The number of solutions for the equation `2 sin^(-1)(sqrt(x^(2) - x + 1)) + cos^(-1)(sqrt(x^(2) - x) )= (3pi)/(2)` is

Options

A. 1
B. 2
C. 3
D. infinite

Answer

To solve the equation

\[ 2 \sin^{-1}(\sqrt{x^2 - x + 1}) + \cos^{-1}(\sqrt{x^2 - x}) = \frac{3\pi}{2}, \]

we will follow these steps:

### Step 1: Determine the domain of the functions involved

The expressions inside the inverse trigonometric functions must be valid.

1. For \(\sin^{-1}(y)\), \(y\) must be in the range \([-1, 1]\).
2. For \(\cos^{-1}(y)\), \(y\) must also be in the range \([0, 1]\).

Thus, we need:

\[ \sqrt{x^2 - x + 1} \leq 1 \quad \text{and} \quad \sqrt{x^2 - x} \leq 1. \]

### Step 2: Solve the inequalities

#### Inequality 1:

\[ \sqrt{x^2 - x + 1} \leq 1 \]

Squaring both sides gives:

\[ x^2 - x + 1 \leq 1 \implies x^2 - x \leq 0 \implies x(x - 1) \leq 0. \]

This inequality holds for \(0 \leq x \leq 1\).

#### Inequality 2:

\[ \sqrt{x^2 - x} \leq 1 \]

Squaring both sides gives:

\[ x^2 - x \leq 1 \implies x^2 - x - 1 \leq 0. \]

To solve \(x^2 - x - 1 = 0\), we use the quadratic formula:

\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2}. \]

The roots are approximately \(x \approx 1.618\) and \(x \approx -0.618\). The inequality \(x^2 - x - 1 \leq 0\) holds between the roots:

\[ -\frac{1 + \sqrt{5}}{2} \leq x \leq \frac{1 + \sqrt{5}}{2}. \]

### Step 3: Find the intersection of the intervals

From the first inequality, we have \(0 \leq x \leq 1\). From the second inequality, we have:

\[ -\frac{1 + \sqrt{5}}{2} \leq x \leq \frac{1 + \sqrt{5}}{2}. \]

The intersection of these intervals is:

\[ 0 \leq x \leq 1. \]

### Step 4: Check the endpoints

Now we will check the values at the endpoints \(x = 0\) and \(x = 1\).

1. **For \(x = 0\)**:

\[ 2 \sin^{-1}(\sqrt{0^2 - 0 + 1}) + \cos^{-1}(\sqrt{0^2 - 0}) = 2 \sin^{-1}(1) + \cos^{-1}(0) = 2 \cdot \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2}. \]

This satisfies the equation.

2. **For \(x = 1\)**:

\[ 2 \sin^{-1}(\sqrt{1^2 - 1 + 1}) + \cos^{-1}(\sqrt{1^2 - 1}) = 2 \sin^{-1}(1) + \cos^{-1}(0) = 2 \cdot \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2}. \]

This also satisfies the equation.

### Conclusion

Both \(x = 0\) and \(x = 1\) are solutions. Therefore, the number of solutions for the equation is:

\[ \boxed{2}. \]