The extreme value of f(x) obtained after substitution is:

f( 1/a)= 1+ln(a)= 1

The solution is a = 1

(2) G(x)=a*sin( 1-x) + ln(x)

g '(x)=-a * cos( 1-x)+ 1/x

When x∈(0, 1), G'(x) ≥ 0.

According to:-a * cos (1-x)+1/x ≥ 0.

= = = = = & gt； a* x * cos( 1-x)≤ 1(x ∈( 0， 1))

= = = = = & gt； a ≦ 1

(3)∑sin[ 1/(k+ 1)^2](k = 1，2，..n)

≤∑ 1/(k+ 1)^2(k = 1，2，..n)

& lt 1/4+ 1/9+ 1/ 16+ 1/25+∑ 1/[k(k+ 1)(k = 5，6，..n)

= 1/4+ 1/9+ 1/ 16+ 1/25+∑( 1/k- 1/(k+ 1))(k = 5，6，..n)

= 1/4+ 1/9+ 1/ 16+ 1/25+ 1/5- 1/(n+ 1)

& lt 1/4 + 1/9+ 1/ 16 + 1/25 + 1/5

& lt second edition