CUET UG 2025 — Mathematics Calculus

To determine whether $f(x) = x - \frac{1}{x}$f(x)=x−x1​ is increasing or decreasing, find the derivative and check its sign.

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$$f(x) = x - \frac{1}{x}$$f(x)=x−x1​

Rewrite as:

$$f(x) = x - x^{-1}$$f(x)=x−x−1

Differentiate:

$$f'(x) = 1 - (-1)x^{-2}$$f′(x)=1−(−1)x−2

$$f'(x) = 1 + \frac{1}{x^2}$$f′(x)=1+x21​

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For $f'(x) = 1 + \frac{1}{x^2}$f′(x)=1+x21​:

Since $x^2$x2 is always positive for $x \neq 0$x=0, then $\frac{1}{x^2}$x21​ is always positive.

Therefore:

$$f'(x) = 1 + \frac{1}{x^2} > 1 > 0$$f′(x)=1+x21​>1>0

Since $f'(x) > 0$f′(x)>0 for all $x \neq 0$x=0, the function is increasing.

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The function $f(x) = x - \frac{1}{x}$f(x)=x−x1​ is not defined at $x = 0$x=0 (division by zero).

The domain is $x \in (-\infty, 0) \cup (0, \infty)$x∈(−∞,0)∪(0,∞).

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Therefore, the function is increasing for all $x \in (-\infty, 0) \cup (0, \infty)$x∈(−∞,0)∪(0,∞).