已知函数f（X）=asin2x+cos2x,且f（π/3）=√3-1/2

显示全部楼层 · 2012-9-9 09:25:53

解:1、已知f(x)=asin2x+cos2x且f(π/3)=√3-1/2√3-1/2=asin(2π/3)+cos(2π/3)√3-1/2=a*√3/2-1/2a=2y=f(x)=2sin2x+cos2xy-2sin2x=cos2x=√[1-(sin2x)^2]y^2+4(sin2x)^2-4y*sin2x=1-(sin2x)^25(sin2x)^2-4y*sin2x+y^2-1=0上方程未知数为(sin2x)的判别式△≥0,即(4y)^2-4*5*(y^2-1)≥0y^2≤5-√5≤y≤√5即： a=2,f(x)最大值=√5 2、f(x)=2sin2x+cos2xf(x)=√5sin（...

千问 · 2012-9-9 09:25:53

(1)f(π/3)=(√3-1)/2asin(2π/3)+cos(2π/3)=(√3-1)/2(a根号3)/2-1/2=(√3-1)/2a=1辅助角=arctan (1)=π/4f(x)=根号2 sin(2x+π/4)最大值为根号2(2)减函数看图像[2kπ+π/2,2kπ+3π/2]2kπ+π/2<=2x+π/4...

千问 · 2012-9-9 09:25:53

f（π/3）=asin2π/3+cos2π/3=a√3/2-1/2=(√3-1)/2所以a=1,f（x）=sin2x+cos2x=√2sin(2x+π/4),因为sin(2x+π/4)最大值为1,所以f(x)的最大值为√2若要f(x）为减函数，则π/2+2kπ＜2x+π/4＜3π/2+2kπ,解得π/8+kπ＜x＜5π/8+kπ,其中k为整数，...

千问 · 2012-9-9 09:25:53

函数f（X）=asin2x+cos2x,且f（π/3）=√3-1/2解：f（π/3）=asin（2π/3）+cos(2π/3)
=asin（π/3）-cos(π/3)
=√3a/2-1/2=√3-1/2所以
a=2.
所以f（X）=2...