girish3535 wrote:...

A,B,C are bounded only on 1 side.
D has 2 line segments
2 <= 3x +4 <= 6
or -2 <= 3x <= 2
or -2/3 <= x <= 2/3. This is a finite line segment

vishal.pathak wrote:

girish3535 wrote:...

A,B,C are bounded only on 1 side.
D has 2 line segments
2 <= 3x +4 <= 6
or -2 <= 3x <= 2
or -2/3 <= x <= 2/3. This is a finite line segment

Yup.. That should be correct. The number line for |x| will be two different line segments, one for negative values of x and one for positive values of x.

The question here is whether the solution will have two options

Answer choice A - the even exponent hides the sign, thus x could be positive or negative, won't be a single continuous line.

B) the odd exponent means that x will keep it's sign and since x^3 <27 we know that x <3 - this will be a continuous line from positive three to negative infinity.

C) same problem as A, x coudl be greater than or equal to 4 OR less than or equal to -4

D) Same problem - absolute value hides the sign and thus the x can be positive OR negative.

E. This answer does not contain a hidden sign, as there is no even exponent or absolute value, however this line will not be infinite because it is bounded on both sides x will be between -2/3 and 2/3.