Correct Answer - Option 3 : (-3, 2)
Concept:
Solution Set: The set of all possible values of x.
When (a > 0 and b < 0) or (a < 0 and b > 0) then \(\rm \frac a b <0\)
When (a > 0 and b > 0) or (a < 0 and b < 0) then \(\rm \frac a b >0\)
Calculation:
Here, \(\rm \frac{x-2}{x+3}<0\)
Case 1: When x - 2 > 0 and x + 3 < 0
⇒ x > 2 and x < -3
This is not possible, as we can never find a real number which is greater than 2 and less than -3.
Case 2: When x - 2 < 0 and x + 3 > 0
⇒ x < 2 and x > -3
⇒ -3 < x < 2
⇒x ∈ (-3, 2)
∴ Solution set = (-3, 2)
Hence, option (3) is correct.
