JEE Main 2021 (Online) 22nd July Evening ShiftQuestion 1
Let $[x]$ denote the greatest integer less than or equal to $x$. Then, the values of $x\in\mathbf{R}$ satisfying the equation $\left[e^{x}\right]^{2}+\left[e^{x}+1\right]-3=0$ lie in the interval :
Functions
Solution
With $n=[e^x]\ge0$, $[e^x+1]=n+1$: $n^2+n-2=0$, so $n=1$ (as $n\ge0$). $1\le e^x<2$ gives $0\le x<\ln2$.