CAT 2020

Slot 2

QA

Question & Solution

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

There can be a solution when $\left(x^2-5x+7\right)=1$ or $x^2-5x\ +6=0$

or x=3 and x=2

There can also be a solution when x+1 = 0 or x=-1

Hence three possible solutions exist.