If `cos^2x-(c-1)cosx+2cgeq6` for every `x in R ,` then the true set of values of `c` is (a) `(2,oo)` (b) `(4,oo)` (c) `(-oo,-2)` (d) `(-oo,-4)`

Options

A. `[2,oo)`
B. `[4,oo)`
C. (-oo,-2]`
D. (-oo,-4]`

Explanation

To solve the inequality \( \cos^2 x - (c - 1) \cos x + 2c \geq 6 \) for every \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Rearranging the Inequality
Start with the given inequality:
\[ \cos^2 x - (c - 1) \cos x + 2c \geq 6 \]
Rearranging gives:
\[ \cos^2 x - (c - 1) \cos x + 2c - 6 \geq 0 \]

### Step 2: Define a Quadratic in Terms of \( \cos x \)
Let \( y = \cos x \). The inequality can be rewritten as:
\[ y^2 - (c - 1)y + (2c - 6) \geq 0 \]
This is a quadratic inequality in \( y \).

### Step 3: Determine the Conditions for the Quadratic
For the quadratic \( ay^2 + by + c \) to be non-negative for all \( y \), the discriminant must be less than or equal to zero:
\[ D = b^2 - 4ac \leq 0 \]
Here, \( a = 1 \), \( b = -(c - 1) \), and \( c = 2c - 6 \).

Calculating the discriminant:
\[ D = (-(c - 1))^2 - 4 \cdot 1 \cdot (2c - 6) \leq 0 \]
\[ D = (c - 1)^2 - 4(2c - 6) \leq 0 \]

### Step 4: Simplifying the Discriminant
Expanding the discriminant:
\[ (c - 1)^2 - (8c - 24) \leq 0 \]
\[ c^2 - 2c + 1 - 8c + 24 \leq 0 \]
\[ c^2 - 10c + 25 \leq 0 \]

### Step 5: Factoring the Quadratic
This can be factored as:
\[ (c - 5)^2 \leq 0 \]
The only solution to this inequality is:
\[ c - 5 = 0 \implies c = 5 \]

### Step 6: Analyzing the Range of \( c \)
Since we need the quadratic to be non-negative for all \( y \), we also consider the maximum value of \( \cos x \), which is 1. Thus, we need:
\[ c - 3 \geq 1 \implies c \geq 4 \]

### Conclusion
Combining the conditions, we find that \( c \) must be greater than or equal to 4. Therefore, the true set of values of \( c \) is:
\[ c \in [4, \infty) \]

From the options provided, the correct answer is:
(b) \( (4, \infty) \)