已知函數(shù)f(x)=2sin(2x+π/6)+a+1(a∈Ra為常數(shù))(1)若x∈R求f(x)的最小正周期(2)若f(x)在[-π/6,π/6]上最大
已知函數(shù)f(x)=2sin(2x+π/6)+a+1(a∈Ra為常數(shù))(1)若x∈R求f(x)的最小正周期(2)若f(x)在[-π/6,π/6]上最大
值與最小值之和為3,求a的值(3)求在(2)條件下f(x)的單調(diào)減區(qū)間
優(yōu)質(zhì)解答
(1)正周期顯然為2π/ω=2π/2=π
(2)單看2sin(2x+π/6)這個函數(shù)在[-π/6,π/6]上的極值,解-0.5π<2x+π/6<0.5π
可知-π/3所以f(x)最大值為f(π/6),最小值為f(-π/6),算出來分別為a+3和a,所以a顯然為0
(3)單調(diào)遞增區(qū)間已經(jīng)知道了,+π/2就是一個單調(diào)遞減區(qū)間(因?yàn)槭?x),注意要加上kπ
結(jié)果就是π/6+kπ