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Re: Is [m]x>3[/m] ? [#permalink]
29 Aug 2010, 17:42
This post received KUDOS
Is x>3 ? 1. (x-3)(x-2)(x-1) > 0 2. x > 1
Use the following approach with your all such inequality questions.
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Arrange the roots of the equal in the increasing order and create separators in the form of curves as shown. Start from the right most i.e. x>0 to be +ve and put alternate -ve , +ve signs as you move along the left side of the number line.
If the inequality says p(x) > 0 then the domain of the inequality is in +ve curve. If the inequality says p(x) < 0 then the domain of the inequality is in -ve curve.
For the given question in the statement 1 - p(x) > 0 => consider +ve curve i.e. x>3 and 2>x>1 This is not sufficient as 2>x>1 is also there and we can not the question whether x>3 or not.
Consider the 2nd statement. x>1 does not answer the question x>3 as x>1 could be 2 or 4. 2 will give the answer "No" to the given question whereas 4 will give "yes". Thus not sufficient.
Take both the statements together. we have 2>x>1, x>3 and x>1
When we combine all the given three inequalities we still can not answer as x=1.5 and x = 4 will give different answer to the question.
S1 gives three roots of equation and is true in two conditions: either all are positive or two of them are negative. If x is positive then equations are x > 1 or x > 2 or x > 3 => x > 3 for two negative => 1 <x<2 insufficient s2 says x> 1 insuff
combining together s2 does not give extra information and hence e
I did it the pleonasm way!However could somebody explain me the concept of roots cos i didn't get the sign changing around values thing..
The polynomial has roots (x-2)(x-3)(x-1) . They are distinct, which means that the polynomial changes its sign around the roots. If is greater than 3, then it is positive. If is between 2 and 3 then it is negative, between 1 and 2, positive, and below 1, negative. x is therefore limited to (1,2)U(3,infinity)
Hey , here is my explanation. it might help you to understand. here we go,
the question is x>3 ? statement 1:(x-1)(x-2)(x-3) >0
statement 2: x>1
starting with statement 1 : FOR (x-1)(x-2)(x-3) >0 there are 3 condition for which this eq will be +ve case 1: (x-1)>0 , (x-2) >0, (x-3) >0 ; on plotting the point on number line, we will get x>3 ( chk with no 4) case 2: (x-1)<0, (x-2) <0, (x-3) >0 ;; on plotting the point on number line, we will get 2<x<1 ( chk with no 1.5) case 3: (x-1)<0, (x-2) > 0, (x-3) <0; ; on plotting the point on number line, we will get 2<x<1 ( chk with no 1.5) case 4: (x-1)>0, (x-2) < 0, (x-3) <0; ; on plotting the point on number line, we will get 3<x<2 ( chk with no 2.5)
so, from here we cant say that x>3.
statement 2: x>1; doesn't say anything at all. (chk with x=1.5,2.5)
on combining these two statements we cant find anything as we will get the same case as said in CASE 4.
so, either of the statement cant answer this question alone or on combination.
so, final ans is E.
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