If f(x)=`-x^(3)-3x^(2)-2x+a,a in R` then the real values of x satisfying `f(x^(2)+1)gtf(2x^(2)+2x+3)` will be

Options

A. `(-oo,oo)`
B. `(0,oo)`
C. `-(oo,0)`
D. `phi`

Explanation

To solve the inequality \( f(x^2 + 1) > f(2x^2 + 2x + 3) \) where \( f(x) = -x^3 - 3x^2 - 2x + a \) and \( a \in \mathbb{R} \), we will follow these steps:

### Step 1: Define the function
The function is given as:
\[ f(x) = -x^3 - 3x^2 - 2x + a \]

### Step 2: Find the derivative of the function
To analyze the monotonicity of \( f(x) \), we need to find its derivative:
\[ f'(x) = \frac{d}{dx}(-x^3 - 3x^2 - 2x + a) = -3x^2 - 6x - 2 \]

### Step 3: Analyze the derivative
The derivative \( f'(x) = -3x^2 - 6x - 2 \) is a quadratic function that opens downwards (since the coefficient of \( x^2 \) is negative). To find the roots of this quadratic, we can use the discriminant:
\[ D = b^2 - 4ac = (-6)^2 - 4(-3)(-2) = 36 - 24 = 12 \]
Since the discriminant is positive, there are two real roots. We can find the roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{12}}{-6} = \frac{6 \pm 2\sqrt{3}}{-6} = -1 \mp \frac{\sqrt{3}}{3} \]
This means \( f'(x) < 0 \) for all \( x \) in the intervals outside the roots, indicating that \( f(x) \) is decreasing in those intervals.

### Step 4: Set up the inequality
We need to solve:
\[ f(x^2 + 1) > f(2x^2 + 2x + 3) \]
Since \( f(x) \) is decreasing, this inequality holds if:
\[ x^2 + 1 < 2x^2 + 2x + 3 \]

### Step 5: Simplify the inequality
Rearranging the inequality:
\[ x^2 + 1 < 2x^2 + 2x + 3 \]
\[ 0 < 2x^2 + 2x + 3 - x^2 - 1 \]
\[ 0 < x^2 + 2x + 2 \]

### Step 6: Analyze the quadratic inequality
The quadratic \( x^2 + 2x + 2 \) can be analyzed using its discriminant:
\[ D = 2^2 - 4(1)(2) = 4 - 8 = -4 \]
Since the discriminant is negative, \( x^2 + 2x + 2 \) has no real roots and is always positive for all \( x \in \mathbb{R} \).

### Conclusion
Thus, the inequality \( f(x^2 + 1) > f(2x^2 + 2x + 3) \) holds for all real values of \( x \).

### Final Answer
The real values of \( x \) satisfying the inequality are:
\[ \text{All } x \in \mathbb{R} \]