Given: \(f(x) - f(y) \geq \ln x - \ln y + x - y\)
Rewriting:
\(f(x) - f(y) = \frac{\sqrt{x} - \sqrt{y}}{x - y} \cdot \frac{\sqrt{x} + \sqrt{y}}{x - y}\)
- Case 1: Let \( x \to y \)
\(\lim_{y \to x^-} f' \left( \frac{1}{x} \right) \geq \frac{1}{x} + 1 \quad \text{...(1)}\)
- Case 2: Let \( x \neq y \)
\(\lim_{y \to x^+} f' \left( x^+ \right) \leq \frac{1}{x} + 1 \quad \text{...(2)}\)
Thus, \(f'(x) = \frac{1}{x+1}\).
Now, substitute \( f' \left( \frac{1}{x} \right) = n + 1 \) into the sum:
\(\sum_{n=1}^{20} f' \left( \frac{1}{x} \right) = \sum_{n=1}^{20} \left( n^2 + 1 \right)\)
Calculating:
\(\sum_{n=1}^{20} n^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870, \quad \sum_{n=1}^{20} 1 = 20\)
Therefore:
\(\sum_{n=1}^{20} f' \left( \frac{1}{x} \right) = 2870 + 20 = 2890\)
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)