Let \(f : [a, \infty) \to [2, \infty)\) be defined by \(f(x) = x^{2} - 6x + 11\). The value of a for which f is both one-one and onto is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3
Completing the square,
\(f(x) = x^{2} - 6x + 9 + 2\)
\(f(x) = (x - 3)^{2} + 2\)
So the least value of f is 2, taken only at \(x = 3\), and f increases for \(x \ge 3\).
One-one needs \(a \ge 3\). If \(a \lt 3\), the domain contains two points at the same distance from 3, for example \(3 - h\) and \(3 + h\) with a small \(h \gt 0\), and
\(f(3 - h) = h^{2} + 2 = f(3 + h)\)
Onto needs the value 2 to be taken, and 2 is taken only at \(x = 3\). So 3 must be in the domain, which needs \(a \le 3\).
Both together give \(a = 3\). Then for any \(y \ge 2\), \(x = 3 + \sqrt{y - 2}\) lies in \([3, \infty)\) and \(f(x) = y\).
Why the other options are incorrect
Option (1) 2 — Incorrect. f is not one-one on \([2, \infty)\), since \(f(2) = 3 = f(4)\).
Option (2) 6 — Incorrect. On \([6, \infty)\) the least value is \(f(6) = 11\), so values such as 2 are never taken and f is not onto.
Option (4) 11 — Incorrect. On \([11, \infty)\) the least value is \(f(11) = 66\), so f is not onto.