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Is |n| < 1? (1) (n^x) – n < 0 (2) x–1 = –2 OA later...

Should be E.

from 1:
n^x – n < 0
n^x < n

if n = 4, and x = 1/2, (n^x) < (n) and |n| is grater than 1
if n = 2, and x = -1, (n^x) < (n) and |n| is grater than 1
if n = 0.25, and x = 2, (n^x) < (n) and |n| is smaller than 1. so nsf.

from 2:
x – 1 = –2
x = – 1. also nsf.

From 1 and 2:

if x = -1 and n = 2, (n^x) < (n) and |n| is grater than 1
if x = -1 and n = -0.5, (n^x) < (n) and |n| is smaller than 1

Try st2 first:
x-1 = -2
x = -1 --> nothing about n, insufficient.

try st1:
If x is even, say 2, then:
n^2-n < 0 if n is a positive integer
n^2-n < 0 if n is a positive fraction

If x is odd, say 3, then:
n^3-n < 0 if n is a negative integer
n^3-n < 0 if n is a positive fraction

We don't know status of n, insufficient.

Using both:
1/n - n <0> 0 --> out
If n = positive integer, say 3, then 1/n-n <0> possible
If n = negative fraction, say -1/3, then 1/n-n <0> possible
If n = negative integer, say -3, then 1/n-n > 0 --> out

Stmt1: n^x -n < 0 n[n^(x-1) - 1] < 0 n <0> 0 or n > 0 and n^(x-1) - 1 < 0

Lets say n <0> 0 n = -4 , x =3 ===> |n| > 1 n=-1/2, x =-1 ====> |n| <1> 0 and n^(x-1) - 1 <0>|n| > 1 n=1/2 , x=5 ====>|n| <1> 0 and n^(-3/2) < 1 or n<0> 1 for n> 0, n * sqrt(n) > 1 |n| > 1

For n < 0 NOT POSSIBLE as sqrt(n) requires n to be +ve.

Hence SUFF

(1)n^x<n
(2)x=-1
(1)+(2): n^(-1)<n 1/n<n
case 1:if 0<n; 1<n^2; 1<n
case 2:if n<0; n^2<1; -1<n<0
Answer is E.

The two cases listed below will call answer C into question.
case 1: n=10, x=-1; comply (1) and (2), 1<|n|
case 2: n=-0.5, x=-1; comply (1) and (2), |n|<1

Stmt1: n^x -n < 0 n[n^(x-1) - 1] < 0 n <0> 0 or n > 0 and n^(x-1) - 1 < 0

Lets say n <0> 0 n = -4 , x =3 ===> |n| > 1 n=-1/2, x =-1 ====> |n| <1> 0 and n^(x-1) - 1 <0>|n| > 1 n=1/2 , x=5 ====>|n| <1> 0 and n^(-3/2) < 1 or n<0> 1 for n> 0, n * sqrt(n) > 1 |n| > 1

For n < 0 NOT POSSIBLE as sqrt(n) requires n to be +ve.

Hence SUFF

(1)n^x<n (2)x=-1 (1)+(2): n^(-1)<n 1/n<n case 1:if 0<n; 1<n^2; 1<n case 2:if n<0; n^2<1; -1<n<0 Answer is E.

The two cases listed below will call answer C into question. case 1: n=10, x=-1; comply (1) and (2), 1<|n| case 2: n=-0.5, x=-1; comply (1) and (2), |n|<1

n can't be -ve as n^-3/2 will be undefined for -ve values of n.