Inequalities

[rishab1988]

Is n negative?

1) [1-(n^2)] <0
2) n^2-n-2<0

Should the answer be C or E?

[abhishekg21]

is square bracket in [1-(n^2)] representing abs value ?

[Rahul@gurome]

Is n negative?

1) [1-(n^2)] <0
2) n^2-n-2<0

Statement 1: (1 - n²) < 0
Implies n² > 1 => Either n < -1 or n > 1

Not sufficient.

Statement 2: (n² - n - 2) < 0
Implies (n - 2)(n + 1) < 0 => -1 < n < 2

Not sufficient.

1 & 2 Together: Combining both of the inequalities we get only acceptable region of values is 1 < n < 2. Thus n is not negative.

Sufficient.

The correct answer is C.

Note: Combining both the inequalities means n must satisfy both of them. Thus the possible region of values for n is the intersection of the both regions.

[rishab1988]

Rahul.Seems I made a mistake in statement 2.

I interpreted it as (x-2)(x+1)<0

Therefore, either x-2 <0 -> x<2 or x+1<0 -> x<-1

So,should it not be x<2 or x<-1

Please shed some more light on this issue.I treated statement 2 in the same way as I did 1

[Rahul@gurome]

Rahul.Seems I made a mistake in statement 2.

I interpreted it as (x-2)(x+1)<0
Therefore, either x-2 <0 -> x<2 or x+1<0 -> x<-1
So,should it not be x<2 or x<-1

Please shed some more light on this issue.I treated statement 2 in the same way as I did 1

Careful!
When you are saying x < -1, then x is also less than 2. Thus the product becomes positive.

In this kind of cases, draw a number line, mark the critical points (here -1 and 2) on it and try to see which region satisfies the inequality by picking some easy integers (like 0) and plugging it into the inequality.

[rishab1988]

Rahul.Seems I made a mistake in statement 2.

I interpreted it as (x-2)(x+1)<0
Therefore, either x-2 <0 -> x<2 or x+1<0 -> x<-1
So,should it not be x<2 or x<-1

Please shed some more light on this issue.I treated statement 2 in the same way as I did 1

Careful!
When you are saying x < -1, then x is also less than 2. Thus the product becomes positive.

In this kind of cases, draw a number line, mark the critical points (here -1 and 2) on it and try to see which region satisfies the inequality by picking some easy integers (like 0) and plugging it into the inequality.

brilliant!

You solved the problem.I will keep these points in mind.