PS.. Equation of line

[jeevan.Gk]

Pls explain..

[DanaJ]

Eliminate the first three options quickly:
A. x^4 > 1 is valid for any number greater than 1 or smaller than -1, so we have an infinity of solutions
B. x^3 < 27 is valid for any number smaller than 3 (3^3 = 27), including all negative numbers. So this one also has an infinity of solutions
C. x^2>=16 also has an infinity of solutions, since all numbers greater than 4 or smaller than -4 are consistent with this inequality.

My hunch is that you got confused by the last two options.
D. 2 <= |x| <= 5 has two cases, depending on the sign of x:
a. x is negative makes |x| = -x, making the inequality 2 <= -x <= 5. Multiply by -1 and you get that -2 >= x >= -5, or that x is between -5 and -2. This is a finite segment.
b. x is positive, with |x| = x, giving us that x is between 2 and 5. This is also a finite segment.
But if you put together the two cases, you get that the solution is actually two finite segments, so it's not what you're looking for.

I would advise just stopping here and picking E, since it's the only solution left. But we can demonstrate that:
2 <= 3x + 4 <=6
-2 <= 3x <= 2
-2/3 <= x < 2/3 ----- finite segment between -4/3 and 2/3.


Edited it.

[sureshbala]

Eliminate the first three options quickly:
A. x^4 > 1 is valid for any number greater than 1 or smaller than -1, so we have an infinity of solutions
B. x^3 < 27 is valid for any number smaller than 3 (3^3 = 27), including all negative numbers. So this one also has an infinity of solutions
C. x^2>=16 also has an infinity of solutions, since all numbers greater than 4 or smaller than -4 are consistent with this inequality.

My hunch is that you got confused by the last two options.
D. 2 <= |x| <= 5 has two cases, depending on the sign of x:
a. x is negative makes |x| = -x, making the inequality 2 <= -x <= 5. Multiply by -1 and you get that -2 >= x >= -5, or that x is between -5 and -2. This is a finite segment.
b. x is positive, with |x| = x, giving us that x is between 2 and 5. This is also a finite segment.
But if you put together the two cases, you get that the solution is actually two finite segments, so it's not what you're looking for.

I would advise just stopping here and picking E, since it's the only solution left. But we can demonstrate that:
2 <= 3x + 4 <=6
6 <= 3x <= 10
2 <= x < 10/3 ----- finite segment between 2 and 10/3.

Perfectly fine....other than the above bold statements