Let `f : R to [0, pi/2)` be defined by `f ( x) = tan^(-1) ( 3x^(2) + 6x + a)". If " f(x)` is an onto function . then the value of a si

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to determine the value of \( a \) such that the function \( f(x) = \tan^{-1}(3x^2 + 6x + a) \) is an onto function from \( \mathbb{R} \) to \( [0, \frac{\pi}{2}) \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = \tan^{-1}(3x^2 + 6x + a) \) maps real numbers to the interval \( [0, \frac{\pi}{2}) \). For \( f(x) \) to be onto, the expression inside the arctangent must cover all values from \( 0 \) to \( +\infty \).

2. **Finding the Quadratic Expression**:
The expression \( 3x^2 + 6x + a \) is a quadratic function in \( x \). We can analyze its behavior by determining its minimum value.

3. **Finding the Vertex**:
The vertex of a quadratic \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 3 \) and \( b = 6 \):
\[ x = -\frac{6}{2 \cdot 3} = -1 \]

4. **Calculating the Minimum Value**:
Substitute \( x = -1 \) back into the quadratic:
\[ 3(-1)^2 + 6(-1) + a = 3 - 6 + a = a - 3 \]
The minimum value of \( 3x^2 + 6x + a \) is \( a - 3 \).

5. **Setting the Minimum Value**:
For \( f(x) \) to be onto, the minimum value \( a - 3 \) must be \( 0 \) (since \( f(x) \) must reach \( 0 \)):
\[ a - 3 = 0 \implies a = 3 \]

6. **Conclusion**:
The value of \( a \) that makes \( f(x) \) onto is:
\[ \boxed{3} \]