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(D) Given the limit: $\mathop {\lim }\limits_{t 	o x} \frac{{{t^2}f(x) - {x^2}f(t)}}{{t - x}} = 1$. Applying $L$'Hopital's rule with respect to $t$:

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$\frac{31}{18}$

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(D) Given the limit: $\mathop {\lim }\limits_{t 	o x} \frac{{{t^2}f(x) - {x^2}f(t)}}{{t - x}} = 1$. Applying $L$'Hopital's rule with respect to $t$: $\mathop {\lim }\limits_{t 	o x} \frac{2t f(x) - x^2 f'(t)}{1} = 1$. Substituting $t = x$,we get the differential equation: $2x f(x) - x^2 f'(x) = 1$. Rearranging: $f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{1}{x^2}$. The integrating factor $I.F. = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$. Multiplying by $I.F.$: $\frac{d}{dx} [f(x) \cdot \frac{1}{x^2}] = -\frac{1}{x^4}$. Integrating both sides: $\frac{f(x)}{x^2} = \int -x^{-4} dx = \frac{1}{3x^3} + C$. So,$f(x) = \frac{1}{3x} + Cx^2$. Using $f(1) = 1$: $1 = \frac{1}{3} + C \implies C = \frac{2}{3}$. Thus,$f(x) = \frac{1}{3x} + \frac{2x^2}{3}$. For $x = \frac{3}{2}$: $f(\frac{3}{2}) = \frac{1}{3(3/2)} + \frac{2(3/2)^2}{3} = \frac{2}{9} + \frac{2}{3} \cdot \frac{9}{4} = \frac{2}{9} + \frac{3}{2} = \frac{4 + 27}{18} = \frac{31}{18}$.

If $f(x)$ is a differentiable function in the interval $(0, \infty)$ such that $f(1) = 1$ and $\mathop {\lim }\limits_{t \to x} \frac{{{t^2}f(x) - {x^2}f(t)}}{{t - x}} = 1$ for each $x > 0$,then $f(\frac{3}{2})$ is equal to

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