Let f(x) be a function defined by
`f(x) =(ab -a^2-2)x -underset(0)overset(x)(cos^4 t + sin^2t-2)dt `
If (x) is a decreasing function for all `x in R ` and a in R where a is independent of x, then

Options

A. a. `be in (1,oo)`
B. b. `b in (-1,1)`
C. c. `b in (-oo,]`
D. d. non of these

Explanation

To determine the conditions under which the function \( f(x) \) is decreasing for all \( x \in \mathbb{R} \), we will analyze the given function step by step.

### Step 1: Define the Function
The function is given by:
\[ f(x) = (ab - a^2 - 2)x - \int_0^x (\cos^4 t + \sin^2 t - 2) \, dt \]

### Step 2: Differentiate the Function
To find when \( f(x) \) is decreasing, we need to find the derivative \( f'(x) \):
\[ f'(x) = ab - a^2 - 2 - (\cos^4 x + \sin^2 x - 2) \]
Here, we used the Fundamental Theorem of Calculus to differentiate the integral.

### Step 3: Simplify the Derivative
Now, we simplify \( f'(x) \):
\[ f'(x) = ab - a^2 - 2 - \cos^4 x - \sin^2 x + 2 \]
This simplifies to:
\[ f'(x) = ab - a^2 - \cos^4 x - \sin^2 x \]

### Step 4: Set the Condition for Decreasing Function
For \( f(x) \) to be a decreasing function, we need:
\[ f'(x) < 0 \]
Thus, we require:
\[ ab - a^2 - \cos^4 x - \sin^2 x < 0 \]

### Step 5: Analyze the Trigonometric Terms
We know that \( \cos^4 x + \sin^2 x \) can be bounded. The maximum value of \( \cos^4 x \) is 1 (when \( \cos x = 1 \)) and the maximum value of \( \sin^2 x \) is also 1 (when \( \sin x = 1 \)). Therefore:
\[ \cos^4 x + \sin^2 x \leq 1 + 1 = 2 \]

### Step 6: Substitute the Bound into the Inequality
Substituting this bound into our inequality gives:
\[ ab - a^2 - 2 < 0 \]
This simplifies to:
\[ ab - a^2 < 2 \]

### Step 7: Rearranging the Inequality
Rearranging gives us:
\[ ab < a^2 + 2 \]

### Step 8: Conclusion
Thus, the condition for \( f(x) \) to be a decreasing function for all \( x \in \mathbb{R} \) is:
\[ ab < a^2 + 2 \]