JEE Main 2024 — Definite Integration Question with Solution

\begin{aligned} & \text { Let } \sqrt{e^x-1}=t \\ & e^x-1=t^2 \\ & e^x=1+t^2 \\ & e^x=0+2 t-\frac{d t}{d x} \\ & \frac{d t}{d x}=\frac{e^x}{2 t}=\frac{t^2+1}{2 t} \\ & I=\int \frac{2 t}{t\left(1+t^2\right)} d t=2 \tan ^{-1} t \\ & \Rightarrow \quad I=\int_\limits\alpha^{\log ^4} \frac{d x}{\sqrt{e^x-1}} \\ & I=\left.2 \tan ^{-1} \sqrt{e^x-1}\right|_\alpha ^{\log 4} \\ & =2\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{e^\alpha-1}\right)=\frac{\pi}{6} \\ & \Rightarrow \frac{\pi}{3}-\tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{12} \\ & \tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{4} \\ & \Rightarrow e^\alpha-1=1 \\ & e^\alpha=2 \Rightarrow e^{-\alpha}=\frac{1}{2} \end{aligned}

$$\therefore$$ Quadratic equation whose roots are $$e^a$$ & $$e^{-\alpha }$$ is

$$\begin{array}{rl}& x^2-\left(e^{\alpha }+e^{-\alpha }\right)x+e^{\alpha }\times e^{-\alpha }=0\\ & x^2-\left(2+\frac{1}{2}\right)x+1=0\\ & 2x^2-5x+2=0\end{array}$$