If range of the function `f(x) = tan^(-1) ( 3x^(2) + bx + 3) , x in R " is " [0, pi/2)`, then square of sum of all possible values of b will be

Options

A. 0
B. 18
C. 72
D. None of these

Answer

To solve the problem, we need to ensure that the function \( f(x) = \tan^{-1}(3x^2 + bx + 3) \) has a range of \([0, \frac{\pi}{2})\). This means that the expression inside the inverse tangent must be non-negative and must approach infinity as \( x \) approaches some value.

### Step 1: Analyze the function
The function \( f(x) \) is defined as:
\[ f(x) = \tan^{-1}(3x^2 + bx + 3) \]
For the range to be \([0, \frac{\pi}{2})\), the argument \( 3x^2 + bx + 3 \) must be non-negative for all \( x \in \mathbb{R} \) and must approach infinity as \( x \) approaches infinity.

### Step 2: Ensure non-negativity
The quadratic \( 3x^2 + bx + 3 \) must not have any real roots, which means its discriminant must be less than or equal to zero:
\[ D = b^2 - 4ac = b^2 - 4 \cdot 3 \cdot 3 = b^2 - 36 \]
Setting the discriminant less than or equal to zero gives:
\[ b^2 - 36 \leq 0 \]
This simplifies to:
\[ b^2 \leq 36 \]
Thus, we have:
\[ -6 \leq b \leq 6 \]

### Step 3: Find the sum of possible values of \( b \)
The possible integer values of \( b \) within the range \([-6, 6]\) are:
\[ -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 \]
The sum of these values can be calculated as follows:
\[ \text{Sum} = (-6) + (-5) + (-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 + 6 \]
Grouping the terms:
\[ = (-6 + 6) + (-5 + 5) + (-4 + 4) + (-3 + 3) + (-2 + 2) + (-1 + 1) + 0 = 0 \]

### Step 4: Calculate the square of the sum
The square of the sum of all possible values of \( b \) is:
\[ 0^2 = 0 \]

### Final Answer
The square of the sum of all possible values of \( b \) is:
\[ \boxed{0} \]