Question

一道数学题，急需，在线等

若函数f(x)=2sin(2x+π/6)+a+1,x∈[0,π/2])时，f(x)的最大值为4，则a=，最小值为=

Answer 1

因为sin(2x+π/6) <= 1，当x=π/6时取等号，所以
f(x) <= 2+a+1 = a+3 = 4
因此a = 1
f(x) = 2sin(2x+π/6)+2
当x∈[0,π/2]时，π/6<=2x+π/6<=7π/6，
所以sin(2x+π/6)的最小值为x=π/2时，sin(2x+π/6) = sin(7π/6) = -1/2
此时f(x)有最小值1。

Answer 2

a=0 最小值=2

Answer 3

2x+π/6∈[π/6 , 7π/6]
sin(2x+π/6)最大值为1 a=1
sin(2x+π/6)最小值为 -sinπ/6=-1/2
f(x)最小值=1

Answer 4

x∈[0,π/2]) 得到 2x+π/6∈[π/6,7π/6])
当2x+π/6=π/2取得最大值此时sin(2x+π/6)=sinπ/2=1 f(x)=2+a+1=4 a=1
当2x+π/6=7π/6时去最小值 sin(2x+π/6)=sin7π/6=-1/2 f(x)=-1+a+1=1

Answer 5

sin(2x+π/6）=1，即x=π/6,所以a=1,x=π/2时最小值为1

Answer 6

a=1 最小值为1 解：x∈[0,π/2] 2x∈[0,π] 2x+π/6∈[π/6,7π/6] 函数f(x)在[0,5π/12]上为增函数 在[5π/12，π/2]为减函数 x=5π/12 时函数f(x)取得最大值 最大值为a+3=4 故a=1 f(0)=3 f(π/2)=1 故最小值为1