Question

Let $A=diag(x^2,6x,1)$ and $B=diag(\frac{1}{x^2},\frac{1}{x},1)$ be two diagonal matrices. A function $f$ is defined as $f\left(x\right)=tr\left(\right(A^{-1}B{)}^{-1}).$ If $\int \frac{f\left(x\right)}{x^2+1}dx=g\left(x\right)$ where $g\left(0\right)=0,$ then the number of roots of $g\left(x\right)=0$ is

$\left[tr\right(P)$ denotes the trace of matrix $P]$

Answer 1

$f\left(x\right)=tr\left(\right(A^{-1}B{)}^{-1})=tr(B^{-1}A)$
Now, $B^{-1}=diag(x^2,x,1)$
$A=diag(x^2,6x,1)\Rightarrow B^{-1}A=diag(x^4,6x^2,1)$
$\therefore f\left(x\right)=x^4+6x^2+1$

Now, $\int \frac{f\left(x\right)}{x^2+1}dx$
$=\int \frac{x^4+6x^2+1}{x^2+1}dx=\int \frac{(x^2+5)(x^2+1)-4}{x^2+1}dx=\int (x^2+5-\frac{4}{x^2+1})dx=\frac{x^3}{3}+5x-4{\tan }^{-1}x+C\Rightarrow g\left(x\right)=\frac{x^3}{3}+5x-4{\tan }^{-1}x(\because g(0)=0)$
Differentiating w.r.t. $x$, we get
$g^{\prime }\left(x\right)=x^2+5-\frac{4}{x^2+1}$

Assuming $x^2+1=t$, then
$g^{\prime }\left(x\right)=\frac{t^2+4t-4}{t}=\frac{(t+2{)}^2-8}{t}$
As $x^2\ge 0\Rightarrow t\ge 1$
$\Rightarrow (t+2{)}^2\ge 9\Rightarrow (t+2{)}^2-8\ge 1\therefore g^{\prime }\left(x\right)>0$
$\Rightarrow g\left(x\right)$ is monotonically increasing function.
Hence, $g\left(x\right)=0$ has only one solution i.e., $x=0$