If `c=(1)/(2) and f(x)=2x-x^(2)`, then interval of x in which LMVT is applicable, is

Options

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

Explanation

To determine the interval of \( x \) in which the Mean Value Theorem (MVT) is applicable for the function \( f(x) = 2x - x^2 \) and \( c = \frac{1}{2} \), we will follow these steps:

### Step 1: Verify the conditions of MVT
The Mean Value Theorem states that if a function \( f \) is continuous on the 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:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

### Step 2: Find the derivative of \( f(x) \)
First, we need to find the derivative \( f'(x) \):
\[ f(x) = 2x - x^2 \]
Differentiating \( f(x) \):
\[ f'(x) = 2 - 2x \]

### Step 3: Calculate \( f'(c) \) at \( c = \frac{1}{2} \)
Now, we will evaluate \( f' \left( \frac{1}{2} \right) \):
\[ f' \left( \frac{1}{2} \right) = 2 - 2 \left( \frac{1}{2} \right) = 2 - 1 = 1 \]

### Step 4: Check the given intervals
We need to check the provided intervals to see if MVT holds true.

#### Option 1: Interval \( [1, 2] \)
- \( a = 1 \), \( b = 2 \)
- \( f(1) = 2(1) - (1)^2 = 2 - 1 = 1 \)
- \( f(2) = 2(2) - (2)^2 = 4 - 4 = 0 \)
- Calculate \( \frac{f(b) - f(a)}{b - a} \):
\[ \frac{f(2) - f(1)}{2 - 1} = \frac{0 - 1}{1} = -1 \]
Since \( f' \left( \frac{1}{2} \right) = 1 \) does not equal \(-1\), MVT does not apply.

#### Option 2: Interval \( [-1, 1] \)
- \( a = -1 \), \( b = 1 \)
- \( f(-1) = 2(-1) - (-1)^2 = -2 - 1 = -3 \)
- \( f(1) = 1 \)
- Calculate \( \frac{f(b) - f(a)}{b - a} \):
\[ \frac{f(1) - f(-1)}{1 - (-1)} = \frac{1 - (-3)}{2} = \frac{4}{2} = 2 \]
Since \( f' \left( \frac{1}{2} \right) = 1 \) does not equal \( 2 \), MVT does not apply.

#### Option 3: Interval \( [0, 1] \)
- \( a = 0 \), \( b = 1 \)
- \( f(0) = 2(0) - (0)^2 = 0 \)
- \( f(1) = 1 \)
- Calculate \( \frac{f(b) - f(a)}{b - a} \):
\[ \frac{f(1) - f(0)}{1 - 0} = \frac{1 - 0}{1} = 1 \]
Since \( f' \left( \frac{1}{2} \right) = 1 \) equals \( 1 \), MVT applies.

#### Option 4: Interval \( [2, 1] \)
This interval is invalid as \( a \) cannot be greater than \( b \).

### Conclusion
The interval in which the Mean Value Theorem is applicable is:
\[ \text{Interval: } [0, 1] \]