已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x属于[0,π/2]时,-5≤f(x)≤1.求f(x)的单调区间

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/03 02:37:07
已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x属于[0,π/2]时,-5≤f(x)≤1.求f(x)的单调区间
xRn@+RDg~Fq6QƯĔ&"bUA6 P6?T IꊇM73wsg_O|C Gn3DB%mM_~o28 z=r9& 8e[܂ ҭΕŹɏh]0)2ϓ|"nduH$-P&bH^AN[S=6 #4DR7_̾| v#jeIN3&6JĤ]v:B51x!9HQ4$"Igd$5.K6wggewNsw;VVt画ܘel׆1 |Z//s)9۬ CFxdglYFlw t 5ϯ5`R~\iͳi>IT] !xպAM,bS%Y)늅%RA[VHضSĸ08`SbLs#-kQZ0BEk-k-K\K]Uֈ:ľwoó)

已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x属于[0,π/2]时,-5≤f(x)≤1.求f(x)的单调区间
已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x属于[0,π/2]时,-5≤f(x)≤1.求f(x)的单调区间

已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x属于[0,π/2]时,-5≤f(x)≤1.求f(x)的单调区间
(1)因为,x∈[0,π/2],
2x+π/6∈[π/6,7π/6],
sin(2x+π/6)∈[-1/2,1],
又 a>0
所以,-2a+2a+b=-5
a+2a+b=1
解得:a=2,b=-5
(2) 由(1)知,f(x)=-4sin(2x+π/6)-1
由题意 g(x)=f(x+π/2)
=-4sin(2x+π+π/6)-1
=4sin(2x+π/6)-1>1
即 sin(2x+π/6)>1/2
所以 2x+π/6∈(2kπ+π/6,2kπ+5π/6)
单调增区间满足 2x+π/6∈(2kπ+π/6,2kπ+π/2]
单调减区间满足 2x+π/6∈[2kπ+π/2,2kπ+5π/6)
解得 g(x)的单调增区间为 (kπ,kπ+π/6]
单调减区间为 [kπ+π/6,kπ+π/3]