Let f(x) = int frac{7x^{10} + 9x^{8}}{(1 + x^ | vedclass

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(D) First,simplify the integrand: $f(x) = \int \frac{x^{18}(7x^{-8} + 9x^{-10})}{(x^9(x^{-9} + x^{-7} + 2))^2} dx = \int \frac{7x^{-8} + 9x^{-10}}{(x^

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$4$

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(D) First,simplify the integrand: $f(x) = \int \frac{x^{18}(7x^{-8} + 9x^{-10})}{(x^9(x^{-9} + x^{-7} + 2))^2} dx = \int \frac{7x^{-8} + 9x^{-10}}{(x^{-9} + x^{-7} + 2)^2} dx$.Let $t = x^{-9} + x^{-7} + 2$,then $dt = (-9x^{-10} - 7x^{-8}) dx$,so $-(7x^{-8} + 9x^{-10}) dx = dt$.Thus,$f(x) = \int -t^{-2} dt = t^{-1} + C = \frac{1}{x^{-9} + x^{-7} + 2} + C = \frac{x^9}{1 + x^2 + 2x^9} + C$.Given $\lim_{x 	o 0} f(x) = 0$,we find $C = 0$. Also $f(1) = \frac{1}{1+1+2} = \frac{1}{4}$,which is consistent.Now,$f'(x) = \frac{9x^8(1+x^2+2x^9) - x^9(2x + 18x^8)}{(1+x^2+2x^9)^2}$.At $x=1$,$f'(1) = \frac{9(4) - 1(20)}{4^2} = \frac{36-20}{16} = 1$.Matrix $A = \begin{bmatrix} 0 & 0 & 1 \ 1/4 & 1 & 1 \ \alpha^2 & 4 & 1 \end{bmatrix}$.$|A| = 1(\frac{1}{4} 	imes 4 - \alpha^2 	imes 1) = 1 - \alpha^2$.Given $|B| = |	ext{adj}(	ext{adj } A)| = |A|^{(n-1)^2} = |A|^4 = 81$,so $|A| = \pm 3$.$1 - \alpha^2 = 3 \Rightarrow \alpha^2 = -2$ (not possible for real $\alpha^2$) or $1 - \alpha^2 = -3 \Rightarrow \alpha^2 = 4$.

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