CAT 1995 — QA Question 35

If x = 1, we have min(3, 3) = 3.
If x = 2, we have min(6, 0) = 0.
If x = 3, we have min(11, –3) = –3.
If x = 0.5, we have min(2.25, 4.5) = 2.25.
If x = 0.3, we have min(2.09, 5.1) = 2.09.
Thus, we find that as x increases above 1 and when it decreases below 1, the value of the function decreases.
It is maximum at x = 1 and the corresponding value = 3.

Hint: Please note that the highest value of the given fraction will be at a point where (2 + x²) = (6 – 3x), as
even if one of the values increases beyond this, the other value will be the minimum value.
If we equate the two, we get x² + 3x – 4 = 0. Solving this, we get x = 1 or x = –4.
Since x > 0, it has to be 1 and hence the result.