VeritasPrepKarishma wrote:
|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.
When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4
We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.
Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4
The only point is that we don't include the transition points twice.
Hope the role of '=' sign is clear.
Quote:
In GMATCLUB Maths book it is given that to solve any Modulus Questions we need to follow 3 step process. First is by opening the Modulus and exposing to signs. In the problem given below I am not able to deduce, as how to proceed for opening the mod and exposing to signs (both +ve and -ve)
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Q: Mod (x+3) - Mod (4-x)= Mod (8+x). How many solutions does the equation have?
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As a general approach, the terms in MOD are put in brackets and then solved by putting positive and then negative signs.
The process given in the book is perfectly fine but don't expect it to be 100% mechanical. You will need to evaluate the question every time. You CANNOT just make equations taking +ve and -ve value for every mod and get 2*2*2 = 8 equations. Understand the reason for this.
By definition,
|x+8| = x+8 when x+8 >= 0 i.e. when x >= -8
|x+8| = -(x+8) when x+8 < 0 i.e. when x < -8
Similarly,
|x+3| = x+3 when x+3 >= 0 i.e. when x >= -3
|x+3| = -(x+3) when x+3 < 0 i.e. when x < -3
|x-4| = x-4 when x-4 >= 0 i.e. when x >= 4
|x-4| = -(x-4) when x-4 < 0 i.e. when x < 4
Now think, is it possible that |x+8| = -(x+8) but |x+3| = x+3? No because this will happen only when x < -8 AND x >= -3. There is no such value of x.
As mentioned in the post above - you will have 3 transition points: -8, -3 and 4. In between these ranges, the signs of the 3 mods change.
x < -8 ------ Here, |x+8| = -(x+8), |x+3| = -(x+3), |x-4| = -(x-4). So you make an equation using these.
-8 <= x < -3 ------- Here, |x+8| = (x+8), |x+3| = -(x+3), |x-4| = -(x-4). So you make an equation using these.
-3 <= x < 4 ------- |x+8| = (x+8), |x+3| = (x+3), |x-4| = -(x-4). So you make an equation using these.
x >=4 -------- |x+8| = (x+8), |x+3| = (x+3), |x-4| = (x-4). So you make an equation using these.
The transition points decide the number of equations you will get. If you have 3 transition points, you will get 4 equations. If you have 4 transition points, you will get 5 equations and so on...