We need to evaluate
∫ α ln 4 d x e x − 1 = π 6 . \int_{\alpha}^{\ln 4} \frac{dx}{\sqrt{e^x-1}}=\frac{\pi}{6}. ∫ α l n 4 e x − 1 d x = 6 π .
Use the substitution
e x − 1 = tan 2 θ . e^x-1=\tan^2\theta. e x − 1 = tan 2 θ . Then
e x = 1 + tan 2 θ = sec 2 θ ⇒ x = ln ( sec 2 θ ) . e^x=1+\tan^2\theta=\sec^2\theta
\quad\Rightarrow\quad x=\ln(\sec^2\theta). e x = 1 + tan 2 θ = sec 2 θ ⇒ x = ln ( sec 2 θ ) . Differentiate:
e x d x = 2 tan θ sec 2 θ d θ . e^x\,dx=2\tan\theta\sec^2\theta\,d\theta. e x d x = 2 tan θ sec 2 θ d θ . Since e x = sec 2 θ e^x=\sec^2\theta e x = sec 2 θ
sec 2 θ d x = 2 tan θ sec 2 θ d θ ⇒ d x = 2 tan θ d θ . \sec^2\theta\,dx=2\tan\theta\sec^2\theta\,d\theta
\quad\Rightarrow\quad dx=2\tan\theta\,d\theta. sec 2 θ d x = 2 tan θ sec 2 θ d θ ⇒ d x = 2 tan θ d θ . Also,
e x − 1 = tan θ . \sqrt{e^x-1}=\tan\theta. e x − 1 = tan θ . Hence,
d x e x − 1 = 2 tan θ d θ tan θ = 2 d θ . \frac{dx}{\sqrt{e^x-1}}=\frac{2\tan\theta\,d\theta}{\tan\theta}=2\,d\theta. e x − 1 d x = tan θ 2 tan θ d θ = 2 d θ .
Change limits.
e x = 4 ⇒ e x − 1 = 3 ⇒ tan 2 θ = 3. e^x=4 \Rightarrow e^x-1=3 \Rightarrow \tan^2\theta=3. e x = 4 ⇒ e x − 1 = 3 ⇒ tan 2 θ = 3. So
θ = π 3 \theta=\frac{\pi}{3} θ = 3 π (in the principal range).
When x = α x=\alpha x = α θ α \theta_\alpha θ α
Thus,
∫ α ln 4 d x e x − 1 = ∫ θ α π / 3 2 d θ = 2 ( π 3 − θ α ) . \int_\alpha^{\ln 4} \frac{dx}{\sqrt{e^x-1}}
=\int_{\theta_\alpha}^{\pi/3} 2\,d\theta
=2\left(\frac{\pi}{3}-\theta_\alpha\right). ∫ α l n 4 e x − 1 d x = ∫ θ α π /3 2 d θ = 2 ( 3 π − θ α ) . Given this equals π 6 \frac{\pi}{6} 6 π
2 ( π 3 − θ α ) = π 6 . 2\left(\frac{\pi}{3}-\theta_\alpha\right)=\frac{\pi}{6}. 2 ( 3 π − θ α ) = 6 π . Therefore,
π 3 − θ α = π 12 ⇒ θ α = π 4 . \frac{\pi}{3}-\theta_\alpha=\frac{\pi}{12}
\quad\Rightarrow\quad
\theta_\alpha=\frac{\pi}{4}. 3 π − θ α = 12 π ⇒ θ α = 4 π .
Now find e α e^\alpha e α
e α − 1 = tan 2 θ α = tan 2 π 4 = 1 , e^\alpha-1=\tan^2\theta_\alpha=\tan^2\frac{\pi}{4}=1, e α − 1 = tan 2 θ α = tan 2 4 π = 1 , we get
e α = 2. e^\alpha=2. e α = 2. Therefore,
e − α = 1 2 . e^{-\alpha}=\frac{1}{2}. e − α = 2 1 .
The required quadratic has roots 2 2 2 1 2 \frac12 2 1
( x − 2 ) ( x − 1 2 ) = 0. (x-2)\left(x-\frac12\right)=0. ( x − 2 ) ( x − 2 1 ) = 0. Expanding,
x 2 − 5 2 x + 1 = 0. x^2-\frac{5}{2}x+1=0. x 2 − 2 5 x + 1 = 0. Multiplying by 2 2 2
2 x 2 − 5 x + 2 = 0. 2x^2-5x+2=0. 2 x 2 − 5 x + 2 = 0.
Compare with options:
A: 2 x 2 − 5 x + 2 = 0 2x^2-5x+2=0 2 x 2 − 5 x + 2 = 0
B: x 2 − 2 x − 8 = 0 x^2-2x-8=0 x 2 − 2 x − 8 = 0
C: 2 x 2 − 5 x − 2 = 0 2x^2-5x-2=0 2 x 2 − 5 x − 2 = 0
D: x 2 + 2 x − 8 = 0 x^2+2x-8=0 x 2 + 2 x − 8 = 0
Hence the correct option is A .