函数f={cos(3/2)πx(0≤x≤1){log1/3[x](x>1) ,若a,b,c互不相等,且f=f=f,则a+b+c的取值范围是A (4/3,3)B [4/3,3)C (4/3,4)D (7/3,13/3)当0≤x≤1时,f=cos(3/2)πx当x>1时,f=log1/3[x]【1/3是底数,x是真数】
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![函数f={cos(3/2)πx(0≤x≤1){log1/3[x](x>1) ,若a,b,c互不相等,且f=f=f,则a+b+c的取值范围是A (4/3,3)B [4/3,3)C (4/3,4)D (7/3,13/3)当0≤x≤1时,f=cos(3/2)πx当x>1时,f=log1/3[x]【1/3是底数,x是真数】](/uploads/image/z/8976281-41-1.jpg?t=%E5%87%BD%E6%95%B0f%3D%7Bcos%283%2F2%29%CF%80x%280%E2%89%A4x%E2%89%A41%EF%BC%89%7Blog1%2F3%5Bx%5D%28x%3E1%29+%2C%E8%8B%A5a%2Cb%2Cc%E4%BA%92%E4%B8%8D%E7%9B%B8%E7%AD%89%2C%E4%B8%94f%3Df%3Df%2C%E5%88%99a%2Bb%2Bc%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AFA+%284%2F3%2C3%29B+%5B4%2F3%2C3%29C+%284%2F3%2C4%29D+%287%2F3%2C13%2F3%29%E5%BD%930%E2%89%A4x%E2%89%A41%E6%97%B6%EF%BC%8Cf%3Dcos%283%2F2%29%CF%80x%E5%BD%93x%3E1%E6%97%B6%EF%BC%8Cf%3Dlog1%2F3%5Bx%5D%E3%80%901%2F3%E6%98%AF%E5%BA%95%E6%95%B0%EF%BC%8Cx%E6%98%AF%E7%9C%9F%E6%95%B0%E3%80%91)
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函数f={cos(3/2)πx(0≤x≤1){log1/3[x](x>1) ,若a,b,c互不相等,且f=f=f,则a+b+c的取值范围是A (4/3,3)B [4/3,3)C (4/3,4)D (7/3,13/3)当0≤x≤1时,f=cos(3/2)πx当x>1时,f=log1/3[x]【1/3是底数,x是真数】
函数f={cos(3/2)πx(0≤x≤1)
{log1/3[x](x>1) ,若a,b,c互不相等,且f=f=f,则a+b+c的取值范围是
A (4/3,3)
B [4/3,3)
C (4/3,4)
D (7/3,13/3)
当0≤x≤1时,f=cos(3/2)πx
当x>1时,f=log1/3[x]【1/3是底数,x是真数】
函数f={cos(3/2)πx(0≤x≤1){log1/3[x](x>1) ,若a,b,c互不相等,且f=f=f,则a+b+c的取值范围是A (4/3,3)B [4/3,3)C (4/3,4)D (7/3,13/3)当0≤x≤1时,f=cos(3/2)πx当x>1时,f=log1/3[x]【1/3是底数,x是真数】
可以画出图像,
当cos3πx/2=0时(0≤x≤1)
x=1/3或x=1
这是前一段的图像
因为若a,b,c互不相等,且f<a>=f<b>=f<c>
由三角函数对称性来说
设前两个数为a,b
所以a,b必定关于直线x=2/3对称
所以a+b=4/3
还有一个数在(1,3)上
所以a+b+c的取值范围为(7/3,13/3)
选D
啊 D