1

y=(tanx-1)^2

[0,π/6]

tanx<1

函数为减

f(min)=f(π/6)=2(1-√3)/3

2

y=-cosx^2+acosx-a/2-1/2

令cosx=t

y=-t^2+at-a/2-1/2=-(t-a/2)^2+(a^2-2a-2)/4

t=a/2时y最大

a^2-2a-2=1

(a-3)(a+1)=0

a=3或a=-1

3

(1)定义域x/2-x/3=x/6属于[-π/2+kπ,π/2+kπ]

x属于[-3π+6kπ,3π+6kπ]

周期为6kπ

单调区间[-3π+6kπ,3π+6kπ]为增

(2)π/4+kπ<=x/6<=π/3+kπ

3π/2+6kπ<=x<=2π+6kπ

4

没有x