CAT Exam 2021 Slot 1 Solved Paper

${\mathrm{x}}^0=1$
${\mathrm{x}}^1=2$
${\mathrm{x}}^2=\frac{\left(1+{\mathrm{x}}^1\right)}{{\mathrm{x}}^0}=\frac{\left(1+2\right)}{1}=3$
${\mathrm{x}}^3=\frac{\left(1+{\mathrm{x}}^2\right)}{\mathrm{x}1}=\frac{\left(1+3\right)}{2}=2$
${\mathrm{x}}^4=\frac{\left(1+{\mathrm{x}}^3\right)}{{\mathrm{x}}^2}=\frac{\left(1+2\right)}{3}=1$
${\mathrm{x}}^5=\frac{\left(1+{\mathrm{x}}^4\right)}{{\mathrm{x}}^3}=\frac{\left(1+1\right)}{2}=1$
${\mathrm{x}}^6=\frac{\left(1+{\mathrm{x}}^5\right)}{{\mathrm{x}}^4}=\frac{\left(1+1\right)}{1}=2$ Hence, the series begins to repeat itself after every 5 terms. Terms whose number is of the form 5n are $1,5\mathit{n}+1$ are 2... and so on, where $\mathit{n}=0,1,2,3,....$
2021 is of the form $5\mathit{n}+1$.
Hence, its value will be 2.