If the interval in which f(x) = x/4 + 4/x, x ≠ 0 is strictly increasing is (-∞, a) ∪ (b, ∞), then Option 1: a = b Option 2: a = -b Option 3: a = 2b Option 4: a = -2b

Answer: a = -b. Solution: To find where f(x) = x/4 + 4/x is strictly increasing, we need f'(x) > 0. Rewriting the function: f(x) = x/4 + 4x^(-1) Differentiating term by term: f'(x) = 1/4 - 4x^(-2) f'(x) = 1/4 - 4/x^2 For the function to be strictly increasing: 1/4 - 4/x^2 > 0 1/4 > 4/x^2 Multiplying both sides by x^2 (which is always positive): x^2/4 > 4 x^2 > 16 The inequality x^2 > 16 gives:, x, > 4 This means either x > 4 or x Comparing with the given form (-∞, a) (b, ∞): a = -4 b = 4 Therefore, a = -b.

Calculus

Application of Derivatives

Medium

common

If the interval in which f(x) = $\frac{x}{4}$ + $\frac{4}{x}$ , x ≠ 0 is strictly increasing is (-∞, a) ∪ (b, ∞), then

a = b

a = -b

a = 2b

a = -2b

✅ Correct Option: 2

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

Rewriting the function:

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

Differentiating term by term:

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

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

For the function to be strictly increasing:

$\frac{1}{4} - \frac{4}{x^2} > 0$

$\frac{1}{4} > \frac{4}{x^2}$

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

$\frac{x^2}{4} > 4$

$x^2 > 16$

The inequality $x^2 > 16$ gives:

$|x| > 4$

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

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

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

$a = -4$

$b = 4$

Therefore, $a = -b$ .

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Choose the correct answer from the options given below:

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If the interval in which f(x) = x4\frac{x}{4}4x​ + 4x\frac{4}{x}x4​, x ≠ 0 is strictly increasing is (-∞, a) ∪ (b, ∞), then

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If the interval in which f(x) = $\frac{x}{4}$ + $\frac{4}{x}$ , x ≠ 0 is strictly increasing is (-∞, a) ∪ (b, ∞), then