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 $a_i<a_{i+1}\forall i=1,2,\dots .8.$ If the matrix $A=\left[\begin{array}{ccc}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right]$ and $B^3-p{B}^2+qB-rI=0$ , (where $B=adjA$ ), then $\left(2r+p\right)=$

Answer 1

Answer: 5

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