Question

已知t=|x|*根号下(1-x^2),则实数t的取值范围

Answer 1

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