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Source: — Data Sufficiency |
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by clock60 » Sun Oct 24, 2010 12:42 pm
looks like A
(1) x^2-4x+3<0, (x-1)(x-3)<0 given is valid for x 1<x<3, and the only integer inside is 2. so x=2 suff
(2) x^2+4x+3>0, (x+1)(x+3)>0 valid for x<-3, x>-1 many integers are possible insuff
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by nasir » Sun Oct 24, 2010 7:25 pm
clock60 wrote:looks like A
(1) x^2-4x+3<0, (x-1)(x-3)<0 given is valid for x 1<x<3, and the only integer inside is 2. so x=2 suff
/quote]


x-1 <0 => x<1
x-3<0 => x<3

how you made the relation 1<x<3
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by Rahul@gurome » Sun Oct 24, 2010 9:28 pm
Consider first (1) alone.
x^2 - 4x + 3 < 0 implies (x-3)(x-1) < 0.
This means case (1) (x-3) > 0 and (x-1) < 0 which is x > 3 and x < 1.
This is not possible since no real number can be greater than 3 and less than 1. This case is hence eliminated.
Or
case (2) (x-3) < 0 and (x-1) > 0 which is x<3 and x > 1.
This is same as 1<x<3.
We accept case (2).
If x is an integer, the only possible value is x = 2.
Hence (1) alone is sufficient to answer the question.
Next consider (2) alone.
x^2 + 4x + 3 > 0.
This means (x+1)(x+3) > 0.
So we have case (1) x+1 > 0 and x+3>0 which is x> -1 and x> -3.
So combining we have x > -1.
Or
case (2) x+1 < 0 and x+3 < 0 which is x<-1 and x<-3.
On combining we have that x< -3 .
Hence from (2) we get that either x< -3 or x> -1.
There can be infinite many values of x if x is an integer.
So (2) alone is not sufficient to answer the question.

The correct answer is (A).
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by goyalsau » Mon Oct 25, 2010 4:44 am
nasir wrote:
clock60 wrote:looks like A
(1) x^2-4x+3<0, (x-1)(x-3)<0 given is valid for x 1<x<3, and the only integer inside is 2. so x=2 suff
/quote]


x-1 <0 => x<1
x-3<0 => x<3

how you made the relation 1<x<3
Guys please explain , We both Don't know this rule.........
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by kvcpk » Mon Oct 25, 2010 6:32 am
nasir wrote:If x is an integer, what is the value of x?

(1) x2 - 4x + 3 < 0
(2) x2 + 4x +3 > 0

Please explain

(1) x2 - 4x + 3 < 0

(x-1)(x-3)<0

Rule to remember:
Whenever (x-a)(x-b) <0 then x lies between a and b.
Whenever (x-a)(x-b)>0 then x does not lie between a and b.

From above, we can see that x is between 1 and 3.
Integer lying between 1 and 3 is only 2.
Hence x should be equal to 2.
SUFF

(2) x2 + 4x +3 > 0
(x+1)(x+3)>0

This means that x does not lie between -1 and -3.
Hence x can be any value >-1 or <-3
Hence INSUFF

pick A.

Hope this helps!!

Let me know if you have any troubles understanding this.
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by goyalsau » Mon Oct 25, 2010 7:40 am
kvcpk wrote:
(1) x2 - 4x + 3 < 0

(x-1)(x-3)<0

Rule to remember:
Whenever (x-a)(x-b) <0 then x lies between a and b.

if equation is like this. (x+a)(x+b) <0 Then x will lie in between a and b a < x < b

and if equation is like this. (x-a)(x+b) <0 Then x will lie in between -a < x < b
kvcpk wrote: Whenever (x-a)(x-b)>0 then x does not lie between a and b.
if equation is like this. (x+a)(x+b) >0 Then x will not lie in between a and b a> x, x< b

and if equation is like this. (x-a)(x+b) >0 Then x will lie in between -a > x, x < b

Please correct me if i am wrong.
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by shovan85 » Mon Oct 25, 2010 8:14 am
goyalsau wrote:
if equation is like this. (x+a)(x+b) <0 Then x will lie in between a and b a < x < b

Please correct me if i am wrong.
Grrr.... Man why do you need a Concrete formula always. Just give yourself sometime and analyse it (IMHO) ;)

(x+a)(x+b) <0

Let me take a = 2 and b = 3.

Thus, (x+2) (x+3) < 0
So x+2 < 0 and x+3 > 0 OR x+2 > 0 and x+3 < 0
So x < -2 and x > -3 OR x > -2 and x < -3

Consider x<-2 and x>-3. Which decimal will satisfy this? all in between -3 and -2 say x = -2.5

What happened here?

-3 < x < -2 but what is your a and b? a and b are 2 and 3 respectively.

In general what happened... -b < x < -a (Not what you have thought of "a < x < b")

Thus we can say the solution will be -b < x < -a OR a < -x < b.

One question to you why did I select only x < -2 and x > -3, NOT x > -2 and x < -3 ?

Please Analyse the other cases what ever you have mentioned in your earlier post ;)

PS: And also I hope I m correct. I have never encountered this so if it is wrong Sourabh please bear with me ;)
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by kvcpk » Mon Oct 25, 2010 8:35 am
goyalsau wrote:
if equation is like this. (x+a)(x+b) <0 Then x will lie in between a and b a < x < b

and if equation is like this. (x-a)(x+b) <0 Then x will lie in between -a < x < b
Hi Saurabh,

I think you got me wrong. You missed on the sign before the values a and b.
If (x-a)(x-b)<0
then a<x<b
[Notice that signs before a and b are negative signs]

Ok, So what if (x+a)(x+b)<0??
We can rewrite it is (x- (-a))(x- (-b))<0
So x lies between -a and -b
-a<x<-b

If (x+a)(x-b)<0 then
-a<x<b
[Notice the positive sign before a in x+a]

If (x-a)(x+b)<0 then
-b<x<a
[Notice the positive sign before b in x+b]


Helpful?? If not let me know.

When >0, then the above funda remains same. Only change is that x does not lie in that range now.
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by jeetu_vishnoi » Mon Oct 25, 2010 2:11 pm
Answer is A.

first equation can be written as (x-3)(x-1)<0 which means that (x-3)<0 and (x-1)>0
it shows that x<3 and x>1 and it is given that x is integer hence x=2.

first equation can be written as (x+3)(x+1)<0 which means that either {(x+3)<0 and (x+1)<0} or {(x+3)>0 and (x+1)>0}
which clearly shows that it's not possible to decide tha value of x.
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