If f (x) is odd, G (x) is even, and f (x) - G (x) = x2 + 2x + 3, then f (x) + G (x)= It's the square of x plus 2x + 3

If f (x) is odd, G (x) is even, and f (x) - G (x) = x2 + 2x + 3, then f (x) + G (x)= It's the square of x plus 2x + 3

The function f (x) is odd and G (x) is even
So f (- x) = - f (x), G (- x) = g (x)
So f (x) + G (x) = - f (- x) + G (- x)
=-[f(-x)-g(-x)]
=-[(-x)^2+2(-x)+3]
=-x^2+2x-3

If f (x) = x2 + (A-1) x is even, then the monotone decreasing interval of G (x) = ax2-2x-1 is
If f (x) = x2 + (A-1) x is even function, does it mean that a = 1?

F (x) = x2 + (A-1) x is an even function
The odd coefficient is equal to 0
So A-1 = 0
a=1

2. If f (x) - G (x) = x2 + 2x + 3, then f (x) + G (x) = (a) - x2 + 2x-3
It is known that f (x) is an odd function on R and G (x) is an even function on R. if f (x) - G (x) = x ^ 2 + 2x + 3, then f (x) + G (x)=
（A）-x^2+2x-3\x05（B）x^2+2x-3\x05（C）-x^2-2x+3\x05（D）x^2-2x+3
Method of solving

A,f(-x)=- f(x),g(-x)= g(x) ,f(-x)-g(-x)=（-x）^2+2（-x)+3,- f(x)-g(x)=x^2-2x+3,
f (x)+g(x)=--x^2+2x - 3