If `f:R^(+)rarrA`, where `A={x:-5ltxltoo}` is defined by f(x) = `x^(2)` - 5 and if
`f^(-1)(13)={-lambdasqrt((lambda-1)),lambdasqrt((lambda-1))}`, the value of `lambda` is

Explanation

To solve the problem, we need to find the value of \( \lambda \) given the function \( f(x) = x^2 - 5 \) and the condition \( f^{-1}(13) = \{-\lambda \sqrt{\lambda - 1}, \lambda \sqrt{\lambda - 1}\} \).

### Step 1: Set up the equation
We start by finding the inverse function value \( f^{-1}(13) \). Since \( f(x) = x^2 - 5 \), we set it equal to 13:
\[ x^2 - 5 = 13 \]

### Step 2: Solve for \( x \)
Now, we solve for \( x \):
\[ x^2 = 13 + 5 \]
\[ x^2 = 18 \]
Taking the square root of both sides, we get:
\[ x = \pm \sqrt{18} \]
This can be simplified to:
\[ x = \pm 3\sqrt{2} \]

### Step 3: Express in terms of \( \lambda \)
According to the problem, we have:
\[ f^{-1}(13) = \{-\lambda \sqrt{\lambda - 1}, \lambda \sqrt{\lambda - 1}\} \]
This means we can equate the values we found to the expressions involving \( \lambda \):
\[ 3\sqrt{2} = \lambda \sqrt{\lambda - 1} \]

### Step 4: Square both sides
To eliminate the square root, we square both sides:
\[ (3\sqrt{2})^2 = (\lambda \sqrt{\lambda - 1})^2 \]
This gives:
\[ 18 = \lambda^2 (\lambda - 1) \]

### Step 5: Rearrange the equation
Expanding the right side:
\[ 18 = \lambda^3 - \lambda^2 \]
Rearranging gives us:
\[ \lambda^3 - \lambda^2 - 18 = 0 \]

### Step 6: Solve the cubic equation
Now we can try to find the roots of the cubic equation \( \lambda^3 - \lambda^2 - 18 = 0 \). We can use the Rational Root Theorem or trial and error to find a suitable root. Testing \( \lambda = 3 \):
\[ 3^3 - 3^2 - 18 = 27 - 9 - 18 = 0 \]
Thus, \( \lambda = 3 \) is a root.

### Conclusion
The value of \( \lambda \) is:
\[ \lambda = 3 \]