Question

已知函数f(x)=sin^2x+sinxcosx (1)求f( x)的最大值及取得最大值时对应的X的值 (2)求该函数的单调递增区间

Answer 1

f(x)=sin^2x+sinxcosx
=1/2(2sin^2x+2sinxcosx)
=1/2(1-cos2x+sin2x)
=1/2(sin2x-cos2x)+1/2
=√2/2sin(2x-π/4)+1/2
f(x)max=√2/2+1/2,当2x-π/4=2kπ+π/2, 即：x=kπ+3π/8
(2)求该函数的单调递增区间
2kπ-π/2 ≤2x-π/4≤2kπ+π/2
kπ-π/8 ≤x≤kπ+3π/8

Answer 2

f(x)=-1/2cos2x+1/2sin2x+1/2=-√2/2sin（2x-π/4）+1/2
所以当2x-π/4=2nπ+π/2时，即x=nπ+3π/8（n为整数）时，有最小值-√2/2+1/2
当2x-π/4=2nπ+3π/2时，即x=nπ+7π/8（n为整数）时，有最大值√2/2+1/2
2）
对于sinx的单调递增区间是（2nπ-π/2,2nπ+π/2）
即有2nπ-π/2<2x-π/4＜2nπ+π/2
解得
nπ-π/8＜x＜nπ+3π/8
所以f(x)=sin^2x+sinxcosx 的单调递增区间是 （nπ-π/8，nπ+3π/8）n为整数

Answer 3

(1)f(x)=sin^2x+sinxcosx
=(1-cos2x )/2+1/2sin2x
=1/2sin2x-1/2cos2x+1/2
=sin(2X-π/4)+1/2
所以当X=Kπ+3/8π时f(x)=3/2最大
（2）求f(x)=sin(2X-π/4)+1/2单调区间