Let [_^ _e 4 d xe^x-1=()/(6)].

Question

Let $$\int_\limits\alpha^{\log _e 4} \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=\frac{\pi}{6}$$. Then $$\mathrm{e}^\alpha$$ and $$\mathrm{e}^{-\alpha}$$ are the roots of the equation :

Accepted Answer

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

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

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

Solutions · 1