For x gt 0 , the minimum value of displaystylex/(log_e x) is

The correct answer is Option (3) → \(e\)

Let \(f(x) = \frac{x}{\log_e x}\), defined for \(x \gt 0\), \(x \ne 1\). By the quotient rule:

    \( \displaystyle f'(x) = \frac{\log_e x - x \cdot \frac{1}{x}}{(\log_e x)^2} = \frac{\log_e x - 1}{(\log_e x)^2} \)

    \(f'(x) = 0 \Rightarrow \log_e x = 1 \Rightarrow x = e\)

The denominator is positive, so the sign of \(f'\) is the sign of \(\log_e x - 1\):

    \(1 \lt x \lt e\): \(f'(x) \lt 0\), f decreasing
    \(x \gt e\): \(f'(x) \gt 0\), f increasing

So f has a local minimum at x = e, and

    \( \displaystyle f(e) = \frac{e}{\log_e e} = e \)

This is the minimum value meant here: the least value of f for \(x \gt 1\). (For \(0 \lt x \lt 1\), \(\log_e x \lt 0\), so f is negative there and has no least value, as it falls without limit when x approaches 1 from the left.)