Well, from x^n = x^(n+2) we get that x^(n+2) - x^n = x^n(x^2 - 1) = 0. This is equivalent to x^n(x - 1)(x + 1) = 0. This happens only when x^n = 0 (so x = 0), x = -1 or x = 1.
1. This equation is equivalent to x^2 - x -2 = 0. And this one basically means that (x - 2)(x + 1) = 0. This is only true when x is either 2 or -1. Since only x = -1 fits the inital restriction, then x is indeed -1, which is smaller than 0. So 1 is sufficient.

2. 2x < x^5 is equivalent to x^5 - 2x > 0. Now let's just use the numbers we've narrowed down in our intial restriction.
a. x = 0, then x^5 - 2x = 0, which is not consistent with x^5 - 2x > 0. So 0 is out.
b. x = 1. x^5 - 2x = -1, which again is not greater than 0. So -1 is out as well.
c. x = -1 and x^5 - 2x = 1 > 0 . So -1 is the only one that fits.
2 is sufficient as well.

So the answer is D

Given x^n = x^n+2.

i.e x^n = x^n . x^2

i.e x^n(x^2-1) = 0

i.e x^n = 0 or x^2-1 = 0

So x =0 or x^2 = 1

Thus from the given question itself we know that x =0 or 1 or -1.

Statement 1: x= x^2-2 . Since we have the value of x as 0 or 1 or -1, only -1 satisfies this equation. Hence x=-1. Thus x<0

Hence statement 1 alone is sufficient.

Statement 2: 2x<x^5. Since from the question we have x = 0 or 1 or -1, this condition is satisfied only when x = -1. Thus x<0.

Hence statement 2 alone is sufficient.

So C must be the answer

Mr sureshbala,
in statement 1 ---- 2 also satisfies the equation .......
how it could be right....
hence it is insufficient right...coz once x<0 and once x>0

detailed explanation will be appreciated .....

If you solve x^n = x^(n+2), u get two solutions. Either x^n=0 (means x=0) or x^2=1 (means x could be 1 or -1). We have 3 possible solutions here -1,0 and 1.

1) x = (x^2) - 2
if u solve this equation you get x=-1 or x=2
but there is only one value that satisfies both the equations x = (x^2) - 2 and x^n = x^(n+2) and that value is -1. So x=-1 meaning x<0.

A alone is sufficient.

2) 2x<x^5
The above equation holds true for -1 and all the numbers >= 2.
But again there is only one value that satisfies both the equations 2x<x^5 and x^n = x^(n+2) and that value is -1.

So B alone is also sufficient.

So Answer is D.

naaga: Solving the initial equation of x^n = x^(n+2) yields 3 answers: 0, 1 and -1. It is true that in stmt 1 you have two solutions, 2 and -1, but since only -1 is also in the list of solutions for the inital euqtion, then this means that only -1 fits.

DanaJ wrote:Well, from x^n = x^(n+2) we get that x^(n+2) - x^n = x^n(x^2 - 1) = 0. This is equivalent to x^n(x - 1)(x + 1) = 0. This happens only when x^n = 0 (so x = 0), x = -1 or x = 1.
1. This equation is equivalent to x^2 - x -2 = 0. And this one basically means that (x - 2)(x + 1) = 0. This is only true when x is either 2 or -1. Since only x = -1 fits the inital restriction, then x is indeed -1, which is smaller than 0. So 1 is sufficient.

2. 2x < x^5 is equivalent to x^5 - 2x > 0. Now let's just use the numbers we've narrowed down in our intial restriction.
a. x = 0, then x^5 - 2x = 0, which is not consistent with x^5 - 2x > 0. So 0 is out.
b. x = 1. x^5 - 2x = -1, which again is not greater than 0. So -1 is out as well.
c. x = -1 and x^5 - 2x = 1 > 0 . So -1 is the only one that fits.
2 is sufficient as well.

So the answer is D

Dana - Would statement 2 fail if x=0 were one of the solutions for 2x < x^5?

beater wrote:

DanaJ wrote:Well, from x^n = x^(n+2) we get that x^(n+2) - x^n = x^n(x^2 - 1) = 0. This is equivalent to x^n(x - 1)(x + 1) = 0. This happens only when x^n = 0 (so x = 0), x = -1 or x = 1.
1. This equation is equivalent to x^2 - x -2 = 0. And this one basically means that (x - 2)(x + 1) = 0. This is only true when x is either 2 or -1. Since only x = -1 fits the inital restriction, then x is indeed -1, which is smaller than 0. So 1 is sufficient.

2. 2x < x^5 is equivalent to x^5 - 2x > 0. Now let's just use the numbers we've narrowed down in our intial restriction.
a. x = 0, then x^5 - 2x = 0, which is not consistent with x^5 - 2x > 0. So 0 is out.
b. x = 1. x^5 - 2x = -1, which again is not greater than 0. So -1 is out as well.
c. x = -1 and x^5 - 2x = 1 > 0 . So -1 is the only one that fits.
2 is sufficient as well.

So the answer is D

Dana - Would statement 2 fail if x=0 were one of the solutions for 2x < x^5?

x^n= x^(n+2) --> (1)
st 1:

x=x^2-2

say n=0 from (1) x^2=x^0 =1

x= 1-2=-1
sufficient

st 2:

from equ (1) x^5 = x^3 = x

2x < x^5 --> 2x<x --> x<0

sufficient.


D

Yes, if 2 were an equation that had 0 as a root as well as -1, then we would not be able to make a choice.