Is n negative? (1) (1 n^2)

Question

Is n negative?

(1)  (1 – n^2) < 0

(2)  n^2 – n – 2 < 0

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IshanSaini · Answer

According to my understanding.

Question Stem : Is n negative?

(1) (1–n2)<0 
can be written as 
(1+n)(1-n)<0
now, using in-equality properties, we can say that the equation will be 0 at n = -1 and n = +1.
Now note that this equation at n=+2 and n=-2 will be negative, however at n=0 will be positive so we get the sign ranges where A,C are -ve while B is +ve.
 .......-1...........+1.......
---A---|-----B----|---C--
we need <0 neg so n can take values in the range (-inf,-1) U (1,inf).
Hence n can be +ve or negative.
 Strike A,D

(2) n2–n–2<0 
solving this equation, we get (n+1)(n-2) < 0
From here we get n = -1 and n = +2 as the =0 points.
Now take n = 0 we get a negative result, so the range of the above equation is n belongs to (-1,2) or  
Therefore from this we can say that :
B is enough to tell that x is not negative.
 Strike C, E. 
 B is the answer.

Archit3110 · Answer

target is n -ve
#1
 (1 – n^2) < 0

test with n -2, -3/2,-1/2,0,1/2,1, 3/2, 2
we get yes at n = 2 , -2 , -3/2 , 3/2
insufficient
#2
 n^2 – n – 2 < 0

test with n -2, -3/2,-1/2,0,1/2,1, 3/2, 2

n can be 0 , 3/2 
insufficient
from 1 &2
n = +3/2
option C

Archit3110 · Answer

IshanSaini 
check fractions as well when testing values of n

8.toastcrunch · Answer

Statement 1 -  (1–n2)<0 , n can be 2 and -2 .  Not Sufficient

Statement 2 -  n2–n–2<0 , n=1 and -3/2 .  Not Sufficient

Together 1 and 2  - n is positive .  Sufficient

C is the answer

IshanSaini · Answer

Ahh, I see where my reasoning fell short, I forgot to consider values like -0.11, -0.9 that fell between -1 and 0.
Thanks,  Archit3110 , for pointing this out. 
Much appreciated.

fdfd97 · Answer

Can someone post the official explanation please? Thank you.

singhall · Answer

So we need to find if n<0.

A) (1–n^2)<0

(a^2-b^2)= (a+b)(a-b)
(1-n^2) = (1-n)(1+n) < 0
multiply by -1
(n-1)(n+1)>0

zero points, n=1, n=-1
this means n should be greater than zero for values of n less than -1 and greater than +1.

n can be positive or negative. NOT SUFFICIENT

B) n^2–n–2<0
n^2-2n+n-2<0
n(n-2)+1(n-2)<0
(n+1)(n-2)<0

Zero Points=> -1,2

n should lie on the number line at points less than 2 and greater than -1.
n can still be positive or negative. NOT SUFFICIENT.

A + B
if we plot this on the number line, A and B would coincide only for points between 1 and 2.
so n lies between 1 and 2.
n is positive. SUFFICIENT.

Answer is C.

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Please offer suggestions if anyone finds a flaw in the solution or the approach.

bv8562 · Answer

Is n negative? (1) (1-n^2) < 0 (2) n^2-n-2 < 0 Solution : (1) (1-n^2) < 0 (1-n)(1+n)<0 either: 1-n>0 & 1+n<0 n<1 & n<-1 Therefore, n<-1 or: 1-n<0 & 1+n>0 n>1 & n>-1 Therefore, n>1 we get: n<-1 or n>1 , therefore, not sufficient. Eliminate A & D. (2) n^2-n-2 < 0 n^2-2n+n-2 < 0 n(n-2)+1(n-2) < 0 (n-2)(n+1) < 0 either: n-2>0 & n+1<0 n>2 & n<-1 This is not a valid inequality. or: n-2<0 & n+1>0 n<2 & n>-1 Therefore, -11 AND -1

Avneeshyadav01 · Answer

(1) N^2>1 => |n|>1. that means n>1 ot n<-1
Range of n = [-infinite, -1)U(1 to infinite]. N can be =ve or -ve. -----> NOT SUFFICIENT

(2) (n-2)(n+1)<0
range of n = (-1, +2). Hence n can be +ve or -ve.---->NOT SUFFICIENT

(1)+(2): Overlapped ranfe of n = (1,2). thet means n is +ve.------> SUFFICIENT.

OA: C

Is n negative? (1) (1 n^2) < 0 (2) n^2 n 2 < 0

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