Question

8．設(shè)函數(shù)$f（x）=\left\{\begin{array}{l}{2^{x+1}}，x≤0\\{log_2}x，x＞0\end{array}\right.$，若關(guān)于x的方程[f（x）]²-af（x）=0恰有三個(gè)不同的實(shí)數(shù)解，則實(shí)數(shù)a的取值范圍是（0，2]．

Answer 1

分析 作函數(shù)$f（x）=\left\{\begin{array}{l}{2^{x+1}}，x≤0\\{log_2}x，x＞0\end{array}\right.$的圖象，而[f（x）]²-af（x）=0得f（x）=0或f（x）=a，從而可得f（x）=a有兩個(gè)解，從而判斷．

解答 解：作函數(shù)$f（x）=\left\{\begin{array}{l}{2^{x+1}}，x≤0\\{log_2}x，x＞0\end{array}\right.$的圖象如下，
，
∵[f（x）]²-af（x）=0，
∴f（x）=0或f（x）=a，
∴f（x）=0的解為x=1，
∴f（x）=a有兩個(gè)解，
∴0＜a≤2；
故答案為：（0，2]．

點(diǎn)評(píng) 本題考查了分段函數(shù)的應(yīng)用及數(shù)形結(jié)合的思想應(yīng)用．