高一数学20题
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f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1
=2√3sinxcosx+2cos²x-1+2
=√3sin2x+cos2x+2
=2(sin2x*√3/2+cos2x*1/2)+2
=2(sin2xcosπ/6+cos2xsinπ/6)+2
=2sin(2x+π/6)+2
所以T=2π/2=π
-π/6<=2x+π/6<=2π/3
所以-1/2<=sin(2x+π/6)<=1
所以最大值=2×1+2=4
最小值=2×(-1/2)+2=1
(满意请采纳)
=2√3sinxcosx+2cos²x-1+2
=√3sin2x+cos2x+2
=2(sin2x*√3/2+cos2x*1/2)+2
=2(sin2xcosπ/6+cos2xsinπ/6)+2
=2sin(2x+π/6)+2
所以T=2π/2=π
-π/6<=2x+π/6<=2π/3
所以-1/2<=sin(2x+π/6)<=1
所以最大值=2×1+2=4
最小值=2×(-1/2)+2=1
(满意请采纳)
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x∈[-π/6,π/4],2x+π/6∈[-π/6,2π/3],f(x)max=2,f(x)min=-1
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