Stuart ,
Thanks man that was really an excellent reply .
@crackthetest , buddy you are again going the same path that I have walked , the answers dont have a typo
.
But here is the catch and thanks to Stuart I now understand why I have arrived at the solution 5/3 < x and 3/5 <x which eventually means 5/3<x .
Here is the catch :
The basic rule when you remove the mod operator is that you have to consider 4 cases :
positive/positive , positive/negative , negative/positive , and negative / negative.
But if the problem you have is an equation
ex : |x| = |y|
Then the cases will be
(1) x= y
(2) x= -y
(3) -x= y
(4)- x= -y
Now if you compare case 2 and 3 they complement each other and if we multiply case 2 by -1 we get case 3 .
Similarly if we compare case 1 and 4 they too complement each other and if we multiply case 4 by -1 we get case 1 .
NOTE : since we have an "=" sign we dont have to bother much about flipping the sign and basically case we need only to solve just two cases
(1) Same sign x=y
(2) Different sign x= -y
and this will give us the final answer .
But is the problem is and inequality with variables watch out . Reason being
Ex :
(1) x> y
(2) x> -y
(3) -x>y
(4)- x> -y
case 1 and 4 are not complementary now , if we multiply case 4 by - 1 we get x < y and not x > y that is case 1.
(flip the sign when multiplying or dividing by -ve number)
Similarly case 2 and 3 are not complementary .
So when faced with inequality always solve for 4 cases .
NOTE : this rule is to applied when you remove mod operator from both sides of inequality . If you have mod operator on just one side of inequality then solve for just two cases .
So coming back to the problem we were discussing we need to consider 4 cases :
1) Positive / positive case:
x+1/x-1 < 4
x+1<4x-4
0<3x-5
5/3 < x
2) Positive/ negative case case:
x+1/x-1 < - 4
x+1 <-4x+4
5x<3
x<3/5
3) Negative/Positive case:
- (x+1/x-1) < 4
-(x+1)< 4x-4
3<5x
3/5 < x
4) Negative/ Negative case:
-(x+1)/x-1 < -4
-x-1< -4x+4
3x < 5
x < 5/3
So we have now 4 values of x , they are
x < 5/3 and x < 3/5 ; x> 3/5 and x>5/3
Now in x<5/3 and x<3/5 , x<3/5 is more limiting . Also in x> 3/5 and x>5/3 , x>5/3 is more limiting .
So we have two limiting values for x as x < 3/5 and x >5/3 .
As you can see this is a long process and chances of making a mistake are many so better method will be to square both sides as suggested by sanju09
(x +1)^2/(x - 1)^2 < 16
x^2 + 2x + 1 < 16(x^2 - 2x + 1) =16x^2 -32x +16
15x^2 - 34x + 15 > 0
(15x - 9)(15x - 25)/15 > 0
(5x - 3)(3x - 5) > 0
But the point to note here , as Stuart explains
if the product of the two terms is positive, then there are two options - either both brackets are positive or both brackets are negative.
Case 1 (both positive):
5x -3 > 0 AND 3x - 5 > 0
5x > 3 AND 3x > 5
x > 3/5 AND x > 5/3
When we have two inequalities facing in the same direction, and they must both be true, we only need to pay attention to the more limiting (i.e. extreme) inequality.
So, from Case 1 we know that x > 5/3.
Case 2 (both negative):
x -3 < 0 AND 3x - 5 < 0
5x < 3 AND 3x < 5
x < 3/5 AND x < 5/3
Again, we have two inequalities facing in the same direction, and both must be true; accordingly, we only pay attention to the more limiting inequality.
So, from Case 2, we know that x < 3/5.
Since the two cases are ALTERNATIVES, we conclude that:
x < 3/5 OR x > 5/3 ... choose (B).
Because this type of problem has a lot of places to make mistakes, it's a great idea to double check your solution by plugging back into the original inequality. For example, if papgust (sorry to pick on you!) had plugged in x = 1.1 (a number inside his range), he would have gotten:
|2.1/.1| < 4
21 < 4
which is clearly not true, showing that there was an error.
Of course, we could have also attacked this question by working directly with the choices and eliminating those that violated the original condition.
Well then that sumps up every thing , thanks a lot to you all for your contribution specially Stuart for your input , I hope this helps others too.
