Question

已知函數(shù)f(x)=(x＞0).

（1）當(dāng)x₁＞0,x₂＞0且f(x₁)·f(x₂)=1時(shí)，求證：x₁·x₂≥3+2；

（2）若數(shù)列{a_n}滿足a₁=1,a_n＞0,a_n+1=f(a_n)(n∈N^*)，求數(shù)列{a_n}的通項(xiàng)公式.

Answer 1

(1)證明:∵x₁＞0,x₂＞0,f(x₁)·f(x₂)=1,

∴·=1.

∴(x₁x₂)²=(2x₁+1)(2x₂+1)=4x₁x₂+2(x₁+x₂)+1≥4x₁x₂+4+1=(2+1)².

∴x₁x₂≥2+1.∴(-1)²≥2.∴-1≥或-1≤-(舍去).

∴≥+1.∴x₁x₂≥(+1)²=3+2.

(2)解:∵a₁=1,a_n＞0,a_n+1=f(a_n)=,

∴==+=(1+)²-1.∴1+=(1+)².

∴l(xiāng)g(1+)=lg(1+)²=2lg(1+).

∴數(shù)列{lg(1+)}是首項(xiàng)為lg(1+)=lg2,公比為2的等比數(shù)列.

∴l(xiāng)g(1+)=2^n-1·lg2=.

∴1+=.∴a_n=.