Given that \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\) where \(ab\neq{0}\) and we need to find the sum of all solutions of the equation \(x@2 = \frac{x@(-1)}{2}\)

Let's start by finding the value of x@2 first
To find the value of x@2, compare x@2 with a@b
=> a=x and b=2
So, to find x@2 substitute a=x and b=2 in \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\)
=> \(x@2 = |\frac{x+1}{x}| - \frac{2+1}{2}\) =\( |\frac{x+1}{x}| - \frac{3}{2}\)

Similarly, \(x@(-1) = |\frac{x+1}{x}| - \frac{-1+1}{-1}\) =\(| \frac{x+1}{x}| - 0\) = \(|\frac{x+1}{x}|\)

Given that \(x@2 = \frac{x@(-1)}{2}\)
=> \( |\frac{x+1}{x}| - \frac{3}{2}\) = \(|\frac{x+1}{x}|\) / 2
=> \(| \frac{x+1}{x}| - |\frac{x+1}{x}|\) / 2 =\( \frac{3}{2}\)
Multiply both the sides by 2 we get
\( 2*|\frac{x+1}{x}| - |\frac{x+1}{x}|\) = 3
=> \(|\frac{x+1}{x}|\) = 3

We will get two cases (Watch this video to know about the basic of Absolute Value)
Case 1: \(\frac{x+1}{x }\)>= 0 (0r x+1 > 0 => x > -1 and x≠0
=> \(|\frac{x+1}{x}|\) = \(\frac{x+1}{x }\)
=> \(\frac{x+1}{x }\) = 3
=> x + 1 = 3x
=> 2x = 1 => x = \(\frac{1}{2}\) which is > -1 so this is one solution

Case 2: \(\frac{x+1}{x }\)< 0
=> \(|\frac{x+1}{x}|\) = - \(\frac{x+1}{x }\)
=> -\(\frac{x+1}{x }\) = 3
=> -x - 1 = 3x => 4x = -1
or x = \(\frac{-1}{4}\)
Let's check if this is a right answer by substituting the value in \(\frac{x+1}{x }\)< 0
((-1/4) + 1) / (-1/4) = (3/4) / (-1/4) < 0 so this is also a solution

=> Sum of values of the solution = \(\frac{1}{2}\) + \(\frac{-1}{4}\) = \(\frac{1}{4}\) = 0.25

So, Answer will be D
Hope it helps!

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