题目
f(x1)-f(x2) |
x1-x2 |
A.f(3)<f(-2)<f(1) | B.f(1)<f(-2)<f(3) | C.f(-2)<f(1)<f(3) | D.f(3)<f(1)<f(-2) |
答案
∴
f(x2)-f(x1) |
x2-x1 |
又f(x)是偶函数,故f(x)在x1,x2∈(-∞,0](x1≠x2)单调递减.
且满足n∈N*时,f(-2)=f(2),3>2>1>0,
得f(1)<f(-2)<f(3),
故选B.
f(x1)-f(x2) |
x1-x2 |
A.f(3)<f(-2)<f(1) | B.f(1)<f(-2)<f(3) | C.f(-2)<f(1)<f(3) | D.f(3)<f(1)<f(-2) |
f(x2)-f(x1) |
x2-x1 |