)f(x)是幂函数.
分析 (1)由f(x)是正比例函数,可得$\left\{\begin{array}{l}{{m}^{2}-m-1>0}\\{-5m-3=1}\end{array}\right.$,解得m即可得出;
(2))由f(x)是反比例函数,可得$\left\{\begin{array}{l}{{m}^{2}-m-1≠0}\\{-5m-3=-1}\end{array}\right.$,解得m即可得出;
(3)由f(x)是二次例函数,可得$\left\{\begin{array}{l}{{m}^{2}-m-1≠0}\\{-5m-3=2}\end{array}\right.$,解得m即可得出;
(4)由f(x)是幂函数,可得$\left\{\begin{array}{l}{{m}^{2}-m-1=1}\\{-5m-3≠0}\end{array}\right.$,解得m即可得出.
解答 解:f(x)=(m2-m-1)x-5m-3,
(1)∵f(x)是正比例函数,∴$\left\{\begin{array}{l}{{m}^{2}-m-1>0}\\{-5m-3=1}\end{array}\right.$,解得m=-$\frac{4}{5}$,此时f(x)=$\frac{11}{25}$x;
(2))∵f(x)是反比例函数,∴$\left\{\begin{array}{l}{{m}^{2}-m-1≠0}\\{-5m-3=-1}\end{array}\right.$,解得m=-$\frac{2}{5}$,此时f(x)=$-\frac{11}{25x}$;
(3)∵f(x)是二次例函数,∴$\left\{\begin{array}{l}{{m}^{2}-m-1≠0}\\{-5m-3=2}\end{array}\right.$,解得m=-1,此时f(x)=x2;
(4)∵f(x)是幂函数,∴$\left\{\begin{array}{l}{{m}^{2}-m-1=1}\\{-5m-3≠0}\end{array}\right.$,解得m=2或-1,此时f(x)=x-13或f(x)=x2.
点评 本题考查了正比例函数、反比例函数、二次函数、幂函数的定义、不等式的解法,考查了推理能力与计算能力,属于中档题.
A. | $\frac{\sqrt{2}}{4}$a2 | B. | a2 | C. | 2$\sqrt{2}$a2 | D. | 2a2 |
A. | 1 | B. | 2 | C. | -1 | D. | -2 |