In a problem of differentiation of `(f(x))/(g(x))`, one student writes the derivative of `(f(x))/(g(x))` as `(f'(x))/(g'(x))` and he finds the correct result . If `g(x)=x^(2)and lim_(xtooo) f(x)=4` On the basis of above information , answer the following questions :
The function `f(x)` is

Options

A. a.`(4x)/((x-2))`
B. b.`(4x^(2))/((2-x)^(2))`
C. c.`(4x^(2))/(x^(2)+x+1)`
D. d.`(4x^(2))/((4-x)^(2))`

Answer

To find the function \( f(x) \) based on the given information, we will follow these steps:

### Step 1: Understand the given information
We know that:
- \( g(x) = x^2 \)
- \( \lim_{x \to \infty} f(x) = 4 \)

### Step 2: Apply the Quotient Rule for differentiation
The derivative of \( \frac{f(x)}{g(x)} \) is given by the quotient rule:
\[ \left( \frac{f(x)}{g(x)} \right)' = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \]
Substituting \( g(x) = x^2 \) and \( g'(x) = 2x \), we have:
\[ \left( \frac{f(x)}{x^2} \right)' = \frac{x^2 f'(x) - f(x) \cdot 2x}{(x^2)^2} = \frac{x^2 f'(x) - 2x f(x)}{x^4} \]

### Step 3: Analyze the student's incorrect differentiation
The student incorrectly wrote the derivative as:
\[ \frac{f'(x)}{g'(x)} = \frac{f'(x)}{2x} \]
This means they equated:
\[ \frac{x^2 f'(x) - 2x f(x)}{x^4} = \frac{f'(x)}{2x} \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives:
\[ 2x(x^2 f'(x) - 2x f(x)) = x^4 f'(x) \]
Expanding this results in:
\[ 2x^3 f'(x) - 4x^2 f(x) = x^4 f'(x) \]
Rearranging leads to:
\[ 2x^3 f'(x) - x^4 f'(x) = 4x^2 f(x) \]
Factoring out \( f'(x) \):
\[ f'(x)(2x^3 - x^4) = 4x^2 f(x) \]

### Step 5: Rearranging the equation
We can write:
\[ f'(x) = \frac{4x^2 f(x)}{x^4 - 2x^3} \]
This simplifies to:
\[ f'(x) = \frac{4 f(x)}{x(2 - x)} \]

### Step 6: Separate variables and integrate
Rearranging gives:
\[ \frac{f'(x)}{f(x)} = \frac{4}{x(2 - x)} \]
Integrating both sides:
\[ \int \frac{f'(x)}{f(x)} \, dx = \int \frac{4}{x(2 - x)} \, dx \]
The left side integrates to \( \ln |f(x)| \).

### Step 7: Solve the right side integral
Using partial fractions:
\[ \frac{4}{x(2 - x)} = \frac{A}{x} + \frac{B}{2 - x} \]
Solving gives \( A = 2 \) and \( B = 2 \):
\[ \int \left( \frac{2}{x} + \frac{2}{2 - x} \right) dx = 2 \ln |x| - 2 \ln |2 - x| + C \]

### Step 8: Combine results
Thus,
\[ \ln |f(x)| = 2 \ln |x| - 2 \ln |2 - x| + C \]
Exponentiating gives:
\[ f(x) = e^C \frac{x^2}{(2 - x)^2} \]
Letting \( e^C = k \), we have:
\[ f(x) = k \frac{x^2}{(2 - x)^2} \]

### Step 9: Use the limit to find \( k \)
Using the limit condition \( \lim_{x \to \infty} f(x) = 4 \):
\[ \lim_{x \to \infty} k \frac{x^2}{(2 - x)^2} = 4 \]
As \( x \to \infty \), \( f(x) \) approaches \( k \) since \( \frac{x^2}{x^2} \to 1 \):
\[ k = 4 \]

### Final Result
Thus, the function is:
\[ f(x) = 4 \frac{x^2}{(2 - x)^2} \]