The image of the function f (x) = | x + 1 | + | x-a | (x ∈ R, a is a constant) is symmetric about the y-axis The image of the function f (x) = ┃ x + 1 ┃ + ┃ x-a ┃ (x ∈ R, a is a constant) is symmetric about the Y axis. (1) find the value of a (2) Let G (x) = f (x-t) - f (x + T), judge the parity of G (x), and give the proof

The image of the function f (x) = | x + 1 | + | x-a | (x ∈ R, a is a constant) is symmetric about the y-axis The image of the function f (x) = ┃ x + 1 ┃ + ┃ x-a ┃ (x ∈ R, a is a constant) is symmetric about the Y axis. (1) find the value of a (2) Let G (x) = f (x-t) - f (x + T), judge the parity of G (x), and give the proof

（1）
Solution 1: how to solve the equation f (1) = f (- 1) 2 + | 1-A | = | 1 + a |,
Solution 1: honest method, discuss a to absolute value
Solution 2: of course, it can be understood that the distance from a to - 1 minus the distance from a to 1 on the number axis is equal to 2. Then a > = - 1
F (2) = f (- 2) 3 + | 2-A | = 1 + | 2 + a |, then a = 1
This solution can only use f (2). The reason why f (1) is also included is that I first thought of 1
Solution 2: F (x) = f (- x)
|The approximation of X + 1 | + | x-a | = | - x + 1 | + | x + a |, makes me want to change the right x coefficient to positive
=|X + a | + | X-1 |, it is not difficult to find that a = 1
This method is not rigorous enough, a may have other possible values
Solution 3: from the number axis, f (x) represents the sum of the distances from X to - 1 and A. since it is an axisymmetric figure, the symmetric point of - 1 is 1, then a = 1