问题
填空题
若函数f(x)=loga(x+
|
答案
∵函数f(x)=loga(x+
)是奇函数,x2+2a2
∴f(x)+f(-x)=0
即loga(x+
)+loga(-x+x2+2a2
)=0x2+2a2
∴loga(x+
)×(-x+x2+2a2
)=0x2+2a2
∴x2+2a2-x2=1,即2a2=1,
∴a=±2 2
又a对数式的底数,a>0
∴a=2 2
故应填2 2
若函数f(x)=loga(x+
|
∵函数f(x)=loga(x+
)是奇函数,x2+2a2
∴f(x)+f(-x)=0
即loga(x+
)+loga(-x+x2+2a2
)=0x2+2a2
∴loga(x+
)×(-x+x2+2a2
)=0x2+2a2
∴x2+2a2-x2=1,即2a2=1,
∴a=±2 2
又a对数式的底数,a>0
∴a=2 2
故应填2 2