The function f(x) = x/3 + 3/x is increasing in the interval: Option 1: (-∞, -3) (3, ∞) Option 2: (-3, 0) (0, 3) Option 3: (-∞, ∞) Option 4: (0, ∞)

Answer: (-∞, -3) (3, ∞). Solution: A function is increasing in an interval when its derivative is positive. Given: f(x) = x/3 + 3/x Finding the derivative: f'(x) = 1/3 - 3/x^2 For the function to be increasing, f'(x) > 0: 1/3 - 3/x^2 > 0 1/3 > 3/x^2 x^2/3 > 3 x^2 > 9 Taking the square root:, x, > 3 This gives x > 3 or x Therefore, the function is increasing in (-∞, -3) (3, ∞).

Calculus

Application of Derivatives

Medium

applied

The function $f(x) = \frac{x}{3} + \frac{3}{x}$ is increasing in the interval:

$(-\infty, -3) \cup (3, \infty)$

$(-3, 0) \cup (0, 3)$

$(-\infty, \infty)$

$(0, \infty)$

✅ Correct Option: 1

A function is increasing in an interval when its derivative is positive.

Given: $f(x) = \frac{x}{3} + \frac{3}{x}$

Finding the derivative:

$f'(x) = \frac{1}{3} - \frac{3}{x^2}$

For the function to be increasing, $f'(x) > 0$ :

$\frac{1}{3} - \frac{3}{x^2} > 0$

$\frac{1}{3} > \frac{3}{x^2}$

$\frac{x^2}{3} > 3$

$x^2 > 9$

Taking the square root:

$|x| > 3$

This gives $x > 3$ or $x < -3$

In interval notation: $(-\infty, -3) \cup (3, \infty)$

Therefore, the function is increasing in $(-\infty, -3) \cup (3, \infty)$ .

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Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The function $f(x) = \frac{x}{3} + \frac{3}{x}$ is increasing in the interval: