P= 1-t^2+t

(2)t=√2sin(θ-π/4) from θ to [0, π] t to [(-√2)/2, 1].

p= 1-t^2+t=-[t-( 1/2)]^2+(5/4)

When t= 1/2, P max =5/4.

When t=-√2/2, the minimum value of p =( 1-√2)/2.

2.f( 12k+ 1)= 1/2 f( 12k+3)= 1 f( 12k+5)= 1/2 f(65438

So f (1) f (3) f (5) ... f (101) = (1/2) *1* (1/2) * (.

3. Angle A = 90, so vector AB is multiplied by vector BC=0, that is, 2k+3=0 k=-3/2.

4. Domain 2 sin (2x+π/3)+1>; 0 to get 2kπ-π/6.

Value range 0

Monotonicity: The function consists of y= ㏒0.2(t) and t=2sin(2x+π/3)+ 1.

Therefore, when 2x+π/3∈(2kπ-π/6, 2kπ+π/2), that is, x ∈ (kπ-π/4, kπ+π/ 12), the internal function t increases and the external function y decreases, so the external function y decreases.

When 2x+π/3∈[2kπ+π/2, 2kπ+7π/6], that is, x ∈ [kπ+π/ 12, kπ+5π/ 12), the internal function T decreases and the external function Y decreases.

Periodicity: T=2π/2=π

Maximum value: y minimum value =㏒0.2(3) At this time 2x+π/3= 2kπ+π/2, that is, x= kπ+π/ 12.