Let f(x) is differentiable function in [2, 5] such that `f(2)=(1)/(5)` and `f(5)=(1)/(2)`, then the exists a number `c, 2 lt c lt 5` for which f '(c) equals :-

Options

A. `(1)/(2)`
B. `(1)/(5)`
C. `(1)/(10)`
D. `(1)/(15)`

Explanation

To solve the problem, we will apply the Mean Value Theorem (MVT). Here are the steps:

### Step 1: Identify the function and the interval
We have a function \( f(x) \) that is differentiable on the interval \([2, 5]\). We know:
- \( f(2) = \frac{1}{5} \)
- \( f(5) = \frac{1}{2} \)

### Step 2: Check the conditions for the Mean Value Theorem
The Mean Value Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
In our case, \( a = 2 \) and \( b = 5 \). Since \( f(x) \) is given to be differentiable on \([2, 5]\), it is also continuous on this interval.

### Step 3: Apply the Mean Value Theorem
Using the values given:
\[ f'(c) = \frac{f(5) - f(2)}{5 - 2} \]
Substituting the known values:
\[ f'(c) = \frac{\frac{1}{2} - \frac{1}{5}}{5 - 2} \]

### Step 4: Simplify the expression
First, we need to find \( f(5) - f(2) \):
\[ f(5) - f(2) = \frac{1}{2} - \frac{1}{5} \]
To subtract these fractions, we need a common denominator:
\[ \frac{1}{2} = \frac{5}{10}, \quad \frac{1}{5} = \frac{2}{10} \]
Thus,
\[ f(5) - f(2) = \frac{5}{10} - \frac{2}{10} = \frac{3}{10} \]
Now substituting back into the equation for \( f'(c) \):
\[ f'(c) = \frac{\frac{3}{10}}{3} = \frac{3}{10} \cdot \frac{1}{3} = \frac{1}{10} \]

### Conclusion
Therefore, there exists a number \( c \) in the interval \( (2, 5) \) such that:
\[ f'(c) = \frac{1}{10} \]