Let f(x)=∫ (7 x^10+9 x^8)/((1+x^2+2 x^9)^2) d x, x>0, lim_x → 0 f(x)=0 and f(1)=1/4. If A = [0, 0, 1; 1/4, f^′(1), 1; α^2, 4, 1] and B = adj ( adj A…

Let $\displaystyle f(x)=\int \frac{7 x^{10}+9 x^8}{\left(1+x^2+2 x^9\right)^2} d x, x>0, \lim _{x \rightarrow 0} f(x)=0$ and $\displaystyle f(1)=\frac{1}{4}$. If $\displaystyle \mathrm{A}=\left[\begin{array}{ccc}0 & 0 & 1 \\ \frac{1}{4} & f^{\prime}(1) & 1 \\ \alpha^2 & 4 & 1\end{array}\right]$ and $\displaystyle \mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})$ be such that $\displaystyle |\mathrm{B}|=81$, then $\displaystyle \alpha^2$ is equal to