定义在R上的偶函数f(x)满足:对任意的x1,x2∈(-∞,0](x1≠x2),有(x2-x1)(f(x2)-f(x1))>0.则当n∈N*时,有( )
A. f(-n)<f(n-1)<f(n+1)
B. f(n-1)<f(-n)<f(n+1)
C. f(n+1)<f(-n)<f(n-1)
D. f(n+1)<f(n-1)<f(-n)
人气:215 ℃ 时间:2020-05-20 18:20:01
解答
x1,x2∈(-∞,0](x1≠x2),有(x2-x1)(f(x2)-f(x1))>0∴x2>x1时,f(x2)>f(x1)∴f(x)在(-∞,0]为增函数∵f(x)为偶函数∴f(x)在(0,+∞)为减函数而n+1>n>n-1>0,∴f(n+1)<f(n)<f...
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