If $x$ is real,the expression $\frac{x + 2}{2x^2 + 3x + 6}$ takes all values in the interval

Let $a \in \mathbb{R}$ and let $\alpha, \beta$ be the roots of the equation $x^2+60^{\frac{1}{4}} x+a=0$. If $\alpha^4+\beta^4=-30$,then the product of all possible values of $a$ is $......$

Let $\alpha$ and $\beta$ be the roots of $x^{2}-3x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6x+q=0$. If $\alpha, \beta, \gamma, \delta$ form a geometric progression,then the ratio $(2q+p):(2q-p)$ is:

The population of cattle in a farm increases such that the difference between the population in year $n+2$ and that in year $n$ is proportional to the population in year $n+1$. If the populations in years $2010, 2011$ and $2013$ were $39, 60$ and $123$ respectively,then the population in $2012$ was:

If $x$ is real,then the maximum and minimum values of the expression $\frac{x^2 - 3x + 4}{x^2 + 3x + 4}$ are

Suppose $a, b, c$ are three distinct real numbers. Let $P(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$. When simplified,$P(x)$ becomes: