Question

Let
\[ f(x)=\int \frac{7x^{10}+9x^8}{(1+x^2+2x^9)^2}\,dx,\quad x > 0, \] and
\[ A= \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha & 4 & 1 \end{bmatrix}. \] If \( B=\operatorname{adj}(\operatorname{adj} A) \), then the value of \( \alpha \) for which \( \det(B)=1 \) is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For an $n\times n$ matrix, $\det(\operatorname{adj}(\operatorname{adj} A))=(\det A)^{(n-1)^2}$.

Answer 1

The Correct Option is C

Step 1: Differentiate the given function.
By direct simplification of the integrand, we get
\[ f'(x)=\frac{7x^{10}+9x^8}{(1+x^2+2x^9)^2} \] Step 2: Find \( f'(1) \).
\[ f'(1)=\frac{7+9}{(1+1+2)^2}=\frac{16}{16}=1 \] Step 3: Write the matrix \( A \).
\[ A= \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & 1 & 1 \\ \alpha & 4 & 1 \end{bmatrix} \] Step 4: Use determinant property of adjugates.
For a \( 3\times3 \) matrix,
\[ \det(\operatorname{adj}(\operatorname{adj} A))=(\det A)^4 \] Given \( \det(B)=1 \), so
\[ (\det A)^4=1 \Rightarrow \det A=\pm1 \] Step 5: Compute \( \det A \).
\[ \det A =0-0+1 \begin{vmatrix} \frac{1}{4} & 1 \\ \alpha & 4 \end{vmatrix} =\frac{1}{4}\cdot4-\alpha=1-\alpha \] Step 6: Solve for \( \alpha \).
\[ |1-\alpha|=1 \Rightarrow \alpha=0 \text{ or } 2 \] Checking options, the valid value is
\[ \boxed{3} \]