If the interval in which the function \(f(x)=\displaystyle\frac{x}{x^2+1}\) is strictly increasing is \((-a, a)\), then a is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
A function is strictly increasing where \(f'(x) \gt 0\). By the quotient rule,
\(\displaystyle f'(x) = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2}\)
\(\displaystyle f'(x) = \frac{1 - x^2}{(x^2 + 1)^2}\)
The denominator is always positive, so \(f'(x) \gt 0\) exactly when
\(1 - x^2 \gt 0\), that is, \(-1 \lt x \lt 1\)
So f is strictly increasing on (−1, 1), and a = 1.