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设函数F(x)=根号3sinxcosx+cos^2x+a,当x∈[-π/6,π/3]时,此函数的最大值与最小值的和为3/2求解析式
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设函数F(x)=根号3sinxcosx+cos^2x+a,当x∈[-π/6,π/3]时,此函数的最大值与最小值的和为3/2求解析式
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答案和解析
易知F(x)=√3sinxcosx+cos^2x+a
=√3/2sin2x+1/2cos2x+a+1/2
=(cosπ/6sin2x+sinπ/6cos2x)+a+1/2
=sin(2x+π/6)+a+1/2
因x∈[-π/6,π/3]
则2x+π/6∈[-π/6,5π/6]
显然Fmin=sin(-π/6)+a+1/2=a
Fmax=sinπ/2+a+1/2=a+3/2
而Fmin+Fmax=3/2
即2a+3/2=3/2
所以a=0
于是F(x)=√3sinxcosx+cos^2x
=√3/2sin2x+1/2cos2x+a+1/2
=(cosπ/6sin2x+sinπ/6cos2x)+a+1/2
=sin(2x+π/6)+a+1/2
因x∈[-π/6,π/3]
则2x+π/6∈[-π/6,5π/6]
显然Fmin=sin(-π/6)+a+1/2=a
Fmax=sinπ/2+a+1/2=a+3/2
而Fmin+Fmax=3/2
即2a+3/2=3/2
所以a=0
于是F(x)=√3sinxcosx+cos^2x
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