定义在R上的函数f(x)满足(x-1)f’(x)≤0,且y=f(x+1)为偶函数,当|x1-1|＜|x2-1|时,有a.f(2-x1)≥f(2-x2)b.f(2-x1)=f(2-x2)c.f(2-x1)＜f(2-x2)d.f(2-x1)≤f(2-x2)求详解

Question

定义在R上的函数f(x)满足(x-1)f’(x)≤0,且y=f(x+1)为偶函数,当
|x1-1|＜|x2-1|时,有 a.f(2-x1)≥f(2-x2) b.f(2-x1)=f(2-x2) c.f(2-x1)＜f(2-x2) d.f(2-x1)≤f(2-x2) 求详解

Answer 1

当x≥1时,f'(x)≤0,f(x)在[1,+∞)上单调递增(不一定是严格单调递增),
当x≤1时,f'(x)≥0,f(x)在(-∞,1]上单调递减(不一定是严格单调递减),
y=f(x+1)为偶函数,所以f(x+1)=f(-x+1),
即f(x)图像关于直线x=1对称,
又|x1-1|＜|x2-1|,所以f(x1)≥f(x2),
而f(2-x1)=f(1+1-x1)=f(1-(1-x1))=f(x1),
f(2-x2)=f(1+1-x2)=f(1-(1-x2))=f(x2),
所以f(2-x1)≥f(2-x2).
答案选A