If `g(x)` is a differentiable function such that `int_(1)^(sinalpha)x^(2)g(x)dx=(sin alpha-1), AA alpha in (0,(pi)/(2))`, then the value of `g((1)/(3))` is equal to

Options

A. 4
B. `(4)/(3)`
C. `(sqrt3)/(2)`
D. 9

Answer

To solve the problem, we need to find the value of \( g\left(\frac{1}{3}\right) \) given the condition:

\[ \int_{1}^{\sin \alpha} x^2 g(x) \, dx = \sin \alpha - 1 \quad \text{for } \alpha \in \left(0, \frac{\pi}{2}\right) \]

### Step 1: Differentiate both sides with respect to \( \alpha \)

Using Leibniz's rule for differentiation under the integral sign, we differentiate the left-hand side:

\[ \frac{d}{d\alpha} \left( \int_{1}^{\sin \alpha} x^2 g(x) \, dx \right) = \sin^2 \alpha \cdot g(\sin \alpha) \cdot \cos \alpha \]

The right-hand side differentiates to:

\[ \frac{d}{d\alpha} (\sin \alpha - 1) = \cos \alpha \]

### Step 2: Set the derivatives equal to each other

From the differentiation, we have:

\[ \sin^2 \alpha \cdot g(\sin \alpha) \cdot \cos \alpha = \cos \alpha \]

Assuming \( \cos \alpha \neq 0 \) (which is valid since \( \alpha \) is in \( (0, \frac{\pi}{2}) \)), we can divide both sides by \( \cos \alpha \):

\[ \sin^2 \alpha \cdot g(\sin \alpha) = 1 \]

### Step 3: Solve for \( g(\sin \alpha) \)

Rearranging gives us:

\[ g(\sin \alpha) = \frac{1}{\sin^2 \alpha} \]

### Step 4: Evaluate \( g\left(\frac{1}{3}\right) \)

To find \( g\left(\frac{1}{3}\right) \), we set \( \sin \alpha = \frac{1}{3} \). Thus, we have:

\[ g\left(\frac{1}{3}\right) = \frac{1}{\left(\frac{1}{3}\right)^2} = \frac{1}{\frac{1}{9}} = 9 \]

### Conclusion

The value of \( g\left(\frac{1}{3}\right) \) is:

\[ \boxed{9} \]