Question

高一数学题要求有详细的解答过程

Answer 1

13.解答：f(x)=sqrt(3)sin x cos x-(cos x)^2+1/2=sqrt(3)sin(2x)/2-(1-cox(2x))/2+1/2=sin(2x)cos(π/6)-cos(2x)sin(π/6)=sin(2x-π/6)，显然周期T=2π/2=π
当x∈[0,π/2],2x-π/6∈[-π/6,5π/6],sin(2x-π/6)∈[-1/2,1]。(注：sqrt表示根号)

14.解答:（PS：该题与上面一题方法很接近）
（1）f(x)=-sin(wx)cos(wx)+cos(wx)*cos(wx)=-sin(2wx)/2+(1+cos(2wx))/2
=-1/sqrt(2)*(sin(2wx)cos(π/4)-cos(2wx)sin(π/4))+1/2
=-1/sqrt(2)*sin(2wx-π/4)+1/2
T=2π/(2w)=π，则w=1
（2）即令z=x/2，则有x=2z，故g(x)==-1/sqrt(2)*sin(4x-π/4)+1/2
当x∈[0,π/16],4x-π/4∈[-π/4,0],sin(4x-π/4)∈[-1/sqrt(2),0],g(x)(min)=1/2