Question:
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:
Correct Option: , 4
Solution:
$\left[e^{x}\right]^{2}+\left[e^{x}+1\right]-3=0$
$\Rightarrow\left[e^{x}\right]^{2}+\left[e^{x}\right]+1-3=0$
Let $\left[e^{x}\right]=t$
$\Rightarrow \mathrm{t}^{2}+\mathrm{t}-2=0$
$\Rightarrow \mathrm{t}=-2,1$
$\left[e^{x}\right]=-2$ (Not possible)
or $\left[e^{x}\right]=1 \quad \therefore 1 \leq e^{x}<2$
$\Rightarrow \quad \ln (1) \leq x<\ln (2)$
$\Rightarrow \quad 0 \leq x<\ln (2)$
$\Rightarrow \quad x \in[0, \ln 2)$