For real values of $x$,the range of $\frac{x^2+2x+1}{x^2+2x-1}$ is

If $\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of the equation $a x^4+b x^3+c x^2+d x+e=0$ and $\alpha_2, \beta_2, \gamma_2, \delta_2$ are the roots of the equation $e x^4+d x^3+c x^2+b x+a=0$ such that $0 < \alpha_1 < \beta_1 < \gamma_1 < \delta_1$,$0 < \alpha_2 < \beta_2 < \gamma_2 < \delta_2$,$\alpha_1-\delta_2=2$,$\beta_1-\gamma_2=2$,$\gamma_1-\beta_2=4$,and $\delta_1-\alpha_2=4$,then $a+b+c+d+e=$

Both equations $x^2 + b^2 = 1 - 2bx$ and $x^2 + a^2 = 1 - 2ax$ have exactly one root each,and they share the same root. Then:

The probability of selecting integers $a \in [-5, 30]$ such that $x^{2}+2(a+4)x-5a+64 > 0$ for all $x \in \mathbb{R}$ is:

If $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$ for all $x \in R$,then the interval of maximum length in which $y$ lies is

If $\frac{1}{2} \leq \frac{x^2+x+a}{x^2-x+a} \leq 2$ for all $x \in R$,then $a=$