The number of real solutions of the following equation is : $tan^{-1}\sqrt{x(x+1)}+sin^{-1}\sqrt{(x^{2}+x+1)}=\frac{\pi}{2}$

Question

Accepted Answer

To determine the number of real solutions for the equation $ \tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{(x^2+x+1)} = \frac{\pi}{2} $, we need to analyze the behavior of each term in the equation and how they interact.

### Explanation Supporting Option 3:

1. **Understanding the Equation:**
   - The equation is $ \tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{(x^2+x+1)} = \frac{\pi}{2} $.
   - The range of $ \tan^{-1}(y) $ is $ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $ and the range of $ \sin^{-1}(y) $ is $ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $.

2. **Analyzing the Terms:**
   - $ \tan^{-1}\sqrt{x(x+1)} $ is defined for $ x(x+1) \geq 0 $, which implies $ x \leq -1 $ or $ x \geq 0 $.
   - $ \sin^{-1}\sqrt{(x^2+x+1)} $ is defined for $ 0 \leq \sqrt{(x^2+x+1)} \leq 1 $, which is always true since $ x^2+x+1 \geq 1 $.

3. **Behavior of the Equation:**
   - For the equation to hold, the sum of the two inverse trigonometric functions must equal $ \frac{\pi}{2} $.
   - Consider the identity $ \tan^{-1}(a) + \sin^{-1}(b) = \frac{\pi}{2} $ when $ a = \sqrt{1-b^2} $.

4. **Finding Solutions:**
   - Set $ \sqrt{x(x+1)} = \sqrt{1 - (x^2+x+1)} $.
   - Simplifying gives $ x(x+1) = 1 - (x^2+x+1) $, leading to $ x(x+1) = -x^2-x $.
   - Solving $ x(x+1) = -x^2-x $ results in $ x^2 + x = 0 $, which simplifies to $ x(x+1) = 0 $.
   - The solutions to this are $ x = 0 $ and $ x = -1 $.

5. **Verification:**
   - For $ x = 0 $: $ \tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} $.
   - For $ x = -1 $: $ \tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} $.

Thus, there are two real solutions, $ x = 0 $ and $ x = -1 $, confirming that Option 3 is correct.

### Explanation for Incorrect Options:

- **Option 1 (Zero):** This is incorrect because we found two valid solutions, $ x = 0 $ and $ x = -1 $, that satisfy the equation.
- **Option 2 (One):** This is incorrect because there are two distinct solutions, not just one.
- **Option 4 (Infinite):** This is incorrect because the equation only holds for the specific values $ x = 0 $ and $ x = -1 $, not infinitely many values.

Therefore, the correct answer is indeed Option 3, as there are exactly two real solutions to the given equation.