Question

4．已知函數(shù) f（x）=log₃$\frac{2{x}^{2}+bx+c}{{x}^{2}+1}$的值域?yàn)閇0，1]，求b和c的值．

Answer 1

分析 直接利用對(duì)數(shù)的性質(zhì)轉(zhuǎn)化為求解二次函數(shù)的問(wèn)題．

解答 解析：因?yàn)閒（x）的值域?yàn)閇0，1]，即：0≤log₃$\frac{2{x}^{2}+bx+c}{{x}^{2}+1}$≤1
所以：log₃1≤log₃$\frac{2{x}^{2}+bx+c}{{x}^{2}+1}$≤log₃3．
∵底數(shù)3＞1，y=log₃x是增函數(shù)，
∴$\left\{\begin{array}{l}{\frac{2{x}^{2}+bx+c}{{x}^{2}+1}≥1}\\{\frac{2{x}^{2}+bx+c}{{x}^{2}+1}≤3}\end{array}\right.$⇒$\left\{\begin{array}{l}{{x}^{2}+bx+c-1≥0}\\{{x}^{2}-bx+3-c≥0}\end{array}\right.$⇒$\left\{\begin{array}{l}{{△}_{1}=^{2}-4（c-1）≥0}\\{{△}_{2}=^{2}-4（3-c）≥0}\end{array}\right.$
當(dāng)且僅當(dāng)$\left\{\begin{array}{l}{{△}_{1}=0}\\{{△}_{2}=0}\end{array}\right.$時(shí)，則有0≤log₃$\frac{2{x}^{2}+bx+c}{{x}^{2}+1}$≤1取等號(hào)．
解方程組：可得：$\left\{\begin{array}{l}{b=2}\\{c=2}\end{array}\right.$或$\left\{\begin{array}{l}{b=-2}\\{c=2}\end{array}\right.$．
故b和c的值為：$\left\{\begin{array}{l}{b=2}\\{c=2}\end{array}\right.$或$\left\{\begin{array}{l}{b=-2}\\{c=2}\end{array}\right.$．

點(diǎn)評(píng) 本題考查了對(duì)數(shù)函數(shù)的性質(zhì)的運(yùn)用，轉(zhuǎn)化為二次方程成立的問(wèn)題．屬于中檔題．