Let `g (x) =x ^(2),f (x) = sqrtx, and alpha, beta ( alpha lt beta)` be the roots of the quadratic equation `18 x ^(2) - 9pix + pi^(2) =0.` Then the area (is sq. units) bounded by curve `y = (gof) (x)` and the lines `x =alpha, x = betaand y =0` is :

Options

A. `(pi^2)/24`
B. `(pi^2)/9`
C. `(pi^2)/36`
D. `(pi^2)/72`

Answer

To solve the problem, we need to find the area bounded by the curve \( y = (g \circ f)(x) \), the lines \( x = \alpha \), \( x = \beta \), and \( y = 0 \). Let's break down the solution step by step.

### Step 1: Define the Functions
We are given:
- \( g(x) = x^2 \)
- \( f(x) = \sqrt{x} \)

### Step 2: Find the Composition of Functions
We need to find \( g(f(x)) \):
\[ g(f(x)) = g(\sqrt{x}) = (\sqrt{x})^2 = x \]
Thus, the curve is represented by the equation \( y = x \).

### Step 3: Identify the Roots of the Quadratic Equation
We need to find the roots of the quadratic equation:
\[ 18x^2 - 9\pi x + \pi^2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 18 \), \( b = -9\pi \), and \( c = \pi^2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-9\pi)^2 - 4 \cdot 18 \cdot \pi^2 = 81\pi^2 - 72\pi^2 = 9\pi^2 \]
Now, applying the quadratic formula:
\[ x = \frac{9\pi \pm 3\pi}{36} \]
This gives us two roots:
\[ x_1 = \frac{12\pi}{36} = \frac{\pi}{3}, \quad x_2 = \frac{6\pi}{36} = \frac{\pi}{6} \]
Thus, we have:
\[ \alpha = \frac{\pi}{6}, \quad \beta = \frac{\pi}{3} \]

### Step 4: Set Up the Integral for Area Calculation
The area \( A \) bounded by the curve \( y = x \) and the lines \( x = \alpha \), \( x = \beta \), and \( y = 0 \) can be calculated using the integral:
\[ A = \int_{\alpha}^{\beta} x \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} x \, dx \]

### Step 5: Compute the Integral
Calculating the integral:
\[ A = \left[ \frac{x^2}{2} \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{1}{2} \left( \left(\frac{\pi}{3}\right)^2 - \left(\frac{\pi}{6}\right)^2 \right) \]
Calculating each term:
\[ \left(\frac{\pi}{3}\right)^2 = \frac{\pi^2}{9}, \quad \left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{36} \]
Now substituting back:
\[ A = \frac{1}{2} \left( \frac{\pi^2}{9} - \frac{\pi^2}{36} \right) \]
Finding a common denominator (36):
\[ \frac{\pi^2}{9} = \frac{4\pi^2}{36} \]
Thus:
\[ A = \frac{1}{2} \left( \frac{4\pi^2}{36} - \frac{\pi^2}{36} \right) = \frac{1}{2} \left( \frac{3\pi^2}{36} \right) = \frac{3\pi^2}{72} = \frac{\pi^2}{24} \]

### Final Result
The area bounded by the curve and the lines is:
\[ \boxed{\frac{\pi^2}{24}} \text{ square units} \]