Correct Answer - Option 4 : (2, 11/3)
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, \(\rm \frac{5}{x-2}>3\)
Adding -3 to both sides, we get
⇒\(\rm \frac{5}{x-2}-3>3-3\)
⇒\(\rm \frac{5-3x+6}{x-2}>0\)
⇒\(\rm \frac{11-3x}{x-2}>0\)
Case 1: When 11 - 3x > 0 and x - 2 > 0
⇒3x < 11 and x > 2
⇒x < 11/3 and x > 2
⇒ 2 < x < 11/3
⇒x ∈ (2, 11/3)
Case 2: When 11 - 3x < 0 and x - 2 < 0
⇒3x > 11 and x < 2
⇒x > 11/3 and x < 2
This is not possible, as we can never find a real number which is greater than 11/3 and less than 2.
∴ Solution set = (2, 11/3)
Hence, option (4) is correct.
