The set of values of `a` for which `x^2+ax+sin^(-1)(x^2-4x+5)+cos^(-1)(x^2-4x+5)=0` has at least one real root is given by

Options

A. `(-oo, - sqrt(2 pi)] uu[sqrt(2 pi), oo)`
B. `(-oo, -sqrt(2 pi))uu(sqrt(2 pi), oo)`
C. R
D. None of these

Answer

To solve the equation

\[ x^2 + ax + \sin^{-1}(x^2 - 4x + 5) + \cos^{-1}(x^2 - 4x + 5) = 0, \]

we start by using the property of inverse trigonometric functions:

\[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}. \]

### Step 1: Simplifying the Equation

Using this property, we can rewrite the equation as:

\[ x^2 + ax + \frac{\pi}{2} = 0. \]

### Step 2: Finding the Discriminant

For the quadratic equation \(x^2 + ax + \frac{\pi}{2} = 0\) to have at least one real root, its discriminant must be greater than or equal to zero. The discriminant \(D\) of a quadratic equation \(Ax^2 + Bx + C = 0\) is given by:

\[ D = B^2 - 4AC. \]

In our case, \(A = 1\), \(B = a\), and \(C = \frac{\pi}{2}\). Thus, the discriminant is:

\[ D = a^2 - 4 \cdot 1 \cdot \frac{\pi}{2} = a^2 - 2\pi. \]

### Step 3: Setting Up the Inequality

For the quadratic to have at least one real root, we require:

\[ D \geq 0 \implies a^2 - 2\pi \geq 0. \]

This can be rearranged to:

\[ a^2 \geq 2\pi. \]

### Step 4: Solving the Inequality

The inequality \(a^2 \geq 2\pi\) can be factored as:

\[ (a - \sqrt{2\pi})(a + \sqrt{2\pi}) \geq 0. \]

### Step 5: Analyzing the Sign of the Factors

To find the intervals where this inequality holds, we identify the critical points:

- \(a = -\sqrt{2\pi}\)
- \(a = \sqrt{2\pi}\)

We can use a sign chart or the wavy curve method to analyze the intervals:

1. For \(a < -\sqrt{2\pi}\), both factors are negative, so the product is positive.
2. For \(-\sqrt{2\pi} < a < \sqrt{2\pi}\), one factor is negative and the other is positive, so the product is negative.
3. For \(a > \sqrt{2\pi}\), both factors are positive, so the product is positive.

### Step 6: Conclusion

The solution to the inequality \(a^2 \geq 2\pi\) is:

\[ a \in (-\infty, -\sqrt{2\pi}] \cup [\sqrt{2\pi}, \infty). \]

Thus, the set of values of \(a\) for which the original equation has at least one real root is:

\[ \boxed{(-\infty, -\sqrt{2\pi}] \cup [\sqrt{2\pi}, \infty)}. \]