Step 1: Find the derivative of \( f(x) \)
The given function is \( f(x) = \frac{\log x}{x} \). Differentiate using the quotient rule: \[ f'(x) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}. \]
Step 2: Find critical points
For \( f'(x) = 0 \): \[ 1 - \log x = 0 \implies \log x = 1 \implies x = e. \]
Step 3: Determine intervals of increase and decrease
For \( x \in (0, e) \): \[ 1 - \log x>0 \implies f'(x)>0 \quad {(strictly increasing)}. \] For \( x \in (e, \infty) \): \[ 1 - \log x<0 \implies f'(x)<0 \quad {(strictly decreasing)}. \]
Step 4: Conclude the result
The function is strictly increasing on \( (0, e) \) and strictly decreasing on \( (e, \infty) \).
Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]When observed over a long period of time, a time series data can predict trends that can forecast increase, decrease, or stagnation of a variable under consideration. The table below shows the sale of an item in a district during 1996–2001:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Year} & 1996 & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \textbf{Sales (in lakh ₹)} & 6.5 & 5.3 & 4.3 & 6.1 & 5.6 & 7.8 \\ \hline \end{array} \]