Question

高一数学

Answer 1

log2(20)=log2（2）+log2(10)=1+log2(10)

由f(1+x)=f(1-x)
f[log2(20)]=f[1+log2(10)]=f[1-log2(10)]
奇函数
=-f[log2(10)-1]
=-f[log2(10/2)]
=-f[log2(2*2.5)]
=-f[1+log2(2.5)]
=-f[1-log2(2.5)]
=f[log2(2.5)-1]
=f[log2(1.25)]
1<1.25<2
所以0<log2(1.25)<1
-1<-log2(1.25)<0
所以f[-log2(1.25)]=2^[-log2(1.25)]+1/5
=2^[log2(1/1.25)]+1/5
=1/1.25+1/5
=1
奇函数则f[log2(1.25)]=-f[-log2(1.25)]
所以原式=-f[-log2(1.25)]=-1

追问

噢不 能给手写版吗