Let $$y = y(x)$$ be the solution of the differential equation $$\frac{dy}{dx} = (1 + x + x^2)(1 - y + y^2)$$, $$y(0) = \frac{1}{2}$$. Then $$(2y(1) - 1)$$ is equal to :

The given differential equation is$$\frac{dy}{dx}=(1+x+x^2)(1-y+y^2),\qquad y(0)=\frac12.$$

Separate the variables:$$\frac{dy}{1-y+y^2}=(1+x+x^2)\,dx.$$

Integrate both sides:$$\int\frac{dy}{1-y+y^2}=\int(1+x+x^2)\,dx.$$

Left integral
Write the quadratic in completed-square form:$$1-y+y^2=y^2-y+1=\left(y-\frac12\right)^2+\frac34.$$

Use the standard formula $$\int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\!\left(\frac{u}{a}\right)+C.$$
Here $$u=y-\frac12,\;a=\frac{\sqrt3}{2},\;du=dy.$$ Thus

$$\int\frac{dy}{1-y+y^2}=\frac2{\sqrt3}\tan^{-1}\!\left(\frac{2(y-\frac12)}{\sqrt3}\right)+C_1.$$

Right integral
$$\int(1+x+x^2)\,dx=x+\frac{x^2}{2}+\frac{x^3}{3}+C_2.$$

Combining the two antiderivatives and absorbing constants into a single constant $$C$$, we obtain

$$\frac2{\sqrt3}\tan^{-1}\!\left(\frac{2(y-\frac12)}{\sqrt3}\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+C.$$

Use the initial condition
At $$x=0,\;y=\frac12$$, the left side is $$\frac2{\sqrt3}\tan^{-1}(0)=0$$. Hence $$C=0$$.

Therefore the relation between $$x$$ and $$y$$ is

$$\frac2{\sqrt3}\tan^{-1}\!\left(\frac{2(y-\frac12)}{\sqrt3}\right)=x+\frac{x^2}{2}+\frac{x^3}{3}.$$(1)

Find $$y(1)$$
Substitute $$x=1$$ in (1):

Right side: $$1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}.$$

Left side: $$\frac2{\sqrt3}\tan^{-1}\!\left(\frac{2(y(1)-\frac12)}{\sqrt3}\right).$$

Hence$$\frac2{\sqrt3}\tan^{-1}\!\left(\frac{2y(1)-1}{\sqrt3}\right)=\frac{11}{6}.$$

Multiply by $$\frac{\sqrt3}{2}$$:

$$\tan^{-1}\!\left(\frac{2y(1)-1}{\sqrt3}\right)=\frac{11\sqrt3}{12}.$$

Take the tangent of both sides:

$$\frac{2y(1)-1}{\sqrt3}=\tan\!\left(\frac{11\sqrt3}{12}\right).$$

Therefore$$2y(1)-1=\sqrt3\,\tan\!\left(\frac{11\sqrt3}{12}\right).$$

Hence $$(2y(1)-1)=\sqrt3\tan\left(\frac{11\sqrt3}{12}\right).$$ This matches Option C.