7.設(shè)函數(shù)f(x)的定義域為D,記f(X)={y|y=f(x),x∈X⊆D},f-1(Y)={x|f(x)∈Y,x∈D},若f(x)=2sin(ωx+\frac{5π}{6})(ω>0)且f(f-1([0,2]))=[0,2],則ω的取值范圍是ω>0.
分析 可求得f-1([0,2])=[\frac{1}{ω}(kπ-\frac{5π}{6}),\frac{1}{ω}(kπ+\frac{π}{6})],(k∈Z),從而求得2sin(ωx+\frac{5π}{6})∈[0,2],從而解得.
解答 解:由題意得,
f(x)=2sin(ωx+\frac{5π}{6})(ω>0)的定義域為R,
f-1([0,2])={x|f(x)∈[0,2],x∈R},
故2sin(ωx+\frac{5π}{6})∈[0,2];
故kπ≤ωx+\frac{5π}{6}≤kπ+π,k∈Z;
故\frac{1}{ω}(kπ-\frac{5π}{6})≤x≤\frac{1}{ω}(kπ+\frac{π}{6}),k∈Z;
即f-1([0,2])=[\frac{1}{ω}(kπ-\frac{5π}{6}),\frac{1}{ω}(kπ+\frac{π}{6})],(k∈Z);
故f(f-1([0,2]))
=f([\frac{1}{ω}(kπ-\frac{5π}{6}),\frac{1}{ω}(kπ+\frac{π}{6})])
={y|y=f(x),x∈[\frac{1}{ω}(kπ-\frac{5π}{6}),\frac{1}{ω}(kπ+\frac{π}{6})]},
故f(x)=2sin(ωx+\frac{5π}{6})在[\frac{1}{ω}(kπ-\frac{5π}{6}),\frac{1}{ω}(kπ+\frac{π}{6})]上的值域為[0,2];
∵x∈[\frac{1}{ω}(kπ-\frac{5π}{6}),\frac{1}{ω}(kπ+\frac{π}{6})],
∴ωx∈[(kπ-\frac{5π}{6}),(kπ+\frac{π}{6})],
∴ωx+\frac{5π}{6}∈[(kπ,(kπ+π)],
∴2sin(ωx+\frac{5π}{6})∈[0,2].
故答案為:ω>0.
點評 本題考查了對應(yīng)關(guān)系的應(yīng)用及函數(shù)的定義域與值域的關(guān)系應(yīng)用.