A derivable function `f : R^(+) rarr R` satisfies the condition `f(x) - f(y) ge log((x)/(y)) + x - y, AA x, y in R^(+)`. If g denotes the derivative of f, then the value of the sum `sum_(n=1)^(100) g((1)/(n))` is

Options

A. (a)5050
B. (b)5510
C. (c)5150
D. (d)1550

Answer

To solve the problem step by step, we start with the condition given for the function \( f \):

### Step 1: Understand the given inequality
The condition states:
\[ f(x) - f(y) \geq \log\left(\frac{x}{y}\right) + (x - y) \quad \forall x, y \in \mathbb{R}^+ \]

### Step 2: Set \( y = 1 \)
Let’s set \( y = 1 \):
\[ f(x) - f(1) \geq \log(x) + (x - 1) \]

### Step 3: Rearranging the inequality
This can be rearranged to express \( f(x) \):
\[ f(x) \geq f(1) + \log(x) + (x - 1) \]

### Step 4: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). The left-hand side gives us \( f'(x) \) (denote it as \( g(x) \)), and the right-hand side gives:
\[ g(x) \geq \frac{1}{x} + 1 \]
Thus, we have:
\[ g(x) \geq \frac{1 + x}{x} = \frac{x + 1}{x} \]

### Step 5: Evaluate \( g\left(\frac{1}{n}\right) \)
Now, substituting \( x = \frac{1}{n} \):
\[ g\left(\frac{1}{n}\right) \geq \frac{\frac{1}{n} + 1}{\frac{1}{n}} = n + 1 \]

### Step 6: Sum up \( g\left(\frac{1}{n}\right) \) from 1 to 100
We need to calculate:
\[ \sum_{n=1}^{100} g\left(\frac{1}{n}\right) \geq \sum_{n=1}^{100} (n + 1) \]
This simplifies to:
\[ \sum_{n=1}^{100} g\left(\frac{1}{n}\right) \geq \sum_{n=1}^{100} n + \sum_{n=1}^{100} 1 \]

### Step 7: Calculate the sums
The first sum is the sum of the first 100 natural numbers:
\[ \sum_{n=1}^{100} n = \frac{100 \cdot 101}{2} = 5050 \]
The second sum is simply:
\[ \sum_{n=1}^{100} 1 = 100 \]

### Step 8: Combine the results
Thus, we have:
\[ \sum_{n=1}^{100} g\left(\frac{1}{n}\right) \geq 5050 + 100 = 5150 \]

### Step 9: Conclusion
Since \( g\left(\frac{1}{n}\right) \) is greater than or equal to \( n + 1 \), the sum \( \sum_{n=1}^{100} g\left(\frac{1}{n}\right) \) is at least 5150. Thus, the value of the sum is:
\[ \sum_{n=1}^{100} g\left(\frac{1}{n}\right) = 5150 \]

### Final Answer
The value of the sum \( \sum_{n=1}^{100} g\left(\frac{1}{n}\right) \) is \( 5150 \).

---