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解:由于：f(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=<x<=kπ+π/8故单调递增区间为：[kπ-3π/8,kπ+π/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=3b=4故：a=2-2√2,b=4 |
