第二问
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第二问
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![第二问](/uploads/image/z/5223976-16-6.jpg?t=%E7%AC%AC%E4%BA%8C%E9%97%AE+%26nbsp%3B)
f(x)=2sin(x-π/3)cos(x-π/3)+2√3cos^2(x-π/3)-√3
=sin(2x-2π/3)+√3cos(2x-2π/3)
=2[sin(2x-2π/3)cosπ/3+cos(2x-2π/3sinπ/3]
=2sin(2x-2π/3+π/3)+1-√3
=2sin(2x-π/3)
故 f(2x)-a=2sin(4x-π/3)-a
当4x-π/3=π/2+kπ 时,y取最大值,此时x=5π/24+kπ(k∈Z).
又∵在区间[0,π/4]中
∴对称轴为x=5π/24
∵两零点x1和x2关于x=5π/24对称.
∴x1+x2=5π/12
故tan(x1+x2)=tan(5π/12)
=tan(π/4+π/6)
=(1+√3 /3)/(1-√3 /3)
=2+√3
=sin(2x-2π/3)+√3cos(2x-2π/3)
=2[sin(2x-2π/3)cosπ/3+cos(2x-2π/3sinπ/3]
=2sin(2x-2π/3+π/3)+1-√3
=2sin(2x-π/3)
故 f(2x)-a=2sin(4x-π/3)-a
当4x-π/3=π/2+kπ 时,y取最大值,此时x=5π/24+kπ(k∈Z).
又∵在区间[0,π/4]中
∴对称轴为x=5π/24
∵两零点x1和x2关于x=5π/24对称.
∴x1+x2=5π/12
故tan(x1+x2)=tan(5π/12)
=tan(π/4+π/6)
=(1+√3 /3)/(1-√3 /3)
=2+√3