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 determine 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 \( \left(0, \frac{\pi}{2}\right] \).

### Step 1: Understand the function and its range
The function \( f(x) \) maps real numbers \( x \) to the interval \( \left(0, \frac{\pi}{2}\right] \). For \( f(x) \) to be surjective, the range of \( f(x) \) must equal the co-domain \( \left(0, \frac{\pi}{2}\right] \).

### Step 2: Set up the equation
Let \( y = f(x) \). Then we have:
\[ y = \sin^{-1}\left(\frac{40}{x^2 + x + \lambda}\right) \]
Taking the sine of both sides gives:
\[ \sin y = \frac{40}{x^2 + x + \lambda} \]
Rearranging this, we find:
\[ x^2 + x + \lambda = \frac{40}{\sin y} \]

### Step 3: Analyze the quadratic equation
Rearranging gives us a quadratic equation in \( x \):
\[ x^2 + x + \left(\lambda - \frac{40}{\sin y}\right) = 0 \]
For this quadratic to have real solutions for all \( y \in \left(0, \frac{\pi}{2}\right] \), the discriminant must be non-negative.

### Step 4: Calculate the discriminant
The discriminant \( D \) of the quadratic \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
For our equation:
- \( a = 1 \)
- \( b = 1 \)
- \( c = \lambda - \frac{40}{\sin y} \)

Thus, the discriminant is:
\[ D = 1^2 - 4 \cdot 1 \cdot \left(\lambda - \frac{40}{\sin y}\right) = 1 - 4\lambda + \frac{160}{\sin y} \]
For \( f(x) \) to be surjective, we require:
\[ D \geq 0 \]
This leads to:
\[ 1 - 4\lambda + \frac{160}{\sin y} \geq 0 \]

### Step 5: Find the critical points
Rearranging gives:
\[ \frac{160}{\sin y} \geq 4\lambda - 1 \]
This can be rewritten as:
\[ \sin y \leq \frac{160}{4\lambda - 1} \]
The maximum value of \( \sin y \) is 1 (when \( y = \frac{\pi}{2} \)), hence:
\[ 1 \leq \frac{160}{4\lambda - 1} \]
This implies:
\[ 4\lambda - 1 \geq 160 \]
Solving for \( \lambda \):
\[ 4\lambda \geq 161 \quad \Rightarrow \quad \lambda \geq \frac{161}{4} = 40.25 \]

### Conclusion
Thus, the value of \( \lambda \) must be:
\[ \lambda = 40.25 \]