CUET UG 2025 — Mathematics Calculus

To find where $f(x) = \frac{x}{4} + \frac{4}{x}$f(x)=4x​+x4​ is strictly increasing, we need $f'(x) > 0$f′(x)>0.

Rewriting the function:

$$f(x) = \frac{x}{4} + 4x^{-1}$$f(x)=4x​+4x−1

Differentiating term by term:

$$f'(x) = \frac{1}{4} - 4x^{-2}$$f′(x)=41​−4x−2

$$f'(x) = \frac{1}{4} - \frac{4}{x^2}$$f′(x)=41​−x24​

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For the function to be strictly increasing:

$$\frac{1}{4} - \frac{4}{x^2} > 0$$41​−x24​>0

$$\frac{1}{4} > \frac{4}{x^2}$$41​>x24​

Multiplying both sides by $x^2$x2 (which is always positive):

$$\frac{x^2}{4} > 4$$4x2​>4

$$x^2 > 16$$x2>16

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The inequality $x^2 > 16$x2>16 gives:

$$|x| > 4$$∣x∣>4

This means either $x > 4$x>4 or $x < -4$x<−4.

The function is strictly increasing on $(-\infty, -4) \cup (4, \infty)$(−∞,−4)∪(4,∞).

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Comparing with the given form $(-\infty, a) \cup (b, \infty)$(−∞,a)∪(b,∞):

$$a = -4$$a=−4

$$b = 4$$b=4

Therefore, $a = -b$a=−b.