The minimum positive period of the function f (x) = 2cos2x2 + SiNx is______ ．

The minimum positive period of the function f (x) = 2cos2x2 + SiNx is______ ．

F (x) = 2cos2x2 + SiNx = cosx + SiNx + 1 = 2Sin (x + π 4) + 1  t = 2 π 1 = 2 π, so the answer is: 2 π

The minimum positive period of F (x) = 2cos & # 178; X / 2 + SiNx is
Why is "2" not "2"

f(x)=[2cos²(x/2)－1]＋sinx＋1
=cosx＋sinx＋1
=√2[(√2/2)sinx＋(√2/2)cosx]＋1
=√2[sinxcos45°＋cosxsin45°]＋1
=√2sin(x＋45)°＋1
The minimum positive period is 2 π / 1 = 2 π

Given the function f (x + y) + F (X-Y) = 2F (x), and f (0) ≠ 0, it is proved that f (x) is an even function
Urgent need

Let x = y and substitute f (x + y) + F (X-Y) = 2F (x) to get f (2x) + F (0) = 2F (x)
That is, f (2x) - 2F (x) = f (0) ①
∴f(x)-2f(2x)=f(0)②
F (x) = - f (0)
The definition field is r, f (- x) = f (x) = - f (0) (f (0) ≠ 0)
F (x) is an even function

Given that the function f (x) satisfies f (x + y) + F (X-Y) = 2F (x) · f (y) & nbsp; (x ∈ R, y ∈ R), and f (0) ≠ 0, we try to prove that f (x) is an even function

It is proved that: let x = y = 0 ∵ f (x + y) + F (X-Y) = 2F (x) · f (y) ∵ f (0) ∵ f (0) ≠ 0, ∵ f (0) = 1, let x = 0 ∵ f (x + y) + F (X-Y) = 2F (x) · f (y) ∵ f (y) + F (- y) = 2F (0) · f (y) ∵ f (- y) = f (y), that is, f (x) is even function

It is known that the function y = f (x) is an even function on R and an increasing function on (- ∞, 0). It is proved that y = f (x) is a decreasing function on (0, + ∞)

Set at (- ∞, 0) and take any two numbers x2 > X10,
In this interval is an increasing function,
f(x2)>f(x1),
Even function again,
f(-x2)=f(x2),
f(-x1)=f(x1),
f(-x2)>f(-x1),
Because 0 > x2 > x1, X2 ∈ (- ∞, 0], x1 ∈ (- ∞, 0],
Then 0 〈 - x2

Given the even function f (x) defined on [- 5,5], when x ∈ (- ∞, 0), f (x) = - x2 + 2x + 1, find the analytic expression of F (x) when x ≥ 0

_ x 2_ 2x +1

The even function FX defined on R satisfies the following conditions: for any x1, X2 ∈ [0, positive infinity), and x1 ≠ x2
If f (x1) - f (x2) / (x1-x2) > 0, then f (3) f (- 2) f (1)

F (- 2) x2
f(x1)>f(x2)
F (x) is an increasing function
When x1

Let f (x) be an odd function, G (x) be an even function, and f (x) - G (x) = 1 / X & # 178; + X, find the expression of F (x)

f(x)-g(x)=1/x²+x①
F (x) is an odd function and G (x) is an even function
f(-x)-g(-x)
=-f(x)-g(x)
=1/(-x)²+(-x)
=1/x²-x②
① - 2
2f(x)=1/x²+x-1/x²+x
2f(x)=2x
f(x)=x
① + 2
g(x)=-1/x²

Let f (x) = sin2 (x + π 4), if a = f (lg5), B = f (LG15), then ()
A. a+b=0B. a-b=0C. a+b=1D. a-b=1

F (x) = sin2 (x + π 4) = 1 − cos (2x + π 2) 2 = 1 + sin2x2, a = f (lg5), B = f (LG15) = f (- lg5), ∩ a + B = 1 + sin2lg52 + 1 − sin2lg52 = 1, A-B = 1 + sin2lg52-1 − sin2lg52 = sin2lg5, so option C is correct, so select C

Let f (x) = Sin & # 178; (x + 5pai / 4), if a = f (Lg3), B = f (LG1 / 3), then a, a+
Let f (x) = Sin & # 178; (x + 5pai / 4), if a = f (Lg3), B = f (LG1 / 3), then
A、a+b=0
B、a+b=1
C、a-b=0
D、a-b=1

f(x)=sin²(x+5pai/4),
=(1-cos(2x+5pai/2))/2
=(1+sin2x)2
=1/2+1/2sin2x
Lg3 and LG1 / 3 are opposite to each other
So sin (- x) = - f (x)
So a + B = 1