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The operator @ is defined by the following expression: a@b = |(a + 1)| Updated on: 22 Oct 2025, 08:58
Originally posted by Bunuel on 27 Jun 2018, 21:14.
Last edited by Bunuel on 22 Oct 2025, 08:58, edited 1 time in total.
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The operator \(@\) is defined by the following expression: \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\) where \(ab\neq{0}\). What is the sum of the solutions to the equation \(x@2 = \frac{x@(-1)}{2}\) ?

A. -1
B. -0.75
C. -0.25
D. 0.25
E. 0.75
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D
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The operator @ is defined by the following expression: a@b = |(a + 1)| 28 Jun 2018, 05:08
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Bunuel
The operator \(@\) is defined by the following expression: \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\) where \(ab\neq{0}\). What is the sum of the solutions to the equation \(x@2 = \frac{x@(-1)}{2}\) ?


A. -1

B. -0.75

C. -0.25

D. 0.25

E. 0.75
Show more

Before we go to actual calculations, let's look at what we have...
1) We have x on each side, with function of x on left side and \(\frac{(function of x)}{2}\) on right side..
So final will be \(\frac{(function of x)}{2}\)on left side...
2) @-1 will leave 0 as -1+1 is zero

So \(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......
\(|\frac{x+1}{x}|=3\)....
Two cases..
1) \(\frac{(x+1)}{x}=3......x+1=3x....x=\frac{1}{2}\)
2) \(\frac{(x+1)}{x}=-3.....x+1=-3x.....x=-\frac{1}{4}\)
Sum = \(\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\) or 0.25

even if you solve it completely
\(x@2 = |\frac{x+1}{x}| - \frac{2+1}{2}=|\frac{x+1}{x}| - \frac{3}{2}\).....
\(\frac{x@-1}{2} = \frac{1}{2}*|\frac{x+1}{x}| - \frac{-1+1}{1}=\frac{1}{2}*|\frac{x+1}{x}|\)
so \(|\frac{x+1}{x}| - \frac{3}{2}=\frac{1}{2}*|\frac{x+1}{x}|........................\frac{1}{2}|\frac{x+1}{x}| = \frac{3}{2}...................|\frac{x+1}{x}| = 3\)
Two cases..
1) \(\frac{(x+1)}{x}=3......x+1=3x....x=\frac{1}{2}\)
2) \(\frac{(x+1)}{x}=-3.....x+1=-3x.....x=-\frac{1}{4}\)

D
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The operator @ is defined by the following expression: a@b = |(a + 1)| 18 Oct 2019, 01:16
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So \(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......
\(|\frac{x+1}{x}|=3\)....

Can anyone explain how \(|\frac{x+1}{x}|=3\)?
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The operator @ is defined by the following expression: a@b = |(a + 1)| 19 Oct 2019, 09:00
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So \(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......
\(|\frac{x+1}{x}|=3\)....

Can anyone explain how \(|\frac{x+1}{x}|=3\)?
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\(\frac{1}{2}*(|\frac{x+1}{x}|)-|\frac{2+1}{2}|=0\)......

=> \(\frac{1}{2}*(|\frac{x+1}{x}|) = \frac{3}{2}\)

Multiply L.H.S. and R.H.S. by 2

\(|\frac{x+1}{x}|=3\)
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The operator @ is defined by the following expression: a@b = |(a + 1)| 12 Aug 2021, 09:57
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|(X+1)/X|=3

If (X+1)/X < 0, then X<-1
On that basis I started solving
(X+1)/X = -3
X + 1 = -3X
X = -1/4 or -0.25

this solution did not match the criteria as -0.25 is not less than -1
So I rejected it. Where am I going wrong? Bunuel
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The operator @ is defined by the following expression: a@b = |(a + 1)| 12 Aug 2021, 19:34
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|(X+1)/X|=3

If (X+1)/X < 0, then X<-1
On that basis I started solving
(X+1)/X = -3
X + 1 = -3X
X = -1/4 or -0.25

this solution did not match the criteria as -0.25 is not less than -1
So I rejected it. Where am I going wrong? Bunuel
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If \(|\frac{x+1}{x}|=3\)

\(|\frac{x+1}{x}|<0\)

So, both x+1 and x are of opposite sign.

Two cases
1) x>0, then x+1<0 or x<-1…….NO
2) x<0, then x+1>0 or x>-1…..YES
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The operator @ is defined by the following expression: a@b = |(a + 1)| 16 Dec 2021, 06:40
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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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The operator @ is defined by the following expression: a@b = |(a + 1)| 25 Jul 2023, 11:58
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Bunuel
The operator \(@\) is defined by the following expression: \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\) where \(ab\neq{0}\). What is the sum of the solutions to the equation \(x@2 = \frac{x@(-1)}{2}\) ?


A. -1

B. -0.75

C. -0.25

D. 0.25

E. 0.75
Show more


a@b = |(a+1)/a| - (b+1)/b
We need to solve: x@2 = 1/2 * [x@(-1)]
i.e. 2 * x@2 = x@(-1)

We have: 2 * x@2 = 2 * [|(x+1)/x| - (2+1)/2] = 2 * |(x+1)/x| - 3
Also: x@(-1) = |(x+1)/x| - (-1+1)/2 = |(x+1)/x| - 0

Thus, we have: 2 * |(x+1)/x| - 3 = |(x+1)/x|
=> |(x+1)/x| = 3
=> (x+1)/x = 3 or -3
=> x+1 = 3x or -3x
=> x = 1/2 or -1/4

Sum of the solutions = 1/2 + (-1/4) = 1/4

Answer D
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The operator @ is defined by the following expression: a@b = |(a + 1)| 16 Sep 2024, 19:51
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The operator \(@\) is defined by the following expression: \(a@b = |\frac{a+1}{a}| - \frac{b+1}{b}\) where \(ab\neq{0}\). What is the sum of the solutions to the equation \(x@2 = \frac{x@(-1)}{2}\) ?

\(x@2 = |\frac{x+1}{x}| - \frac{3}{2}\)
\(x@(-1) = |\frac{x+1}{x}|- 0 = |\frac{x+1}{x}|\)
\(x@(-1)/2 = |\frac{x+1}{2x}|\)

\(|\frac{x+1}{x}| - \frac{3}{2} = |\frac{x+1}{2x}|\)
\(|\frac{x+1}{2x}| = \frac{3}{2}\)
\(|\frac{x+1}{x}| = 3\)
\(x = \frac{1}{2} or \frac{-1}{4}\)

Sum of the solutions = 1/2 - 1/4 = 1/4 = 0.25

IMO D
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The operator @ is defined by the following expression: a@b = |(a + 1)| 08 Oct 2025, 12:35
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