Question

已知函数f(x)=(2x^2+bx+c)/(x^2+1)(b＜0)的值域[1,3]

(1)求b.c的值
(2)判断该函数在[-1,1]上的单调性,并给出证明
(3)若t∈R,求f(丨t-1/6丨-丨t+1/6丨)

Answer 1

(1)y=(2x^2+bx+c)/(x^2+1)
(y-2)x^2-bx+y-c=0
Δ=b^2-4(y-2)(y-c)
y=3、y=1时分别代入得
b=-2 c=2
(2)f(-1)=3 f(1)=1 判断单调递减
证明：y=(2x^2+2-2x)/(x^2+1)=2-2x/(x^2+1)=2-2/(x+1/x)
由于g(x)=x+1/x 在[-1,0)和(0,1]上单调递减 故y=2-2/(x+1/x)在[-1,1]上是减函数
(3)①t<-1/6 f[|t-1/6|-|t+1/6|]=f(1/3)=1.4
②-1/6≤t≤1/6 f[|t-1/6|-|t+1/6|]=f(-2t)=(8t^2+4t+2)/(4t^2+1)
③t>1/6 f[|t-1/6|-|t+1/6|]=f(-1/3)=2.6