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when x< 0 : -x=3x-2 --> x = 1/2 , this is not a valid solution since we assumed x<0 therefore the soulution is invalid. when x>0 then x=3x-2 ---> x = 1 , this is a valid solution since we assumed that x > 0 , therefore Conditon B gives us a unique answer i.e. x= 1. Hence the answer is B.
i dont think that we need a condition " x must be an integer to get the answer B because statement 1 is surely not sufficient. statement 2 could be case first: if x> 0 so x is 1 case secondL if x< so so x is 1/2 . this case should be eliminated becos x must be negative so if x=1/2 this case will be deleted.... thus x=1
You are missing an important point - let me explain.
We all know that if we have mods, we take two cases - positive and negative - and then solve the equation. The point is - why do we do that? If you remember, this is how we define mods
|x| = x if x is positive and -x if x is negative
So basically, |x| takes different forms depending on whether x is positive or negative. When I want to solve |x|=3x-2, I can't solve with |x|.
So I split it into two cases: Case 1: x is positive I get x = 3x - 2 x = 1 I accept this value of x since x has to be +ve and satisfy given equation. It does both. Case 2: x is negative -x = 3x - 2 x = 1/2 I reject this value since x should be negative for the equation to look like this. So x can only take 1 value i.e. x = 1.
Remember, when you split |x| into two cases, you have to check that the value you get lies in the region in which you are expecting it to lie. _________________
Now in option B people are confused whether 1/2 can be a value of the x or not But people ....try putting x=1/2 in the option B equation .... l x l = 3x-2 therefore l 1/2 l = 3*1/2 -2 here l 1/2 l will be +ve ...i.e. 1/2 therefore 1/2 = 3/2 - 2 = (3-4)/2 hence 1/2 = - 1/2 ...which is not possible
similarly we even take x as -1/2 then also the resulting values come as 1/2 = - 7/2 ...again unequal ....
in case if x is negative then we cannot say -x=3x-2 we should say -x= -3x -2 2x=-2 so x= -1
In case x is negative, 3x does not change to -3x. When we write 'x', the negative sign is already included. But when x is negative, |x| becomes -x. The reason is that |x| can never be negative. If x is negative, -x makes it positive. e.g. If x = -4 |x| = |-4| = 4 which is actually -x = -(-4) = 4 Hence |x| = -x when x is negative. When x is positive, |x| = x _________________
"I reject this value since x should be negative for the equation to look like this. So x can only take 1 value i.e. x = 1."
im sorry but could you please explaine why you reject 1/2... its also positive as 1
Let me highlight the main points: |x| = x if x is positive and [highlight]|x| = -x if x is negative[/highlight] (As explained in the reply above)
Case 1: x is positive |x|=3x-2 becomes x=3x-2 When I solve this, I get x = 1. This is acceptable since here I am working on the case where x is positive.
[highlight]Case 2: x is negative[/highlight] [highlight]|x|[/highlight]=3x-2 becomes [highlight]-x[/highlight]= 3x - 2 Now I solve and get a value of x. This value will be acceptable only if it is negative since I am working on the case where x is negative. But when I solve, I get x = 1/2. It is not negative so I reject it.
To test, try and put x = 1 in |x|=3x-2 It satisfies.
Put x = 1/2 in |x|=3x-2 It doesn't satisfy. _________________
Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...