The domain of the function \(f(x) = \cos^{-1}\left(\displaystyle\frac{x^{2} + 1}{2x}\right)\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\{-1, 1\}\)
We need \(x \ne 0\) and the argument in \([-1, 1]\), that is,
\( \displaystyle \left|\frac{x^{2} + 1}{2x}\right| \le 1 \)
Multiplying by the positive number \(2|x|\):
\(x^{2} + 1 \le 2|x|\)
Since \(x^{2} = |x|^{2}\), this is
\(|x|^{2} - 2|x| + 1 \le 0\)
\(\left(|x| - 1\right)^{2} \le 0\)
A square is never negative, so it must be 0:
\(|x| = 1\), so \(x = 1\) or \(x = -1\)
Check: at \(x = 1\) the argument is 1, and at \(x = -1\) it is \(-1\); both are allowed. Hence the domain is \(\{-1, 1\}\).
Why the other options are incorrect
Option (1) \([-1, 1]\) — Incorrect. At \(x = \frac{1}{2}\) the argument is \(\frac{5}{4}\), which is greater than 1 (and \(x = 0\) is not allowed at all).
Option (2) \(\{1\}\) — Incorrect. It misses \(x = -1\), where the argument is \(\frac{2}{-2} = -1\) and \(\cos^{-1}(-1) = \pi\) is defined.
Option (4) \(\mathrm{R} - \{0\}\) — Incorrect. At \(x = 2\) the argument is \(\frac{5}{4}\), outside \([-1, 1]\).