Let $f:R\rightarrow[0,\frac{\pi}{2})$ defined by $f(x)=tan^{-1}(x^{2}+x+2a),$ then f will be onto function, if :

Question

Accepted Answer

To determine why Option 4 is the correct answer, we need to analyze the function $ f(x) = \tan^{-1}(x^2 + x + 2a) $ and understand the conditions under which it is onto from $ \mathbb{R} $ to $[0, \frac{\pi}{2})$.

### Explanation:

1. **Understanding the Function:**
   - The function $ f(x) = \tan^{-1}(x^2 + x + 2a) $ maps real numbers $ x $ to the interval $[0, \frac{\pi}{2})$.
   - For $ f $ to be onto, for every $ y $ in $[0, \frac{\pi}{2})$, there must exist an $ x \in \mathbb{R} $ such that $ f(x) = y $.

2. **Range of the Argument of $\tan^{-1}$:**
   - The function $\tan^{-1}(u)$ covers the interval $[0, \frac{\pi}{2})$ as $ u $ ranges from $[0, \infty)$.
   - Therefore, $ x^2 + x + 2a $ must cover $[0, \infty)$ for $ f(x) $ to be onto.

3. **Analyzing $ x^2 + x + 2a $:**
   - The expression $ x^2 + x + 2a $ is a quadratic function in $ x $.
   - The minimum value of this quadratic occurs at the vertex, which is at $ x = -\frac{1}{2} $.
   - Substituting $ x = -\frac{1}{2} $ into the quadratic gives the minimum value as:
     $$
     \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 2a = \frac{1}{4} - \frac{1}{2} + 2a = 2a - \frac{1}{4}
     $$

4. **Condition for the Quadratic to Cover $[0, \infty)$:**
   - For the quadratic to start from 0 and extend to infinity, its minimum value must be 0 or less.
   - Therefore, we set:
     $$
     2a - \frac{1}{4} \geq 0
     $$
   - Solving this inequality gives:
     $$
     2a \geq \frac{1}{4} \quad \Rightarrow \quad a \geq \frac{1}{8}
     $$

5. **Conclusion:**
   - The condition $ a > \frac{1}{8} $ ensures that the quadratic $ x^2 + x + 2a $ starts from a non-negative value and covers $[0, \infty)$, making $ f(x) $ onto $[0, \frac{\pi}{2})$.

### Why Other Options are Incorrect:

- **Option 1: $ a \in (-\frac{1}{4}, \infty) $**
  - This range includes values of $ a $ less than $\frac{1}{8}$, which would not ensure the quadratic starts from a non-negative value.

- **Option 2: $ a \in [-\frac{1}{4}, \infty) $**
  - Similar to Option 1, this includes values of $ a $ that do not satisfy the condition $ a > \frac{1}{8} $.

- **Option 3: $ a \in (-\frac{1}{8}, \infty) $**
  - This range includes values of $ a $ less than $\frac{1}{8}$, which would not ensure the quadratic starts from a non-negative value.

Thus, Option 4 is the correct choice as it precisely satisfies the condition needed for $ f(x) $ to be onto.