1.f(x)=a cos(2x+π/3)+a/2+b (a>0),定义域[-π/3,π/6],值域[-1,5],求a,b.2.若对任意实数a,函数y=5sin((2k+1)/3πx-π/6)(k属于N)在区间[a,a+3]上的值5/4出现不少于4次且不多于8次,则K的值是( )A.2 B.4 C.3或4 D.
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![1.f(x)=a cos(2x+π/3)+a/2+b (a>0),定义域[-π/3,π/6],值域[-1,5],求a,b.2.若对任意实数a,函数y=5sin((2k+1)/3πx-π/6)(k属于N)在区间[a,a+3]上的值5/4出现不少于4次且不多于8次,则K的值是( )A.2 B.4 C.3或4 D.](/uploads/image/z/14330006-62-6.jpg?t=1.f%28x%29%3Da+cos%282x%2B%CF%80%2F3%29%2Ba%2F2%2Bb+%28a%3E0%29%2C%E5%AE%9A%E4%B9%89%E5%9F%9F%5B-%CF%80%2F3%2C%CF%80%2F6%5D%2C%E5%80%BC%E5%9F%9F%5B-1%2C5%5D%2C%E6%B1%82a%2Cb.2.%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8F%E5%AE%9E%E6%95%B0a%2C%E5%87%BD%E6%95%B0y%3D5sin%EF%BC%88%EF%BC%882k%2B1%29%2F3%CF%80x-%CF%80%2F6%EF%BC%89%EF%BC%88k%E5%B1%9E%E4%BA%8EN%EF%BC%89%E5%9C%A8%E5%8C%BA%E9%97%B4%5Ba%2Ca%2B3%5D%E4%B8%8A%E7%9A%84%E5%80%BC5%2F4%E5%87%BA%E7%8E%B0%E4%B8%8D%E5%B0%91%E4%BA%8E4%E6%AC%A1%E4%B8%94%E4%B8%8D%E5%A4%9A%E4%BA%8E8%E6%AC%A1%2C%E5%88%99K%E7%9A%84%E5%80%BC%E6%98%AF%EF%BC%88+%EF%BC%89A.2+B.4+C.3%E6%88%964+D.)
1.f(x)=a cos(2x+π/3)+a/2+b (a>0),定义域[-π/3,π/6],值域[-1,5],求a,b.2.若对任意实数a,函数y=5sin((2k+1)/3πx-π/6)(k属于N)在区间[a,a+3]上的值5/4出现不少于4次且不多于8次,则K的值是( )A.2 B.4 C.3或4 D.
1.f(x)=a cos(2x+π/3)+a/2+b (a>0),定义域[-π/3,π/6],值域[-1,5],求a,b.
2.若对任意实数a,函数y=5sin((2k+1)/3πx-π/6)(k属于N)在区间[a,a+3]上的值5/4出现不少于4次且不多于8次,则K的值是( )
A.2 B.4 C.3或4 D.2或3
3.函数f(x)=Msin(wx+b)(w>0)在区间[a,b]上是增函数,且f(a)=-M,f(b)=M,则函数g(x)=Mcos(wx+b)在[a,b]上( )
A.是增函数 B.是减函数 C.可以取最大值M D.可以取最小值-M
1.f(x)=a cos(2x+π/3)+a/2+b (a>0),定义域[-π/3,π/6],值域[-1,5],求a,b.2.若对任意实数a,函数y=5sin((2k+1)/3πx-π/6)(k属于N)在区间[a,a+3]上的值5/4出现不少于4次且不多于8次,则K的值是( )A.2 B.4 C.3或4 D.
1.定义域[-π/3,π/6],则(2x+π/3)属于[-π/3,2π/3],cos(2x+π/3)属于[cos(2/3π),1],(画图可以看出来),那么有a cos(2π/3)+a/2+b =-1
a ×1+a/2+b=5,求得a,b
2.另R=((2k+1)/3πx-π/6),则有y=5sinR,又sin曲线看,一个周期就会出现两个1/4,T=2π/w,w=(2k+1)/3π,所以由题可以知道:2T>3,4T
[1];a=4 b=-1
[2].d.
[3].c.
1,定义域[-π/3,π/6]时,2x+π/3的范围是[-π/3,2π/3],所以 cos(2x+π/3)的值域为[-0.5,1]
因为a>0,所以cos(2x+π/3)=-0.5时有最小值-1,cos(2x+π/3)=1时有最大值5,代入解得a=4,b=-1
2,不少于4次且不多于8次即跨越的周期应该在2至4间
所以 2《[(2k+1)/3π*(...
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1,定义域[-π/3,π/6]时,2x+π/3的范围是[-π/3,2π/3],所以 cos(2x+π/3)的值域为[-0.5,1]
因为a>0,所以cos(2x+π/3)=-0.5时有最小值-1,cos(2x+π/3)=1时有最大值5,代入解得a=4,b=-1
2,不少于4次且不多于8次即跨越的周期应该在2至4间
所以 2《[(2k+1)/3π*(a+3-a)]/2π《4
解得 2《k《3 选D
3,由题目,sin(wa+b)=-1,sin(wb+b)=1 ,且单调递增。
类似于-π/2到π/2的区间,所以选C
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