Q29 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
Let $f(x)=\int \frac{7 x^{10}+9 x^{8}}{\left(1+x^{2}+2 x^{9}\right)^{2}} d x, x>0, \lim _{x \rightarrow 0} f(x)=0$ and $f(1)=\frac{1}{4}$. If $\mathrm{A}=\left[\begin{array}{ccc}0 & 0 & 1 \\ \frac{1}{4} & f^{\prime}(1) & 1 \\ \alpha^{2} & 4 & 1\end{array}\right]$ and $\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})$ be such that $|\mathrm{B}|=81$, then $\alpha^{2}$ is equal to
- A2
- B1
- C3
- Correct answerD4
Solution
Divide numerator and denominator by $x^{18}$, let $t = x^{-9} + x^{-7} + 2$, then $dt = -(9x^{-10}+7x^{-8})dx$.
$f(x) = \frac{1}{t} + C = \frac{x^9}{1+x^2+2x^9} + C$.
$\lim_{x \to 0} f(x) = 0 \Rightarrow C = 0$. $f(1) = \frac{1}{4}$ (verified).
$f'(x) = \frac{9x^8(1+x^2+2x^9) - x^9(2x+18x^8)}{(1+x^2+2x^9)^2}$.
$f'(1) = \frac{36-20}{16} = 1$.
$|A| = \begin{vmatrix} 0 & 0 & 1 \\ 1/4 & 1 & 1 \\ \alpha^2 & 4 & 1 \end{vmatrix} = 1 - \alpha^2$.
$|B| = |\text{adj}(\text{adj } A)| = |A|^{(3-1)^2} = (1-\alpha^2)^4 = 81 = 3^4$.
$|1-\alpha^2| = 3 \Rightarrow \alpha^2 = 4$.
$f(x) = \frac{1}{t} + C = \frac{x^9}{1+x^2+2x^9} + C$.
$\lim_{x \to 0} f(x) = 0 \Rightarrow C = 0$. $f(1) = \frac{1}{4}$ (verified).
$f'(x) = \frac{9x^8(1+x^2+2x^9) - x^9(2x+18x^8)}{(1+x^2+2x^9)^2}$.
$f'(1) = \frac{36-20}{16} = 1$.
$|A| = \begin{vmatrix} 0 & 0 & 1 \\ 1/4 & 1 & 1 \\ \alpha^2 & 4 & 1 \end{vmatrix} = 1 - \alpha^2$.
$|B| = |\text{adj}(\text{adj } A)| = |A|^{(3-1)^2} = (1-\alpha^2)^4 = 81 = 3^4$.
$|1-\alpha^2| = 3 \Rightarrow \alpha^2 = 4$.
Hence, the correct answer is option: 'D'.