The minimum value of $\frac{9 \cdot 3^{2x} + 6 \cdot 3^x + 4}{9 \cdot 3^{2x} - 6 \cdot 3^x + 4}$ is

Let $2 \sin^2 x + 3 \sin x - 2 > 0$ and $x^2 - x - 2 < 0$ ($x$ is measured in radians). Then $x$ lies in the interval

The curves $y=x^2+9x+20$ and $y=x^2+bx+c$ intersect the $X$-axis at the points $(\alpha_i, 0)$ for $i=1, 2, 3, 4$. If $\alpha_1 < \alpha_2 < \alpha_3 < \alpha_4$ are such that $|\alpha_1-\alpha_3|=|\alpha_2-\alpha_4|=8$,then the sum of all possible values of $b$ and $c$ is:

If $x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right)$,then the value of $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$ is equal to

If $x$ is real,then the maximum and minimum values of $\frac{x^2+14x+9}{x^2+2x+3}$ are respectively

For $x \in R$, the least value of $\frac{x^2-6x+5}{x^2+2x+1}$ is