已知函数f(x)=√2/2 sin2x+√2 (cosx)^2-(√2+1)/21)当x∈[0,π/2],求函数f(x)的单调递增区间2)若不等式a

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/06 22:02:18
已知函数f(x)=√2/2 sin2x+√2 (cosx)^2-(√2+1)/21)当x∈[0,π/2],求函数f(x)的单调递增区间2)若不等式a
x͔n@_%V>!x*'&1+,9DEEU!BDBDB"wN{W`M 7fr/N?;㲹F*tCj^~#,'&<FcXNν~ ?? hZM_GYuC(P8K|ɔĖPث̸ :dȜͬB".I!5aΩBϕ2<{-ͺwХ*'w߻1'7R°c'q/:IIȈdno{} @9gjuε%cĝ~`QqiKSQGÈwp2X zlwkA4bN&gG<= k˚RD[F7n9W\8%PlZS*2դVeSG5p \9n^V8fKeq r)EN!+H *xNMPS tQBohˊ4L^JB,cU 2D"*RENT!uA.`[D?kn

已知函数f(x)=√2/2 sin2x+√2 (cosx)^2-(√2+1)/21)当x∈[0,π/2],求函数f(x)的单调递增区间2)若不等式a
已知函数f(x)=√2/2 sin2x+√2 (cosx)^2-(√2+1)/2
1)当x∈[0,π/2],求函数f(x)的单调递增区间
2)若不等式a

已知函数f(x)=√2/2 sin2x+√2 (cosx)^2-(√2+1)/21)当x∈[0,π/2],求函数f(x)的单调递增区间2)若不等式a
f(x)=√2/2sin2x+√2cos&sup2;x-(√2+1)/2=√2/2sin2x+√2/2(2cos&sup2;x-1)+√2/2-(√2+1)/2=√2/2(sin2x+cos2x)-1/2=sin(2x+π/4)-1/2
即     f(x)=sin(2x+π/4)-1/2
(1)x∈[0,π/2],所以 t=(2x+π/4)∈[π/4,5π/4]     (令t=2x+π/4)
函数 f(t)=sint-1/2 在t∈[[π/4,π/2]上单调递增,即是函数f(x)在(2x+π/4)∈[π/4,π/2]上单调递增,则可求得函数f(x)的单调递增区间为[0,π/8]
(2)f(t)=sint-1/2   (t∈[π/4,5π/4])  在定义域内可求得值域为 [-(√2+1)/2,1/2] ,所以,欲使不等式 f(x)>a 恒成立,则 a<-(2+1)/2 即可.