The value of 'a' for which $x^{2}+ax+sin^{-1}(x^{2}-4x+5)+cos^{-1}(x^{2}-4x+5)=0$ has at least one solution, is:

Question

Accepted Answer

To determine the value of 'a' for which the equation $x^{2} + ax + \sin^{-1}(x^{2} - 4x + 5) + \cos^{-1}(x^{2} - 4x + 5) = 0$ has at least one solution, we need to analyze the components of the equation.

### Explanation:

1. **Understanding the Inverse Trigonometric Functions:**
   - The expressions $\sin^{-1}(y)$ and $\cos^{-1}(y)$ are defined such that $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$ for any $y$ in the domain $[-1, 1]$.
   - Therefore, for $\sin^{-1}(x^{2} - 4x + 5) + \cos^{-1}(x^{2} - 4x + 5)$ to be defined, $x^{2} - 4x + 5$ must lie within the interval $[-1, 1]$.

2. **Analyzing the Quadratic Expression:**
   - The quadratic expression $x^{2} - 4x + 5$ can be rewritten as $(x-2)^2 + 1$.
   - The minimum value of $(x-2)^2$ is 0, which occurs when $x = 2$.
   - Thus, the minimum value of $x^{2} - 4x + 5$ is 1, and it occurs at $x = 2$.

3. **Domain Consideration:**
   - Since the minimum value of $x^{2} - 4x + 5$ is 1, the expression $\sin^{-1}(x^{2} - 4x + 5) + \cos^{-1}(x^{2} - 4x + 5)$ is only defined when $x^{2} - 4x + 5 = 1$.
   - Solving $x^{2} - 4x + 5 = 1$ gives $x^{2} - 4x + 4 = 0$, which simplifies to $(x-2)^2 = 0$.
   - Therefore, $x = 2$ is the only value that satisfies this condition.

4. **Substituting Back into the Original Equation:**
   - Substitute $x = 2$ into the original equation: 
     $$
     2^2 + a \cdot 2 + \sin^{-1}(1) + \cos^{-1}(1) = 0
     $$
   - Simplifying, we get:
     $$
     4 + 2a + \frac{\pi}{2} = 0
     $$
   - Solving for $a$, we find:
     $$
     2a = -4 - \frac{\pi}{2}
     $$
     $$
     a = -2 - \frac{\pi}{4}
     $$

5. **Conclusion:**
   - Therefore, the value of $a$ for which the equation has at least one solution is $-2 - \frac{\pi}{4}$, which corresponds to Option 3.

### Why Other Options are Incorrect:

- **Option 1: $\sqrt{2\pi}$**
  - This value does not satisfy the equation when substituted back, as it does not account for the necessary balance of terms to achieve zero.

- **Option 2: $-2 + \pi$**
  - This value results in a positive term when substituted, which cannot satisfy the equation equating to zero.

- **Option 4: $-\frac{\pi}{4}$**
  - This value does not correctly balance the equation when substituted, leading to a non-zero result.

Thus, Option 3 is the only value that satisfies the condition for the equation to have at least one solution.