Let `f(x)=lim_(nto oo) 1/n((x+1/n)^(2)+(x+2/n)^(2)+……….+(x+(n-1)/n)^(2))`
Then the minimum value of `f(x)` is

Options

A. `1//4`
B. `1//6`
C. `1//9`
D. `1//12`

Explanation

To find the minimum value of the function \( f(x) \) defined as

\[ f(x) = \lim_{n \to \infty} \frac{1}{n} \left( \left( x + \frac{1}{n} \right)^2 + \left( x + \frac{2}{n} \right)^2 + \ldots + \left( x + \frac{n-1}{n} \right)^2 \right), \]

we will follow these steps:

### Step 1: Rewrite the expression
We can rewrite the sum inside the limit:

\[ f(x) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left( x + \frac{k}{n} \right)^2. \]

### Step 2: Expand the square
Expanding the square gives:

\[ f(x) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left( x^2 + 2x \frac{k}{n} + \left( \frac{k}{n} \right)^2 \right). \]

### Step 3: Separate the terms
We can separate the sum:

\[ f(x) = \lim_{n \to \infty} \left( \frac{1}{n} \sum_{k=1}^{n} x^2 + \frac{2x}{n} \sum_{k=1}^{n} \frac{k}{n} + \frac{1}{n} \sum_{k=1}^{n} \left( \frac{k}{n} \right)^2 \right). \]

### Step 4: Calculate each term
1. The first term:
\[ \frac{1}{n} \sum_{k=1}^{n} x^2 = x^2. \]

2. The second term:
\[ \frac{2x}{n} \sum_{k=1}^{n} \frac{k}{n} = \frac{2x}{n^2} \cdot \frac{n(n+1)}{2} \to x \quad \text{as } n \to \infty. \]

3. The third term:
\[ \frac{1}{n} \sum_{k=1}^{n} \left( \frac{k}{n} \right)^2 = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \to \frac{1}{3} \quad \text{as } n \to \infty. \]

### Step 5: Combine the results
Putting all these together:

\[ f(x) = x^2 + x + \frac{1}{3}. \]

### Step 6: Find the minimum value
To find the minimum value of \( f(x) = x^2 + x + \frac{1}{3} \), we can complete the square:

\[ f(x) = \left( x + \frac{1}{2} \right)^2 - \frac{1}{4} + \frac{1}{3} = \left( x + \frac{1}{2} \right)^2 + \frac{1}{12}. \]

The minimum value occurs when \( \left( x + \frac{1}{2} \right)^2 = 0 \), which gives \( x = -\frac{1}{2} \).

Thus, the minimum value of \( f(x) \) is:

\[ f\left(-\frac{1}{2}\right) = 0 + \frac{1}{12} = \frac{1}{12}. \]

### Final Answer:
The minimum value of \( f(x) \) is \( \frac{1}{12} \).

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