Parity judgment of mathematical function in senior one.

1.∫ The function f (x) = a-2/(2 x+ 1) is odd function.

∴ There is f(x)=-f(-x)

Let x= 1, then f( 1)=-f(- 1).

f( 1)=a-2/2^ 1+ 1=a-2/3

-f(- 1)=-(a-2/2^- 1+ 1)=-a+4/3

∴a-2/3=-a+4/3, and the answer is a= 1.

2.f (x) is an even function.

∴f(x)=f(-x)。 f( 1)= f(- 1)=- 1

When x=- 1,

f(2)= f(- 1+3)= f(- 1)=- 1

When x=2,

f(5)=f(2+3)=f(2)=- 1

∴f(5)+f( 1)=- 1+(- 1)=-2

F(x) is the odd function of the world,

So there is f(x)=-f(-x)

f(3)=-f(-3)

f(- 1)=-f( 1)

∵f(3)& lt； f(- 1)

∴-f(-3)<； -f( 1)

∴f(-3)>f( 1)

There is no answer to the original question.

Because the increase or decrease of the original function is unknown, ABCD may not hold.

According to the known conditions, only F (-3) > F (1) must hold.