Solution of inequailty `x^(2)+x+|x|+1le 0` is

Options

A. (1,2)
B. (0,1)
C. No solution
D. None of these

Answer

To solve the inequality \( x^2 + x + |x| + 1 \leq 0 \), we will consider two cases based on the definition of the absolute value function.

### Step 1: Case 1 - \( x \geq 0 \)
In this case, \( |x| = x \). Therefore, the inequality becomes:
\[ x^2 + x + x + 1 \leq 0 \]
This simplifies to:
\[ x^2 + 2x + 1 \leq 0 \]
Factoring the quadratic expression:
\[ (x + 1)^2 \leq 0 \]
The square of any real number is non-negative, so \( (x + 1)^2 = 0 \) is the only solution. Thus, we have:
\[ x + 1 = 0 \implies x = -1 \]
However, since we are in the case where \( x \geq 0 \), there are no valid solutions from this case.

### Step 2: Case 2 - \( x < 0 \)
In this case, \( |x| = -x \). Therefore, the inequality becomes:
\[ x^2 + x - x + 1 \leq 0 \]
This simplifies to:
\[ x^2 + 1 \leq 0 \]
Since \( x^2 \) is always non-negative for any real \( x \), \( x^2 + 1 \) is always greater than 0. Thus, there are no solutions in this case as well.

### Conclusion
Since both cases yield no valid solutions, we conclude that the inequality \( x^2 + x + |x| + 1 \leq 0 \) has no real solutions.

### Final Answer
The solution of the inequality is: **No solution**.