Let `f:R in R` be a continuous function such that f(1)=2. If `lim_(x to 1) int_(2)^(f(x)) (2t)/(x-1)dt=4`, then the value of f'(1) is

Options

A. 1
B. 2
C. 4
D. none of these

Answer

To solve the problem step by step, we start with the given limit involving an integral.

### Step 1: Set up the integral
We are given:
\[ \lim_{x \to 1} \int_{2}^{f(x)} \frac{2t}{x-1} dt = 4 \]
We can rewrite the integral by factoring out \(\frac{1}{x-1}\):
\[ \int_{2}^{f(x)} \frac{2t}{x-1} dt = \frac{1}{x-1} \int_{2}^{f(x)} 2t \, dt \]

### Step 2: Evaluate the integral
Now, we need to evaluate the integral \(\int_{2}^{f(x)} 2t \, dt\):
\[ \int 2t \, dt = t^2 + C \]
Thus,
\[ \int_{2}^{f(x)} 2t \, dt = \left[ t^2 \right]_{2}^{f(x)} = f(x)^2 - 2^2 = f(x)^2 - 4 \]

### Step 3: Substitute back into the limit
Substituting this back into our limit gives:
\[ \lim_{x \to 1} \frac{f(x)^2 - 4}{x - 1} = 4 \]

### Step 4: Apply L'Hôpital's Rule
As \(x \to 1\), both the numerator and denominator approach 0, resulting in a \(0/0\) form. We can apply L'Hôpital's Rule:
\[ \lim_{x \to 1} \frac{f(x)^2 - 4}{x - 1} = \lim_{x \to 1} \frac{d}{dx}(f(x)^2 - 4) \bigg/ \frac{d}{dx}(x - 1) \]
Differentiating the numerator:
\[ \frac{d}{dx}(f(x)^2 - 4) = 2f(x)f'(x) \]
And the denominator:
\[ \frac{d}{dx}(x - 1) = 1 \]
Thus, we have:
\[ \lim_{x \to 1} 2f(x)f'(x) = 4 \]

### Step 5: Substitute \(f(1)\)
We know \(f(1) = 2\):
\[ 2 \cdot 2 \cdot f'(1) = 4 \]
This simplifies to:
\[ 4f'(1) = 4 \]

### Step 6: Solve for \(f'(1)\)
Dividing both sides by 4 gives:
\[ f'(1) = 1 \]

### Final Answer
The value of \(f'(1)\) is:
\[ \boxed{1} \]