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:
\(\sum_{n=1}^{20} f' \left( \frac{1}{x} \right) = 2870 + 20 = 2890\)
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)