CUET UG 2025 — Mathematics Calculus

To find the value of $a$a such that the function $f(x) = \frac{x}{x^2+1}$f(x)=x2+1x​ is strictly increasing in the interval $(-a, a)$(−a,a), we need to find where $f'(x) > 0$f′(x)>0.

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

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

---

For the function to be strictly increasing:

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

The denominator $(x^2+1)^2$(x2+1)2 is always positive.

Therefore, we need:

$$1 - x^2 > 0$$1−x2>0

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

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

---

The inequality $x^2 < 1$x2<1 gives:

$$-1 < x < 1$$−1<x<1

The interval is $(-1, 1)$(−1,1).

---

Comparing with the given interval $(-a, a)$(−a,a):

$$a = 1$$a=1