The range of the function f(x) = cot^-1≤ft(x^2 - 2x + 2) , x real, is

The correct answer is Option (3) → \(\left(0, \displaystyle\frac{\pi}{4}\right]\)

Let \(t = x^{2} - 2x + 2\). By completing the square,

    \(t = (x - 1)^{2} + 1\)

So t takes every value in \([1, \infty)\), with \(t = 1\) at \(x = 1\).

\(\cot^{-1}\) is decreasing, so the largest value of f comes from the smallest t:

    \( \displaystyle \cot^{-1}(1) = \frac{\pi}{4} \) (taken at \(x = 1\))

As t becomes large, \(\cot^{-1} t\) comes closer and closer to 0, but never equals 0, because \(\cot 0\) is not defined.

Hence the range is \(\left(0, \frac{\pi}{4}\right]\).

Why the other options are incorrect

• Option (1) \(\left[\frac{\pi}{4}, \frac{\pi}{2}\right)\) — Incorrect. This is the range of \(\tan^{-1}\left(x^{2} - 2x + 2\right)\); for \(\cot^{-1}\), \(t = 2\) gives a value below \(\frac{\pi}{4}\).

• Option (2) \(\left(0, \frac{\pi}{4}\right)\) — Incorrect. It leaves out \(\frac{\pi}{4}\), which is taken at \(x = 1\).

• Option (4) \(\left[0, \frac{\pi}{4}\right]\) — Incorrect. It includes 0, but \(\cot^{-1} t\) is never 0 for any real t.