The function \(f(x) = \displaystyle\frac{e^{x}}{1 + x^{2}}\), \(x \in R\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → strictly increasing on R
By the quotient rule:
\( \displaystyle f'(x) = \frac{e^{x}(1 + x^{2}) - e^{x}(2x)}{(1 + x^{2})^{2}} \)
\( \displaystyle = \frac{e^{x}(x - 1)^{2}}{(1 + x^{2})^{2}} \)
Here \(e^{x} \gt 0\), \((x - 1)^{2} \ge 0\) and the denominator is positive. So \(f'(x) \gt 0\) for every x ≠ 1, and f′(1) = 0 only at that single point, with no change of sign.
So f is strictly increasing on R.
Why the other options are incorrect
Option (1) strictly decreasing on R — Incorrect. f′(x) is never negative; for example f(0) = 1 and f(2) = \(\frac{e^{2}}{5} \approx 1.48\).
Option (3) strictly increasing in \((-\infty, 1)\) and strictly decreasing in \((1, \infty)\) — Incorrect. x = 1 is a critical point, but \((x - 1)^{2}\) stays positive on both sides, so f keeps increasing after x = 1.
Option (4) strictly decreasing in \((-1, 1)\) and strictly increasing in \((-\infty, -1)\) and in \((1, \infty)\) — Incorrect. f′(0) = 1 > 0, so f is increasing near 0, inside (−1, 1).