If the range of the function `F(x)=tan^-1(3x^2+bx+c)` is `[0,pi/2)` ; (domain in R) then :

Options

A. `b ^(2) =3c`
B. `b ^(2) =4c`
C. `b ^(2) =12c`
D. `b ^(2) =8c`

Answer

To solve the problem, we need to analyze the function \( F(x) = \tan^{-1}(3x^2 + bx + c) \) and determine the conditions under which its range is \( [0, \frac{\pi}{2}) \).

### Step-by-Step Solution:

1. **Understanding the Range**:
The range of \( F(x) \) is given as \( [0, \frac{\pi}{2}) \). This implies that:
\[ 0 \leq F(x) < \frac{\pi}{2} \]

2. **Applying the Arctangent Function**:
Since \( F(x) = \tan^{-1}(3x^2 + bx + c) \), we can express the inequalities in terms of the tangent function:
\[ \tan(0) \leq 3x^2 + bx + c < \tan\left(\frac{\pi}{2}\right) \]
This simplifies to:
\[ 0 \leq 3x^2 + bx + c < \infty \]

3. **Analyzing the Quadratic Expression**:
The quadratic \( 3x^2 + bx + c \) must be non-negative for all \( x \) in \( \mathbb{R} \). For a quadratic function \( ax^2 + bx + c \) to be non-negative for all \( x \), the following conditions must be satisfied:
- The coefficient of \( x^2 \) (which is 3 here) must be positive.
- The discriminant must be less than or equal to zero.

4. **Condition on Coefficient**:
Since the coefficient of \( x^2 \) is 3 (which is positive), this condition is satisfied.

5. **Finding the Discriminant**:
The discriminant \( D \) of the quadratic \( 3x^2 + bx + c \) is given by:
\[ D = b^2 - 4ac = b^2 - 4(3)(c) = b^2 - 12c \]
For the quadratic to be non-negative, we require:
\[ D \leq 0 \implies b^2 - 12c \leq 0 \]
This leads to:
\[ b^2 \leq 12c \]

6. **Conclusion**:
The final condition we derived is:
\[ b^2 \leq 12c \]
This means that for the function \( F(x) \) to have the range \( [0, \frac{\pi}{2}) \), the values of \( b \) and \( c \) must satisfy this inequality.