CAT 2020 Set-2 | Question: 71

Question

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

• A. $2$
• B. $3$
• C. $5$
• D. $4$

Answer 1

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$: \text{B}$