Question

高一数学求解！如图，第21题，在线等，谢谢了！

Answer 1

Answer 2

21、f(x)=ab=cos²ωx+√3sinωxcosωx
=(1+cos2ωx)/2+√3/2sin2ωx
=1/2+sin(2ωx+π/3)
(1) T=2π 则 2π/2ω=2π ω=1/2
单调递增：2kπ-π/2<x+π/3<2kπ+π/2
即 2kπ-5π/6<x<2kπ+π/6
(2) f(x)=1/2+sin(2ωx+π/3)
=1/2+sin(2ω(x+π/6ω))
-π/6ω=π/6
得 ω=-1
22、f(x)=px²+qx
f(x-1)=f(x)+x-1
p(x-1)²+q(x-1)=px²+qx+x-1
px²-(2p-q)x+p²-q=px²+(q+1)x-1
系数相等，得
-(2p-q)=q+1
p²-q=-1
得 p=1/2 q=5/4
(1) f(x)=x²/2+5x/4
(2) 令f(x)=0 得 x=0 x=-5/2
f(x)<0 得 -5/2<x<0

Answer 3