The range of k for which the inequality `kcos^2x-kcosx+1>=0AAxepsilon(-oo,oo)` is:

Options

A. `k gt 1/2`
B. `k gt 4`
C. `1/2 le k le 4`
D. `1/2 le k le 5`

Answer

To solve the inequality \( k \cos^2 x - k \cos x + 1 \geq 0 \) for \( x \in (-\infty, \infty) \), we will analyze the quadratic expression in terms of \( \cos x \).

### Step 1: Rewrite the inequality
We start with the inequality:
\[ k \cos^2 x - k \cos x + 1 \geq 0 \]
Let \( y = \cos x \). The inequality becomes:
\[ k y^2 - k y + 1 \geq 0 \]

### Step 2: Identify the quadratic coefficients
The quadratic can be expressed as:
\[ ay^2 + by + c = 0 \]
where \( a = k \), \( b = -k \), and \( c = 1 \).

### Step 3: Determine the discriminant
For the quadratic \( ay^2 + by + c \) to be non-negative for all \( y \), the discriminant must be less than or equal to zero:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = (-k)^2 - 4(k)(1) = k^2 - 4k \]
We require:
\[ k^2 - 4k \leq 0 \]

### Step 4: Factor the discriminant
Factoring the inequality gives:
\[ k(k - 4) \leq 0 \]

### Step 5: Solve the inequality
To solve \( k(k - 4) \leq 0 \), we find the roots:
- \( k = 0 \)
- \( k = 4 \)

The critical points divide the number line into intervals:
1. \( (-\infty, 0) \)
2. \( (0, 4) \)
3. \( (4, \infty) \)

We test each interval:
- For \( k < 0 \): Choose \( k = -1 \) → \( (-1)(-1 - 4) = -1 \cdot -5 = 5 > 0 \) (not valid)
- For \( 0 < k < 4 \): Choose \( k = 2 \) → \( (2)(2 - 4) = 2 \cdot -2 = -4 \leq 0 \) (valid)
- For \( k > 4 \): Choose \( k = 5 \) → \( (5)(5 - 4) = 5 \cdot 1 = 5 > 0 \) (not valid)

### Step 6: Conclusion
The solution to the inequality \( k(k - 4) \leq 0 \) is:
\[ k \in [0, 4] \]

### Final Answer
Thus, the range of \( k \) for which the inequality holds is:
\[ \boxed{[0, 4]} \]