Question
Let $f(x) = \int_{}^{}\ \frac{7x^{10} + 9x^{8}}{\left( 1 + x^{2} + 2x^{9} \right)^{2}}dx,\ x\ > 0$,
$\lim_{x \rightarrow 0}\mspace{2mu} f(x) = 0\ $ and $\ f(1) = \frac{1}{4}.\ $ If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^{2} & 4 & 1 \end{bmatrix}$ and $B = adj(adjA)$ be such that $|B| = 81$, then $\alpha^{2}$ is equal to
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Solution
$f(x) = \frac{\int_{}^{}\ \left( \frac{7}{x^{8}} + \frac{9}{x^{10}} \right)}{\left( \frac{1}{x^{9}} + \frac{1}{x^{7}} + 2 \right)^{2}}dx$
Put $t = \frac{1}{x^{9}} + \frac{1}{x^{7}} + 2 \Rightarrow \frac{dt}{dx} = \frac{- 9}{x^{10}} - \frac{7}{x^{8}}$
$${f(x) = \int\frac{- dt}{t^{2}} = \frac{1}{t} + C }{f(x) = \frac{1}{\frac{1}{x^{9}} + \frac{1}{x^{7}} + 2} + C }{= \frac{x^{9}}{1 + x^{2} + 2x^{9}} + C }$$Given $f(1) = \frac{1}{4} = \frac{1}{4} + C \Rightarrow C = 0$
$${f(x) = \frac{x^{9}}{1 + x^{2} + 2x^{9}} }{f'(x) = \frac{\left( 1 + x^{2} + 2x^{9} \right) - 9x^{8} - x^{9}\left( 2x + 18x^{8} \right)}{\left( 1 + x^{2} + 2x^{9} \right)^{2}} }{f'(x) = \frac{36 - 20}{16} = 1 }{A = \begin{pmatrix} 0 & 0 & 1 \\ 4 & 1 & 1 \\ \alpha^{2} & \frac{1}{4} & 1 \end{pmatrix} }{|A| = \left| 1 - \alpha^{2} \right| = 3 }{1 - \alpha^{2} = 3, - 3 \Rightarrow \alpha^{2} = - 2,4 }$$Value of $\alpha^{2} = 4$
$${B = adj(adjA) }{|B| = 81 = |A|^{4} \Rightarrow |\text{ }A| = 3}$$
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