y=arccos2x/(1+x∧)的值域

Question

y=arccos2x/(1+x∧)的值域

Answer 1

首先,x=0时,y = arccos0 = π/2

当x＜0时,2x/(1+x²) = -2/{-1/x-x} = -2/{1/√(-x)-√(-x)]²+2}
-1≤2x/(1+x²)＜0,
π/2＜arccos[2x/(1+x²)]≤π

当x＞0时,2x/(1+x²) = 2/{1/x+x} = 2/{1/√x-√x]²+2}
0＜2x/(1+x²)≤1
0≤arccos[2x/(1+x²)]＜π/2

综上,值域【0,π】