Question

Consider, $f\left(x\right)=\sqrt{\frac{\pi }{2} - \left(tan\right)^{- 1} \sqrt{\frac{- x^{2}}{\left(x^{2} - 9\right) \left(x - 7\right)^{2} \left(x - 9\right) \left(x - 3\right)}}}$ , $a_{i}$ are the integral values of $x$ for which $f\left(x\right)$ is defined and $a_{i} < a_{i + 1}\forall i=1, \, 2, \, \ldots .8.$ If the matrix $A=\begin{bmatrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{bmatrix}$ and $B^{3}-pB^{2}+qB-rI=0$ , (where $B=adjA$ ), then $\left(2 r + p\right)=$

Answer 1

$\frac{- x^{2}}{\left(x^{2} - 9\right) \left(x - 7\right)^{2} \left(x - 9\right) \left(x - 3\right)}\geq 0$
$\Rightarrow x\in \left(- 3 , \, 3\right)\cup\left(3 , \, 7\right)\cup\left(7 , \, 9\right)$
So $a_{i}=-2,-1, \, 0,1,2,4,5,6,8$
$A=\begin{bmatrix} -2 & -1 & 0 \\ 1 & 2 & 4 \\ 5 & 6 & 8 \end{bmatrix}$
$B=adjA=\begin{bmatrix} -8 & 8 & -4 \\ 12 & -16 & 8 \\ -4 & 7 & -3 \end{bmatrix}$
$Tr\left(B\right)=-27\&\left|B\right|=16$
Now, $B^{3}-pB^{2}+qB-rI=0$
Here, sum of roots of above equation $p=Tr\left(B\right)=-27$
and product of roots $r=\left|B\right|=16$
So, $2r+p=2\left(16\right)-27=5$