If `[x]` denotes the greatest integer function and `[e^x]^2+[e^x+1]-3=0` then `x in`

Options

A. `(1,e)`
B. `(1,1/e)`
C. `(log_e2,log_e3)`
D. `[0,log_e2)`

Answer

To solve the equation \([e^x]^2 + [e^x + 1] - 3 = 0\), where \([x]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ [e^x]^2 + [e^x + 1] - 3 = 0 \]
Let \( [e^x] = n \), where \( n \) is an integer. Then we can rewrite the equation as:
\[ n^2 + (n + 1) - 3 = 0 \]

### Step 2: Simplify the Equation
Now, simplify the equation:
\[ n^2 + n + 1 - 3 = 0 \implies n^2 + n - 2 = 0 \]

### Step 3: Factor the Quadratic
Next, we can factor the quadratic equation:
\[ n^2 + n - 2 = (n - 1)(n + 2) = 0 \]
This gives us two possible solutions for \( n \):
\[ n - 1 = 0 \implies n = 1 \]
\[ n + 2 = 0 \implies n = -2 \]

### Step 4: Analyze the Solutions
Now we analyze the solutions for \( n \):
1. If \( n = 1 \):
\[ [e^x] = 1 \implies 1 \leq e^x < 2 \]
Taking the natural logarithm:
\[ 0 \leq x < \ln(2) \]

2. If \( n = -2 \):
\[ [e^x] = -2 \]
However, since \( e^x \) is always positive, this case is not possible.

### Step 5: Conclusion
Thus, the only valid solution comes from \( n = 1 \):
\[ 0 \leq x < \ln(2) \]
Therefore, the solution set for \( x \) is:
\[ x \in [0, \ln(2)) \]

### Final Answer
Thus, the final solution is:
\[ x \in [0, \ln(2)) \]