因 1-x^2>=0
解得 x的取值范围是 -1≤x ≤1
由基本不等式,a,b∈R+,a+b ≥2√(ab)
则 t=|x|*√(1-x^2) = √x^2(1-x^2)≤ [x^2 + (1-x^2)]/2 = 1/2
所以,t的取值范围是 0≤t≤1/2