1. Find the derivative: \[ f(x) = \frac{\log x}{x}, \quad f'(x) = \frac{1 \cdot x - \log x \cdot 1}{x^2} = \frac{x - \log x}{x^2}. \]
2. Critical points: The critical points are obtained by solving \( f'(x) = 0 \): \[ x - \log x = 0 \quad \Rightarrow \quad x = \log x. \] Let \( x = e^k \). The equality holds for \( x = e \).
3. Sign of \( f'(x) \): - For \( x \in (0, e) \), \( x - \log x > 0 \), so \( f'(x) > 0 \): \( f(x) \) is increasing. - For \( x \in (e, \infty) \), \( x - \log x<0 \), so \( f'(x)<0 \): \( f(x) \) is decreasing.
Intervals: \[ f(x) { is strictly increasing on } (0, e) { and strictly decreasing on } (e, \infty). \]
Show that \( R \) is an equivalence relation. Also, write the equivalence class \([2]\).
List-I | List-II |
(A) Absolute maximum value | (I) 3 |
(B) Absolute minimum value | (II) 0 |
(C) Point of maxima | (III) -5 |
(D) Point of minima | (IV) 4 |