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已知函数f(x)=x2+mx-|1-x2|(m∈R),若f(x)在区间(-2,0)上有且只有1个零点,则实数m的取值范围

已知函数f(x)=x2+mx-|1-x2|(m∈R),若f(x)在区间(-2,0)上有且只有1个零点,则实数m的取值范围是
已知函数f(x)=x2+mx-|1-x2|(m∈R),若f(x)在区间(-2,0)上有且只有1个零点,则实数m的取值范围是______.
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-1≤x<0时,f(x)=2x2+mx-1,
-2<x<-1时,f(x)=mx+1,
∴当x=-1时,f(-1)=1-m,
当1-m=0,即m=1时,符合题意,
当1-m>0时,f(x)在(-1,0)有零点,
∴f(-2)=-2m+1≥0,解得:m≤ 1年前

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