已知函数f(x)=(sin2x+cos2x+1)/2cosx,
2个回答
分式有意义,cosx≠0
f(x)=[sin(2x)+cos(2x)+1]/(2cosx)
=(2sinxcosx+cos²x-sin²x+cos²x+sin²x)/(2cosx)
=(2sinxcosx+2cos²x)/(2cosx)
=2cosx(sinx+cosx)/(2cosx)
=sinx+cosx
=√2sin(x+π/4)
x∈[0,π/3],则x+π/4∈[π/4,7π/12],当5π/12≤x+π/4≤7π/12时,对于同一正弦值,x有两不同解.
cos(5π/6)=-cos(π/6)=-√3/2
sin(5π/12)=√{[1-cos(5π/6)]/2}=√[(4+2√3)/8]=(√6+√2)/4
此时,sin(x+π/4)∈[(√6+√2)/4,1]
f(x)=a
√2sin(x+π/4)=a
sin(x+π/4)=a/√2
(√6+√2)/4≤a/√2≤1
(√3+1)/2≤a≤√2
a的取值范围为[(√3 +1)/2,√2]
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