Let $\mathop {\lim }\limits_{x \to 2} \frac{(\tan (x-2))\left(\mathrm{r} x^2+(\mathrm{p}-2) x-2 \mathrm{p}\right)}{(x-2)^2}=5$ for some $\mathrm{r}, \mathrm{p} \in \boldsymbol{R}$. If the set of all possible values of q , such that the roots of the equation $\mathrm{r} x^2-\mathrm{p} x+\mathrm{q}=0$ lie in $(0,2)$, be the interval $(\alpha, \beta]$, then $4(\alpha+\beta)$ equals :

If the quadratic equation $(\lambda+2) x^2-3 \lambda x+4 \lambda=0, \lambda \neq-2$, has two positive roots, then the number of possible integral values of $\lambda$ is :

If the set of all solutions of $\left|x^2+x-9\right|=|x|+\left|x^2-9\right|$ is $[\alpha, \beta] \cup[\gamma, \infty)$, then ( $\alpha^2+\beta^2+\gamma^2$ ) 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 :