If `f(x)` is differentiable function in the interval `(0,oo)` such that f(1) = 1 and `lim_(trarrx) (t^(2)f(x)-x^(2)(t))/(t-x)=1` for each `x gt 0`, then `f((3)/(2))` is equal tv

Options

A. `(13)/(6)`
B. `(23)/(18)`
C. `(25)/(9)`
D. `(31)/(18)`

Explanation

To solve the problem, we need to find the value of \( f\left(\frac{3}{2}\right) \) given the conditions provided.

### Step-by-Step Solution:

1. **Understanding the Limit Condition**:
We are given:
\[ \lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1 \]
This limit suggests that we can apply L'Hôpital's Rule since it is in the form \( \frac{0}{0} \) when \( t = x \).

2. **Applying L'Hôpital's Rule**:
Differentiate the numerator and denominator with respect to \( t \):
- The numerator:
\[ \frac{d}{dt}(t^2 f(x) - x^2 f(t)) = 2t f(x) - x^2 f'(t) \]
- The denominator:
\[ \frac{d}{dt}(t - x) = 1 \]
Thus, we have:
\[ \lim_{t \to x} (2t f(x) - x^2 f'(t)) = 1 \]

3. **Substituting \( t = x \)**:
Substituting \( t = x \) into the limit gives:
\[ 2x f(x) - x^2 f'(x) = 1 \]
Rearranging this, we get:
\[ x^2 f'(x) = 2x f(x) - 1 \]

4. **Rearranging the Differential Equation**:
Dividing through by \( x^2 \):
\[ f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2} \]
This is a first-order linear differential equation in standard form \( f' + p(x)f = q(x) \) where \( p(x) = -\frac{2}{x} \) and \( q(x) = -\frac{1}{x^2} \).

5. **Finding the Integrating Factor**:
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int -\frac{2}{x} dx} = e^{-2 \ln |x|} = |x|^{-2} = \frac{1}{x^2} \]

6. **Multiplying through by the Integrating Factor**:
Multiply the entire differential equation by \( \frac{1}{x^2} \):
\[ \frac{f(x)}{x^2} = \int -\frac{1}{x^2} \cdot \frac{1}{x^2} dx + C \]
This simplifies to:
\[ \frac{f(x)}{x^2} = \int -\frac{1}{x^4} dx + C \]

7. **Integrating the Right Side**:
The integral is:
\[ -\frac{1}{3x^3} + C \]
Thus:
\[ f(x) = x^2 \left(-\frac{1}{3x^3} + C\right) = -\frac{1}{3x} + Cx^2 \]

8. **Using the Initial Condition**:
We know \( f(1) = 1 \):
\[ 1 = -\frac{1}{3 \cdot 1} + C \cdot 1^2 \implies 1 = -\frac{1}{3} + C \]
Solving for \( C \):
\[ C = 1 + \frac{1}{3} = \frac{4}{3} \]

9. **Final Form of \( f(x) \)**:
Therefore, we have:
\[ f(x) = -\frac{1}{3x} + \frac{4}{3}x^2 \]

10. **Finding \( f\left(\frac{3}{2}\right) \)**:
Substitute \( x = \frac{3}{2} \):
\[ f\left(\frac{3}{2}\right) = -\frac{1}{3 \cdot \frac{3}{2}} + \frac{4}{3}\left(\frac{3}{2}\right)^2 \]
Simplifying:
\[ = -\frac{2}{9} + \frac{4}{3} \cdot \frac{9}{4} = -\frac{2}{9} + 3 \]
Converting \( 3 \) to a fraction:
\[ = -\frac{2}{9} + \frac{27}{9} = \frac{25}{9} \]

### Final Answer:
Thus, \( f\left(\frac{3}{2}\right) = \frac{25}{9} \).