Question

16．已知函數(shù)y=$\frac{a}{{a}^{2}-2}$（a^x-a^-x）（a＞0，a≠1）在（-∞，+∞）上遞增，求a的取值范圍．

Answer 1

分析 根據(jù)函數(shù)單調(diào)性的性質(zhì)對(duì)a進(jìn)行討論即可．

解答 解：設(shè)f（x）=a^x-a^-x，則當(dāng)a＞1時(shí)，f（x）為增函數(shù)，當(dāng)0＜a＜1時(shí)，f（x）為減函數(shù)，
∴若函數(shù)y=$\frac{a}{{a}^{2}-2}$（a^x-a^-x）（a＞0，a≠1）在（-∞，+∞）上遞增，
則等價(jià)為①$\left\{\begin{array}{l}{\frac{a}{{a}^{2}-2}＞0}\\{a＞1}\end{array}\right.$或②$\left\{\begin{array}{l}{\frac{a}{{a}^{2}-2}＜0}\\{0＜a＜1}\end{array}\right.$，
即$\left\{\begin{array}{l}{a＞1}\\{{a}^{2}-2＞0}\end{array}\right.$或$\left\{\begin{array}{l}{0＜a＜1}\\{{a}^{2}-2＜0}\end{array}\right.$，
則$\left\{\begin{array}{l}{a＞1}\\{a＞\sqrt{2}或a＜-\sqrt{2}}\end{array}\right.$或$\left\{\begin{array}{l}{0＜a＜1}\\{-\sqrt{2}＜a＜\sqrt{2}}\end{array}\right.$，
即a＞$\sqrt{2}$或0＜a＜1．

點(diǎn)評(píng) 本題主要考查函數(shù)單調(diào)性的應(yīng)用，根據(jù)復(fù)合函數(shù)單調(diào)性之間的關(guān)系是解決本題的關(guān)鍵．