CAT 2020 Quant Question [Slot 2] with Solution 21

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

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