高一数学急急急~~~ 要具体步骤

显示全部楼层 · 2010-8-27 15:39:18

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

千问 · 2010-8-27 15:39:18

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/22kπ-3π/4≤2x≤2kπ+π/4kπ-3π/8≤x≤kπ+π/8x∈[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=4a=-√2/2,b=7/2

千问 · 2010-8-27 15:39:18

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

千问 · 2010-8-27 15:39:18

自己做的错了别喷令cos2x+sinxcosx=tt=(1+cos2x)/2 + 1/2sin2x =1/2+cosx+sinx =1/2+1/2(cos2x+sin2x) cos2x+sin2x 一化后面会了吧