Question

(理)已知奇函數(shù)f(x)=

(1)求實(shí)數(shù)m的值,并在直角坐標(biāo)系中畫出y=f(x)的圖象;

(2)若函數(shù)f(x)在區(qū)間［-1,|a|-2］上單調(diào)遞增,試確定a的取值范圍.

(文)設(shè)函數(shù)f(x)=(sinωx+cosωx)cosωx(其中0＜ω＜2).

(1)若f(x)的周期為π,求當(dāng)-≤x≤時(shí),f(x)的值域;

(2)若函數(shù)f(x)的圖象的一條對(duì)稱軸方程為x=,求ω的值.

Answer 1

答案：(理)解:(1)當(dāng)x＜0時(shí),-x＞0,f(-x)=-(-x)²+2(-x)=-x²-2x.

又f(x)為奇函數(shù),∴f(-x)=-f(x)=-x²-2x.∴f(x)=x²+2x.∴m=2.y=f(x)的圖象如上圖所示.

(2)由(1)知f(x)=由圖象可知,f(x)在[-1,1]上單調(diào)遞增,要使f(x)在[-1,|a|-2]上單調(diào)遞增,只需解之,得-3≤a＜-1或1＜a≤3.

(文)解:f(x)=sinωxcosωx+cos²ωx=sin2ωx+=sin(2ωx+)+.

(1)f(x)周期為π,∴=π.∴ω=1.∴f(x)=sin(2x+)+.∵-≤x≤,∴-≤2x+≤.

∴-≤sin(2x+)≤1.∴0≤f(x)≤.

(2)令2ωx+=kπ+(k∈Z),得ωx=+(k∈Z).當(dāng)x=時(shí),得ω=(k∈Z),0＜ω＜2且k∈Z,∴k=0.∴ω=.