If `a_(1), a_(2),…….a_(n)` are real numbers and the function `f(x)=(x-a_(1))^(2)+(x-a_(2))^(2)+…..+(x-a_(n))^(2)` attains its minimum value for some x = p, then p is

Options

A. harmonic mean of `a_(1), a_(2), ……….., a_(n)`
B. geometric mean of `a_(1), a_(2), …a_(n)`
C. arithmetic mean of `a_(1), a_(2), ……., a_(n)`
D. none of these

Answer

To solve the problem, we need to find the value of \( p \) for which the function

\[ f(x) = (x - a_1)^2 + (x - a_2)^2 + \ldots + (x - a_n)^2 \]

attains its minimum value.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) \) is a sum of squares of the differences between \( x \) and the real numbers \( a_1, a_2, \ldots, a_n \). This function is continuous and differentiable everywhere.

2. **Finding the First Derivative**:
We differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = 2(x - a_1) + 2(x - a_2) + \ldots + 2(x - a_n) \]
Simplifying this, we have:
\[ f'(x) = 2\left[(x - a_1) + (x - a_2) + \ldots + (x - a_n)\right] \]
\[ f'(x) = 2\left[nx - (a_1 + a_2 + \ldots + a_n)\right] \]

3. **Setting the First Derivative to Zero**:
To find the critical points, we set \( f'(x) = 0 \):
\[ 2\left[nx - (a_1 + a_2 + \ldots + a_n)\right] = 0 \]
This simplifies to:
\[ nx = a_1 + a_2 + \ldots + a_n \]
Therefore, we can solve for \( x \):
\[ x = \frac{a_1 + a_2 + \ldots + a_n}{n} \]

4. **Identifying the Minimum**:
The value we found, \( x = \frac{a_1 + a_2 + \ldots + a_n}{n} \), is the arithmetic mean of the numbers \( a_1, a_2, \ldots, a_n \).

5. **Second Derivative Test**:
To confirm that this point is indeed a minimum, we can find the second derivative:
\[ f''(x) = 2n \]
Since \( f''(x) = 2n > 0 \) for all \( n > 0 \), this indicates that the function is concave up at this point, confirming that it is a minimum.

### Conclusion:
Thus, the value of \( p \) at which the function \( f(x) \) attains its minimum value is:

\[ p = \frac{a_1 + a_2 + \ldots + a_n}{n} \]

This is the arithmetic mean of the numbers \( a_1, a_2, \ldots, a_n \).