已知f(x)=asinx+bcosx,定积分（0到π/2)f(x)dx=4,定积分（0到π/6）f(

已知f(x)=asinx+bcosx,定积分（0到π/2)f(x)dx=4,定积分（0到π/6）f(

f(x)=asinx+bcosx∫【x=0→π/2】f(x)dx=∫【x=0→π/2】(asinx+bcosx)dx=a∫【x=0→π/2】sinxdx+b∫【x=0→π/2】cosxdx=a【x=0→π/2】-cosx+b【x=0→π/2】sinxdx=a[-cos(π/2)+cos0]+b(sin(π/2)-sin0)=a+b依题意,有：a+b=4……………………………………………………(1)∫【x=0→π/6】f(x)dx=∫【x=0→π/6】(asinx+bcosx)dx=a∫【x=0→π/6】sinxdx+b∫【x=0→π/6】cosxdx=a【x=0→π/6】-cosx+b【x=0→π/6】sinxdx=a[-cos(π/6)+cos0]+b(sin(π/6)-sin0)=a(2-√3)/2+(b√3)/2=a+√3(b-a)/2依题意,有：a+√3(b-a)/2=(7-3√3)/2……………………………………(2)由(1)得：a=4-b……………………………………………(3)代(3)入(2),有：4-b+√3[b-(4-b)]/2=(7-3√3)/24-b+√3(2b-4)/2=(7-3√3)/2(√3-1)b+4-2√3=7/2-(3/2)√3(√3-1)b=(√3-1)/2b=1/2代入(3),有：a=4-1/2解得：a=7/2所以：f(x)=(7/2)sinx+(1/2)cosxf(x)=(1/2)(7sinx+cosx)f(x)=[√(7^2+1^2)/2]{[7/√(7^2+1^2)]sinx+[1/√(7^2+1^2)]cosx}f(x)=[(√50)/2][(7/√50)sinx+(1/√50)cosx]不妨设：7/√50=cosα,则：1/√50=sinα代入上式,有：f(x)=[(√50)/2](cosαsinx+sinαcosx)f(x)=[(√50)/2]sin(x+α)f(x)=[(5√2)/2]sin(x+α)因为：-1≤sin(x+α)≤1所以：-(5√2)/2≤[(√50)/2]sin(x+α)≤(5√2)/2即：-(5√2)/2≤f(x)≤(5√2)/2因此：f(x)的最大值是(5√2)/2,最小值是-(5√2)/2.

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