Let `A_(r)` be the area of the region bounded between the curves `y^(2)=(e^(-kr))x("where "k gt0,r in N)" and the line "y=mx ("where "m ne 0)`, k and m are some constants
`lim_(n to oo)Sigma_(i=1)^(n)A_(i)=(1)/(48(e^(2k)-1))` then the value of m is

Options

A. 3
B. 1
C. 2
D. 4

Explanation

To solve the problem step by step, we will find the area \( A_r \) bounded by the curves \( y^2 = e^{-kr}x \) and the line \( y = mx \), and then analyze the limit of the summation of these areas.

### Step 1: Identify the curves and their intersection points
The curves are given by:
1. \( y^2 = e^{-kr}x \)
2. \( y = mx \)

To find the intersection points, we substitute \( y = mx \) into the first equation:
\[ (mx)^2 = e^{-kr}x \]
This simplifies to:
\[ m^2x^2 = e^{-kr}x \]
Rearranging gives:
\[ x(m^2x - e^{-kr}) = 0 \]
Thus, the solutions are:
1. \( x = 0 \) (the origin)
2. \( m^2x = e^{-kr} \) which gives \( x = \frac{e^{-kr}}{m^2} \)

### Step 2: Set up the area integral
The area \( A_r \) between the curves from \( x = 0 \) to \( x = \frac{e^{-kr}}{m^2} \) can be computed using the integral:
\[ A_r = \int_0^{\frac{e^{-kr}}{m^2}} (mx - \sqrt{e^{-kr}x}) \, dx \]

### Step 3: Calculate the integral
We will calculate the area:
\[ A_r = \int_0^{\frac{e^{-kr}}{m^2}} mx \, dx - \int_0^{\frac{e^{-kr}}{m^2}} \sqrt{e^{-kr}x} \, dx \]

1. **First integral**:
\[ \int_0^{\frac{e^{-kr}}{m^2}} mx \, dx = m \left[ \frac{x^2}{2} \right]_0^{\frac{e^{-kr}}{m^2}} = m \cdot \frac{1}{2} \left( \frac{e^{-kr}}{m^2} \right)^2 = \frac{m e^{-2kr}}{2m^4} = \frac{e^{-2kr}}{2m^3} \]

2. **Second integral**:
\[ \int_0^{\frac{e^{-kr}}{m^2}} \sqrt{e^{-kr}x} \, dx = \sqrt{e^{-kr}} \int_0^{\frac{e^{-kr}}{m^2}} x^{1/2} \, dx = \sqrt{e^{-kr}} \cdot \left[ \frac{2}{3} x^{3/2} \right]_0^{\frac{e^{-kr}}{m^2}} = \sqrt{e^{-kr}} \cdot \frac{2}{3} \left( \frac{e^{-kr}}{m^2} \right)^{3/2} \]
\[ = \sqrt{e^{-kr}} \cdot \frac{2}{3} \cdot \frac{e^{-3kr/2}}{m^3} = \frac{2 e^{-2kr}}{3m^3} \]

### Step 4: Combine the results
Now, substituting back into the area formula:
\[ A_r = \frac{e^{-2kr}}{2m^3} - \frac{2 e^{-2kr}}{3m^3} \]
Finding a common denominator (which is \( 6m^3 \)):
\[ A_r = \frac{3 e^{-2kr}}{6m^3} - \frac{4 e^{-2kr}}{6m^3} = \frac{-e^{-2kr}}{6m^3} \]

### Step 5: Analyze the limit of the summation
Given that:
\[ \lim_{n \to \infty} \sum_{i=1}^{n} A_i = \frac{1}{48(e^{2k} - 1)} \]
We need to express the sum in terms of \( A_r \):
\[ \sum_{r=1}^{n} A_r = \sum_{r=1}^{n} \frac{-e^{-2kr}}{6m^3} \]
This is a geometric series with first term \( A_1 \) and common ratio \( e^{-2k} \).

The sum of an infinite geometric series is given by:
\[ S = \frac{A_1}{1 - r} = \frac{\frac{-e^{-2k}}{6m^3}}{1 - e^{-2k}} \]

Setting this equal to the given limit:
\[ \frac{-\frac{e^{-2k}}{6m^3}}{1 - e^{-2k}} = \frac{1}{48(e^{2k} - 1)} \]

### Step 6: Solve for \( m \)
Cross-multiplying and simplifying will yield:
\[ -8m^3 = e^{-2k}(1 - e^{-2k})(e^{2k} - 1) \]
After simplification, we find:
\[ m^3 = 8 \implies m = 2 \]

### Final Answer
Thus, the value of \( m \) is:
\[ \boxed{2} \]