Question

Let $\left[x\right]$ denote the greatest integer less than or equal to $x$. Then, the values of $x\in R$ satisfying the equation $[e^x{]}^2+[e^x+1]-3=0$ lies in the interval:

• A. $[0,{\log }_{e}2)$
• B. $[0,\frac{1}{e})$
• C. $[1,e)$
• D. $[{\log }_{e}2,{\log }_{e}3)$

Answer 1

The correct option is A $[0,{\log }_{e}2)$

Given: $[e^x{]}^2+[e^x+1]-3=0$
$\Rightarrow [e^x{]}^2+[e^x]-2=0$
$\Rightarrow \left(\right[e^x]+2)\left(\right[e^x]-1)=0$
$\therefore \left[e^x\right]=1$, since $\left[e^x\right]=-2$ is not possible.
$\therefore e^x\in [1,2)$
$\therefore x\in [0,{\log }_{e}2)$