Question

已知函数f(x)=2x^2,g(x)=alnx(a>0),若不等式f(x)>=g(x)恒成立,,(1)求a的取值范围

(2)求证ln2^4/2^4+......+lnn^4/n^4<2/e

Answer 1

已知函数f(x)=2x^2,g(x)=alnx(a>0),若不等式f(x)>=g(x)恒成立,,(1)求a的取值范围
解析：∵函数f(x)=2x^2,g(x)=alnx(a>0),不等式f(x)>=g(x)恒成立
设h(x)=f(x)-g(x)=2x^2-alnx
h’(x)=4x-a/x=0==>(4x^2-a)/x=0==>x=√a/2
h(x)在x=√a/2处取极小值a/2-aln(√a/2)
令a/2-aln(√a/2)>=0==>a<=4e
∵a>0
∴a的取值范围为0<a<=4e

求证ln2^4/2^4+......+lnn^4/n^4<2/e
证明：设f(x)=lnx/x
令f’(x)=(1-lnx)/x^2=0==>x=e
f(x)在x=e处取极大的值1/e
∴x>2时，lnx＜x/2==>lnx^2＜x==>lnx^2/x^2＜1/x
ln2^4/2^4+ln3^4/3^4+….+lnn^4/n^4＜1/2^2+1/3^2+1/4^2……+1/n^2
<1/4+1/9+1/(3•4)+….+1/[(n-1)n]=1/4+1/9+(1/3-1/4)+(1/4-1/5)+…+[1/(n-1)-1/n]
﹦1/4+1/9+1/3-1/n＜25/36＜2/e