Consider the following statements:
(i) When 0 < x < 1, then $$\dfrac{1}{1+x} < 1 - x + x^{2}$$.
(ii) When 0 < x < 1, then $$\dfrac{1}{1+x} > 1 - x + x^{2}$$.
(iii) When -1 < x < 0, then $$\dfrac{1}{1+x} < 1 - x + x^{2}$$.
(iv) When -1 < x < 0, then $$\dfrac{1}{1+x} > 1 - x + x^{2}$$.

$$\dfrac{1}{1+x} < 1 - x + x^{2}$$

$$\dfrac{1}{1+x}-\left(1-x+x^2\right)<0$$

$$\dfrac{1-\left(1+x\right)\left(1-x+x^2\right)}{1+x}<0$$

$$\dfrac{1-\left(1+x^3\right)}{1+x}<0$$

$$\dfrac{1-1-x^3}{1+x}<0$$

$$\dfrac{x^3}{1+x}>0$$

The solution for the above equation is $$x<-1\ or\ x>0$$

$$\dfrac{1}{1+x} > 1 - x + x^{2}$$

$$\dfrac{1}{1+x}-\left(1-x+x^2\right)>0$$

$$\dfrac{1-\left(1+x\right)\left(1-x+x^2\right)}{1+x}>0$$

$$\dfrac{1-\left(1+x^3\right)}{1+x}>0$$

$$\dfrac{1-1-x^3}{1+x}>0$$

$$\dfrac{x^3}{1+x}<0$$

The solution for the above equation is $$-1<x<0$$

Thus, $$\dfrac{1}{1+x} < 1 - x + x^{2}$$ is satisfied when $$x<-1\ or\ x>0$$

and, $$\dfrac{1}{1+x} > 1 - x + x^{2}$$ is satisfied when $$-1<x<0$$

Thus, statements (i) and (iv) are correct.