Tough question on absolute value & inequalities

[vinni.k]

Do people have any idea in solving the following question ?

S = x^3 + 3^x
Is S > 0?

(1) x < 0

(2) | x | > 1

Answer C

[spoiler]I only face problem when i combine the two statements and don't know which way to go from there. Please explain with the help of examples.[/spoiler]

Thanks & Regards
Vinni

[pemdas]

Do people have any idea in solving the following question ?

S = x^3 + 3^x
Is S > 0?

(1) x < 0

(2) | x | > 1

Answer C

[spoiler]I only face problem when i combine the two statements and don't know which way to go from there. Please explain with the help of examples.[/spoiler]

Thanks & Regards
Vinni

all what the question is asking: whether x<0 and how much x is less than 0. Obviously the only way S<0 is when x<0 but how far to the left from zero?

st(1) doesn't tell the distance on the number line, so Insuff
st(2) suggests the absolute value of x, Insuff too

combined we know that x<-1 hence Suff

[GMATGuruNY]

Do people have any idea in solving the following question ?

S = x^3 + 3^x
Is S > 0?

(1) x < 0
(2) | x | > 1

Answer C

[spoiler]I only face problem when i combine the two statements and don't know which way to go from there. Please explain with the help of examples.[/spoiler]

Thanks & Regards
Vinni

Statement 1: x<0
If x=-2, then S = (-2)^3 + 3^(-2) = -8 + (1/9), a sum less than 0.
If x=-1/2, then S = (-1/2)^3 + 3^(-1/2) = -1/8 + 1/√3 ≈ -1/8 + 1/1.7 ≈ -2/16 + 10/16 = 1/2, which is greater than 0.
INSUFFICIENT.

Statement 2: |x| > 1
If x=-2, then S = (-2)^3 + 3^(-2) = -8 + (1/9), a sum less than 0.
If x=2, then S = 2³ + 3² = 17, which is greater than 0.
INSUFFICIENT.

Combined:
Only values less than -1 satisfy both x<0 and |x|>1.

x³:
Cubing a value less than -1 yields a value LESS than the original value:
(-2)³ = -8.
(-3)³ = -27.
And so on.

3^x:
Raising 3 to a power less than -1 yields a POSITIVE FRACTION between 0 and 1:
3^(-2) = 1/9.
3^(-3) = 1/27.
And so on.

Thus, S = (value less than -1) + (positive fraction between 0 and 1), a sum less than 0.
SUFFICIENT.

The correct answer is C.

[vinni.k]

Statement 1: x<0
If x=-2, then S = (-2)^3 + 3^(-2) = -8 + (1/9), a sum less than 0.
If x=-1/2, then S = (-1/2)^3 + 3^(-1/2) = -1/8 + 1/√3 ≈ -1/8 + 1/1.7 ≈ -2/16 + 10/16 = 1/2, which is greater than 0.
INSUFFICIENT.

Statement 2: |x| > 1
If x=-2, then S = (-2)^3 + 3^(-2) = -8 + (1/9), a sum less than 0.
If x=2, then S = 2³ + 3² = 17, which is greater than 0.
INSUFFICIENT.

Combined:
Only values less than -1 satisfy both x<0 and |x|>1.

x³:
Cubing a value less than -1 yields a value LESS than the original value:
(-2)³ = -8.
(-3)³ = -27.
And so on.

3^x:
Raising 3 to a power less than -1 yields a POSITIVE FRACTION between 0 and 1:
3^(-2) = 1/9.
3^(-3) = 1/27.
And so on.

Thus, S = (value less than -1) + (positive fraction between 0 and 1), a sum less than 0.
SUFFICIENT.

The correct answer is C.

Mitch, thanks for your reply.
Very nicely explained. It now seems that the question isn't very difficult.
However, one doubt in statement (2).
(2) | x | > 1 :- when I open the absolute value of x, it gives
x > 1 or x < -1
After combining the two statements :- x < 0 and x > 1 or x < -1
As you have explained in your earlier post

Combined:
Only values less than -1 satisfy both x<0 and |x|>1.

