If denote the set of all real x for which
`(x^(2)+x+1)^(x) lt 1`, then S =

Options

A. `(1,oo)`
B. `(-1,oo)`
C. `(-oo,-1)`
D. None of these

Explanation

To solve the inequality \((x^2 + x + 1)^x < 1\), we will analyze the expression step by step.

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ (x^2 + x + 1)^x < 1 \]
Taking the logarithm on both sides, we have:
\[ \log((x^2 + x + 1)^x) < \log(1) \]
Since \(\log(1) = 0\), this simplifies to:
\[ x \log(x^2 + x + 1) < 0 \]

### Step 2: Analyze the Logarithm
The inequality \(x \log(x^2 + x + 1) < 0\) suggests that the product of \(x\) and \(\log(x^2 + x + 1)\) must be negative. This can happen in two scenarios:
1. \(x > 0\) and \(\log(x^2 + x + 1) < 0\)
2. \(x < 0\) and \(\log(x^2 + x + 1) > 0\)

### Step 3: Case 1: \(x > 0\)
For \(x > 0\):
- We need \(\log(x^2 + x + 1) < 0\), which implies:
\[ x^2 + x + 1 < 1 \]
This simplifies to:
\[ x^2 + x < 0 \]
Factoring gives:
\[ x(x + 1) < 0 \]
The roots of the equation \(x(x + 1) = 0\) are \(x = 0\) and \(x = -1\). The intervals to test are:
- \(x < -1\)
- \(-1 < x < 0\)
- \(x > 0\)

Since we are considering \(x > 0\), this case does not yield any valid solutions.

### Step 4: Case 2: \(x < 0\)
For \(x < 0\):
- We need \(\log(x^2 + x + 1) > 0\), which implies:
\[ x^2 + x + 1 > 1 \]
This simplifies to:
\[ x^2 + x > 0 \]
Factoring gives:
\[ x(x + 1) > 0 \]
The roots are again \(x = 0\) and \(x = -1\). The intervals to test are:
- \(x < -1\)
- \(-1 < x < 0\)

In the interval \(x < -1\), both factors \(x < 0\) and \(x + 1 < 0\) are negative, thus \(x(x + 1) > 0\) is satisfied.

### Conclusion
The solution set \(S\) for the inequality \((x^2 + x + 1)^x < 1\) is:
\[ S = (-\infty, -1) \]