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

1. $2$
2. $3$
3. $5$
4. $4$

Given that, $(x^{2} – 5x + 7)^{x+1} = 1$

We know that, for $a^{b} = 1$

• If $b = 0 \Rightarrow a^{0} = 1$
• If $a = 1 \Rightarrow 1^{b} = 1$
• If $a = -1, b = \text{even} \Rightarrow (-1)^{\text{even}} = 1$

$\textbf{Case 1} : \; x+1 = 0$

$\Rightarrow \boxed{ x = – 1 \;\textsf{(This case will be accepted)} }$

$\textbf{Case 2} : \; x^{2} – 5x + 7 = 1$

$\Rightarrow x^{2} – 5x + 6 = 0$

$\Rightarrow x^{2} – 3x – 2x + 6 = 0$

$\Rightarrow x (x – 3) – 2 (x-3) = 0$

$\Rightarrow (x-2)(x-3) = 0$

$\Rightarrow \boxed{x = 2,3 \;\textsf{(This case will be accepted)} }$

$\textbf{Case 3}: \; x^{2} -5x + 7 = -1$

$\Rightarrow x^{2} – 5x + 8 = 0$

$\Rightarrow x = \dfrac{-(-5) \pm \sqrt{25-32}}{2}$

$\Rightarrow x = \dfrac{5 \pm \sqrt{-7}}{2}$

$\Rightarrow \boxed{x = \frac{5 \pm 7i}{2} \;\textsf{(This case will be rejected)}} \qquad [\because \sqrt{-1} = i ]$

Because we don’t want value of $x$ is complex number.

$\therefore$ The number of integers satisfies the equation is $3.$

Correct Answer$: \text{B}$

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