CAT 1995 — QA Question 35
Direction: Answer the questions based on the following information.
Four sisters — Suvarna, Tara, Uma and Vibha are playing a game such that the loser doubles the money of each of the other players from her share. They played four games and each sister lost one game in alphabetical order. At the end of fourth game, each sister had Rs.32.
Largest value of min(2 + x2, 6 – 3x), when x > 0, is
Answer & solution
- A
1
- B
2
3
- D
4
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.