If `f:R rarr (0, pi//2], f(x)=sin^(-1)((40)/(x^(2)+x+lambda))` is a surjective function, then the value of `lambda` is equal to

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the function \( f(x) = \sin^{-1}\left(\frac{40}{x^2 + x + \lambda}\right) \) is surjective from \( \mathbb{R} \) to \( (0, \frac{\pi}{2}] \).

### Step-by-step Solution:

1. **Understanding Surjectivity**:
A function \( f: A \to B \) is surjective (onto) if for every element \( b \in B \), there exists at least one \( a \in A \) such that \( f(a) = b \). In this case, we need every value in the interval \( (0, \frac{\pi}{2}] \) to be achieved by \( f(x) \).

2. **Range of the Function**:
The range of \( f(x) \) is determined by the expression \( \frac{40}{x^2 + x + \lambda} \). For \( f(x) \) to cover the entire interval \( (0, \frac{\pi}{2}] \), the expression \( \frac{40}{x^2 + x + \lambda} \) must take values in \( (0, 1] \) because \( \sin^{-1}(y) \) is defined for \( y \in [0, 1] \).

3. **Setting Up Inequalities**:
To ensure \( f(x) \) is surjective, we need:
\[ 0 < \frac{40}{x^2 + x + \lambda} \leq 1 \]
This implies:
\[ x^2 + x + \lambda \geq 40 \]

4. **Finding the Minimum Value**:
The expression \( x^2 + x + \lambda \) is a quadratic function in \( x \). The minimum value of a quadratic \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = 1 \), so:
\[ x = -\frac{1}{2} \]
Substituting \( x = -\frac{1}{2} \) into the quadratic:
\[ \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + \lambda = \frac{1}{4} - \frac{1}{2} + \lambda = \lambda - \frac{1}{4} \]

5. **Setting the Minimum Value**:
For \( f(x) \) to be surjective, we need:
\[ \lambda - \frac{1}{4} \geq 40 \]
This leads to:
\[ \lambda \geq 40 + \frac{1}{4} = 40.25 \]

6. **Conclusion**:
The minimum value of \( \lambda \) that makes \( f(x) \) surjective is:
\[ \lambda = 40.25 \]

### Final Answer:
The value of \( \lambda \) is \( 40.25 \).