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Doubt in Inequality

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by rijul007 » Sat Nov 19, 2011 10:42 am
1. If x^2 > 1, then what are the limits of that inequality?

x^2 > 1
x^2 - 1 > 0
(x+1)(x-1) > 0

Take cases

Case 1: x > 1
(x+1)(x-1) will be positive

Case 2 -1<x<1
(x+1)(x-1) will be ngative

Case 3 x<-1
(x+1)(x-1) will be positive

Hence
x^2 > 1 for x<-1 and x>1

2. If (5X - 2)(3X + 1) > 0, then what are the limits of that inequality?
similarly take cases in these too..

you'll get the ans
x<-1/2 and x>2/5
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by kanwar86 » Sat Nov 19, 2011 10:46 am
IMO
1) x<-1 or x>1
2) x<-1/3 or x>2/5
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by zooki » Sat Nov 19, 2011 11:14 am
for Q 2: x=-1/3 does not satisfy the inequality.
neither does x=2/5
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by vishal.pathak » Sat Nov 19, 2011 11:36 am
seema19 wrote:1. If x^2 > 1, then what are the limits of that inequality?
2. If (5X - 2)(3X + 1) > 0, then what are the limits of that inequality?
1.if x^2 > 1 then -infinity < x < -1 and 1 < x < infinity
2. if (5x - 2) (3x + 1) > 0
then
either both (5x-2) and (3x+1) have to be +ve or both have to be -ve for the final product to b +ve (>0)
when both +ve x > 2/5 (5x-2 > 0), x>-1/3(3x+1 > 0)
so we see that x > 2/5 satisfies both x > 2/5 and x>-1/3
when both -ve x < 2/5, x < -1/3
so we see that x < -1/3 satisfies both x<2/5 and x<-1/3
so the range is x < -1/3 , x > 2/5
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by seema19 » Sat Nov 19, 2011 11:42 am
X^2 > 1
X^2 - 1 > 0
(X + 1)(X - 1) > 0
Thus, X > -1 and X > +1.
However, according to the answer,this is wrong. Can someone plz explain why this is wrong.
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by kanwar86 » Sat Nov 19, 2011 11:57 am
seema19 wrote:X^2 > 1
X^2 - 1 > 0
(X + 1)(X - 1) > 0
Thus, X > -1 and X > +1.
However, according to the answer,this is wrong. Can someone plz explain why this is wrong.
For expression (X-1)(X+1) to be greater than zero either X should be >1 or <-1
In your case for 1>X>-1, the term (X+1)>0 but (X-1)<0, which implies the product, X^2-1, is less than zero. Hence, -1<x<1 can't be the solution.
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by vishal.pathak » Sat Nov 19, 2011 12:06 pm
seema19 wrote:X^2 > 1
X^2 - 1 > 0
(X + 1)(X - 1) > 0
Thus, X > -1 and X > +1.
However, according to the answer,this is wrong. Can someone plz explain why this is wrong.
for (X + 1)(X - 1) > 0 either both ((X + 1) & (X - 1)) have to be +ve or both have to be -ve so that the final product is +ve
+ve*+ve = +ve
-ve*-ve = -ve
if both are +ve then x+1>0 and x-1>0
or x > -1 and x>1
So u can see that x>1 satisfies both x > -1 and x>1 (so range from this is x>1, lets call it eqn (1))
Now if both ((X + 1) & (X - 1)) are -ve
then x<-1 and x<1
So u can see that x<-1 satisfies both x<-1 and x<1 (so range from this is x<-1, lets call it eqn (2))
From eqn (1) and eqn(2)
x<-1 and x >1
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by pemdas » Sat Nov 19, 2011 12:09 pm
likewise |x|>|1| you solve by squaring both sides you can consider the roots here
for |x|>|1| the only solutions are mode must be greater than the absolute value of 1 regardless of signs, hence x<-1 and x>1
seema19 wrote:X^2 > 1
X^2 - 1 > 0
(X + 1)(X - 1) > 0
Thus, X > -1 and X > +1.
However, according to the answer,this is wrong. Can someone plz explain why this is wrong.
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by kanwar86 » Sat Nov 19, 2011 12:20 pm
Please note that the solution to 1)is X<-1 OR X>1 instead of X<-1 AND X>1 (which is not possible as both sets are mutually exclusive)
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by GMATGuruNY » Sun Nov 20, 2011 5:32 am
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