STATEMENT- 1: The area bounded by the curves `y = x^(2)` and `y = (2)/(1+ x^(2))` is `2 ((pi)/(2) - (1)/(3))` sq. units
and
STATEMENT-2 : The curves meet at a point whose abscissae are `pm2`.

Options

A. (a)Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-1
B. (b)Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-1
C. (c)Statement-1 is True , Statement-2 is False
D. (d)Statement-1 is False , Statement-2 is True

Answer

To solve the problem of finding the area bounded by the curves \( y = x^2 \) and \( y = \frac{2}{1+x^2} \), we will follow these steps:

### Step 1: Find the points of intersection
To determine the area between the curves, we first need to find the points where they intersect. We set the equations equal to each other:

\[ x^2 = \frac{2}{1+x^2} \]

Multiplying both sides by \( 1 + x^2 \) to eliminate the fraction gives:

\[ x^2(1 + x^2) = 2 \]

This simplifies to:

\[ x^4 + x^2 - 2 = 0 \]

Let \( u = x^2 \). Then the equation becomes:

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

### Step 2: Solve the quadratic equation
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \]

This gives us:

\[ u = 1 \quad \text{or} \quad u = -2 \]

Since \( u = x^2 \), we only consider \( u = 1 \):

\[ x^2 = 1 \implies x = \pm 1 \]

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 1 \) is given by:

\[ A = \int_{-1}^{1} \left( \frac{2}{1+x^2} - x^2 \right) \, dx \]

### Step 4: Calculate the integral
We can split the integral:

\[ A = \int_{-1}^{1} \frac{2}{1+x^2} \, dx - \int_{-1}^{1} x^2 \, dx \]

The first integral can be calculated as:

\[ \int \frac{2}{1+x^2} \, dx = 2 \tan^{-1}(x) \]

Evaluating from \(-1\) to \(1\):

\[ \left[ 2 \tan^{-1}(x) \right]_{-1}^{1} = 2 \left( \tan^{-1}(1) - \tan^{-1}(-1) \right) = 2 \left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = 2 \cdot \frac{\pi}{2} = \pi \]

The second integral is:

\[ \int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3} \]

### Step 5: Combine the results
Now, substituting back into the area formula:

\[ A = \pi - \frac{2}{3} \]

### Step 6: Final expression for the area
To express the area in the required form:

\[ A = \frac{3\pi}{3} - \frac{2}{3} = \frac{3\pi - 2}{3} \]

However, the problem states that the area is \( 2 \left( \frac{\pi}{2} - \frac{1}{3} \right) \):

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

Thus, **Statement 1 is true**.

### Step 7: Verify Statement 2
From our earlier calculations, the points of intersection were \( x = \pm 1 \), not \( x = \pm 2 \). Therefore, **Statement 2 is false**.

### Conclusion
- **Statement 1**: True
- **Statement 2**: False