Two curves `C_(1)equiv[f(y)]^(2//3)+[f(x)]^(1//3)=0 and C_(2)equiv[f(y)]^(2//3)+[f(x)]^(2//3)=12,` satisfying the relation `"(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^(2)-y^(2))`
The area bounded by the curve `C_(2) and |x|+|y|=sqrt(12)` is

Options

A. (a) `12pi-24` sq. units
B. (b) `6-sqrt(12)` sq. units
C. (c) `2sqrt(12)-6` sq. units
D. (d) None of these

Explanation

To find the area bounded by the curves \( C_2 \) and \( |x| + |y| = \sqrt{12} \), we will follow these steps:

### Step 1: Identify the curves
The curves given are:
1. \( C_1: f(y)^{2/3} + f(x)^{1/3} = 0 \)
2. \( C_2: f(y)^{2/3} + f(x)^{2/3} = 12 \)

From the relation provided, we can deduce that:
- \( f(x+y) = (x+y)^3 \)
- \( f(x-y) = (x-y)^3 \)

This leads us to conclude that:
- \( f(x) = x^3 \)
- \( f(y) = y^3 \)

### Step 2: Formulate the equations for the curves
Using the functions derived:
- For \( C_1 \):
\[ y^{2/3} + x^{1/3} = 0 \quad \text{(not relevant for area calculation)} \]
- For \( C_2 \):
\[ y^{2/3} + x^{2/3} = 12 \]

### Step 3: Rewrite \( C_2 \)
Rearranging \( C_2 \):
\[ x^{2/3} + y^{2/3} = 12 \]
This can be rewritten as:
\[ \left( \frac{x}{2\sqrt{3}} \right)^{2/3} + \left( \frac{y}{2\sqrt{3}} \right)^{2/3} = 1 \]
This represents a superellipse.

### Step 4: Identify the area of the circle
The equation \( |x| + |y| = \sqrt{12} \) describes a square with vertices at \( (\sqrt{12}, 0), (0, \sqrt{12}), (-\sqrt{12}, 0), (0, -\sqrt{12}) \).

### Step 5: Calculate the area of the circle
The area of the circle defined by \( C_2 \) can be calculated using the formula for the area of a circle:
\[ \text{Area} = \pi r^2 \]
Where \( r = 2\sqrt{3} \):
\[ \text{Area} = \pi (2\sqrt{3})^2 = \pi \cdot 12 = 12\pi \]

### Step 6: Calculate the area of the square
The area of the square formed by \( |x| + |y| = \sqrt{12} \):
The side length of the square is \( \sqrt{12} \), thus:
\[ \text{Area of square} = (\sqrt{12})^2 = 12 \]

### Step 7: Find the bounded area
The area bounded by the curves \( C_2 \) and \( |x| + |y| = \sqrt{12} \) is given by:
\[ \text{Bounded Area} = \text{Area of Circle} - \text{Area of Square} \]
\[ \text{Bounded Area} = 12\pi - 12 \]

### Final Answer
The area bounded by the curve \( C_2 \) and \( |x| + |y| = \sqrt{12} \) is:
\[ \boxed{12\pi - 12} \]