# CAT 2020 Quant Question [Slot 2] with Solution 21

Question

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
Option: 2
Solution:

${\left( {{x^2} - 5x + 7} \right)^{x + 1}} = 1$

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

$- a = - 1$ then $b$ is even.

$- a = 1$ then $b$ is any number

$- a > 0$ then $b = 0$

Case 1: $x + 1 = 0 \Rightarrow x = - 1$

Case 2: ${x^2} - 5x + 7 = 1 \Rightarrow {x^2} - 5x + 6 = 0 \Rightarrow x = 2{\text{ or }}3$

Case 3: ${x^2} - 5x + 7 = - 1 \Rightarrow {x^2} - 5x + 8 = 0$

but x is not an integer

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

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