CUET UG 2025 — Mathematics Calculus

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

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Given: $f(x) = \frac{x}{2} + \frac{2}{x}$f(x)=2x​+x2​

Differentiating term by term:

Derivative of $\frac{x}{2}$2x​ is $\frac{1}{2}$21​

Derivative of $\frac{2}{x} = 2x^{-1}$x2​=2x−1 is $2 \times (-1) \times x^{-2} = -\frac{2}{x^2}$2×(−1)×x−2=−x22​

Therefore:

$$f'(x) = \frac{1}{2} - \frac{2}{x^2}$$f′(x)=21​−x22​

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For the function to be increasing:

$$\frac{1}{2} - \frac{2}{x^2} > 0$$21​−x22​>0

$$\frac{1}{2} > \frac{2}{x^2}$$21​>x22​

$$\frac{x^2}{2} > 2$$2x2​>2

$$x^2 > 4$$x2>4

$$|x| > 2$$∣x∣>2

This means: $x > 2$x>2 or $x < -2$x<−2

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Checking the options:

(A) $(-\infty, -2)$(−∞,−2) represents $x < -2$x<−2 — the function is increasing here

(B) $(-2, 2)$(−2,2) represents $-2 < x < 2$−2<x<2 — the function is not increasing here

(C) $(2, \infty)$(2,∞) represents $x > 2$x>2 — the function is increasing here

(D) $(-1, 1)$(−1,1) represents $-1 < x < 1$−1<x<1 — the function is not increasing here

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The function is increasing on $(-\infty, -2)$(−∞,−2) and $(2, \infty)$(2,∞).

The correct answer is (A) and (C).