Let `f:R rarr [0, (pi)/(2))` be a function defined by `f(x)=tan^(-1)(x^(2)+x+a)`. If f is onto, then a is equal to

Options

A. 0
B. 1
C. `1//2`
D. `1//4`

Answer

To solve the problem, we need to determine the value of \( a \) such that the function \( f(x) = \tan^{-1}(x^2 + x + a) \) is onto, meaning it covers the entire range from \( 0 \) to \( \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = \tan^{-1}(x^2 + x + a) \) maps real numbers \( x \) to the interval \( [0, \frac{\pi}{2}) \). The output of the function approaches \( \frac{\pi}{2} \) as its argument approaches infinity.

2. **Condition for Onto Function**:
For \( f \) to be onto, the expression \( x^2 + x + a \) must be non-negative for all \( x \) in \( \mathbb{R} \). This ensures that \( \tan^{-1}(x^2 + x + a) \) can take all values from \( 0 \) to \( \frac{\pi}{2} \).

3. **Finding the Discriminant**:
The quadratic expression \( x^2 + x + a \) can be analyzed using its discriminant:
\[ D = b^2 - 4ac = 1^2 - 4(1)(a) = 1 - 4a \]
For the quadratic to be non-negative for all \( x \), the discriminant must be less than or equal to zero:
\[ D \leq 0 \implies 1 - 4a \leq 0 \]

4. **Solving the Inequality**:
Rearranging the inequality gives:
\[ 1 \leq 4a \implies a \geq \frac{1}{4} \]

5. **Conclusion**:
The minimum value of \( a \) that satisfies the condition for \( f \) to be onto is \( a = \frac{1}{4} \).

Thus, the value of \( a \) is:
\[ \boxed{\frac{1}{4}} \]