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求答案图片表情包(求答案)

导读 大家好,我是小十,我来为大家解答以上问题。求答案图片表情包,求答案很多人还不知道,现在让我们一起来看看吧!1、1、解:把y = sin(ω...

大家好,我是小十,我来为大家解答以上问题。求答案图片表情包,求答案很多人还不知道,现在让我们一起来看看吧!

1、

1、解:把y = sin(ωx + φ)的图像向左平移π/3个单位,得到如图所示的图像,那么新图像的最小正周期是(7π/12 – π/3)*4 = (3π/12)*4 = π,这也是原函数y = sin(ωx + φ)的最小正周期,由已知ω > 0,所以2π/|ω| = 2π/ω = π,可得ω = 2,因此y = sin(2x + φ),向左平移π/3个单位,得到的函数的解析式为g(x) = sin[2(x + π/3) + φ] = sin(2x + 2π/3 + φ),由图像g(π/3) = 0,所以sin(2*(π/3) + 2π/3 + φ) = 0,即sin(4π/3 + φ) = 0,可得4π/3 + φ = kπ,k∈Z,所以φ = kπ – 4π/3,k∈Z,由已知|φ| < π/2,所以当且仅当k = 1时,对应的φ = π – 4π/3 = -π/3,代入可得原函数的解析式为y = sin(2x – π/3) 。

2、 

3、f(x) = 2cos2x + √3sin2x + a,

4、解:(1)由已知f(x) = 2cos2x + √3sin2x + a = 1 + cos2x + √3sin2x + a = 2sin(2x + arctan(√3/3)) + a + 1 = 2sin(2x + π/6) + a + 1,所以f(x)的最小正周期为2π/2 = π,函数f(x) = 2sin(2x + π/6) + a + 1在(2x + π/6)∈[2kπ – π/2,2kπ + π/2],k∈Z上单调递增,即(2x)∈[2kπ – 2π/3,2kπ + π/3],k∈Z,即x∈[kπ – π/3,kπ + π/6],k∈Z,所以函数f(x)的单调递增区间是[kπ – π/3,kπ + π/6],k∈Z ;

5、(2)在(1)中令k = 0,可得函数f(x)在[-π/3,π/6]上单调递增,所以f(x)在[-π/6,π/6]上单调递增,因此在[-π/6,π/6]上,f(x)max = f(π/6) = 2sin(2π/6 + π/6) + a + 1 = a + 3;而且f(x)min = f(-π/6) = 2sin(-2π/6 + π/6) + a + 1 = a;由题意(a + 3) + a = 2a + 3 = 3,那么2a = 0,所以a = 0 ;

6、综上所述,常数a的值是0 。

本文到此讲解完毕了,希望对大家有帮助。