JEE Main 2026 — 24 January, Evening Shift — Question 1

$f(x)=\frac{\int\left(\frac{7}{x^{8}}+\frac{9}{x^{10}}\right)}{\left(\frac{1}{x^{9}}+\frac{1}{x^{7}}+2\right)^{2}} d x$

Put $t=\frac{1}{x^{9}}+\frac{1}{x^{7}}+2 \Rightarrow \frac{d t}{d x}=\frac{-9}{x^{10}}-\frac{7}{x^{8}}$

$\mathrm{f}(\mathrm{x})=\int \frac{-\mathrm{dt}}{\mathrm{t}^{2}}=\frac{1}{\mathrm{t}}+\mathrm{C}$

$f(x)=\frac{1}{\frac{1}{x^{9}}+\frac{1}{x^{7}}+2}+C$

$=\frac{x^{9}}{1+x^{2}+2 x^{9}}+C$

Given $\mathrm{f}(1)=\frac{1}{4}=\frac{1}{4}+\mathrm{C} \Rightarrow \mathrm{C}=0$

$f(x)=\frac{x^{9}}{1+x^{2}+2 x^{9}}$

$f^{\prime}(x)=\frac{\left(1+x^{2}+2 x^{9}\right)-9 x^{8}-x^{9}\left(2 x+18 x^{8}\right)}{\left(1+x^{2}+2 x^{9}\right)^{2}}$

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{36-20}{16}=1$

$\mathrm{A}=\left(\begin{array}{ccc}0 & 0 & 1 \\4 & 1 & 1 \\\alpha^{2} & \frac{1}{4} & 1\end{array}\right)$

$\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})$

$|\mathrm{B}|=81=|\mathrm{A}|^{4} \Rightarrow|\mathrm{~A}|=3$

$|\mathrm{A}|=\left|1-\alpha^{2}\right|=3$

$1-\alpha^{2}=3,-3 \Rightarrow \alpha^{2}=-2,4$

Value of $\alpha^{2}=4$