The function y= x/x³ is strictly increasing function for

Correct Answer is : 0<x<e130< x< e¹/3}0<x<e31

Correct Answer is : 0<x<e130< x< e¹/3}0<x<e31

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COMEDK 2025 Evening Solved Paper

Question : 162 of 180

Marks: +1, -0

The function

y=\frac{\log x}{x^3}

is strictly increasing function for

Solution:

Concept:

To determine where a function is strictly increasing, find its derivative and set it greater than zero.

Explanation:

Given

y = \frac{\log x}{x^3}

with

x > 0

Compute derivative using quotient rule: let

u = \log x

v = x^3

Then

u' = \frac{1}{x}

v' = 3x^2

y' = \frac{u'v - uv'}{v^2} = \frac{\frac{1}{x} \cdot x^3 - \log x \cdot 3x^2}{x^6} = \frac{x^2 - 3x^2 \log x}{x^6} = \frac{1 - 3\log x}{x^4}

For

x > 0

x^4 > 0

, so the sign of

y'

matches the numerator

1 - 3\log x

Set

y' > 0

1 - 3\log x > 0 \implies \log x < \frac{1}{3} \implies x < e^{1/3}

Since domain requires

x>0

, function is strictly increasing for

0 < x < e^{1/3}

Answer:

Option A:

0 < x < e^{\frac{1}{3}}

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