Th range of k for which the inequaliity `kcos^2x-kcosx+1>=0 AA x in(-oo,oo) is`

Options

A. `k lt ( -1)/(2)`
B. `k lt 4`
C. `(-1)/(2) le k le 4`
D. `1/2 le k le 5`

Answer

To find the range of \( k \) for which the inequality \( k \cos^2 x - k \cos x + 1 \geq 0 \) holds for all \( x \in (-\infty, \infty) \), we can follow these steps:

### Step 1: Rewrite the inequality
We start with the inequality:
\[ k \cos^2 x - k \cos x + 1 \geq 0 \]
This can be treated as a quadratic in \( \cos x \):
\[ k y^2 - k y + 1 \geq 0 \]
where \( y = \cos x \).

### Step 2: Determine the discriminant
For the quadratic \( k y^2 - k y + 1 \) to be non-negative for all values of \( y \) (which ranges from -1 to 1), the discriminant must be less than or equal to zero. The discriminant \( D \) of the quadratic \( ay^2 + by + c \) is given by:
\[ D = b^2 - 4ac \]
In our case:
- \( a = k \)
- \( b = -k \)
- \( c = 1 \)

Thus, the discriminant is:
\[ D = (-k)^2 - 4(k)(1) = k^2 - 4k \]

### Step 3: Set the discriminant less than or equal to zero
We require:
\[ k^2 - 4k \leq 0 \]
Factoring gives:
\[ k(k - 4) \leq 0 \]

### Step 4: Solve the inequality
The solutions to \( k(k - 4) \leq 0 \) can be found by determining the intervals where the product is non-positive. The critical points are \( k = 0 \) and \( k = 4 \).

Using a sign chart:
- For \( k < 0 \): \( k(k - 4) > 0 \)
- For \( 0 < k < 4 \): \( k(k - 4) < 0 \)
- For \( k > 4 \): \( k(k - 4) > 0 \)

Thus, the solution to the inequality is:
\[ 0 \leq k \leq 4 \]

### Step 5: Combine with the condition for \( k \)
Since we also need to ensure that the quadratic opens upwards (i.e., \( k > 0 \)), we combine this with our previous result:
\[ 0 < k \leq 4 \]

### Final Result
The range of \( k \) for which the inequality holds true is:
\[ k \in (0, 4] \]