Let `x"*"y = x^(2) +y^(3) and (x "*"1)_("*")1=x_("*")(1_("*")1)` Then a value of `2 sin^(-1) ((x^(4)+x^(2)-2)/(x^(4)+x^(2)+2))` is

Options

A. `pi/4`
B. `pi/3`
C. `pi/2`
D. `pi/6`

Answer

To solve the problem step by step, we start with the given operation and equation:

1. **Define the operation**:
\[ x * y = x^2 + y^3 \]

2. **Set up the equation**:
We need to evaluate the expression:
\[ (x * 1) * 1 = x * (1 * 1) \]
First, calculate \( x * 1 \):
\[ x * 1 = x^2 + 1^3 = x^2 + 1 \]
Now, calculate \( (x * 1) * 1 \):
\[ (x^2 + 1) * 1 = (x^2 + 1)^2 + 1^3 = (x^2 + 1)^2 + 1 \]
Expanding \( (x^2 + 1)^2 \):
\[ (x^2 + 1)^2 = x^4 + 2x^2 + 1 \]
Therefore,
\[ (x * 1) * 1 = x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 + 2 \]

3. **Calculate \( 1 * 1 \)**:
\[ 1 * 1 = 1^2 + 1^3 = 1 + 1 = 2 \]
Now, calculate \( x * (1 * 1) \):
\[ x * 2 = x^2 + 2^3 = x^2 + 8 \]

4. **Set the two expressions equal**:
\[ x^4 + 2x^2 + 2 = x^2 + 8 \]

5. **Rearranging the equation**:
\[ x^4 + 2x^2 + 2 - x^2 - 8 = 0 \]
Simplifying gives:
\[ x^4 + x^2 - 6 = 0 \]

6. **Let \( z = x^2 \)**:
The equation becomes:
\[ z^2 + z - 6 = 0 \]

7. **Factor the quadratic**:
\[ (z - 2)(z + 3) = 0 \]
This gives us:
\[ z = 2 \quad \text{or} \quad z = -3 \]
Since \( z = x^2 \), we discard \( z = -3 \) and take \( z = 2 \):
\[ x^2 = 2 \]

8. **Substitute \( x^2 \) into the expression**:
We need to evaluate:
\[ 2 \sin^{-1} \left( \frac{x^4 + x^2 - 2}{x^4 + x^2 + 2} \right) \]
First, calculate \( x^4 \):
\[ x^4 = (x^2)^2 = 2^2 = 4 \]
Now substitute \( x^4 \) and \( x^2 \):
\[ 2 \sin^{-1} \left( \frac{4 + 2 - 2}{4 + 2 + 2} \right) = 2 \sin^{-1} \left( \frac{4}{8} \right) = 2 \sin^{-1} \left( \frac{1}{2} \right) \]

9. **Evaluate \( \sin^{-1} \left( \frac{1}{2} \right) \)**:
We know that:
\[ \sin^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6} \]
Therefore,
\[ 2 \sin^{-1} \left( \frac{1}{2} \right) = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} \]

10. **Final answer**:
The value is:
\[ \frac{\pi}{3} \]