Question

求答案，要有过程

Answer 1

f(x)=a(cos²x+sinxcosx)+b=a/2(cos2x+sin2x+1)+b=√2/2asin(2x+π/4)+a/2+b

追问

答案？

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追答

f(x)=a(cos²x+sinxcosx)+b=a/2(cos2x+sin2x+1)+b=√2/2asin(2x+π/4)+a/2+b

1) ∵a>0
∴2kπ-π/2≤2x+π/4≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8
递增区间：[kπ-3π/8,kπ+π/8]k∈Z

2)∵a<0
∴{√2/2a+a/2+b=3
{-√2/2a+a/2+b=4
解得，a=-√2/2,b=7/2+√2/4

第二题看错了：
2)∵x∈[0,π/2]
∴函数在[0,π/8]单调递增，在[π/8,π/2]单调递减
{√2/2asin(2*π/2+π/4)+a/2+b=3
{√2/2asin(2*π/8+π/4)+a/2+b=4
解得，a=2-2√2,b=4

追问

谢谢