Let $f\left(x\right)=\int \frac{7{x}^{10}+9{x}^{8}}{{(1+{x}^{2}+2{x}^{9})}^{2}}dx,x0$, $\lim _{x\to 0}f\left(x\right)=0$ and $f\left(1\right)=\frac{1}{4}$. If $A=\left[\begin{matrix}0 & 0 & 1 \ \frac{1}{4} & {f}^{'}(1) & 1 \ {\alpha }^{2} & 4 & 1\end{matrix}\right]$ and $B=adj(adjA)$ be such that $|B|=81$, then ${\alpha }^{2}$ is equal to [JEE Main 2026, 24 Jan (Shift 2)]

Question

Let $f\left(x\right)=\int \frac{7{x}^{10}+9{x}^{8}}{{(1+{x}^{2}+2{x}^{9})}^{2}}dx,x>0$, $\lim _{x\to 0}f\left(x\right)=0$ and $f\left(1\right)=\frac{1}{4}$. If $A=\left[\begin{matrix}0 & 0 & 1 \ \frac{1}{4} & {f}^{'}(1) & 1 \ {\alpha }^{2} & 4 & 1\end{matrix}\right]$ and $B=adj(adjA)$ be such that $|B|=81$, then ${\alpha }^{2}$ is equal to [JEE Main 2026, 24 Jan (Shift 2)]

YESPYQ · Accepted Answer

Correct answer: b) 4. $f\left(x\right)=\int \left(\frac{\frac{7}{{x}^{8}}+\frac{9}{{x}^{10}}}{{\left(\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2\right)}^{2}}\right)dx$ Put $t=\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2$ $\Rightarrow \frac{dt}{dx}=\frac{−9}{{x}^{10}}−\frac{7}{{x}^{8}}$ $f\left(x\right)=\int \frac{−dt}{{t}^{2}}=\frac{1}{t}+C$ $f\left(x\right)=\frac{1}{\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2}+C$ $=\frac{{x}^{9}}{1+{x}^{2}+2{x}^{9}}+C$ Given $f\left(1\right)=\frac{1}{4}=\frac{1}{4}+C\Rightarrow C=0$ $f\left(x\right)=\frac{{x}^{9}}{1+{x}^{2}+2{x}^{9}}$ ${f}^{'}\left(x\right)=\frac{\left(1+{x}^{2}+2{x}^{9}\right)9{x}^{8}−{x}^{9}\left(2x+18{x}^{8}\right)}{{\left(1+{x}^{2}+2{x}^{9}\right)}^{2}}$ ${f}^{'}\left(1\right)=\frac{36−20}{16}=1$ $\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})$ $|\mathrm{B}|=81=|\mathrm{A}|^4 \Rightarrow|\mathrm{~A}|=\pm 3$ $A=\left[\begin{matrix}0 & 0 & 1 \ \frac{1}{4} & 1 & 1 \ {\alpha }^{2} & 4 & 1\end{matrix}\right]$ $|\mathrm{A}|=\left(1-\alpha^2\right)$ $1-\alpha^2=3,-3$ $\Rightarrow \alpha^2=-2,4$ value of \({

Let \(f\left(x\right)=\int \frac{7{x}^{10}+9{x}^{8}}{{(1+{x}^{2}+2{x}^{9})}^{2}}dx,x>0\), \(\lim _{x\to 0}f\left(x\right…

Practice more JEE Maths PYQs