Animated Solution for Mathematics - Functions: Let $[x]$ denote the greatest integer less than or equal to $x$. Then, the values of $x\in \mathbf{R}$ satisfying the equation $[e^x{]}^2+[e^x+1]-3=0$ lie in the interval :

We start with the equation $[e^x{]}^2+[e^x+1]-3=0$. At first glance, it looks like a complex arrangement of brackets and exponentials.

One of the most elegant properties of the Greatest Integer Function (GIF) is that if you add an integer $n$ inside the brackets, you can pull it out: $[x+n]=[x]+n$. Since $1$ is an integer, we can rewrite our equation as:

This simplifies to the quadratic form:

To simplify the algebra, let us substitute $t=[e^x]$. The equation transforms into the familiar quadratic:

We factorize this by splitting the middle term:

This leads us to the factored form:

Thus, we have two potential candidates for $t$: $t=-2$ and $t=1$.

We must now consider the domain of our function. We defined $t=[e^x]$.

For any real number $x$, the value of $e^x$ is always strictly positive ($e^x>0$). Consequently, its greatest integer value $[e^x]$ must be non-negative.

This immediately disqualifies $t=-2$. It is a mathematical ghost—a root that exists in the algebra but not in the physical reality of our function. We are left with the only valid survivor: $t=1$.

Now we return to our original variable. We have $[e^x]=1$.

By the definition of the GIF, if the greatest integer of a number is $1$, that number must be at least $1$ but strictly less than $2$. This gives us the inequality:

To isolate $x$, we apply the natural logarithm, $\ln$, to all parts of the inequality. Because the logarithm is a strictly increasing function, the inequality signs remain unchanged:

Since $\ln (1)=0$ and $\ln (e^x)=x$, we arrive at the final result:

You have successfully navigated the staircase of the GIF. By understanding the constraints of the system, you have reduced a complex equation to a clean, logical interval.