Previous Year Paper
CAT-2006-Previous Years Paper
Let f(x) = max (2x + 1, 3 – 4x), where x is any real number. Then the minimum possible value of f(x) is
To find out min of f(x) = max (2x + 1, 3 – 4x), we should be taking the point of intersection of (2x + 1) and (3 – 4x). [Since one of these equations is increasing and other one is decreasing]
2x + 1 = 3 – 4x, or, x = 1/3
2x + 1 = 3 – 4x = 5/3
Hence option (e) is the answer.