已知函数f(x)=(bx+c)/(ax^2 +1)(a,c∈R,a＞0,b是自然数）是奇函数,f(x)有最大值1/2,且

(1) f(-x) = (-bx + c)/(ax² + 1) = -f(x) = (-bx- c)/(ax² + 1)
c = -c = 0
x >0时, f(x) = bx/(ax² + 1) = b/(ax + 1/x) ≤ b/[2√(ax)*√(1/x)] = b/(2√a) = 1/2
a = b²
f(x) = bx/(b²x² + 1)
f(1) = b/(b² + 1) >2/5
2b² - 5b + 2 < 0
(2b -1)(b - 2) < 0
1/2 < b < 2
此范围内的自然数只有1, b= 1
a = 1
f(x) = x/(x² + 1)
(2)
P(p, p/(p² + 1)), Q(q, q/(q² + 1))
(1, 0)为PQ的中点:
(p + q)/2 = 1, p + q = 2 (i)
[p/(p² + 1) + q/(q² + 1)]/2 = 0
pq² + p + p²q + q = (pq + 1)(p + q) = 2(pq + 1) = 0
pq = -1 (ii)
p, q为方程x² - 2x - 1 = 0的根
x = 1 ±√2
P(1 + √2, √2/4), Q(1 - √2, -√2/4)
PQ的斜率k = (√2/4 + √2/4)/(1 + √2 - 1 + √2) = 1/4
直线l的方程: y - 0 = k(x - 1)
y = (x - 1)/4