反正弦、反余弦函数的定义域都为[-1,1],故有
{-1≤1-x≤1,-1≤2x≤1}
→0≤x≤1/2.
结合它们的单调性,知
y=arcsin(1-x)和y=arccos2x在[0,1/2]上都单调递减.
∴f(0)≥f(x)≥f(1/2)
从而,所求函数的值域为[π/6,π].
答：设y=x^2/(x^2-x+1),则 y(x^2-x+1)=x^2, (y-1)x^2-yx+y=0, ∵x∈R, ∴y^2-4y(y-1)>=0, 即y(y-...详情>>
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