Question

The function \( y=\frac{\log x}{x^3} \) is strictly increasing function for

• A. \( 0<x<e^{\frac{1}{3}} \)
• B. \( x>e^{\frac{1}{3}} \)
• C. \( x<2 \)
• D. \( x<e^{\frac{1}{3}} \)

Hint:

For functions of type \( \frac{\log x}{x^n} \), simplify derivative firstAlways separate numerator sign from denominator for inequality solving.

Answer 1

The Correct Option is A

Step 1: Write the function.
\[ y=\frac{\log x}{x^3} \]

Step 2: Differentiate using quotient rule.
\[ \frac{dy}{dx} = \frac{x^3\cdot \frac{1}{x} - \log x \cdot 3x^2}{x^6} \]
\[ = \frac{x^2 - 3x^2\log x}{x^6} \]

Step 3: Simplify derivative.
\[ \frac{dy}{dx} = \frac{x^2(1-3\log x)}{x^6} = \frac{1-3\log x}{x^4} \]

Step 4: Determine increasing condition.
Function is increasing when:
\[ \frac{dy}{dx} > 0 \]
Since \( x^4>0 \) for \( x>0 \), we get:
\[ 1-3\log x > 0 \]

Step 5: Solve inequality.
\[ 1 > 3\log x \]
\[ \log x < \frac{1}{3} \]

Step 6: Convert to exponential form.
\[ x < e^{\frac{1}{3}} \]
Also domain: \( x>0 \).

Step 7: Final conclusion.
\[ \boxed{0<x<e^{\frac{1}{3}}} \]