Let `f:[1/2,1]->R` (the set of all real numbers) be a positive, non-constant, and differentiable function such that `f^(prime)(x)<2f(x))a n df(1/2)=1` . Then the value of `int_(1/2)^1f(x)dx` lies in the interval (a) `(2e-1,2e)` (b) `(3-1,2e-1)` `((e-1)/2,e-1)` (d) `(0,(e-1)/2)`

Options

A. `(2e-1,2e)`
B. `(e-1,2e-1)`
C. `((e-1)/2,e-1)`
D. `(0,(e-1)/2)`

Explanation

To solve the problem, we need to analyze the given conditions and derive the integral of the function \( f(x) \) over the interval \([1/2, 1]\).

### Step 1: Understand the given conditions
We are given that:
- \( f: [1/2, 1] \to \mathbb{R} \) is a positive, non-constant, and differentiable function.
- The condition \( f'(x) < 2f(x) \) holds for all \( x \) in the interval.
- \( f(1/2) = 1 \).

### Step 2: Analyze the inequality \( f'(x) < 2f(x) \)
This inequality suggests that the growth rate of \( f(x) \) is less than twice its current value. We can rewrite this as:
\[ \frac{f'(x)}{f(x)} < 2 \]
Integrating both sides will help us understand the behavior of \( f(x) \).

### Step 3: Integrate the inequality
Integrating \( \frac{f'(x)}{f(x)} < 2 \) gives:
\[ \int \frac{f'(x)}{f(x)} \, dx < \int 2 \, dx \]
This leads to:
\[ \ln |f(x)| < 2x + C \]
Where \( C \) is a constant of integration. Exponentiating both sides yields:
\[ f(x) < e^{2x + C} \]

### Step 4: Determine the constant \( C \)
To find \( C \), we can use the initial condition \( f(1/2) = 1 \):
\[ 1 < e^{2(1/2) + C} \implies 1 < e^{1 + C} \implies 0 < 1 + C \implies C > -1 \]

### Step 5: Express \( f(x) \)
We can express \( f(x) \) in terms of \( e^{2x} \):
\[ f(x) < e^{2x - 1} \]
This gives us an upper bound for \( f(x) \).

### Step 6: Integrate \( f(x) \) over \([1/2, 1]\)
Now we can integrate \( f(x) \):
\[ \int_{1/2}^{1} f(x) \, dx < \int_{1/2}^{1} e^{2x - 1} \, dx \]
Calculating the integral on the right:
\[ \int e^{2x - 1} \, dx = \frac{1}{2} e^{2x - 1} \]
Evaluating this from \( 1/2 \) to \( 1 \):
\[ \left[ \frac{1}{2} e^{2x - 1} \right]_{1/2}^{1} = \frac{1}{2} e^{1} - \frac{1}{2} e^{0} = \frac{1}{2} e - \frac{1}{2} \]

### Step 7: Final bounds for the integral
Thus, we have:
\[ \int_{1/2}^{1} f(x) \, dx < \frac{1}{2} (e - 1) \]
Since \( f(x) \) is positive and non-constant, we also know that:
\[ \int_{1/2}^{1} f(x) \, dx > 0 \]

### Conclusion
Therefore, the value of \( \int_{1/2}^{1} f(x) \, dx \) lies in the interval:
\[ (0, \frac{1}{2}(e - 1)) \]
This corresponds to option (d) \( (0, (e-1)/2) \).