It is known that function f (x) satisfies f (a) + F (b) = f (a + b) + 2 for any real number a and B, and if a > 0, f (a) > 2 holds. 1. Find the value of F (0): 2. Prove the increasing function of function f (x) on R

It is known that function f (x) satisfies f (a) + F (b) = f (a + b) + 2 for any real number a and B, and if a > 0, f (a) > 2 holds. 1. Find the value of F (0): 2. Prove the increasing function of function f (x) on R

It is known that f (x) satisfies f (a) + F (b) = f (a + b) + 2 for any real number a and B, and if a > 0, f (a) > 2 holds
1. Find the value of F (0)
f(0)+f(0)=f(0+0)+2
2f(0)=f(0)+2
f(0)=2
2. Proof: the increasing function of function f (x) on R
Let: n > M
n-m>0
f（n-m）>2
f(n-m)+f(m)=f(n)+2
f(n)-f(m)=f(n-m)-2>0
Namely
f(n)-f(m)>0
f(n)>f(m)
So: the function f (x) is an increasing function on R

If the period of the function f (x) = 2Sin (KX / π + π / 4) is within (2 / 3,3 / 4), then all positive integer values of K are?

T=2π/ω
=2π/(k/π)
=2*π^2*k
From t ∈ (2 / 3,3 / 4)
The solution is k ∈ (8 / 3 π ^ 2,3 π ^ 2)
8/3*3.14^2=26.29
3*3.14^2=29.58
∴k∈{27,28,29}

If the period of function f (x) = sin (KX + Pai / 3) is t and t belongs to (1,3), then the positive integer k=

Firstly, when k = 1, the period of this function is 2 π; if k = 2, then it is π, that is:
Period = 2 π / K
The problem is to find the value of 2 π / K in (1,3), K can be 3 and 4,

If the image of the function y = 2Sin (8x + φ) + 2 is symmetric with respect to the straight line x = π / 6, then the number of zeros of the function on [0,2 π]
See someone write φ = - 6 / π + π / 2 = π / 3

F (x) = f (π / 3-x)
We can solve the problem by taking it in
Then let f (x) = 0, find out the number of X and draw a picture
On the axis of symmetry, if a function is symmetric with respect to x = a, then f (x) = f (2a-x)
I hope it will help you