Solution set of `1/(x^2-x+1) >= 4` is

Options

A. ` x in R`
B. `x in (0,oo)`
C. `phi`
D. None of these

Answer

To solve the inequality \( \frac{1}{x^2 - x + 1} \geq 4 \), we can follow these steps:

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ \frac{1}{x^2 - x + 1} \geq 4 \]
To eliminate the fraction, we can multiply both sides by \( x^2 - x + 1 \) (noting that \( x^2 - x + 1 > 0 \) for all \( x \)):
\[ 1 \geq 4(x^2 - x + 1) \]

### Step 2: Rearrange the Inequality
Now, we rearrange the inequality:
\[ 1 \geq 4x^2 - 4x + 4 \]
This can be rewritten as:
\[ 0 \geq 4x^2 - 4x + 3 \]
or
\[ 4x^2 - 4x + 3 \leq 0 \]

### Step 3: Analyze the Quadratic
Next, we need to analyze the quadratic \( 4x^2 - 4x + 3 \). To do this, we will calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 4 \cdot 3 = 16 - 48 = -32 \]
Since the discriminant is negative (\( D < 0 \)), the quadratic \( 4x^2 - 4x + 3 \) does not intersect the x-axis and is always positive.

### Step 4: Conclusion
Since \( 4x^2 - 4x + 3 > 0 \) for all \( x \), the inequality \( 4x^2 - 4x + 3 \leq 0 \) has no solutions. Therefore, the solution set for the original inequality is:
\[ \text{The solution set is } \emptyset \]