CUET UG 2025 — Mathematics Algebra

A function increases where its derivative is positive, i.e., $f'(x) > 0$f′(x)>0.

The function is $f(x) = \frac{x}{x^2 + 1}$f(x)=x2+1x​

Using the quotient rule where $u = x$u=x and $v = x^2 + 1$v=x2+1:

$$f'(x) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2}$$f′(x)=(x2+1)2(1)(x2+1)−(x)(2x)​

$$f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2}$$f′(x)=(x2+1)2x2+1−2x2​

$$f'(x) = \frac{1 - x^2}{(x^2 + 1)^2}$$f′(x)=(x2+1)21−x2​

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For the function to increase:

$$\frac{1 - x^2}{(x^2 + 1)^2} > 0$$(x2+1)21−x2​>0

The denominator $(x^2 + 1)^2$(x2+1)2 is always positive since $x^2 + 1 \geq 1$x2+1≥1 for all real $x$x.

Therefore, the sign of $f'(x)$f′(x) depends only on the numerator:

$$1 - x^2 > 0$$1−x2>0

$$1 > x^2$$1>x2

$$x^2 < 1$$x2<1

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The inequality $x^2 < 1$x2<1 holds when:

$$-1 < x < 1$$−1<x<1

Therefore, the function increases in the interval $(-1, 1)$(−1,1).