If `f(x)=abs(x-1)+abs(x-4)+abs(x-9)+…+abs(x-2500)AA x inR`, where m and n are respectively the number of integral points at which f(x) is non-differentiable and f(x) has absolute minimum, then (m,n) is

Options

A. `(50,56)`
B. `(50,52)`
C. `(49,55)`
D. none of these

Answer

To solve the problem, we need to analyze the function \( f(x) = |x - 1| + |x - 4| + |x - 9| + \ldots + |x - 2500| \).

### Step 1: Identify the points of non-differentiability
The function \( f(x) \) consists of absolute value terms. Each term \( |x - a| \) is non-differentiable at \( x = a \). The points where \( f(x) \) is non-differentiable are the points where the arguments of the absolute values are zero.

The sequence of points from which we take absolute values is \( 1, 4, 9, \ldots, 2500 \). These points correspond to the squares of integers from \( 1^2 \) to \( 50^2 \) (since \( 2500 = 50^2 \)).

Thus, the points of non-differentiability are:
\[ 1^2, 2^2, 3^2, \ldots, 50^2 \]
This gives us a total of \( 50 \) points.

### Step 2: Determine the number of integral points where \( f(x) \) has an absolute minimum
The function \( f(x) \) will have an absolute minimum at the median of the points of non-differentiability. Since we have \( 50 \) points, the median will be between the \( 25^{th} \) and \( 26^{th} \) points.

The \( 25^{th} \) point is \( 25^2 = 625 \) and the \( 26^{th} \) point is \( 26^2 = 676 \).

Since \( f(x) \) is constant between these two points, the integral points where \( f(x) \) has an absolute minimum are all the integers from \( 625 \) to \( 676 \).

To find the number of integral points, we calculate:
\[ 676 - 625 + 1 = 52 \]
Thus, \( n = 52 \).

### Step 3: Conclusion
We have determined that:
- \( m \) (the number of integral points where \( f(x) \) is non-differentiable) is \( 50 \).
- \( n \) (the number of integral points where \( f(x) \) has an absolute minimum) is \( 52 \).

Therefore, the answer is \( (m, n) = (50, 52) \).

### Final Answer
\[ (m, n) = (50, 52) \]