I am not able to understand this particular part. Are you taking the same inequality signs as common
i.e x < 0 and x < -1 ?

I am not able to get through this part. Absolute value of |x| > 1 gives two inequalities. One less and another greater. Shouldn't we consider all the inequality signs from both statements ?

Regards
Vinni

[FLUID]

Statement 1: x<0
If x=-2, then S = (-2)^3 + 3^(-2) = -8 + (1/9), a sum less than 0.
If x=-1/2, then S = (-1/2)^3 + 3^(-1/2) = -1/8 + 1/√3 ≈ -1/8 + 1/1.7 ≈ -2/16 + 10/16 = 1/2, which is greater than 0.
INSUFFICIENT.

Statement 2: |x| > 1
If x=-2, then S = (-2)^3 + 3^(-2) = -8 + (1/9), a sum less than 0.
If x=2, then S = 2³ + 3² = 17, which is greater than 0.
INSUFFICIENT.

Combined:
Only values less than -1 satisfy both x<0 and |x|>1.

x³:
Cubing a value less than -1 yields a value LESS than the original value:
(-2)³ = -8.
(-3)³ = -27.
And so on.

3^x:
Raising 3 to a power less than -1 yields a POSITIVE FRACTION between 0 and 1:
3^(-2) = 1/9.
3^(-3) = 1/27.
And so on.

Thus, S = (value less than -1) + (positive fraction between 0 and 1), a sum less than 0.
SUFFICIENT.

The correct answer is C.

Mitch, thanks for your reply.
Very nicely explained. It now seems that the question isn't very difficult.
However, one doubt in statement (2).
(2) | x | > 1 :- when I open the absolute value of x, it gives
x > 1 or x < -1
After combining the two statements :- x < 0 and x > 1 or x < -1
As you have explained in your earlier post

Combined:
Only values less than -1 satisfy both x<0 and |x|>1.

I am not able to understand this particular part. Are you taking the same inequality signs as common
i.e x < 0 and x < -1 ?

I am not able to get through this part. Absolute value of |x| > 1 gives two inequalities. One less and another greater. Shouldn't we consider all the inequality signs from both statements ?

Regards
Vinni

Hey Vinni,
2) | x | > 1 :- when I open the absolute value of x, it gives
x > 1 or x < -1

Above statement is correct

After combining the two statements :- x < 0 and x > 1 or x < -1

=> You need to ignore x > 1 Since statement 1 is saying X < 0

[vinni.k]

Hey Vinni,
2) | x | > 1 :- when I open the absolute value of x, it gives
x > 1 or x < -1

Above statement is correct

After combining the two statements :- x < 0 and x > 1 or x < -1

=> You need to ignore x > 1 Since statement 1 is saying X < 0

Thanks for your reply, but please explain why there is need to ignore x > 1.
When i consider all the inequalities, i get E, and it is incorrect. I know that it is incorrect. All i want to understand the ignorance of x > 1.
In DS, we have to satisfy all the statements. I am bit confused on this part.

Regards
Vinni

[GMATGuruNY]

Hey Vinni,
2) | x | > 1 :- when I open the absolute value of x, it gives
x > 1 or x < -1

Above statement is correct

After combining the two statements :- x < 0 and x > 1 or x < -1

=> You need to ignore x > 1 Since statement 1 is saying X < 0

Thanks for your reply, but please explain why there is need to ignore x > 1.
When i consider all the inequalities, i get E, and it is incorrect. I know that it is incorrect. All i want to understand the ignorance of x > 1.
In DS, we have to satisfy all the statements. I am bit confused on this part.

Regards
Vinni

x>1 is not a viable range because statement 1 requires that x be NEGATIVE.

When COMBINING the two statements, we may consider only values that satisfy BOTH STATEMENTS.
Statement 1 requires that x be NEGATIVE.
Statement 2 requires that |x|>1.
The only NEGATIVE values of x such that |x|>1 are x<-1.