If the interval in which the function $f(x)=\frac{x}{x^2+1}$ is strictly increasing is, $(-a, a)$, then a is equal to

1

The correct answer is Option (1) → 1

Given: $f(x) = \frac{x}{x^2 + 1}$

Differentiate using quotient rule:

$f'(x) = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}$

Function is strictly increasing where $f'(x) > 0$:

$\frac{1 - x^2}{(x^2 + 1)^2} > 0 \Rightarrow 1 - x^2 > 0$

$\Rightarrow x^2 < 1 \Rightarrow -1 < x < 1$

Thus, function is strictly increasing in the interval $(-1, 1)$.

Hence, $a = 1$.