x,n

[daretodream]

If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?

1) x < 1

2) n is even.

[ajith]

If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?

1) x < 1

2) n is even.

1) is not sufficient say x = -0.5 and n = 3, x^n is not greater than x^(n+1), x = -0.5 and n =4 x^n is greater than x^(n+1)
2) n is even is not sufficient either -

Combined sufficient to prove that x^n > x^(n+1)

[thephoenix]

If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?

1) x < 1
x can be between 0 and 1 or x can be negative
If x is b/w 0 and 1, then x^n > x^(n+1)
if x is negative, x^n > x^(n+1) - here it depends on n
not suff


2) n is even
not suff because no information on x

Together

If x is b/w 0 and 1 and n is even then x^n > x^(n+1)
ex 0.1 -->x^2=0.01
0.1 -->x^2+1=x^3=0.001
0.01>0.001

if x is negative and n is even then x^n > x^(n+1)
ex x=-2-->x^2=4
-2 -->x^2+1=x^3=-8

hence C

[shashank.ism]

If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?

1) x < 1

2) n is even.

for x^n > x^(n+1) , --> 0<x<1
St.1) x<1 not suff.
St.2) if n is even then also condition on x is not fullfilled

combined : if n is even then its possible as for -ve value of x^n will be +ve and x^(n+1 ) will be -ve ...

also 0<x<1 is applicable for all positive integer x


Hence , Ans C