Let `A={x: 0 le x lt pi//2} and f:R to A` be an onto function given by `f(x)=tan^(-1)(x^(2)+x+lambda)`, where `lambda` is a constant. Then,

Options

A. `lambda gt 0`
B. `lambda ge 1//4`
C. `lambda lt 1//4`
D. `0 le lambda le 1`

Explanation

To solve the problem, we need to find the value of the constant \(\lambda\) such that the function \(f(x) = \tan^{-1}(x^2 + x + \lambda)\) is onto from \(\mathbb{R}\) to the interval \(A = [0, \frac{\pi}{2})\).

### Step 1: Understanding the function and its range
The function \(f(x) = \tan^{-1}(x^2 + x + \lambda)\) maps real numbers to the interval \([0, \frac{\pi}{2})\). For \(f\) to be onto, the expression \(x^2 + x + \lambda\) must cover all values from \(0\) to \(\infty\) as \(x\) varies over \(\mathbb{R}\).

### Step 2: Analyzing the expression \(x^2 + x + \lambda\)
The quadratic expression \(x^2 + x + \lambda\) is a parabola that opens upwards (since the coefficient of \(x^2\) is positive). We need to ensure that this expression is non-negative for all \(x \in \mathbb{R}\).

### Step 3: Finding the minimum value of the quadratic
To find the minimum value of the quadratic \(x^2 + x + \lambda\), we can use the vertex formula. The \(x\)-coordinate of the vertex is given by:
\[ x = -\frac{b}{2a} = -\frac{1}{2} \]
Substituting \(x = -\frac{1}{2}\) into the quadratic gives:
\[ \text{Minimum value} = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + \lambda = \frac{1}{4} - \frac{1}{2} + \lambda = \lambda - \frac{1}{4} \]

### Step 4: Ensuring the minimum value is non-negative
For the quadratic to be non-negative for all \(x\), we need:
\[ \lambda - \frac{1}{4} \geq 0 \]
This simplifies to:
\[ \lambda \geq \frac{1}{4} \]

### Step 5: Conclusion
Thus, the value of \(\lambda\) must be at least \(\frac{1}{4}\) for the function \(f\) to be onto from \(\mathbb{R}\) to \([0, \frac{\pi}{2})\).

### Final Answer
\[ \lambda \geq \frac{1}{4} \]