Largest value of min(2 + x2, 6 – 3x), when x > 0, is
Explanation:
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.
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.