Question

The function \(f(x) = \dfrac{x}{\log x}\) is increasing in the interval
(consider \(\log_e x = \log x\))

• A. \((1, \infty)\)
• B. \((e, \infty)\)
• C. \((-\infty, e)\)
• D. \((-\infty, -1)\)

Hint:

Find \(f'(x) = \dfrac{\log x - 1}{(\log x)^2}\) and see when it is positive.

Answer 1

The Correct Option is B

Step 1: Find the domain:
\(\log x\) exists only for \(x > 0\), and the function needs \(\log x \neq 0\), so \(x \neq 1\). Hence the domain is \((0,1) \cup (1,\infty)\).

Step 2: Differentiate:
Use the quotient rule:
\[ f'(x) = \frac{\log x \cdot 1 - x \cdot \frac{1}{x}}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} \]

Step 3: Find where f' is positive:
The denominator \((\log x)^2\) is positive on the domain. So \(f'(x) > 0\) when \(\log x - 1 > 0\), that is \(\log x > 1\), that is \(x > e\).

Step 4: Check the options:
Option 1, \((1, \infty)\), includes \((1, e)\) where \(\log x < 1\) and \(f'\) is negative, so it is wrong. Options 3 and 4 include negative numbers, outside the domain. Only option 2 is correct.

Final Answer:
The function increases on \((e, \infty)\), option 2. \[ \boxed{(e, \infty)} \]