Let f(x) = Max. `{x^2, (1 - x)^2, 2x(1 - x)}` where `x in [0, 1]` If Rolle's theorem is applicable for f(x) on largestpossible interval [a, b] then the value of `2(a+b+c)` when `c in [a, b]` such that f'(c) = 0, is

Options

A. (a)`(2)/(3)`
B. (b)`(1)/(3)`
C. (c)`(1)/(2)`
D. (d)`(3)/(2)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \max\{x^2, (1-x)^2, 2x(1-x)\} \) over the interval \( [0, 1] \) and determine where Rolle's theorem can be applied.

### Step 1: Identify the points of intersection
We first need to find the points where the three functions intersect within the interval \( [0, 1] \).

1. Set \( x^2 = (1-x)^2 \):
\[ x^2 = 1 - 2x + x^2 \implies 2x = 1 \implies x = \frac{1}{2} \]

2. Set \( x^2 = 2x(1-x) \):
\[ x^2 = 2x - 2x^2 \implies 3x^2 - 2x = 0 \implies x(3x - 2) = 0 \implies x = 0 \text{ or } x = \frac{2}{3} \]

3. Set \( (1-x)^2 = 2x(1-x) \):
\[ 1 - 2x + x^2 = 2x - 2x^2 \implies 3x^2 - 4x + 1 = 0 \]
The roots of this quadratic can be found using the quadratic formula:
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]
This gives us:
\[ x = 1 \quad \text{and} \quad x = \frac{1}{3} \]

### Step 2: Determine the intervals
The critical points we found are \( x = 0, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 1 \). We will evaluate \( f(x) \) in the intervals \( [0, \frac{1}{3}] \), \( [\frac{1}{3}, \frac{1}{2}] \), \( [\frac{1}{2}, \frac{2}{3}] \), and \( [\frac{2}{3}, 1] \).

- For \( x \in [0, \frac{1}{3}] \): \( f(x) = (1-x)^2 \)
- For \( x \in [\frac{1}{3}, \frac{1}{2}] \): \( f(x) = 2x(1-x) \)
- For \( x \in [\frac{1}{2}, \frac{2}{3}] \): \( f(x) = (1-x)^2 \)
- For \( x \in [\frac{2}{3}, 1] \): \( f(x) = x^2 \)

### Step 3: Check differentiability
Rolle's theorem requires that \( f \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \). We need to check the points of transition:

- At \( x = \frac{1}{3} \) and \( x = \frac{1}{2} \), we need to ensure that the left-hand derivative equals the right-hand derivative.

### Step 4: Calculate the derivatives
1. For \( f(x) = (1-x)^2 \):
\[ f'(x) = -2(1-x) \]

2. For \( f(x) = 2x(1-x) \):
\[ f'(x) = 2 - 4x \]

### Step 5: Find \( c \) where \( f'(c) = 0 \)
Set \( f'(x) = 0 \):
- From \( f'(x) = 2 - 4x \):
\[ 2 - 4c = 0 \implies c = \frac{1}{2} \]

### Step 6: Identify \( a \) and \( b \)
The largest interval where \( f \) is continuous and differentiable is \( [\frac{1}{3}, \frac{2}{3}] \).

### Step 7: Calculate \( 2(a + b + c) \)
Here, \( a = \frac{1}{3}, b = \frac{2}{3}, c = \frac{1}{2} \):
\[ 2(a + b + c) = 2\left(\frac{1}{3} + \frac{2}{3} + \frac{1}{2}\right) = 2\left(1 + \frac{1}{2}\right) = 2 \cdot \frac{3}{2} = 3 \]

Thus, the final answer is:
\[ \boxed{3} \]