已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f
已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f(x)≤1(1)求常数a,b的值(2)设g(x)=f(x+π/2)且lg...
已知a>0,函数f(x)=-2asin(2x+π/6)+2a+b,当x∈[0,π/2]时,-5≤f(x)≤1 (1)求常数a,b的值 (2)设g(x)=f(x+π/2)且lgg(x)>0,求g(x)的单调 我算了好几遍第一问,a+b<=fx<=3a+b,可看网上人家回答得都是b<=fx<=3a+b,谁能把这题解答一下?谢谢
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(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]
打字不易,如满意,望采纳。
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]
打字不易,如满意,望采纳。
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第4行,当x=6/派的时候,y不等于1呀
对不起,打错了
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