Question

18．已知函數(shù)f（x）=$\left\{\begin{array}{l}{（\frac{1}{2}）^{x}-3\\;x≤0}\\{{x}^{\frac{1}{2}}\\;x＞0}\end{array}\right.$，若f（a）＞1，則實(shí)數(shù)a的取值范圍是a＜-2或a＞1．

Answer 1

分析 由題意原不等式等價(jià)于$\left\{\begin{array}{l}{a≤0}\\{（\frac{1}{2}）^{a}-3＞1}\end{array}\right.$或$\left\{\begin{array}{l}{a＞0}\\{{a}^{\frac{1}{2}}＞1}\end{array}\right.$，解不等式組可得．

解答 解：∵函數(shù)f（x）=$\left\{\begin{array}{l}{（\frac{1}{2}）^{x}-3\\;x≤0}\\{{x}^{\frac{1}{2}}\\;x＞0}\end{array}\right.$，
∴f（a）＞1等價(jià)于$\left\{\begin{array}{l}{a≤0}\\{（\frac{1}{2}）^{a}-3＞1}\end{array}\right.$或$\left\{\begin{array}{l}{a＞0}\\{{a}^{\frac{1}{2}}＞1}\end{array}\right.$，
分別解關(guān)于a的不等式組可得a＜-2或a＞1，
故答案為：a＜-2或a＞1．

點(diǎn)評(píng) 本題考查分段函數(shù)不等式的解法，化為不等式組是解決問題的關(guān)鍵，屬基礎(chǔ)題．