If Rolle's theorems applicable to the function `f(x)=ln x/x, (x> 0)` over the interval `[a, b]` where `a in I, b in I` ,then value of `a^2+ b^2` can be:

Options

A. `20`
B. `25`
C. `45`
D. `10`

Answer

To solve the problem, we need to determine the values of \( a \) and \( b \) such that Rolle's Theorem is applicable to the function \( f(x) = \frac{\ln x}{x} \) over the interval \([a, b]\) where \( a \) and \( b \) are integers. We will also find the value of \( a^2 + b^2 \).

### Step 1: Check the conditions for Rolle's Theorem
Rolle's Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), and if \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

### Step 2: Determine continuity and differentiability
The function \( f(x) = \frac{\ln x}{x} \) is continuous and differentiable for \( x > 0 \). Therefore, it is continuous and differentiable on any interval \([a, b]\) where \( a > 0 \) and \( b > 0 \).

### Step 3: Set up the equation for Rolle's Theorem
Since \( f(a) = f(b) \), we have:
\[ \frac{\ln a}{a} = \frac{\ln b}{b} \]
Cross-multiplying gives:
\[ b \ln a = a \ln b \]

### Step 4: Rearranging the equation
Rearranging the equation, we can write:
\[ \frac{\ln a}{a} = \frac{\ln b}{b} \]
Taking the exponential of both sides gives:
\[ \frac{a}{b} = \frac{b}{a} \]
This implies:
\[ a^2 = b^2 \]
Thus, \( a = b \) or \( a = -b \). Since \( a \) and \( b \) are both positive integers, we have \( a = b \).

### Step 5: Finding integer solutions
To satisfy the condition \( b \ln a = a \ln b \) with \( a \neq b \), we can try specific integer values.

Let’s assume \( a = 2 \) and \( b = 4 \):
\[ 2 \ln 4 = 4 \ln 2 \]
Calculating both sides:
\[ 2 \cdot 1.386 = 4 \cdot 0.693 \quad \text{(approximately)} \]
Both sides equal \( 2.772 \), confirming that \( a = 2 \) and \( b = 4 \) satisfy the equation.

### Step 6: Calculate \( a^2 + b^2 \)
Now, we calculate:
\[ a^2 + b^2 = 2^2 + 4^2 = 4 + 16 = 20 \]

### Final Answer
Thus, the value of \( a^2 + b^2 \) is \( 20 \).