Let $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, $x > 0$, $\lim_{x \rightarrow 0} f(x) = 0$ 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 $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to
Finding the Integral Function $f(x)$
We need to find the integral $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, for $x > 0$. Consider the function $g(x) = \frac{x^9}{1 + x^2 + 2x^9}$. Its derivative is:
$ \frac{d}{dx} \left( \frac{x^9}{1 + x^2 + 2x^9} \right) = \frac{9x^8(1 + x^2 + 2x^9) - x^9(2x + 18x^8)}{(1 + x^2 + 2x^9)^2} $
$ = \frac{9x^8 + 9x^{10} + 18x^{17} - 2x^{10} - 18x^{17}}{(1 + x^2 + 2x^9)^2} = \frac{9x^8 + 7x^{10}}{(1 + x^2 + 2x^9)^2} $
This matches the integrand. So, $f(x) = \frac{x^9}{1 + x^2 + 2x^9} + C$. Using the condition $\lim_{x \rightarrow 0} f(x) = 0$: $ \lim_{x \rightarrow 0} \left( \frac{x^9}{1 + x^2 + 2x^9} + C \right) = \frac{0}{1 + 0 + 0} + C = 0 \implies C = 0 $. Thus, $f(x) = \frac{x^9}{1 + x^2 + 2x^9}$. This is consistent with the given $f(1) = \frac{1}{1+1^2+2(1)^9} = \frac{1}{1+1+2} = \frac{1}{4}$.
Calculating the Derivative $f'(1)$
The derivative $f'(x)$ is the integrand itself:
$ f'(x) = \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} $
Evaluating $f'(x)$ at $x=1$:
$ f'(1) = \frac{7(1)^{10} + 9(1)^8}{(1 + (1)^2 + 2(1)^9)^2} = \frac{7 + 9}{(1 + 1 + 2)^2} = \frac{16}{4^2} = \frac{16}{16} = 1 $
Matrix Properties and Determinant Calculation
The matrix $A$ is constructed using $f'(1) = 1$:
$ A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & 1 & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix} $
The determinant of $A$, denoted $|A|$, is calculated as:
$ |A| = 0 \cdot \det(\dots) - 0 \cdot \det(\dots) + 1 \cdot \begin{vmatrix} \frac{1}{4} & 1 \\ \alpha^2 & 4 \end{vmatrix} $
$ |A| = 1 \cdot \left( \frac{1}{4}(4) - 1(\alpha^2) \right) = 1 - \alpha^2 $
We are given $B = \text{adj}(\text{adj } A)$ and $|B| = 81$. For an $n \times n$ matrix $A$, a key property is $\text{adj}(\text{adj } A) = |A|^{n-2} A$. Since $A$ is a $3 \times 3$ matrix ($n=3$), we have: $ B = |A|^{3-2} A = |A| A $. The determinant of $B$ is $|B| = | |A| A |$. Using the property $|kA| = k^n |A|$ for a scalar $k$ and $n \times n$ matrix $A$:
$ |B| = |A|^3 |A| = |A|^4 $
Solving for $\alpha^2$
We are given $|B| = 81$. Substituting $|B| = |A|^4$: $ |A|^4 = 81 $ Taking the fourth root, we get: $ |A| = \pm \sqrt[4]{81} = \pm 3 $.
We now use $|A| = 1 - \alpha^2$ and consider the two possible values for $|A|$:
- Case 1: $|A| = 3$
$ 1 - \alpha^2 = 3 $ $ \alpha^2 = 1 - 3 = -2 $
This result is typically not considered valid in this context as the options are positive integers. - Case 2: $|A| = -3$
$ 1 - \alpha^2 = -3 $ $ \alpha^2 = 1 - (-3) = 1 + 3 = 4 $
The value $\alpha^2 = 4$ corresponds to option D.
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