Question

The number of integers that satisfy the equality \((x^2-5x+7)^{x+1} = 1\) is

• A. 2
• B. 3
• C. 5
• D. 4

Answer 1

The Correct Option is B

Considering \( a^b = 1 \), the following conditions apply:

• If \( a = 1 \), this holds for all \( b \).
• If \( a = -1 \), this holds only when \( b \) is an even number.
• If \( b = 0 \), this holds for all \( a \) except \( a = 0 \).

Evaluation Process:

Condition 1: \( x + 1 = 0 \Rightarrow x = -1 \)

Any non-zero base raised to the power of 0 equals 1. We verify the base: \[ (x^2 - 5x + 7) = (-1)^2 + 5 + 7 = 1 + 5 + 7 = 13 eq 0 \] Thus, this is valid. ✅ \( x = -1 \) is a solution.

Condition 2: \( x^2 - 5x + 7 = 1 \)

Solving the equation: \[ x^2 - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0 \Rightarrow x = 2, \; x = 3 \] Both solutions are valid integers.
✅ \( x = 2 \) and \( x = 3 \) are solutions.

Condition 3: \( x^2 - 5x + 7 = -1 \)

\[ x^2 - 5x + 8 = 0 \Rightarrow \text{Discriminant} = 25 - 32 = -7 < 0 \] This equation has no real roots. ❌ No integer solution.

Conclusion:

The total count of integer solutions is: \[ \boxed{3} \] Correct Option: (B)