Try again. . .
1, from f(π/6+x)=f(π/6-x), the straight line x=π/6 is the symmetry axis of f(x). By observing the image, we can know whether the function value corresponding to the symmetry axis is a peak or a valley, so f (π/6) = 2 (Note: if F (t+)
2, y = cosxtanx = cosx (sinx/cosx) = sinx, because in the original function, x≠π/2+kπ, y ≠ 1, that is, the interval is an open interval (-1).
3.① When sin θ=0, θ=0 or π, f (x) = 2x+ 1, which is obviously not always greater than 0.
② When sin θ≠0, that is, θ≠0 and θ ≠ π, if the function is always greater than zero, the opening of the function image is upward, that is, sin θ >; 0, there are 0.
At the same time △ < 0, [2(cosθ)]? -6sinθ& lt； 0,
2(cosθ)？ -3 sinθ& lt； 0, let sinθ=t, then t? = 1-(cosθ)？ ，(cosθ)？ = 1-t？
2( 1-t？ )-3t & lt； 0, simplification and factorization result in (2t- 1)(t+2)>0, i.e. t> 1/2, π/6 <; θ& lt； 5π/6
4.( 1) f(x) has a minimum value, x∈R, then the image opening of f(x) is upward, and a >；; 0
If the minimum value is 0, there is only one intersection with the X axis, that is, △=0=b? -4a，b？ =4a (Note: here, the formula method is used to calculate the minimum value =0, and b? =4a)
F(- 1)=a-b+ 1=0, b=a+ 1, and substitute b? =4a，a= 1，b=2。
That is, f(x)=(x+ 1)? , F(x) I won't write here.
(2)g(x)=x？ +(2-k)x+ 1, and g(x) is also a quadratic function.
(1) g(x) monotonically increases in the definition domain, then the symmetry axis of g(x) is on the left side of x=-2, that is, -(2-k)/2≤-2, and the solution is k≤-2.
(2) g(x) decreases monotonously in the definition domain, then the symmetry axis of g(x) is on the right side of x=2, that is, -2-k)/2≥2, and the solution is k≥6.
To sum up, k∈(-∞, -2]∩[6, +∞)
(3) If f (x) is an even function, then f(x)=f(-x), so that b=0 and f(x)=ax? + 1
M>0, so F(m)=am? + 1; N<0, so F(n)=-an? - 1
F(m)+F(n)=a(m？ -n？ )=a(m+n)(m-n)
M>0, n<0, so there are m>0>n, M-n > 0; We also know that m+n >; 0，a & gt0
Therefore, f (m)+f (n) >: 0.