Question

Let \[ f(x)=\int \frac{7x^{10}+9x^8}{(1+x^2+2x^9)^2}\,dx \] and $f(1)=\frac14$. Given that 

and $B=\operatorname{adj}(\operatorname{adj}A)$, if $|B|=81$, find the value of $\alpha^2$ (where $\alpha\in\mathbb{R}$). 
 

• A. 2
• B. 4
• C. 6
• D. 8

Hint:

For $n\times n$ matrices, remember: $|\operatorname{adj}(\operatorname{adj}A)|=|A|^{(n-1)^2}$.

Answer 1

The Correct Option is D

Step 1: Use property of adjoint. 
For a $3\times3$ matrix, \[ |\operatorname{adj}(\operatorname{adj}A)|=|A|^{(3-1)^2}=|A|^4 \] Given $|B|=81$, \[ |A|^4=81 \Rightarrow |A|=3 \] Step 2: Evaluate $f(1)$. 
\[ f'(x)=\frac{7x^{10}+9x^8}{(1+x^2+2x^9)^2} \] Let $u=1+x^2+2x^9$ 
\[ f(x)=-\frac{1}{u}+C \] Using $f(1)=\frac14$: \[ -\frac{1}{1+1+2}+C=\frac14 \Rightarrow C=\frac12 \] Step 3: Compute determinant of $A$. 
\[ =0-0+1\left(4\cdot\frac14-\alpha^2\cdot\frac14\right) \] \[ =1-\frac{\alpha^2}{4} \] Step 4: Use $|A|=3$. 
\[ 1-\frac{\alpha^2}{4}=3 \Rightarrow \alpha^2=8 \] Final conclusion. 
The value of $\alpha^2$ is 8.