The values of $\frac{x^2-2x+1}{x^2+x-1}$ do not lie in the interval

Consider the following statements: For any integer $n$,
$I.$ $n^2+3$ is never divisible by $17$.
$II.$ $n^2+4$ is never divisible by $17$.
Then,

Let $p$ and $q$ be roots of the equation $x^2-2x+A=0$ and let $r$ and $s$ be the roots of the equation $x^2-18x+B=0$. If $p < q < r < s$ are in $A.P.$,then $A$ and $B$ are

Let $x$ be a real number. Match the following:

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$,$\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$,$\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

All the roots of the equation $x^5+15x^4+94x^3+305x^2+507x+353=0$ are increased by some real number $k$ in order to eliminate the $4^{th}$ degree term from the equation. Now,the coefficient of $x$ in the transformed equation is

If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3=$