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CUET · MATHS · PYQ PAPER 2025

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. A 1
  2. B 2
  3. C 3
  4. D 4
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Answer & Solution

Correct Answer

(A) 1

Step-by-step Solution

Detailed explanation

\(f'(x) = \frac{(1)(x^2+1) - (x)(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}\) \(f'(x) > 0 \implies \frac{1-x^2}{(x^2+1)^2} > 0\) \(1-x^2 > 0\) \(x^2 \(-1 Interval is \((-1, 1)\). Comparing with \((-a, a)\), \(a=1\).
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