CUET UG 2025 — Mathematics Algebra

A function is increasing when its derivative $f'(x) > 0$f′(x)>0.

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Given $f(x) = \frac{x - 2}{x + 1}$f(x)=x+1x−2​

Using the quotient rule with $u = x - 2$u=x−2 and $v = x + 1$v=x+1:

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

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

$$f'(x) = \frac{3}{(x+1)^2}$$f′(x)=(x+1)23​

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For $f'(x) > 0$f′(x)>0:

$$\frac{3}{(x+1)^2} > 0$$(x+1)23​>0

The numerator $3$3 is always positive.

The denominator $(x+1)^2$(x+1)2 is always positive for all $x \neq -1$x=−1.

Therefore $f'(x) > 0$f′(x)>0 for all $x \neq -1$x=−1.

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The function is increasing for $x \in \mathbb{R} - \{-1\}$x∈R−{−1}.

This represents all real numbers except $-1$−1.