(1) Find the monotonic increasing interval of f(x);
(2) if the inequality | f (x)-m | < 2 is constant on x∈[π/4, π/2], find the range of the number m.
Solution: (1). f(x)= 1-cos[2(π/4+x)]-(√3)cos2x- 1 =-cos(π/2+2x)-(√3)cos2x = sin2x。
= 2[( 1/2)sin2x-(√3/2)cos2x]= 2[sin 2 xcos(π/3)-cos2xsin(π/3)]= 2 sin(2x-π/3)
Use -π/2+2kπ
-π/ 12+kπ& lt； x & lt5π/ 12+kπ。
(2) When π/4≦x≦π/2, π/2≦2x≦π/6≦2x-π/3≦2π/3, 0 ≦ f (x) ≦.
If f (x)-m < 2, -2 is obtained.
So the range of m is: (0,2).
2. Known quadratic function f(x)=x? +x, if the solution set of inequality f(-x)+f(x)≤2|x| is C.
(1) solution set c;
(2) if the equation f (a x)-a (x+1) = 5 (a >; 0, a≠1) has a solution on c, and the range of the number a is realistic;
(3) if the equation g(x)=f(x)-x+ 1+|x-a|, x∈C, find the minimum value of the function g(x).
Solution: (1) f(-x)+f(x)=x? -x+x？ +x=2x？ ≦ 2 𔱑 X 𔱑, that is, X? ≦ x, so the solution set c = {x-1≦ x ≦1}.
(2)f(a^x)+a^(x+ 1)=a^(2x)+a^x+a×a^x=a^2x+(a+ 1)a^x=5
That's (an X)? +(a- 1) a x-5 = 0 has a solution on c,
Have a rest and do it after a nap. )