Question

Let P be the sum of all possible determinants of order 2 having 0, 1, 2, & 3 as their four elements. Then find the common root $\alpha$ of the equations $\begin{array}{c}{x}^2+ax+[m+1]=0\\ x^2+bx+[m+4]=0\\ x^2-cx+[m+15]=0\end{array}$ such that $\alpha$ > p, where $a+b+c=0$ and $m=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{2n}-\frac{r}{\sqrt{n^2+r^2}}$ (where [.] denotes the greatest integer function)

Answer 1

Let $\alpha$ be the common root, then
${\alpha }^2+a\alpha +[m+1]=0$ ...(i)
${\alpha }^2+b\alpha +[m+1]=0$ ...(ii)
${\alpha }^2+c\alpha +[m+1]=0$ ...(iii)
From (i)+(ii)-(iii), we get
$a^2+\left[m\right]-10=0$ ...(iv)
But $m=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}$
$=\underset{n\to \infty }{lim}\frac{l}{n}\sum _{r=1}^{2n}\frac{\frac{r}{n}}{\sqrt{1+{\left(\frac{r}{n}\right)}^2}}={\int }_0^2\frac{x}{\sqrt{1+x^2}}dx$
$={\left[\sqrt{1+x^2}\right]}_0^2=\sqrt{5}-1$
Now $\left[m\right]=[\sqrt{5}-1]=1$
From equation (iv) ${\alpha }^2+1-10=0$

$\Rightarrow \alpha =\pm 3$