If $\left|\frac{x^2+kx+1}{x^2+x+1}\right| < 3$ for all real $x$,then $k$ is in the interval

Suppose,$\alpha$ is the minimum value of $x^2+bx+5$ and $\beta$ is the maximum value of $-x^2+ax+5$. If $[\alpha, \beta]$ is the interval of maximum length for $x$ in which $x^2-10x+24 \leq 0$,then $a^2b^2$ is equal to

Let $\alpha, \beta$ be the roots of the equation $x^2 - 3x + r = 0$, and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2 + 3x + r = 0$. If the roots of the equation $x^2 + 6x = m$ are $2\alpha + \beta + 2r$ and $\alpha - 2\beta - \frac{r}{2}$, then $m$ is equal to:

If $x$ is complex,the expression $\frac{x^2+34x-71}{x^2+2x-7}$ takes all values which lie in the interval $(a, b)$,find the value of $a$ and $b$.

If the minimum value of $f(x) = x^2 + 2bx + 2c^2$ is greater than the maximum value of $g(x) = -x^2 - 2cx + b^2$ for all real $x$,then:

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