Let $f:[1, \infty) \rightarrow \mathbf{R}$ be a differentiable function defined as $f(x)=\int_1^x f(\mathrm{t}) \mathrm{dt}+(1-x)\left(\log _{\mathrm{e}} x-1\right)+\mathrm{e}$.

Then the value of $f(f(1))$ is :

Let $y=y(x)$ be the solution of the differential equation :

$$ \frac{d y}{d x}+\left(\frac{6 x^2+\left(3 x^2+2 x^3+4\right) e^{-2 x}}{\left(x^3+2\right)\left(2+e^{-2 x}\right)}\right) y=2+e^{-2 x} $$

$x \in(-1,2)$, satisfying $y(0)=\frac{3}{2}$. If $y(1)=\alpha\left(2+e^{-2}\right)$, then $\alpha$ is equal to :

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

Let $x = x(y)$ be the solution of the differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$, $y > 1$, $x(e) = e$.

Then $x(e^2)$ is equal to: