已知函数f(x)=a(cos2x+sinxcosx)+b (1)当a>0时,求f(x)的单调递增区间(2)当a<0且x∈[0,π/2]时,f(
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![已知函数f(x)=a(cos2x+sinxcosx)+b (1)当a>0时,求f(x)的单调递增区间(2)当a<0且x∈[0,π/2]时,f(](/uploads/image/z/4321203-51-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Da%28cos2x%2Bsinxcosx%29%2Bb+%EF%BC%881%EF%BC%89%E5%BD%93a%EF%BC%9E0%E6%97%B6%2C%E6%B1%82f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%EF%BC%882%EF%BC%89%E5%BD%93a%EF%BC%9C0%E4%B8%94x%E2%88%88%5B0%2C%CF%80%2F2%5D%E6%97%B6%2Cf%28)
已知函数f(x)=a(cos2x+sinxcosx)+b (1)当a>0时,求f(x)的单调递增区间(2)当a<0且x∈[0,π/2]时,f(
已知函数f(x)=a(cos2x+sinxcosx)+b (1)当a>0时,求f(x)的单调递增区间(2)当a<0且x∈[0,π/2]时,f(
已知函数f(x)=a(cos2x+sinxcosx)+b (1)当a>0时,求f(x)的单调递增区间(2)当a<0且x∈[0,π/2]时,f(
f(x)的值域是[3,4],求a,b的值?
(x)=a[cos^2(x)+sinxcosx]+b
=a[(1+cos2x)/2+(1/2)(2sinxcosx)]+b
=a[(1/2)sin2x+(1/2)cos2x+1/2]+b
=a[(1/2)(sin2x+cos2x)]+(a+2b)/2
=(√2a/2)sin(2x+π/4)+(a+2b)/2
则:
(1)
由于:a>0
则:
当f(x)单调递增时,
2kπ-π/2=<2x+π/4<=2kπ+π/2
即:
kπ-3π/8=
(2)由于:x属于[0,π/2]
则:2x+π/4属于[π/4,5π/4]
则:sin(2x+π/4)属于[-√2/2,1]
由于:a<0
则:f(x)属于:[(√2+1)a/2+b,b]
又:f(x)的值域是[3,4]
则:
(√2+1)a/2+b=3
b=4
故:
a=2-2√2,b=4