If $$\int_0^x f(t)dt = x^2 + \int_x^1 t^2 f(t)dt$$, then $$f'\left(\frac{1}{2}\right)$$ is:

We start with the given functional equation

$$\int_{0}^{x} f(t)\,dt \;=\; x^{2} \;+\; \int_{x}^{1} t^{2}\,f(t)\,dt.$$

To obtain an explicit expression for $$f(x)$$, we differentiate both sides with respect to the variable $$x$$. We recall the standard results:

1. If $$F(x)=\displaystyle \int_{0}^{x} g(t)\,dt$$, then $$\dfrac{dF}{dx}=g(x).$$

2. If $$G(x)=\displaystyle \int_{x}^{1} g(t)\,dt,$$ then $$\dfrac{dG}{dx}=-g(x)$$ (because the lower limit depends on $$x$$).

Applying these, the left-hand side differentiates to $$f(x).$$ On the right-hand side, the derivative of $$x^{2}$$ is $$2x,$$ and the derivative of $$\int_{x}^{1} t^{2} f(t)\,dt$$ is $$-x^{2} f(x).$$ Hence

$$f(x) \;=\; 2x \;-\; x^{2}\,f(x).$$

We collect the $$f(x)$$ terms on one side:

$$f(x) \;+\; x^{2}\,f(x) \;=\; 2x,$$

so

$$f(x)\,(1 + x^{2}) \;=\; 2x.$$

Dividing by $$1 + x^{2}$$ gives

$$f(x) \;=\; \frac{2x}{\,1 + x^{2}\,}.$$

Next, we need $$f'(x).$$ We differentiate $$f(x)=\dfrac{2x}{1+x^{2}}$$ using the quotient rule. For a quotient $$\dfrac{u}{v},$$ the derivative is $$\dfrac{u'v - uv'}{v^{2}}.$$ Here $$u=2x$$ with $$u'=2$$ and $$v=1+x^{2}$$ with $$v'=2x.$$ Therefore

$$f'(x)=\frac{\,2(1+x^{2}) \;-\; 2x(2x)\,}{(1+x^{2})^{2}} =\frac{\,2+2x^{2}-4x^{2}\,}{(1+x^{2})^{2}} =\frac{\,2-2x^{2}\,}{(1+x^{2})^{2}} =\frac{\,2(1 - x^{2})\,}{(1 + x^{2})^{2}}.$$

We now substitute $$x=\dfrac{1}{2}$$:

$$1 - x^{2}=1-\frac{1}{4}=\frac{3}{4}, \qquad 1 + x^{2}=1+\frac{1}{4}=\frac{5}{4}.$$

Hence

$$f'\!\left(\frac12\right)=\frac{\,2\left(\frac34\right)\,}{\left(\frac54\right)^{2}} =\frac{\,\frac32\,}{\frac{25}{16}} =\frac{3}{2}\times\frac{16}{25} =\frac{48}{50} =\frac{24}{25}.$$

Hence, the correct answer is Option B.