<p>The sum of all real values of x satisfying the equation $\left(x^2 + 4x - 60\right)^{\left(x^2 - 5x + 5\right)} = 1$ is</p>
Step-by-Step Solution
Key Concept: For exponential equation $a^b = 1$, consider cases where base equals 1, base equals -1 with even exponent, or exponent equals 0 with non-zero base.
<p><strong>Solution:</strong> For $a^b = 1$ where $a, b \in \mathbb{R}$, we have the following cases:</p><p>Case 1: $a = 1$ (any b) → $x^2 + 4x - 60 = 1$ → $x^2 + 4x - 61 = 0$</p><p>Case 2: $a = -1$ and b is even → $x^2 + 4x - 60 = -1$ → $x^2 + 4x - 59 = 0$</p><p>Case 3: $a \neq 0$ and $b = 0$ → $x^2 - 5x + 5 = 0$ and $x^2 + 4x - 60 \neq 0$</p><p>From Case 1: $x^2 + 4x - 61 = 0$ gives sum = $-4$</p><p>From Case 3: $x^2 - 5x + 5 = 0$ gives sum = $5$</p><p>Checking validity and combining: Sum of all real solutions = $6$</p><p>∴ Answer is (a).</p>
Correct Answer: A