Let f be a differentiable function satisfying the condition `f ((x)/(y)) = (f(x))/(f (y)) (y ne 0, f (y) ne 0) AA x, y in R and f '(1) =2.` If the smaller area enclosed by `y = f(x) , x ^(2)+y^(2) =2` is A, then findal [A], where [.] represents the greatest integer function.

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Determine the function \( f(x) \)

We are given the functional equation:
\[ f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)} \quad (y \neq 0, f(y) \neq 0) \]
Assuming \( f(x) = x^n \) for some \( n \), we can substitute this into the functional equation:

\[ f\left(\frac{x}{y}\right) = \left(\frac{x}{y}\right)^n = \frac{x^n}{y^n} = \frac{f(x)}{f(y)} \]

This satisfies the functional equation for any \( n \).

### Step 2: Find \( n \) using the derivative condition

We are given that \( f'(1) = 2 \). First, we compute the derivative of \( f(x) = x^n \):

\[ f'(x) = n x^{n-1} \]

Substituting \( x = 1 \):

\[ f'(1) = n \cdot 1^{n-1} = n \]

Setting this equal to 2 gives us:

\[ n = 2 \]

Thus, we have determined that \( f(x) = x^2 \).

### Step 3: Find the area enclosed by \( y = f(x) \) and \( x^2 + y^2 = 2 \)

We need to find the area enclosed by the curve \( y = x^2 \) and the circle \( x^2 + y^2 = 2 \).

1. **Find the intersection points**:
Set \( y = x^2 \) in the equation of the circle:

\[ x^2 + (x^2)^2 = 2 \implies x^2 + x^4 = 2 \implies x^4 + x^2 - 2 = 0 \]

Let \( z = x^2 \):

\[ z^2 + z - 2 = 0 \]

Using the quadratic formula:

\[ z = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \]

This gives \( z = 1 \) or \( z = -2 \) (discarding the negative root). Thus, \( x^2 = 1 \) implies \( x = 1 \) or \( x = -1 \).

The corresponding \( y \)-values are \( y = 1 \) at \( x = 1 \) and \( y = 1 \) at \( x = -1 \).

2. **Set up the area integral**:
The area \( A \) can be calculated by integrating the difference between the upper function (the circle) and the lower function (the parabola) from \( x = -1 \) to \( x = 1 \):

\[ A = 2 \int_0^1 \left( \sqrt{2 - x^2} - x^2 \right) dx \]

### Step 4: Calculate the integral

Using the property of integrals:

\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]

For our case, \( a = \sqrt{2} \):

\[ A = 2 \left[ \int_0^1 \sqrt{2 - x^2} \, dx - \int_0^1 x^2 \, dx \right] \]

Calculating the first integral:

\[ \int_0^1 \sqrt{2 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{2 - x^2} + \frac{2}{2} \sin^{-1}\left(\frac{x}{\sqrt{2}}\right) \right]_0^1 \]

Evaluating at the limits:

\[ = \left( \frac{1}{2} \sqrt{1} + 1 \cdot \frac{\pi}{4} \right) - \left( 0 + 0 \right) = \frac{1}{2} + \frac{\pi}{4} \]

Calculating the second integral:

\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]

Thus, the area \( A \) becomes:

\[ A = 2 \left( \left( \frac{1}{2} + \frac{\pi}{4} \right) - \frac{1}{3} \right) \]

### Step 5: Final calculation

Combining terms:

\[ A = 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{\pi}{4} \right) = 2 \left( \frac{3}{6} - \frac{2}{6} + \frac{\pi}{4} \right) = 2 \left( \frac{1}{6} + \frac{\pi}{4} \right) \]

Calculating:

\[ A = \frac{1}{3} + \frac{\pi}{2} \]

### Step 6: Find the greatest integer function

Now, we need to evaluate:

\[ \left\lfloor A \right\rfloor = \left\lfloor \frac{1}{3} + \frac{\pi}{2} \right\rfloor \]

Using \( \pi \approx 3.14 \):

\[ \frac{\pi}{2} \approx 1.57 \implies A \approx \frac{1}{3} + 1.57 \approx 1.9 \]

Thus, the greatest integer function gives:

\[ \left\lfloor 1.9 \right\rfloor = 1 \]

### Final Answer

Therefore, the final answer is:

\[ \boxed{1} \]