Question

高一数学急急急~~~ 要具体步骤

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属于[0,π]时，f(x)的值域为[3,4]，求实数a,b的值

Answer 1

f(x)=a/2(2cos²x+2sinxcosx)+b
=a/2(cos2x+sin2x)+b
=a√2/2sin(2x+π/4)+b
(1).2kπ-π/2≤2x+π/4≤2kπ+π/2
2kπ-3π/4≤2x≤2kπ+π/4
kπ-3π/8≤x≤kπ+π/8
x∈[kπ-3π/8,kπ+π/8],k∈Z 递增。

(2)x∈[0,π]
2x+π/4∈[π/4，9π/4]
sin(2x+π/4)∈[-1,1]
a√2/2+b=3
-a√2/2+b=4
a=-√2/2,b=7/2

Answer 2

⑴f(x)=a(cos²x+sinxcosx)+b
=a(0.5cos2x+0.5+0.5sin2x)+b
=0.5a(cos2x+sin2x)+0.5a+b
= 根号2/2asin(2x+ π/4 )+a/2+b
所以单调增区间：-π/2+2kπ<2x+π/4<π/2+2kπ
-3π/8+kπ<x<π/8+kπ
⑵因为x属于[0,π]，所以2x+a/4属于[π/4,9π/4]
因为—根号2/2*a+a/2 +b=4且—根号2/2*a+a/2 +b=3
所以a=-根号2/2 b=9/2-根号2/4

Answer 3

自己做的错了别喷

令cos²x+sinxcosx=t
t=(1+cos2x)/2 + 1/2sin2x
=1/2+cosx+sinx
=1/2+1/2(cos2x+sin2x)

cos2x+sin2x 一化
后面会了吧