Let `f(x)=x^2 + 3x-3,x leq 0`. If n points `x_1, x_2, x_3,.....,x_n` are so chosen on the x-axis such that 1/n summationf^-1(xi)=f(1/n summation i=1 to n(xi)) (2) summationi=1 to n f^-1(xi)=summation i=1 to n (xi,)where f^-1 denotes the inverse of f, Then the AM of xi's is a)1 b)2 c)3 d)4

Answer

To solve the problem, we need to analyze the given function and the conditions provided. Let's break it down step by step.

### Step 1: Understand the Function
The function given is:
\[ f(x) = x^2 + 3x - 3 \]
for \( x \leq 0 \).

### Step 2: Find the Inverse Function \( f^{-1}(x) \)
To find the inverse, we first set \( y = f(x) \):
\[ y = x^2 + 3x - 3 \]
Rearranging gives:
\[ x^2 + 3x - (y + 3) = 0 \]
This is a quadratic equation in \( x \). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 3, c = -(y + 3) \).

Calculating the discriminant:
\[ b^2 - 4ac = 3^2 - 4(1)(-(y + 3)) = 9 + 4(y + 3) = 4y + 21 \]

Thus, the inverse function is:
\[ f^{-1}(y) = \frac{-3 \pm \sqrt{4y + 21}}{2} \]
Since we are considering \( x \leq 0 \), we take the negative root:
\[ f^{-1}(y) = \frac{-3 - \sqrt{4y + 21}}{2} \]

### Step 3: Apply the Given Conditions
We have two conditions:
1. \( \frac{1}{n} \sum_{i=1}^{n} f^{-1}(x_i) = f\left(\frac{1}{n} \sum_{i=1}^{n} x_i\right) \)
2. \( \sum_{i=1}^{n} f^{-1}(x_i) = \sum_{i=1}^{n} x_i \)

Let \( t = \frac{1}{n} \sum_{i=1}^{n} x_i \). Then from the first condition, we have:
\[ \frac{1}{n} \sum_{i=1}^{n} f^{-1}(x_i) = f(t) \]

### Step 4: Calculate \( f(t) \)
Substituting \( t \) into the function:
\[ f(t) = t^2 + 3t - 3 \]

### Step 5: Substitute into the First Condition
From the first condition:
\[ \frac{1}{n} \sum_{i=1}^{n} f^{-1}(x_i) = f(t) \]
This implies:
\[ \sum_{i=1}^{n} f^{-1}(x_i) = n f(t) \]

### Step 6: Substitute into the Second Condition
From the second condition:
\[ \sum_{i=1}^{n} f^{-1}(x_i) = \sum_{i=1}^{n} x_i \]
Thus, we have:
\[ n f(t) = \sum_{i=1}^{n} x_i \]

### Step 7: Equate the Two Expressions
Setting the two expressions for \( \sum_{i=1}^{n} f^{-1}(x_i) \) equal gives:
\[ n f(t) = n t \]
This simplifies to:
\[ f(t) = t \]

### Step 8: Solve for \( t \)
Substituting \( f(t) \):
\[ t^2 + 3t - 3 = t \]
Rearranging gives:
\[ t^2 + 2t - 3 = 0 \]

### Step 9: Factor the Quadratic
Factoring:
\[ (t - 1)(t + 3) = 0 \]
Thus, \( t = 1 \) or \( t = -3 \). Since \( t \) represents the average of \( x_i \) and must be non-negative, we have:
\[ t = 1 \]

### Conclusion
The arithmetic mean of the \( x_i \)'s is:
\[ \text{AM} = t = 1 \]

### Final Answer
The arithmetic mean of \( x_i \)'s is \( \boxed{1} \).