Question

看看下面这个导数题 谢谢啦

导数题 麻烦写一下过程

Answer 1

1) f'(x) = lnx +1 - a = 0
x = e^(a-1)
单调减：f'(x) < 0, when x < e^(a-1); 单调增：f'(x) > 0, when x > e^(a-1);
By the first derivative test, f min = f( e^(a-1)) = (a-1)e^(a-1) - ae^(a-1) + 3 = 2
=> e^(a-1) = 1 => a = 1
2) 用数学归纳法：(ATTN: ln(1+x) > 1-x^2/2, where x = 1/i, i = 1, 2, ..., n)
a1 = [ln(1+1)]^2 > 1/4 > 1/7
Assume a1+a2+...+an = [ln(1+1)]^2 + [ln(1+1/2)]^2 + ... + [[ln(1 + 1/n)]^2 > n/(2n+4)
a1+a2+...+an+a(n+1) > n/(2n+4) + [[ln(1 + 1/(n+1))]^2 > n/(2n+4) + {1 - 1/[2(n+1)^2]}^2
> n/(2n+4) + 1 - 2/[2(n+1)^2] > n/(2n+4) + (2n^2+4n)/[2(n+1)^2]
> n/(2n+4) + 2/(2n+2) > n/(2n+6) + 2/(2n+6) > (n+1)/(2n+6)
QED

Answer 2